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Tamilnadu Samacheer Kalvi 10th Science Solutions Chapter 18 Heredity

Kickstart your preparation by using this Tamilnadu State Board Solutions for Class 18th Science Chapter 18 Heredity Questions and Answers and get the max score in the exams. You can cover all the topics of Chapter 18 easily after studying the Tamilnadu State Board Class 18th Science Textbook solutions pdf. Download the Tamilnadu State Board Science Chapter 18 Heredity solutions of Class 18th by accessing the links provided here and ace up your preparation.

Samacheer Kalvi 10th Science Heredity Textual Evaluation Solved

I. Choose the Correct Answer.

Question 1.
According to Mendel, alleles have the following character ______.
(a) Pair of genes
(b) Responsible for character
(c) Production of gametes
(d) Recessive factors.
Answer:
(a) Pair of genes

Question 2.
9 : 3 : 3 : 1 ratio is due to:
(a) Segregation
(b) Crossing over
(c) Independent assortment
(d) Recessiveness
Answer:
(c) Independent assortment

Question 3.
The region of the chromosome where the spindle fibres get attached during cell division ______.
(a) Chromomere
(b) Centrosome
(c) Centromere
(d) Chromonema.
Answer:
(c) Centromere

Question 4.
The centromere is found at the centre of the ……… chromosome.
(a) Telocentric
(b) Metacentric
(c) Sub-metacentric
(d) Acrocentric
Answer:
(b) Metacentric

Question 5.
The ______ units form the backbone of the DNA.
(a) 5 carbon sugar
(b) Phosphate
(c) Nitrogenous bases
(d) Sugar phosphate.
Answer:
(c) Nitrogenous bases

Question 6.
Okasaki fragments are joined together by:
(a) Helicase
(b) DNA polymerase
(c) RNA primer
(d) DNA ligase
Answer:
(d) DNA ligase

Question 7.
The number of chromosomes found in human beings are ______.
(a) 22 pairs of autosomes and 1 pair of allosomes.
(b) 22 autosomes and 1 allosome.
(c) 46 autosomes.
(d) 46 pairs of autosomes and 1 pair of allosomes.
Answer:
(a) 22 pairs of autosomes and 1 pair of allosomes.

Question 8.
The loss of one or more chromosome in a ploidy is called:
(a) Tetraploidy
(b) Aneuploidy
(c) Euploidy
(d) polyploidy
Answer:
(b) Aneuploidy

II. Fill in the blanks.

Question 1.
The pairs of contrasting character (traits) of Mendel are called ______.
Answer:
Alleles or allelomorphs.

Question 2.
The physical expression of a gene is called ______.
Answer:
Phenotype.

Question 3.
The thin thread-like structures found in the nucleus of each cell are called ______.
Answer:
Chromosomes.

Question 4.
DNA consists of two _______ chains.
Answer:
Polynucleotide.

Question 5.
An inheritable change in the amount or the structure of a gene or a chromosome is called ______.
Answer:
Mutation.

III. Identify whether the statement is True or False. Correct the False Statement.

Question 1.
A typical Mendelian dihybrid ratio of F2 generation is 3 : 1.
Answer:
False.
Correct statement: A typical Mendelian dihybrid ratio of F2 generation is 9 : 3 : 3 : 1.

Question 2.
A recessive factor is altered by the presence of a dominant factor.
Answer:
True.

Question 3.
Each gamete has only one allele of a gene.
Answer:
True.

Question 4.
hybrid is an offspring from a cross between genetically different parent.
Answer:
True.

Question 5.
Some of the chromosomes have an elongated knob-like appendage known as a telomere.
Answer:
False.
Correct statement: Some of the chromosomes have an elongated knob – like appendage known as the satellite.

Question 6.
New nucleotides are added and a new complementary strand of DNA is formed with the help of enzyme DNA polymerase.
Answer:
True.

Question 7.
Down’s syndrome is a genetic condition with 45 chromosomes.
Answer:
False.
Correct statement: Down’s syndrome is the genetic condition with 21 chromosomes.

IV. Match the following.

Question 1.

Column A Column B
1. Autosomes (a) Trisomy 21
2. Diploid condition (b) 9 : 3 : 3 : 1
3. Allosome (c) 22 pair of chromosome
4. Down’s syndrome (d) 2n
5. Dihybrid ratio (e) 23rd pair of chromosome

Answer:
1. (c) 22 pair of chromosome
2. (d) 2n
3. (e) 23rd pair of chromosome
4. (a) Trisomy 21
5. (b) 9 : 3 : 3 : 1.

V. Answer in a Sentence.

Question 1.
What is a cross in which inheritance of two pairs of contrasting characters is studied?
Answer:
Dihybrid cross.

Question 2.
Name the conditions when both the alleles are identical.
Answer:
Homozygous

Question 3.
A garden pea plant produces axial white flowers. Another of the same species produced terminal violet flowers. Identify the dominant trait?
Answer:

  • Axial position – a dominant trait
  • White flowers – a recessive trait
  • Terminal position – a recessive trait
  • Violet flower – dominant trait.

Question 4.
What is the name given to the segments of DNA, which are responsible for the inheritance of a particular character?
Answer:
Genes

Question 5.
Name the bond which binds the nucleotides in a DNA.
Answer:
The hydrogen bonds bind the nucleotides in a DNA.

VI. Short Answer Questions.

Question 1.
Why did Mendel select pea plant for his experiments?
Answer:

  • The pea plant is self-pollinating and so it is very easy to raise pure breeding individuals.
  • It has a short life span, as it is an annual.
  • It is easy to cross-pollinate.
  • It has deeply defined contrasting characters.
  • The flowers are bisexual.

Question 2.
What do you understand by the term phenotype and genotype?
Answer:
Phenotype is the outward appearance or morphological character of an organism. The expression of gene or the genetic make up of an individual for a particular trait is called genotype.

Question 3.
What are allosomes?
Answer:
Out of 23 pairs of chromosomes, 22 pairs are autosomes and the 23rd pair is the allosome or sex chromosome.

Question 4.
What are Okazaki fragments?
Answer:
For the synthesis of new DNA, two things are required one is RNA primer and the enzyme primase. The DNA polymerase moves along the newly formed RNA primer nucleotides, which leads to the elongation of DNA. In the other strand, DNA is synthesis in small fragments called okazaki fragments. These fragments are linked by the enzyme called ligase.

Question 5.
Why is euploidy considered to be advantageous to both plants and animals?
Answer:
In euploidy condition, the individual bears more than the usual number of diploid chromosomes. The triploid plants and animals are sterile. The tetraploid plants, often result in increased fruit and flower size.

Question 6.
A pure tall plant (TT) is crossed with pure dwarf plant (tt), what would be the F1 and F2 generations? Explain.
Answer:
When a pure tall plant (TT) is crossed with pure dwarf plant (tt), In F1 generation all the plants will be tall. In F2 generation by selling the F1 monohybrid the tall and dwarf plant will be in the ratio of 3 : 1. In F2 generation phenotypic ratio will be 3 : 1. genotypic ratio will be 1 : 2 : 1

Question 7.
Explain the structure of a chromosome.
Answer:
The chromosomes are thin long and thread-like structures, consisting of two identical strands called sister chromatids. They are held together by a centromere. Each chromatid is made up of a spirally coiled thin structure called chromonema. The bead-like structures along the length are called chromomeres. The chromosomes are made up of DNA, RNA, chromosomal proteins (histones and non-histones) and certain metallic ions. These proteins give structural support to the chromosome.

Question 8.
Label the parts of the DNA in the diagram given below. Explain the structure briefly. Structure of DNA.
Answer:
DNA is the hereditary material as it contains the genetic information. DNA is a large molecule, consisting of millions of nucleotides. Each nucleotide consists of three compounds.
Samacheer Kalvi 10th Science Solutions Chapter 18 Heredity 1
(a) A sugar molecule – Deoxy Ribose sugar

(b) A nitrogenous base [Purines and Pyrimidines]

  • Purines (Adenine and Guanine)
  • Pyrimidines (Cytosine and Thymine)

(c) A phosphate group Nucleoside and Nucleotide:
Nucleoside = Nitrogen base + sugar
Nucleotide = Nucleoside + Phosphate
The nucleotides are formed according to the purines and pyrimidines present in them.

VII. Long Answer Questions.

Question 1.
Explain with an example of the inheritance of the dihybrid cross. How is it different from a monohybrid cross?
Answer:
The dihybrid cross involves the inheritance of two pairs of contrast characteristics, round – yellow seeds and wrinkled – green seeds. When pea plants having round – yellow seeds cross – bred with pea plants having wrinkled – green seeds, in the first generation (F1), only round yellow seeds were produced.
No wrinkled – green seeds were obtained. Round yellow colour seeds were dominant and wrinkled-green seeds were recessive.

When round – yellow seeds were cross-bred by self-pollination, four types of seeds having different combinations of shape and colour were obtained in the F2 generation. They were round- yellow, round-green, wrinkled-yellow and wrinkled – green seeds.
Samacheer Kalvi 10th Science Solutions Chapter 18 Heredity 2
A dihybrid cross produced four types of F2 offsprings in the ratio of 9 with two dominant traits, 3 with one dominant trait and one recessive trait, 3 with another dominant trait and another recessive trait and one with two recessive traits. The new combinations of traits with round green and wrinkled yellow had appeared in the dihybrid cross (F2 generation). The ratio of each phenotype of seeds in the F2 generation is 9 : 3 : 3 : 1. This is known as the Dihybrid ratio.

Difference between a monohybrid cross and dihybrid cross:
Monohybrid cross:
Monohybrid cross is a genetic cross, that involves a single pair of genes, which is responsible for one trait.
Parents differ by a single trait.
Monohybrid ratio in F2 generation is 3 : 1.

Dihybrid cross:
Dihybrid cross is a genetic cross, that involves two pairs of genes, which are responsible for two traits,
The parents have two different independent traits.
The dihybrid ratio in the F2 generation is 9 : 3 : 3 : 1.

Question 2.
How is the structure of DNA organised? What is the biological significance of DNA?
Answer:
DNA is the hereditary material, as it contains the genetic information. It is a large molecule consisting of millions of nucleotides, so it is called a polynucleotide. Each nucleotide consists of three components.
(a) A sugar molecule – Deoxyribose sugar

(b) A nitrogenous base – There are two types of the nitrogenous base in DNA they are

  • Purines (Adenine and Guanine)
  • Pyrimidines (Cytosine and Thymine)

(c) A phosphate group – The polynucleotide chains from a double helix. Nitrogenous bases in the centre are linked to sugar – phosphate units, which form the backbone of the DNA. Pairing between the nitrogenous bases is very specific and is always between purine and pyrimidine, linked by hydrogen bonds.

Adenine (A) links Thymine (T) with two hydrogen bonds [A=T]. Cytosine (C) links Guanine (G) with three hydrogen bonds (C = G). The hydrogen bonds between the nitrogenous bases make the DNA molecule stable. The nucleotides in a helix are joined together by phosphodiester bonds.
Samacheer Kalvi 10th Science Solutions Chapter 18 Heredity 3
The biological significance of DNA:

  • It is responsible for the transmission of heredity information from one generation to the next generation.
  • It contains the information required for the formation of proteins.
  • It controls the developmental process and life activities of an organism.

Question 3.
The sex of the newborn child is a matter of chance and neither of the parents may be considered responsible for it. What would be the possible fusion of gametes to determine the sex of the child?
Answer:
Out of 23 pairs of chromosomes, 22 pairs are autosomes and one pair (23rd pair) is the sex chromosome. The female gametes or the eggs formed are similar in their chromosome type [22 + XX], So human females are homogametic. The male gametes or sperms produced are of two types. They are produced in equal proportions. The sperm bearing [22 + X] chromosomes and the sperm bearing (22 + Y) chromosomes. So human males are called heterogametic.

It is a chance, as to which category of sperm fuses with the egg. If the egg [X] is fused by the X – bearing sperm an [XX] individual (female) is produced. If the egg [X] is fused by the Y – bearing sperm an [XY] individual (male) is produced. The sperm produced by the father determines the sex of the child. The mother is not responsible for determining the sex of the child.
Samacheer Kalvi 10th Science Solutions Chapter 18 Heredity 4

VIII. Higher Order Thinking Skills (HOTS) Questions

Question 1.
Flowers of the garden pea are bisexual and self-pollinated. Therefore, it is difficult to perform hybridization experiment by crossing a particular pistil with the specific pollen grains. How Mendel made it possible in his monohybrid and dihybrid crosses?
Answer:
As the garden pea is self-pollinating plant the parent plant were emasculated to prevent self pollination.
The anthers were collected from male parent and dusted on the female parent and the stigma was bagged.

Question 2.
Pure-bred tall pea plants are first crossed with pure-bred dwarf pea plants. The pea – plants obtained in the F1 generation are then cross-bred to produce F2 generation of pea plants.

  1. What do the plants of the F1 generation look like?
  2. What is the ratio of tall plants to dwarf plants in the F2 generation?
  3. Which type of plants was missing in F1, generation but reappeared in the F2 generation?

Answer:

  1. Tall
  2. 1 : 2 : 1
  3. Dwarf plants.

Question 3.
Kavitha gave birth to a female baby. Her family members say that she can give birth to only female babies because of her family history. Is the statement given by her family members true. Justify your answer.
Answer:
No, the statement is not true.
Sex determination is a chance of probability as to which category of sperm fuses with the eggs. If the egg(x) is fused by the x-bearing sperm, then
individual is female. If the egg (x) is fused by the y-bearing sperm then the individual is male. The sperm produced by the father only determines the sex of the child.

IX. Value-Based Questions

Question 1.
Under which conditions does the law of independent assortment hold good and why?
Answer:
Mendel gave this law based on his dihybrid cross experiment. Here the total number of individuals is F2 will be sixteen which occur in a ratio of 9 : 3 : 3 : 1 where two parental classes and two new combination will be produced.

Samacheer Kalvi 10th Science Heredity Additional Questions Solved

I. Fill in the blanks.

Question 1.
The branch of biology that deals with the genes, genetic variation and heredity of living organisms is called ______.
Answer:
Genetics.

Question 2.
The chromatids are held together by ______.
Answer:
Centromere.

Question 3.
______ is the enzyme, which separates the double helix of DNA, above the replication fork.
Answer:
Topoisomerase.

Question 4.
_____ mutation is the changes occurring in the nucleotide sequence of a gene.
Answer:
Gene or point.

Question 5.
The end of the chromosome is called ______.
Answer:
Telomere.

Question 6.
The constriction of the chromosome at any point is called ______.
Answer:
Nuclear Zone.

Question 7.
The nucleotides in a helix are joined together by ______ bonds.
Answer:
Phosphodiester.

II. Match the following:

Question 1.

1. Homozygous (a) polynucleotide
2. Genes (b) human male
3. DNA (c) occurs in pairs and alike
4. Heterozygous (d) segments of DNA
5. Heterogametic (e) bind DNA to the origin of replication site
6. Autosome (f) occurs in pairs and unlike
7. Helicase (g) somatic characters

Answer:

  1. (c) occurs in pairs and alike
  2. (d) segments of DNA
  3. (a) polynucleotide
  4. (f) occurs in pairs and unlike
  5. (b) Human male
  6. (g) somatic characters
  7. (e) bind DNA to the origin of replication site.

III. Choose the correct answer.

Question 1.
The thin long thread-like structures consisting of two identical strands ______.
(a) Hybrid
(b) Chromosomes
(c) Genes
(d) DNA and RNA.
Answer:
(b) Chromosomes

Question 2.
A cross between a tall plant (TT) and short pea plant (tt) resulted in progeny that were all tall plants because:
(a) Tallness is the dominant trait
(b) Shortness in the dominant trait
(c) Tallness in the recessive trait
(d) Height of pea plant is not governed by gene ‘T’ or ‘t’
Answer:
(a) Tallness is the dominant trait

Question 3.
The number, size and shape of chromosomes in the cell nucleus of an organism ______
(a) Karyotype
(b) Heterozygous
(c) Autosome
(d) Nucleotide
Answer:
(a) Karyotype

Question 4.
In human males all the chromosomes are paired perfectly except one. The unpaired chromosomes are:
(a) large chromosome
(b) small chromosome
(c) Y- chromosome
(d) X- chromosome
Answer:
(c) Y- chromosome

Question 5.
The addition or deletion in the number of chromosomes present in a cell is called _______
(a) Homogametic
(b) Polymerase
(c) Ploidy
(d) Nucleoside
Answer:
(c) Ploidy

IV. Write true or false for the statements. Correct the false statement.

Question 1.
The character which expresses itself is called the dominant condition and that which is masked is called recessive condition. Answer:
True

Question 2.
The chromosomes of body cells of an organism are haploid in condition and the single set of chromosomes in gametes are diploid in condition.
Answer:
False
Correct statement: The chromosomes of body cells of an organism are diploid in condition and the single set of chromosomes in gametes are haploid in condition.

Question 3.
The mother is responsible for determining the sex of the child. The sperm produced by the father does not determine the sex of the child.
Answer:
False
Correct statement: The mother is not responsible for determining the sex of the child. The sperm produced by the father determines the sex of the child.

Question 4.
The human female is called Homogametic. The human males are called heterogametic.
Answer:
True

Question 5.
Adenine always links with Guanine with three hydrogen bonds and cytosine always links with thymine with two hydrogen bonds.
Answer:
False
Correct statement: Adenine always links with thymine with two hydrogen bonds. Cytosine always links with Guanine with three hydrogen bonds.

V. Match the following persons with the correct statements.

Question 1.

1. Waldeyar (a) Role of chromosome in heredity
2. James Watson and Crick (b) Father of genetics
3. Hugo de Vries (c) Down’s syndrome
4. Langdon Down (d) Chromosomes
5. T.H Morgan (e) Mutation
6. Gregor Johan Mendel (f) Model of DNA

Answer:

  1. (d) chromosomes
  2. (f) Model of DNA
  3. (e) Mutation
  4. (c) Down’s syndrome
  5. (a) Role of chromosome in heredity
  6. (b) Father of genetics

VI. Answer the following in a word or with a sentence.

Question 1.
Who is the father of genetics?
Answer:
Gregor Johann Mendel.

Question 2.
Name the plant on which Mendel performed his experiment.
Answer:
Pisum sativum (Garden Pea).

Question 3.
What is locus?
Answer:
Each gene is present at a specific position on a chromosome called its locus.

Question 4.
Who proposed the double helical model of DNA?
Answer:
Watson and Crick

Question 5.
What does the dihybrid cross involve?
Answer:
The dihybrid cross involves the inheritance of two pairs of contrasting characteristics.

Question 6.
What are the components of chromosome?
Answer:
The components of chromosomes are DNA, RNA, chromosomal proteins like histones and non histones and certain metallic ions.

Question 7.
What are Nucleoside and Nucleotide?
Answer:
Nucleoside = Nitrogen base + sugar
Nucleotide = Nucleoside + phosphate

VII. Answer the following briefly.

Question 1.
Write the Law of purity of gametes.
Answer:
When a pair of contrasting alleles are brought together in a heterozygote, the two members of the allelic pair remain together without mixing and when gametes are formed, the two separate out, so that only one enters each gamete.

Question 2.
What are chromonema and chromomeres?
Answer:
Each chromatid is made up of a spirally coiled thin structure called chromonema. The chromonema has a number of bead – like structures along its length which are called chromomeres.

Question 3.
Explain the types of chromosome-based on the position of the centromere.
Answer:
Samacheer Kalvi 10th Science Solutions Chapter 18 Heredity 5
Based on the position of the centromere, the chromosomes are classified as follows:

  • Telocentric: The centromere is at the proximal end. The chromosomes are rod – shaped.
  • Acrocentric: The centromere is found at the end with a short arm and a long arm. They are rod-shaped chromosomes.
  • Sub metacentric: The centromere is found near the centre of the – chromosome. Thus forming two unequal arms. They are J shaped or L shaped chromosomes.
  • Metacentric: The centromere occurs in the centre of the chromosome and forms two equal arms. They are V – shaped chromosomes.

Question 4.
Give reason for the appearance of new combination of characters in F2 progene.
Answer:
At the time of gamete formation in the F1 Hybrid genes character asserted independently resulted in the appearance of new combination of character in the F2 progene.

Question 5.
What is sickle cell anaemia?
Answer:
Sickle cell anaemia is caused by the mutation of a single gene. This alteration in gene brings a change in the structure of protein part of Haemoglobin molecule. The red blood cells, that carry the haemoglobin is sickle-shaped and oxygen-carrying capacity reduces, causing sickle cell anaemia.

Question 6.
Represent the phenotypic, genotypic ratio of both monohybrid cross and dihybrid cross in 72 generations of pea plants.
Answer:
Monohybrid cross:
Phenotypic ratio – 3 : 1
Genotypic ratio – 1 : 2 : 1

Dihybrid cross:
Phenotypic ratio – 9 : 3 : 3 : 1
Genotypic ratio – 1 : 2 : 2 : 1 : 4 : 1 : 2 : 2 : 1.

Question 7.
How do chromosomes take part in the formation of the male and female child?
Answer:
Fertilization of the egg (22 + X) with a sperm [22 + X] will produce a female child [44 + XX] The fertilization of the egg (22 + X) with a sperm [22 + Y] will give rise to a male child [44 + XY].

VIII. Answer the following in detail.

Question 1.
Explain the regions of chromosomes with a neat labelled diagram.
Answer:
The chromosomes are thin, long and thread-like structures, with two identical strands called chromatids. They are held together by a centromere. The chromatid is made up of spirally coiled, a thin structure called chromonema, which has a number of bead-like structures along its length called chromomeres.

A chromosome consists of the following regions:

  1. Primary constriction: The two arms of a chromosome meet at a point called primary constriction or centromere. The centromere is the region, where spindle fibres attach to the chromosome during cell division.
  2. Secondary constriction: Some chromosomes have a secondary constriction at any point of the chromosome, called the nuclear zone or nucleolar organizer. (Formation of the nucleolus in the nucleus).
  3. Telomere: The end of the chromosome is called telomere. Each extremity of the chromosome has a polarity and prevents it from joining the adjacent chromosome. It maintains and provides stability to the chromosomes.
  4. Satellite: Some of the chromosomes have an elongated knob-like appendage at one end of the chromosome, known as a satellite. The chromosomes with satellites are called as sat- chromosomes.
    Samacheer Kalvi 10th Science Solutions Chapter 18 Heredity 6

Question 2.
Given an account of the Laws of Mendel.
Answer:
Mendel proposed three important laws which are now called as Mendel’s Laws of Heredity.
(i) Law of Dominance : “When two homozygous individuals with one or more sets of contrasting characters are crossed, the characters that appear in the F1 hybrid are dominant and those that do not appear in F1 are recessive characters”.

(ii) Law of Segregation or Law of purity of gametes : “When a pair of contrasting factors or genes or allelomorphs are brought together in a heterozygote or hybrid, the two members of the allelic pair remain together without mixing and when gametes are formed, the two separate out, so that only one enters each gamete.”

(iii) Law of independent assortment : “In case of inheritance of two or more pairs of characters simultaneously, the factors or genes of one pair assort out independently of the other pair.”

Question 3.
Explain the monohybrid cross, with a diagram and describe the interpretation of Mendel oh monohybrid cross.
Answer:
Crosses involving the inheritance of only one pair of contrasting characters are called monohybrid cross. It is a cross between two forms of a single trait like a cross between tall and dwarf plant. In a monohybrid cross, a pure breeding tall plant and a pure breeding dwarf plant, results, tall and monohybrids in the F1 generation.

In the F2 generation, selfing of the F1 monohybrids resulted in tall and dwarf plants in the ratio of 3 : 1. The external expression of a particular trait is known as the phenotype. So the phenotypic ratio is 3 : 1.
In the F2 generations, 3 different types were obtained.
Tall Homozygous – TT (pure) – 1
Tall Heterozygous – Tt – 2
Dwarf Homozygous – tt – 1
So the genotypic ratio 1 : 2 : 1.
A genotype is the genetic expression of an organism.

Mendel’s interpretation on monohybrid cross:
Factors are passed on from one generation to another, factors are referred to as genes. Tallness and dwarfism are determined by a pair of contrasting factors tall plant T, a dominant character and a plant is dwarf’s’, recessive character in pairs. Pure breeding tall plants (TT) and pure dwarf plants (tt) are called homozygous. If they are unlike, (Tt) they are referred to as heterozygous.
Samacheer Kalvi 10th Science Solutions Chapter 18 Heredity 7
(i) Two factors make up a pair of contrasting characters are called alleles or allelomorphs. One member of each pair is contributed by one parent.

(ii) When two factors of a trait are brought together, by fertilization, only one expresses itself (tallness), masking the expression of the other (dwarfness). The character which expresses itself is called dominant and the character, which is masked is called recessive.

(iii) The factors for tallness [T] and dwarfness (t) are separate entities, and in a gamete either T or t is present. When F1 hybrids are self crossed the two entities separate and then unite independently, forming tall and dwarf plants.

Question 4.
Explain Mendel’s laws of heredity with the results of a dihybrid cross.
Answer:
Mendel crossed pea plants having round yellow seeds (dominant) with pea plants having wrinkled green seeds. In the F1 generation, round shape, yellow colour of the seeds were dominant over the wrinkled green colour seeds. When the hybrids of the F1 generation were cross-bred by self-pollination, the dihybrid cross, produced four types of F2 offsprings in the ratio of 9, with two dominant traits, 3 with one dominant trait and one recessive trait, 3 with another dominant trait and another recessive trait and 1 with two recessive traits.

Two new combinations of traits with round green and wrinkled yellow had appeared in the dihybrid cross. [F2 generation]
Mendel’s law of heredity:
(i) Law of dominance: When two homozygous individuals with contrasting characters are crossed, the characters, that appear in the F1 hybrid are dominant and those do not appear in F1 are recessive characters.

(ii) Law of segregation or Law of purity of gametes: When a pair of contrasting factors or genes are brought together, in a heterozygote the contrasting pair remain together without, mixing and when gametes are formed, the two separate out so that only one enters each gamete.

(iii) Law of Independent Assortment: In case of inheritance of two or more pairs of characters simultaneously, the factors or genes of one pair assort out independently of the other pair.

Question 5.
Explain the Watson and Crick model of DNA.
Answer:
Watson and Crick model of DNA:
DNA molecules consist of two polynucleotide chains. These chains form a double helix. Structure, with two strands, which run anti-parallel to one another. Nitrogenous bases in the centre are linked to sugar – phosphate units, which form the backbone of the DNA. Pairing between the nitrogenous bases is very specific and is always between purine and pyrimidine linked by hydrogen bonds.

Adenine [A] links Thymine [T] with two hydrogen bonds (A = T)
Cytosine [C] links Guanine [G] with three hydrogen bonds (C = G). This is called complementary base pairing.
Hydrogen bonds between the nitrogenous bases make the DNA molecule stable.
Each turn of the double helix is 34 A° (3.4 nm). There are ten base pairs in a complete turn.
Nucleotides in a helix are joined together by phosphodiester bonds.
Samacheer Kalvi 10th Science Solutions Chapter 18 Heredity 8
Samacheer Kalvi 10th Science Solutions Chapter 18 Heredity 9

Question 6.
Explain in detail, the various steps of DNA replication.
Answer:
Replication of DNA:
Replication of DNA occurs within a cell. DNA molecule produces exact copies of its own structure during replication.
The two strands of a DNA molecule have complementary base pairs, the nucleotides of each strand, provide the information needed to produce its new strand. The two resulting daughter cells contain exactly the same genetic information as the parent cell.
Samacheer Kalvi 10th Science Solutions Chapter 18 Heredity 10
DNA replication involves the following steps:
(a) Origin of replication: The points on the DNA, where replication begins, is the site of origin of replication. The two strands open and separate at this point, forming the replication fork.

(b) An unwinding of DNA molecule: The enzyme, helicase, bind to the origin of a replication site. Helicase separates the two strands of DNA. The enzyme called topoisomerase separates the double helix above the replication fork and removes the twists formed during the unwinding process.

(c) Formation of RNA primer: An RNA primer is a short segment of RNA nucleotides. The primer is synthesized by the DNA template, close to the origin of a replication site.

(d) Synthesis of the new complementary strand from the parent strand: After the formation of RNA primer, nucleotides are added with the help of an enzyme DNA polymerase, a new complementary strand of DNA is formed from each of the parent strands. The daughter strand is synthesized as a continuous strand, which is called the leading strand.

The short segments of DNA are synthesized, in the other strand and called lagging strand. The short segments of DNA are called Okazaki fragments. The enzyme DNA ligase joins the fragments. The replication fork of the two sides meets at a site called terminus, which is stimulated opposite to origin of a replication site.

IX. Higher Order Thinking Skills [HOTS] Questions

Question 1.
What do you understand about DNA, gene and chromosome in short?
Answer:
Deoxy ribonucleic acid [DNA] is the material located in the cells and nucleus, that makes up the chromosomes and genes. Its molecule is in the shape of a double helix.

A gene is a segment of DNA that is passed down from parents to children and confers a trait to the offspring. Genes are organised and packaged in units called chromosomes.

Chromosomes are thin, long thread-like structures contain two identical strands, called chromatids, held together by a centromere. Humans have 23 pairs of chromosomes.

Question 2.
What is the difference between a gene mutation and a chromosomal mutation?
Answer:

Gene mutation Chromosomal mutation
1. A change in the nucleotide sequence, in a particular gene. 1. A change in several genes in the chromosome.
2. Gene mutation is only a structural alteration. 2. The chromosomal mutation is either numerical or structural changes in the entire DNA strand.

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