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Tamilnadu Samacheer Kalvi 12th Commerce Solutions Chapter 6 Money Market

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Samacheer Kalvi 12th Commerce Money Market Textbook Exercise Questions and Answers

I. Choose the Correct Answer:

Question 1.
The money invested in the call money market provides high liquidity with _____
(a) Low Profitability
(b) High Profitability
(c) Limited Profitability
(d) Medium Profitability
Answer:
(a) Low Profitability

Question 2.
A major player in the money market is the _____
(a) Commercial Bank
(b) Reserve Bank of India
(c) State Bank of India
(d) Central Bank
Answer:
(a) Commercial Bank

Question 3.
Money Market provides _____
(a) Medium-term Funds
(b) Short-term Funds
(c) Long-term Funds
(d) Shares
Answer:
(b) Short-term Funds

Question 4.
Money Market Institutions are _____
(a) Investment Houses
(b) Mortgage Banks
(c) Reserve Bank of India
(d) Commercial Banks and Discount Houses
Answer:
(d) Commercial Banks and Discount Houses

Question 5.
Risk in the Money Market is _____
(a) High
(b) Market Risk
(c) Low Credit and Market Risk
(d) Medium Risk
Answer:
(c) Low Credit and Market Risk

Question 6.
Debt Instruments are issued by Corporate Houses are raising short-term financial resources from the money market are called _____
(a) Treasury Bills
(b) Commercial Paper
(c) Certificate of Deposit
(d) Government Securities
Answer:
(b) Commercial Paper

Question 7.
The market for buying and selling of Commercial Bills of Exchange is known as a _____
(a) Commercial Paper Market
(b) Treasury Bill Market
(c) Commercial Bill Market
(d) Capital Market
Answer:
(c) Commercial Bill Market

Question 8.
A marketable document of title to a time deposit for a specified period may be referred to as a _____
(a) Treasury Bill
(b) Certificate of Deposit
(c) Commercial Bill
(d) Government Securities
Answer:
(b) Certificate of Deposit

Question 9.
Treasury Bills commands _____
(a) High Liquidity
(b) Low Liquidity
(c) Medium Liquidity
(d) Limited Liquidity
Answer:
(a) High Liquidity

Question 10.
Government Securities are issued by agencies such as _____
(a) Central Government
(b) State Governments
(c) Semi-government Authorities
(d) All of the above
Answer:
(d) All of the above

II. Very Short Answer Questions

Question 1.
Define the term “Money Market”.
Answer:
According to Crowther, ’’the money market is the collective name given to the various firms and institutions that deal in the various grades of near money”.

Question 2.
What is commercial bill market?
Answer:
The Commercial Bill is an instrument drawn by a seller of goods on a buyer of goods. A bill of exchange issued by a commercial organization to raise money for short-term needs. These bills are of 30 days, 60 days and 90 days maturity.

Question 3.
What is a CD market?
Answer:
Certificate of Deposits are short-term deposit instruments issued by banks and financial institutions to raise large sums of money. The Certificate of Deposit is transferable from one party to another. Due to their negotiable feature, they are also known as negotiable certificate of deposit.

Question 4.
What is Government Securities Market?
Answer:
A market whereby the Government or gilt-edged securities can be bought and sold is called ‘Government Securities Market’.

Question 5.
What are the Instruments of Money Market?
Answer:

1.  Treasury Bills in the Treasury Market
2. Money at Call and Short Notice in the Call Loan Market
3. Commercial Bills and Promissory Notes in the Bill Market

Now in addition to the above, the following new instruments come into existence:

1. Commercial Papers
2. Certificate of Deposits
3. Inter-Bank participation Certificates
4. Repo Instruments

Question 6.
Explain the two oldest money markets.
Answer:

1. Treasury Bills- These are very popular and enjoy a higher degree of liquidity since they are issued by the Government.
2. Commercial Bills- It is an instrument drawn by a seller of goods on a buyer of goods.

Question 7.
What do you mean by Auctioning?
Answer:
A method of trading whereby merchants bid against one another and where the securities are sold to the highest bidder is known as‘auctioning’.

Question 8.
What do you mean by Switching?
Answer:
The purchase of one security against the sale of another security carried out by the RBI in the secondary market as part of its open market operations is described as ‘Switching’.

III. Short Answer Questions

Question 1.
What are the features of Treasury Bills?
Answer:

1. Issuer
2. Finance Bills
3. Liquidity
4. Vital Source
5. Monetary Management

Question 2.
Who are the participants of Money Market?
Answer:

1. Government of different countries
2. Central Banks of different countries
3. Private and Public Banks
4. Mutual Funds Institutions
5. Insurance Companies
6. Non-Banking Financial Institutions
7. RBI and SBI
8. Commercial Banks
9. State Governments
10. Public

Question 3.
Explain the types of Treasury Bills.
Answer:
Treasury Bills are issued to the public and other financial institutions for meeting the short term financial requirements of the Central Government.
Treasury Bills may be classified into three!
They are:

1. 91 days Treasury Bills
2. 182 days Treasury Bills
3. 364 days Treasury Bills

Question 4.
What are the features of Certificate of Deposit?
Answer:

1. Document of title to time deposit
2. It is unsecured negotiable instruments
3. It is freely transferable by endorsement and delivery
4. It is issued at discount to face value
5. It is repayable on a fixed date without grace days

Question 5.
What are the types of Commercial Bill?
Answer:

1. Demand and Usance Bills
2. Clean bills and documentary Bills
3. Inland bills and Foreign Bills
4. Indigeneous Bills
5. Accommodation and supply Bills

IV. Long Answer Questions

Question 1.
Define Money Market and Capital Market. Explain the difference between the Money -Market and Capital Market.
Answer:
(i) Money Market Definition: According to Crowther, “the money market is the collective name given to the various firms and institutions that deal in the various grades of near money.”
(ii) Capital Market Definition: According to Aran K. Datta, capital market may be defined as “a complex of institutions investment and practices with established links between the demand for and supply of different types of capital gains”.

Question 2.
Explain the characteristics of Money Market.
Answer:

1. Short-term Funds: It is a market purely for short-term funds or financial assets called near money.
2. Maturity Period: It deals with financial assets having a maturity period upto one year only.
3. Conversion of Cash: It deals with only those assets which can be converted into cash readily without loss and with minimum transaction cost.
4. No Formal Place: Generally, transactions take place through phone, i.e., oral communication. Relevant documents and written communications can be exchanged subsequently.
5. Sub-markets: It is not a single homogeneous market. It comprises of several sub-markets each specialising in a particular type of financing.
6. Role of Market: The components of a money market are the Central Bank, Commercial Banks. Commercial banks generally play a dominant role in this market.
7. Highly Organized Banking System: The Commercial Banks are the nerve centre of the whole money market. They are the principal suppliers of short-term funds.
8. Existence of Secondary Market: There should be an active secondary market for these instruments.
9. Demand and Supply of Funds: There should be a large demand and supply of short-term funds.
10. Wholesale Market: It is a wholesale market and the volume of funds or financial assets traded in the market is very large.
11. Flexibility: Due to greater flexibility in the regulatory framework, there are constant endeavours for introducing new instruments.
12. Presence of a Central Bank: The central bank keeps their cash reserves and provides them financial accommodation in difficulties by discounting their eligible securities.

Question 3.
Explain the Instruments of Money Market.
Answer:

(i) Treasury Bills: Treasury bills are very popular and enjoy a higher degree of liquidity since they are issued by the Government. A Treasury bill is nothing but a promissory note issued for a specified period stated therein. The Government promises to pay the specified amount mentioned therein to the bearer of the instrument on the due date. The period does not exceed a period of one year.

(ii) Certificate of Deposits: Certificate of Deposits are short-term deposit instruments issued by banks and financial institutions to raise large sums of money. The Certificate of Deposit is transferable from one party to another. Due to their negotiable feature, they are also known as negotiable certificate of deposit.

(iii) Commercial Bills: The Commercial Bill is an instrument drawn by a seller of goods on a buyer of goods. It possesses the advantages like self-liquidating in nature, recourse to two parties, knowing exact date of transactions, transparency of transactions etc.,

Question 4.
Explain the features and types of Commercial Bills.
The features of the Commercial Bills are as follows:
Answer:

1. Drawer
2. Acceptor
3. Payee
4. Discounter
5. Endorser
6. Assessment
7. Maturity
8. Credit Rating

Types:

1. Demand and Usance Bills: A demand bill is one wherein no specific time of payment is mentioned. So, demand bills are payable immediately when they are presented to the drawee.
2. Clean Bills and Documentary Bills: Bills that are accompanied by documents of title to goods are called documentary bills. Clean bills are drawn without accompanying any document.
3. Inland Bills and Foreign Bills: Bills that are drawn and payable in India on a person who is resident in India are called inland bills.
4. Indigeneous Bills: The drawing and acceptance of indigenous bills are governed by native custom or usage of trade.
5. Accommodation and Supply Bills: Accommodation bills are those wrhich do not arise out of genuine trade of transactions.

Question 5.
What are the features of Government Securities?
Answer:

1. Agencies: Government securities are issued by agencies such as Central Government State Governments, semi-government authorities like local Government authorities.
2. RBI Special Role: RBI takes a special and an active role in the purchase and sale of these securities as part of its monetary management exercise.
3. Nature of Securities: Securities offer a safe avenue of investment through guaranteed payment of interest and repayment of principal by the Government.
4. Liquidity Profile: The liquidity profile of gilt-edged securities varies. Accordingly liquidity profile of securities issued by Central Government is high.
5. Tax Rebate: A striking feature of these securities is that they offer wide-range of tax incentives to investors.
6. Market: As each sale and purchase has to be negotiated separately, the Gilt-Edged Market is an Over-The-Counter Market.
7. Forms: The securities of Central and State Government take such forms as inscribed stock or stock certificate, promissory note and bearer bond.
8. Participants: The participants in Government securities market include the Government sector comprising Central and State Governments
9. Trading: Small and less active, banks and corporate holders who purchase and sell Government securities on the stock exchanges participate in trading.
10. Issue Mechanism: The Public Debt Office (PDO) of the RBI undertakes to issue government securities.
11. Issue opening: A notification for the issue of the securities is made a few days before the public subscription is open.
12. Grooming Gradual: It is the acquisition of securities nearing maturity through the stock exchanges by the RBI.
13. Switching: It is the purchase of one security against the sale of another security carried out by the RBI in the secondary market as part of its open market operations.
14. Auctioning: A method of trading whereby merchants bid against one another and where the securities are sold to the highest bidder.

Samacheer Kalvi 12th Commerce Money Market Additional Questions and Answers

I. A. Choose the Correct Answer

Question 1.
Money market is a market for purely ______
(a) Short-term funds
(b) Long-term funds
(c) Medium-term funds
(d) None of these
Answer:
(a) Short-term funds

Question 2
______ deals with the financial assets and securities whose maturity period does not exceed one year.
(a) Money market
(b) Capital market
(c) Stock exchange
(d) Government bonds
Answer:
(a) Money market

Question 5.
Treasury Bills is an example of ______
(a) Capital market
(b) Money market
(c) Stock exchange
(d) None of these
Answer:
(b) Money market

Question 4
______ is a focal point for Central Bank intervention for influencing liquidity in the company.
(a) Capital market
(b) Money market
(c) Stock exchange
(d) Central Bank
Answer:
(b) Money market

Question 5.
Which is dealt with only those assets which can be converted into cash readily?
(a) Capital market
(b) Money market
(c) Stock exchange
(d) Bank
Answer:
(b) Money market

Question 6.
The issuers of certificate of deposits are ______
(i) commercial banks
(ii) cooperative banks
(iii) private company
(iv) financial institutions
(a) (i) and (ii)
(b) (i) and (iii)
(c) (i) and (iv)
(d) (ii) and (iii)
Answer:
(c) (i) and (iv)

B. Fill in the blanks:

1. A market for the purchase and sale of Treasury Bills is known as _______
2. Bills which are drawn and payable in India on a person who is resident of India are called _______

Answers:

1. Treasury Bills Market
2. Inland bills

II. Very Short Answer Questions

Question 1.
What is Grooming Gradual?
Answer:
Acquisition of securities nearing maturity through the stock exchanges by the RBI in order to facilitate redemption is described as ‘grooming’.

Question 2.
What is Indigenous Bills ?
Answer:
The drawing and acceptance of indigenous bills are governed by native custom or usage of trade.

Question 3.
What is Liquidity Profile?
Answer:
The liquidity profile of gilt-edged securities varies. Accordingly liquidity profile of securities issued by Central Government is high.

Question 4.
Who are Participants?
Answer:
The participants in Government securities market include the Government sector comprising Central and State Governments whose holdings represent governmental transfer of resources.

Question 5.
What is Wholesale Market?
Answer:
It is a wholesale market and the volume of funds or financial assets traded in the market is very large.

Question 6.
Who will subscribe certificate of deposits?
Answer:
Certificate of deposits are subscribed by individuals, corporations, trusts, associations and NRIs. It is a document of title to a time deposit. .

Question 7.
What are the components of money market?
Answer:
The components of a money market are the Central Bank, Commercial Banks, Non-Banking Financial Companies, Discount Houses and Acceptance Houses.

III. Short Answer Questions

Question 1.
What is sub-markets?
Answer:
It is not a single homogeneous market. It comprises of several sub-markets each specialising in a particular type of financing. E.g, Call Money Market, Acceptance Market, Bill Market.

Question 2.
Write any three differences between money market and capital market.
Answer:

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Tamilnadu Samacheer Kalvi 12th Bio Zoology Solutions Chapter 4 Principles of Inheritance and Variation

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Samacheer Kalvi 12th Bio Zoology Principles of Inheritance and Variation Text Book Back Questions and Answers

Question 1.
Haemophilia is more common in males because it is a ____________
(a) Recessive character carried by Y-chromosome
(b) Dominant character carried by Y-chromosome
(c) Dominant trait carried by X-chromosome
(d) Recessive trait carried by X-chromosome
Answer:
(d) Recessive trait carried by X-chromosome

Question 2.
ABO blood group in man is controlled by ____________
(a) Multiple alleles
(b) Lethal genes
(c) Sex linked genes
(d) Y-linked genes
Answer:
(a) Multiple alleles

Question 3.
Three children of a family have blood groups A, AB and B. What could be the geno types of their parents?
(a) I^A I^B and ii
(b) I^A 1^O and I^B I^O
(c) I^B I^B and I^A I^A
(d) I^A I^A and ii
Answer:
(b) I^A I^O and I^B I^O

Question 4.
Which of the following is not correct?
(a) Three or more alleles of a trait in the population are called multiple alleles.
(b) A normal gene undergoes mutations to form many alleles
(c) Multiple alleles map at different loci of a chromosome
(d) A diploid organism has only two alleles out of many in the population
Answer:
(c) Multiple alleles map at different loci of a chromosome

Question 5.
Which of the following phenotypes in the progeny are ____________
(a) A and B only
(b) A,B only and AB
(c) AB only
(d) A,B,AB and O
Answer:
(d) A,B,AB and O

Question 6.
Which of the following phenotypes is not possible in the progeny of the parental genotypic combination I^AI^O × l^Al^B ?
(a) AB
(b) O
(c) A
(d) B
Answer:
(b) O

Question 7.
Which of the following is true about Rh factor in the offspring of a parental combination DdXDd (both Rh positive)?
(a) All will be Rh positive
(b) Half will be Rh positive
(c) About 3/4 will be Rh negative
(d) About one fourth will be Rh negative
Answer:
(d) About one fourth will be Rh negative

Question 8.
What can be the blood group of offspring when both parents have AB blood group?
(a) AB only
(b) A, B and AB
(c) A, B, AB and O
(d) A and B only
Answer:
(b) A, B and AB

Question 9.
If the childs blood group is ‘O’ and fathers blood group is ‘A’ and mother’s blood group is ‘B’ the genotype of the parents will be _________
(a) I^AI^A and I^AI^O
(b) I^AI^O and I^BI^O
(c) I^AI^O and I^OI^O
(d) I^OI^O and I^BI^B
Answer:
(b) I^AI^O and I^BI^O

Question 10.
XO type of sex determination and XY type of sex determination are examples of _________
(a) Male heterogamety
(b) Female heterogamety
(c) Male homogamety
(d) Both (b) and (c)
Answer:
(a) Male heterogamety

Question 11.
In an accident there is great loss of blood and there is no time to analyse the blood group which blood can be safely transferred?
(a) ‘O’ and Rh negative
(b) ‘O’ and Rh positive
(c) ‘B’ and Rh negative
(d) ‘AB’ and Rh positive
Answer:
(a) ‘O’ and Rh negative

Question 12.
Father of a child is colour blind and mother is carrier for colour blindness, the probability of the child being colour blind is _________
(a) 25%
(b) 50%
(c) 100%
(d) 75%
Answer:
(b) 50%

Question 13.
A marriage between a colour blind man and a normal woman produces _________
(a) All carrier daughters and normal sons
(b) 50% carrier daughters, 50% normal daughters
(c) 50% colour blind sons, 50% normal sons
(d) All carrier off springs
Answer:
(a) All carrier daughters and normal sons

Question 14.
Mangolism is a genetic disorder which is caused by the presence of an extra chromosome number.
(a) 20
(b) 21
(C) 4
(d) 23
Answer:
(b) 21

Question 15.
Klinefelters’ syndrome is characterized by a karyotype of _________
(a) XYY
(b) XO
(c) XXX
(d) XXY
Answer:
(d) XXY

Question 16.
Females with Turners’syndrome have _________
(a) Small uterus
(b) Rudimentary ovaries
(c) Underdeveloped breasts
(d) All of these
Answer:
(d) All of these

Question 17.
Pataus’ syndrome is also referred to as _________
(a) 13-Trisomy
(b) 18-Trisormy
(c) 21-Trisormy
(d) None of these
Answer:
(a) 13-Trisomy

Question 18.
Who is the founder of Modem Eugenics movement?
(a) Mendel
(b) Darwin
(c) Fransis Galton
(d) Karl pearson
Answer:
(c) Fransis Galton

Question 19.
Improvement of human race by encouraging the healthy persons to marry early and produce large number of children is called _________
(a) Positive eugenics
(b) Negative eugenics
(c) Positive euthenics
(d) Positive euphenics
Answer:
(a) Positive eugenics

Question 20.
The _________ deals with the control of several inherited human diseases especially inborn errors of metabolism.
(a) Euphenics
(b) Eugenics
(c) Euthenics
(d) All of these
Answer:
(a) Euphenics

Question 21.
“Universal Donor” and “Universal Recipients” blood group are _________ and _________ respectively.
(a) AB, O
(b) O, AB
(c) A, B
(d) B, A
Answer:
(b) O, AB

Question 22.
ZW-ZZ system of sex determination occurs in _________
(a) Fishes
(b) Reptiles
(c) Birds
(d) All of these
Answer:
(d) All of these

Question 23.
Co-dominant blood group is _________
(a) A
(b) AB
(c) B
(d) O
Answer:
(b) AB

Question 24.
Which of the following is incorrect regarding ZW-ZZ type of sex determination?
(a) It occurs in birds and some reptiles
(b) Females are homogametic and males are heterogametic
(c) Male produce two types of gametes
(d) It occurs in gypsy moth
Answer:
(b) Females are homogametic and males are heterogametic

Question 25.
What is haplodiploidy?
Answer:
In haplodiploidy, the sex of the offspring is determined by the number of sets of chromosomes it receives. Fertilized eggs develop into females (Queen or Worker) and unfertilized eggs develop into males (drones) by parthenogenesis. It means that the males have half the number of chromosomes (haploid) and the females have double the number (diploid).

Question 26.
Distinguish between heterogametic and homogametic sex determination systems
Answer:
Heterogametic Sex:

1. Organisms producing two different types of gametes.
2. Example : Human male.
3. Sperm with X chromosome Sperm with Y chromosome

Heterogametic Sex:

1. Organisms producing two different types of gametes.
2. Example: Human female.
3. Every egg produced contain X chromosomes.

Question 27.
What is Lyonisation?
Answer:
Lyonisation is a process of inactivation of one of the X chromosomes in some females.

Question 28.
What is criss-cross inheritance?
Answer:

1. Inheritance of genes from a male parent to female child and then to male grandchild or female parent to male child and then to female grandchild.
2. E.g., X-linked gene inheritance.

Question 29.
Why are sex linked recessive characters more common in the male human beings?
Answer:
Sex linked inherited traits are more common in males than females because, males are hemizygous and therefore express the trait when they inherit one mutant allele.

Question 30.
What are holandric genes?
Answer:
The genes present in the differential region of Y chromosome are called Y- linked or holandric genes. The Y linked genes have no corresponding allele in X chromosome.

Question 31.
Mention the symptoms of Phenylketonuria.
Answer:
Severe mental retardation, light pigmentation of skin and hair. Phenylpyruvic acid is excreted in the urine.

Question 32.
Mention the symptoms of Down’s syndrome.
Answer:
Severe mental retardation, defective development of the central nervous system, increased separation between the eyes, flattened nose, ears are malformed, mouth is constantly open and the tongue protrudes.

Question 33.
Differentiate Intersexes from Supersexes.
Answer:
Intersexes:

1. Intersexes refers to the individuals having the characteristics of both female and male sexes and their sexual anatomy doesnot seem to fit the typical definition of male or female.

Supersexes:

1. Supersexes ar formed as a result of abnormal combination of sex chromosomes.
2. Example : Super males in humans humanbeings have 44+XYY chromosomes.

Question 34.
Explain the genetic basis of ABO blood grouping in man.
Multiple allele inheritance of ABO blood groups
Answer:
Blood differs chemically from person to person. When two different incompatible blood types are mixed, agglutination (clumping together) of erythrocytes (RBC) occurs. The basis of these chemical differences is due to the presence of antigens (surface antigens) on the membrane of RBC and epithelial cells. Karl Landsteiner discovered two kinds of antigens called antigen ‘A’ and antigen ‘B’ on the surface of RBC’s of human blood. Based on the presence or absence of these antigens three kinds of blood groups, type ‘A’, type ‘B’, and type ‘O’ (universal donor) were recognized. The fourth and the rarest blood group ‘AB’ (universal recipient) was discovered in 1902 by two of Landsteiner’s students Von De Castelle and Sturli.

Bernstein in 1925 discovered that the inheritance of different blood groups in human beings is determined by a number of multiple allelic series. The three autosomal alleles located on chromosome 9 are concerned with the determination of blood group in any person. The gene controlling blood type has been labeled as ‘L’ (after the name of the discoverer, Landsteiner) or I (from isoagglutination). The I gene exists in three allelic forms, IA, IB and 1°. IA specifies A antigen. IB allele determines B antigen and 1° allele specifies no antigen. Individuals who possess these antigens in their fluids such as the saliva are called secretors.

Each allele (IA and IB) produces a transferase enzyme. IA allele produces N-acetyl galactose transferase and can add N-acetyl galactosamine (NAG) and IB allele encodes for the enzyme galactose transferase that adds galactose to the precursor (i.e. H substances). In the case of I^O/I^O allele no terminal transferase enzyme is produced and therefore called “null” allele and hence cannot add NAG or galactose to the precursor.

From the phenotypic combinations it is evident that the alleles I^A and I^B are dominant to 1°, but co-dominant to each other (I^A = I^B). Their dominance hierarchy can be given as (I^A=I^B> 1^O). A child receives one of three alleles from each parent, giving rise to six possible genotypes and four possible blood types (phenotypes). The genotypes are I^AI^A, I^AI^O, I^BI^B, I^BI^O, I^AI^B and I^O I^O

Question 35.
How is sex determined in human
Answer:

Genes determining sex in human beings are located on two sex chromosomes, called allosomes. In mammals, sex determination is associated with chromosomal differences between the two sexes, typically XX females and XY males. 23 pairs of human chromosomes include 22 pairs of autosomes (44A) and one pair of sex chromosomes (XX or XY). Females are homogametic producing only one type of gametes (egg), each containing one X chromosome while the males are heterogametic producing two types of sperms with X and Y chromosomes. An independently evolved XX: XY system of sex chromosomes also exist in Drosophila.

Question 36.
Explain male heterogamety.
Answer:
Male heterogamety (XY males) is a type of sex determination in which males produce two different types of gametes. For example, human males produce two kinds of sperms that is sperm with X-chromosome and sperms with Y-chromosome.

Question 37.
Brief about female heterogamety.
Answer:
Female heterogamety (ZO females) refers to the condition, where female produces two types of egg cells. Some with Z chromosome and some without Z chromosome.

Question 38.
Give an account of genetic control of Rh factor?
Genetic control of Rh factor
Answer:

Fisher and Race hypothesis: Rh factor involves three different pairs of alleles located on three different closely linked loci on the chromosome pair. This system is more commonly in use today, and uses the ‘Cde’ nomenclature. In the given figure, three pairs of Rh alleles (Cc, Dd and Ee) occur at 3 different loci on homologous chromosome pair-1.

The possible genotypes will be one C or c, one D or d, one E or e from each chromosome. For e.g. CDE/cde; CdE/cDe; cde/cde; CDe/CdE etc. All genotypes carrying a dominant ‘D’ allele will produce Rh+positive phenotype and double recessive genotype ‘dd’ will give rise to Rh negative phenotype.

Wiener Hypothesis:
Wiener proposed the existence of eight alleles (R¹, R², R⁰, R^Z, r, r¹, r¹¹, r^y) at a single Rh locus. All genotypes carrying a dominant ‘R allele’ (R¹, R² ,R⁰ ,R^z) will produce ‘Rh-positive’ ^phenotype and double recessive genotypes (rr, rr¹, rr¹¹, rr^y) will give rise to Rh-negative phenotype.

Question 39.
Explain the mode of sex determination in honeybees.
Haplodiploidv in Honeybees:
Answer:
In hymenopteran insects such as honeybees, ants and wasps, a mechanism of sex determination called haplodiploidy mechanism of sex determination is common. In this system, the sex of the offspring is determined by the number of sets of chromosomes it receives. Fertilized eggs develop into females (Queen or Worker) and unfertilized eggs develop into males (drones) by parthenogenesis. It means that the males have half the number of chromosomes (haploid) and the females have double the number (diploid), hence the name haplodiplody for this system of sex determination.

This mode of sex determination facilitates the evolution of sociality in which only one diploid female becomes a queen and lays the eggs for the colony. All other females which are diploid having developed from fertilized eggs help to raise the queen’s eggs and so contribute to the queen’s reproductive success and indirectly to their own, a phenomenon known as Kin Selection. The queen constructs their social environment by releasing a hormone that suppresses fertility of the workers.

Question 40.
Discuss the genic balance mechanism of sex determination with reference to Drosophila?
Answer:
XX-XY type (Lygaeus Type) sex determination is seen in Drosophila. The females are homogametic with XX chromosome, while the males are heterogametic with X and Y

chromosome. Homogametic females produce only one kind of egg, each with one X chromosome, while the heterogametic males produce two kinds of sperms some with X chromosome and some with Y chromosome.

The sex of the embryo depends on the fertilizing sperm. An egg fertilized by an ‘X’ bearing sperm produces a female, if fertilized by a ‘Y’ bearing sperm, a male is produced.

Question 41.
What are the applications of Karyotyping?
Answer:

1. Karyotyping helps in gender identification.
2. It is used to detect the chromosomal aberrations like deletion, duplication, translocation, non-disjunction of chromosomes.
3. It helps to identify the abnormalities of chromosomes like aneuploidy.
4. It is also used in predicting the evolutionary relationships between species.
5. Genetic diseases in human beings can be detected by this technique.

Question 42.
Explain the inheritance of sex linked characters in human being.
Answer:
Haemophilia is commonly known as bleeder’s disease, which is more common in men than women. This hereditary disease was first reported by John Cotto in 1803. Haemophilia is caused by a recessive X-linked gene. A person with a recessive gene for haemophilia lacks a normal clotting substance (thromboplastin) in blood, hence minor injuries cause continuous ’bleeding, leading to death. The females are carriers of the disease and would transmit the disease to 50% of their sons even if the male parent is normal. Haemophilia follows the characteristic criss-cross pattern of inheritaitce.

Question 43.
What is extra chromosomal inheritance? Explain with an example.
Answer:
The cytoplasmic extra nuclear genes have a characteristic pattern of inheritance which does not resemble genes of nuclear chromosomes and are known as Extrachromosomal/ Cytoplasmic inheritance.

Question 44.
Comment on the methods of Eugenics.
Answer:
Eugenics refers to the study of the possibility of improving the qualities of human population. Methods of Eugenics:

1. Sex-education in school and public forums.
2. Promoting the uses of contraception.
3. Compulsory sterilization for mentally retarded and criminals.
4. Egg donation.
5. Artificial insemination by donors.
6. Prenatal diagnosis of genetic disorders and performing MTP
7. Gene therapy
8. Cloning
9. Egg/sperm donation of healthy individuals.

Samacheer Kalvi 12th Bio Zoology Principles of Inheritance and Variation Additional Questions and Answers

1 – Mark Questions

Question 1.
If a colorblind female marries a normal male, their sons will be _________
(a) All normal visioned
(b) All color blinded
(c) One half normal visioned other half colorblind
(d) Three fourth colorblind one fourth normal
Answer:
(c) One half normal visioned other half colorblind

Question 2.
Excess hair growth on pinna is a feature noticed only in males because _________
(a) Males produce more testosterone
(b) gene responsible for the character is located in Y-chromosome
(c) Estrogen suppresses the character in females
(d) females act only as a carriers for this character
Answer:
(b) gene responsible for the character is located in Y-chromosome

Question 3.
ABO blood group is a classical example for _________
(a) Multiple allelism
(b) Pleotropism
(c) Incomplete dominance
(d) Polygenic mechanism
Answer:
(a) Multiple allelism

Question 4.
Unit of heredity is _________
(a) allele
(b) allelomorph
(d) gene
Answer:
(d) gene

Question 5.
Identify the proper dominance hierarchy.
(a) I^B = 1° > I^B
(b) I^A = I^B > 0
(c) I^O = I^B > 
(d) I^B = I^A > O
Answer:
(b) I^A = I^B > 0

Question 6.
Haemophilia is more common in human males than human females. The reason is due to ______
(а) X-linked dominant gene
(b) X-linked recessive gene
(c) Y-linked recessive gene
(d) Allosomal abnormality
Answer:
(b) X-linked recessive gene

Question 7.
Identify the correct statement.
(a) Homozygous sex chromosome (XX) produce males in Drosophila
(b) Homozygous sex chromosome (ZZ) determine female sex in birds
(c) Heterozygous sex chromosome (XO) determine male sex in grasshopper
(d) Heterozygous sex chromosome (ZW) determine male sex in gypsy moth
Answer:
(c) Heterozygous sex chromosome (XO) determine male sex in grasshopper

Question 8.
Which blood group doesnot possess antibodies?
(a) I^AI^B
(b) I^OI^O
(c) I^AO
(d) I^BI^B
Answer:
(a) I^AI^B

Question 9.
Assertion (A): On diagnosis, Ramu is reported to have under developed testis and gynaecomastia.
Reason (R): His karyotype reveals XXY condition.
(а) A is right but R is wrong
(b) R explains A
(c) Both A and R are wrong
(d) Both and R are right but R is not the correct explanation of A
Answer:
(b) R explains A

Question 10.
Pick out the odd man out.
(a) Klinefelter’s syndrome
(b) Turner’s syndrome
(c) Huntington’s chorea
(d) 13-Trisomy
Answer:
(c) Huntington’s chorea

Question 11.
Pick out the odd one out regarding Mendelian disorder.
(a) Thalassemia
(b) phenylketonuria
(c) Albinism
(d) Huntington’s chorea
Answer:
(d) Huntington’s chorea

Question 12.

Answer:
(a) A – iv, B – ii, D – i, D – iii

Question 13.
Identify the proper ratio of normal visioned individuals against colorblind individuals, if colorblind carrier female marries a normal male.
(a) 1 : 1
(b) 3 : 1
(c) 1 : 3
(d) All four are normal visioned
Answer:
(c) 1 : 3

Question 14.
Pick out the correct statement.
(i) Karyotyping helps in gender identification
(ii) Holandric genes are located on X-chromosome
(iii) Trisomy-21 is an allosomal abnormality
(iv) Cooley’s anaemia is an autosomal recessive disorder
(a) i, iii, iv are correct
(c) i and iv are correct
Answer:
(c) i and iv are correct

Question 15.
DOPA stands for _________
(a) 3,4 – dihydroxy phenyl acetate
(b) 3,4 – dihydroxy phenyle alanine
(c) 3,4 – dihydroxy phenyl asparate
(d) 3,4 – dihydroxy phenyle aldehyde
Answer:
(b) 3,4 – dihydroxy phenyle alanine

Question 16.
The type of antibody generated against Rh antigen is _________
(a) IgE
(b) IgG
(c) IgA
Answer:
(b) IgG

Question 17.
Which of the following symbol is used in pedigree analysis to represent unspecified sex?

Answer:

Question 18.
A colorblind man marries a woman with normal sight who has no history of color blindness in her family. What is the probability of their grandson being colorblind?
(a) 1/4
(b) 3/4
(c) 2/4
(d) 4/4
Answer:
(a) 1/4

Question 19.
Multiple alleles are located _________
(a) at different loci on homologous chromosome
(b) at same locus on homologous chromosome
(c) at different loci on non-homologous chromosome
(d) at different chromosomes
Answer:
(b) at same locus on homologous chromosome

Question 20.
Identify the incorrect statement regarding haplodiploidy.
(g) Haplodiploidy is noticed in honeybees and drosophila
(b) Unfertilized eggs develop into drones
(c) Fertilized eggs develop into queen and worker bees
(d) Males have half the total chromosomal number
Answer:
(a) Haplodiploidy is noticed in honeybees and drosophila

Question 21.
IA and IB genes of ABO blood group are _________
(a) Co-dominant
(b) Pleotropic
(c) Dominant and recessive
(d) Epistatic
Answer:
(a) Co-dominant

Question 22.
Which one of the following crosses show 3 : 1 ratio of normal visioned versus carrier blind?
(a) X^C X^C × X⁺Y
(b) X⁺ X^C × X^C Y
(c) X⁺X^C × X⁺Y
(d) X⁺X⁺× X^C Y
Answer:
(c) X⁺X^C × X⁺Y

2 – Mark Questions

Question 1.
Define multiple allelism.
Answer:
When three or more alleles of a gene that control a particular trait occupy the same locus on the homologous chromosome of an organism, they are called multiple alleles and their inheritance is called multiple allelism.

Question 2.
Name the discoverers of antigen A, B and AB.
Answer:
Antigens A and Antigen B was discovered by Karl Landsteiner. Antigen AB was discovered by Von De Castelle and Sturli.

Question 3.
What happens if type A blood is injected to a person having B blood group? Explain the reason.
Answer:
When two different incompatible blood types are mixed, agglutination (clumping together) of erythrocytes (RBC) occurs. The basis of these chemical differences is due to the presence of antigens (surface antigens) on the membrane of RBC and epithelial cells.

Question 4.
State the allelic forms of I gene and mention its chromosomal location.
Answer:
The I gene exists in three forms: IA, IB and I°. The alleles are located on chromosome 9.

Question 5.
Write the possible genotypes for a person having B-blood group.
Answer:
The possible genotypes of a B-blood group person are IBIB or IBI°.

Question 6.
State Wiener Hypothesis on Rh-factor.
Answer:
Wiener proposed the existence of eight alleles (R¹, R², R⁰, R^z, r, r¹, r¹¹, r^y) at a single Rh locus. All genotypes carrying a dominant ‘R allele’ (R¹, R² ,R⁰ ,R^z) will produce ‘Rh-positive’ phenotype and double recessive genotypes (rr, rr¹, rr¹¹, rr^y) will give rise to Rh-negative phenotype.

Question 7.
Distinguish between homogametic and heterogametic condition with example.
Answer:
Homogametic organism:

1. Organism producing only one type of gametes.
2. e.g. Human female (Only X)

Heterogametic organism:

1. Organism producing two different types of gametes.
2. e.g. Human Male (X and Y)

Question 8.
Name any four organism expressing ZW-ZZ type of sex determination.
Answer:
Gypsy moth, fishes, reptiles and birds.

Question 9.
Expand

1. SRY
2. TDF

Answer:

1. SRY – Sex Determining region
2. TDF – Testes Determining Factor

Question 10.
Define Barr body.
Answer:
In 1949, Barr and Bertram first observed a condensed body in the nerve cells of female cat which was absent in the male. This condensed body was called sex chromatin by them and was later referred as Barr body.

Question 11.
Based on Lyon’s hypothesis, mention the number of Barr bodies in XXY males, XO females.
Answer:

1. XXY males – One Barr body.
2. XO females – No Barr body.

Question 12.
State Lyon’s hypothesis.
Answer:
Lyon’s hypothesis states that in mammals the necessary dosage compensation is accomplished by the inactivation of one of the X chromosome in females so that both males and females have only one functional X chromosome per cell.

Mary Lyon suggested that Barr bodies represented an inactive chromosome, which in females becomes tightly coiled into a heterochromatin, a condensed and visible form of chromatin (Lyon’s hypothesis). The number of Barr bodies observed in cell was one less than the number of X-Chromosome. XO females have no Barr body, whereas XXY males have one Barr body.

Question 13.
Mention few X-linked inherited diseases.
Answer:
Red-green colour blindness or daltonism, haemophilia and Duchenne’s muscular dystrophy.

Question 14.
Define Karyotyping.
Answer:
Karyotyping is a technique through which a complete set of chromosomes is separated from a cell and the chromosomes are arranged in pairs. An idiogram refers to a diagrammatic representation of chromosomes.

Question 15.
Explain the inheritance pattern of Y-linked genes with example.
Answer:
Genes in the non-homologous region of the Y-chromosome are inherited directly from male to male. In humans, the Y-linked or holandric genes for hypertrichosis (excessive development of hairs on pinna of the ear) are transmitted directly from father to son, because males inherit the Y chromosome from the father. Female inherits only X chromosome from the father and are not affected.

Question 16.
Observe the symbol used in pedigree analysis and give the proper terms they represent.

Answer:

Question 17.
Write a brief note on pedigree analysis.
Answer:
Pedigree is a “family tree”, drawn with standard genetic symbols, showing the inheritance pathway for specific phenotypic characters. Pedigree analysis is the study of traits as they have appeared in a given family line for several past generations.

Question 18.
What do you mean by ‘Mendelian disorder’.
Answer:
Alteration or mutation in a single gene causes Mendelian disorders. These disorders are transmitted to the off springs on the same line as the Mendelian pattern of inheritance.
E.g., Thalassemia.

Question 19.
Name any four Mendelian disorders.
Answer:

1. Thalassemia
2. Albinism
3. sickle cell anaemia
4. Huntington’s chorea

Question 20.
What is the phenotype of

1. I^AI^O
2. I^OI^O

Answer:

1. I^AI^O – A blood group person
2. I^OI^O – O blood group person

Question 21.
On which chromosomes does HBA1 gene and HBB genes are located?
Answer:

1. HBA1 gene is located on chromosome 16.
2. HBB gene is located on chromosome 11.

Question 22.
Complete the equation.

Answer:
(a) A = Phenylalanine hydroxylase
(b) B = Tyrosinase

Question 23.
Write a note on Huntington’s chorea.
Answer:
Huntington’s chorea is inherited as an autosomal dominant lethal gene in man. It is characterized by involuntary jerking of the body and progressive degeneration of the nervous system, accompanied by gradual mental and physical deterioration. The patients with this disease usually die between the age of 35 and 40.

Question 24.
Comment on Trisomy-21.
Answer:
Trisomic condition of chromosome – 21 results in Down’s syndrome. It is characterized by severe mental retardation, defective development of the central nervous system, increased separation between the eyes, flattened nose, ears are malformed, mouth is constantly open and the tongue protrudes.

Question 25.
Mention the genetic makeup of Turner’s syndrome person and Klinefelter’s syndrome , person.
Answer:
Klinefelter’s syndrome – 44AA+XXY Turner’s syndrome – 44AA+XO

Question 26.
List out any four clinical symptoms of Klinefelter’s syndrome.
Answer:
Gynaecomastia, high pitched voice, under developed genetalia and tall with long limbs.

3 – Mark Questions

Question 27.
Write the types of sex-determination mechanisms does the following crosses as shown. Give an example for each.

1. Female XX with Male XO
2. Female ZW with Male ZZ

Answer:

1. Male heterogamety. e.g., Human beings.
2. Female heterogamety. e.g., Birds.

Question 28.
What are the enzymes encoded by the alleles I^A, I^B and I°?
Answer:
I^A allele produces N-acetyl galactose transferase and can add N-acetyl galactosamine (NAG) and I^B allele encodes for the enzyme galactose transferase that adds galactose to the precursor (i.e. H substances). In the case of I^O/I^O allele no terminal transferase enzyme is produced and therefore called “null” allele and hence cannot add NAG or galactose to the precursor.

Question 29.
Draw a tabular column representing various types of blood group in human beings, their phenotypes, genotypes, antigens and respective antibodies.
Answer:
Genetic basis of the human ABO blood groups:

Question 30.
Give an account on Rhesus factor.
Answer:
Rhesus or Rh – Factor: The Rh factor or Rh antigen is found on the surface of erythrocytes. It was discovered in 1940 by Karl Landsteiner and Alexander Wiener in the blood of rhesus monkey, Macaca rhesus and later in human beings. The term ‘Rh factor’ refers to “immunogenic D antigen of the Rh blood group system. An individual having D antigen are Rh D positive (Rh+) and those without D antigen are Rh D negative (Rh”)”. Rhesus factor in the blood is inherited as a dominant trait. Naturally occurring Anti D antibodies are absent in the plasma of any normal individual. However if an Rh” (Rh negative) person is exposed to Rh+ (Rh positive) blood cells (erythrocytes) for the first time, anti D antibodies are formed in the blood of that individual. On the other hand, when an Rh positive person receives Rh negative blood no effect is seen.

Question 31.
How Erythroblastosis foetalis can be prevented?
Answer:
If the mother is Rh negative and foetus is Rh positive, anti D antibodies should be administered to the mother at 28th and 34th week of gestation as a prophylactic measure. If the Rh negative mother delivers Rh positive child then anti D antibodies should be administered to the mother soon after delivery. This develops passive immunity and prevents the formation of anti D antibodies in the mothers blood by destroying the Rh foetal RBC before the mother’s immune system is sensitized. This has to be done whenever the woman attains pregnancy.

Question 32.
Explain XX-XO type of sex determination.
Answer:

XX-XO method of sex determination is seen in bugs, some insects such as cockroaches and grasshoppers. Pi The female with two X chromosomes are homogametic Gametes (XX) while the males with only one X chromosome are heterogametic (XO). The presence of an unpaired X chromosomes determines the male sex. The males PI Generation with unpaired ‘X’ chromosome produce two types of sperms, one half with X chromosome and other half without X chromosome. The sex of the offspring depends upon the sperm that fertilizes the egg.

Question 33.
Name the type of sex-determination mechanism of the following organisms.

1. Gypsy moth
2. Human beings
3. Butterflies

Answer:

1. Gypsy moth -ZW – ZZ type (ZW-females, ZZ – males)
2. Human beings – XX – XY type (XX-females, XY – males)
3. Butterflies – ZO – ZZ type (ZO-females, ZZ – males)

Question 34.
Complete the following cross.
AAZW × AAZZ (female) (male)
Answer:

Question 35.
Role of Y- chromosome is crucial for maleness – Justify.
Answer:
Current analysis of Y chromosomes has revealed numerous genes and regions with potential genetic function; some genes with or without homologous counterparts are seen on the X. Present at both ends of the Y chromosome are the pseudoautosomal regions (PARs) that are similar with regions on the X chromosome which synapse and recombine during meiosis. The remaining 95% of the Y chromosome is referred as the Non-combining Region of the Y (NRY).

The NRY is divided equally into functional genes (euchromatic) and non-functional genes (heterochromatic). Within the euchromatin regions, is a gene called Sex determining region Y (SRY). In humans, absence of Y chromosome inevitably leads to female development and this SRY gene is absent in X chromosome. The gene product of SRY is the testes determining factor (TDF) present in the adult male testis.

Question 36.
Color blindness is a perfect example for criss-cross of inheritance – Justify the statement.
Answer:
A marriage between a colour blind man and a normal visioned woman will produce normal visioned male and female individuals in F₁ generation but the females are carriers. The marriage between a F₁ normal visioned carrier woman and a normal visioned male will produce one normal visioned female, one carrier female, one normal visioned male and one colour blind male. Colour blind trait is inherited from the male parent to his grandson through carrier daughter, which is an example of criss-cross pattern of inheritance.
image 12

Question 37.
How the Karyotype of lymphocytes was prepared by Tjio and Levan?
Answer:
Preparation of Karyotype Tjio and Levan (1960) described a simple method of culturing lymphocytes from the human blood. Mitosis is induced followed by addition of colchicine to arrest cell division at metaphase stage and the suitable spread of metaphase chromosomes is photographed. The individual chromosomes are cut from the photograph and are arranged in an orderly fashion in homologous pairs. This arrangement is called a karyotype. Chromosome banding permits structural definitions and differentiation of chromosomes.

Question 38.
What is a genetic disorder? Mention its types?
Answer:
A genetic disorder is a disease or syndrome that is caused by an abnormality in an individual -DNA. Abnormalities can range from a small mutation in a single gene to the addition or subtraction of an entire chromosome or even a set of chromosomes. Genetic disorders are of two types namely, Mendelian disorders and chromosomal disorders.

Question 39.
Explain the genetic basis of Phenylketonuria.
Answer:
Phenylketonuria is an inborn error of Phenylalanine metabolism caused due to a pair of autosomal recessive genes. It is caused due to mutation in the gene PAH (phenylalanine hydroxylase gene) located on chromosome 12 for the hepatic enzyme “phenylalanine hydroxylase”.

This enzyme is essential for the conversion of phenylalanine to tyrosine. Affected individual lacks this enzyme, so phenylalanine accumulates and gets converted to phenylpyruvic acid and other derivatives. It is characterized by severe mental retardation, light pigmentation of skin and hair. Phenylpyruvic acid is excreted in the urine.

Question 40.
Give an account on Patau’s syndrome.
Answer:
Trisomic condition of chromosome 13 results in Patau’s syndrome. Meiotic non-disjunction is thought to be the cause for this chromosomal abnormality. It is characterized by multiple and severe body malformations as well as profound mental deficiency. Small head with small eyes, cleft palate, malformation of the brain and internal organs are some of the symptoms of this syndrome.

Question 41.
Define aneuploidy.
Answer:
Failure of chromatids to segregate during cell division resulting in the gain or loss of one or more chromosomes is called aneuploidy. It is caused by non-disjunction of chromosomes.

Question 42.
What do you mean by “syndrome”? Give two examples.
Answer:
Group of signs and symptoms that occur together and characterize a particular abnormality is called a syndrome, e.g., Down’s syndrome and Turner’s syndrome.

5 – Mark Questions

Question 42.
Explain in detail about Erythroblastosis foetalis.
Answer:
Rh incompatability has great significance in child birth. If a woman is Rh negative and the man is Rh positive, the foetus may be Rh positive having inherited the factor from its father. The Rh negative mother becomes sensitized by carrying Rh positive foetus within her body. Due to damage of blood vessels, during child birth, the mother’s immune system recognizes the Rh antigens and gets sensitized. The sensitized mother produces Rh antibodies. The antibodies are IgG type which are small and can cross placenta and enter the foetal circulation. By the time the mother gets sensitized and produce anti ‘D’ antibodies, the child is delivered.

Usually no effects are associated with exposure of the mother to Rh positive antigen during the first child birth, subsequent Rh positive children carried by the same mother, may be exposed to antibodies produced by the mother against Rh antigen, which are carried across the placenta into the foetal blood circulation. This causes haemolysis of foetal RBCs resulting in haemolytic jaundice and anaemia. This condition is known as Erythroblastosis foetalis or Haemolytic disease of the new bom (HDN).

Question 43.
Decribe female heterogamy and its types.
Answer:
Heterogametic Females:
In this method of sex determination, the homogametic male possesses two ‘X’ chromosomes as in certain insects and certain vertebrates like fishes, reptiles and birds producing a single type of gamete; while females produce dissimilar gametes. The female sex consists of a single ‘X’ chromosome or one ‘X’ and one ‘Y’ chromosome. Thus the females are heterogametic and produce two types of eggs. Heterogametic females are of two types, ZO-ZZ type and ZW-ZZ type.

ZO-ZZ Type:
This method of sex determination is seen in certain moths, butterflies and domestic chickens. In this type, the female possesses
single ‘Z’ chromosome in its body cells and is heterogametic (ZO) producing two kinds of eggs some with ‘Z’ chromosome and some without ‘Z’ chromosome, while the male possesses two ‘Z’ chromosomes and is homogametic (ZZ).

ZW-ZZ type:
This method of sex determination occurs in certain insects (gypsy moth) and in vertebrates such as fishes, reptiles and birds. In this method the female has one ‘Z’ and one ‘ W’ chromosome (ZW) producing two types of eggs, some carrying the Z chromosomes and some carry the W chromosome. The male sex has two ‘Z’ chromosomes and is homogametic (ZZ) producing a single type of sperm.

Question 44.
Write elaborately about the following Mendelian disorders.
Answer:
(a) Thalassemia
(b) Albinism
Answer:
(a) Thalassemia: Thalassemia is an autosomal recessive disorder. It is caused by gene mutation resulting in excessive destruction of RBC’s due to the formation of abnormal haemoglobin molecules. Normally haemoglobin is composed of four polypeptide chains, two alpha and two beta globin chains. Thalassemia patients have defects in either the alpha or beta globin chain causing the production of abnormal haemoglobin molecules resulting in anaemia.

Thalassemia is classified into alpha and beta based on which chain of haemoglobin molecule _ is affected. It is controlled by two closely linked genes HBA1 and HBA2 on chromosome 16. Mutation or deletion of one or more of the four alpha gene alleles causes Alpha Thalassemia. In Beta Thalassemia, production of beta globin chain is affected. It is controlled by a single gene (HBB) on chromosome 11. It is the most common type of Thalassemia and is also known as Cooley’s anaemia. In this disorder, the alpha chain production is increased and damages the membranes of RBC.

(b) Albinism: Albinism is an inborn error of metabolism, caused due to an autosomal recessive gene. Melanin pigment is responsible for skin colour. Absence of melanin results in a condition called albinism. A person with the recessive allele lacks the tyrosinase enzyme system, which is required for the conversion of dihydroxyphenyl alanine (DOPA) into melanin pigment inside the melanocytes. In an albino, melanocytes are present in normal numbers in their skin, hair, iris, etc., but lack melanin pigment.

Question 45.
Discuss any two Allosomal anomalies in human.
Allosomal abnormalities in human beings
Answer:
Mitotic or meiotic non-disjunction of sex chromosomes causes allosomal abnormalities. Several sex chromosomal abnormalities have been detected.
E.g. Klinefelter’s syndrome and Turner’s syndrome.

1. Klinefelter’s Syndrome (XXY Males): This genetic disorder is due to the presence of an additional copy of the X chromosome resulting in a karyotype of 47,XXY. Persons with this syndrome have 47 chromosomes (44AA+XXY). They are usually sterile males, tall, obese, with long limbs, high pitched voice, under developed genetalia and have feeble breast (gynaecomastia) development.

2. Turner’s Syndrome (XO Females): This genetic disorder is due to the loss of a X chromosome resulting in a karyotype of 45,X. Persons with this syndrome have 45 chromosomes (44 autosomes and one X chromosome) (44AA+XO) and are sterile females. Low stature, webbed neck, under developed breast, rudimentary gonads lack of menstrual cycle during puberty, are the main symptoms of this syndrome.

Higher Order Thinking Skills (HOTs) Questions

Question 1.
On analysis, a person’s karyotype reveals an extra one chromosome of twenty first pair. What does this condition represents? which type of symptoms can be noticed in the person?
Answer:

1. Trisomy-21 or Down’s syndrome.
2. Symptoms – Mental retardation, malformed ears, protruded tongue, mouth is constantly open etc.

Question 2.
A female whose blood group is AB- got conceived and later it is diagnoised that her – foetus possess B+. What measures would be taken to prevent the foetus from Haemolytic
disease of Newborn (HDN)
Answer:
If the mother is Rh negative and foetus is Rh positive, anti D antibodies should be administered to the mother at 28th and 34th week of gestation as a prophylactic measure. If the Rh negative mother delivers Rh positive child then anti D antibodies should be administered to the mother soon after delivery. This develops passive immunity and prevents the formation of anti D antibodies in the mothers blood by destroying the Rh foetal RBC before the mother’s immune system is sensitized. This has to be done whenever the woman attains pregnancy.

Question 3.
The following table shows the genotypes for ABO blood grouping and this phenotypes. Complete the table by filling the gaps.
Answer:

Question 4.
Give one example for each of the following group of drugs,

1. Stimulants
2. Analgesic
3. Hallucinogens

Answer:

1. Stimulants – Eg : Nicotine
2. Analgesic – Eg : Opium
3. Hallucinogens – Phencyclidine

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Tamilnadu Samacheer Kalvi 12th Bio Zoology Solutions Chapter 3 Reproductive Health

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Samacheer Kalvi 12th Bio Zoology Reproductive Health Text Book Back Questions and Answers

Question 1.
Which of the following is correct regarding HIV, hepatitis B, gonorrhoea and trichomoniasis?
(a) Gonorrhoea is a STD whereas others are not.
(b) Trichomoniasis is a viral disease whereas others are bacterial.
(c) HIV is a pathogen whereas others are diseases.
(d) Hepatitis B is eradicated completely whereas others are not.
Answer:
(c) HIV is a pathogen whereas others are diseases.

Question 2.
Which one of the following groups includes sexually transmitted diseases caused by bacteria . only?
(a) Syphilis, gonorrhoea and candidiasis
(b) Syphilis, chlamydiasis and gonorrhoea
(c) Syphilis, gonorrhoea and trichomoniasis
(d) Syphilis, trichomoniasis and pediculosis
Answer:
(b) Syphilis, chlamydiasis and gonorrhoea

Question 3.
Identify the correct statements from the following:
(a) Chlamydiasis is a viral disease.
(b) Gonorrhoea is caused by a spirochaete bacterium, Treponema palladium.
(c) The incubation period for syphilis is 2 to 14 days in males and 7 to 21 days in females.
(d) Both syphilis and gonorrhoea are easily cured with antibiotics.
Answer:
(d) Both syphilis and gonorrhoea are easily cured with antibiotics.

Question 4.
A contraceptive pill prevents ovulation by
(a) blocking fallopian tube
(b) inhibiting release of FSH and LH
(c) stimulating release of FSH and LH
(d) causing immediate degeneration of released ovum.
Answer:
(b) inhibiting release of FSH and LH

Question 5.
The approach which does not give the defined action of contraceptive is

Answer:
(b) Vasectomy – Prevents spermatogenesis

Question 6.
Read the given statements and select the correct option.
Statement 1: Diaphragms, cervical caps and vaults are made of rubber and are inserted into the female reproductive tract to cover the cervix before coitus.
Statement 2: They are chemical barriers of conception and are reusable.
(a) Both statements 1 and 2 are correct and statement 2 is the correct explanation of statement 1.
(b) Both statements 1 and 2 are correct but statement 2 is not the correct explanation of statement 1.
(c) Statement 1 is correct but statement 2 is incorrect.
(d) Both statements 1 and 2 are incorrect.
Answer:
(c) Statement 1 is correct but statement 2 is incorrect.

Question 7.
Match column I with column II and select the correct the correct option from option the code given below

Answer:
(d) A – (iv), B – (i), C – (ii), D – (iii)

Question 8.
Select the incorrect action of hormonal contraceptive pills from the following
(a) Inhibition of spermatogenesis.
(b) Inhibition of ovulation.
(c) Changes in cervical mucus impairing its ability to allow passage and transport of sperms.
(d) Alteration in uterine endometrium to make it unsuitable for implantation.
Answer:
(a) Inhibition of spermatogenesis.

Question 9.
What is amniocentesis? Why a statutory ban is imposed on this technique?
Answer:
Amniocentesis is a prenatal technique used to detect any chromosomal abnormalities in the foetus and it is being often misused to determine the sex of the foetus. Once the sex of the foetus is known, there may be a chance of female foeticide. Hence, a statutory ban on amniocentesis is imposed

Question 10.
Select the correct term from the bracket and complete the given branching tree

(Barriers, Lactational amonerrhoea, Tubectomy, CuT)
Answer:

Question 11.
Correct the following statements

1. Transfer of an ovum collected from donor into the fallopian tube is called ZIFT.
2. Transfering of an embryo with more than 8 blastomeres into uterus is called GIFT,
3. Multiload 375 is a hormone releasing IUD.

Answer:

1. Transfer of 8 celled blastomere collected from donor into the fallopian tube is called ZIFT.
2. Transfering of an embryo with more than 8 blastomeres into uterus is called IUT.
3. Multiload 375 is a copper releasing IUD.

Question 12.
Which method do you suggest the couple to have a baby, if the male partner fails to inseminate the female or due to very low sperm count in the ejaculate?
Answer:
Micro-testicular sperm extraction (TESE):
Microsurgical sperm retrieval from the testicle involves a small midline incision in the scrotum, through which one or both testicles can be seen. Under the microscope, the seminiferous tubules are dilated and small amount of testicular tissue in areas of active sperm production are removed and improved for sperm yield compared to traditional biopsy techniques.

Question 13.
Expand the following: (a) ZIFT (b) ICSI
Answer:
Zygote intra-fallopian transfer Intra-cytoplasmic sperm injection

Question 14.
What are the strategies to be implemented in India to attain total reproductive health?
Answer:

• Creating awareness and providing medical assistance to build a healthy society.
• Introducing sex education in schools to provide information about adolescence and adolescence related changes.
• Educating couples and those in the marriageable age groups about the available birth control methods and family planning norms.
• Creating awareness about care for pregnant women, post-natal care of mother and child and the importance of breast feeding.
• Encouraging and supporting governmental and non-governmental agencies to identify new methods and/or to improve upon the existing methods of birth control.

Question 15.
Differentiate foeticide and infanticide.
Answer:
Female foeticide refers to ‘aborting the female in the mother’s womb’.
Female infanticide is ‘killing the female child after her birth’.

Question 16.
Describe the major STDs and their symptoms.
Answer:

Cervical cancer:
Cervical cancer is caused by a sexually transmitted virus called Human Papilloma virus (HPV). HPV may cause abnormal growth of cervical cells or cervical dysplasia. The most common symptoms and signs of cervical cancer are pelvic pain, increased vaginal discharge and abnormal vaginal bleeding. The risk factors for cervical cancer include

• Having multiple sexual partners
• Prolonged use of contraceptive pills

Cervical cancer can be diagnosed by a Papanicolaou smear (PAP smear) combined with an HPV test. X-Ray, CT scan, MRI and a PET scan may also be used to determine the stage of cancer. The treatment options for cervical cancer include radiation therapy, surgery and chemotherapy.

Modem screening techniques can detect precancerous changes in the cervix. Therefore screening is recommended for women above 30 years once in a year. Cervical cancer can be prevented with vaccination. Primary prevention begins with HPV vaccination of girls aged 9-13 years, before they become sexually active. Modification in lifestyle can also help in preventing cervical cancer. Healthy diet, avoiding tobacco usage, preventing early marriages, practicing monogamy and regular exercise minimize the risk of cervical cancer.

Question 17.
How are STDs transmitted?
Answer:
Sexually transmitted diseases (STD) are called as Sexually transmitted infections (STI). Normally STI are transmitted from person to person during intimate sexual contact with an infected partner. Infections like Hepatitis-B and HIV are transmitted sexually as well as by sharing of infusion needles, surgical instruments, etc with infected people, blood transfusion or from infected mother to baby.

Question 18.
Write the preventive measures of STDs.
Answer:
Prevention of STDs:

1. Avoid sex with unknown partner/ multiple partners
2. use condoms
3. In case of doubt, consult a doctor for diagnosis and get complete treatment.

Question 19.
The procedure of GIFT involves the transfer of female gametes into the fallopain tube, can gametes be transferred to the uterus to achieve the same result? Explain.
Answer:
Gamete intra-fallopian transfer (GIFT): Transfer of an ovum collected from a donor into the fallopian tube. In this the eggs are collected from the ovaries and placed with the sperms in one of the fallopian tubes. The zygote travels toward the uterus and gets implanted in the inner lining of the uterus.

Question 20.
Amniocentesis, the foetal sex determination test, is banned in our country. Is it necessary? Comment.
Answer:
Amniocentesis is a prenatal technique used to detect any chromosomal abnormalities in the foetus and it is being often misused to determine the sex of the foetus. Once the sex of ‘ the foetus is known, there may be a chance of female foeticide. Hence, a statutory ban on amniocentesis is imposed.

Question 21.
Open Book Assessment ‘Healthy reproduction, legally checked birth control measures and proper family planning programmes are essential for the survival of mankind’ Justify.
Answer:
Reproductive health and proper family planning programmes are highly essential for survival of mankind. Reproductive health refers to the state of physical, psycological and social well being merely in absence of illness in all matter related to reproductive system and its proper functioning. Family planning has the following health and social benefits.

1. Protecting the health of women by reducing frequent pregnancies.
2. Reducing abortions.
3. Providing stable population growth.
4. Assuming the infants well being by providing adequate time lapse between successive pregnancies.

Samacheer Kalvi 12th Bio Zoology Reproductive Health Additional Questions and Answers

1 – Mark Questions

Question 1.
Which of the following is Not a natural contraceptive?
(a) Rhythm method
(b) Lactational amenorrhoea
(c) Progestasert
(d) Continuous abstinence
Answer:
(c) Progestasert

Question 2.
Identify the fungal STD(s)
(i) Trichomoniasis
(ii) Genital herps
(iii) Candidiasis
(a) Only (i)
(b) Only (z’zz)
(c) Only (zv)
Answer:
(A) Only (iii)

Question 3.
Match the following.

Answer:
(a) – iii, (b) – i, (c) – iv, (d) – ii

Question 4.
Pick out the incorrect statement regarding the character of an good contraceptive.
(a) It should be user friendly
(b) should not affect sexual drive
(c) side effects must be least
(d) should not be easily available
Answer:
(d) should not be easily available

Question 5.
Select the proper hormonal composition of oral contraceptive pills
(a) FSH & Prolactin (b) Prolactin & TSH
(c) TSH & FSH (d) FSH & LH
Answer:
(d) FSH & LH

Question 6.
Assertion (A): IUD’s are inserted in the ovary.
Reason (R): IUD’s Increases phagocytosis of the sperm.
(a) Both A and R are correct
(b) Both A and R are incorrect
(c) A is correct R is incorrect
(d) A is incorrect R is correct
Answer:
(d) A is incorrect R is correct

Question 7.
Identify the mismatched pair.

Answer:
(c) Candidias (iii) Albugo candida

Question 8.
In India, family planning programme was initiated in
(a) 1961
(b) 1981
(c) 1951
(d) 1971
Answer:
(c) 1951

Question 9.
Assertion (A): Amniocentesis helps to diagnose the chromosomal aberrations in foetus. Reason (R): Amniocentesis is legalized is our country.
(a) Both A and R are wrong
(b) A is right and R is wrong
(c) R explains A
(d) A is wrong R is right
Answer:
(b) A is right and R is wrong

Question 10.
Legalized marriageable age of female in India is
(a) 19 years
(b) 20 years
(c) 18 years
(d) 21 years
Answer:
(c) 18 years

Question 11.
Identify the correct statement.
(a) Lactational amenorrhea is a permanent birth control method
(b) Condoms are made of polyethylene glycol and lambskin
(c) LNG -20 is a copper – releasing IUD
(d) Diaphragm covers the cervix there by preventing sperm entry
Answer:
(d) Diaphragm covers the cervix there by preventing sperm entry

Question 12.
According to WHO, India is the largest HIV affected country.
(a) first
(b) second
(c) third
(d) seventh
Answer:
(e) third

Question 13.
Identify the correct statement.
(a) MTP is the voluntary killing of infant.
(b) MTP is legalized in India from 1974.
(c) Performing MTP during second trimester is more risky.
(d) It is a surgical – based abortion.
Answer:
(c) Performing MTP during second trimester is more risky.

Question 14.
Saheli, contains a non-steroidal preparation called
Answer:
Centchroman

Question 15.
The average foetal heart beat rate is between beats per minute.
Answer:
120-160

Question 16.
World AIDS day is observed on
Answer:
11th July

Question 17.
Indian Government legalized MTP in
Answer:
1971

Question 18.
In chorionic villus sampling test, the tissue sample is taken from
(a) amniotic fluid
(b) placental tissue
(c) Intestinal villi
(d) foetal liver
Answer:
(b) placental tissue

Question 19.
In ZIFT technique, the zygote is transferred at the stage of
(a) 16 blastomere
(b) morula
(c) 12 blastomere
(d) 8 blastomere
Answer:
(d) 8 blastomere

Question 20.
Given below are the basic steps in IVF treatment cycle. Select the proper sequence.
(i) Ovarian stimulation
(ii)Egg retrieval
(iii) fertilization
(iv) Embryo culture
(v) Embryo transfer
(a) (ii) → (ii) → (v) → (i) → (iii)
(b) (i) → (iii) → (ii) → (v) → (iv)
(c) (i) → (ii) → (ii) → (iv) → (v)
(d) (ii) → (i) → (iii) → (v) → (iv)
Answer:
(c) (i) → (ii) → (ii) → (iv) → (v)

Question 21.
Enactment of _____ banned the identification of sex and to prevent the prenatal abortion.
(a) POGSOAct
(b) POTAAct
(c) PCPNDTAct
(d) GOONDAAct
Answer:
(c) PCPNDT Act

Question 22.
Which is NOT a national health care programme?
(a) Pradhan Mantri Surakshit Matritva Abhiyan
(b) Pradhan Mantri Fiscal BhimaYojana
(c) RMNCH + A approach
(d) Janani Shishu Suraksha Karyakaram
Answer:
(b) Pradhan Mantri Fiscal Bhima Yoj ana

Question 23.
______ is Known as anti-sterility vitamin.
Answer:
Vitamin-E

2 – Mark Questions

Question 1.
Name any four national level health care programmes run by Indian Government.
Answer:

1. Janani Suraksha Yojana
2. Shishu Suraksha Karyakaram
3. RMNCH+A Programme
4. Pradhan Mantri Surakshit Matritva Abhiyan

Question 2.
Comment on ‘Saheli’
Answer:
Saheli is an oral contraceptive pill provided by Central Drug Research Institute in Lucknow, – India. It contains a non-steroidal hormone preparation called centchroman.

Question 3.
What is Mayer – Rokitansky Syndrome?
Answer:
All women are bom with ovaries, but some do not have functional uterus. This condition is called Mayer-Rokitansky syndrome.

Question 4.
Point out any four STD’s caused by viruses.
Answer:

1. Genital herpes
2. Hepatitis-B
3. Genital warts
4. AIDS

Question 5.
Define Surrogacy.
Answer:
Surrogacy is a method of assisted reproduction or agreement whereby a woman agrees to carry a pregnancy for another person, who will become the newborn child’s parent after birth. Through In Vitro Fertilization (IVF), embryos are created in a lab and are transferred into the surrogate mother’s uterus.

Question 6.
Name the causative organism of

1. Syphilis
2. Genital warts

Answer:

1. Syphilis is caused by Treponema palladium
2. Genital warts caused by Human Papilloma vims.

Question 7.
Define Infertility.
Answer:
Inability to conceive or produce children even after unprotected sexual cohabitation is called infertility. That is, the inability of a man to produce sufficient numbers or quality of sperm to impregnate a woman or inability of a woman to become pregnant or maintain a pregnancy.

Question 8.
Expand the acronyms:

1. GIFT
2. ICSI

Answer:

1. GIFT – Gamete Intra – Fallopian Transfer.
2. ICSI – Intra Cytoplasmic Sperm Injection.

Question 9.
What is amniocentesis?
Answer:
Amniocentesis involves taking a small sample of the amniotic fluid that surrounds the foetus to diagnose for chromosomal abnormalities.

Question 10.
How copper IUD’s provide contraception?
Answer:
Copper IUD’s release free copper and copper salts into the uterus and suppress the sperm motility. They remain in uterus for 5-10 years.
Eg: NovaT, Cu T-380.

Question 11.
What do you mean by the term – coitus interruptus?
Answer:
Coitus interruptus is a oldest family planning method, where the male partner withdraws his penis before ejaculation, there by preventing the deposition of semen into vagina.

Question 12.
What is MTP?
Answer:
Medical Termination of Pregnancy (MTP) : Medical method of abortion is a voluntary or intentional termination of pregnancy in a non-surgical or non-invasive way. Early medical termination is extremely safe upto 12 weeks (the first trimester) of pregnancy and generally ,has no impact on a women’s fertility. Abortion during the second trimester is more risky as the foetus becomes intimately associated with the maternal tissue.

Question 13.
Which type of women are benefited by In Vitro Fertilization?
Answer:
IVF is used to treat women with blocked, damaged or absent fallopian tubes.

3 – Mark Questions

Question 1.
What are the steps taken by Government to overcome population explosion?
Answer:
To overcome the problem of population explosion, birth control is the only available solution.
People should be motivated to have smaller families by using various contraceptive devices. Advertisements by the Government in the media as well as posters/bills, etc., with a slogan Naam iruvar namakku iruvar (we two, ours two) and Naam iruvar namakku oruvar (we two, ours one) have also motivated to control population growth in Tamil Nadu. Statutory rising of marriageable age of the female to 18 years and that of males to 21 years and incentives given to pouples with small families are the other measures taken to control population growth in our country.

Question 2.
Lactational amenorrhoea – Comment.
Answer:
Menstrual cycles resume as early as 6 to 8 weeks from parturition. However, the reappearance of normal ovarian cycles may be delayed for six months during breastfeeding. This delay in ovarian cycles is called lactational amenorrhoea. It serves as a natural, but an unreliable form of birth control.

Question 3.
Write a note on sterilization procedure in the male.
Answer:
Vasectomy is the surgical procedure for male sterilisation. In this procedure, both vas deferens are cut and tied through a small incision on the scrotum to prevent the entry of sperm into the urethra. Vasectomy prevents sperm from heading off to penis as the discharge has no sperms in it.

Question 4.
Define Tubectomy.
Answer:
Tubectomy is the surgical sterilisation in women. In this procedure, a small portion of both fallopian tubes are cut and tied up through a small incision in the abdomen or through vagina. This prevents fertilization as well as the entry of the egg into the uterus.

Question 5.
Complete the table by filling the gaps.

Answer:
A – Neisseria Gonorrhoea
B – AIDS
C- Jaundice

Question 6.
List various natural methods of birth control.
Answer:

1. Periodic abstinence
2. Coitus interruptus
3. Lactational amenorrhea

Question 7.
What does ICSI stands for? Describe the technique.
Answer:
ICSI stands for Intra-Cytoplasmic Sperm Injection. In this method, a sperm is carefully injected into the cytoplasm of the egg and the zygote formed is allowed to divide till 8 celled stage and transferred to uterus. Fertilization occurs in 75-85% of eggs injected with sperm.

Question 8.
How LNG-20 act as contraceptive?
Answer:
LNG-20 is a Intra Uterine System of contraceptive method. It increases the viscosity of cervical mucus and thereby preventing the sperms from entering the cervix.

Question 9.
MTP is legalized in our country. Yes or No? Why?
Answer:
Yes. Government oflndia legalized MTP in 1971 for medical necessity and social consequences ’with certain restrictions like sex discrimination and illegal female foeticides to avoid its misuse. MTP performed illegally by unqualified quacks is unsafe and could be fatal. MTP of the first conception may have serious psychological consequences

Question 10.
Write the hormonal composition of oral contraceptive pills. Also explain their action mode.
Answer:
Oral contraceptive pills are enriched with synthetic progesterone and oestrogen hormones. These pills prevent the ovulation by inhibiting the secretion of LH and FSH hormones.

Question 11.
Suggest some methods to help infertile couples to have children.
Answer:

1. Invitro fertilisation
2. Zygote intra-fallopian transfer (ZIFT)
3. Gamete intra-fallopian transfer (GIFT)
4. Surrogacy

Question 12.
What are the characteristics of a good contraceptive?
Answer:
An ideal contraceptive should be user friendly, easily available, with least side effects and should not interfere with sexual drive.

Question 13.
Write is a note on foetoscope.
Answer:
Foetoscope is used to monitor the foetal heart rate and other functions during late pregnancy and labour. The average foetal heart rate is between 120 and 160 beats per minute. An abnormal foetal heart rate or pattern may mean that the foetus is not getting enough oxygen and it indicates other problems. A hand-held doppler device is often used during prenatal visits to count the foetal heart rate. During labour, continuous electronic foetal monitoring is often used.

Question 14.
How the technique of amniocentesis is performed?
Answer:
Amniocentesis is generally performed in a pregnant woman between the 15th and 20th weeks of pregnancy by inserting a long, thin needle through the abdomen into the amniotic sac to withdraw a small sample of amniotic fluid. The amniotic fluid contains cells shed from the foetus.

Question 15.
Mention the role of prolactin in lactational amenorrhoea.
Answer:
Suckling by the baby during breast-feeding stimulates the pituitary to secrete increased prolactin hormone in order to increase milk production. This high prolactin concentration in the mother’s blood may prevent menstrual cycle by suppressing the release of GnRH (Gonadotropin Releasing Hormone) from hypothalamus and gonadotropin secretion from the pituitary.

Question 16.
Name any one

1. fungal STI
2. Bacterial STI
3. Protozoan STI

Answer:

1. Fungal STI – Candidiasis
2. Bacterial STI -Gonorrhoea
3. Protozoan STI – Trichomoniasis

Question 17.
Why ultrasonography is performed for a carrying women?
Answer:
Ultrasonography is usually performed in the first trimester for dating, determination of the number of foetuses, and for assessment of early pregnancy complications.

Question 18.
Suggest any two simple precautions to avoid contracting RTFs.
Answer:

1. Avoiding coitus with unknown/multiple partners.
2. Use of condoms during coitus.

Question 19.
Name the most effective long-acting reversible contraceptive methods.
Answer:
Intrauterine devices and contraceptive implants.

5 – Mark Question

Question 1.
Give an detailed account on various natural methods of contraception.
Answer:
Natural method is used to prevent meeting of sperm with ovum, i.e., Rhythm method (safe period), coitus interruptus, continuous abstinence and lactational amenorrhoea.

a. Periodic abstinence/rhythm method: Ovulation occurs at about the 14th day of the menstrual cycle. Ovum survives for about two days and sperm remains alive for about 72 hours in the female reproductive tract. Coitus is to be avoided during this time.

b. Continuous abstinence is the simplest and most reliable way to avoid pregnancy is not to have coitus for a defined period that facilitates conception.

c. Coitus interruptus is the oldest family planning method. The male partner withdraws his penis before ejaculation, thereby preventing deposition of semen into the vagina.

d. Lactational amenorrhoea : Menstrual cycles resume as early as 6 to 8 weeks from parturition. However, the reappearance of normal ovarian cycles may be delayed for six months during breast-feeding. This delay in ovarian cycles is called lactational amenorrhoea. It serves as a natural, but an unreliable form of birth control.

Suckling by the baby during breast-feeding stimulates the pituitary to secrete increased prolactin hormone in order to increase milk production. This high prolactin concentration in the mother’s blood may prevent menstrual cycle by suppressing the release of GnRH (Gonadotropin Releasing Hormone) from hypothalamus and gonadotropin secretion from the pituitary.

Question 2.
What are IUD’s? Explain its way of functioning. Also describe their types.
Answer:
Intrauterine Devices (IUDs) are inserted by medical experts in the uterus through the vagina. These devices are available as copper releasing IUDs, hormone releasing IUDs and non- medicated IUDs. IUDs increase phagocytosis of sperm within the uterus. IUDs are the ideal contraceptives for females who want to delay pregnancy.

It is one of the popular methods of contraception in India and has a success rate of 95 to 99%. Copper releasing IUDs differ from each other by the amount of copper. Copper IUDs such as Cu T-380 A, Nova T, Cu 7, Cu T 380 Ag, Multiload 375, etc. release free copper and copper salts into the uterus and suppress sperm motility. They can remain in the uterus for five to ten years. Hormone-releasing IUDs such as Progestasert and LNG – 20 are often called as intrauterine systems (IUS). They increase the viscosity of the cervical mucus and thereby prevent sperms from entering the cervix. Non-medicated IUDs are made of plastic or stainless steel. Lippes loop is a double S-shaped plastic device.

Question 3.
Write in detail about cervical cancer.
Answer:
Cervical cancer is caused by a sexually transmitted virus called Human Papilloma virus (HPV). HPV may cause abnormal growth of cervical cells or cervical dysplasia.
The most common symptoms and signs of cervical cancer are pelvic pain, increased vaginal discharge and abnormal vaginal bleeding. The risk factors for cervical cancer include

1. Having multiple sexual partners
2. Prolonged use of contraceptive pills

Cervical cancer can be diagnosed by a Papanicolaou smear (PAP smear) combined with an HPV test. X-Ray, CT scan, MRI and a PET scan may also be used to determine the stage of cancer. The treatment options for cervical cancer include radiation therapy, surgery and chemotherapy. Modem screening techniques can detect precancerous changes in the cervix. Therefore screening is recommended for women above 30 years once in a year.

Cervical cancer can be prevented with vaccination. Primary prevention begins with HPV vaccination of girls aged 9-13 years, before they become sexually active. Modification in lifestyle can also help in preventing cervical cancer. Healthy diet, avoiding tobacco usage, preventing early marriages, practicing monogamy and regular exercise minimize the risk of cervical cancer.

Question 4.
List out the causes of fertility in human.
Answer:
Causes for male infertility:

1. Undescended tests and swollen veins in scrotum
2. Underdeveloped testes
3. Tight clothing increases the temperature in scrotum and affect sperm production.
4. Autoimmune response against own sperm.
5. Usage of alcohol, tobacco, marijuana drugs etc.

Causes for female infertility:

1. Malformation of cervix or fallopian tubes.
2. Inadequate nutrition at puberty.
3. Low body fat (anorexia = psychological eating disorder due to fear of gaining weight)
4. Pelvic Inflammatory Disease (PID), uterine disorders, endometriosis.
5. underdeveloped ovaries.
6. Developing antibodies against sperm.

Common Causes for both the sexes:

1. Tumours in pituitary or sex organs
2. Inherited mutation of hormone synthesizing genes
3. Long term stress
4. Ingestion of toxins (Cadmium) & drugs
5. Injuries to gonads
6. Ageing

Higher Order Thinking Skills (HOTs) Questions

Question 1.
Dr. Sheela is a famous Gynaecologist at Poes garden. However illegally she performed amniocentesis for several pregnant illiterates and did MTP if identified as female foetus, on request. Under which act will she get arrested, if a complaint is filed against her.
Answer:
Dr. Sheela will be be arrested under PCPNDT Act – preconception and prenatal diagnostic technique Act-1994.

Question 2.
Identify the picture and describe the surgical procedure.
Answer:
The picture describes tubectomy.

1. Tubectomy is the surgical sterilization in woman.
2. In this a portion of oviduct is cut and tied through vagina or minor incision in abdomen.
3. It prevents fertilization and also the entry of egg to uterus.

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Tamilnadu Samacheer Kalvi 12th Bio Zoology Solutions Chapter 5 Molecular Genetics

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Samacheer Kalvi 12th Bio Zoology Molecular Genetics Text Book Back Questions and Answers

Question 1.
Hershey and Chase experiment with bacteriophage showed that
(a) Protein gets into the bacterial cells
(b) DNA is the genetic material
(c) DNA contains radioactive sulphur
(d) Viruses undergo transformation
Answer:
(b) DNA is the genetic material

Question 2.
DNA and RNA are similar with respect to
(a) Thymine as a nitrogen base
(b) A single-stranded helix shape
(c) Nucleotide containing sugars, nitrogen bases and phosphates
(d) The same sequence of nucleotides for the amino acid phenyl alanine
Answer:
(c) Nucleotide containing sugars, nitrogen bases and phosphates

Question 3.
A mRNA molecule is produced by
(a) Replication
(b) Transcription
(c) Duplication
(d) Translation
Answer:
(b) Transcription

Question 4.
The total number of nitrogenous bases in human genome is estimated to be about
(a) 3.5 million
(b) 35000
(c) 35 million
(d) 3.1 billion
Answer:
(d) 3.1 billion

Question 5.
E. coli cell grown on 15N medium are transferred to 14N medium and allowed to grow for two generations. DNA extracted from these cells is ultracentrifuged in a cesium chloride density gradient. What density distribution of DNA would you expect in this experiment?
(a) One high and one low density band
(b) One intermediate density band
(c) One high and one intermediate density band
(d) One low and one intermediate density band
Answer:
(d) One low and one intermediate density band

Question 6.
What is the basis for the difference in the synthesis of the leading and lagging strand of DNA molecules?
(a) Origin of replication occurs only at the 5’ end of the molecules
(b) DNA ligase works only in the 3’ → 5’ direction
(c) DNA polymerase can join new nucleotides only to the 3 ’ end of the growing stand
(d) Helicases and single-strand binding proteins that work at the 5’ end
Answer:
(d) Helicases and single-strand binding proteins that work at the 5’ end

Question 7.
Which of the following is the correct sequence of event with reference to the central dogma?
(a) Transcription, Translation, Replication
(b) Transcription, Replication, Translation
(c) Duplication, Translation, Transcription
(d) Replication, Transcription, Translation
Answer:
(d) Replication, Transcription, Translation

Question 8.
Which of the following statements about DNA replication is not correct?
(a) Unwinding of DNA molecule occurs as hydrogen bonds break
(b) Replication occurs as each base is paired with another exactly like it
(c) Process is known as semi – conservative replication because one old strand is conserved in the new molecule
(d) Complementary base pairs are held together with hydrogen bonds
Answer:
(b) Replication occurs as each base is paired with another exactly like it

Question 9.
Which of the following statements is not true about DNA replication in eukaryotes?
(a)) Replication begins at a single origin of replication.
(b) Replication is bidirectional from the origins.
(c) Replication occurs at about 1 million base pairs per minute.
(d) There are numerous different bacterial chromosomes, with replication occurring in each at the same time.
Answer:
(d) There are numerous different bacterial chromosomes, with replication occurring in each at the same time.

Question 10.
The first codon to be deciphered was which codes for
(a) AAA, proline
(b) GGG, alanine
(c) UUU, Phenylalanine
(d) TTT, arginine
Answer:
(c) UUU, Phenylalanine

Question 11.
Meselson and Stahl’s experiment proved __________
(a) Transduction
(b) Transformation
(c) DNA is the genetic material
(d) Semi-conservative nature of DNA replication
Answer:
(d) Semi-conservative nature of DNA replication

Question 12.
Ribosomes are composed of two subunits; the smaller subunit of a ribosome has a binding site for and the larger subunit has two binding sites for two
Answer:
mRNA, tRNA

Question 13.
Anoperonisa:
(a) Protein that suppresses gene expression
(b) Protein that accelerates gene expression
(c) Cluster of structural genes with related function
(d) Gene that switched other genes on or off
Answer:
(d) Gene that switched other genes on or off

Question 14.
When lactose is present in the culture medium:
(a) Transcription of lacy, lac z, lac a genes occurs
(b) Repressor is unable to bind to the operator
(c) Repressor is able to bind to the operator
(d) Both (a) and (b) are correct
Answer:
(d) Both (a) and (b) are correct

Question 15.
Give reasons: Genetic code is ‘universal’.
Answer:
The genetic code is universal. It means that all known living systems use nucleic acids and the same three base codons (triplet codon) direct the synthesis of protein from amino acids. For example, the mRNA (UUU) codon codes for phenylalanine in all cells of all organisms. Some exceptions are reported in prokaryotic, mitochondrial and chloroplast genomes. However similarities are more common than differences.

Question 16.
Name the parts marked ‘A’ and ‘B’ in the given transcription unit:
Answer:

Question 17.
Differentiate – Leading strand and lagging strand
Answer:

1. DNA polymerase I Involved DNA repair mechanism
2. DNA polymerase II Involved DNA repair mechanism
3. DNA polymerase III Involved in DNA replicaton

Question 18.
Differentiate – Template strand and coding strand.
Answer:

1. Template Strand: During replication, DNA strand having the polarity 3’ → 5’ act as template strand.
2. Coding Strand: During replication, DNA strand having the polarity 5’ → 3’ act as coding strand.

Question 19.
Mention any two ways in which single nucleotide polymorphism (SNPs) identified in human genome can bring revolutionary change in biological and medical science.
Answer:
Scientists have identified about 1.4 million locations, where single base DNA differences (SNPs – Single nucleotide polymorphism – pronounced as ‘snips’) occur in humans. Identification of ‘SNIPS’ is helpful in finding chromosomal locations for disease associated sequences and tracing human history.

Question 20.
State any three goals of the human genome project.
Answer:

1. Identify all the genes (approximately 30000) in human DNA.
2. Determine the sequence of the three billion chemical base pairs that makeup the human DNA.
3. To store this information in databases.

Question 21.
In E.coli, three enzymes 0- galactosidase, permease and transacetylase are produced in the presence of lactose. Explain why the enzymes are not synthesized in the absence of lactose.
Answer:
In the absence of lactose, the repressor protein binds to the operator and prevents the transcription of structural gene by RNA polymerase, hence the enzymes are not produced. However, there will always be a minimal level of lac operon expression even in absence of lactose.

Question 22.
Distinguish between structural gene, regulatory gene and operator gene.
Answer:
Structure of the operon: Each operon is a unit of gene expression and regulation and consists of one or more structural genes and an adjacent operator gene that controls transcriptional, activity of the structural gene.

1. The structural gene codes for proteins, rRNA and tRNA required by the cell.
2. Promoters are the signal sequences in DNA that initiate RNA synthesis. RNA polymerase binds to the promoter prior to the initiation of transcription.
3. The operators are present between the promoters and structural genes. The repressor protein binds to the operator region of the operon.

Question 23.
A low level of expression of lac operon occurs at all the time in E-coli. Justify the statement.
Answer:
One of the enzyme synthesized by lac operon is permease which is involved in the transport of lactose into the cells. If the lac operon gets inactivated, permease is not synthesized hence lactose cannot enter the cell. Lactose acts as a inducer, binding to the repressor protein and switch on the operator to initiate gene expression.

Question 24.
Why the human genome project is called a mega project?
Answer:
The international human genome project was launched in the year 1990. It was a mega project and took 13 years to complete. The human genome is about 25 times larger than the genome of any organism sequenced to date and is the first vertebrate genome to be completed. Human genome is said to have approximately 3 >109 bp. HGP was closely associated with the rapid development of a new area in biology called bioinformatics.

Question 25.
From their examination of the structure of DNA, What did Watson and Crick infer about the probable mechanism of DNA replication, coding capability and mutation?
Answer:
Inference of Watson and Crick on DNA replication: They concluded that each of the DNA strand in a helix act as template during DNA replication leading to formation of new daughter DNA molecules, which are complementary to parental strand, (i.e., Semi-conservative method of replication) Inference on coding capability: During transcription, the genetic information in the DNA strand is coded to mRNA as complementary bases, (except for uracil in place of thymine in RNA) Inference on mutation: Any changes in the nucleotide sequence of DNA leads to corresponding alteration in aminoacid sequence of specific protein thus confirming the validity of genetic code.

Question 26.
Why tRNA is called an adapter molecule?
Answer:
The transfer RNA, (tRNA) molecule of a cell acts as a vehicle that picks up the amino acids scattered through the cytoplasm and also reads specific codes of mRNA molecules. Hence it is called an adapter molecule. This term was postulated by Francis Crick.

Question 27.
What are the three structural differences between RNA and DNA?
Answer:
DNA:

1. Sugar is deoxyribose sugar.
2. Double stranded structure.
3. Nitrogen bases are Adenine, Guanine, Cytosine and Thymine.

RNA:

1. Sugar is ribose sugar.
2. Single stranded molecule.
3. Nitrogen bases are Adenine, Guanine, Cytosine and Uracil.

Question 28.
Name the anticodon required to recognize the following codons:
AAU, CGA, UAU, and GCA.
Answer:
UUA, GCU, AUA and CGU.

Question 29.
(a) Identify the figure given below
(b) Redraw the structure as a replicating fork and label the parts
(c) Write the source of energy for this replication and name the enzyme involved in this process.
(d) Mention the differences in the synthesis of protein, based on the polarity of the two template strands.
Answer:
(a) Replication fork
(b)

(c) Deoxy nucleotide, triphosphate acts as a energy source for replication. DNA polymerase is used for replication
(d) mRNA contacting information for protein synthesis will developed from DNA strand having polariy 5’ → 3’

Question 30.
If the coding sequence in a transcription unit is written as follows:
5’ TGCATGCATGCATGCATGCATGCATGC 3’
Write down the sequence of mRNA.
Answer:
mRNA sequence is 3’ACGUACGUACGUUCGUACGUACGUACG5’

Question 31.
How is the two stage process of protein synthesis advantageous?
Answer:
The split gene feature of eukaryotic genes is almost entirely absent in prokaryotes. Originally each exon may have coded for a single polypeptide chain with a specific function. Since exon arrangement and intron removal are flexible, the exon coding for these polypeptide subunits act as domains combining in various ways to form new genes. Single genes can produce different functional proteins by arranging their exons in several different ways through alternate splicing patterns, a mechanism known to play an important role in generating both protein and functional diversity in animals. Introns would have arosen before or after the evolution of eukaryotic gene.

If introns arose late how did they enter eukaryotic gene? Introns are mobile DNA sequences that can splice themselves out of, as well as into, specific ‘target sites’ acting like mobile transposon-like elements (that mediate transfer of genes between organisms – Horizontal Gene Transfer – HGT). HGT occurs between lineages of prokaryotic cells, or from prokaryotic to eukaryotic cells and between eukaryotic cells. HGT is now hypothesized to have played a major role in the evolution of life on Earth.

Question 32.
Why did Hershey and Chase use radioactively labelled phosphorous and sulphur only? Would they have got the same result if they use radiolabelled carbon and nitrogen?
Answer:
Generally proteins contain sulphur but not phosphorous and nucleic acid (DNA) contains , phosphorous but not sulphur. Hence Hershey – Chase used radioactive isotopes of sulphur (³⁵S) and phosphorus (³²P) to keep separate track of viral protein and nucleic acid in culture medium. The expected result cannot be achieved, if radioactive carbon and nitrogen is used, since these molecules are present in both DNA and proteins.

Question 33.
Explain the formation of a nucleosome.
Answer:
Komberg proposed a model for the nucleosome, in which 2 molecules of the four histone proteins H2A, H2B, H3 and H4 are organized to form a unit of eight molecules called histone octamere.

The negatively charged DNA is wrapped around the positively charged histone octamere to form a structure called nucleosome. A typical nucleosome contains 200 bp of DNA helix. The histone octameres are in close contact and DNA is coiled on the outside of nucleosome.

Question 34.
It is established that RNA is the first genetic material. Justify giving reasons.
Answer:
Three molecular biologists in the early 1980’s (Leslie Orgel, Francis Brick and Carl Woese) independently proposed the ‘RNA world’ as the first stage in the evolution of life, a stage when RNA catalysed all molecules necessary for survival and replication. The term ‘RNA world’ first used by Walter Gilbert in 1986, hypothesizes RNA as the first genetic on Earth. There is now enough evidence to suggest that essential life processes (such as metabolism, translation and splicing etc.,) evolved around RNA. RNA has the ability to act as both genetic material and catalyst. There are several biochemical reactions in living systems that are catalysed by RNA. This catalytic RNA is known as ribozyme. But, RNA being a catalyst was reactive and hence unstable.

This led to evolution of a more stable form of DNA, with certain chemical modifications. Since DNA is a double stranded molecule having complementary strand, it has resisted changes by evolving a process of repair. Some RNA molecules function as gene regulators by binding to DNA and affect gene expression. Some viruses use RNA as the genetic material. Andrew Fire and Craig Mellow (recipients of Nobel Prize in 2006) were of the opinion that RNA is an active ingredient in the chemistry of life.

Samacheer Kalvi 12th Bio Zoology Molecular Genetics Additional Questions and Answers

1 – Mark Questions

Question 1.
The term‘gene’was coined by ___________
Answer:
Wilhelm Johannsen

Question 2.
Whose experiment finally provided convincing evidence that DNA is the genetic material?
(a) Griffith experiment
(b) Avery, Macleod and McCarty’s experiment
(c) Hershey-Chase experiment
(d) Urey-Miller’s experiment
Answer:
(c) Hershey-Chase experiment

Question 3.
In Hershey – Chase experiment, the DNA of T² phase was made radioactive by using ___________
(a) ³²P
(b) ³²S
(c) ³⁵P
(d) ³²S
Answer:
(a) ³²P

Question 4.
A nucleoside is composed of ___________
(a) Sugar and Phosphate
(b) Nitrogen base and Phosphate
(c) Sugar and Nitrogen base
(d) Sugar, Phosphate and Nitrogenous base
Answer:
(c) Sugar and Nitrogen base

Question 5.
Identify the incorrect statement
(a) a base is a substance that accepts H+ ion
(b) Both DNA and RNA have four bases
(c) Purines have single carbon-nitrogen ring
(d) Thymine is unique for DNA
Answer:
(c) Purines have single carbon-nitrogen ring

Question 6.
Watson and Crick proposed their double helical DNA model based on the X-ray diffraction analysis of ___________
(a) Erwin Chargaff
(b) Meselson and Stahl
(c) Wilkins and Franklin
(d) Griffith
Answer:
(c) Wilkins and Franklin

Question 7.
The term ‘RNA world’ was first used by ___________
Answer:
Walter Gilbert

Question 8.
The distance between two consecutive base pairs in DNA is ___________
(a) 0.34 nm
(b) 3.4 nm
(c) 0.034 nm
(d) 34 nm
Answer:
(a) 0.34 nm

Question 9.
If the length of E. coli DNA is 1.36 mm, the number of base pairs is ___________
(a) 0.36 × 10⁶m
(b) 4 × 10⁶m
(c) 0.34 × 10^-9nm
(d) 4 × 10^-9m
Answer:
(b) 4 × 10⁶m

Question 10.
Identify the proper sequence in the organisation of eukaryotic chromosome.
(a) Nucleosome – Solenoid – Chromatid
(b) Chromatid – Nucleosome – Solenoid
(c) Solenoid – chromatin – DNA
(d) Nucleosome – solenoid – genophore
Answer:
(a) Nucleosome – Solenoid – Chromatid

Question 11.
Assertion (A) : Genophore is noticed in prokaryotes.
Reason (R) : Bacteria possess circular DNA without chromatin organisation.
(a) Both A and R are correct
(b) A is correct R is incorrect
(c) R explains A
(d) A is incorrect R is correct
Answer:
(c) R explains A

Question 12.
Assertion (A): Heterochromatin is transcriptionally active.
Reason (R): Tightly packed chromatin which stains dark.
(a) Both A and R are correct
(b) A is correct R is incorrect
(c) R explains A
(d) A is incorrect R is correct
Answer:
(d) A is incorrect R is correct

Question 13.
Assertion (A) : Semi-conservative model was proposed by Hershey and Chase.
Reason (R) : The daughter DNA contains only new strands.
(a) Both A and R are incorrect
(b) A is correct R is incorrect
(c) R explains A
(d) A is incorrect R is correct
Answer:
(a) Both A and R are incorrect

Question 14.
Komberg enzyme is called as _____
Answer:
DNA polymerase I

Question 15.
Replication of DNA occurs at __________ phase of cell cycle.
(a) M
(b) S
(c) G₁
(d) G₂
Answer:
(b) S

Question 16.
Semi-conservative model of replication was proved by __________
(a) Hershey and Chase
(b) Griffith
(c) Meselson and Stahl
(d) Macleod and McCarty
Answer:
(c) Meselson and Stahl

Question 17.
How many types of DNA polymerases does an eukaryotic cell possess?
(a) two
(b) three
(c) four
(d) five
Answer:
(d) Five

Question 18.
Identify the incorrect statement
(a) Replication occurs at ori – site of DNA
(b) Deoxy nucleotide triphosphate acts as a substrate
(c) Unwinding of DNA strand is carried out by topoisomerase
(d) DNA polymerase catalyses the polymerization at 3-OH
Answer:
(c) Unwinding of DNA strand is carried out by topoisomerase

Question 19.
The discontinuously synthesized fragments of lagging strand are called ________
Answer:
Okazaki fragments

Question 20.
Retroviruses possess ________ as genetic material.
Answer:
RNA

Question 21.
Which is NOT a part of transcription unit?
(a) Promoter
(b) Operator
(c) Structural gene
(d) Terminator
Answer:
(b) Operator

Question 22.
Goldberg – Hogness box of eukaryotes is equivalent to ________ of prokaryotes.
Answer:
Pribnow box

Question 23.
Okazaki fragments are joined by the enzyme ________ during DNA replication.
Answer:
DNA ligase

Question 24.

Answer:
(a) A – iv, B – i, C – ii, D – iii

Question 25.
The RNA polymerase of prokaryotes binds with factor to initiate polymerization.
(a) rho
(b) theta
(c) sigma
(d) psi
Answer:
(c) sigma

Question 26.

(a) Capping
(b) Tailing
(c) Splicing
(d) Transcribing
Answer:
(c) Splicing

Question 27.
Which of the following feature is absent in prokaryotes?
(a) Prokaryotes possess three major types of RNAs
(b) Structural genes are polycistronic
(c) Initiation process of transcription requires ‘P’ factor
(d) Split gene feature
Answer:
(d) Split gene feature

Question 28.
Which of the following sequence has completely translated?
(i) AGA, UUU, UGU, AGU, UAG
(ii) AUG, UUU, AGA, UAC, UAA
(iii) AAA, UUU, UUG, UGU, UGA
(iv) AUG,AAU,AAC,UAU,UAG
(a) i and ii
(b) ii only
(c) i and iii
(d) ii and iv
Answer:
(d) ii and iv

Question 29.
Capping of mRNA occurs using __________
(a) Poly A residues
(b) Methyl guanosine triphosphate
(c) Deoxy ribonucleotide triphosphate
(d) Ribonucleotide triphosphate
Answer:
(b) Methyl guanosine triphosphate

Question 30.
One of the aspect is not a feature of genetic code?
(a) Specific
(b) Degenerate
(c) Universal
(d) Ambiguous
Answer:
(d) Ambiguous

Question 31.
Which of the triplet codon is not a code of proline?
(i) CCU
(ii) CAU
(iii) CCG
(iv) CAA
(a) i only
(b) ii and iv
(c) iii only
(d) all the above
Answer:
(b) ii and iv

Question 32.
Coding sequences found in split genes are called.
(a) Operons
(b) Introns
(c) Exons
(d) Cistron
Answer:
(c) Exons

Question 33.
Which of the following mRNA yields 6 aminoacids after translation?
(i) UCU UAU AGU CGA UGC AGU UGA AAA UUU
(ii) UGA AGA UAG GAG CAU CCC UAC UAU GAU
(iii) GUC UGC UGG GCU GAU UAA AGG AGC AUU
(iv) AUG UAC CAU UGC UGA UGC AGG AGC CCG
Answer:
(i) UCU UAU AGU CGA UGC AGU UGA AAA UUU

Question 34.
The transcription termination factor associated with RNA polymerase in prokaryotes is
(a) θ
(b) σ
(c) ρ
(d) ∑
Answer:
(c) ρ

Question 35.
In a DNA double strand, if guanine is of 30%, what will be the percentage of thymine?
(a) 100%
(b) 20%
(c) 10%
(d) 70%
Answer:
(b) 20%

Question 36.
Identify the triplet pairs that code for Tyrosine
(a) UUU,UUC
(b) UAU, UAU
(c) UGC, UGU
(d) CAU, CAC
Answer:
(b) UAU, UAU

Question 37.

Answer:
A – ii B – i C – iv D – iii

Question 38.
AUG code is for __________
(a) Arginine
(b) Tyrosine
(c) Tryptophan
(d) Methionine
Answer:
(d) Methionine

Question 39.
The sequence of bases in coding strand of DNA is G A G T  T A G C A G G C, then the sequence of codons in primary transcript is
(a) C U C A U A C G C C C G
(b) C U C A A U C G U C C G
(c) U C A G A U C U G C G C
(d) U U C A A U C G U G C G
Answer:
(b) C U C A A U C G U C C G

Question 40.
The promoter region of eukaryote is __________
(a) TATAA
(b) AUGUT
(c) UUUGA
(d) AAAAU
Answer:
(a) TATAA

Question 41.
Match the following:
(A) AUG – (i) Tyrosine
(B) UGA – (ii) Glycine
(C) UUU – (iii) Methionine
(D) GGG – (iv) Phenylalanine
(a) A – iii B – i C – iv D – ii
(b) A – iii B – ii C – i D – iv
(c) A – iv B – i C – iii D – ii
(d) A – ii B – i C – iv D – iii
Answer:
(d) A – ii B – i C – iv D – iii

Question 42.
__________ number of codons, codes for cystine.
Answer:
Two

Question 43.
In sickle cell anaemia, the __________ codon of β – globin gene is modified.
(a) Eighth
(b) Seventh
(c) Sixth
(d) Nineth
Answer:
(c) Sixth

Question 44.
Pick out the incorrect statement.
(a) tRNA acts as a adapter molecule
(b) Stop codons donot have tRNA’s
(c) Addition of aminoacid leads to hydrolysis of tRNA
(d) tRNA has four major loops
Answer:
(c) Addition of aminoacid leads to hydrolysis of tRNA

Question 45.
Which of the following antibiotic inhibits the interaction between tRNA and mRNA?
(a) Neomycin
(b) Streptomycin
(c) Tetracycline
(d) Chloramphenicol
Answer:
(a) Neomycin

Question 47.
The cluster of genes with related function is called _________
(a) Cistron
(b) Operon
(c) Muton
(d) Recon
Answer:
(b) Operon

Question 48.
Repressor protein of Lac operon binds to __________ of operon.
(a) Promoter region
(b) Operator region
(c) terminator region
(d) inducer region
Answer:
(b) Operator region

Question 49.
Lac Z gene codes for __________
(a) Permease
(b) transacetylase
(c) β -galactosidase
(d) Aminoacyl transferase
Answer:
(c) β -galactosidase

Question 50.
Lac operon model was proposed by __________
Answer:
Jacob and Monod

Question 51.
Approximate count of base pair in human genome is __________
Answer:
3 × 10⁹ bp

Question 52.
Automated DNA sequences are developed by.
Answer:
Frederick Sanger

Question 53.
Which of the chromosome has higher gene density?
(a) Chromosome 20
(b) Chromosome 19
(c) Chromosome 13
(d) Chromosome Y
Answer:
(b) Chromosome 19

Question 54.
Number of genes located in chromosome Y is __________
(a) 2968
(b) 213
(c) 2869
(d) 231
Answer:
(d) 231

Question 55.
How many structural genes are located in lac operon of E.Coli?
(a) 4
(b) 3
(c) 2
(d) 1
Answer:
(b) 3

Question 56.
DNA finger printing technique was developed by
(a) Jacob and Monod
(b) Alec Jeffreys
(c) Frederick Sanger
Answer:
(b) Alec Jeffreys

Question 57.
In DNA fingerprinting, separation of DNA fragments is done by __________
(a) Centrifugation
(b) Electrophoresis
(c) X-ray diffraction
(d) denaturation
Answer:
(b) Electrophoresis

Question 58.
SNP stands for
(a) Single nucleotide Polymorphism
(b) Single Nucleoside Polypeptide
(c) Single nucleotide Polymorphism
(d) Single nucleotide polymer
Answer:
(a) Single nucleotide Polymorphism

Question 59.
Specific sequences of mRNA that are not translated are __________
Answer:
UnTranslated Regions (UTR)

Question 60.
Non-coding or intervening DNA sequence is called __________

Question 61.
_______ Intron is the monomer of DNA.
Answer:
Nucleotide

Question 62.
Which one of the following is wrongly matched?
(a) Transcription – Copying information from DNA to RNA
(b) Translation – Decoding information from mRNA to protein
(c) Replication – Making of DNA copies
(d) Splicing – Joining of exons with introns
Answer:
(d) Splicing – Joining of exons with introns

2- Mark Questions

Question 1.
Who proposed One gene – One enzyme hypothesis? Define it.
Answer:
George Beadle and Edward Tatum proposed One gene – One enzyme hypothesis which states that one gene controls the production of one enzyme.

Question 2.
Differentiate nucleoside from nucleotide.
Answer:

1. Nucleoside: Nucleoside subunit is composed of nitrogenous bases linked to a pentose sugar molecule.
2. Nucleotide: Nucleotide subunit is composed of nitrogenous bases, a pentose sugar and a phosphate group.

Question 3.
State the key differences between DNA and RNA.
Answer:
DNA:

1. DNA is made of deoxyribose sugar.
2. Nitrogenous bases of DNA are Adenine, Guanine, Cytosine and Thymine.

RNA:

1. RNA is made of ribose sugar.
2. Nitrogenous bases of RNA are Adenine, Guanine, Cytosine and Uracil.

Question 4.
Point out the nitrogenous bases of RNA.
Answer:
Adenine, Guanine, Cytosine and Uracil.

Question 5.
What makes the DNA and RNA as acidic molecules?
Answer:
The phosphate functional group (PO₄) gives DNA and RNA the property of an acid at physiological pH, hence the name nucleic acid.

Question 6.
Which type of bond is formed

1. between a purine and pyrimidine base?
2. between the pentose sugar and adjacent nucleotide?

Answer:

1. Purine and pyrimidine bases are linked by hydrogen bonds.
2. Pentose sugar is linked to adjacent nucleotide by phosphodiester bonds.

Question 7.
DNA acts as genetic material for majority of living organisms and not the RNA. Give reasons to support the statement.
Answer:

1. RNA was reactive and hence highly unstable.
2. Some RNA molecules acts as gene regulators by binding to DNA and affect gene expression.
3. Uracil of RNA is less stable than thymine of DNA.

Question 8.
Name any two viruses whose genetic material is RNA.
Answer:

1. Tobacco Mosaic Virus (TMV)
2. Bacteriophage 0B

Question 9.
What are the properties that a molecule must possess to act as genetic material?
Answer:

1. Self replication
2. Information storage
3. Stability
4. Variation through mutation

Question 10.
How many base pairs are present in one complete turn of DNA helix? What is the distance between two consecutive base pairs?
Answer:
There are ten base pairs in each turn with a distance of 0.34 x 109m between two adjacent base pairs.

Question 11.
What is a genophore?
Answer:
In prokaryotes such as E. coli though they do not have defined nucleus, the DNA is not scattered throughout the cell. DNA (being negatively charged) is held with some proteins (that have positive charges) in a region called the nucleoid. The DNA as a nucleoid is organized into large loops held by protein. DNA of prokaryotes is almost circular and lacks chromatin organization, hence termed genophore.

Question 12.
Whqt is nucleosome? How many base pairs are there in a typical nucleosome?
Answer:
The negatively charged DNA is wrapped around the positively charged histone octamere to form a structure called nucleosome. A typical nucleosome contains 200 bp of DNA helix.

Question 13.
Expand and define NHC
Answer:

1. NHC : Non-histone Chromosomal protein.
2. In eukaryotes, apart from histone proteins, additional set of proteins are required for packing of chromatin at higher level and are referred as non – histone chromosomal proteins.

Question 14.
Differentiate between Heterochromatin and Euchromatin.
Answer:
Heterochromatin:

1. Region of nucleus where the chromatin are loosely packed and stains light are called Heterochromatin.
2. Transcriptionally inactive.

Euchromatin:

1. Region of nucleus where the chromatin are tightly packed and stains dark are called Euchromatin.
2. Transcriptionally active.

Question 15.
Which is the widely accepted model of DNA replication? Who has proved it?
Answer:
Semi-conservative replication model. It was proved by Meselson and Stahl in 1958.

Question 16.
Name the chemical substance which is called by the name

1. Kornberg Enzyme
2. Ochoa’s enzyme

Answer:

1. DNA polymerase I is also known as Komberg enzyme.
2. Polynucleotide phosphorylase is also known as Ochoa’s enzyme.

Question 17.
Name the various types of prokaryotic DNA polymerase. State their role in replication process.
Answer:

1. DNA Polymerase I Involver in DNA repair mechanism
2. DNA Polymerase II Involver in DNA repair mechanism
3. DNA Polymerase III Involver in DNA replication

Question 18.
What is the function of Deoxy nucleotide triphosphate in replication?
Answer:
Deoxy nucleotide triphosphate acts as substrate and also provides energy for polymerization reaction.

Question 19.
Given below are some events of eukaryotic replication. Name the enzymes involved in the process.

1. Unwinding of DNA
2. Joining of Okazaki fragments
3. Addition of nucleotides to new strand
4. Correcting the repair

Answer:

1. Helicase
2. DNA ligase
3. DNA polymerase
4. Nuclease

Question 20.
Differentiate leading strand from lagging strand
Answer:
Leading strand:

1. Leading strand has the polarity 3’ → 5’.
2. Replication is continuous.

Lagging strand:

1. Lagging strand has the polarity 5’ → 3’.
2. Replication is discontinuous.

Question 21.
What are Okazaki fragments?
Answer:
The discontinuously synthesized fragments of the lagging strand are called the Okazaki fragments are joined by the enzyme DNA ligase.

Question 22.
What is a replication fork?
Answer:
At the point of origin of replication, the helicases and topoisomerases (DNA gyrase) unwind and pull apart the strands, forming a Y-Shaped structure called the replication fork. There are two replication forks at each origin.

Question 23.
Apart from DNA polymerase, name any other four enzymes which were involved in DNA replication of eukaryotic cell.
Answer:
DNA ligase, Topoisomerase (DNA gyrase), Helicase and Nuclease.

Question 24.
Who proposed the central dogma? Write its concept.
Answer:
Francis Crick proposed the Central dogma in molecular biology which states that genetic information flows as follows:

Question 25.
Define transcription and name the enzyme involved in this process.
Answer:
The process of copying genetic information from one strand of DNA into RNA is termed transcription. This process takes place in presence of DNA dependent RNA polymerase.

Question 26.
What is TATA box? State its function.
Answer:
In eukaryotes, the promoter has AT rich regions called TATA box or Goldberg-Hogness box. It acts as a binding site for RNA polymerase.

Question 27.
Structural gene of eukaryotes differ from prokaryotes. How?
Answer:
In eukaryotes, the structural gene is monocistronic coding for only one protein whereas in prokaryotes the structural gene is polycistronic coding for many proteins.

Question 28.
What are the two major components of prokaryotic RNA polymerase? How do they act?
Answer:
Bacterial (prokaryotic) RNA polymerase consists of two major components, the core enzyme and the sigma subunit. The core enzyme (β1, β, and α) is responsible for RNA synthesis ” whereas a sigma subunit is responsible for recognition of the promoter.

Question 29.
Distinguish between exons and introns.
Answer:

1. Exons: Expressed sequences (Coding sequences) of an eukaryotic gene
2. Introns: Interveining sequences (non-coding sequences) of an eukaryotic gene

Question 30.
Define splicing.
Answer:
The process of removing introns from hnRNA is called splicing.

Question 31.
What is capping and tailing?
Answer:
In capping an unusual nucleotide, methyl guanosine triphosphate is added at the 5’ end of hnRNA, whereas adenylate residues (200-300) (Poly A) are added at the 3’ end in tailing.

Question 32.
If a double stranded DNA has 20% of cytosine, calculate the percentage of adenine in DNA.
Answer:
Cytdsine = 20, hence Guanine = 20
As per ChargafFs rule (A+T) = (G+C) =100
Percent of Thymine + Adenine = 20 + 20 = 100
(T + A) = (20 + 20) =100
(T + A)=100-(20 + 20)
T +A = 100 – 40
T + A = 60
Therefore the percent of Adenine will be 60/2 = 30%.

Question 33.
Mention the dual functions of AUG.
Answer:
AUG has dual functions. It acts as a initiator codon and also codes for the amino acid methionine.

Question 34.
How many codons are involved in termination of translation. Name them.
Answer:
Three codons terminate translation process. They are UAA, UAG and UGA.

Question 35.
Degeneracy of codon – comment.
Answer:
A degenerate code means that more than one triplet codon could code for a specific amino acid. For example, codons GUU, GUC, GUA and GUG code for valine.

Question 36.
Point out the exceptional categories to universality of genetic code.
Answer:
Exceptions to universal nature of genetic code is noticed in prokaryotic mitochondrial and chloroplast genomes.

Question 37.
What are non-sense codons?
Answer:
UGA, UAA and UAG are the non-sense codons, which terminates translation.

Question 38.
Name the triplet codons that code for

1. Tyrosine
2. Histidine

Answer:

1. Tyrosine – UAU, UAC
2. Histidine – CAU, CAC

Question 39.
Why hnRNA has to undergo splicing?
Answer:
Since hnRNA contains both coding sequences (exons) and non-coding sequences (introns) it has to undergo splicing to remove introns.

Question 40.
State the role of following codons in translation process

1. AUG
2. UAA

Answer:

1. AUG is the initiator codon and also codes for methionine.
2. UAA is a terminator codon.

Question 41.
Given below is mRNA sequence. Mention the aminoacids sequence that is formed after its translation.
Answer:
3’AUGAAAGAUGGGUAA5’
Methionine – Lysine – Aspartic acid – Glycine

Question 42.
Name the four codons that codes valine.
Answer:
GUU, GUC, GUA and GUG.

Question 43.
The base sequence in one of the DNA strand is TAGC ATGAT. Mention the base sequence in its complementary strand.
Answer:
The complementary strand has ATCGTACTA.

Question 44.
Why t-RNA is called as adapter molecule?
Answer:
The transfer RNA (tRNA) molecule of a cell acts as a vehicle that picks up the amino acids scattered through the cytoplasm and also reads specific codes of mRNA molecules. Hence it is called as adapter molecule.

Question 45.
What do you mean by charging of tRNA? Name the enzyme involved in this process.
Answer:
The process of addition of amino acid to tRNA is known as aminoacylation or charging and the resultant product is called aminoacyl- tRNA (charged tRNA). Aminoacylation is catalyzed by an enzyme aminoacyl – tRNA synthetase.

Question 46.
What are UTR’s?
Answer:
mRNA also have some additional sequences that are not translated and are referred to as Untranslated Regions (UTR). UTRs are present at both 5’ end (before start codon) and at 3’ end (after stop codon).

Question 47.
What is S – D sequence?
Answer:
The 5’ end of the mRNA of prokaryotes has a special sequence which precedes the initial AUG start codon of mRNA. This ribosome binding site is called the Shine – Dalgamo sequence or S-D sequence. This sequences base-pairs with a region of the 16Sr RNA of the small ribosomal subunit facilitating initiation.

Question 48.
Define translation unit.
Answer:
A translation unit in mRNA is the sequence of RNA that is flanked by the start codon on 5’ end and stop codon on 3’ end and codes of polypeptide.

Question 49.
Mention the inhibitory role of tetracycline and streptomycin in bacterial translation.
Answer:
Tetracycline inhibits binding between aminoacyl tRNA and mRNA.Streptomycin inhibits initiation of translation and causes misreading.

Question 50.
At what stage, does the gene expression is regulated?
Answer:
Gene expression can be controlled or regulated at transcriptional or translational levels.

Question 51.
What is a operon? Give example.
Answer:
The cluster of genes with related functions is called operon.
E.g: lac operon in E.coli.

Question 52.
Considering the lac operon of E.coli, name the products of the following genes.

1. i gene
2. lac Z gene
3. lac Y gene
4. lac a gene

Answer:

1. i gene – repressor protein
2. lac Z gene – fS-galactosidase
3. Lac Y gene – Permease
4. lac a gene – transacetylase

Question 53.
Expand

1. ETS
2. YAC.

Answer:

1. ETS : Expressed Sequence Tags
2. YAC : Yeast Artificial Chromosomes

Question 54.
Name the human chromosome that has

1. most number of genes
2. least number of genes

Answer:

1. Chromosome 1 has maximum number of genes (2968 genes)
2. Chromosome Y has least genes (231 genes)

Question 55.
What are SNPs? Mention its uses.
Answer:
SNPs : Single nucleotide polymorphism. It helps to find chromosomal locations for disease associated sequences and tracing human history.

Question 56.
Mention any four areas where DNA fingerprinting can be used.
Answer:

1. Forensic analysis
2. Pedigree analysis
3. Conservation of wild life
4. Anthropological studies

3 – Mark Questions

Question 57.
Classify nucleic acid based on sugar molecules.
Answer:
There are two types of nucleic acids depending on the type of pentose sugar. Those containing deoxyribose sugar are called Deoxyribo Nucleic Acid (DNA) and those with ribose sugar are known as Ribonucleic Acid (RNA). The only difference between these two sugars is that there is one oxygen atom less in deoxyribose.

Question 58.
Both purines and pyrimidines are nitrogen bases yet they differ. How?
Answer:
Both purines and pyrimidines are nitrogen bases. The purine bases Adenine and Guanine have double carbon – nitrogen ring, whereas cytosine and thymine bases have single carbon nitrogen ring.

Question 59.
How 5’ of DNA differ from its 3’?
Answer:
The 5’ of DNA refers to the carbon in the sugar to which phosphate (P04V) functional group is attached. The 3’ of DNA refers to the carbon in the sugar to which a hydroxyl (OH) group is attached.

Question 60.
State Chargaff’s rule.
Answer:
According to Erwin Chargaff,

1. Adenine pairs with Thymine with two hydrogen bonds.
2. Guanine pairs with Cytosine with three hydrogen bonds.

Question 61.
Chemically DNA is more stable than RNA – Justify.
Answer:
In DNA, the two strands being complementary, if separated (denatured) by heating can come together (renaturation) when appropriate condition is provided. Further 2 OH group present at every nucleotide in RNA is a reactive group that makes RNA liable and easily degradable. RNA is also known to be catalytic and reactive. Hence, DNA is chemically more stable and chemically less reactive when compared to RNA. Presence of thymine instead of uracil in DNA confers additional stability to DNA.

Question 62.
Write in simple about semi-conservative mode of DNA replication.
Answer:
Semi-conservative replication was proposed by Watson and Crick in 1953. This mechanism of replication is based on the DNA model. They suggested that the two polynucleotide strands of DNA molecule unwind and start separating at one end. During this process, covalent hydrogen bonds are broken. The separated single strand then acts as template for the synthesis of a new strand. Subsequently, each daughter double helix carries one polynucleotide strand from the parent molecule that acts as a template and the other strand is newly synthesised and complementary to the parent strand.

Question 63.
Draw a simplified diagram of nucleosome and label it.
Answer:

Question 64.
What is a primer?
Answer:
A primer is a short stretch of RNA. It initiates the formation of new strand. The primer produces 3’-OH end on the sequence of ribonucleotides, to which deoxyribonucleotides are added to form a new strand.

Question 65.
Both strands of DNA are not copied during transcription. Give reason.
Answer:
Both the strands of DNA are not copied during transcription for two reasons.

1. If both the strands act as a template, they would code for RNA with different sequences. This in turn would code for proteins with different amino acid sequences. This would result in one segment of DNA coding for two different proteins, hence complicate the genetic information transfer machinery.
2. If two RNA molecules were produced simultaneously, double stranded RNA complementary to each other would be formed. This would prevent RNA from being translated into proteins.

Question 66.
What do you mean by a template strand and coding strand?
Answer:
DNA dependent RNA polymerase catalyses the polymerization in only one direction, the strand that has the polarity 3’→ 5’ acts as a template, and is called the template strand. The other strand which has the polarity 5’→ 3’ has a sequence same as RNA (except thymine instead of uracil) and is displaced during transcription. This strand is called coding strand.

Question 67.
Name the factors that are responsible for initiation and termination of transcription in prokaryotes.
Answer:

1. Sigma factor is responsible for initiation of transcription.
2. Rho factor is responsible for termination of transcription.

Question 68.
Name the major RNA types of prokaryotes and mention their role.
Answer:
In prokaryotes, there are three major types of RNAs: mRNA, tRNA, and rRNA. All three \ RNAs are needed to synthesize a protein in a cell. The mRNA provides the template, tRNA brings amino acids and reads the genetic code, and rRNAs play structural and catalytic role
during translation.

Question 69.
Define genetic code.
Answer:
The order of base pairs along DNA molecule controls the kind and order of amino acids found in the proteins of an organism. This specific order of base pairs is called genetic code.

Question 70.
Explain Wobble hypothesis.
Answer:
Wobble Hypothesis is proposed by Crick (1966) which states that tRNA anticodon has the ability to wobble at its 5’ end by pairing with even non-complementary base of mRNA codon.’ According to this hypothesis, in codon-anticodon pairing the third base may not be complementary.

The third base of the codon is called wobble base and this position is called wobble position. The actual base pairing occurs at first two positions only. The importance of Wobbling hypothesis is that it reduces the number of tRNAs required for polypeptide synthesis and it overcomes the effect of code degeneracy.

Question 71.
Explain the nature of eukaryotic ribosome.
Answer:
The ribosomes of eukaryotes (80 S) are larger, consisting of 60 S and 40 S sub units. ‘S’ denotes the sedimentation efficient which is expressed as Svedberg unit (S). The larger subunit in eukaryotes consist of a 23 S RNA and 5Sr RNA molecule and 31 ribosomal proteins. The smaller eukaryotic subunit consist of 18Sr RNA component and about 33 proteins.

Question 72.
Expand and define ORF.
Answer:
Any sequence of DNA or RNA, beginning with a start codon and which can be translated into a protein is known as an Open Reading Frame (ORF).

Question 73.
What are the components of initiation complex of prokaryotic translation?
Answer:
Initiation of translation in E. coli begins with the formation of an initiation complex, consisting of the 30S subunits of the ribosome, a messenger RNA and the charged N-formyl methionine tRNA (f^met -1 RNA f^met), three proteinaceous initiation factors (IF 1, IF2, IF3), GTP (Guaniner Tri Phosphate) and Mg ²⁺.

Question 74.
Explain the components of operon.
Answer:
Structure of the operon: Each operon is a unit of gene expression and regulation and consists
of one or more structural genes and an adjacent operator gene that controls transcriptional
activity of the structural gene.

1. The structural gene codes for proteins, rRNA and tRNA required by the cell.
2. Promoters are the signal sequences in DNA that initiate RNA synthesis. RNA polymerase I binds to the promoter prior to the initiation of transcription.
3. The operators are present between the promoters and structural genes. The repressor protein binds to the operator region of the operon.

5 – Mark Question

Question 75.
Describe Hershey and Chase experiment. What is concluded by their experiment?
Answer:
Alfred Hershey and Martha Chase (1952) conducted experiments on bacteriophages that infect bacteria. Phage T₂ is a virus that infects the bacterium Escherichia coli. When phages (virus) are added to bacteria, they adsorb to the outer surface, some material enters the bacterium, and then later each bacterium lyses to release a large number of progeny phage. Hershey and Chase wanted to observe whether it was DNA or protein that entered the bacteria. All nucleic acids contain phosphorus, and contain sulphur (in the amino acid cysteine and methionine). Hershey and Chase designed an experiment using radioactive isotopes of Sulphur (³⁵S) and phosphorus (³²P) to keep separate track of the viral protein and nucleic acids during the infection process.

The phages were allowed to infect bacteria in culture medium which containing the radioactive isotopes ³⁵S or ³²P. The bacteriophage that grew in the presence of ³⁵S had labelled proteins and bacteriophages grown in the presence of ³²P had labelled DNA. The differential labelling thus enabled them to identity DNA and proteins of the phage. Hershey and Chase mixed the labelled phages with unlabeled E. coli and allowed bacteriophages to attack and inject their genetic material. Soon after infection (before lysis of bacteria), the bacterial cells were gently agitated in a blender to loosen the adhering phase particles.

It was observed that only ³²P was found associated with bacterial cells and ³⁵S was in the surrounding medium and not in the bacterial cells. When phage progeny was studied for radioactivity, it was found that it carried only ³²P and not ³⁵ S. These results clearly indicate that only DNA and not protein coat entered the bacterial cells. Hershey and Chase thus conclusively proved that it was DNA, not protein, which carries the hereditary information from virus to bacteria.

Question 76.
Explain the properties of DNA that makes it an ideal genetic material.
Answer:
1. Self Replication: It should be able to replicate. According to the rule of base pairing and complementarity, both nucleic acids (DNA and RNA) have the ability to direct duplications. Proteins fail to fulfill this criteria.

2. Stability: It should he stable structurally and chemically. The genetic material should be stable enough not to change with different stages of life cycle, age or with change in physiology of the organism. Stability as one of property of genetic material was clearly evident in Griffith’s transforming principle. Heat which killed the bacteria did not destroy some of the properties of genetic material. In DNA the two strands being complementary.

if separated (denatured) by heating can come together (renaturation) when appropriate condition is provided. Further 2’ OH group present at every nucleotide in RNA is a reactive group that makes RNA liable and easily degradable. RNA is also known to be catalytic and reactive. Hence, DNA is chemically more stable and chemically less reactive when compared to RNA. Presence of thymine instead of uracil in DNA confers additional stability to DNA.

3. Information storage: It should be able to express itself in the form of ‘Mendelian characters’. RNA can directly code for protein synthesis and can easily express the characters. DNA, however depends on RNA for synthesis of proteins. Both DNA and RNA can act as a genetic material, but DNA being more stable stores the genetic information and RNA transfers the genetic information.

4. Variation through mutation: It should be able to mutate. Both DNA and RNA are able to mutate. RNA being unstable, mutates at a faster rate. Thus viruses having RNA genome with shorter life span can mutate and evolve faster. The above discussion indicates that both RNA and DNA can function as a genetic material. DNA is more stable, and is preferred for storage of genetic information.

Question 77.
How the DNA is packed in an eukaryotic cell? ft
Answer:
In eukaryotes, organization is more complex. Chromatin is formed by a series of repeating units called nucleosomes. Komberg proposed a model for the nucleosome, in which 2 molecules of the four histone proteins H2A, H2B, H3 and H4 are organized to form a unit of eight molecules called histone octamere. The negatively charged DNA is wrapped around the positively charged histone octamere to form a structure called nucleosome. A typical nucleosome contains 200 bp of DNA helix. The histone octameres are in close contact and DNA is coiled on the outside of nucleosome. Neighbouring nucleosomes are connected by linker DNA (HI) that is exposed to enzymes.

The DNA makes two complete turns around the histone octameres and the two turns are sealed off by an HI molecule. Chromatin lacking HI has a beads-on-a-string appearance in which DNA inters and leaves the nucleosomes at random places. HI of one nucleosome can interact with 33l of the neighbouring nucleosomes resulting in the further folding of the fibre.

The chrof&atin fiber in interphase nuclei and mitotic chromosomes have a diameter that vary between 200-300 nm and represents inactive chromatin. 30 nm fibre arises from the folding of nucfeosbme, chains into a solenoid structure having six nucleosomes per turn. This structure is stabilized by interaction between different HI molecules. DNA is a solenoid and packed about,%)_folds. The hierarchical nature of chromosome structure is illustrated.

Additional set of pteins are required for packing of chromatin at higher level and are referred to as non-histone chromosomal proteins (NHC). In*,a typical nucleus, some regions of chromatin are Ibosely packed (lightly stained) and are referred to as euchromatin. The chromatin that is,-tightly packed (stained darkly) is called heterochromatin. Euchromatin is transcriptionally active and heterochromatin is transcriptionally inactive.

Question 78.
Meselson and Stahl’s experiment proved the semi-coflBptervation mode of DNA replication. Explain.
Answer:
The mode of DNA replication was determined in 1958 by Meselson and Stahl. They designed an experiment to distinguish between semi-conservative, conservative and dispersive replications. In their experiment, they grew two cultures of E.coli for many generations in separate media. The ‘heavy’ culture was grown in a medium in which the nitrogen source (NH4CI) contained the heavy isotope ¹⁵N and the ‘ light’ culture was grown in a medium in which the nitrogen source contained light isotope ¹⁴H for many generations. At the end of growth, they observed that the bacterial DNA in the heavy culture contained only ¹⁵N and in the light culture only ¹⁴N. The heavy DNA could be distinguished from light DNA (¹⁵N from ¹⁴N) with a technique called Cesium Chloride (CsCl) density gradient centrifugation. In this process, heavy and light DNA extracted from cells in thtytwo cultures settled into two distinct and separate bands (hybrid DNA).

The heavy culture (¹⁵N) was then transferred into a medium that had only NH4CI, and took samples at various definite time intervals (20 minutes duration). After the first replication, they extracted DNA and subjected it to density gradient centrifugation. The DNA settled into a band that was intermediate in position between the previously determined heavy and light bands. After the second replication (40 minutes duration), they again extracted DNA samples,and this time found the DNA settling into two bands, one at the light band position and one at intermediate position. These results confirm Watson and Crick’s semi – conservative replication hypothesis.

Question 79.
Give a detailed account of a transcription unit.
Answer:
A transcriptional unit in DNA is defined by three regions, a promoter, the structural gene and a terminator. The promoter is located towards the 5 ’ end. It is a DNA sequence that provides binding site for RNA polymerase. The presence of promoter in a transcription unit, defines the template and coding strands. The terminator region located towards the 3’ end of the coding strand contains a DNA sequence that causes the RNA polymerase to stop transcribing. In eukaryotes the promoter has AT rich regions called TATA box (Goldberg- Hogness box) ‘ and in prokaryotes this region is called Pribnow box.

Besides promoter, eukaryotes also require an enhancer. The two strands of the DNA in the structural gene of a transcription unit have opposite polarity. DNA dependent RNA polymerase catalyses the polymerization in only one direction, the strand that has the polarity 3’→5’ acts as a template, and is called the template strand. The other strand which has the polarity 5’→ 3’ has a sequence same as RNA (except thymine instead of uracil) and is displaced during transcription. This strand is called coding strand

The structural gene may be monocistronic (eukaryotes) or polycistronic (prokaryotes). In eukaryotes, each mRNA carries only a single gene and encodes information for only a single protein and is called monocistronic mRNA. In prokaryotes, clusters of related genes, known as operon, often found next to each other on the chromosome are transcribed together to give a single mRNA and hence are polycistronic.

Question 80.
Explain the transcription process in prokaryotes with needed diagram.
Answer:
In prokaryotes, there are three major types of RNAs:
mRNA, tRNA, and rRNA. All three RNAs are needed to synthesize a protein in a cell. The mRNA provides the template, tRNA brings amino acids and reads the genetic code, and rRNAs play structural and catalytic role during translation. There is a single DNA-dependent RNA polymerase that catalyses transcription of all types of RNA. It binds to the promoter and initiates transcription (Initiation).

The polymerases binding sites are called promoters. It uses nucleoside triphosphate as substrate and polymerases in a template depended fashion following the rule of complementarity. After the initiation of transcription, the polymerase continues to elongate the RNA, adding one nucleotide after another to the growing RNA chain. Only a short stretch of RNA remains bound to the enzyme, when the polymerase reaches a terminator at the end of a gene, the, nascent RNA falls off, so also the RNA polymerase. The RNA polymerase is only capable of catalyzing the process of elongation. The RNA polymerase associates transiently with initiation factor sigma (a) and termination factor rho (p) to initiate and terminate the transcription, respectively.

Association of RNA with these factors instructs the RNA polymerase either to initiate or terminate the process of transcription. In bacteria, since the mRNA does not require any processing to become active and also since transcription and translation take place simultaneously in the same compartment sincethere is no separation of cytosol and nucleus in bacteria), many times the translation can begin much before the mRNA is fully transcribed. This is because the genetic material is not separated from other cell organelles by a nuclear membrane consequently; transcription and translation can be coupled in bacteria.

Question 81.
Write the salient features of genetic code.
Answer:
The salient features of genetic code are as follows:

1. The genetic codon is a triplet code and 61 codons code for amino acids and 3 codons do not code for any amino acid and function as stop codon (Termination).
2. The genetic code is universal. It means that all known living systems use nucleic acids and the same three base codons (triplet codon) direct the synthesis of protein from amino acids. For example, the mRNA (UUU) codon codes for phenylalanine in all cells of all organisms. Some exceptions are reported in prokaryotic, mitochondrial and chloroplast genomes. However similarities are more common than differences.
3. A non-overlapping codon means that the same letter is not used for two different codons. For instance, the nucleotide sequence GUTJ and GUC represents only two codons.
4. It is comma less, which means that the message would be read directly from one end to the other i.e., no punctuation are needed between two codes.
5. A degenerate code means that more than one triplet codon could code for a specific amino acid. For example, codons GUU, GUC, GUA and GUG code for valine.
6. Non-ambiguous code means that one codon will code for one amino acid.
7. The code is always read in a fixed direction i.e. from 5’→3’ direction called polarity.
8. AUG has dual functions. It acts as a initiator codon and also codes for the amino acid methionine.
9. UAA, UAG (tyrosine) and UGA (tryptophan) codons are designated as termination (stop) codons and also are known as “non-sense” codons.

Question 82.
Mutations on genetic code affects the phenotype. Describe with example.
Answer:
The simplest type of mutation at the molecular level is a change in nucleotide that substitutes one base for another. Such changes are known as base substitutions which may occur spontaneously or due to the action of mutagens. A well studied example is sickle cell anaemia in humans which results from a point mutation of an allele of β-haemoglobin gene (βHb).

A haemoglobin molecule consists of four polypeptide chains of two types, two a chains and two P-chains. Each chain has a heme group on its surface. The heme groups are involved in the binding of oxygen. The jruman blood disease, sickle cell anaemia is due to abnormal haemoglobin. This abnormality in haemoglobin is due to a single base substitution at the sixth codon of the beta globingene from GAG to GTG in p -chain of haemoglobin.

It results in a change of amino acid glufeniic acid to valine at the 6th position of the p -chain. This is the classical example of point mutation that results in the change of amino acids residue glutamic acid to valine. The mutant haemoglobin undergoes polymerisation under oxygen tension causing the change in the shape of the RBC from biconcave to a sickle shaped structure.

Question 83.
Explain the mechanism of AteArperon of the E-coli.
Answer:
The Lac (Lactose) operon: The metabolism of lactose in E.coli requires three enzymes – permease, P-galactosidase (P-gat) and transacetylase. The enzyme permease is needed for entry of lactose into the cell, Pjgglactosidase brings about hydrolysis of lactose to glucose and galactose, while transacety gtransfers acetyl group from acetyl Co A to P-galactosidase. The lac operon consists of one-regulator gene (T gene refers to inhibitor) promoter sites (p), and operator site (o). Besides these, it has three structural genes namely lac z, y and lac a. The lac ‘z’ gene codes for P-gaiaqtttsidase, lac ‘y’ gene codes for permease and ‘a’ gene codes for transacetylase.

Jacob and Monod proposed the classical model of Lac operon to explain gene expression and regulation in E.coli. In lac a polycistronic structural gene is regulated by a common promoter and regulatory genfc When the cell is using its normal energy source as glucose, the ‘i’ gene transcribes a repressor mRNA and after its translation, a repressor protein is produced. It binds to the operator region of the operon and prevents translation, as a result, P-galactosidase is not produced. In the absence of preferred carbon source such as glucose, if lactose is available as an energy source for the bacteria then lactose enters the cell as a result of permease enzyme. Lactose acts as an inducer and interacts with the repressor to inactivate it.

The repressor protein binds to the operator of the operon and prevents RNA polymerase from transcribing the operon. In the presence of inducer, such as lactose or allolactose, the repressor is inactivated by interaction with the inducer. This allows RNA polymerase to bind to the promotor site and transcribe the operon to produce lac mRNA which enables formation of all the required enzymes needed for lactose metabolism. This regulation of lac operon by the repressor is an example of negative control of transcription initiation.

Question 84.
What are the objectives of Human Genome project?
Answer:
The main goals of Human Genome Project are as follows:

1. Identify all the genes (approximately 30000) in human DNA.
2. Determine the sequence of the three billion chemical base pairs that makeup the human DNA.
3. To store this information in databases.
4. Improve tools for data analysis.
5. Transfer related technologies to other sectors, such as industries.
6. Address the ethical, legal and social issues (ELSI) that may arise from the project.

Question 85.
Write the salient features of Human Genome Project.
Answer:

1. Although human genome contains 3 billion nucleotide bases, the DNA sequences that encode proteins make up only about 5% of the genome.
2. An average gene consists of 3000 bases, the largest known human gene being dystrophin with 2.4 million bases.
3. The function of 50% of the genome is derived from transposable elements such as LINE and ALU sequence.
4. Genes are distributed over 24 chromosomes. Chromosome 19 has the highest gene density. Chromosome 13 and Y chromosome have lowest gene densities.
5. The chromosomal organization of human genes shows diversity.
6. There may be 35000-40000 genes in the genome and almost 99.9 nucleotide bases are exactly the same in all people.
7. Functions for over 50 percent of the discovered genes are unknown.
8. Less than 2 percent of the genome codes for proteins.
9. Repeated sequences make up very large portion of the human genome. Repetitive sequences have no direct coding functions but they shed light on chromosome structure, dynamics and evolution (genetic diversity).
10. Chromosome 1 has 2968 genes, whereas chromosome ’Y’ has 231 genes.
11. Scientists have identified about 1.4 million locations, where single base DNA differences (SNPs – Single nucleotidepolymorphism – pronounce as ‘snips’) occur in humans. Identification of ‘SNIPS’ is helpful in finding chromosomal locations for disease associated sequences and tracing human history.

Question 86.
Describe the principle involved in DNA fingerprinting technique.
Answer:
The DNA fingerprinting technique was first developed by Alec Jeffreys in 1985. The DNA of a person and finger prints are unique. There are 23 pairs of human chromosomes with 1.5 million pairs of genes. It is a well known fact that genes are segments of DNA which differ in the sequence of their nucleotides. Not all segments of DNA code for proteins, some DNA segments have a regulatory function, while others are intervening sequences (introns) and still others are repeated DNA sequences. In DNA fingerprinting, short repetitive nucleotide sequences are specific for a person. These nucleotide sequences are called as variable number tandem repeats (VNTR). The VNTRs of two persons generally show variations and are useful as genetic markers.

DNA finger printing involves identifying differences in some specific regions in DNA sequence called repetitive DNA, because in these sequences, a small stretch of DNA is repeated many times. These repetitive DNA are separated from bulk genomic DNA as different peaks during density gradient centrifugation. The bulk DNA forms a major peak and the other small peaks are referred to as satellite DNA. Depending on base composition (A: T rich or G : C rich), length of segment and number of repetitive units, the satellite DNA is classified into many sub – categories such as micro-satellites and mini satellites, etc.

These sequences do not code for any proteins, but they form a large portion of human genome. These sequences show high degree of polymorphism and form the basis of DNA fingerprinting. DNA isolated from blood, hair, skin cells, or other genetic evidences left at the scene of a crime can be compared through VNTR patterns, with the DNA of a criminal suspect to determine guilt or innocence. VNTR patterns are also useful in establishing the identity of a homicide victim, either from DNA found as evidence or from the body itself.

Question 87.
Draw a flow chart depicting the steps of DNA finger printings technique
Answer:

Higher Order Thinking Skills (HOTs) Questions

Question 1.
A mRNA strand has a series of triplet codons of which the first three codons are given below
(a) AUG
(b) UUU
(c) UGC
(i) Name the amino acid encoded by these triplet codons.
(ii) Mention the DNA sequence from which these triplet codons would have transcribed?
Answer:
(i) AUG codes for Methionine
UUU codes for Phenylalanine
UGC codes for Cysteine
(ii) TAC sequence of DNA is transcribed to AUG
AAA sequence of DNA is transcribed to UUU
ACG sequence of DNA is transcribed to UGC

Question 2.
Given below are the structures of tRNA molecules which are involved in translation process. In one tRNA, codon is mentioned but not the amino acid. In another tRNA molecule, amino acid is named and not the codon. Complete the figure by mentioning the respective amino acids and codons.
Answer:

Question 3.
A DNA fragment possesses 32 adenine bases and 32 cytosine bases. How many total number of nucleotides does that DNA fragment contains? Explain.
Answer:
128 nucleotides. Adenine always pair Thymine base. If there are 32 adenine bases then there must be 32 Thymine bases. Similarly cytosine pairs with guanine. If cytosine bases are 32 in number the guanine bases will be equal to cytosine. So it make a total of 128 nucleotides.

Question 4.
Following is a DNA sequence representing a part of gene TAC TCG CCC TAT UAA CCC AAA ACC TCT using this derive A.

1. The RNA transcript
2. The spliced mRNA (consider all the codons with two Aderine bases are introns)
3. The total number of aminoacids coded by the mRNA

Answer:

1. RNA transcript: AUG UGC GGG AUA GGG UUU UGG AGA
2. Spliced mRNA : AUG UGC GGG GGG UUU UGG
3. 6 aminoacids are coded by mRNA

Question 5.
Complete the molecular processes by naming them

1. DNA → DNA
2. mRNA → Protein
3. RNA transcript → mRNA

Answer:

1. Replication
2. Translation
3. Splicing

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Samacheer Kalvi 12th Bio Botany Economically Useful Plants and Entrepreneurial Botany Text Book Back Questions and Answers

Question 1.
Consider the following statements and choose the right option:
(i) Cereals are members of grass family.
(ii) Most of the food grains come from monocotyledon.
(a) (i) is correct and (ii) is wrong
(b) Both (i) and (ii) are correct
(c) (i) is wrong and (ii) is correct
(d) Both (i) and (ii) are wrong
Answer:
(b) Both (i) and (ii) are correct

Question 2.
Assertion: Vegetables are important part of healthy eating.
Reason: Vegetables are succulent structures of plants with pleasant aroma and flavours.
(a) Assertion is correct, Reason is wrong
(b) Assertion is wrong, Reason is correct
(c) Both are correct and reason is the correct explanation for assertion.
(d) Both are correct and reason is not the correct explanation for assertion.
Answer:
(d) Both are correct and reason is not the correct explanation for assertion.

Question 3.
Groundnut is native of _____________
(a) Philippines
(b) India
(c) North America
(d) Brazil
Answer:
(d) Brazil

Question 4.
Statement A: Coffee contains caffeine.
Statement B: Drinking coffee enhances cancer.
(a) A is correct, B is wrong
(b) A and B – Both are correct
(c) A is wrong, B is correct
(d) A and B – Both are wrong
Answer:
(a) A is correct, B is wrong

Question 5.
Tectona grandis is coming under family _____________
(a) Lamiaceae
(b) Fabaceae
(c) Dipterocaipaceae
(d) Ebenaceae
Answer:
(a) Lamiaceae

Question 6.
Tamarindus indica is indigenous to _____________
(a) Tropical African region
(b) South India, Sri Lanka
(c) South America, Greece
(d) India alone
Answer:
(a) Tropical African region

Question 7.
New world species of cotton _____________
(a) Gossipium arboretum
(b) G.herbaceum
(c) Both a and b
(d) G.barbadense
Answer:
(d) G.barbadense

Question 8.
Assertion: Turmeric fights various kinds of cancer.
Reason: Curcumin is an anti-oxidant present in turmeric.
(a) Assertion is correct, Reason is wrong
(b) Assertion is wrong, Reason is correct
(c) Both are correct
(d) Both are wrong
Answer:
(c) Both are correct

Question 9.
Find out the correctly matched pair:
(a) Rubber Shorea robusta
(b) Dye
(c) Timber Cyperus papyrus
(d) Pulp Hevea brasiliensis
Answer:
(b) Dye

Question 10.
Observe the following statements and pick out the right option from the following:
Statement I – Perfumes are manufactured from essential oils.
Statement II – Essential oils are formed at different parts of the plants.
(a) Statement I is correct
(b) Statement II is correct
(c) Both statements are correct
(d) Both statements are wrong
Answer:
(c) Both statements are correct

Question 11.
Observe the following statements and pick out the right option from the following:
Statement I: The drug sources of Siddha include plants, animal parts, ores and minerals.
Statement II: Minerals are used for preparing drugs with long shelf-life.
(a) Statement I is correct
(b) Statement II is correct
(c) Both statements are correct
(d) Both statements are wrong
Answer:
(c) Both statements are correct

Question 12.
The active principle trans-tetra hydro canabial is present in _____________
(a) Opium
(b) Curcuma
(c) Marijuana
(d) Andrographis
Answer:
(c) Marijuana

Question 13.
Which one of the following matches is correct?
(a) Palmyra – Native of Brazil
(b) Saccharun – Abundant in Kanyakumari
(c) Steveocide – Natural sweetener
(d) Palmyra sap – Fermented to give ethanol
Answer:
(c) Steveocide – Natural sweetener

Question 14.
The only cereal that has originated and domesticated from the New world.
(a) Oryza sativa
(b) Triticum aestivum
(c) Triticum duram
(d) Zea mays
Answer:
(c) Triticum duram

Question 15.
Write the cosmetic uses of Aloe.
Answer:
Aloe gel are used as skin tonic. It has a cooling effect and moisturizing characteristics and hence used in preparation of creams, lotions, shampoos, shaving creams, after shave lotions and allied products. It is used in gerontological applications for rejuvenation of aging skin. Products prepared from aloe leaves have multiple properties such as emollient, antibacterial, antioxidant, antifungal and antiseptic. Aloe vera gel is used in skin care cosmetics.

Question 16.
What is pseudo cereal? Give an example.
Answer:
The term pseudo-cereal is used to describe foods that are prepared and eaten as a whole grain, but are botanical outliers from grasses. Example: quinoa. It is actually a seed from the Chenopodium quinoa plant, belongs to the family Amaranthaceae.

Question 17.
Discuss which wood is better for making furniture.
Answer:
Teak wood is the ideal type of wood for making household furnitures because, it is highly durable and shows great resistance against the attack of termites and fungi. Moreover it doesnot split or crack and is a carpenter friendly wood.

Question 18.
A person got irritation while applying chemical dye. What would be your suggestion for alternative?
Answer:
If a grey haired person is allergic on using chemical dyes then he can go for natural dyes like Henna. Henna is an organic dye obtained from leaves and young shoots of Lawsonia inermis. The principal colouring matter is Tacosone’ which is harmless and causes no irritation on skin.

Question 19.
Name the humors that are responsible for the health of human beings.
Answer:
Vatam, Pittam and Kapam.

Question 20.
Give definitions for organic farming?
Answer:
Organic farming is an alternative agricultural system in which plants/crops are cultivated – in natural ways by using biological inputs to maintain soil fertility and ecological balance thereby minimizing pollution and wastage.

Question 21.
Which is called as the “King of Bitters”? Mention their medicinal importance.
Answer:
Andrographis paniculata is called as King of Bitters. Andrographis is a potent hepatoprotective agent and is widely used to treat liver disorders. Concoction of Andrographis paniculata and eight other herbs (Nilavembu Kudineer) is effectively used to treat malaria and dengue.

Question 22.
Differentiate bio-medicines and botanical medicines.
Answer:

1. Bio-medicines: Medically useful molecules obtained from plants that are marketed as drugs are called Bio-medicines.
2. Botanical medicines: Parts of medicinal plants which are modified as powers or pills or other forms and marketed. These are called botanical medicines.

Question 23.
Write the origin and area of cultivation of green gram and red gram.
Answer:

Question 24.
What are millets? What are its types? Give example for each type.
Answer:
Millet is the term applied to a variety of very small seeds originally cultivated by ancient people in Africa and Asia. They are gluten-free with less glycemic index.
Types of millet:

1. Major millets – E.g: Ragi (Eleusine coracana)
2. Minor millets – E.g: Foxtail millet (Setaria italica)

Question 25.
If a person drinks a cup of coffee daily it will help him for his health. Is this correct? If it is correct, list out the benefits.
Answer:
Yes, drinking coffee in moderation enhances the health of a person. Caffeine enhances release of acetylcholine in brain, which in turn enhances efficiency. It can lower the incidence of fatty liver diseases, cirrhosis and cancer. It may reduce the risk of type 2 diabetes.

Question 26.
Enumerate the uses of turmeric.
Answer:
Turmeric is one of the most important and ancient Indian spices and used traditionally over thousands of years for culinary, cosmetic, dyeing and for medicinal purposes. It is an important constituent of curry powders. Turmeric is used as a colouring agent in pharmacy, confectionery and food industry. Rice coloured with turmeric (yellow) is considered sacred and auspicious which is used in ceremonies. It is also used for dyeing leather, fibre, paper and toys.

Curcumin extracted from turmeric is responsible for the yellow colour. Curcumin is a very good anti-oxidant which may help fight various kinds of cancer. It has anti-inflammatory, anti- ‘ diabetic, anti-bacterial, anti-fungal and anti-viral activities. It stops platelets from clotting in arteries, which leads to heart attack.

Question 27.
What is TSM? How does it classified and what does it focuses on?
Answer:
TSM stands for Traditional Systems of Medicines India has a rich medicinal heritage. Anumber of Traditional Systems of Medicine (TSM) are practiced in India some of which come from outside India. TSM in India can be broadly classified into institutionalized or documented and non-institutionalized or oral traditions. Institutionalized Indian systems include Siddha and Ayurveda which are practiced for about two thousand years.

These systems have prescribed texts in which the symptoms, disease diagnosis, drugs to cure, preparation of drugs, dosage and diet regimes, daily and seasonal regimens. Non-institutional systems, whereas, do not have such records and or practiced by rural and tribal peoples across India. The knowledge is mostly held in oral form. The TSM focus on healthy lifestyle and healthy diet for maintaining good health and disease reversal.

Question 28.
Write the uses of nuts you have studied.
Answer:
Cashews are commonly used for garnishing sweets or curries, or ground into a paste that forms a base of sauces for curries or some sweets. Roasted and raw kernels are used as snacks.

Question 29.
Give an account on the role of Jasminum in perfuming.
Answer:
The essential jasmine oil is present in the epidermal cells of the inner and outer surfaces of both the sepals and petals. One ton of Jasmine blossom yields about 2.5 to 3 kg of essential oil, comprising 0.25 to 3% of the weight of the fresh flower. Jasmine oil is an essential oil that is valued for its soothing, relaxing and antidepressant qualities.

Question 30.
Give an account of active principle and medicinal values of any two plants you have studied
Answer:

Question 31.
Write the economic importance of rice.
Answer:
Rice is the easily digestible calorie rich cereal food which is used as a staple food in Southern and North East India. Various rice products such as Flaked rice (Aval), Puffed rice / parched rice (Pori) are used as breakfast cereal or as snack food in different parts of India. Rice bran oil obtained from the rice bran is used in culinary and industrial purposes. Husks are used as fuel, and in the manufacture of packing material and fertilizer.

Question 32.
Which TSM is widely practiced and culturally accepted in Tamil Nadu? – explain.
Answer:
Siddha is the most popular, widely practiced and culturally accepted system in Tamil Nadu. It is based on the texts written by 18 Siddhars. There are different opinions on the constitution of 18 Siddhars. The Siddhars are not only from Tamil Nadu, but have also come from other countries. The entire knowledge is documented in the form of poems in Tamil. Siddha is principally based on the Pancabuta philosophy. According to this system three humors namely Vatam, Pittam and Kapam that are responsible for the health of human beings and ‘ any disturbance in the equilibrium of these humors result in ill health.

The drug sources of Siddha include plants, animal parts, marine products and minerals. This system specializes in using minerals for preparing drugs with the long shelf-life. This system uses about 800 herbs as source of drugs. Great stress is laid on disease prevention, health promotion, rejuvenation and cure.

Question 33.
What are psychoactive drugs? Add a note Marijuana and Opium.
Answer:
Phytochemicals or drugs from some of the plants alter an individuals perceptions of mind by producing hallucination are known as psychoactive drugs.

1. Marijuana: Marijuana is obtained from Cannabis sativa. The active principle in Marijuana is trans – tetrahydrocanabinal (TCH). It is used as pain killer and reduce hypertension. It is also used in the treatment of Glaucoma, cancer radiotherapy and asthma, etc.
2. Opium: Opium is obtained from the exudates of the fruits of papaver somniferum (poppy plants). It is used to induce sleep and relieve pain. Opium yields morphine which is used as a strong analgesic in surgeries.

Question 34.
What are the King and Queen of spices? Explain about them and their uses.
Answer:

1. King of Spices: Pepper is one of the most important Indian spices referred to as the “King of Spices” and also termed as “Black Gold of India”. Kerala, Karnataka and Tamil Nadu are the top producers in India. The characteristic pungency of the pepper is due to the presence of alkaloid Pipeline. There are two types of pepper available in the market namely black and white pepper.
2. Uses: It is used for flavouring in the preparation of sauces, soups, curry powder and pickles. It is used in medicine as an aromatic stimulant for enhancing salivary and gastric secretions and also as a stomachic. Pepper also enhances the bio-absorption of medicines.
3. Queen of Spices: Cardamom is called as “Queen of Spices”. In India, it is one of the main cash crops cultivated in the Western Ghats, and North Eastern India.
4. Uses: The seeds have a pleasing aroma and a characteristic warm, slightly pungent taste. It is used for flavouring confectionaries, bakery products and beverages. The seeds are used in the preparation of curry powder, pickles and cakes. Medicinally, it is employed as a stimulant and carminative. It is also chewed as a mouth freshener.

Question 35.
How will you prepare an organic pesticide for your home garden with the vegetables available from your kitchen?
Answer:
Step 1: Mix 120 g of hot chillies with 110 g of garlic or onion. Chop them thoroughly.

Step 2: Blend the vegetables together manually or using an electric grinder until it forms a thick paste.

Step 3: Add the vegetable paste to 500 ml of warm water. Give the ingredients a stir to thoroughly mix them together.

Step 4: Pour the solution into a glass container and leave it undisturbed for 24 hours. If possible, keep the container in a sunny location. If not, at least keep the mixture in a warm place.

Step 5: Strain the mixture. Pom- the solution through a strainer, remove the vegetables and collect the vegetable-infused water and pour into another container. This filtrate is the pesticide. Either discard the vegetables or use it as a compost.

Step 6: Pour the pesticide into a squirt bottle. Make sure that the spray bottle has first been cleaned with warm water and soap to get rid it of any potential contaminants. Use a funnel to transfer the liquid into the squirt bottle and replace the nozzle.

Step 7: Spray your plants with the pesticide. Treat the infected plants every 4 to 5 days with the solution. After 3 or 4 treatments, the pest will be eliminated. If the area is thoroughly covered with the solution, this pesticide should keep bugs away for the rest of the season.

Samacheer Kalvi 12th Bio Botany Economically Useful Plants and Entrepreneurial Botany Additional Questions and Answers

1 – Mark Questions

Question 1.
Paddy, Wheat and Sorghum, etc., comes under the category of cereals. All the members of cereals belong to which of the following family?
(a) Fabaceae
(b) Poaceae
(c) Leguminoneae
(d) Caesalpinaeceae
Answer:
(b) Poaceae

Question 2.
Match the common names of the given plant species with their respective binomials

Answer:
(a) A – iii, B – iv, C – iv, C – ii and D – i

Question 3.
Given below are the plant species and their parts used. Which is the incorrect pair(s)?
(i) Cajanus cajan : Seeds
(ii) Anacardium occidentale : nuts
(iii) Borassus flabellifer : Endosperm
(iv) Capsicum annum : leaves
(a) i and ii
(b) ii and iii
(c) iii only
(d) iv only
Answer:
(d) iv only

Question 4.
Identify the tamil name for flaked rice
(a) Nel
(b) Aval
(c) Pori
(d) Umi
Answer:
(b) Aval

Question 5.
Pigeon pea is the common name for.
(a) Vigna radiata
(b) Vigna mungo
(c) Cajanus cajan
(d) Sorghum vulgare
Answer:
(c) Cajanus cajan

Question 6.
Sorghum is native to ______ continent.
Answer:
Africa

Question 7.
Match the following:

Answer:
(a) 1 – ii, 2 – iii, 3 – iv and 4 – i

Question 8.
_____ is the state tree of Tamil Nadu.
Answer:
Palmyra

Question 9.
Statement 1: Arachis hypogea belongs to Fabaceae Statement 2: It is a native of Brazil.
(a) Statement 1 is correct and Statement 2 is also correct
(b) Statement 1 is correct and Statement 2 is incorrect
(c) Statement 1 is incorrect and Statement 2 is correct
(d) Both the Statements are incorrect
Answer:
(a) Statement 1 is correct and Statement 2 is also correct

Question 10.
Statement 1: Chinese discovered the paper.
Statement 2: Eucalyptus and Casuarina are the widely used tree species for making paper pulp.
(a) Statement 1 is correct and Statement 2 is also correct
(b) Statement 1 is correct and Statement 2 is incorrect
(c) Statement 1 is incorrect and Statement 2 is correct
(d) Both the Statements are incorrect
Answer:
(a) Statement 1 is correct and Statement 2 is also correct

Question 11.
Statement 1: Andrographis paniculata is known as King of Bitters.
Statement 2: The decoction of Andrographis is used against Diabetes mellitus.
(a) Statement 1 is correct and Statement 2 is also correct
(b) Statement 1 is correct and Statement 2 is incorrect
(c) Statement 1 is incorrect and Statement 2 is correct
(d) Both the Statements are incorrect
Answer:
(b) Statement 1 is correct and Statement 2 is incorrect

Question 12.
Statement 1: Aloe vera belongs to the family Asphodelaceae.
Statement 2: Jasminum grandiflorum belongs to the family of Oleaceae.
(a) Statement 1 is correct and Statement 2 is also correct
(b) Statement 1 is correct and Statement 2 is incorrect
(c) Statement 1 is incorrect and Statement 2 is correct
(d) Both the Statements are incorrect
Answer:
(a) Statement 1 is correct and Statement 2 is also correct

Question 13.
Which district of Tamil Nadu is the World’s largest wholesale turmeric dealer?
Answer:
Erode

Question 14.
Assertion (A): Turmeric is used to treat cancer.
Reason (R): Curcumin is biomolecule present in turmeric.
(a) A is right R is wrong
(b) Both A and R are wrong
(c) A is wrong R is right
(d) A and R are right. R explains A.
Answer:
(d) A and R are right. R explains A.

Question 15.
Assertion (A): Black pepper is a spice.
Reason (R): Condiments are flavouring substances, generally added after the cooking of food.
(a) A is right R is wrong
(b) R explains A
(c) Both A and R are right. R is not correct explanation for A.
(d) Both A and R are wrong
Answer:
(c) Both A and R are right. R is not correct explanation for A.

Question 16.
Select the new world species of cotton.
(i) Gossypium hirsutum
(ii) Gossypium arboreum
(iii) Gossypium arboreum
(iv) Gossipium
(a) i and ii only
(b) i and iii only
(c) iii and iv only
(d) ii and iv only
Answer:
(a) i and ii only

Question 17.
Binomial of teak is ________
Answer:
Tectona grandis

Question 18.
The plant source of Marijuana is ________
(a) Andrographis paniculata
(b) Phyllanthus maderspatensis
(c) Cannabis sativa
(d) Papaver somniferum
Answer:
(c) Cannabis sativa

Question 19.
Identify the incorrect statements:
(a) Morphine is used as potent hepatoprotective.
(b) Phyllanthin is used as a strong analgesic in surgery.
(c) Indian Acalypha is used to cure skin diseases.
(d) Cissus quadrangularis is widely used for treating bone fractures.
(i) a and c
(ii) a and d
(iii) b and c
(iv) a and b
Answer:
(iv) a and b

Question 20.
Identify the mismatched pair:
(a) Holy basil – Ocimum sanctum
(b) Indian gooseberry – Phyllanthus amarus
(c) Vilvam – Aegle marmelos
(d) Veldt grape – Cissus quadrangularis
Answer:
(b) Indian gooseberry – Phyllanthus amarus

Question 21.
The active principle in Marijuana is _________
Answer:
Trans-tetrahydrocanabinal (THC).

Question 22.
The second GI tag for Jasmine is ‘Madurai Malli’, which jasmine type has got the first GI tag?
Answer:
‘Mysore Malli’

Question 23.
The principal coloring agent present in the leaf paste of Lawsonia inermis which give orange to dark red when applied on skin is _______
Answer:
Lacosone

Question 24.
Which state of India ranks first in coffee consumption?
Answer:
Tamil Nadu

Question 25.
Name the plant family to which the coffee plant belongs to?
Answer:
Rubiaceae

2 – Mark Questions

Question 1.
Give the binomials of paddy and wheat.
Answer:

1. Paddy : Oryza sativa.
2. Wheat: Triticum aestivum.

Question 2.
What is maida? Mention its culinary purpose.
Answer:
Processed wheat flour, that has little fibre, is called Maida which is used extensively in making parota, naan and bakery products.

Question 3.
Name any four millet varieties.
Answer:
Finger millet, Sorghum, Foxtail millet and Kodo millet.

Question 4.
Ragi rich food helps to overcome bone related ailments – justify.
Answer:
Ragi (Finger millet) is a kind of millet which has less glycemic index and rich in calcium. Hence consuming ragi food items enhances the bone strength.

Question 5.
Eating millets is good for diabetic patients. How?
Answer:
Millets are gluten free and have less glycemic index, hence it is good for diabetic patients.

Question 6.
Write the health benefits of Kodo millet.
Answer:
Kodo millet is a good diuretic and cures constipation. Helps to reduce obesity, blood sugar and blood pressure.

Question 7.
Name the family does the cereals and pulses belong to.
Answer:

1. Cereals belong to the family Poaceae.
2. Pulses belong to the family Fabaceae.

Question 8.
Write the common name for

1. Vigna radiata
2. Cajanus cajatt

Answer:

1. Vigna radiata – Green gram.
2. Cajanus cajan – Red gram or pigeon pea.

Question 9.
Why do we take vegetables in our diet?
Answer:
Vegetables are the important part of healthy eating and provide many nutrients, including potassium, fiber, folic acid and vitamins A, E and C. The nutrients in vegetables are vital for maintenance of our health.

Question 10.
Write the binomial and the family to which lady’s finger belongs to.
Answer:
Binomial of lady’s finger is Abelmoschus esculentus.
Family: Malvaceae.

Question 11.
Which is our National fruit and State tree of Tamil Nadu?
Answer:

1. National fruit of India is Mango.
2. State tree of Tamil Nadu is Palmyra.

Question 12.
Point out the uses of sugarcane.
Answer:
Sugar cane is the raw material for extracting white sugar. Sugarcane supports large number of industries like sugar mills producing refined sugars, distilleries producing liquor grade ethanol and millions of jaggery manufacturing units. Fresh sugarcane juice is a refreshing drink. Molasses is the raw material for the production of ethyl alcohol.

Question 13.
What is toddy?
Answer:
Inflorescence of palmyra is tapped for its sap which is used as health drink. Sap is processed to get palm jaggery or fermented to give toddy.

Question 14.
How coffee plants are introduced to India?
Answer:
Coffea arabica is the prime source of commercial coffee which is native to the tropical Ethiopia. An Indian Muslim saint, Baba Budan introduced coffee from Yemen to Mysore.

Question 15.
Compare Spices and Condiments.
Answer:
Spices are accessory foods mainly used for flavouring during food preparation to improve , their palatability. Spices are aromatic plant products and are characterized by sweet or bitter taste. Spices are added in minimal quantities during the cooking process. For example black pepper. Condiments, on the other hand, are flavouring substances having a sharp taste and are usually added to food after cooking. For example, curry leaves.

Question 16.
Mention any two Jute species.
Answer:

1. Corchorus capsularis
2. Corchorus olitorius

Question 17.
Name the plant species used for making paper pulp.
Answer:
Wood of Melia azadirachta, Neolamarkia chinensis, Casuarina spp and Eucalyptus spp are used for making paper pulp.

Question 18.
What does NCB stands for? Mention its role.
Answer:
The Narcotics Control Bureau (NCB) is the nodal drug law enforcement and intelligence agency of India and is responsible for fighting drug trafficking and the abuse of illegal substances.

3 – Mark Questions

Question 19.
Write a short note on foxtail millet. Foxtail Millet
Answer:

1. Botanical name: Setaria italica This is one of the oldest millet used traditionally in India. Which is domesticated first in China about 6000 years. Rich in protein, carbohydrate, vitamin B and C, Potassium and Calcium.
2. Uses:It supports in strengthening of heart and improves eye sight. Thinai porridge is given to lactating mother.

Question 20.
Write the botanical name and culinary uses of green gram.
Answer:
Botanical name: Vigna radiata Uses It can be used as roasted, cooked and sprouted pulse. Green gram is one of the ingredients of pongal, a popular breakfast dish in Tamil Nadu. Fried dehulled and broken or whole green gram is used as popular snack. The flour is traditionally used as a cosmetic, especially for the skin.

Question 21.
Mention the ways in which mangoes are used in Indian culinary.
Answer:
Mango is the major table fruit of India, which is rich in beta carotenes. It is utilized in many ways, as dessert, canned, dried and preserves in Indian cuisine. Sour, unripe mangoes are used in chutneys, pickles, side dishes, or may be eaten raw with salt and chili. Mango pulp is made into jelly. Aerated and non-aerated fruit juice is a popular soft drink.

Question 22.
Write a note on origin and area of cultivation of cotton.
Answer:
Cotton is one of the oldest cultivated crops of the world. It has been cultivated for about 8000 years both in new world and in old world. Commercial cotton comes from four cotton species: two from the new world and two from the old world.

1. Gossypium hirsutum
2. G.barbadense are the New world species and
3. G. arboretum
4. G. herbaceum are the old world species. In India, cotton is cultivated in Gujarat, Maharashtra, Andhra Pradesh and Tamil Nadu.

Question 23.
What is vulcanization? Who invented it?
Answer:
The heating of the rubber with sulphur under pressure at 150°C is called vulcanization. It was invented by Charles Goodyear.

Question 24.
Write a note on Henna.
Answer:

1. Botanical name: Lawsonia inermis.
2. Family: Lythraceae.
3. Origin and Area of cultivation: It is indigenous to North Africa and South-west Asia. It is grown mostly throughout India, especially in Gujarat, Madhya Pradesh and Rajasthan.
4. Uses: An orange dye ‘Henna’ is obtained from the leaves and young shoots of Lawsonia inermis. The principal colouring matter of leaves Tacosone” is harmless and causes no irritation to the skin. This dye has long been used to dye skin, hair and finger nails. It is used for. colouring leather, for the tails of horses and in hair-dyes.

Question 25.
What are organic pesticides?
Answer:
Pest like aphids, spider and mites can cause serious damage to flowers, fruits, and vegetables. These creatures attack the garden in swarms, and drain the life of the crop and often invite disease in the process. Many chemical pesticides prove unsafe for human and the environment. It turns fruits and vegetables unsafe for consumption. Thankfully, there are many homemade, organic options to turn to war against pests.

5 – Mark Questions

Question 26.
Write the botanical name, origin, cultivational area and uses of Black gram.
Answer:

1. Botanical name : Vigna mungo
2. Origin and Area of cultivation : Black gram is native to India. Earliest archeobotanical evidences record the presence of black gram about 3,500 years ago. It is cultivated as a rain fed crop in drier parts of India. India contributes to 80% of the global production of black gram. Important states growing black gram in India are Uttar Pradesh, Chattisgarh and Karnataka.
3. Uses : Black gram is eaten whole or split, boiled or roasted or ground into fl our. Black gram batter is a major ingredients for the preparation of popular Southern Indian breakfast dishes. Split pulse is used in seasoning Indian curries.

Question 27.
Give a detailed account on any one fibe yielding plant. Jute
Answer:

1. Botanical name : Corchorus app.
2. Family: Malvaceae
3. Origin and Area of cultivation: Jute is derived from the two cultivated species Corchorus capsularis  Colittorius is of African origin whereas C. Capsularis, is believed to Indo- Burmaese origin. It is an important cultivated commercial crop in Gangetic plains of India and Bangladesh.
4. Use : It is one of the largest exported fibre material of India. The jute industry occupies an important place in the national economy of India. Jute is used for ‘safe’ packaging in view of being natural, renewable, bio-degradable and eco-friendly product. It is used in bagging and wrapping textile. About 75% of the jute produced is used for manufacturing sacks and bags. It is also used in manufacture of blankets, rags, curtains etc. It is also being used as a textile fibre in recent years.

Question 28.
Draw a table mentioning binominals, family name, useful part and their medical uses of any five local medicinal plants.
Answer:

Question 29.
Explain the stepwise preparation of Bio-pest repellent from the leaves of Azadirachta indica
Answer:
Botanical pest repellent and insecticide made with the dried leaves of Azadirachta indica
Preparation of Bio-pest repellent

1. Pluck leaves from the neem tree and chop the leaves finely.
2. The chopped up leaves were put in a 50-liter container and fill to half with water; put the lid on and leave it for 3 days to brew.
3. Using another container, strain the mixture which has brewed for 3 days to remove the leaves, through fine mesh sieve. The filtrate can be sprayed on the plants to repel pests.
4. To make sure that the pest repellent sticks to the plants, add 100 ml of cooking oil and the same amount of soap water. (The role of the soap water is to break down the oil, and the role of the oil is to make it stick to the leaves).
5. The stewed leaves from the mixture can be used in the compost heap or around the base of the plants.

Higher Order Thinking Skills (HOTs) Questions

Question 1.
The following are the binomials of economically useful plants. Mention their respective families.
(a) Borassus flabellifer
(b) Mangifera indica
(c) Saccharum officinarum
(d) Curcuma longa
(e) Capsicum annum
(f) Tamarindus indica
Answer:

Question 2.
Name the plant source from which the following products are obtained.

1. Maida
2. Rubber
3. Morphine
4. Coffee

Answer:

1. Maida – Triticum durum
2. Rubber – Hevea brasiliensis
3. Morphine – Papaver somniferum
4. Coffee – Coffea arabica

Question 3.
Mention the plant species or their products which are popularly known as

1. Queen of species
2. Black Gold of India and
3. King of bitters.

Answer:

1. Queen of Species – Cardamom
2. Black Gold of India – Pepper
3. King of bitters – Andrographis paniculata

Question 4.
Complete the statements.

1. Alphonsa is a variety of ______
2. Cayenne pepper is a variety of ______

Answer:

1. Mango
2. Capsicum

Question 5.
Siddha medicine is an age old practice in our nation. According to this medicinal system, what is the major cause for all sort of illness in humans?
Answer:
According to Siddha medicine, three humors namely vatam, pittam and kapam are responsible for the health of human being and any disturbance in the equilibrium of these humors result in ill health.

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Tamilnadu Samacheer Kalvi 12th Bio Botany Solutions Chapter 2 Classical Genetics

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Samacheer Kalvi 12th Bio Botany Classical Genetics Text Book Back Questions and Answers

Question 1.
Extra nuclear inheritance is a consequence of presence of genes in __________
(a) Mitrochondria and chloroplasts
(b) Endoplasmic reticulum and mitrochondria
(c) Ribosomes and chloroplast
(d) Lysosomes and ribosomes
Answer:
(a) Mitrochondria and chloroplasts

Question 2.
In order to find out the different types of gametes produced by a pea plant having the genotype AaBb, it should be crossed to a plant with the genotype __________
(a) aaBB
(b) AaBB
(c) AABB
(d) aabb
Answer:
(d) aabb

Question 3.
How many different kinds of gametes will be produced by a plant having die genotype AABbCC?
(a) Three
(b) Four
(c) Nine
(d) Two
Answer:
(d) Two

Question 4.
Which one of the following is an example of polygenic inheritance?
(a) Flower colour in MirabilisJalapa
(b) Production of male honey bee
(c) Pod shape in garden pea
(d) Skin Colour in humans
Answer:
(d) Skin Colour in humans

Question 5.
In Mendel’s experiments with garden pea, round seed shape (RR) was dominant over wrinkled seeds (rr), yellow cotyledon (YY) was dominant over green cotyledon (yy). What are the expected phenotypes in the F₂ generation of the cross RRYY xrryy?
(a) Only round seeds with green cotyledons
(b) Only wrinkled seeds with yellow cotyledons
(c) Only wrinkled seeds with green cotyledons
(d) Round seeds with yellow cotyledons an wrinkled seeds with yellow cotyledons
Answer:
(d) Round seeds with yellow cotyledons an wrinkled seeds with yellow cotyledons

Question 6.
Test cross involves __________
(a) Crossing between two genotypes with recessive trait
(b) Crossing between two F₁ hybrids
(c) Crossing the F₁ hybrid with a double recessive genotype
(d) Crossing between two genotypes with dominant trait
Answer:
(c) Crossing the F₁ hybrid with a double recessive genotype

Question 7.
In pea plants, yellow seeds are dominant to green. If a heterozygous yellow seed pant is crossed with a green seeded plant, what ratio of yellow and green seeded plants would you expect in generation?
(a) 9:3
(b) 1:3
(c) 3:1
(d) 50:50
Answer:
(d) 50:50

Question 8.
The genotype of a plant showing the dominant phenotype can be determined by __________
(a) Back cross
(b) Test cross
(c) Dihybrid cross
(d) Pedigree analysis
Answer:
(b) Test cross

Question 9.
Select the correct statement from the ones given below with respect to dihybrid cross
(a) Tightly linked genes on the same chromosomes show very few combinations
(b) Tightly linked genes on the same chromosomes show higher combinations
(c) Genes far apart on the same chromosomes show very few recombinations
(d) Genes loosely linked on the same chromosomes show similar recombinations as the tightly I linked ones
Answer:
(a) Tightly linked genes on the same chromosomes show very few combinations

Question 10.
Which Mendelian idea is depicted by a cross in which the F₁ generation resembles both the parents
(a) Incomplete dominance
(b) Law of dominance
(c) Inheritance of one gene
(d) Co-dominance
Answer:
(d) Co-dominance

Question 11.
Fruit colour in squash is an example of __________
(a) Recessive epistasis
(b) Dominant epistasis
(c) Complementary genes
(d) Inhibitory genes
Answer:
(b) Dominant epistasis

Question 12.
In his classic experiments on Pea plants, Mendel did not use __________
(a) Flowering position
(b) Seed colour
(c) Pod length
(d) Seed shape
Answer:
(c) Pod length

Question 13.
The epistatic effect, in which the hybrid cross 9:3:3:1 between AaBb Aabb is modified as
(a) Dominance of one allele on another allele of both loci
(b) Interaction between two alleles of different loci
(c) Dominance of one allele to another alleles of same loci
(d) Interaction between two alleles of some loci
Answer:
(b) Interaction between two alleles of different loci

Question 14.
In a test cross involving F₁ dihybrid flies, more parental type offspring were produced than the recombination type offspring. This indicates __________
(a) The two genes are located on two different chromosomes
(b) Chromosomes failed to separate during meiosis
(c) The two genes are linked and present on the some chromosome
(d) Both of the characters are controlled by more than one gene
Answer:
(c) The two genes are linked and present on the some chromosome

Question 15.
The genes controlling the seven pea characters studied by Mendel are known to be located on h6w many different chromosomes?
(a) Seven
(b) Six
(c) Five
(d) Four
Answer:
(a) Seven

Question 16.
Which of the following explains how progeny can posses the combinations of traits that none
of the parent possessed?
(a) Law of segregation
(b) Chromosome theory
(c) Law of independent assortment
(d) Polygenic inheritance
Answer:
(d) Polygenic inheritance

Question 17.
“Gametes are never hybrid”. This is a statement of __________
(a) Law of dominance
(b) Law of independent assortment
(c) Law of segregation
(d) Law of random fertilization
Answer:
(c) Law of segregation

Question 18.
Gene which suppresses other genes activity but does not lie on the same locus is called as __________
(a) Epistatic
(b) Supplement only
(c) Hypostatic
(d) Codominant
Answer:
(c) Hypostatic

Question 19.
Pure tall plants are crossed with pure dwarf plants. In the F₁ generation, all plants were tall. These tall plants of generation were selfed and the ratio of tall to dwarf plants obtained was 3:1. This is called __________
(a) Dominance
(b) Inheritance
(c) Codominance
(d) Heredity
Answer:
(a) Dominance

Question 20.
The dominant epistatis ratio is _________
(a) 9:3:3:1
(b) 12:3:1
(c) 9:3:4
(d) 9:6:1
Answer:
(b) 12:3:1

Question 21.
Select the period for Mendel’s hybridization experiments.
(a) 1856 – 1863
(b) 1850 – 1870
(c) 1857-1869
(d) 1870 – 1877
Answer:
(a) 1856 – 1863

Question 22.
Among the following characters which one was not considered by Mendel in his experimentation pea?
(a) Stem – Tall or dwarf
(b) Trichomal glandular or non-glandular
(c) Seed – Green or yellow
(d) Pod – Inflated or constricted
Answer:
(b) Trichomal glandular or non-glandular

Question 23.
Name the seven contrasting traits of Mendel.
Answer:
Plant Height, Seed Shape, Cotyledon colour, Flower colour, Pod colour, Pod form, Flower position

Question 24.
What is meant by true breeding or pure breeding lines / strain?
Answer:
A true breeding lines (Pure-breeding strains) means it has undergone continuous self pollination having stable trait inheritance from parent to offspring. Matings within pure breeding lines produce offsprings having specific parental traits that are constant in inheritance and expression for many generations. Pure line breed refers to homozygosity only.

Question 25.
Give the names of the scientists who rediscovered Mendelism.
Answer:
Mendel’s experiments were rediscovered by three biologists, Hugo de Vries of Holland, Car Correns of Germany and Erich von Tschermak of Austria.

Question 26.
What is back cross?
Answer:
Back cross is a cross of F₁ hybrid with any one of the parental genotypes. The back cross is of two types; they are dominant back cross and recessive back cross. It involves the cross between the F₁ off spring with either of the two parents.

Question 27.
Define Genetics.
Answer:
“Genetics” is the branch of biological science which deals with the mechanism of transmission of characters from parents to offsprings. The term Genetics was introduced by W. Bateson in 1906.

Question 28.
What are multiple alleles?
Answer:
Three or more alternative forms of a gene that occupy the same locus and control the expression of a single trait.
E.g : ABO blood group

Question 29.
What are the reasons for Mendel’s successes in his breeding experiment?
Answer:
Mendel was successful because:

1. He applied mathematics and statistical methods to biology and laws of probability to his breeding experiments.
2. He followed scientific methods and kept accurate and detailed records that include quantitative data of the outcome of his crosses.
3. His experiments were carefully planned and he used large samples.
4. The pairs of contrasting characters which were controlled by factor (genes) were present on separate chromosomes.
5. The parents selected by Mendel were pure breed lines and the purity was tested by self
   crossing the progeny for many generations.

Question 30.
Explain the law of dominance in monohybrid cross.
Answer:
Law of dominance states that the offsprings of an individual with contrasting (dissimilar) traits will only express the dominant trait in F₁ generation and both the characters are expressed in F₂ generation. This law also explains the proportion of 3 : 1 ratio in F₂ generation.

Question 31.
Differentiate incomplete dominance and codominance.
Answer:
Incomplete Dominance:

1. In incomplete dominance, neither of the allele is not completely dominant to another allele rather combine and produce new trait
2. New phenotype is formed due to character blending (not alleles)
3. Example : Pink flowers of Mirabilis Jalapa

Co-dominance:

1. In co-dominance, both the alleles in heterozygote are dominant and the traits are equally expressed (joint expression)
2. No formation of new phenotype rather both dominant traits are expressed, conjointly
3. Example: Red and white flowers of camellia

Question 32.
What is meant by cytoplasmic inheritance
Answer:
DNA is the universal genetic material. Genes located in nuclear chromosomes follow Mendelian inheritance. But certain traits are governed either by the chloroplast or mitochondrial genes. This phenomenon is known as extra nuclear inheritance. It is a kind of Non-Mendelian inheritance. Since it involves cytoplasmic organelles such as chloroplast and mitochondrion that act as inheritance vectors, it is also called Cytoplasmic inheritance.

Question 33.
Describe dominant epistasis with an example.
Answer:
Dominant Epistasis – It is a gene interaction in which two alleles of a gene at one locus interfere and suppress or mask the phenotypic expression of a different pair of alleles of another gene at another locus. The gene that suppresses or masks the phenotypic expression of a gene at another locus is known as epistatic.

The gene whose expression is interfered by non-allelic genes and prevents from exhibiting its character is known as hypostatic. When both the genes are present together, the phenotype is determined by the epistatic gene and not by the hypostatic gene.

In the summer squash the fruit colour locus has a dominant allele ‘W’ for white colour and a recessive allele ‘w’ for coloured fruit. ‘W’ allele is dominant that masks the expression of any colour.

Dominant epistasis in summer squash

In another locus hypostatic allele ‘G’ is for yellow fruit and its recessive allele ‘g’ for green fruit. In the first locus the white is dominant to colour where as in the second locus yellow is dominant to green. When the white fruit with genotype WWgg is crossed with yellow fruit with genotype wwGG, the F₁ plants have white fruit and are heterozygous (WwGg). When F₁ heterozygous plants are crossed.

they give rise to F₂ with the phenotypic ratio of 12 white : 3 yellow : 1 green.Since W is epistatic to the alleles ‘G’ and ‘g’, the white which is dominant, masks the effect of yellow or green. Homozygous recessive ww genotypes only can give the coloured fruits (4/16). Double recessive ‘wwgg’ will give green fruit (1/16). The Plants having only ‘G’ in its genotype (wwGg or wwGG) will give the yellow fruit(3/l 6).

Question 34.
Explain polygenic inheritance with an example.
Answer:
Polygenic inheritance – Several genes combine to affect a single trait. A group of genes that together Dark Red determine (contribute) a characteristic of an organism is called polygenic inheritance. It gives explanations to the inheritance of continuous traits which are compatible with Mendel’s Law. The first experiment on polygenic inheritance was demonstrated by Swedish Geneticist H.

Nilsson-Ehle (1909) in wheat kernels. Kernel colour is controlled by two genes each with two alleles, one with red kernel colour was dominant to white. He crossed the two pure breeding wheat varieties dark red and a white. Dark red genotypes F₁ generation R₁R₁R₂R₂ and white genotypes are r₁r₁r₂r₂ – F₁ generation medium red were obtained with the genotype R₁r₁R₂r₂. F₁ wheat plant produces

four types of gametes R₁R₂, R₁r₂, r,₁r₂. The intensity of the red colour is determined by the number of R genes in the F₂ generation. Four R genes: A dark red kernel colour is obtained. Three R genes: Medium – dark red kernel colour is obtained. Two R genes: Medium-red kernel colour is obtained. One R gene: Light red kernel colour is obtained. Absence of R gene: Results in White kernel colour.

The R gene in an additive manner produces the red kernel colour. The number of each phenotype is plotted against the intensity of red kernel colour which produces a bell shaped curve. This represents the distribution of phenotype.

Conclusion: Finally the loci that was studied by Nilsson – Ehle were not linked and the genes assorted independently. Later, researchers discovered the third gene that also affect the kernel colour of wheat. The three independent pairs of alleles were involved in wheat kernel colour. Nilsson – Ehle found the ratio of 63 red : 1 white in F₂ generation – 1 : 6 : 15 : 20 : 15 : 6 : 1 in F₂ generation.

Question 35.
Differentiate continuous variation with discontinuous variation.
Answer:
1. Discontinuous Variation: Within a population there are some characteristics which show a limited form of variation.
Example: Style length in Primula, plant height of garden pea. In discontinuous variation, the characteristics are controlled by one or two major genes which may have two or more allelic forms.

These variations are genetically determined by inheritance factors. Individuals produced by this variation show differences without any intermediate form between them and there is no overlapping between the two phenotypes. The phenotypic expression is unaffected by environmental conditions. This is also called as qualitative inheritance

2. Continuous Variation: This variation may be due to the combining effects of environmental and genetic factors. In a population most of the characteristics exhibit a complete gradation, from one extreme to the other without any break. Inheritance of phenotype is determined by the combined effects of many genes, (polygenes) and environmental factors. This is also known as quantitative inheritance.
Example: Human height and skin color.

Question 36.
Explain with an example how single genes affect multiple traits and alleles the phenotype of an organism.
Answer:
In Pleiotropy, the single gene affects multiple traits and alter the phenotype of the organism. The Pleiotropic gene influences a number of characters simultaneously and such genes are called pleiotropic gene. Mendel noticed pleiotropy while performing breeding experiment with peas (Pisum sativum).

Peas with purple flowers, brown seeds and dark spot on the axils of the leaves were crossed with a variety of peas having white flowers, light coloured seeds and no spot on the axils of the leaves, the three traits for flower colour, seed colour and a leaf axil spot all were inherited together as a single unit. This is due to the pattern of inheritance where the three traits were controlled by a single gene with dominant and recessive alleles. Example: sickle cell anemia.

Question 37.
Bring out the inheritance of chloroplast gene with an example.
Answer:
Chloroplast Inheritance

It is found in 4 ‘O’ Clock plant (Mirabilis jalapa). In this, there are two types of variegated leaves namely dark green leaved plants and pale green leaved plants. When the pollen of dark green leaved plant (male) is transferred to the stigma of pale green leaved plant (female) and pollen of pale green leaved plant is transferred to the stigma of dark green leaved plant, the F₁ generation of both the crosses must be identical as per Mendelian inheritance. But in the reciprocal cross the F₁ plant differs from each other.

In each cross, the F plant reveals the character of the plant which is used as female plant. This inheritance is not through nuclear gene. It is due to the chloroplast gene found in the ovum of the female plant which contributes the cytoplasm during fertilization since the male gamete contribute only the nucleus but not cytoplasm.

Samacheer Kalvi 12th Bio Botany Classical Genetics Additional Questions and Answers

1 – Mark Questions

Question 1.
The term ‘Genetics’ was introduced by __________
(a) Gregor Mendel
(b) Bateson
(c) Hugo de vries
(d) Carl Correns
Answer:
(b) Bateson

Question 2.
Which is not a correct statements?
(A) Variations are the raw materials for evolution
(B) Variations provide genetic material for natural selection
(C) It helps the individual to adapt to changing environment
(D) Variations allow breeders to improve the crop field
(a) A and D
(b) B only
(c) C and D
(d) nono of he above
Answer:
(d) nono of he above

Question 3.
The process of removal of anthers from the flower is called __________
Answer:
Emasculation

Question 4.
An allede is __________
(a) another word for a gene
(b) alternate forms of a gene
(c) morphological expression of a gene
(d) genitic
Answer:
(b) alternate forms of a gene

Question 5.
Gregor Mendel __________
(i) was born in Czechoslovakia
(ii) did his experiments in Pisum fulvum
(iii) was the first systemic researcher in genetics
(iv) Published his results in the paper “Experiments on Plant Hybrids”
(a) All are correct
(b) (ii),(iii), (iv) are correct
(c) (i), (iii),(iv) are correct
(d) (i), (iii),(iv) are correct
Answer:
(c) (i), (iii),(iv) are correct

Question 6.

Answer:
A – (ii) B – (iv) C – (iii) D – (i)

Question 7.
How many characters studied by Mendel in pisum sativum
(a) Three
(b) Five
(c) Seven
(d) Nine
Answer:
(c) Seven

Question 8.
Mendel’s work were rediscovered by __________
(a) Hugo de Vries
(b) Tschermak
(c) Carl Correns
(d) All the above
Answer:
(d) All the above

Question 9.
Crossing of F₁, to any one of the parent refers to __________
(a) selling
(b) back cross
(c) test cross
(d) all of the above
Answer:
(b) back cross

Question 10.
Match the following

Answer:
A – (ii)
B – (iv)
C – (i)
D – (iii)

Question 11.
In an intergenic interaction, the gene that suppresses the pherotype of a gene is said to Crossing of F, to any one of the parent refers to __________
(a) Dominant
(b) Inhibitory
(c) Epistatic
(d) Hypostatic
Answer:
(c) Epistatic

Question 12.
Assertion (A) : Test cross is done between F₂ hybrid with F₁ recessive
Reason (R) : It helps to identify the homozygosity of hybrids
(a) A and R are correct R explains A
(b) A and R are incorrect
(c) A is correct R is incorrect
(d) A is incorrect R is correct
Answer:
(b) A and R are incorrect

Question 13.
Assertion (A) : Codominance is an example for intragenic interaction
Reason (R) : Interaction take place between the alleles of same gene
(a) A and R are correct R explains A
(b) A and R are incorrect
(c) A is correct R is incorrect
(d) A is incorrect R is correct
Answer:
(a) A and R are correct R explains A

Question 14.
Assertion (A) : Pleiotropic gene affects multiple traits
Reason (R) : ABO blood group is an example for Pleiotropism
(a) A and R are correct R explains A
(b) A and R are incorrect
(c) A is correct R is incorrect
(d) A is incorrect R is correct
Answer:
(c) A is correct R is incorrect

Question 15.
Assertion (A) : Cytoplasmic male sterility is a Mendelian inheritance
Reason (R) : The genes for cytoplasmic male sterility in peal maize is located at mitochondrial DNA
(a) A and R are correct R explains A
(b) A and R are incorrect
(c) A is correct R is incorrect
(d) A is incorrect R is correct
Answer:
(d) A is incorrect R is correct

Question 16.
What is the phenotypic ratio in case of incomplete dominance
(a) 9 : 7
(b) 3 : 1
(c) 1 : 2 : 1
(d) 1 : 1 : 1 : 1
Answer:
(c) 1 : 2 : 1

Question 17.
Identify the mismatched pair
(a) Chloroplast inheritance – Gregor Mendel
(b) Polygenic inheritance – H. Nilsson
(c) Lethal genes – E. Baur
(d) Incomplete dominance – Carl Correns
Answer:
(a) Chloroplast inheritance – Gregor Mendel

Question 18.
Statement 1 : Intergenic gene interaction occurs between alleles at same locus
Statement 2 : Co-dominance is an example for intergenic gene interaction
(a) Statement 1 is correct & Statement 2 is incorrect
(b) Statement 1 is incorrect & Statement 2 is correct
(c) Both Statements 1 & 2 are correct
(d) Both Statements 1 & 2 are incorrect
Answer:
(c) Both Statements 1 & 2 are correct

Question 19.
Statement 1 : Test cross is done between F₁ individual with homozygous recessive
Statement 2 : If F₁ individual is homozygous, the rate of a monohybrid cross will be 1:1
(a) Statement 1 is correct & Statement 2 is incorrect
(b) Statement 1 is incorrect & Statement 2 is correct
(c) Both Statements 1 & 2 are correct
(d) Both Statements 1 & 2 are incorrect
Answer:
(a) Statement 1 is correct & Statement 2 is incorrect

Question 20.
Identify the incorrect statement
Answer:
(a) In incomplete dominance, the traits are blended not the genes
(b) Incomplete dominance is noticed in Mirabilis jalapa by Carl Correns
(c) It is a type of Intragenic gene interaction
(d) Incomplete dominance F₂ ratio is 1 : 3 : 1
Answer:
(d) Incomplete dominance F₁ ratio is 1 : 3 :1

Question 21.
In case of co-dominance, monohybrid F_{1 __________} is 1 : 2 : 1
(a) Genotype ratio
(b) Phenotype ratio
(c) Both genotype & Phenotype ratio
(d) Ratio is wrong
Answer:
(c) Both genotype & Phenotype ratio

Question 22.
Identify the wrong statement (s)
(i) Monohybrid cross involve the inhertance of teo alleles of a gene
(ii) The dwarf traits reappeared in F₂
(iii) Law of dominance was proved by monohybrid cross
(iv) F₁ monohybrid was an hererozygous
(a) i and ii
(b) iii and iv
(c) i only
(d) none of the above
Answer:
(d) none of the above

Question 23.
Result of incomplete dominance is __________
(а) Intermediate genotype
(b) Intermediate phenotype
(c) Recessive phenotype
(d) Epistasis
Answer:
(b) Intermediate phenotype

Question 24.
Heterozygous Tall mono hybrid is cross with homozygous dwarf. What will be characteristic of offspring?
(a) 25 % recessive 75% dominant
(b) 75 % recessive 25% dominant
(c) 50 % recessive 50% dominant
(d) All are dominance
Answer:
(c) 50 % recessive 50% dominant

Question 25.
ABO blood group is a classical example for __________
(a) Polygenic inheritance
(b) Incomplete Dominance
(c) Epistasis
(d) Dominance
Answer:
(d) Dominance

Question 26.
RR (Red) flower of Mirabilis is crossed with White (WW) flowers. Resultant offspring are pink RW. This is an example of __________
(a) Epistasis
(b) Co-dominance
(c) Incomplete dominance
(d) Pleiotropism
Answer:
(c) Incomplete dominance

Question 27.
How many genetically different gametes are produced by a plant have genotype TtYyRr?
(a) 2
(b) 4
(c) 6
(d) 8
Answer:
(d) 8

Question 28.
When a single gene influences multiple traits then the phenomenon is called __________
(a) Pleiotropy
(b) Polygenic inheritance
(c) Epistasis
(d) Atavism
Answer:
(a) Pleiotropy

Question 29.
According to Mendel which character shown dominance.
(a) Yellow flower color
(b) Yellow cotyledon color
(c) Wrinkled seeds
(d) Inflated pod
Answer:
(d) Inflated pod

Question 30.
Ratio of recessive epistasis is __________
(a) 12 : 3 : 1
(b) 9 : 7
(c) 9 : 3 : 4
(d) 9 : 6 : 1
Answer:
(c) 9 : 3 : 4

Question 31.
According to Mendel, which is not a dominant trait?
(a) Wrinkled seeds
(b) Purple flower
(c) Inflated pod form
(d) Axial flower portion
Answer:
(a) Wrinkled seeds

Question 32.
Identify the allelic interaction.
(a) Domination epistasis
(b) Co – dominance
(c) Recessive epistasis
(d) Duplicate genes
Answer:
(b) Co – dominance

Question 33.
Gametes are never hybrid’ is concluded by __________
(a) Law of dominance
(b) Law of segregation
(c) Law of independent environment
(d) Law of lethality
Answer:
(b) Law of segregation

Question 34.
Factor hypothesis was proposed by __________
(a) Reginald Punnett
(b) W. Bateson
(c) Gregor Mende
(d) Carl Correns
Answer:
(b) W. Bateson

Question 35.
The 1:2:1 ratio of co-dominance process Mendel’s __________
(a) Law of dominance
(b) Law of recessiveness
(c) Law of segregation
(d) Law of independent assortment
Answer:
(b) Law of recessiveness

Question 36.
Match the following:
Epistatic interaction Example
(A) Complementary genes (i) Seed capsule in xxxxx
(B) Supplementary genes (ii) Leaf color in rice plant
(C) Inhibitory genes (iii) Grain color in maize
(D) Duplicate genes (iv) Flower color in sweet peas
Answer:
A – (iv)
B – (iii)
C – (ii)
D – (i)

2 – Mark Questions

Question 1.
Who coined the term genetics? Also define it.
Answer:
“Genetics” is the branch of biological science which deals with the mechanism of transmission of characters from parents to off springs. The term Genetics was introduced by W. Bateson in 1906.

Question 2.
Name the four major subdisciplines of genetics.
Answer:
(a) Classical genetics
(b) Molecular genetics
(c) Population genetics
(d) Quantitative genetics

Question 3.
Define Heredity and variations.
Answer:
Heredity : Heredity is the transmission of characters from parents to off springs.
Variations : The organisms belonging to the same natural population or species that shows a difference in the characteristics is called variation.

Question 4.
Mendel’s theory is a particulate theory – justify.
Answer:
Mendel’s theory of inheritance, known as the Particulate theory, establishes the existence of minute particles or hereditary units or factors, which are now called as genes.

Question 5.
Which organism was studied by Gregor Mendel? How many traits does he considered on his experiments?
Answer:
Gregor Mendel selected seven pairs of characters in Pisum sativum (garden pea)

Question 6.
Name any four characters of pisum sativum that was studied by Mendel.
Answer:
Seed shape, flower color, flower position & pod color.

Question 7.
Define the terms

1. Emasculation
2. Alleles.

Answer:

1. Emasculation : Removal of anthers from the flower
2. Alleles : Alternate forms of a gene

Question 8.
Name the first and second law of Mendel.
Answer:

1. The Law of Dominance
2. The Law of Segregation

Question 9.
What is genotype & phenotype?
Answer:
genotype & phenotype

1. The term genotype is the genetic constitution of an individual.
2. The term phenotype refers to the observable characteristic of an organism.

Question 10.
Write the phenotypic and genotypic ratio of monohybrid cross.
Answer:
(a) Phenotypic ratio = 3:1.
(b) Genotypic ratio =1 : 2 : 1

Question 11.
What is test cross? Why it is done?
Answer:

1. Test cross is crossing an individual of unknown genotype with a homozygous recessive.
2. Test cross is used to identify whether an individual is homozygous or heterozygous for dominant character.

Question 12.
State the law of independent assortment.
Answer:
When two pairs of traits are combined in a hybrid, segregation of one pair of characters is independent to the other pair of characters. Genes that are located in different chromosomes assort independently during meiosis.

Question 13.
Give the phenotypic ratio of
(a) Dihybrid cross
(b) Dihybrid test cross
Answer:
(a) Dihybrid cross ratio = 9 : 3 : 3 : 1
(b) Dihybrid test cross ratio = 1 : 1 : 1 : 1

Question 14.
RrYyf (F₁ hybrid)  rryy (recessive parent). Name the type of cross. Mention its ratio.
Answer:
Dihybrid test cross and the ratio is 1 : 1 : 1 : 1

Question 15.
How many types of gametes are produced by a dihybrid plant. If the same plant is self fertilized, how many second generation offsprings are developed?
Answer:
Four different gametes are produced by a dihybrid plant and on selfing, it yield 16 off springs.

Question 16.
Write the phenotypic ratio of trihybrid cross.
Answer:
27 : 9 : 9 : 9 : 3 : 3 : 3 : 1

Question 17.
Define gene interaction.
Answer:
A single phenotype is controlled by more than one set of genes, each of which has two or more alleles. This phenomenon is called Gene Interaction.

Question 18.
Classify gene interactions with an example.
The gene interactions may be
(a) Intragenic gene interaction. E.g.: Codominance
(b) Intergenic gene interaction. E.g.: Epistasis

Question 19.
Provide any four intergenic gene interactions.
Answer:
(a) Incomplete dominance
(b) Codominance
(c) Multiple alleles
(d) Pleiotropic genes are common examples for intragenic interaction.

Question 20.
Define intragenic interaction
Answer:
Interactions take place between the alleles of the same gene i.e., alleles at the same locus is called intragenic or intralocus gene interaction.

Question 21.
In which plant does the incomplete dominance was studied by Carl Correns? Write the ratio of the cross.
Answer:
Mirabilis Jalapa (4 o’ clock plant). Incomplete dominance ratio is 1 : 2 : 1

Question 22.
What are lethal alleles? Give example.
Answer:
An allele which has the potential to cause the death of an organism is called a Lethal Allele.
E.g : Recessive lethality in Antirrhinum species.

Question 23.
Give the proper terminologies for the following statement
(a) Single gene affecting multiple traits
(b) Single trait affected by many genes.
Answer:
(a) Pleiotropism
(b) Poly genic inheritance

Question 24.
What is intergenic gene interactions? Give example
Answer:
Interlocus interactions take place between the alleles at different loci i.e. between alleles of different genes.
Eg: Dominant Epistasis

Question 25.
Name any two extranuclear inheritance.
Answer:
(a) Chloroplast inheritance
(b) Mitrochondrial inheritance

Question 26.
What are plasmogenes?
Answer:
Plasmogenes are independent, self-replicating, extra-chromosomal units located in cytoplasmic organelles, chloroplast and mitochondrion

Question 27.
What are extra nuclear inheritance?
Answer:
Certain characters/traits are governed and inherited by genes located in cytoplasmic organelles (chloroplast or mitochondrion) other than nucleus. This is called extra nuclear inheritance.

Question 28.
Why extranuclear inheritance is called as cytoplasmic inheritance.
Answer:
Extra nuclear inheritance is due to genes located on the cytoplasmic organelles such as chloroplast and mitochondrion hence it is called cytoplasmic inheritance.

Question 29.
What is cytoplasmic male sterility?
Answer:
In Sorghum vulgare (Pearl maize), the gene located for the sterility pollens are located in the mitochondrial DNA. This phenomenon is called as cytoplasmic male sterility.

3 – Mark Questions

Question 30.
Point out any three importance of variations.
Answer:

1. They help the individuals to adapt themselves to the changing environment.
2. Variations allow breeders to improve better yield, quicker growth, increased resistance and lesser input.
3. They constitute the raw materials for evolution.

Question 31.
Why Mendel selected pea plants for his experiments.
Answer:
He choose pea plant because,

1. It is an annual plant and has clear contrasting characters that are controlled by a single gene separately.
2. Self-fertilization occurred under normal conditions in garden pea plants. Mendel used both self-fertilization and cross-fertilization.
3. The flowers are large hence emasculation and pollination are very easy for hybridization.

Question 32.
State the law of segregation.
Answer:
The Law of Segregation (Law of Purity of gametes): Alleles do not show any blending. During the formation of gametes, the factors or alleles of a pair separate and segregate from each other such that each gamete receives only one of the two factors. A homozygous parent produces similar gametes and a heterozygous parent produces two kinds of gametes each having one allele with equal proportion. Gametes are never hybrid.

Question 33.
How many types of gametes are produced by heterozygous dihybrid plant with a genotypeRrYy? Write them.
Answer:
Four gametes – RY, Ry, rY, ry

Question 34.
Define trihybrid cross. Mention its F₂ phenotypic ratio.
Answer:
A cross between homozygous parents that differ in three gene pairs (i.e. producing trihybrids) is called trihybrid cross, F₂ Phenotypic ratio -27 : 9 : 9 : 9 : 3 : 3 : 3 : 1

Question 35.
Define co-dominance. How it is proved by using Gossypium species?
Answer:
The phenomenon in which two alleles are both expressed in the heterozygous individual is known as codominance. The codominance was demonstrated in plants with the help of electrophoresis or chromatography for protein or flavonoid substance.

Example: Gossypium hirsutum and Gossypium sturtianum, their F₁ hybrid (amphiploid) was tested for seed proteins i by electrophoresis. Both the parents have different banding patterns for their seed proteins. In hybrids, additive banding pattern was noticed. Their hybrid shows the presence of both the types of proteins similar to their parents.

Question 36.
Give an account on cytoplasmic male sterility.
Answer:
Male sterility found in pearl maize (Sorgum vulgare) is the best example for mitochondrial cytoplasmic inheritance. So it is called cytoplasmic male sterility. In this, male sterility is inherited maternally. The gene for cytoplasmic male sterility is found in the mitochondrial DNA.

Question 37.
Write a short note on Atavism.
Answer:
Atavism is a modification of a biological structure whereby an ancestral trait reappears after having been lost through evolutionary changes in the previous generations. Evolutionary traits that have disappeared phenotypically do not necessarily disappear from an organism’s DNA. The gene sequence often remains, but is inactive.

Such an unused gene may remain in the genome for many generations. As long as the gene remains intact, a fault in the genetic control suppressing the gene can lead to the reappearance of that character again. Reemergence of sexual reproduction in the flowering plant Hieracium pilosella is the best example for Atavism in plants.

5 – Mark Questions

Question 38.
Explain Dihybrid cross in pea plant.
Answer:
The crossing of two plants differing in two pairs of contrasting traits is called dihybrid cross. In dihybrid cross, two characters (colour and shape) are considered at a time. Mendel considered the seed shape (round and wrinkled) and cotyledon colour (yellow & green) as the two characters. In seed shape round (R) is dominant over wrinkled (r); in cotyledon colour yellow (Y) is dominant over green (y).

Hence the pure breeding round yellow parent is represented by the genotype RRYY and the pure breeding green wrinkled parent is represented by the genotype rryy. During gamete formation the paired genes of a character assort out ‘ independently of the other pair. During the F₁ x F, fertilization each zygote with an equal probability receives one of the four combinations from each parent. The resultant gametes thus will be genetically different and they are of the following four types:

(1) Yellow round (YR) – 9/16
(2) Yellow wrinkled (Yr) – 3/16
(3) Green round (yR) – 3/16
(4) Green wrinkled (yr) -1/16

These four types of gametes of F₁ dihybrids unite randomly in the process of fertilization and produce sixteen types of individuals in F₂ in the ratio of 9:3:3:1 as shown in the figure. Mendel’s 9:3:3:1 dihybrid ratio is an ideal ratio based on the probability including segregation, independent assortment and random fertilization. In sexually reproducing organism / plants from the garden peas to human beings, Mendel’s findings laid the foundation for understanding inheritance and revolutionized the field of biology. The dihybrid cross and its result led Mendel to propose a second set of generalisations that we called Mendel’s Law of independent assortment.

Question 39.
How does the wrinkled gene make Mendel’s peas wrinkled? Find out the molecular explanation.
Answer:
The protein called starch branching enzyme (SBEI) is encoded by the wild-type allele of the gene (RR) which is dominant. When the seed matures, this enzyme SBEI catalyzes the formation of highly branched starch molecules. Normal gene (R) has become interrupted by the insertion of extra piece of DNA (0.8 kb) into the gene, resulting in allele. In the homozygous mutant form of the gene (R) which is recessive, the activity of the enzyme SBEI is lost resulting in wrinkled peas.

The wrinkled seed accumulates more sucrose and high water content. Hence Ore osmotic pressure inside the seed rises. As a result, the seed absorbs more water and when it matures it loses water as it dries. So it becomes wrinkled at maturation. When the seed has at least one copy of normal dominant gene heterozygous, the dominant allele helps to synthesize starch, amylopectin an insoluble carbohydrate, with the osmotic balance which minimises the loss of water resulting in smooth structured round seed.

Question 40.
Describe incomplete dominance exhibited by Mirabilis jalapa.
Answer:
The German Botanist Carl Correns’s (1905) Experiment – In 4 O’ clock plant, Mirabilis jalapa when the pure breeding homozygous red (R¹R¹) parent is crossed with homozygous white (R²R²), the phenotype of the F₁ hybrid is heterozygous pink (R¹R²). The F₁ heterozygous phenotype differs from both the parental homozygous phenotype. This cross did not exhibit the character of the dominant parent but an intermediate colour pink. When one allele is not completely dominant to another allele it shows incomplete dominance. Such allelic interaction is known as incomplete dominance. F₁ generation produces intermediate phenotype pink coloured flower.

When pink coloured plants of F₁ generation were interbred in F₂ both phenotypic and genotypic ratios were found to be identical as 1 : 2 :1(1 red: 2 pink: 1 white). Genotypic ratio is 1 R¹ R¹ : 2 R¹R² : 1 R²R². From this we conclude that the alleles themselves remain discrete and unaltered proving the Mendel’s Law of Segregation. The phenotypic and genotypic ratios are the same. There is no blending of genes. In the F¹ generation R¹ and R² genes segregate and recombine to produce red, pink and white in the ratio of 1 : 2 : 1. R¹ allele codes for an enzyme responsible for the formation of red pigment. R² allele codes for defective enzyme.

R¹ and R² genotypes produce only enough red pigments to make the flower pink. Two R¹ R² are needed for producing red flowers. Two R²R² genes are needed for white flowers. If blending had taken place, the original pure traits would not have appeared and all F₂ plants would have pink flowers. It is very clear that Mendel’s particulate inheritance takes place in this cross which is confirmed by the reappearance of original phenotype in F₂.

Higher Order Thinking Skills (HOTs) Questions

Question 1.
A yellow colour flower plant indicated by YY is crossed with white color flower plant denoted by yy.
(a) following the Mendelian inheritance pattern, what would be the flower color is first filial generation?
(b) Which Mendelian principle is illustrated in this cross?
(c) Derive the cross and state the phenotypic ratio of yellow flowers to white flowers in F₂ generation?
Answer:
(a) F₁ plants produce yellow colour flower plants.
(b) Law of dominance and Law of segregation
(c)

Question 2.
Mala is a genetic research student. She was given a plant to identify whether it is a homozygous or heterozygous for a particular trait. How will she proceed further?
Answer:
To identify the plant genotype whether homozygous or heterozygous Mala can perform test cross, where the individual is crossed with homozygous recessive for the trait. If the plant is heterozygous then the resultant progenies would be in the ratio 50:50

Question 3.
In the chart given below, ‘AA’ are the genes located in a chromosome of Pisum sativum.
Answer:

Observe the chart and mention the genetic phenomenon does it indicates.
Pleitrophy – A single gene affecting many traits. Here the single gene AA controls the traits – for flower colour, seed colour and leaf axil spot.

Question 4.
Give the F₂ phenotypic ratio of
(a) Supplementary genes
(b) Complementary genes
(c) Dominant epistasis
Answer:
(a) Supplementary genes – 9 : 3 : 4
(b) Complementary genes – 9 : 7
(c) Dominant epistasis -12 : 3 : 1

Question 5.
Name the respective pattern of inheritance where F₁ phenotype
(a) resembles any one of the two parents
(b) is an intermediate between two parental traits.
Answer:
(a) Dominance
(b) Incomplete dominance

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Samacheer Kalvi 12th Bio Botany Principles and Processes of Biotechnology Text Book Back Questions and Answers

Question 1.
Restriction enzymes are ___________
(a) Not always required in genetic engineering
(b) Essential tools in genetic engineering
(c) Nucleases that cleave DNA at specific sites
(d) both b and c
Answer:
(d) both b and c

Question 2.
Plasmids are ___________
(a) circular protein molecules
(b) required by bacteria
(c) tiny bacteria
(d) confer resistance to antibiotics
Answer:
(d) confer resistance to antibiotics

Question 3.
EcoRI cleaves DNA at
(a) AGGGTT
(b) GTATATC
(c) GAATTC
(d) TATAGC
Answer:
(c) GAATTC

Question 4.
Genetic engineering is ___________
(a) making artificial genes
(b) hybridization of DNA of one organism to that of the others.
(c) production of alcohol by using micro organisms.
(d) making artificial limbs, diagnostic instruments such as ECG and EEG, etc.
Answer:
(b) hybridization of DNA of one organism to that of the others.

Question 5.
Consider the following statements:
i. Recombinant DNA technology is popularly known as genetic engineering is a stream of,biotechnology which deals with the manipulation of genetic materials by man invitro
ii. pBR322 is the first artificial cloning vector developed in 1977 by Boliver and Rodriguez from E.coli plasmid.
iii. Restriction enzymes belongs to a class of enzymes called nucleases. Choose the correct option regarding above statements
(a) i and ii
(b) i and iii
(c) ii and iii
(d) i,ii and iii
Answer:
(d) i,ii and iii

Question 6.
The process of recombinant DNA technology has the following steps
i. Amplication of the gene.
ii. Insertion of recombinant DNA into the host cells.
iii. Cutting of DNA at specific location using restriction enzyme.
iv. Isolation of genetic material (DNA).
Pick out the correct sequence of step for recombinant DNA technology.
(a) ii, iii, iv, and i
(b) iv, ii, iii, and i
(c) i, ii, iii and iv
(d) iv, iii, i, and ii
Answer:
(d) iv, iii, i, and ii

Question 7.
Which one of the following palindromic base sequence in DNA can be easily cut at about the middle by some particular restriction enzymes?
(a) 5′ CGTTCG 3′ ATCGTA5′
(b) 5′ GATATG 3′ CTACTA5′
(c) 5′ GAATTC 3′ CTTAAG 5′
(d) 5′ CACGTA 3′ CTCAGT 5′
Answer:
(c) 5′ GAATTC 3′ CTTAAG 5′

Question 8.
pBR 322, BR stands for
(a) Plasmid Bacterial Recombination
(b) Plasmid Bacterial Replications
(c) Plasmid Boliver and Rodriguez
(d) Plasmid Baltimore and Rodriguez
Answer:
(c) Plasmid Boliver and Rodriguez

Question 9.
Which of the following one is used as a Biosensors?
(a) Electrophoresis
(b) Bioreactors
(c) Vectors
(d) Electroporation
Answer:
(b) Bioreactors

Question 10.
Match the following

(A) a b c d
(B) c d b a
(C) a c b d
(D) c d a b
Answer:
(D) c d a b

Question 11.
In which techniques Ethidium Bromide is used?
(a) Southern Blotting techniques
(b) Western Blotting techniques
(c) Polymerase Chain Reaction
(d) Agarose Gel Electrophoresis
Answer:
(d) Agarose Gel Electrophoresis

Question 12.
Assertion: Agrobacterium tumifaciens is popular in genetic engineering because this bacteriumis associated with the root nodules of all cereals and pulse crops.
Reason: A gene incorporated in the bacterial chromosomal genome gets automatically transferred to the cross with which bacterium is associated.
(a) Both assertion and reason are true. But reason is correct explanation of assertion.
(b) Both assertion and reason are true. But reason is not correct explanation of assertion.
(c) Assertion is true, but reason is false.
(d) Assertion is false, but reason is true.
(e) Both assertion and reason are false.
Answer:
(a) Both assertion and reason are true. But reason is correct explanation of assertion.

Question 13.
Which one of the following is not correct statement?
(a) Ti plasmid causes the bunchy top disease
(b) Multiple cloning site is known as Polylinker
(c) Non-viral method of transfection of Nucleic acid in cell
(d) Polylactic acid is a kind of biodegradable and bioactive thermoplastic.
Answer:
(a) Ti plasmid causes the bunchy top disease

Question 14.
An analysis of chromosomal DNA using the southern hybridisation technique does not use
(a) Electrophoresis
(b) Blotting
(c) Autoradiography
(d) Polymerase Chain Reaction
Answer:
(a) Electrophoresis

Question 15.
An antibiotic gene in a vector usually helps in the selection of
(a) Competent cells
(b) Transformed cells
(c) Recombinant cells
(d) None of the above
Answer:
(a) Competent cells

Question 16.
Some of the characteristics of Bt cotton are
(a) Long fibre and resistant to aphids
(b) Medium yield, long fibre and resistant to beetle pests
(c) high yield and production of toxic protein crystals which kill dipteran pests.
(d) High yield and resistant to ball worms
Answer:
(b) Medium yield, long fibre and resistant to beetle pests

Question 17.
How do you use the biotechnology in modern practice?
Answer:
In modem practice, biotechnology is used in the development of herbicide resistance plants, improved crop varieties, producing pharma products like insulin, developing vaccines, diagnosing genetic diseases and designing drgus etc.

Question 18.
What are the materials used to grow microorganism like Spirulinal
Answer:
Spirulina can be grown easily on materials like waste water from potato processing plants (containing starch), straw, molasses, animal manure and even sewage, to produce large quantities.

Question 19.
You are working in a biotechnology lab with a bacterium namely E.coli. How will you cut the nucleotide sequence? explain it.
Answer:
The DNA nucleotide sequence can be cut using Restriction endonucleases (RE). Restriction endonucleases – EcoRI cuts the DNA at GAATTC seqUence, producing sticky ends.CTTAAG

Question 20.
What are the enzymes you can use to cut terminal end and internal phospho diester bond of nucleotide sequence?
Answer:
Restriction exonuclease are the restriction enzyme used to cut nucleotides from the terminal end of DNA. Whereas, restriction endonucleases cut the internal phospho diester bond with DNA molecule.

Question 21.
Name the chemicals used in gene transfer.
Answer:
Polyethylene Glycol (PEG) and Dextran Sulphate.

Question 22.
What do you know about the word pBR332?
Answer:
pBR 322 plasmid is a reconstructed plasmid and most widely used as cloning vector; it contains 4361 base pairs. In pBR, p denotes plasmid, B and R respectively the names of scientist Roliver and/fodriguez who developed this plasmid. The number is the number of plasmid developed from their laboratory.

It contains ampR and tetR two different antibiotic resistance genes and recognition sites for several restriction enzymes. (Hind III, EcoRI, BamH I, Sal I, Pvu II, Pst I and Cla I), ori and antibiotic resistance genes. Rop codes for the proteins involved in the replication of the plasmid.

Question 23.
Mention the application of biotechnology.
Answer:

1. Biotechnology is one of the most important applied interdisciplinary sciences of the 21st century. It is the trusted area that enables us to find the beneficial way of life.
2. Biotechnology has wide applications in various sectors like agriculture, medicine,environment and commercial industries.
3. This science has an invaluable outcome like transgenic varieties of plants e.g. transgenic cotton (Bt-cotton), rice, tomato, tobacco, cauliflower, potato and banana.
4. The development of transgenics as pesticide resistant, stress resistant and disease resistant varieties of agricultural crops is the immense outcome of biotechnology.
5. The synthesis of human insulin and blood protein in E.coli and utilized for insulin deficiency disorder in human is a breakthrough in biotech industries in medicine.
6. The synthesis of vaccines, enzymes, antibiotics, dairy products and beverages are the products of biotech industries.
7. Biochip based biological computer is one of the successes of biotechnology.
8. Genetic engineering involves genetic manipulation, tissue culture involves aseptic cultivation of totipotent plant cell into plant clones under controlled atmospheric conditions.
9. Single cell protein from Spirulina is utilized in food industries.
10. Production of secondary metabolites, biofertilizers, biopesticides and enzymes.
11. Biomass energy, biofuel, bioremediation and phytoremediation for environmental biotechnology.

Question 24.
What are restriction enzyme. Mention their type with role in biotechnology.
Answer:
Restriction enzymes are the enzymes of bacterial origin which cleaves DNA into fragments at or near specific recognition sites within DNA molecules. This principle is used in biotechnology to cut and insert the desired gene (gene of interest) thereby generating an rDNA with desirable characters.

Question 25.
Is there any possibilities to transfer a suitable desirable gene to host plant without vector? Justify your answer.
Answer:
Yes, it is possible to transfer a suitable desired gene to a host plant using certain chemicals, microinjection method, electroporation or by biolistics.

Question 26.
How will you identify a vector?
Answer:

1. Vectors are able to replicate autonomously to produce multiple copies of them along with their DNA insert in the host cell.
2. It should be small in size and of low molecular weight, less than 10 Kb (kilo base pair) in size so that entry/transfer into host cell is easy.
3. Vector must contain an origin of replication so that it can independently replicate within the host.
4. It should contain a suitable marker such as antibiotic resistance, to permit its detection in transformed host cell.
5. Vector should have unique target sites for integration with DNA insert and should have the ability to integrate with DNA insert it carries into the genome of the host cell. Most of the commonly used cloning vectors have more than one restriction site. These are Multiple Cloning Site (MCS) or polylinker. Presence of MCS facilitates the use of restriction enzyme of choice.

Question 27.
Compare the various types of Blotting techniques.
Answer:

Question 28.
Write the advantages of herbicide tolerant crops.
Answer:
Advantages of Herbicide Tolerant Crops:

• Weed control improves higher crop yields;
• Reduces spray of herbicide;
• Reduces competition between crop plant and weed;
• Use of low toxicity compounds which do not remain active in the soil; and
• The ability to conserve soil structure and microbes.

Question 29.
Write the advantages and disadvantages of Bt cotton.
Answer:
The advantages of Bt cotton are:

1. Yield of cotton is increased due to effective control of bollworms.
2. Reduction in insecticide use in the cultivation of Bt cotton
3. Potential reduction in the cost of cultivation.
4. Cost of Bt cotton seed is high.
5. Effectiveness up to 120 days after that efficiency is reduced.
6. Ineffective against sucking pests like jassids, aphids and whitefly.
7. Affects pollinating insects and thus yield.

Question 30.
What is bioremediation? Give some examples of bioremediation.
Bioremediation:
It is defined as the use of microorganisms or plants to clean up environmental pollution. It is an approach used to treat wastes including wastewater, industrial waste and solid waste. Bioremediation process is applied to the removal of oil, petrochemical residues, pesticides or heavy metals from soil or ground water.

In many cases, bioremediation is less expensive and more sustainable than other physical and chemical methods of remediation. Bioremediation process is a cheaper and eco-friendly approach and can deal with lower concentrations of contaminants more effectively. The strategies for bioremediation in soil and water can be as follows:

1. Use of indigenous microbial population as indicator species for bioremediation process.
2. Bioremediation with the addition of adapted or designed microbial inoculants.
3. Use of plants for bioremediation – green technology.

Question 31.
Write the benefits and risk of Genetically Modified Foods.
Answer:
GM Food – Benefits:

1. High yield without pest.
2. 70% reduction of pesticide usage.
3. Reduce soil pollution problem.
4. Conserve microbial population in soil.

Risks – believed to:

1. Affect liver, kidney function and cancer.
2. Hormonal imbalance and physical disorder.
3. Anaphylactic shock (sudden hypersensitive reaction) and allergies.
4. Adverse effect in immune system because of bacterial protein.
5. Loss of viability of seeds show in terminator seed technology of GM crops.

Samacheer Kalvi 12th Bio Botany Principles and Processes of Biotechnology Additional Questions and Answers

Question 1.
Which of the following person coined the term biotechnology?
(a) Ernst Hoppe
(b) Stanley Cohen
(c) Ian Wilmet
(d) Karl Ereky
Answer:
(d) Karl Ereky

Question 2.
Zymology deals with
(a) Study of yeast fungus and its practical applications.
(b) Study of fermentation and its uses.
(c) Study of Bioreactors and their construction methodology.
(d) Study of zymase producing microbes and its benefits.
Answer:
(b) Study of fermentation and its uses.

Question 3.
Match column I with column II

(a) A – iii, B – i, C – iv, D – ii
(b) A – i, B – iv, C – ii, D – iii
(c) A – iv, B – iii, C – ii D – i
(d) A – ii, B – iv C – i, D – iii
Answer:
(a) A – iii, B -1, C – iv, D – ii

Question 4.
Identify the incorrect statement:
(a) French chemist Louis Pasteur demonstrated the fermentation.
(b) Fermentor is a vessel providing optimal condition for microbial action.
(c) Solvent extraction is an upstream process of fermentation.
(d) Distillation and filtration comes under down stream process.
Answer:
(c) Solvent extraction is an upstream process of fermentation.

Question 5.
Pick out the mismatched pair(s):
(i) Amphotericin-B – Streptomyces notatum
(ii) Penicillin – Penicillum nodosus
(iii) Streptomycin – Streptomyces grises
(iv) Tetracycline – Streptomyces aureofacins
(a) i and ii
(b) ii and iii
(c) iii and iv
(d) i only
Answer:
(a) i and ii

Question 6.
Identify the non-fungal species used in SCP production.
(i) Candida
(ii) Chlorella
(iii) Chlamydomonas
(iv) Cellulomonas
(a) i and ii
(b) ii and iii
(c) ii, iii and iv
(d) All the above
Answer:
(c) ii, iii and iv

Question 7.
Select the correct restriction enzyme which breaks the phosphodiester bond within a DNA
molecule.
(i) Bal 31
(ii) Hind II
(iii) Bam HI
(iv) Pvul
(a) i and iii
(b) i, ii and iii
(c) ii, iii and iv
(d) i only
Answer:
(c) ii, iii and iv

Question 8.
Cohesive ends are _______
(a) Blunt ends
(b) Flush ends
(c) Sticky ends
(d) Symmetric cuts
Answer:
(c) Sticky ends

Question 9.
Self-ligation is prevented by __________
(a) DNA Polymerase
(b) Helicase
(c) Alkaline phosphate
(d) DNA lipase
Answer:
(c) Alkaline phosphate

Question 10.
Observe the diagram and name A and B.

(a) A – Plasmid – B – Vector
(b) A – Nucleoid – B – Plasmid
(c) A – Bacterial chromosome B – Vector
(d) A – Nucleoid – B – x phage DNA
Answer:
(b) A – Nucleoid – B – Plasmid

Question 11.
A vector should __________
(i) contain suitable marker
(ii) contain ori site
(iii) have poly linkess
(iv) be small in size
(a) i, ii and iii
(b) ii, iii and iv
(c) i, ii and iv
(d) all the above
Answer:
(d) all the above

Question 12.
Number of base pairs does pBR 322 plasmid contains __________
(a) 322
(b) 4322
(c) 4361
(d) 3264
Answer:
(c) 4361

Question 13.
__________ is the plasmid present in Agrobacterium.
Answer:
Ti plasmid

Question 14.
ptlC 19 is an example for.
(a) Shuttle vector
(b) Expression vector
(c) Cosmid
(d) Phagemid vector
Answer:
(b) Expression vector

Question 15.
Statement 1: YAC plasmid behaves like a yeast chromosome.
Statement 2: Circular YAC multiplies in bacteria.
(a) Statement 1 is correct and Statement 2 is also correct.
(b) Statement 1 is correct and Statement 2 is incorrect.
(c) Both the statements are incorrect.
(d) Statement 1 is incorrect and Statement 2 is correct.
Answer:
(a) Statement 1 is correct and Statement 2 is also correct.

Question 16.
Statement 1: Liposomes are the artificial lipoprotein vesicles. Statement 2: Liposomes are highly used in gene transfer.
(a) Statement 1 is correct and Statement 2 is also correct.
(b) Statement 1 is correct and Statement 2 is incorrect.
(c) Both the statements are incorrect.
(d) Statement 1 is incorrect and Statement 2 is correct.
Answer:
(d) Statement 1 is incorrect and Statement 2 is correct.

Question 17.
Statement 1: DNA is a hydrophobic molecule.
Statement 2: T-DNA is a part of E-coli plasmid.
(a) Statement 1 is correct and Statement 2 is also correct.
(b) Statement 1 is correct and Statement 2 is incorrect.
(c) Both the statements are incorrect.
(d) Statement 1 is incorrect and Statement 2 is correct.
Answer:
(c) Both the statements are incorrect.

Question 18.
Statement 1: Bioventing procedure increases 02 flow to accelerate degradation of pollutants.
Statement 2: Bioaugmentation uses microbes to recover metal pollutants from contaminated sites.
(a) Statement 1 is correct and Statement 2 is also correct.
(b) Statement 1 is correct and Statement 2 is incorrect.
(c) Both the statements are incorrect.
(d) Statement 1 is incorrect and Statement 2 is correct.
Answer:
(b) Statement 1 is correct and Statement 2 is incorrect.

Question 19.
Assertion (A) : Golden rice helps to overcome childhood blindness.
Reason (R) : It is rich in P carotene.
(a) Both A and R are wrong.
(b) A is right R is wrong.
(c) R explains A.
(d) A and R are right, R does not explain A.
Answer:
(c) R explains A.

Question 20.
Assertion (A): Expression vectors are suitable for expressing foreign proteins.
Reason (R): pBR 322 is an expression vectors.
(а) Both A and R are wrong.
(b) A is right R is wrong.
(c) R explains A.
(d) A and R are right, R does not explain A.
Answer:
(b) A is right R is wrong.

Question 21.
Assertion (A) : Pseudomonas putida is utilized in the production of Biological hydrogen.
Reason (R): During photosynthesis, it releases oxygen.
(a) Both A and R are wrong.
(b) A is right R is wrong.
(c) R explains A.
(d) A and R are right, R does not explain A.
Answer:
(a) Both A and R are wrong.

Question 22.
Assertion (A): DMH -11 is a transgenic mustard.
Reason (R): It is developed by using bamase/ barstar technology.
(a) Both A and R are wrong.
(b) A is right R is wrong.
(c) R explains A.
(d) A and R are right, R does not explain A.
Answer:
(c) R explains A.

Question 23.
Green fluorescent protein (GFP) was isolated from
(a) Aequorea victoria
(b) Arabidopsis thaliana
(c) Agrobacterium tumifaciens
(d) Escherichia coli
Answer:
(a) Aequorea victoria

Question 24.
Tetracycline is obtained from
(a) S.nodosus
(b) S.aureofacins
(c) S.grises
(d) P. chryosogenum
Answer:
(a) S.aureofacins

Question 25.
Today more than restriction enzymes have been isolated.
(a) 800
(b) 900
(c) 1000
(d) 870
Answer:
(6) 900

2. Mark Questions

Question 1.
How modern biotechnology differs from conventional biotechnology?
Answer:
There are two main features of modem biotechnology, that differentiated it from the conventional technology are its:
(i) ability to change the genetic material for getting new products with specific requirement through recombinant DNA technology
(ii) ownership of the newly developed technology and its social impact.

Question 2.
What is a fermentor?
Answer:
Bioreactor (Fermentor) is a vessel or a container that is designed in such a way that it can provide an optimum environment in which microorganisms or their enzymes interact with a substrate to produce the required product. In the bioreactor, aeration, agitation, temperature and pH are controlled.

Question 3.
Define fermentation.
Answer:
Fermentation refers to the metabolic process in which organic molecules (normally glucose) are converted into acids, gases, or alcohol in the absence of oxygen or any electron transport chain.

Question 4.
What are primary metabolites? Give example.
Answer:
Metabolites produced for the maintenance of life process of microbes are known as primary metabolites.
E.g. Ethanol, citric acid, lactic acid and acetic acid.

Question 5.
How microbial enzymes are produced? Mention its significance.
Answer:
When microbes are cultured, they secrete some enzymes into the growth media. These enzymes are industrially used in detergents, food processing, brewing and pharmaceuticals.
E.g. Protease, amylase, isomerase, and lipase.

Question 6.
Mention any two bacterial species used as SCP.
Answer:

1. Cellulomonas
2. Alcaligenes

Question 7.
Name any two fungal species used as SCP.
Answer:

1. Agaricus campestris
2. Saccharomyces cerevisiae

Question 8.
Expand PCR and mention its use.
Answer:
PCR stands for Polymerase Chain Reaction.
PCR is a common lab technique used to make copies of particular region of DNA.

Question 9.
What is Restriction Endonuclease?
Answer:
A restriction enzyme or restriction endonuclease is an enzyme that cleaves DNA into fragments at or near specific recognition sites within the molecule known as restriction sites.

Question 10.
What is a palindrome sequence?
Answer:
Palindrome is a sequence of nucleotide in DNA strands at the site which reads the same in 5′-3′ direction and in the 3′-5′ direction.

Question 11.
Write an palindrome sequence of DNA.
Answer:
5′ … CATTATATAATG … 3′
3′ … GTAATATATTAC … 5′

Question 12.
Differentiate between flush end and cohesive end of DNA.
Answer:
Flush End:
Some restriction enzymes cut the strands of DNA through the centre resulting in blunt end or flush end or symmetric cuts.

Cohesive End:
Some restriction enzymes cut the strands in a way producing protruding and recessed ends known as cohesive end or sticky end or asymmetric cuts.

Question 13.
What is the role of DNA ligase in genetic engineering?
Answer:
DNA ligase enzyme joins the sugar and phosphate molecules of double stranded DNA (dsDNA) with 5’-P0₄ and a 3’-OH in an Adenosine Triphosphate (ATP) dependent reaction. This is isolated from T4 phage.

Question 14.
Define plasmids.
Plasmids are extrachromosomal, self replicating ds circular DNA molecules, found in the bacterial cells in addition to the bacterial chromosome. Plasmids contain Genetic information for their own replication.

Question 15.
Classify vectors and explain them.
Answer:
Vectors are of two types:

1. Cloning Vector
2. Expression Vector.

Cloning vector is used for the cloning of DNA insert inside the suitable host cell. Expression vector is used to express the DNA insert for producing specific protein inside the host.

Question 16.
What are expression vectors?
Answer:
Vectors which are suitable for expressing foreign proteins are called expression vectors. This vector consists of signals necessary for transcription and translation of proteins in the host. This helps the host to produce foreign protein in large amounts.
Example: pUC 19.

Question 17.
Name any four vectors that you know?
Answer:
Cosmid, plasmid, Bacteriophage and Phagemids.

Question 18.
Write a brief note on BAC vector.
Answer:
Bacterial Artificial Chromosome (BAC) Vector is a shuttle plasmid vector, created for cloning large-sized foreign DNA. BAC vector is one of the most useful cloning vector in r-DNA technology they can clone DNA inserts of upto 300 Kb and they are stable and more user-friendly.

Question 19.
What does Blotting refers to?
Answer:
Blotting refers to the process of immobilization of sample nucleic acids on solid support (nitrocellulose or nylon membranes). The blotted nucleic acids are then used as target in the hybridization experiments for their specific detection.

Question 20.
Point out any two disadvantages of Bt cotton?
Answer:
Bt cotton has some limitations:

• Cost of Bt cotton seed is high.
• Effectiveness up to 120 days after that efficiency is reduced.

Question 21.
What are the benefits of Genetically Modified plants?
Answer:
GM Food – Benefits:

1. High yield without pest.
2. 70% reduction of pesticide usage.
3. Reduce soil pollution problem.
4. Conserve microbial population in soil.

Question 22.
Expand:

1. PEG
2. PHB

Answer:

1. PEG – Poly Ethylene Glycol
2. PHB – Poly Hydroxy Butyrate

Question 23.
What is Biopharming?
Answer:
Biopharming also known as molecular pharming is the production and use of transgenic plants genetically engineered to produce pharmaceutical substances for use of human beings. This is also called “molecular farming or pharming”. These plants are different from medicinal plants which are naturally available.

Question 24.
Define the terms

1. Bioventing
2. Bioaugmentation

Answer:

1. Bioventing is the process that increases the oxygen or air flow to accelerate the degradation of environmental pollutants.
2. Bioaugmentation is the addition of selected microbes to speed up degradation process.

Question 25.
How hydrogen is biologically synthsized?
Answer:
The biological hydrogen production with algae is a method of photo biological water splitting. In normal photosynthesis the alga, Chlamydomonas reinhardtii releases oxygen. When it is deprived of sulfur, it switches to the production of hydrogen during photosynthesis and the electrons are transported to ferredoxins. [Fe]-hydrogenase enzymes combine them into the production of hydrogen gas.

Question 26.
Define Biopiracy.
Answer:
Biopiracy can be defined as the manipulation of intellectual property rights laws by corporations to gain exclusive control over national genetic resources, without giving adequate recognition or remuneration to the original possessors of those resources.

Question 27.
What are polylinkers?
Answer:
Mostly cloning vectors have more than one restriction sites. These are called as Multiple Cloning Site (MCS) or polylinkers. Presence of MCS facilitates the use of restriction enzyme of choice.

3-Mark Questions

Question 28.
Mention any three historical events which took place in the 21st century for the development of biotechnology.
Answer:
2002 – First crop plant genome sequenced in Oryza sativa.
2003 – Human genome project is completed, providing information on the locations and 1 sequence of human genes on all 46 chromosomes.
2016 – Stem cells injected into stroke patients re-enable patient to walk – Stem cell therapy.

Question 29.
In the fermentation process, what does upstream and downstream refers to? Explain.
Answer:
Upstream process:
All the process before starting of the fermenter such as sterilization of the fermenter, preparation and sterilization of culture medium and growth of the suitable inoculum are called upstream process.

Downstream process:
All the process after the fermentation process is known as the downstream process. This process includes distillation, centrifuging, filtration and solvent extraction. Mostly this process involves the purification of the desired product.

Question 30.
Provide a stepwise procedure of fermentation process.
Procedure of Fermentation
Answer:

1. Depending upon the type of product, bioreactor is selected.
2. A suitable substrate in liquid media is added at a specific temperature, pH and then diluted.
3. The organism (microbe, animal/plant cell, sub-cellular organelle or enzyme) is added to it.
4. Then it is incubated at a specific temperature for the specified time.
5. The incubation may either be aerobic or anaerobic.
6. Withdrawal of product using downstream processing methods.

Question 31.
What are secondary metabolites? Give two examples.
Answer:
Secondary metabolites are those which are not required for the vital life process of microbes, but have value added nature, this includes antibiotics
e.g. -Amphotericin-B (Streptomyces nodosus) and Penicillin (Penicillium chryosogenum).

Question 32.
What is SCP? Mention its nutritional value.
Answer:
Single Cell Protein (SCP) are dried cells of microorganisms that are used as protein supplement in human foods or animal feeds. SCP are rich in proteins, aminoacids, vitamins, carbohydrates, fats and minerals. It is used as food source by Astronauts and Antarctica expedition scientists.

Question 33.
Mention any three algal species used for SCP production.
Answer:
Spirulina, Chlorella and Chlamydomonas.

Question 34.
Though SCP is a rich protein source, it has not been widely used as food supplement. Point a reason to support this statement.
Answer:
Yes, although SCP is a rich protein source it is not widely used by most of the people in countries due to its higher nucleic acid content and slow digestibility.

Question 35.
Point out few advantages of single cell protein.
Answer:
Applications of Single-Cell Protein:

1. It is used as protein supplement.
2. It is used in cosmetics, products for healthy hair and skin.
3. It is used in poultry as the excellent source of proteins and other nutrients, it is widely used for feeding cattle, birds and fishes, etc.
4. It is used in food industry as aroma carriers, vitamin carrier, emulsifying agents to improve the nutritive value of baked products, in soups, in ready-to-serve-meals, in diet recipes.
5. It is used in industries like paper processing and leather processing as foam stabilizers.

Question 36.
Classify restriction enzymes based on their mode of action.
Answer:
Based on their mode of action restriction enzymes are classified into Exonucleases and Endonucleases.

a. Exonucleases are enzymes which remove nucleotides one at a time from the end of a DNA molecule,
e.g. Bal 31 and Exomiclease III.

b. Endonucleases are enzymes which break the internal phosphodiester bonds within a DNA molecule,
e.g. Hind II, EcoRI, Pvul, BamHI and TaqI.

Question 37.
Which type of restriction enzyme is widely used in rDNA technology? Why?
Answer:
Type II enzyme is preferred for use in recombinant DNA technology as they recognise and cut DNA within a specific sequence typically consisting of 4-8 bp.

Question 38.
Explain the procedure behind the naming of Restriction Enzymes by citing an example.
Answer:
Restriction endonucleases are named by a standard procedure. The first letter of the enzymes indicates the genus name, followed by the first two letters of the species, then comes the strain of the organism and finally a roman numeral indicating the order of discovery. For example, EcoRI is from Escherichia (E) coli (co), strain RY 13 (R) and first endonuclease (I) to be discovered.

Question 39.
Give a short note on Alkaline phosphate.
Answer:
Alkaline phosphate is a DNA modifying enzymes and adds or removes specific phosphate group at 5’ terminus of double stranded DNA (dsDNA) or single stranded DNA (ssDNA) or RNA. Thus it prevents self ligation. This enzyme is purified from bacteria and calf intestine.

Question 40.
What are the features that a vector must possess to facilitate cloning?
Answer:
The following are the features that are required to facilitate cloning into a vector.

1. Origin of replication (ori): This is a sequence from where replication starts and piece of DNA when linked to this sequence can be made to replicate within the host cells.
2. Selectable marker: In addition to ori the vector requires a selectable marker, which helps in identifying and eliminating non-transformants and selectively permitting the growth of the transformants,
3. Cloning sites: In order to link the alien DNA, the vector needs to have very few, preferably single, recognition sites for the commonly used restriction enzymes.

Question 41.
Draw and label Ti plasmid
Answer:

Question 42.
What do you mean by the term “walking genes”? Could you explain?
Answer:
Transposons (Transposable elements or mobile elements) are DNA sequence able to insert itself at a new location in the genome without having any sequence relationship with the target locus and hence transposons are called walking genes or jumping genes. They are used as genetic tools for analysis of gene and protein functions, that produce new phenotype on host cell. The use of transposons is well studied in Arabidopsis thaliana and bacteria such as Escherichia coli.

Question 43.
How does shuttle vectors differ from other types of vectors?
Answer:
The shuttle vectors are plasmids designed to replicate in cells of two different species. These vectors are created by recombinant techniques. The shuttle vectors can propagate in one host and then move into another host without any extra manipulation. Most of the Eukaryotic
vectors are Shuttle Vectors.

Question 44.
Given below are the three different DNA palindrome sequences. Name the respective restriction enzymes which cleaves those sequences and also mention the microbial sources of the enzymes.
Answer:
(a) 5 AGCT3′
(b) 5 GGCC3′
(c) 5 GAATTC3′
Answer:

Question 45.
Why is it difficult for DNA to pass through cell membrane? How the bacterial cells can be made competent to take up DNA?
Answer:
Since the DNA is a hydrophilic molecule, it cannot pass through cell membranes, In order to force bacteria to take up the plasmid, the bacterial cells must first be made competent to take up DNA. This is done by treating them with a specific concentration of a divalent cation such as calcium.

Question 46.
Write a brief note on Biolistics.
Answer:
The foreign DNA is coated onto the surface of minute gold or tungsten particles (1-3 pm) and bombarded onto the target tissue Or cells using a particle gun (also called as gene gun/micro projectile gun/shotgun). Then the bombarded cells or tissues are cultured on selected medium to regenerate plants from the transformed cells.

Question 47.
Agrobacterium – a natural genetic engineer of plants. Justify the statement.
Answer:
Among the various vectors used for plant transformation, the Ti-plasmid from Agrobacterium tumefaciens has been used extensively. This bacterium has a large size plasmid, known as Ti plasmid (Tumor inducing) and a portion of it referred as T-DNA (transfer DNA) is transferred to plant genome in the infected cells and cause plant tumors (crown gall). Since this bacterium has the natural ability to transfer T-DNA region of its plasmid into plant genome, upon infection of cells at the wound site, it is also known as the natural genetic engineer of plants.

Question 48.
Give a brief account on antibiotic resistant marker.
Answer:
An antibiotic resistance marker is a gene that produces a protein that provides cells with resistance to an antibiotic. Bacteria with transformed DNA can be identified by growing on a medium containing an antibiotic. Recombinants will grow on these medium as they contain genes encoding resistance to antibiotics such as amphicillin, chloroamphenicol, tetracycline or kanamycin, etc., while others may not be able to grow in these media, hence it is considered useful selectable marker.

Question 49.
Mention the types of Blotting techniques.
Answer:
Types of Blotting Techniques:

1. Southern Blotting: The transfer of DNA from agarose gels to nitrocellulose membrane.
2. Northern Blotting: The transfer of RNA to nitrocellulose membrane.
3. Western Blotting: Electrophoretic transfer of proteins to nitrocellulose membrane:

Question 50.
Expand CRISPR – Cas9.
Answer:
Clustered Regularly Interspaced Short Palindromic Repeats and CRISPR-associated protein 9.

Question 51.
What is RNA interference (RNAi)? How it is carried out?
Answer:
RNA interference is a biological process in which RNA molecules inhibit gene expression or translation. This is done by neutralising targeted mRNA molecules.

A simplified model for the RNAi pathway is based on two steps, each involving ribonuclease enzyme. In the first step, the trigger RNA (either dsRNA or miRNA primary transcript) is processed into a short interfering RNA (siRNA) by the RNase II enzymes called Dicer and Drosha. In the second step, siRNAs are loaded into the effector complex RNA-induced silencing complex (RISC). The siRNA is unwound during RISC assembly and the single- stranded RNA hybridizes with mRNA target. This RNAi is seen in plant feeding nematodes.

Question 52.
Prepare a protocol for Glyphosphate resistant potato plant development.
Answer:

Question 53.
How Bt cotton crops are developed?
Answer:
Bt cotton is a genetically modified organism (GMO) or genetically modified pest resistant plant cotton variety, which produces an insecticide activity to bollworm. Strains of the bacterium Bacillus thuringiensis produce over 200 different Bt toxins, each harmful to different insects. Most Bt toxins are insecticidal to the larvae of moths and butterflies, beetles, cotton bollworms and gatflies but are harmless to other forms of life.The genes are encoded for toxic crystals in the Cry group of endotoxin.

When insects attack and eat the cotton plant the Cry toxins are dissolved in the insect’s stomach.The epithelial membranes of the gut block certain vital nutrients thereby sufficient regulation of potassium ions are lost in the insects and results in the death of epithelial cells in the intestine .membrane which leads to the death of the larvae.

Question 54.
Comment on Golden rice.
Answer:
Golden rice is a variety of Oryza sativa (rice) produced through genetic engineering of biosynthesized beta-carotene, a precursor ofVitamin-A in the edible parts of rice developed by – Ingo Potrykus and his group. The aim is to produce a fortified food to be grown and consumed in areas with a shortage of dietary Vitamin-A, which kills so many children under five year age.

Golden rice differs from its parental strain by the addition of three beta-carotene biosynthesis , genes namely ‘psy’ (phytoene synthase) from daffodil plant Narcissus pseudonarcissus and ‘crt-1’ gene from the soil bacterium Erwinia auredorora and ‘lyc’ (lycopene cyclase) gene from wild-type rice endosperm. The endosperm of normal rice, does not contain beta-carotene. Golden-rice has been genetically altered so that the endosperm now accumulates Beta-carotene. This has been done using Recombinant DNA technology. Golden rice can control childhood blindness – Xerophthalmia.

Question 55.
Name any 3 bacterial species used to generate polyhydroxybutyrates (PHB).
Answer:
Bacillus megaterium.
Corynebacterium glutamicum.
Alcaligenes eutrophus.

Question 56.
Write short note on Green fluorescent protein.
Answer:
The green fluorescent protein (GFP) is a protein containing 238 amino acid residues of 26.9 kDa that exhibits bright green fluorescence when exposed to blue to ultraviolet range (395 nm). GFP refers to the protein first isolated from the jellyfish Aequorea victoria. GFP is an excellent tool in biology due to its ability to form internal chromophore without requiring any accessory cofactors, gene products, enzymes or substrates other than molecular oxygen. In cell and molecular biology, the GFP gene is frequently used as a reporter of expression. It has been used in modified forms to make biosensors.

Question 57.
How turmeric biopiracy is prevented by Indian Government?
Answer:
The United States Patent and Trademark Office, in the year 1995 granted patent to the method of use of turmeric as an antiseptic agent. Turmeric has been used by the Indians as a home remedy for the quick healing of the wounds and also for purpose of healing rashes. The journal article published by the Indian Medical Association, in the year 1953 wherein this remedy was mentioned.

Therefore, in this way it was proved that the use of turmeric as an antiseptic is not new to the world and is not a new invention, but formed a part of the traditional knowledge of the Indians. The objection in this case US patent and trademark office was upheld and traditional knowledge of the Indians was protected.

5 – Mark Questions

Question 58.
Describe the steps involved in recombinant DNA technology.
Answer:
The steps involved in recombinant DNA technology are:

1. Isolation of a DNA fragment containing a gene of interest that needs to be cloned. This is called an insert.
2. Generation of recombinant DNA (rDNA) molecule by insertion of the DNA fragment into a carrier molecule called a vector that can self-replicate within the host cell.
3. Selection of the transformed host cells that is carrying the rDNA and allowing them to multiply thereby multiplying the rDNA molecule. The
4. entire process thus generates either a large amount of rDNA or a large amount of protein expressed by the insert.
5. Wherever vectors are not involved the desired gene is multiplied by PCR technique. The multiple copies are. injected into the host cell protoplast or it is shot into the host cell protoplast by shot gun method.

Question 59.
Explain in detail about various ty pes of direct gene transfer method,
Answer:
a. Chemical mediated gene transfer: Certain chemicals like polyethylene glycol (PEG) and dextran sulphate induce DNA uptake into plant protoplasts.

b. Microinjection: The DNA is directly injected into the Electric field induces a voltage across cell membrane nucleus using fine tipped glass needle or micro pipette Electroporation Methods of Gene Transfer to transform plant cells. The protoplasts are immobilised on a solid support (agarose on a microscopic slide) or held with a holding pipette under suction.

c. Electroporation Methods of Gene Transfer: Apulse of high voltage is applied to protoplasts, cells or tissues which makes transient pores in the plasma membrane through which uptake of foreign DNA occurs.

d. Liposome mediated method of Gene Transfer: Liposomes the artificial phospholipid vesicles are useful in gene transfer. The gene or DNA is transferred from liposome into vacuole of plant cells. It is carried out by encapsulated DNA into the vacuole. This technique is advantageous because the liposome protects the introduced DNA from being damaged by the acidic pH and protease enzymes present in the vacuole. Liposome and tonoplast of vacuole fusion resulted in gene transfer. This process is called lipofection.

e. Biolistics: The foreign DNA is coated onto the surface of minute gold or tungsten particles (1-3 pm) and bombarded onto the target tissue or cells using a particle gun (also called as gene gun/micro projectile gun/shotgun). Then the bombarded cells or tissues are cultured on selected medium to regenerate plants from the transformed cells.

Question 60.
Describe the procedure of Blue-White colony selection methods.
Answer:
Blue- White Colony Selection Method is a powerful method used for screening of recombinant plasmid. In this method, a reporter gene lacZ is inserted in the vector. The lacZ encodes the enzyme P-galactosidase and contains several recognition sites for restriction enzyme.P-galactosidase breaks a synthetic substrates called X-gal (5-bromo-4-chloroindolyl- P-D- galacto-pyranoside) into an insoluble blue coloured product. If a foreign gene is inserted into lacZ, this gene will be inactivated.

Therefore, no-blue colour will develop (white) because P-galactosidase is not synthesized due to inactivation of lacZ. Therefore, the host cell containing r-DNA form white coloured colonies on the medium contain X-gal, whereas the other cells containing non-recombinant DNA will develop the blue coloured colonies. On the basis of colony colour, the recombinants can be selected.

Question 61.
Write a note on Replica plating technique.
Answer:

A technique in which the pattern of colonies growing on a culture plate is copied. A sterile filter plate is pressed against the culture plate and then lifted. Then the filter is pressed against a second sterile culture plate. This results in the new plate being infected with cell in the same
– relative positions as the colonies in the original plate. Usually, the medium used in the second plate will differ from that used in the first. It may include an antibiotic or without a growth factor. In this way, transformed cells can be selected. Replica plating technique

Question 62.
How Agarose Gel Electrophoresis is performed?
Answer:
1. Agarose Gel Electrophoresis is used mainly for the purification of specific DNA fragments. Agarose is convenient for separating DNA fragments ranging in size from a few hundred to about 20000 base pairs. Polyacrylamide is preferred for the purification of smaller DNA fragments.

2. The gel is complex network of polymeric molecules. DNA molecule is negatively charged molecule – under an electric field DNA molecule migrates through the gel. The electrophoresis is frequently performed with marker DNA fragments of known size which allow accurate size

3. determination of an unknown DNA molecule by interpolation. The advantages of agarose gel electrophoresis are that the DNA bands can be readily detected at high sensitivity. The bands of DNA in the gel are stained with the dye Ethidium Bromide and DNA can be detected

4. as visible fluorescence illuminated in UV light will give orange fluorescence, which can be photographed.

Question 63.
Explain the procedure of Southern Blotting Technique. Southern Blotting Techniques – DNA
Answer:
The transfer of denatured DNA from Agarose gel to Nitrocellulose Blotting or Filter Paper technique was introduced by Southern in 1975 and this technique is called Southern Blotting Technique.

Steps:
The transfer of DNA from agarose gel to nitrocellulose filter paper is achieved by Capillary Action.
A buffer Sodium Saline Citrate (SSC) is used, in which DNA is highly soluble, it can be drawn up through the gel into the Nitrocellulose membrane. By this process ss-DNA becomes ‘Trapped’ in the membrane matrix.

This DNA is hybridized with a nucleic acid and can be detected by autoradiography. Autoradiography – A technique that captures the image formed in a photographic emulsion due to emission of light or radioactivity from a labelled component placed together with unexposed
film.

Higher Order Thinking Skills (HOTs) Questions 

Question 1.
Give the technical terminologies for the following statements.
(а) Autonomous, self-replicating, circular DNA
(b) Molecular scissors
(c) Symmetrical repreated sequence in DNA strands
(d) Mobile genetic elements
Answer:
(a) Plasmid
(b) Restriction Enzymes
(c) Palindrome sequence
(d) Transposons

Question 2.
Observe the given flow chart and complete it.

Answer:
A. Plasmid
B. rDNA/chimeric DNA

Question 3.
Name the products of the following combinations.
(a) Bacterial plasmid + cos – site = ______
(b) Bacterial plasmid + phage DNA = ______
Answer:
(a) Cosmid
(b) Phagemid

Question 4.
Golden rice is a bio-fortified rice developed by rDNA technology. It differes from its parental strain by possessing ‘psy’ gene, ‘crt-1’ gene and ‘lye’ gene which are responsible for beta – carotene synthesis.
(a) Name the sources of the above mentioned genes.
(b) Which disease can be controlled / prevented if a person’s diet has golden rice?
Answer:
(a) ‘psy’ gene is obtained from Daffodil plant (Narcissus pseudonarcissus).
(b) ‘crt-1’ gene is from Erwinia auredorora bacterium.
(c) Tyc gene is from wild-type rice endosperm.
(b) Golden rice can control Xerophthalmia.

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Samacheer Kalvi 12th Bio Botany Plant Breeding Text Book Back Questions and Answers

Question 1.
Assertion: Genetic variation provides the raw material for selection.
Reason: Genetic variations are differences in genotypes of the individuals.
(a) Assertion is right and reason is wrong.
(b) Assertion is wrong and reason is right.
(c) Both reason and assertion is right.
(d) Both reason and assertion is wrong.
Answer:
(c) Both reason and assertion is right.

Question 2.
While studying the history of domestication of various cultivated plants recognized earlier.
(a) Centres of origin
(b) Centres of domestication
(c) Centres of hybrid
(d) Centres of variation
Answer:
(a) Centres of origin

Question 3.
Pick out the odd pair.
(a) Mass selection – Morphological characters
(b) Purline selection – Repeated self pollination
(c) Clonal selection – Sexually propagated
(d) Natural selection – Involves nature
Answer:
(a) Mass selection – Morphological characters

Question 4.

Answer:
(b) i – III, ii – I, iii – IV, iv – II

Question 5.
The quickest method of plant breeding is __________
(a) Introduction
(b) Selection
(c) Hybridization
(d) Mutation breeding
Answer:
(d) Mutation breeding

Question 6.
Desired improved variety of economically useful crops are raised by __________
(a) natural selection
(b) hybridization
(c) mutation
(d) biofertilisers
Answer:
(b) hybridization

Question 7.
Plants having similar genotypes produced by plant breeding are called __________
(a) clone
(b) haploid
(c) autopolyploid
(d) genome
Answer:
(a) clone

Question 8.
Importing better varieties and plants from outside and acclimatising them to local environment is called __________
(a) cloning
(b) heterosis
(c) selection
(d) introduction
Answer:
(d) introduction

Question 9.
Dwarfing gene of wheat is __________
(a) pall
(b) Atomita 1
(c) Norin 10
(d) pelita 2
Answer:
(c) Norin 10

Question 10.
Crosses between the plants of the same variety are called __________
(a) interspecific
(b) intervarietal
(c) intravarietal
(d) intergeneric
Answer:
(c) intravarietal

Question 11.
Progeny obtained as a result of repeat self pollination a cross pollinated crop to called __________
(a) pure line
(b) pedigree line
(c) inbreed line
(d) heterosis
Answer:
(a) pure line

Question 12.
Jaya and Ratna are the semi dwarf varieties of __________
(a) wheat
(b) rice
(c) cowpea
(d) mustard
Answer:
(b) rice

Question 13.
Which one of the following are the species that are crossed to give sugarcane varieties with high sugar, high yield, thick stems and ability to grow in the sugarcane belt of North India?
(a) Saccharum robustum and Saccharum officinarum
(b) Saccharum barberi and Saccharum officinarum
(c) Saccharum sinense and Saccharum officinarum
(d) Saccharum barberi and Saccharum robustum
Answer:
(b) Saccharum barberi and Saccharum officinarum

Question 14.
Match column I (crop) with column II (Corresponding disease resistant variety) and select the correct option from the given codes.

Answer:
(b) ii, i, iii, iv

Question 15.
A wheat variety, Atlas 66 which has been used as a donor for improving cultivated wheat, which is rich in __________
(a) iron
(b) carbohydrates
(c) proteins
(d) vitamins
Answer:
(c) proteins

Question 16.
Which one of the following crop varieties correct matches with its resistance to a disease?

Answer:
(a) Pusa Komal – Bacterial blight

Question 17.
Which of the following is incorrectly paired?
(a) Wheat – Himgiri
(b) Milch breed – Sahiwal
(c) Rice – Ratna
(d) Pusa Komal – Brassica
Answer:
(d) Pusa Komal – Brassica

Question 18.
Match list I with list II:

Answer:
(b) i-d, ii-c, iii-a, iv-b

Question 19.
Differentiate primary introduction from secondary introduction
Answer:

1. Primary Introduction: Primary introduction – When the introduced variety is well adapted to the new environment without any alternation to the original genotype.
2. Secondary Introduction: Secondary introduction – When the introduced variety is subjected to selection to isolate a superior variety and hybridized with a local variety to transfer one or a few characters to them.

Question 20.
How are microbial innocnlants used to increase the soil fertility?
Answer:
Biofertilizers or microbial innoculants are defined as preparations containing living cells or latent cells of efficient strains of microorganisms that help crop plants uptake of nutrients by their interactions in the rhizosphere when applied through seed or soil.

They are efficient in fixing nitrogen, solubilising phosphate and decomposing cellulose. They are designed to improve the soil fertility, plant growth, and also the number and biological activity of beneficial microorganisms in the soil. They are ecofriendly organic agro inputs and are more efficient and cost effective than chemical fertilizers.

Question 21.
What are the different types of hybridization?
Answer:
Types of Hybridization
According to the relationship between plants, the hybridization is divided into.

1. Intravarietal hybridization – The cross between the plants of same variety. Such crosses are useful only in the self-pollinated crops.
2. Intervarietal hybridization – The cross between the plants belonging to two different varieties of the same species and is also known as intraspecific hybridization. This technique has been the basis of improving self-pollinated as well as cross pollinated crops.
3. Interspecific hybridization – The cross between the plants belonging to different species belonging to the same genus is also called intragenic hybridization. It is commonly used for transferring the genes of disease, insect, pest and drought resistance from one species to another.

Question 22.
Explain the best suited type followed by plant breeders at present?
Answer:
Mutation breeding represents a new method of conventional breeding procedures as they have the advantage of improving the defect without losing agronomic and quality character in agriculture and crop improvement. Mutation means the sudden heritable changes in the genotype or phenotype of an organism. Gene mutations are of considerable importance in plant breeding as they provide essential inputs for evolution as well as for re-combination and selection. It is the only method for improving seedless crops.

Question 23.
Write a note on heterosis.
Answer:
The superiority of the F₁ hybrid in performance over its parents is called heterosis or hybrid vigour. Vigour refers to increase in growth, yield, greater adaptability of resistance to diseases, pest and drought. Vegetative propagation is the best suited measure for maintaining hybrid vigour, since the desired characters are not lost and can persist over a period of time. Many breeders believe that its magnitude of heterosis is directly related to the degree of genetic diversity between the two parents. Depending on the nature, origin, adaptability and reproducing ability heterosis can be classified as:

1. Euheterosis- This is the true heterosis which is inherited and is further classified as
2. Mutational Euheterosis – Simplest type of euheterosis and results from the sheltering or eliminating of the deleterious, unfavourable often lethal, recessive, mutant genes by their adaptively superior dominant alleles in cross pollinated crops.
3. Balanced Euheterosis – Well balanced gene combinations which is more adaptive to environmental conditions and agricultural usefulness.
4. Pseudoheterosis – Also termed as luxuriance. Progeny possess superiority over parents in vegetative growth but not in yield and adaptation, usually sterile or poorly fertile.

Question 24.
List out the new breeding techniques involved in developing new traits in plant breeding.
Answer:
New Breeding Techniques (NBT) are a collection of methods that could increase and accelerate the development of new traits in plant breeding. These techniques often involve genome editing, to modify DNA at specific locations within the plants to produce new traits in crop plants. The various methods of achieving these changes in traits include the following.

• Cutting and modifying the genome during the repair process by tools like CRISPR /Cas.
• Genome editing to introduce changes in few base pairs using a technique called Oligonucleotide-directed mutagenesis (ODM).
• Transferring a gene from an identical or closely related species (cisgenesis).
• Organising processes that alter gene activity without altering the DNA itself (epigenetic methods)

Samacheer Kalvi 12th Bio Botany Plant Breeding Additional Questions and Answers

1 – Mark Questions

Question 1.
____________ is the process of bringing a plant species under human control.
(a) Emasculation
(b) Hybridization
(c) Domestication
(d) Acclimatization
Answer:
(c) Domestication

Question 2.
Which of the following scientist developed world’s first cotton hybrid?
(a) Dr. B.P. Pal
(b) C.T. Patel
(c) Dr. K. Ramiah
(d) N.G.P. Rao
Answer:
(b) C.T. Patel

Question 3.
Identify the incorrect statement:
(a) Bio-inoculants are efficient in solubilising phosphate
(A) Bio-inoculants are ecofriendly organic agro outputs
(c) Bio-inoculants are obtained from dead organic matters
(d) Bio-inoculants are designed to improve soil fertility
Answer:
(c) Bio-inoculants are obtained from dead organic matters

Question 4.
Which is not a free-living nitrogen fixing species?
(a) Azotobacter
(b) Clostridium
(c) Nostop
(d) Anabaena
Answer:
(d) Anabaena

Question 5.
Arbuscular mycorrhizae is a symbiotic association between ________
(a) Algae and fungi
(b) Angiosperm roots and fungi
(c) Blue green algae and Azolla fern
(d) Cyanobacteria and corolloid root
Answer:
(b) Angiosperm roots and fungi

Question 6.
Azolla is best suited biofertilizer for ____________
(a) Sugar cane cultivation
(b) Paddy cultivation
(c) Wheat cultivation
(d) Cotton cultivation
Answer:
(A) Paddy cultivation

Question 7.
Assertion (A): SLF promotes vigorous growth and provide resistance against diseases.
Reason (R): SLF is made from kelp containing more than 70 minerals.
(a) Both A and R are true. R explains A.
(A) A is true R is false
(c) A is false R is true
(d) Both A and R are false
Answer:
(a) Both A and R are true. R explains A.

Question 8.
Assertion (A): Pure line varieties show homozygosity.
Reason (R): Pure line species are obtained through cross pollination.
(a) Both A and R are true. R explains A.
(b) A is true R is false
(c) A is false R is true
(d) Both A and R are false
Answer:
(b) A is true R is false

Question 9.
Assertion (A): Hybrids show increased growth and elevated yield.
Reason (R): F₁ hybrids show Heterosis.
(a) Both A and R are true. R explains A.
(b) A is true R is false
(c) A is false R is true
(d) Both A and R are false
Answer:
(a) Both A and R are true. R explains A.

Question 10.
Statement (1): Trichoderma species is a free-living bacteria.
Statement (2): It acts as a potent bio-control agent
(a) Statement 1 is correct and Statement 2 is incorrect
(b) Statement 1 is incorrect and Statement 2 is correct
(c) Both the statements are correct
(d) Both the statements are incorrect
Answer:
(b) Statement 1 is incorrect and Statement 2 is correct

Question 11.
Statement (1): Clonal selection is carried out in asexually propagating plants.
Statement (2): Clones show similar genotypes.
(a) Statement 1 is correct and Statement 2 is incorrect
(b) Statement 1 is incorrect and Statement 2 is correct
(c) Both the statements are correct
(d) Both the statements are incorrect
Answer:
(c) Both statements are correct

Question 12.
Removal of anthers from a flower to overcome self-pollination and the phenomenon is.
Answer:
Emasculation

Question 13.
Identify the proper sequence of hybridisation technique.
(a) Emasculation → Selection → Bagging → Crossing → Harvesting
(b) Harvesting → Selection → Crossing → Emasculation → Bagging
(c) Selection → Harvesting → Crossing → Emasculation → Bagging
(d) Selection → Emasculation → Bagging → Crossing → Harvesting
Answer:
(d) Selection → Emasculation → Bagging → Crossing → Harvesting

Question 14.
Intra specific hybridization is also termed as
(a) Intravarietal hybridization (b) Intervarietal hybridization
(c) Interspecific hybridization (d) Intergeneric hybridization
Answer:
(b) Intervarietal hybridization

Question 15.
The period of opening of a flower is ________
Answer:
Anthesis

Question 16.
Superiority of hybrids over parents only in vegetative growth not in yield. This phenomenon is termed as ________
(a) Euheterosis
(b) Balanced euheterosis
(c) Luxuriance
(d) Mutational heterosis
Answer:
(c) Luxuriance

Question 17.
The term green revolution was coined by ________
(a) William S Gaud
(b) M.S. Swaminathan
(c) Dr. B.P. Pal
(d) Dr. N.E. Borlaug
Answer:
(a) William S Gaud

Question 18.
Which is the second gamma garden in India?
Answer:
Indian Agricultural Research Institute (IARI)

Question 19.
Match the following:

Answer:
(a) A – ii, B – i, C – iv, D – iii

Question 20.
Who is popularly called as the “father of green revolution in India”?
(a) Nel Jeyaraman
(b) Dr. M.S. Swaminathan
(c) Dr. Nammalvar
(d) N.G.P. Rao
Answer:
(a) Dr. M. S. Swaminathan

Question 21.
Pusa swamim variety of Brassica species show resistance to ________
(a) White rust
(b) Leaf curl
(c) Black rot
(d) Hill bunt
Answer:
(a) White rust

Question 22.
The first established Atomic Garden in India was ________
(a) Bhabha Atomic Research Institute
(b) Indira Gandhi Centre for Atomic Research
(c) Indian Agricultural Research Institute
(d) Bose Research Institute
Answer:
(d) Bose Research Institute

Question 23.
Triticale is polyploid breed of ________
(a) Triticum cereale x Secale sativus
(b) Triticum durum x Secale cereale
(c) Triticum cereale x Secale sativus
(d) Triticum sativus x Secale cereale
Answer:
(b) Triticum durum x Secale cereale

Question 24.
Raphanobrassica is an example for ________
(a) Autopolyploid
(b) Allopolyploid
(c) Polyploid
(d) Polysomy
Answer:
(b) Allopolyploid

Question 25.
Atlas 66 is a improved variety of ________
(a) Rice
(b) Maize
(c) Wheat
(d) Spinach
Answer:
(c) Wheat

Question 26.
Pusa Sawani variety of okra is resistant against ________
(a) Aphids
(b) Fruit borers
(c) Shoot and fruit borers
(d) Jassids
Answer:
(c) Shoot and fruit borers

Question 27.
________ was the first person to develop world’s first hybrid of sorghum.
Answer:
N.G.P. Rao

Question 28.
Identify the incorrect statement:
(i) SLF is obtained from Kelps – a brown algae
(ii) Azolla is a fern
(iii) Rhizobium is found in association with root nodules
(iv) AM forms symbiotic relation with angiospermic roots
(a) i only
(b) ii, iii and iv only
(c) none of the above
(d) all the above
Answer:
(c) none of the above

Question 29.
Damping off of tomato is controlled by ________
(a) Beauveria species
(b) Trichoderma species
(c) Acacia species
(d) Pseudomonas species
Answer:
(a) Beauveria species

Question 30.
Match the following:

Answer:
(a) A – ii, B – iv, C – i, D – iii

Question 31.
Identify the correct symbiotic association
(a) Rhizobium × Corolloid roots
(b) Arbuscular Mycorrhizae × Angiospermic roots
(c) Azolla × Amantia
(d) Azospirillum × Azolla
(b) Arbuscular Mycorrhizae × Angiospermic roots
Answer:
(b) Arbuscular Mycorrhizae × Angiospermic roots

Question 32.
Atomita 2 – rice is a product by.
(a) Polyploid breeding
(b) Hybridization
(c) Mutation breeding
(d) Clonal selection
Answer:
(c) Mutation breeding

Question 33.
Luxuriance is the term used on par with ________
(a) Heterosis
(b) Anthesis
(c) Hybrids
(d) Mutant breeds
Answer:
(a) Heterosis

Question 34.
The term pureline was coined by.
Answer:
Johannsen

2 – Mark Questions

Question 1.
What is meant by domestication of plants?
Answer:
Domestication is the process of bringing a plant species under the control of humans and gradually changing it through careful selection,, genetic alteration and handling so that it is more useful to people.

Question 2.
Mention any two free – living nitrogen fixing bacteria.
Answer:
Azotobacter and Clostridium.

Question 3.
What are bio-pesticides? Why they are considered better than synthetic pesticides?
Answer:
Bio-pesticides are biologically based agents used for the control of plant pests. They are in high use due to their non-toxic, cheaper and eco-friendly characteristics as compared to chemical or synthetic pesticides.

Question 4.
Name any four plants used in Green leaf manuring.
Answer:

1. Cassia fistula
2. Sesbania grandiflora
3. Azadirachta indica
4. Pongamia pinnata

Question 5.
What is plant breeding?
Answer:
Plant breeding is the science of improvement of crop varieties with higher yield, better quality, resistance to diseases and shorter durations which are suitable to particular environment.

Question 6.
Define acclimatization.
Answer:
The newly introduced plant has to adapt itself to the new environment. This adjustment or adaptation of the introduced plant in the changed environment is called acclimatization.

Question 7.
Who coined the term “pureline”? Define it.
Answer:
Johannsen in 1903 coined the word pureline. It is a collection of plants obtained as a result of repeated self-pollination from a single homozygous individual.

Question 8.
Define the following terms:
Answer:
(a) Emasculation
(b) Bagging
Emasculation: It is a process of removal of anthers to prevent self pollination before anthesis (period of opening of a flower).
Bagging: The stigma of the flower is protected against any undesirable pollen grains, by covering it with a bag.

Question 9.
What is Heterosis?
Answer:
The superiority of the F₁ hybrid in performance over its parents is called heterosis or hybrid vigour. Vigour refers to increase in growth, yield, greater adaptability of resistance to diseases, pest and drought.

Question 10.
What does the term ‘luxuriance’ stands for in plant breeding? Explain.
Answer:
Pseudoheterosis – Also termed as luxuriance. Progeny possess superiority over parents in vegetative growth but not in yield and adaptation, usually sterile or poorly fertile.

Question 11.
Mutagens are the substances that induces mutation. Name any two physical and chemical mutagens.
Answer:

1. UV short waves, X-rays – Physical mutagens.
2. Nitromethyl, Urea – Chemical mutagens.

Question 12.
Write in brief about Atomic Garden.
Answer:
Atomic Garden: Is a form of mutation breeding where plants are exposed to radioactive sources typically cobalt-60 or caesium-137 in order to generate desirable mutation in crop plants.

Question 13.
State any one advantage and one disadvantage of polyploid breeding.
Answer:

1. Advantage: Polyploidy often exhibit hybrid vigour and increased tolerance to biotic and abiotic stresses.
2. Disadvantage: Polyploidy results in reduced fertility due to meiotic error resulting in seedless varieties.

Question 14.
What are polyploids? Mention its nature.
Answer:
Majority of flowering plants are diploid (2n). The plants which possess more than two sets of chromosome are called polyploids. Polyploidy often exhibit increased hybrid vigour, increased heterozygosity, increase the tolerance to both biotic and abiotic stresses and buffering of deleterious mutations.

Question 15.
Name any two allopolyploid plant species.
Answer:

1. Triticale (Triticum durum x Secale cereale).
2. Raphanobrassica (Brassica oleraceae x Raphanus sativus).

Question 16.
What is Biofortification?
Answer:
Biofortification is breeding crops with higher levels of vitamins and minerals or higher protein and healthier fats. It is the most practical means to improve public health.

Question 17.
Name the insect resistant varieties developed in the following crops.

1. Okra
2. Rapeseed mustard.

Answer:

1. Okra: Pusa Sawani and Pusa A-4 varieties of Okra are resistant to shoot and fruit borers.
2. Rapeseed mustard: Pusa gaurav variety of rapeseed mustard shown

Question 18.
In which plant, and by whom the first natural hybridization was performed?
Answer:
The first natural hybridization was done by Cotton Mather in maize.

3 – Mark Questions

Question 19.
In 1926, Vavilov initially proposed eight main geographic centres of crop origin. Mention any six of them.
Answer:
China, India, South America, Mediterranean, Mesoamerica and South East Asia.

Question 20.
Name any three eminent plant breeders of Indian origin.
Answer:

1. Dr. M.S. Swaminathan
2. C.T. Patel
3. Dr. B.P. Pal

Question 21.
How Rhizobium acts as a efficient bio-fertilizer?
Answer:
Bio-fertilizers containing rhizome bacteria are called rhizome bio-fertilizer culture. Symbiotic bacteria that reside inside the root nodules convert the atmospheric nitrogen into a bio available form to the plants. This nitrogen fixing bacterium when applied to the soil undergoes multiplication in billions and fixes the atmospheric nitrogen in the soil. Rhizobium is best suited for the paddy fields which increase the yield by 15 – 40%.

Question 22.
Azolla increases the yield of paddy crop – support your answer.
Answer:
Azolla is a free-floating water fern that fixes the atmospheric nitrogen in association with nitrogen fixing blue green alga Anabaena azolla. It is used as a bio-fertilizer for wetland rice cultivation and is known to contribute 40 – 60 kg/ha/crop. The agronomic potential of Azolla is quite significant particularly for increasing the yield of rice crop, as it quickly decompose in soil.

Question 23.
Mention any three advantages of Arbuscular mycorrhizal association.
Answer:

1. Arbuscular mycorrhizae have the ability to dissolve the phosphate found in soil.
2. Provides resistance against diseases, germs and unfavourable weather conditions.
3. It assures water availability.

Question 24.
What makes the Trichoderma an effective bio-pesticide?
Answer:
Trichoderma species are free-living fungi that are common in soil and root ecosystem. They have been recognized as bio-control agent for

1. the control of plant disease
2. ability to enhance root growth development
3. crop productivity
4. resistance to abiotic stress and
5. uptake and use of nutrients.

Question 25.
Write a note on Green manuring.
Answer:
Green manuring is defined as the growing of green manure crops and use of these crops directly in the field by ploughing. One of the main objectives of the green manuring is to increase the content of nitrogen in the soil. Also it helps in improving the structure and physical properties of the soil. The most important green manure crops are Crotalaria juncea, Tephrosia purpurea and Indigofera tinctoria.

Question 26.
Point out the objectives of plant breeding.
Answer:

1. To increase yield, vigour and fertility of the crop.
2. To increase tolerance to environmental condition, salinity, temperature and drought.
3. To prevent the premature falling of buds and fruits, etc.
4. To improve synchronous maturity.
5. To develop resistance to pathogens and pests.
6. To develop photosensitive and thermos-sensitive varieties

Question 27.
Give an account on clonal selection.
Answer:
In asexually propagated crop, progenies derived from a plant resemble in genetic constitution with the parent plant as they are mitotically divided. Based on their phenotypic appearance, clonal selection is employed to select improved variety from a mixed population (clones). The selected plants are multiplied through vegetative propagation to give rise to a clone. The genotype of a clone remains unchanged for a long period of time.

Question 28.
Write a short note on autopolyploidy with an example.
Answer:
When chromosome number is doubled by itself in the same plant, is called autopolyploidy.
Example: A triploid condition in sugarbeets, apples and pear has resulted in the increase in vigour and fruit size, large root size, large leaves, flower, more seeds and sugar content in them. It also resulted in seedless tomato, apple, watermelon and orange.

5-Mark Questions

Question 29.
Give a comparative account on Seaweed liquid fertilizer.
Answer:
Seaweed liquid fertilizer (SLF) contains cytokinin, gibberellins and auxin apart from macro and micro nutrients. Most seaweed based fertilizers are made from kelp (brown algae) which grows to length of 150 metres. Liquid seaweed fertilizer is not only organic but also eco-friendly. The alginates in the seaweed that reacts with metals in the soil and form long, cross- linked polymers in the soil. These polymers improve the crumbing in the soil, swell up when they get wet and retain moisture for a long time.

They are especially useful in organic gardening which provides carbohydrates for plants. Seaweed has more than 70 minerals, vitamins and enzymes. It promotes vigorous growth. Improves resistance of plants to frost and disease. Seeds soaked in seaweed extract germinate much rapidly and develop a better root system.

Question 30.
Explain the steps involved in hybridization.
Answer:
Steps involved in hybridization are as follows:

1. Selection of Parents: Male and female plants of the desired characters are selected. It should be tested for their homozygosity.
2. Emasculation: It is a process of removal of anthers to prevent self pollination before anthesis (period of opening of a flower).
3. Bagging: The stigma of the flower is protected against any undesirable pollen grains, by covering it with a bag.
4. Crossing: Transfer of pollen grains from selected male flower to the stigma of the female emasculated flower.
5. Harvesting seeds and raising plants: The pollination leads to fertilization and finally seed formation takes place.
   The seeds are grown into new generation which are called hybrid.

Higher Order Thinking Skills (HOTs) Questions

Question 1.
Given below are the examples of symbiotic association in which one partner was mentioned. Write the name of the mutual co-partners.
(a) Azollafem + ______________
(b) Root nodules of legume plants + ______________
(c) phycomycetous fungi + ______________
(a) Anabaena azolla (blue green alga)
(b) Rhizobium
Answer:
(c) Angiosperm roots

Question 2.
Provide an example for each of the following agricultural components.

1. Bio pesticide
2. Green manure crop
3. Bio-Fertilzier

Answer:

1. Bio pesticide – E.g. Trichoderma species
2. Green manure – E.g. Crotalariajuncea
3. Bio fertilizer – E.g. Azolla

Question 3.
State the objective of using green manuring.
Answer:
Green manuring helps to increase the content of nitrogen soil and also improves the structure and physical property of soil.

Question 4.
Why plant Breeding is carried out by farmers and scientists?
Answer:
Plant breeding is a proposal manipulation of plant species in order to develop improved crop varieties with higher yield, better quality, resistance to disease and quick maturation.

Question 5.
Yesterday, Ramu visited his friend’s Orchard, where in he noticed few flowers of certain guava trees are covered using thin paper bags.

1. Name the process carried out there.
2. Why it was done so?

Answer:

1. The process is referred as bagging in plant hybridization.
2. It is done after emasculation to protect the contact of the stigma of the flower against any other pollen grain.

Question 6.
Who am I?

1. Father of Green Revolution.
2. Father of Indian Green Revolution.

Answer:

1. Dr. Norman E. Borlaug.
2. Dr. M.S. Swaminathan.

Question 7.
A plant breeder developed a hybrid sugarcane by grafting two different varieties with desirable characters. The resultant hybrid showed as excellent growth and productivity with increased sucrose content compared to its parental forms.

1. What does this phenomenon refers to?
2. How this condition can be maintained through further generation?

Answer:

1. Heterosis or Hybrid vigour
2. Vegetative propogation is the best suited measure to maintain the vigourisity of hybrid.

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Samacheer Kalvi 12th Bio Botany Chromosomal Basis of Inheritance Text Book Back Questions and Answers

Question 1.
An allohexaploidy contains ___________
(a) Six different genomes
(b) Six copies of three different genomes
(c) Two copies of three different genomes
(d) Six copies of one genome
Answer:
(b) Six copies of three different genomes

Question 2.
The A and B genes are 10 cM apart on a chromosome. If an AB/ab heterozygote is testcrossed to ab/ab, how many of each progeny class would you expect out of 100 total progeny?
(a) 25 AB, 25 ab, 25 Ab, 25 aB
(b) 10AB, 10ab
(c) 45 AB, 45 ab
(d) 45 AB, 45 ab, 5 Ab, 5aB
Answer:
(b) 10AB, 10ab

Question 3.
Match list I with list II

(a) A-i, B-iii, C-ii, D-iv
(b) A-ii, B-iii, C-iv, D-i
(c) A-ii, B-iii, C-i, D-iv
(d) A -iii, B-ii, C-i, D-iv
Answer:
(c) A-ii, B-iii, C-i, D-iv

Question 4.
Which of the following sentences are correct?
1. The offspring exhibit only parental combinations due to incomplete linkage
2. The linked genes exhibit some crossing over in complete linkage
3. The separation of two linked genes are possible in incomplete linkage
4. grossing over is absent in complete linkage
(a) 1 and 2
(b) 2 and 3
(c) 3 and 4
(d) 1 and 4
Answer:
(c) 3 and 4

Question 5.
Accurate mapping of genes can be done by three point test cross because increases ___________
(a) Possibility of single cross over
(b) Possibility of double cross over
(c) Possibility of multiple cross over
(d) Possibility of recombination frequency
Answer:
(b) Possibility of double cross over

Question 6.
Due to incomplete linkage in maize, the ratio of parental and recombinants are ___________
(a) 50 : 50
(b) 7 : 1 : 1 : 7
(c) 96.4 : 3.6
(d) 1 : 7 : 7 : 1
Answer:
(b) 7 : 1 : 1 : 7

Question 7.
Genes G S L H are located on same chromosome. The recombination percentage is between and G is 15%, S and L is 50% and H and S are 20%. The correct order of genes is ___________
(a) GHSL
(b) SHGL
(c) SGHL
(d) HSLG
Answer:
(b) SHGL

Question 8.
The point mutation sequence for transition, transition, transversion and transversion in DNA are ________
(a) A to T, T to A, C to G and G to C
(b) A to G, C to T, C to G and T to A
(c) C to G, A to G, T to A and G to A
(d) G to C, A to T, T to A and C to G
Answer:
(b) A to G, C to T, C to G and T to A

Question 9.
If haploid number in a cell is 18. The double monosomic and trisomic number will be ___________
(a) 35 and 37
(b) 34 and 37
(c) 37 and 35
(d) 17 and 19
Answer:
(b) 34 and 37

Question 10.
Changing the codon AGC to AGA represents ___________
(a) mis-sense mutation
(b) non-sense mutation
(c) frameshift mutation
(d) deletion mutation
Answer:
(a) mis-sense mutation

Question 11.
Assertion (A): Gamma rays are generally used to induce mutation in wheat varieties.
Reason (R): Because they carry lower energy to non-ionize electrons from atom
(a) A is correct. R is correct explanation of A
(b) A is correct. R is not correct explanation of A
(c) A is correct. R is wrong explanation of A
(d) A and R is wrong
Answer:
(c) A is correct. R is wrong explanation of A

Question 12.
How many map units separate two alleles A and B, if the recombination frequency is 0.09?
(a) 900 cM
(b) 90 cM
(c) 9 cM
(d) 0.9 cM
Answer:
(d) 0.9 cM

Question 13.
When two different genes came from same parent they tend to remain together.

1. What is the name of this phenomenon?
2. Draw the cross with suitable example.
3. Write the observed phenotypic ratio.

Answer:
(i) Linkage
(ii)

(iii) Observed Pherotypic ratio 7 : 1 : 1 : 7

Question 14.
If you cross dominant genotype PV/PV male Drosophila with double recessive female and obtain F₁ hybrid. Now you cross F₁ male with double recessive female.

1. What type of linkage is seen?
2. Draw the cross with correct genotype.
3. What is the possible genotype in F₂ generation?

Answer:
(i) Incomplete linkage
(ii) Parent Garnets

F₁ hybrid

Question 15.

1. What is the name of this test cross?
2. How will you construct gene mapping from the above given data?
3. Find out the correct order of genes.

Answer:
(i) Three point test cross.
(ii) Construction of gene map:
To construct the gene map, the recombinant frequency (RF) of the alleles has to be calculated.
From the given data it is clear that ABC and abc are parental (P) types and the others (Abe, abC, AbC, aBc, ABc) are recombinant (R) type.

Lets analyse the loci of two alleles at a time starting with A and B. Since the genes AB and ab are parental type, the recombinants will be Ab and aB.
Therefore
Recombinant frequency of alleles Ab and aB = \(\frac { No.of recombinant }{ Total progenies }\) × 100
\(\frac { 114+5+4+116 }{ 1200 }\) × 100 = 19.91%

Recombinant frequency for the loci B and C
The parental form are Be and bC and the recombinant are Be and bC.

Recombinant frequency of alleles Be and bC = \(\frac { 4+128+124+5 }{ 1200 }\) × 100 + 21.75%
Since the recombinant frequency of the alleles A and C shown highest frequency, they must be the farthest apart and alleles B must lie in between A and C. So the gene map can be constructed as follows

(iii) The correct gene order is ABC/abc.

Question 16.
What is the difference between mis-sense and nonsense mutation?
Answer:
Mis-sense Mutation:
The mutation where the codon for one amino acid is changed into a codon for another amino acid is called Missense or non-synonymous mutations.

Non-sense Mutation:
The mutations where codon for one amino acid is changed into a termination or stop codon is called Nonsense mutation.

Question 17.
A B C C B D E F G H I
From the above figure identify the type of mutation and explain it.
Answer:
In reverse tandem duplication, the duplicated segment is located immediately after the normal segment but the gene sequence other will be reversed.

Question 18.
Write the salient features of Sutton and Boveri concept.
Answer:
Salient features of the chromosomal theory of inheritance:

1. Somatic cells of organisms are derived from the zygote by repeated cell division (mitosis). These consist of two identical sets of chromosomes. One set is received from female parent (maternal) and the other from male parent (paternal). These two chromosomes constitute the homologous pair.
2. Chromosomes retain their structural uniqueness and individuality throughout the life cycle of an organism.
3. Each chromosome carries specific determiners or Mendelian factors which are now termed as genes.
4. The behaviour of chromosomes during the gamete formation (meiosis) provides evidence to the fact that genes or factors are located on chromosomes.

Question 19.
“Explain the mechanism of crossing over.
Answer:
Crossing over is a precise process that includes stages like synapsis, tetrad formation, cross over and terminalization.

(i) Synapsis: Intimate pairing between two homologous chromosomes is initiated during zygotene stage of prophase I of meiosis I. Homologous chromosomes are aligned side by side resulting in a pair of homologous chromosomes called bivalents. This pairing phenomenon is called synapsis or syndesis. It is of three types:

1. Procentric synapsis: Pairing starts from middle of the chromosome.
2. Proterminal synapsis: Pairing starts from the telomeres.
3. Random synapsis: Pairing may start from anywhere.

(ii) Tetrad Formation: Each homologous chromosome of a bivalent begin to form two identical sister chromatids, which remain held together by a centromere. At this stage each bivalent has four chromatids. This stage is called tetrad stage.

(iii) Cross Over: After tetrad formation, crossing over occurs in pachytene stage. The non-sister chromatids of homologous pair make a contact at one or more points. These points of contact between non¬sister chromatids of homologous chromosomes are called Chiasmata (singular-Chiasma).

At chiasma, cross-shaped or X-shaped structures are formed, where breaking and rejoining of two chromatids occur. This results in reciprocal exchange of equal and corresponding segments between them. A recent study reveals that synapsis and chiasma formation are facilitated by a highly organised structure of filaments called Synaptonemal Complex (SC). This synaptonemal complex formation is absent in some species of male Drosophila, hence crossing over does not takes place.

(iv) Terminalisation: After crossing over, chiasma starts to move towards the terminal end of chromatids. This is known as terminalisation. As a result, complete separation of homologous chromosomes occurs.

Question 20.
Write the steps involved in molecular mechanism of DNA recombination with diagram.
Answer:

1. Homologous DNA molecules are paired side by side with their duplicated copies of DNAs
2. one stand place by the enzyme endonuclease.
3. The cut strands cross and join the homologous strands forming the Holliday structure or Holliday junction.
4. The Holliday junction migrates away from the original site, a process called branch migration, as a result heteroduplex region is formed.
5. DNA strands may cut along through the vertical (V) line or horizontal (H) line.
6. The vertical cut will result in heteroduplexes with recombinants.
7. The horizontal cut will result in heteroduplex with non recombinants.

Question 21.
How is Nicotiana exhibit self-incompatibility. Explain its mechanism.
Answer:
Self-sterility means that the pollen from a plant is unable to germinate on its own stigma and will not be able to bring about fertilization in the ovules of the same plant. East (1925) observed multiple alleles in Nicotiana which are responsible for self-incompatibility or self-sterility. The gene for self-incompatibility can be designated as S, which has allelic series S₁ S₂, S₃, S₄ and S₅. The cross-fertilizing tobacco plants were not always homozygous as S₁ S₁ or S₂ S₂, but all plants were heterozygous as S₁ S₂, S₃ S₄ and S₅  S₆. When crosses were made between different S₁ S₂ plants, the pollen tube did not develop normally. But effective pollen tube development was observed when crossing was made with other than S₁S₂ for example S₃S4.

When crosses were made between seed parents with S₁S₂ and pollen parents with S₂S₃, two kinds of pollen tubes were distinguished. Pollen grains carrying S₂ were not effective, but the pollen grains carrying S₃ were capable of fertilization. Thus, from the cross S₁ S₂ X S₃ S₄, all the pollens were effective and four kinds of progeny resulted: S₁S₃, S₁ S₄, S₂ S₃ and S₂ S₄.

Question 22.
How sex is determined in monoecious plants? Write their genes involved in it.
Answer:
Zea mays (maize) is an example for monoecious, which means male and female flowers are present on the same plant. There are two types of inflorescence. The terminal inflorescence which bears staminate florets that develops from shoot apical meristem called tassel. The lateral inflorescence which develop pistillate florets from axillary bud is called ear or cob. Unisexuality in maize occurs through the selective abortion of stamens in ear florets and pistils in tassel florets. A substitution of two single gene pairs ‘ba’ for barren plant and ‘ts’ for tassel
seed makes the difference between monoecious and dioecious (rare)

maize plants. The allele for barren plant (ba) when homozygous makes the stalk staminate by eliminating silk and ears. The allele for tassel seed
(ts) transforms tassel into a pistillate structure that produce no pollen. The table is the resultant sex expression ‘ based on the combination of these
alleles. Most of these mutations are shown to be defects in gibberellin biosynthesis. Gibberellins play an important role in the suppression of stamens in florets on the ears.

Question 23.
What is gene mapping? Write its uses.
Answer:
The diagrammatic representation of position of genes and related distances between the adjacent genes is called genetic mapping. It is directly proportional to the frequency of recombination between them. It is also called as linkage map.
Uses of genetic mapping:

• It is used to determine gene order, identify the locus of a gene and calculate the distances between genes.
• They are useful in predicting results of dihybrid and trihybrid crosses.
• It allows the geneticists to understand the overall genetic complexity of particular organism.

Question 24.
Draw the diagram of different types of aneuploidy.
Answer:

Question 25.
Mention the name of man-made cereal. How it is formed?
Answer:
Triticale, the successful first man made cereal. Depending on the ploidy level Triticale can be divided into three main groups:

1. Tetraploidy: Crosses between diploid wheat and rye.
2. Hexaploidy: Crosses between tetraploid wheat Triticum durum (macaroni wheat) and rye.
3. Octoploidy: Crosses between hexaploid wheat T. aestivum (bread wheat) and rye. Hexaploidy Triticale hybrid plants

demonstrate characteristics of both macaroni wheat and rye. For example, they combine the high-protein content of wheat with rye’s high content of the amino acid lysine, which is low in wheat.

Samacheer Kalvi 12th Bio Botany Chromosomal Basis of Inheritance Additional Questions and Answers

1 – Mark Questions

Question 1.
Name the scientist(s) who rediscovered the Mendelian work?
(i) Hugo de Vries
(ii) Carl Correns
(iii) Tschermak
(a) i and iv
(b) i, ii and iv
(c) (iii) i, ii and iii
(d) ii, iii and iv
Answer:
(c) i, ii and iii

Question 2.
Which is not a feature of the chromosomal theory of inheritance?
(a) Somatic cells of organisms are derived from zygote by repeated meiosis.
(b) Chromosomes retain their structural uniqueness throughout the life of an organism.
(c) Mendelian factors are located in chromosomes
(d) Sutton and Boveri independently proposed the theory.
Answer:
(a) Somatic cells of organisms are derived from zygote by repeated meiosis.

Question 3.
The following sequence represents the location of genes in a chromosome. A-B-C-M-R – S – Y – Z. Which of the gene pairs will have least chance of getting inherited together?
(a) A and M
(b) S and Y
(c) M and Z
(d) A and Y
Answer:
(d) A and Y

Question 4.
Match the column I with column II:

(a) 1 – iii, 2 – i, 3 – iv, 4 – ii
(b) 1 – ii, 2 – i, 3 – iv, 4 – iii
(c) 1 – iv, 2 – iii, 3 – ii, 4 – i
(d) 1 – ii, 2 – iii, 3 – i, 4 – iv
Answer:
(a) 1 – iii, 2 – i, 3 – iv, 4 – ii

Question 5.
Number of chromosomes (2n) in Ophioglossum is ________
(a) 1226
(b) 1622
(c) 1262
(d) 2126
Answer:
(c) 1262

Question 6.
Identify the syntenic gene from the given genes sequence of a chromosome G-H-I-J-K-L-M-A-B
(a) G and H
(b) J, K and L
(c) G and B
(d) A and B
Answer:
(c) G and B

Question 7.
Incomplete linkage was reported by Hutchinson in
(a) Drosophila
(b) Maize
(c) Neurospora
(d) Lathyrus odoratus
Answer:
(b) Maize

Question 8.
Mechanism of crossing over involves the following stages. Select the correct sequence.
(а) Tetrad stage → Synapsis → Bivalent stage → cross over
(b) Syndesis → Tetrad → Crossing over → Terminalisation
(c) Terminalisation → Tetrad → Bivalent → Cross over
(d) Cross over → Bivalent → Tetrad → Terminalisation
Answer:
(b) Syndesis → Tetrad → Cross over → Terminalisation

Question 9.
During cross over, chiasma occurs between
(a) Sister chromatids of non-homologous chromosomes
(b) Non-sister chromatids of non- homologous chromosomes
(c) Non-sister chromatids of homologous chromosomes
(d) Sister chromatids of homologous chromosomes
Answer:
(c) Non-sister chromatids of homologous chromosomes

Question 10.
The term crossing over was coined by _________
Answer:
T.H. Morgan

Question 11.
At which stage of meiosis, does the chromosomes undergo recombination process?
(a) Leptotene stage of prophase I
(b) Zygotene stage of prophase I
(c) Diplotene stage of prophase I
(d) Pachytene stage of prophase I
Answer:
(d) Pachytene stage of prophase I

Question 12.
Which of the following statement(s) is/are wrong with respect to Recombination process?
(i) Mitotic crossing over occurs rarely in somatic cells.
(ii) Syndesis refers to pairing of non-homologous chromosome.
(iii) Procentric synapsis starts from telomeres.
(iv) A Bivalent has four chromatids.
(a) i and iv
(b) ii and i
(c) ii and iii
(d) All the above
Answer:
(c) ii and iii

Question 13.
Recombination frequency (RF) is equal to
(a) \(\frac { No. of recombinants }{ No. of recombinants }\) × 100
(b) \(\frac { No. of recombinants }{ No. of patental strains }\) × 100
(c) \(\frac { No. of recombinants }{ No. of offsprings }\) × 100
(d) \(\frac { No. of recombinants }{ No. of Parental strains }\) × 100
Answer:
(c) \(\frac { No. of recombinants }{ No. of offsprings }\) × 100

Question 14.
In a population of 250 progenies produced, only 120 resemble the parental forms. Calculate the recombinant frequency.
(a) 66%
(b) 52%
(c) 59%
(d) 49%
Answer:
(b) 52%

Question 15.
The unit of distance in genetic map is called as __________
Answer:
Map unit or Centimorgan

Question 16.
The “2n” condition of Carica papaya is __________
Answer:
36

Question 17.
Mutation theory was proposed by. __________
(a) T. H. Morgan
(b) Hugo de Vries
(c) Alfred variety
(d) Erectiferm
Answer:
(b) Hugo de Vries

Question 18.
Identity the mutant variety of castor.
(a) Sharbathi Sonora variety
(b) Aruna variety
(c) Reimei variety
(d) Erectiferm variety
Answer:
(b) Aruna variety

Question 19.
Which is not a non-ionizing radiation?
(a) X-rays
(b) Gamma rays
(c) Alpha rays
(d) UV rays
Answer:
(d) UV rays

Question 20.
Transition type of gene mutation is caused when __________
(a) AC is replaced by GT
(b) AG is replaced by TC
(c) AC is replaced by TG
(d) TC is replaced by AG
Answer:
(a) AC is replaced by GT

Question 21.
Pick out the co-mutagen from the following:
(a) Eosin
(b) Mustard gas
(c) Ascorbic acid
(d) Nitrous acid
Answer:
(c) Ascorbic acid

Question 22.
Sharbati Sonara is a mutant wheat variety which is developed by irradiating the seeds with __________
(a) Thermal neutrons
(b) Gamma radiation
(c) X-rays
(d) UV radiations
Answer:
(b) Gamma radiation

Question 23.
Which one of the following ploidy is irrelevant to others?
(a) Monosomy
(b) Trisomy
(c) Tetrasomy
(d) Pentasomy
Answer:
(a) Monosomy

Question 24.
Match with correct

(a) 1 – ii, 2 – iii, 3 – iv, 4 – i
(b) 1 – ii, 2 – i, 3 – iv, 4 – iii
(c) 1 – iv, 2 – iii, 3 – ii, 4 – i
(d) 1 – ii, 2 – iii, 3 – i, 4 – iv
Answer:
(a) 1 – ii, 2 – iii, 3 – iv, 4 – i

Question 25.
Statement 1: Euploidy involves entire sets of chromosomes
Statement 2: Aneuploidy involves individual chromosomes within a diploid net.
(a) Statement 1 is correct and Statement 2 is incorrect
(b) Statement 1 is incorrect and Statement 2 is correct
(c) Both the statements are correct
(d) Both the statements are incorrect
Answer:
(c) Both the statements are correct

Question 26.
Statement 1: In transversion mutation, single purine is changed to pyrimidine.
Statement 2: In transition mutation, a purine replaced by another purine.
(a) Statement 1 is correct and Statement 2 is incorrect
(b) Statement 1 is incorrect and Statement 2 is correct
(c) Both the statements are correct
(d) Both the statements are incorrect
Answer:
(a) Statement 1 is correct and Statement 2 is incorrect

Question 27.
Statement 1: Pairing of homologous chromosome is called as syndesis.
Statement 2: Proterminal synapsis occurs from telomeres.
(a) Statement 1 is correct and Statement 2 is incorrect
(b) Statement 1 is incorrect and Statement 2 is correct
(c) Both the statements are correct
(d) Both the statements are incorrect
Answer:
(c) Both the statements are correct

Question 28.
Statement 1: The widely accepted DNA replication model is Holliday’s hybrid DNA model.
Statement 2: The vertical cut in the DNA results in heteroduplex with non-recombinants.
(a) Statement 1 is correct and Statement 2 is incorrect
(b) Statement 1 is incorrect and Statement 2 is correct
(c) Both the statements are correct
(d) Both the statements are incorrect
Answer:
(d) Both the statements are incorrect.

Question 29.
Statement 1: Self-sterility in Nicotiana is controlled by multiple alleles.
Statement 2: Multiple alleles are always responsible for the same character.
(a) Statement 1 is correct and Statement 2 is incorrect
(b) Statement 1 is incorrect and Statement 2 is correct
(c) Both the statements are correct
(d) Both the statements are incorrect
Answer:
(c) Both the statements are correct

Question 30.
Sex determination in plants was first discovered by.
Answer:
C.E. Allen

Question 31.
One of the following is not the kind of euploidy
(a) Diploidy
(b) Polyploidy
(c) Hyperploidy
(d) Autoploidy
Answer:
(c) Hyperploidy

Question 32.
The chromosomal condition 2n -2 represents __________
(a) Monosomy
(b) Nullisomy
(b) Nullisomy
(c) Trisomy
Answer:
(d) Tetrasomy

Question 33.
(a) Potato
(b) Coffee
(c) Groundnut
(d) Apple
Answer:
(d) Apple

Question 34.
Assertion (A): Polyploidy is common in plants.
Reason (R): Polyploids possess more than 2 basic sets of chromosomes.
(a) A is true R is false
(b) Both A and R are false
(c) A is true, R is not correct explanation for A
(d) R explains A
Answer:
(d) R explains A

Question 35.
Assertion (A): Complete linkage is noticed in male species of Drosophila.
Reason (R): Completely linked genes show some crossing over.
(a) A is true R is false
(b) Both A and R are false
(c) A is true, R is not correct explanation for A
(d) R explains A
Answer:
(a) A is true R is false

Question 36.
Assertion (A): Self-sterility is observed in Nicotiana species.
Reason (R): Because the genes are located on chromosome.
(a) A is true R is false
(b) Both A and R are false
(c) A is true, R is not correct explanation for A
(d) R explains A
Answer:
(c) A is true, R is not correct explanation for A

Question 37.
The first man made cereal is _______
Answer:
Triticale

Question 38.
Observe the gene sequence and identify the types of aberration A B C B C D E F?
(a) Tandem duplication
(b) Simple duplication
(c) Reverse tandem duplication
(d) Displaced tandem duplication
Answer:
(a) Tandem duplication

2 – Mark Questions

Question 1.
Write any one contribution of the following scientists to the field of molecular biology,
(a) Hugo de Vries
(b) Sutton and Boveri
Answer:
(a) Hugo de Vries proposed Mutation theory.
(b) Sutton and Boveri proposed chromosomal theory of inheritance.

Question 2.
Define Linkage. Mention its types.
Answer:
Tendency of genes to remain together during separation of chromosomes is called linkage. Linkage are of 2 types – complete linkage and incomplete linkage.

Question 3.
What does the condition synteny refers to?
Answer:
The two genes that are sufficiently far apart on the same chromosome are called unlinked genes or syntenic genes. Such condition is known as synteny.

Question 4.
What are linked genes?
Answer:
Genes located close together on the same chromosome and inherited together are called linked genes.

Question 5.
Who coined the term crossing over? When does the crossing over occurs in a cell?
Answer:
The term ‘crossing over ’ was coined by Morgan (1912). It takes place during pachytene stage of prophase I of meiosis.

Question 6.
Crossing over occurs only in germinal cells. Yes or no. Support your answer.
Answer:
No. Though crossing over is a common process in germinal cells rarely it also occurs in somatic cells during mitosis. Such crossing over is called mitotic crossing over or somatic crossing over.

Question 7.
Mention the major stages involved in crossing over.
Answer:
Crossing over is a precise process that includes stages like synapsis, tetrad formation, cross over and terminalization.

Question 8.
What are bivalents? When does this condition is noticed in a cell?
Answer:
During zygotene stage of prophase I of meiosis I, homologous chromosomes are aligned side by side resulting in a pair of homologous chromosomes called bivalents.

Question 9.
What is meant by synaptonemal complex?
Answer:
Synaptonemal complex is a highly organised structure or filaments that facilitates the synapsis and chiasma formation during crossing over mechanism.

Question 10.
What will be the result if there is a failure in the formation of synaptonemal complex. Give one example of organism where such condition is noticed?
Answer:
Failure in the formation of synaptonemal complex leads to the absence of crossing over to take place. It is noticed in certain male species of Drosophila.

Question 11.
Define terminalization.
Answer:
After crossing over, chiasma starts to move towards the terminal end of chromatids. This is known as terminalization. As a result, complete separation of homologous chromosomes occurs.

Question 12.
Write the formula to calculate recombination frequency.
Answer:
\(\frac { No. of recombinants }{ No. of offsprings }\) × 100

Question 13.
What is genetic mapping?
Answer:
The diagrammatic representation of position of genes and related distances between the adjacent genes is called genetic mapping. It is directly proportional to the frequency of recombination between them. It is also called as linkage map.

Question 14.
Define the terms
(a) locus
(b) centimorgan.
Answer:
(a) Locus: A specific location of genes on a chromosome.
(b) Centimorgan: Unit of distance in a genetic map.

Question 15.
What are multiple alleles?
Answer:
When any of the three or more allelic forms of a gene occupy the same locus in a given pair of homologous chromosomes, they are said to be called multiple alleles.

Question 16.
Does environment determines the sex of a plant? Explain in brief with an example.
Answer:
Yes, the equisetum (horsetail plant), which grow under good conditions develop as female plants those grown under stressed condition develops as male then supporting the fact environment can also determines the sex of plant in few species.

Question 17.
Why do we call papaya a dimorphic plant.
Answer:
Papaya is a dimorphic plant since their sexes are separate i.e., male plant produces flowers with stamens and female plants produces flowers with carpels.

Question 18.
Define the term mutation. Who coined the term?
Answer:
A sudden change in the genetic material of an organisms is called mutation. The term mutation was introduced by Hugo de Vries.

Question 19.
Compare point mutation with chromosomal mutation.
Answer:
Mutational events that take place within individual genes are called gene mutations or point mutation, whereas the changes occur in structure and number of chromosomes is called chromosomal mutation.

Question 20.
Based on the effect on translation, classify mutations.
Answer:
(a) Silent mutation
(b) Mis-sense mutation
(c) Non-sense mutation
(d) Frameshift mutation

Question 21.
What does indel mutation refers to?
Answer:
Addition or deletion mutations are actually additions or deletions of nucleotide pairs and also called base pair addition or deletions. Collectively, they are termed as indel mutations.

Question 22.
Define mutagens and mention its types.
Answer:
The factors which cause genetic mutation are called mutagenic agents or mutagens. Mutagens are of two types, physical mutagen and chemical mutagen.

Question 23.
Point out any four physical mutagens.
Answer:
Temperature, X-rays, Gamma rays and UV rays.

Question 24.
Write a brief note on Castor Aruna variety.
Answer:
Castor Aruna is mutant variety of castor which is developed by treatment of seeds with thermal neutrons in order to induce very early maturity (120 days instead of 270 days as original variety).

Question 25.
How Sharbati Sonora was developed by Dr. M.S. Swaminathan etal?
Answer:
Sharbati Sonora is a mutant variety of wheat, which is developed from Mexican variety (Sonora 64) by irradiating of gamma rays.

Question 26.
Name any four chemical mutagens.
Answer:
(a) Ethyl methane sulphonate (EMS)
(b) Mustard gas
(c) Magnous salt
(d) Formaldehyde

Question 27.
Nitrous oxide is a potent mutagen – comment.
Answer:
Nitrous oxide alters the nitrogen bases of DNA and disturb the replication and transcription that leads to die formation of incomplete and defective polypeptide during translation.

Question 28.
What is ploidy?
Answer:
The chromosome number of somatic cells changes due to addition or elimination of individual chromosome or basic set of chromosomes. This condition in known as numerical chromosomal aberration or ploidy.

Question 29.
Differentiate Aneuploidy from Euploidy.
Answer:
Aneuploidy:

1. Ploidy involving individual chromosomes within a diploid set.
2. E.g: Trisomy.

Euploidy:

1. Ploidy involving entire sets of chromosomes.
2. E.g: Polyploidy.

Question 30.
Comment on the chromosomal condition; 2n – 2.
Answer:
Loss of pair of homologous chromosome from the diploid set is called Nullisomy. Selling of monosomic plants produces Nullisomics. They are generally lethal.

Question 31.
Name any 3 auto tetraploids and one natural autotriploid plant species.
Answer:
(a) Autotetraploids : Rye, potato and coffee.
(b) Natural autotriploid : Cyanodon dactylon (common doob grass).

Question 32.
What are deficiency loops?
Answer:
Deletions are observable during meiotic pachytene stage and polytene chromosome. The unpaired loop formed in the normal chromosomal part at the time of chromosomal pairing. Such loops are called as deficiency loops and it can be seen in meiotic prophase.

Question 33.
Given below are the gene sequences on the chromosome. Compare them with the normal chromosome and identify the type of structural chromosomal aberrations.
Normal Chromosome : A-B-C-D-E-F-G-H-I.
Chromosome 1 : A-B-C-B-C-D-E-F-G-H-I.
Chromosome 2 : A-B-C-D-F-G-H-I
Answer:
Chromosome 1 : Tandem duplication.
Chromosome 2 : Intercalary deletion.

3 – Mark Questions

Question 34.
Point out any three salient features of chromosomal theory of inheritance.
Answer:
Salient features of the chromosomal theory of inheritance:

1. Somatic cells of organisms are derived from die zygote by repeated cell division (mitosis). These consist of two identical sets of chromosomes. One set is received from female parent (maternal) and the other from male parent (paternal). These two chromosomes constitute the homologous pair.
2. Chromosomes retain their structural uniqueness and individuality throughout the life cycle of an organism.
3. Each chromosome carries specific determiners or Mendelian factors which are now termed as genes.

Question 35.
Compare Mendelian factors with chromosome
Answer:
Mendelian factors:

1. Alleles of a factor occur in pair.
2. Similar or dissimilar alleles of a factor separate during the gamete formation.
3. Mendelian factors can assort independently.

Chromosomes behaviour:

1. Chromosomes occur in pairs.
2. The homologous chromosomes separate during meiosis.
3.  The paired chromosomes can separate independently during meiosis but the linked genes in the same chromosome normally do not assort independently.

Question 36.
State Coupling and Repulsion theory.
Answer:
The two dominant alleles or recessive alleles occur in the same homologous chromosomes, tend to inherit together into same gamete are called coupling or configuration. If dominant or recessive alleles are present on two different, but homologous chromosomes they inherit apart into different gamete are called repulsion or trans configuration.

Question 37.
Give a short note on incomplete linkage.
Answer:
If two linked genes are sufficiently apart, the chances of their separation are possible. As a result, parental and non-parental combinations are observed. The linked genes exhibit some crossing over. This phenomenon is called incomplete linkage. This was observed in maize. It was reported by Hutchinson.

Question 38.
How crossing over differs from linkage?
Answer:
Linkage:

1. The genes present on chromosome stay close together.
2. It involves same chromosome of homologous chromosome.
3. It reduces new gene combinations.

Crossing over:

1. It leads to separation of linked genes.
2. It involves exchange of segments between non-sister chromatids of homologous chromosome.
3. It increases variability by forming new gene combinations. Leading to formation of new organism.

Question 39.
What is Synapsis? Explain its types.
Answer:
Pairing of homologous chromosomes during zygotene stage of prophase I of meiosis I is called synapsis. Synapsis is of three types:

1. Procentric synapsis: Pairing starts from middle of the chromosome.
2. Proterminal synapsis: Pairing starts from the telomeres.
3. Random synapsis: Pairing may start from anywhere.

Question 40.
How and where chiasma is formed?
Answer:
After tetrad formation, crossing over occurs in pachytene stage. The non-sister chromatids of homologous pair make a contact at one or more points. These points of contact between non sister chromatids of homologous chromosomes are called Chiasmata (singular-Chiasma). At chiasma, cross-shaped or X-shaped structures are formed, where breaking and rejoining of two chromatids occur. This results in reciprocal exchange of equal and corresponding segments between them.

Question 41.
Classify cross over.
Answer:

1. Single cross over: Formation of single chiasma and involves only two chromatids out of four.
2. Double cross over: Formation of two chiasmata and involves two or three or all four strands.
3. Multiple cross over: Formation of more than two chiasmata and crossing over frequency is extremely low.

Question 42.
What is recombination? Which is the widely accepted model of DNA recombination?
Answer:
Crossing over results in the formation of new combination of characters in an organism called recombinants. In this, segments of DNA are broken and recombined to produce new combinations of alleles. This process is called recombination.
The widely accepted model of DNA recombination during crossing over is Holliday’s hybrid DNA model.

Question 43.
Which type of test cross provides the data to construct an efficient genetic map? Explain.
Answer:
A more efficient mapping technique is to construct based on the results of three-point test cross. It refers to analyzing the inheritance patterns of three alleles by test crossing a triple recessive heterozygote with a triple recessive homozygote. It enables to determine the distance between the three alleles and the order in which they are located on the chromosome. Double cross overs can be detected which will provide more accurate map distances.

Question 44.
Enumerate the uses of Genetic mapping.
Answer:

1. It is used to determine gene order, identify the locus of a gene and calculate the distances between genes.
2. They are useful in predicting results of dihybrid and trihybrid crosses.
3. It allows the geneticists to understand the overall genetic complexity of particular organism.

Question 45.
List any three characteristic features of multiple allele.
Answer:

1. Multiple alleles of a series always occupy the same locus in the homologous chromosome. Therefore, no crossing over occurs within the alleles of a series.
2. Multiple alleles are always responsible for the same character.
3. The wild type alleles of a series exhibit dominant character whereas mutant type will influence dominance or an intermediate phenotypic effect.

Question 46.
Explain the sex determination mechanism in Carica papaya.
Answer:
Carica papaya, 2n=36 (Papaya) has 17 pairs of autosomes and one pair of sex chromosomes. Male papaya plants have XY and female plants have XX. Unlike human sex chromosomes, papaya sex chromosomes look like autosomes and it is evolved from autosome. The sex chromosomes are functionally distinct because the Y chromosome carries the genes for male organ development and X bears the female organ developmental genes. In papaya sex determination is controlled by three alleles. They are m, Mj and M2 of a single gene.

Sex chromosome of papaya

Question 47.
Differentiate Mis-sense mutation from Non-sense mutation.
Answer:
Mis-sense Mutation:
The mutation where the codon for one amino acid is changed into a codon for another amino acid is called Mis-sense or non-synonymous mutations.

Non-sense Mutation:
The mutations where codon for one amino acid is changed into a termination or stop codon is called Non-sense mutation.

Question 48.
What is frameshift mutation?
Answer:
Mutations that result in the addition or deletion of a single base pair of DNA that changes the reading frame for the translation process as a result of which there is complete loss of normal protein structure and function are called Frameshift mutations.

Question 49.
How temperature induces mutation?
Answer:
Increase in temperature increases the rate of mutation. While rise in temperature, breaks the hydrogen bonds between two DNA nucleotides which affects the process of replication and transcription.

Question 50.
Explain co-mutagens with examples.
Answer:
The compounds which are not having own mutagenic properties but can enhance the effects of known mutagens are called co-mutagens.
Example: Ascorbic acid increase the damage caused by hydrogen peroxide. Caffeine increase the toxicity of methotrexate.

Question 51.
Name the following chromosomal conditions.
(a) 2n + 2 + 2
(b) 2n – 1 – 1
(c) x
(e) 2n + n + n
(f) 2n + 1
Answer:
(a) Double tetrasomy
(b) Double monosomy
(c) Monoploidy
(e) Polyploidy
(f) Trisomy

Question 52.
Give an account on colchicine.
Answer:
Colchicine, an alkaloid is extracted from root and conns of Colchicum autumnale, when applied in low concentration to the growing tips of the plants it will induce polyploidy. Surprisingly it does not affect the source plant Colchicum, due to presence of anticolchicine.

Question 53.
Explain the three types of duplication with diagram.
Answer:

1. Tandem duplication: The duplicated segment is located immediately after the normal segment of the chromosome in the same order.
2. Reverse tandem duplication The duplicated segment is located immediately after the normal segment but the gene sequence order will be reversed.
3. Displaced duplication: The duplicated segment is located in the same chromosome, but away from the normal segment.

5 – Mark Questions

Question 54.
Why crossing over is important?
Answer:

1. Exchange of segments leads to new gene combinations which plays an important role in evolution.
2. Studies of crossing over reveal that genes are arranged linearly on the chromosomes.
3. Genetic maps are made based on the frequency of crossing over.
4. Crossing over helps to understand the nature and mechanism of gene action.
5. If a useful new combination is formed it can be used in plant breeding.

Question 55.
Draw a flow chart depicting the various types of ploidy.
Answer:

Question 56.
Explain hyperploidy with its types.
Answer:
Hyperploidy
Addition of one or more chromosomes to diploid sets are called hyperploidy. Diploid set of chromosomes represented as Disomy. Hyperploidy can be divided into three types. They are as follows:

(a) Trisomy: Addition of single chromosome to diploid set is called Simple trisomy (2n+l). Trisomics were first reported by Blackeslee (1910) in Datura stramonium (Jimson weed). But later it was reported in Nicotiana, Pisum and Oenothera. Sometimes addition of two individual chromosome from different chromosomal pairs to normal diploid sets are called Double trisomy (2n+l+l).

(b) Tetrasomy: Addition of a pair or two individual pairs of chromosomes to diploid set is called tetrasomy (2n+2) and Double tetrasomy (2n+2+2) respectively. All possible tetrasomics are available in Wheat.

(c) Pentasomy: Addition of three individual chromosome from different chromosomal pairs to normal diploid set are called pentasomy (2n+3).

Question 57.
List out the significance of ploidy.
Answer:

1. Many polyploids are more vigorous and more adaptable than diploids.
2. Many ornamental plants are autotetraploids and have larger flower and longer flowering duration than diploids.
3. Autopolyploids usually have increase in fresh weight due to more water content.
4. Aneuploids are useful to determine the phenotypic effects of loss or gain of different chromosomes.

Question 58.
Explain the Translocation type of chromosomal aberration. Translocation
Answer:
The transfer of a segment of chromosome to a non- homologous chromosome is called translocation. Translocation should not be confused with crossing over, in which an exchange of genetic material between homologous chromosome takes place. Translocation occurs as a result of interchange of chromosome segments in non-homologous cliromosomes.
There are three types:

1. Simple translocation
2. Shift translocation
3. Reciprocal translocation

i. Simple translocation : A single break is made in only one chromosome. The broken segment gets attached to one end of a non-homologous chromosome. It occurs very rarely in nature.

ii. Shift translocation : Broken segment of one chromosome gets inserted interstitially in a non-homologous chromosome.

iii. Reciprocal translocations : It involves mutual exchange of chromosomal segments between two non-homologous chromosomes. It is also called illegitimate crossing over. It is further divided into two types.

a. Homozygous translocation: Both the chromosomes of two pairs are involved in translocation. Two homologous of each translocated chromosomes are identical.

b. Heterozygous translocation: Only one of the chromosome from each pair of two homologous are involved in translocation, while the remaining chromosome is normal. Translocations play a major role in the formation of species.

Higher Order Thinking (HOTs) Questions

Question 1.
Given below is a sequence of alphabets representing the genes of chromosome. Observe it and answer the questions.
A-B-C-D-E-F-G-H-I-J-K.
(a) Write the sequence of genes after the chromosome undergoes terminal deletion of single gene.
(b) What will be the gene sequence, if the genes E and F undergoes tandem duplication?
(c) Consider the centromere is located between the genes F and G and write a gene sequence, after paracentric inversion occurs in between the genes C, D and E.
Answer:
(a) B – C – D – E – F – G – H -I – J – K (or) A- B – C – D – E – F – G – H -I – J.
(b) A – B – C – D – E – F – E – F – G – H – I – J – K.
(c) A – B – E – D – C – F – G – H -1 – J – K.

Question 2.
In Drosophila melanogaster, there are four pairs of chromosomes. If there occurs chromosomal aberrations resulting in trisomic and monosomic condition, what will be the chromosomal count? Write the correct chromosomal count against respective chromosomal aberration.
Answer:
Normal chromosome of Drosophila melanogaster (2n) = 8
Frisomic condition (2n+1) = 9
Monosomic condition (2n – 1) = 7

Question 3.
Study the figures given below and answer the questions.
Answer:

Which type of cross produces higher recombinant percentage? Give reason in support of your answer.
(a) Cross B produces more recombinants.
(b) The gene e and f are located sufficiently apart leading to crossing over resulting in recombinant varieties.

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Samacheer Kalvi 12th Bio Botany Ecosystem Text Book Back Questions and Answers

Question 1.
Which of the following is not a abiotic component of the ecosystem?
(a) Bacteria
(b) Humus
(c) Organic compounds
(d) Inorganic compounds
Answer:
(b) Humus

Question 2.
Which of the following is / are not a natural ecosystem?
(a) Forest ecosystem
(b) Rice field
(c) Grassland ecosystem
(d) Desert ecosystem Pond is a type
Answer:
(b) Rice field

Question 3.
Pond is a type of __________
(a) Forest ecosystem
(b) grassland ecosystem
(c) marine ecosystem
(d) fresh water ecosystem Pond ecosystem
Answer:
(d) fresh water ecosystem Pond ecosystem

Question 4.
Pond ecosystem is __________
(a) not self sufficient and self regulating
(b) partially self sufficient and self regulating
(c) self sufficient and not self regulating
(d) self sufficient and self regulating
Answer:
(d) self sufficient and self regulating

Question 5.
Profundal zone is predominated by heterotrophs in a pond ecosystem, because of __________
(a) with effective light penetration
(b) no effective light penetration
(c) complete absence of light
(d) a and b
Answer:
(d) a and b

Question 6.
Solar energy used by green plants for photosynthesis is only __________
(a) 2 -8%
(b) 2-10%
(c) 3-10%
(d) 2-9%
Answer:
(b) 2-10%

Question 7.
Which of the following ecosystem has the highest primary productivity?
(a) Pond ecosystem
(b) Lake ecosystem
(c) Grassland ecosystem
(d) Forest ecosystem
Answer:
(c) Grassland ecosystem

Question 8.
Ecosystem consists of __________
(a) decomposers
(b) producers
(c) consumers
(d) all of the above
Answer:
(d) all of the above

Question 9.
Which one is in descending order of a food chain?
(a) Producers → Secondary consumers → Primary consumers → Tertiary consumers
(b) Tertiary consumers → Primary consumers → Secondary consumers → Producers
(c) Tertiary consumers → Secondary consumers → Primary consumers → Producers
(d) Tertiary consumers → Producers → Primary consumers → Secondary consumers Significance of food web is / are
Answer:
(c) Tertiary consumers → Secondary consumers → Primary consumers → Producers

Question 10.
Significance of food web is / are __________
(a) it does not maintain stability in nature
(b) it shows patterns of energy transfer
(c) it explains species interaction
(d) b and c
Answer:
(d) b and c

Question 11.
The following diagram represents __________

(a) pyramid of number in a grassland ecosystem
(b) pyramid of number in a pond ecosystem
(c) pyramid of number in a forest ecosystem
(d) pyramid of biomass in a pond ecosystem
Answer:
(c) pyramid of number in a forest ecosystem

Question 12.
Which of the following is / are not the mechanism of decomposition?
(a) Eluviation
(b) Catabolism
(c) Anabolism
(d) Fragmentation
Answer:
(c) Anabolism

Question 13.
Which of the following is not a sedimentary cycle?
(a) Nitrogen cycle
(b) Phosphorous cycle
(c) Sulphur cycle
(d) Calcium cycle
Answer:
(a) Nitrogen cycle

Question 14.
Which of the following are not regulating services of ecosystem services?
(i) Genetic resources
(ii) Recreation and aesthetic values
(iii) Invasion resistance
(iv) Climatic regulation
(a) i and iii
(b) ii and iv
(c) i and ii
(d) i and iv
Answer:
(c) i and ii

Question 15.
Productivity of profundal zone will be low. Why?
Answer:
The producers of the pond ecosystem depends on phytoplankton through photosynthesis. Profundal zone lies below the limnetic zone with no effective light penetration, hence productivity rate is very low.

Question 16.
Discuss the gross primary productivity is more efficient than net primary productivity.
Answer:
Gross primary productivity:

1. It refers to the total amount of food energy or organic matter produced in an ecosystem by autotrophs.
2. GPP = NPP + Respiration

Net primary productivity:

1. It refers to the amount of energy that remain in autotrophs after respiration loss.
2. NPP = GPP – Respiration

Question 17.
Pyramid of energy is always upright. Give reasons.
Answer:
The energy pyramid represents a successive energy flow at each trophic level in an ecosystem. There is a gradual decrease in energy transfer at successive tropic levels from producers to higher levels, hence the pyramid of energy is always upright.

Question 18.
What will happen if all producers are removed from ecosystem?
Answer:
Producers are the autotrophs which occupy the first tropic level in an ecosystem. The energy produced by them is utilized by the herbivores and then by carnivores, thereby maintaining the stability of ecosystem. If producers are removed from an ecosystem, it would lead to starvation and death of herbivores and subsequently the carnivores, thus terminating the entire food web.

Question 19.
Construct the food chain with the following data.
Hawk, plants, frog, snake, grasshopper.
Answer:
Plants → Grasshopper → Frog → Snake → Hawk

Question 20.
Name of the food chain which is generally present in all type of ecosystem. Explain and write their significance.
Answer:
Detritus food chain is common in all type of ecosystem. In detritus food chain, the dead remains of plant and animals or their excreta are broken down by detrivores and the organic and inorganic substances are returned back to environment. Thus maintaining die company of various biogeochemical cycles. Also Microbes growing on detritus makes the soil nutritious for consumers.

Question 21.
Shape of pyramid in a particular ecosystem is always different in shape. Explain with example.
Answer:
In a forest ecosystem the pyramid of number is spindle in shape, it is because the base (T₁) of the pyramid occupies large sized trees (Producer) which are lesser in number. Herbivores (T₂) (Fruit eating birds, elephant and deer) occupying second trophic level, are more in number than the producers. In final trophic level (T₄), .tertiary consumers (lion) are lesser in number than the secondary consumer (T₃) (fox and snake).

Question 22.
Generally human activities are against to the ecosystem, where as you a student how will you help to protect ecosystem?
Answer:

1. Buying and using only ecoffiendly products and recycle them.
2. Growing more trees.
3. Choosing sustained farm products (vegetables, fruits and greens, etc.)
4. Reducing the use of natural resources.
5. Recycling the waste and reduce the amount of waste you produce.
6. Reducing consumption of water and electricity.
7. Reducing or eliminating the use of house-hold chemicals and pesticides.
8. Maintaining your cars and vehicles properly to reduce carbon emission.
9. Creating awareness and educate about ecosystem protection among your friends and family members and ask them to find out solution to minimise this problem.

Question 23.
Generally in summer the forest are affected by natural fire. Over a period of time it recovers itself by the process of successions. Find out the types of succession and explain. Secondary succession.
Answer:
The development of a plant community in an area where an already developed community has been destroyed by natural causes is known as secondary succession. This type of succession takes less time to occur.

Question 24.
Draw a pyramid from following details and explain in brief.
Answer:
Quantities of organisms are given-Hawks-50, plants-1000, rabbit and mouse-250 +250, pythons and lizard – 100 + 50 respectively

T₄ (50) T₃ (100) T₂ (550) T₁ (1000)
The pyramid produced is an upright pyramid of numbers where there is a gradual decrease in number of organisms at each trophic level from T₁ to T₄. This is an example for grassland ecosystem.

Question 25.
Various stages of succession are given bellow. From that rearrange them accordingly. Find out the type of succession and explain in detail.
Answer:
Reed-swamp stage, phytoplankton stage, shrub stage, submerged plant stage, forest stage, submergedfree floating stage and marsh medow stage.

(1) Phytoplankton stage – It is the first stage of succession consisting of the pioneer community like blue green algae, green algae, diatoms, bacteria, etc., The colonization of these organisms enrich the amount of organic matter and nutrients of pond due to their life activities and death. This favors the development of the next serai stages.

(2) Submerged plant stage – As the result of death and decomposition of planktons, silt brought from land by rain water, lead to a loose mud formation at the bottom of the pond. Hence, the rooted submerged hydrophytes begin to appear on the new substratum.
Example: Vallisneria and Hydrilla etc. The death and decay of these plants will build up the substratum of pond to become shallow.

(3) Submerged free floating stage – During this stage, the depth of the pond will become almost 2-5 feet Hence, the rooted hydrophytic plants and with floating large leaves start colonising the pond.
Example: Rooted floating plants like Nelumbo, Nymphaea and some free floating species like Azolla, and Pistia are also present in this stage. By death and decomposition of these plants, further the pond becomes more shallow.

(4) Reed-swamp stage – It is also called an amphibious stage. During this stage, rooted floating plants are replaced by plants which can live successfully in aquatic as well as aerial environment.
Example: Typha, Phragmites, Sagittaria and Scirpus etc. At the end of this stage, water level is very much reduced, making it unsuitable for the continuous growth of amphibious plants.

(5) Marsh meadow stage – When the pond becomes swallowed due to decreasing water level, species of Cyperaceae and Poaceae colonise the area. They form a mat-like vegetation with the help of their much branched root system. This leads to an absorption and loss of large quantity of water. At the end of this stage, the soil becomes dry and the marshy vegetation disappears gradually and leads to shurb stage.

(6) Shrub stage – Here areas are invaded by terrestrial plants like shrubs (Salix and Comus) and trees (Populus and Alnus). These plants absorb large quantity of water and make the habitat dry. Further, the accumulation of humus with a rich flora of microorganisms produce minerals in the soil, ultimately favouring the arrival of new tree species in the area.

(7) Forest stage – It is the climax community of hydrosere. A variety of trees invade the area and develop any one of the diverse type of vegetation.
Example.Temperate mixed forest (Ulmus,Acer and Quercus), Tropical rain forest (Artocaipus and Cinnamomum ) and Tropical deciduous forest (Bamboo and Tectona).

Samacheer Kalvi 12th Bio Botany Ecosystem Additional Questions and Answers

1 – Mark Question

Question 1.
Ecosystem is the structural and functional unit of ecology. This statement was given by ___________
(a) Tansley
(b) Odum
(c) Charles Elton
(d) Edwin
Answer:
(b) Odum

Question 2.
Identify the incorrect option among the following component sequence.
(a) air, water, sunlight and temperature
(b) latitude, altitude, direction of mountain and aptitude
(c) soil air, pH of soil, saltwater and soil moisture
(d) carbohydrate, protein, lipids and humic substances
Answer:
(b) latitude, altitude, direction of mountain and aptitude

Question 3.
Pick out the edaphic factor among the following.
(a) Rainfall
(b) Temperature
(c) Soil pH
(d) Latitude
Answer:
(c) Soil pH

Question 4.

Answer:
a -ii, b – iii, c – iv, d-i

Question 5.
Which is not a macro consumer?
(a) Herbivore
(b) Carnivore
(c) Ominivore
(d) Decomposer
Answer:
(d) Decomposer

Question 6.
Photosynthetically Active Radiation ranges between the wavelength of.
(a) 400 – 600 nm
(b) 600 – 700 nm
(c) 400 – 500 nm
(d) 400 – 700 nm
Answer:
(d) 400 – 700 nm

Question 7.
Who coined the term Ecosystem?
Answer:
A.G. Tansley

Question 8.
Identify the incorrect statement.
(a) Carbon stored in oil is referred as Grey carbon
(b) Carbon stored in atmosphere is referred as Blue carbon
(c) Carbon stored in industrialized forests is referred as Green carbon
(d) Carbon emitted from gas, died engine is referred as Black carbon
Answer:
(c) Carbon stored in industrialized forests is referred as Green carbon

Question 9.
Which group of organism occupies the third tropic level in an ecosystem?
(a) Primary consumers
(b) Secondary consumers
(c) Secondary carnivores
(d) Omnivores
Answer:
(b) Secondary consumers

Question 10.
Which is irrelevant to the first law of thermodynamics?
(a) Energy can be transmitted from one system to other in many forms.
(b) Energy transformation results in reduction of free energy.
(c) Energy can neither be created nor destroyed.
(d) Energy in the universe is constant.
Answer:
(b) Energy transformation results in reduction of free energy.

Question 11.
If 1200 Joules of solar energy is trapped by producers, how much of Joules of energy does the organism in the third trophic level will receive?
(a) 120 Joules
(b) 12 Joules
(c) 1.2 Joules
(d) 0.12 Joules
Answer:
(c) 1.2 Joules

Question 12.
Which of the following food chain is in improper sequence?
(a) Plants → snake → rabbit → lizard → eagle
(b) Plants → grasshopper → lizard → snake → hawk
(c) Plants → lizard → rabbit → snake → eagle
(d) Plants → rabbit → lizard → hawk → eagle
Answer:
(b) Plants → grasshopper → lizard → snake → hawk

Question 13.
Which one of the following is not a functional unit of an ecosystem?
(a) Productivity
(b) Conductivity
(c) Energy flow
(d) Decomposition
Answer:
(b) Conductivity

Question 14.
The upright pyramid is not a feature of. __________
(a) Pond ecosystem
(b) Grassland ecosystem
(c) Forest ecosystem
(d) Terminal ecosystem
Answer:
(a) Pond ecosystem

Question 15.
The type of ecosystem with maximum net primary productivity is __________
(a) Desert ecosystem
(b) Deciduous forest ecosystem
(c) Tropical rain forest ecosystem
(d) Grassland ecosystem
Answer:
(c) Tropical rain forest ecosystem

Question 16.
Pyramid of numbers with broad base indicates __________
(a) High population of old individuals
(b) Low population of young individuals
(c) High population of young individuals
(d) Low population of old individuals
Answer:
(b) Low population of young individuals

Question 17.
Spindle shaped pyramid is a character of __________
(a) Pond ecosystem
(b) Grassland ecosystem
(c) Parasite ecosystem
(d) Forest ecosystem
Answer:
(d) Forest ecosystem

Question 18.
Read the statement and select the correct terminology for the same:
“Carrying away of inorganic compounds of soil by water”.
(a) Eluviation
(b) Fragmentation
(c) Humification
(d) Mineralisation
Answer:
(a) Eluviation

Question 19.
Complete the food chain by filling the link X:
Paddy → Grassopper → Frog → X → Hawk
(a) King cobra
(b) Gorilla
(c) Rabbit
(d) Tasmanial wolf
Answer:
(a) King cobra

Question 20.
Which of the following is abundant in rock deposits and guano?
(a) Nitrogen
(b) Phosphorous
(c) Oxygen
(d) Calcium
Answer:
(b) Phosphorous

Question 21.
The bottom most zone of a pond is termed as
(a) Limnetic zone
(b) Littoral zone
(c) Benthic zone
(d) Profundal zone
Answer:
(c) Benthic zone

Question 22.
Observe the figures and select the correct type of pyramid of numbers

Answer:
(a) (A) Grassland ecosystem (B) Forest Echosystem (c) Parasite ecosystem

Question 23.
Lotic ecosystem refers to ___________
(a) Open water ecosystem
(b)Running water ecosystem
(c) Standing water
(d) Ocean water ecosystem
Answer:
(b)Running water ecosystem

Question 24.
Identify the correct sequence of various zones from surface to depth in a pond ecosystem.
(a) Profundal, limnetic, littoral and benthic
(b) Benthic, littoral, profundal and limnetic
(c) Limnetic, profundal, littoral and benthic
(d) Littoral, limnetic, profundal and benthic
Answer:
(d) Littoral, limnetic, profundal and benthic

Question 25.
Which type of ecosystem service does the genetic resources comes under?
(a) Provisioning services
(b) Supporting services
(c) Regulating services
(d) Cultural services
Answer:
(a) Provisioning services

Question 26.
Assertion (A): Pyramid of energy is upright.
Reason (R): During the energy transfer at successive trophic levels from producers there will be a gradual decrease
(a) Both A and R are wrong
(b) A is right R is wrong
(c) R explains A
(d) A is right R is not the correct explanation for A
Answer:
(c) R explains A

Question 27.
Assertion (A): In forest ecosystem, the pyramid of number is spindle shaped.
Reason (R): Tropical level (T₁) of the pyramid occupies large trees which are maxium in number
(a) Both A and R are wrong
(b) A is right R is wrong
(c) R explains A
(d) A is right R is not the correct explanation for A
Answer:
(b) A is right R is wrong

Question 28.
Species that indicate the health of the ecosystem are called as __________
Answer:
Flagship species

Question 29.
Succession initiating on a sand referred as
(a) Hydrosere
(b) Psammosere
(c) Halosere
(d) Lithosere
Answer:
(b) Psammosere

Question 30.
Match the column I with column II

Answer:
(a) a – iii, b – iv, c – i, d – ii

Question 31.
Statement (I): Allogenic succession occurs as a result of abiotic factors.
Statement (II): Autogenic succession occurs as result of biotic factors.
(a) Statement I is correct; Statement II is incorrect.
(b) Statement I is incorrect; Statement II is correct.
(c) Both Statements I and II are correct.
(d) Both Statements I and II are incorrect.
Answer:
(c) Both Statements I and II are correct.

Question 32.
Statement (I): The first invaded plants in a barren area are called as pioneers.
Statement (II): Marsh meadow stage of hydrosere succession is also called as amphibious stage.
(a) Statement I is correct; Statement II is incorrect.
(b) Statement I is incorrect; Statement II is correct.
(c) Both Statements I and II are correct.
(d) Both Statements I and II are incorrect.
Answer:
(a) Statement I is correct; Statement II is incorrect.

Question 33.
_______ is the climax community of hydrosere.
(a) Reed swamp stage
(b) Marsh medow stage
(c) Shrub stage
(d) Forest stage
Answer:
(d) Forest stage

2-Mark Questions 

Question 1.
According to A.G. Tansley, what is an ecosystem?
Answer:
A.G. Tansley (1935), who defined ecosystem as ‘the system resulting from the integration of all the living and non-living factors of the environment.

Question 2.
Mention any two climatic factors and edaphic factors of an ecosystem.
Answer:

1. Climatic factors: Light and Air.
2. Edaphic factors: Soilwaer and pH of soil.

Question 3.
Define ‘Standing state’ with regard to ecosystem.
Answer:
The total inorganic substances present in any ecosystem at a given time is called standing quality (or) standing state.

Question 4.
What are biotic components?
Answer:
Biotic (living) components includes all living organisms like plants, animals, fungi and bacteria. They form the trophic structures of any ecosystem.

Question 5.
Name the macro consumers and micro consumers.
Answer:

1. Macro consumers are herbivores, carnivores and omnivores.
2. Micro consumers are decomposers.

Question 6.
How will you define decomposers?
Answer:
Decomposers are organisms that decompose the dead plants and animals to release organic and inorganic nutrients into the environment which are again reused by plants.
Example: Bacteria, Actinomycetes and Fungi.

Question 7.
What is standing crop?
Answer:
The amount of living materials present in a population at any given time is known as standing crop, which may be expressed in terms of number or biomass per unit area.

Question 8.
What do you mean by PAR? Mention its significance.
Answer:
The amount of light available for photosynthesis of plants is called Photosynthetically Active Radiation (PAR) which is between the range of 400-700 mm wave length.

Question 9.
Pointout the factors that affects the photosynthetically active radiation.
Answer:
PAR is not always constant because of clouds, tree shades, air, dust particles, seasons, latitudes and length of the daylight availability.

Question 10.
Define Grey carbon and Black carbon.
Answer:
Grey carbon – carbon stored in fossil fuel (coal, oil and biogas deposits in the lithosphere). Black carbon – carbon emitted from gas, diesel engine and coal fired power plants.

Question 11.
What is meant by productivity of an ecosystem?
Answer:
The rate of biomass production per unit area in a unit time is called productivity. It can be expressed in terms of gm /m²/year or Kcal/m²/ year.

Question 12.
How Net Primary Productivity can be derived?
Answer:
Net Primary Productivity (NPP) is derived by the difference between Gross primary productivity (GPP) and respiration.
NPP = GPP – Respiration

Question 13.
Expand GPP and define it.
Answer:
Gross Primary Productivity (GPP) is the total amount of food energy or organic matter or biomass produced in an ecosystem by autotrophs through the process of photosynthesis is called gross primary productivity.

Question 14.
Write the name of four important aspects of ecosystem.
Answer:

1. Productivity
2. Energy flow
3. Decomposition
4. Nutrient cycling