Category: Class 11

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Tamilnadu Samacheer Kalvi 11th Bio Botany Solutions Chapter 11 Transport in Plants

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Samacheer Kalvi 11th Bio Botany Transport in Plants Text Book Back Questions and Answers

I. Choose the correct answers.
Question 1.
In a fully turgid cell:
(a) DPD = 10 atm; OP = 5 atm; TP = 10 atm
(b) DPD = 0 atm; OP =10 atm; TP = 10 atm
(c) DPD = 0 atm; OP = 5 atm; TP = 10 atm
(d) DPD = 20 atm; OP = 20 atm; TP = 10 atm
Answer:
(b) DPD = 0 atm; OP =10 atm; TP = 10 atm

Question 2.
Which among the following is correct?
(i) Apoplast is fastest and operate in nonliving part
(ii) Trahsmembrane route includes vacuole
(iii) Symplast interconnect the nearby cell through plasmadesmata
(iv) Symplast and transmembrane route are in living part of the cell

(a) (i) and (ii) only
(b) (ii) and (iii) only
(c) (iii) and (iv) only
(d) All of these
Answer:
(c) (iii) and (iv) only

Question 3.
What type of transpiration is possible in the xerophyte Opuntia?
(a) Stomatal
(b) Lenticular
(c) Cuticular
(d) All the above
Answer:
(b) Lenticular

Question 4.
Stomata of a plant open due to:
(a) Influx of K⁺
(b) Efflux of K⁺
(c) Influx of Cl^–
(d) Influx of OH^–
Answer:
(a) Influx of K⁺

Question 5.
Munch hypothesis is based on:
(a) translocation of food due to TP gradient and imbibition force
(b) ranslocation of food due to TP
(c) translocation of food due to imbibition force
(d) None of the above
Answer:
(b) ranslocation of food due to TP

Question 6.
If the concentration of salt in the soil is too high and the plants may wilt even if the field is thoroughly irrigated. Explain.
Answer:
The salts present in the soil dissolve in the irrigated water and form hypertonic solution outside the root hairs of the plant and the root hairs cannot absorb water from hypertonic solution, since water molecules cannot move from hypertonic solution to hypotonic solution in the cells of root hair. Hence the plants become wilt even the field is irrigated.

Question 7.
How phosphorylase enzyme open the stomata in starch sugar interconversion theory?
Answer:
The discovery of enzyme phosphorylase in guard cells by Hanes (1940) greatly supports the starch – sugar interconversion theory. The enzyme phosphorylase hydrolyses starch into sugar and high pH followed by endosmosis and the opening of stomata during light. The vice versa takes place during the night.

Question 8.
List out the non photosynthetic parts of a plant that need a supply of sucrose?
Answer:
The non photosynthetic parts of a plant that need a supply of sucrose:

1. Roots
2. Tubers
3. Developing fruits and
4. Immature leaves.

Question 9.
What are the parameters which control water potential?
Answer:
Water potential (Ψ) can be controlled by,

1. Solute concentration or Solute potential (Ψ_s)
2. Pressure potential (Ψ_p).

By correlating two factors, water potential is written as, Ψ_w = Ψ_s + Ψ_p.
Water Potential = Solute potential + Pressure potential.

Question 10.
An artificial cell made of selectively permeable membrane immersed in a beaker (in the figure). Read the values ans answer the following questions?

Ψ_w = 0, Ψ_s = 2, Ψ_p = 0.
(a) Draw an arrow to indicate the direction of water movement.
(b) Is the solution outside the cell isotonic, hypotonic or hypertonic?
(c) Is the cell isotonic, hypotonic or hypertonic?
(d) Will the cell become more flaccid, more turgid or stay in original size?
(e) With reference to artificial cell state, the process is endosmosis or exosmosis? Give reasons.
Answer:
(a) An arrow to indicate the direction of water movement:

(b) Outside solution in hypotonic.
(c) The cell is hypertonic.
(d) The cell become more turgid.
(e) The process is endo – osmosis because the solvent (water) moves inside the cell.

Samacheer Kalvi 11th Bio Botany Transport in Plants Additional Questions & Answers

I. Choose the correct answer (1 Mark)
Question 1 .
In plants, cell to cell transport is aided by:
(a) diffusion alone
(b) osmosis alone
(c) imbibition alone
(d) all the three above
Answer:
(d) all the three above

Question 2.
In passive transport:
(a) no energy expenditure is required
(b) energy expenditure is required
(c) no involvement of physical forces like gravity
(d) no involvement of osmosis
Answer:
(a) no energy expenditure is required

Question 3.
Which of the following statements are correct?
(i) Cell membranes allow water and non polar molecules to permeate by simple diffusion.
(ii) Polar molecules like amino acids can also diffuse through membrane.
(iii) Smaller molecules diffuse faster than larger molecules.
(iv) Larger molecules diffuse faster than smaller molecules.

(a) (i) and (iv) only
(b) (i) and (iii) only
(c) (i) and (ii) only
(d) (ii) and (iv) only
Answer:
(b) (i) and (iii) only

Question 4.
In co – transport across membrane:
(a) two different molecules are transported in opposite direction.
(b) two types of molecules are transported the same direction.
(c) three types of molecules are transported in opposite direction.
(d) two types of molecules are transported in all directions.
Answer:
(b) two types of molecules are transported the same direction.

Question 5.
The swelling of dry seeds is due to phenomenon called:
(a) osmosis
(b) transpiration
(c) imbibition
(d) none of the above
Answer:
(c) imbibition

Question 6.
The concept of water potential was introduced by:
(a) Slatyer and Mosses
(b) Slatyer and Taylor
(c) Armusten and Taylor
(d) Mosses and Robert
Answer:
(b) Slatyer and Taylor

Question 7.
At standard temperature the water potential pure water is:
(a) 1.0
(b) -1.0
(c) 0.5
(d) zero
Answer:
(d) zero

Question 8.
Addition of solute to pure water:
(a) increases water potential
(b) does not change water potential
(c) decreases water potential
(d) does not change the gradient of water potential
Answer:
(b) does not change water potential

Question 9.
Osmotic pressure is increased with:
(a) decrease of dissolved solutes in the solution
(b) increase of dissolved solutes in the solution.
(c) increase of solvent in a solution
(d) isotonic condition of the solution
Answer:
(b) increase of dissolved solutes in the solution.

Question 10.
Diffusion Pressure Deficit (DPD) was termed by Meyer in:
(a) 1928
(b) 1828
(c) 1936
(d) 1938
Answer:
(d) 1938

Question 11.
The root hairs are:
(a) unicellular extensions of epidermal cells with cuticle
(b) Unicellular extensions of xylem parenchyma cells without cuticle
(c) Unicellular extensions of epidermal cells without cuticle
(d) None of the above
Answer:
(c) Unicellular extensions of epidermal cells without cuticle

Question 12.
Kramer (1949) recognised two distinct mechanisms, which independently operate in the absorption of water in plants are:
(a) osmosis and diffusion
(b) imbibition and diffusion
(c) diffusion and absorption
(d) active absorption and passive absorption
Answer:
(d) active absorption and passive absorption

Question 13.
Indicate the correct statements:
(i) the cell sap concentration in xylem is always high.
(ii) the cell sap concentration in xylem is not always high.
(iii) root pressure is not universal in all plants.
(iv) root pressure is universal in all plants.

(a) (i) and (iv) only
(b) (ii) and (iii) only
(c) (i) and (iii) only
(d) (ii) and (iv) only
Answer:
(b) (ii) and (iii) only

Question 14.
When respiratory inhibitors like KCN, chloroform are applied:
(a) there is a decrease in the rate of respiration and increase in the rate of absorption of water.
(b) there is an increase in the rate of respiration and decrease in the rate of absorption of water.
(c) there is a decrease in the rate of respiration and also decrease in the rate of absorption of water.
(d) there is an increase in the rate of respiration and also in the rate of absorption of water.
Answer:
(c) there is a decrease in the rate of respiration and also decrease in the rate of absorption of water.

Question 15.
Relay pump theory was proposed by:
(a) J.C. Bose
(b) Godlewski
(c) Stoking
(d) Strasburger
Answer:
(b) Godlewski

Question 16.
Pulsation theory was proposed by:
(a) Strasburger
(b) Godsey
(c) J.C. Bose
(d) C.V. Raman
Answer:
(c) J.C. Bose

Question 17.
The term ‘root pressure’ was coined by:
(a) Strasburger
(b) Stephen Hales
(c) Amstrong
(d) Overton
Answer:
(b) Stephen Hales

Question 18.
Indicate the correct statements:
(i) Root pressure is absent in gymnosperms.
(ii) Root pressure in totally absent in angiosperms.
(iii) There is a relationship between the ascent of sap and root pressure.
(iv) There is no relationship between the ascent of sap and root pressure.

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

Question 19.
The capillary theory was suggested by:
(a) Unger
(b) J.C. Bose
(c) Boehm
(d) Sachs
Answer:
(c) Boehm

Question 20.
Cohesion and transpiration pull theory was originally proposed by:
(a) Unger and Sachs
(b) Xavier and Dixon
(c) Boehm and Jolly
(d) Dixon and Jolly
Answer:
(d) Dixon and Jolly

Question 21.
Loss of water from mesophyll cells causes:
(a) increase in water potential
(b) decrease in water potential
(c) does not change in water potential
(d) hone of the above events
Answer:
(b) decrease in water potential

Question 22.
The water may move through the xylem at the rate as fast as:
(a) 65 cm / min
(b) 85 cm / min
(c) 75 cm / min
(d) 45 cm / min
Answer:
(c) 75 cm / min

Question 23.
The length and breadth of stomata is:
(a) about 10 – 30μ and 2 – 10μ respectively
(b) about 10 – 14μ and 3 – 10μ respectively
(c) about 10 – 40μ and 3 – 10μ respectively
(d) about 5 – 30μ and 5 – 10μ respectively
Answer:
(c) about 10 – 40μ and 3 – 10μ respectively

Question 24.
The opening and closing of stomata depends upon the change in pH of guard cells. This is observed by:
(a) Loftfield
(b) Sayre
(c) Von Mohl
(d) Amstrong
Answer:
(b) Sayre

Question 25.
Who did observe that stomata open in light and close in the night:
(a) Unger
(b) Sachs
(c) Boehm
(d) Von Mohl
Answer:
(d) Von Mohl

Question 26.
The phosphorylase enzyme in guard cells supports the starch – sugar inter conversion theory. The above reaction is:
(a) oxidation reaction
(b) hydrolyses reaction
(c) reduction reaction
(d) none of the above
Answer:
(b) hydrolyses reaction

Question 27.
Low pH and a shortage of water in the guard cell activate the stress hormone namely:
(a) Ascorbic acid
(b) Malic acid
(c) Abscisic acid
(d) Salisilic acid
Answer:
(c) Abscisic acid

Question 28.
Accumulation of CO₂ in plant cell during dark:
(a) increases the pH level
(b) decreases the pH level
(c) does not alter pH
(d) decreases in H⁺ ion concentration
Answer:
(b) decreases the pH level

Question 29.
Phenyl Mercuric Acetate (PMA), when applied as a foliar spray to plants:
(a) induces partial stomatal closure for two weeks.
(b) induces partial stomatal opening for two weeks.
(c) induces partial stomatal closure for four weeks.
(d) induces stomatal closure permanently
Answer:
(a) induces partial stomatal closure for two weeks.

Question 30.
The transpiration in plants is a “necessary evil” as stated by:
(a) Steward
(b) Sayre
(c) Curtis
(d) Meyer
Answer:
(c) Curtis

Question 31.
Sink in plants, which receives food from source is:
(a) tubers
(b) developing fruits
(c) roots
(d) all the three above
Answer:
(d) all the three above

Question 32.
Activated diffusion theory was first proposed by:
(a) Fenson and Spanner
(b) Mason and Masked
(c) Crafts and Munch
(d) Hanes and Robert
Answer:
(b) Mason and Masked

Question 33.
From sieve elements sucrose is translocated into sink organs such as root, tubers etc and this process is termed as:
(a) Xylem unloading
(b) Xylem uploading
(c) Phloem unloading
(d) Phloem uploading
Answer:
(c) Phloem unloading

Question 34.
In which plant, the petioles are flattened and widened, to become phyllode:
(a) Asparagus
(b) Acacia melanoxylon
(c) Vinca rosea
(d) Delonix regia
Answer:
(b) Acacia melanoxylon

Question 35.
Match the following:

(i) Opuntia	(a) Cladode
(ii) Acacia	(b) Guttation
(iii) Asparagus	(c) Phyllode
(iv) Alocasia	(d) Phylloclade

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

Question 36.
Hydathodes are generally present in plants that grow in:
(a) dry places
(b) moist and shady places
(c) sunny places
(d) deserts
Answer:
(b) moist and shady places

Question 37.
Ganongs potometer is used to measure:
(a) the rate of photosynthesis
(b) the rate of gaseous exchange
(c) the rate of water transport
(d) the rate of transpiration
Answer:
(d) the rate of transpiration

Question 38.
Indicate the correct statement:
(a) Anti – transpirants increases the loss of water by transpiration.
(b) Anti – transpirants do not alter the rate of transpiration.
(c) Anti – transpirants do not decrease the loss water by transpiration in cross plants.
(d) Anti – transpirants reduce the enormous loss of wafer by transpiration in crop plants.
Answer:
(d) Anti – transpirants reduce the enormous loss of wafer by transpiration in crop plants.

Question 39.
The liquid coming out of hydathode of grasses is:
(a) pure water
(b) not pure water
(c) a solution containing a number of dissolved substances
(d) salt water
Answer:
(c) a solution containing a number of dissolved substances

Question 40.
A dry cobalt chloride strip, when hydrated, turns:
(a) white
(b) red
(c) green
(d) pink
Answer:
(d) pink

II. Answer the following (2 Marks)

Question 1.
What is the need for transport of materials in plants?
Answer:
Water absorbed from roots must travel up to leaves by xylem for food preparation by photosynthesis. Likewise, food prepared from leaves has to travel to all parts of the plant including roots.

Question 2.
What are the types of transport based on the distance travelled by the materials?
Answer:
Based on the distance travelled by water (sap) or food (solute) they are classified as

1. Short distance (cell to cell transport)
2. Long distance transport.

Question 3.
Define the term diffusion.
Answer:
The net movement of molecules from a region of their higher concentration to a region of their lower concentration along a concentration gradient until an equilibrium is attained.

Question 4.
Define the term semipermeable.
Answer:
Semipermeable allow diffusion of solvent molecules but do not allow the passage of solute molecule. eg: Parchment paper.

Question 5.
What is meant by Porin?
Answer:
Porin is a large transporter protein found in the outermembrane of plastids, mitochondria and bacteria which facilitates smaller molecules to pass through the membrane.

Question 6.
Define symport or co – transport?
Answer:
The term symport is used to denote an integral membrane protein that simultaneously transports two types of molecules across the membrane in the same direction.

Question 7.
Explain the term counter transport.
Answer:
An antiport is an integral membrane transport protein that simultaneously transports two different molecules, in opposite directions, across the membrane.

Question 8.
What is the difference between co – transport and counter transport?
Answer:
In co – transport, two molecules are transported together whereas, in counter transport two molecules are transported in opposite direction to each other.

Question 9.
Define the term Imbibition.
Answer:
Colloidal systems such as gum, starch, proteins, cellulose, agar, gelatin when placed in water, will absorb a large volume of water and swell up. These substances are called imbibants and the phenomenon is imbibition.

Question 10.
Give two examples for the phenomenon of Imbibition.
Answer:
two examples for the phenomenon of Imbibition:

1. The swelling of dry seeds.
2. The swelling of wooden windows, tables, doors due to high humidity during the rainy season.

Question 11.
Define the term osmotic potential.
Answer:
Osmotic potential is defined as the ratio between the number of solute particles and the number of solvent particles in a solution.

Question 12.
What is transpiration?
Answer:
The loss of excess of water in the form of vapour from various aerial parts of the plant is called transpiration.

Question 13.
What is meant by osmotic pressure?
Answer:
When a solution and its solvent (pure water) are separated by a semipermeable membrane, a pressure is developed in the solution, due to the presence of dissolved solutes. This is called osmotic pressure (OP).

Question 14.
Explain the term wall pressure exerted by the cell wall.
Answer:
The cell wall reacts to this turgor pressure with equal and opposite force, and the counter – pressure exerted by the cell wall towards cell membrane is wall pressure (WP).

Question 15.
Define the term osmosis.
Answer:
Osmosis (Latin: Osmos – impulse, urge) is a special type of diffusion. It represents the movement of water or solvent molecules through a selectively permeable membrane from the place of its higher concentration (high water potential) to the place of its lower concentration (low water potential).

Question 16.
What is meant by isotonic solution?
Answer:
Isotonic (Iso = identical; tonic = soute): It refers to two solutions having same concentration. In this condition the net movement of water molecule will be zero.

Question 17.
What are the three types of plasmolysis?
Answer:
Three types of plasmolysis occur in plants:

1. Incipient plasmolysis
2. Evident plasmolysis
3. Final plasmolysis.

Question 18.
Explain briefly about root hairs.
Answer:
Root hairs are unicellular extensions of epidermal cells without cuticle. Root hairs are extremely thin and numerous and they provide a large surface area for absorption.

Question 19.
Define active absorption of water.
Answer:
The mechanism of water absorption due to forces generated in the root itself is called active absorption. Active absorption may be osmotic or non – osmotic.

Question 20.
Explain briefly the term stomatal transpiration.
Answer:
Stomata are microscopic structures present in high number on the lower epidermis of leaves. This is the most dominant form of transpiration and being responsible for most of the water loss (90 – 95%) in plants.

Question 21.
Give any two objections to starch – sugar inter conversion theory.
Answer:
Two objections to starch – sugar inter conversion theory:

1. In monocots, guard cell does not have starch.
2. There is no evidence to show the presence of sugar at a time when starch disappears and stomata open.

Question 22.
Briefly explain plant anti – transpirants.
Answer:
The term anti – transpirant is used to designate any Material applied to plants for the purpose of retarding transpiration. An ideal anti – transpirant checks the transpiration process without disturbing the process of gaseous exchange.

Question 23.
Mention any two uses of anti – transpirants.
Answer:
Two uses of anti – transpirants:

1. Anti – transpirants reduce the enormous loss of water by transpiration in crop plants.
2. Useful for seedling transplantations in nurseries.

Question 24.
What is meant by translocation of organic solutes.
Answer:
The phenomenon of food transportation from the site of synthesis to the site of utilization is known as translocation of organic solutes. The term solute denotes food material that moves in a solution.

Question 25.
Define the term Ion – Exchange.
Answer:
Ions of external soil solution are exchanged with same charged (anion for anion or cation for cation) ions of the root cells.

III. Answer the following (3 Marks)

Question 1.
Briefly explain the term aquaporin.
Answer:
Aquaporin Is a water channel protein embedded in the plasma membrane. It regulates the massive amount of water transport across the membrane. Plants contain a variety of aquaporins. Over 30 types of aquaporins are known from maize.

Currently, they are also recognised to transport substrates like glycerol, urea, CO₂, NH₃, rhetalloids, and reactive oxygen species (ROS) in addition to water. They increase the permeabi lity of the membrane to water. They confer drought and salt stress tolerance.

Question 2.
What is carrier protein? Mention the. three types of carrier proteins?
Answer:
Carrier protein acts as a vehicle to carry molecules from outside of the membrane to inside the cell and vice versa. Due to association with molecules to be transported, the structure of carrier protein gets modified until the dissociation of the molecules.

There are three types of carrier proteins classified on the basis of handling of molecules and direction of transport. They are:

1. Uniport
2. Symport
3. Antiport.

Question 3.
Explain osmotic potential.
Answer:
Solute potential, otherwise known as osmotic potential denotes the effect of dissolved solute on water potential. In pure water, the addition of solute reduces its free energy and lowers the water potential value from zero to negative. Thus the value of solute potential is always negative. In a solution at standard atmospheric pressure, water potential is always equal to solute potential (Ψ_w = Ψ_s).

Question 4.
What are the types of osmosis based on the direction of the movement of water? Explain briefly.
Answer:
Based on the direction of movement of water or solvent in an osmotic system, two types of osmosis can occur, they are Endosmosis and Exosmosis:

1. Endosmosis: Endosmosis is defined as the osmotic entry of solvent into a cell or a system when it is placed in a pure water or hypotonic solution. eg: dry raisins (high solute and low solvent) placed in the water, it swells up due to turgidity.
2. Exosmosis: Exosmosis is defined as the osmotic withdrawal of water from a cell or system when it is placed in a hypertonic solution. Exosmosis in a plant cell leads to plasmolysis.

Question 5.
Describe the method of demonstration of endo – osmosis by potato Osmoseope.
Answer:
The method of demonstration of endo – osmosis by potato Osmoscope:

1. Take a peeled potato tuber and make a cavity inside with the help of a knife.
2. Fill the cavity with concentrated sugar solution and mark the initial level.
3. Place this setup in a beaker of pure water.
4. After 10 minutes observe the sugar solution level and record your findings.
5. With the help of your teacher discuss the results.
   
   

Instead of potato use beetroot or bottleguard and repeat the above experiment. Compare and discuss the results.

Question 6.
Explain the term reverse osmosis.
Answer:
Reverse Osmosis follows the same principles of osmosis, but in the reverse direction. In this process movement of water is reversed by applying pressure to force the water against a concentration gradient of the solution.

In regular osmosis, the water molecules move from the higher concentration (pure water = hypotonic) to lower concentration (salt water = hypertonic). But in reverse osmosis, the water molecules move from the lower concentration (salt water = hypertonic) to higher concentration (pure water = hypotonic) through a selectively permeable membrane.

Uses:  Reverse osmosis is used for purification of drinking water and desalination of seawater.

Question 7.
Give details of symplast route of water movement.
Answer:
The symplast (Greek: sym = within; plast = cell) consists of the entire mass of cytosol of all the living cells in a plant, as well as the plasmodesmata, the cytoplasmic channel that interconnects them. In the symplastic route, water has to cross plasma membrane to enter the cytoplasm of outer root cell; then it will move within adjoining cytoplasm through plasmodesmata around the vacuoles without the necessity to cross more membrane, till it reaches xylem.

Question 8.
Describe the non – osmotic active absorption theory proposed by Bennet – Clark in 1936.
Answer:
Bennet – Clark (1936), Thimann (1951) and Kramer (1959) observed absorption of water even if the concentration of cell sap in the root hair is lower than that of the soil water. Such a movement requires an expenditure of energy released by respiration (ATP). Thus, there is a link between water absorption and respiration. It is evident from the fact that when respiratory inhibitors like KCN, Chloroform are applied there is a decrease in the rate of respiration and also the rate of absorption of water.

Question 9.
Mention the objections to vital force theory of Ascent of sap.
Answer:
The objections to vital force theory of Ascent of sap:

1. Strasburger (1889) and Overton (1911) experimentally proved that living cells are not mandatory for the ascent of sap. For this, he selected an old oak tree trunk which when immersed in picric acid and subjected to excessive heat killed all the living cells of the trunk. The trunk when dipped in water, the ascent of sap took place.
2. Pumping action of living cells should be in between two xylem elements (vertically) and not on lateral sides.

Question 10.
Explain the capillary theory of Boehm (1809).
Answer:
Capillary theory: Boehm (1809) suggested that the xylem vessels work like a capillary tube. This capillarity of the vessels under normal atmospheric pressure is responsible for the ascent of sap. This theory was rejected because the magnitude of capillary force can raise water level only up to a certain height. Further, the xylem vessels are broader than the tracheid which actually conducts more water and against the capillary theory.

Question 11.
Give a brief account of Lenticular transpiration.
Answer:
In stems of woody plants and trees, the epidermis is replaced by periderm because of secondary growth. In order to provide gaseous exchange between the living cells and outer atmosphere, some pores which looks like lens – shaped raised spots are present on the surface of the stem called Lenticels. The loss of water from lenticels is very insignificant as it amounts to only 0.1% of the total.

Question 12.
Explain the theory of photosynthesis in guard cells observed by Von Mohl with its demerits.
Answer:
Von Mohl (1856) observed that stomata open in light and close in the night. According to him, chloroplasts present in the guard cells photosynthesize in the presence of light resulting in the production of carbohydrate (Sugar) which increases osmotic pressure in guard cells. It leads to the entry of water from other cell and stomatal aperture opens. The above process vice versa in night leads to closure of stomata.

Demerits:

1. Chloroplast of guard cells is poorly developed and incapable of performing photosynthesis.
2. The guard cells already possess much amount of stored sugars.

Question 13.
What are the three types of wilting in plants? Explain them briefly.
Answer:
In general, there are three types of wilting as follows:

1. Incipient wilting: Water content of plant cell decreases but the symptoms are not visible.
2. Temporary wilting: On hot summer days, the freshness of herbaceous plants reduces turgor pressure at the day time and regains it at night.
3. Permanent wilting: The absorption of water virtually ceases because the plant cell does not get water from any source and the plant cell passes into a state of permanent wilting.

Question 14.
Define guttation. Explain it with examples.
Answer:
During high humidity in the atmosphere, the rate of transpiration is much reduced. When plants absorb water in such a condition root pressure is developed due to excess water within the plant. Thus excess water exudates as liquid from the edges of the leaves and is called guttation. eg: Grasses, tomato, potato, brinjal and Alocasia.

Guttation occurs through stomata like pores called hydathodes generally present in plants that grow in moist and shady places. Pores are present over a mass of loosely arranged cells with large intercellular spaces called epithem. This mass of tissue lies near vein endings (xylem and Phloem). The liquid coming out of hydathode is not pure water but a solution containing a number of dissolved substances.

Question 15.
What is the significance of transpiration in plants?
Answer:
Transpiration leads to loss of water, as 95% of absorbed water is lost in transpiration. It seems to be an evil process to plants. However, number of process like absorption of water, ascent of sap and mineral absorption – directly relay on the transpiration. Moreover plants withstand against scorching sunlight due to transpiration. Hence the transpiration is a “necessary evil” as stated by Curtis.

Question 16.
What do you understand by the source and sink organ of plant?
Answer:
The source organ: Source is defined as any organ in plants which are capable of exporting food materials to the areas of metabolism or to the areas of storage. eg: Mature leaves, germinating seeds.

Sink organ: Sink is defined as any organ in plants which receives food from source. eg: Roots, tubers, developing fruits and immature leaves.

Question 17.
Why plants transport sugars as sucrose and not as starch or glucose or fructose?
Answer:
Glucose and Fructose are simple monosaccharides, whereas, Sucrose is a disaccharide composed of glucose and fructose. Starch is a polysaccharide of glucose. Sucrose and starch are more efficient in energy storage when compared to glucose and fructose, but starch is insoluble in water. So it cannot be transported via phloem and the next choice is sucrose, being water soluble and energy efficient, sucrose is chosen as the carrier of energy from leaves to different parts of the plant.

Sucrose has low viscosity even at high concentrations and has no reducing ends which makes it inert than glucose or fructose. During photosynthesis, starch is synthesized and stored in the chloroplast stroma and sucrose is synthesized in the leaf cytosol from which it diffuses to the rest of the plant.

Question 18.
What is meant by phloem unloading?
Answer:
From sieve elements sucrose is translocated into sink organs such as roots, tubers, flowers and fruits and this process is termed as phloem unloading. It consists of three steps:

1. Sieve element unloading: Sucrose leave from sieve elements.
2. Short distance transport: Movement of sucrose to sink cells.
3. Storage and metabolism: The final step when sugars are stored or metabolized in sink cells.

Question 19.
Explain the term Donnam equilibrium.
Answer:
Within the cell, some of the ions never diffuse out through the membrane. They are trapped within the cell and are called fixed ions. But they must be balanced by the ions of opposite charge. Assuming that a concentration of fixed anions is present inside the membrane, more cations would be absorbed in addition to the normal exchange to maintain the equilibrium. Therefore, the cation concentration would be greater in the internal than in the external solution. This electrical balance or equilibrium controlled by electrical as well as diffusion phenomenon is known as the Donnan equilibrium.

IV. Answer the following (5 Marks)

Question 1.
Define the term osmosis. Give details of the types of osmosis in plants.
Answer:
1. Osmosis (Latin: Osmos – impulse, urge) is a special type of diffusion. It represents the movement of water or solvent molecules through a selectively permeable membrane from the place of its higher concentration (high water potential) to the place of its lower concentration (low water potential). Types of Solutions based on concentration:

• Hypertonic (Hyper = High; tonic = solute): This is a strong solution (low solvent / high solute / low Ψ) which attracts solvent from other solutions.
•  Hypotonic (Hypo – low; tonic = solute): This is a weak solution (high sol vent / low or zero solute/ high Ψ) and it diffuses water out to other solutions.
• Isotonic (Iso = identical; tonic = soute): It refers to two solutions having same concentration. In this condition the net movement of water molecule will be zero.

The term hyper, hypo and isotonic are relative terms which can be used only in comparison with another solution.

2. Types of osmosis: Based on the direction of movement of water or solvent in an osmotic system, two types of osmosis can occur, they are Endosmosis and Exosmosis.

• Endosmosis: Endosmosis is defined as the osmotic entry of solvent into a cell or a system when it is placed in a pure water or hypotonic solution. eg: dry raisins . (high solute and low solvent) placed in the water, it swells up due to turgidity.
• Exosmosis: Exosmosis is defined as the osmotic withdrawal of water from a cell or system when it is placed in a hypertonic solution. Exosmosis in a plant cell leads to plasmolysis.

Question 2.
Give an account of active absorption theories with their demerits.
Answer:
The mechanism of water absorption due to forces generated in the root itself is called active absorption. Active absorption may be osmotic or non – osmotic.
1. Osmotic active absorption: The theory of osmotic active absorption was postulated by Atkins (1916) and Preistley (1923). According to this theory, the first step in the absorption is soil water imbibed by cell wall of the root hair followed by osmosis. The soil water is hypotonic and cell sap is hypertonic. Therefore, soil water diffuses into root hair along the concentration gradient (endosmosis).

When the root hair becomes fully turgid, it becomes hypotonic and water moves osmotically to the outer most cortical cell. In the same way, water enters into inner cortex, endodermis, pericycle and finally reaches protoxylem. As the sap reaches the protoxylem a pressure is developed known as root pressure. This theory involves the symplastic movement of water.

2. Objections to osmotic theory:

• The cell sap concentration in xylem is not always high
• Root pressure is not universal in all plants especially in trees.

3. Non – Osmotic active absorption: Bennet – Clark (1936), Thimann (1951) and Kramer (1959) observed absorption of water even if the concentration of cell sap in the root hair is lower than that of the soil water. Such a movement requires an expehditure of energy released by respiration (ATP). Thus, there is a link between water absorption and respiration. It is evident from the fact that when respiratory inhibitors like KCN, Chloroform are applied there is a decrease in the rate of respiration and also the rate of absorption of water.

Question 3.
Explain in detail about the cohesion tension theory proposed by Dixon and Jolly (1894).
Answer:
(i) Strong cohesive force or tensile strength of water: Water molecules have the strong mutual force of attraction called cohesive mutual force of attraction called cohesive force due to which they cannot be easily separated from one another. Further, the attraction between a water molecule and the wall of the xylem element is called adhesion. These cohesive and adhesive force works together to form an unbroken continuous water column in the xylem. The magnitude of the cohesive force is much high (350 atm) and is more than enough to ascent sap in the tallest trees.

(ii) Continuity of the water column in the plant: An important factor which can break the water column is the introduction of air bubbles in the xylem. Gas bubbles expanding and displacing water within the xylem element is called cavitation or embolism. However, the overall continuity of the water column remains undisturbed since water diffuses into the adjacent xylem elements for continuing ascent of sap.

(iii) Transpiration pull or Tension in the unbroken water column: The unbroken water column from leaf to root is just like a rope. If the rope is pulled from the top, the entire rope will move upward. In plants, such a pull is generated by the process of transpiration which is known as transpiration pull. Water vapour evaporates from mesophyll cells to the intercellular spaces near stomata as a result of active transpiration.

The water vapours are then transpired through the stomatal pores. Loss of water from mesophyll cells causes a decrease in water potential. So, water moves as a pull from cell to cell along the water potential gradient. This tension, generated at the top (leaf) of the unbroken water column, is transmitted downwards from petiole, stem and finally reaches the roots. The cohesion theory is the most accepted among the plant physiologists today.

Question 4.
Describe the theory of K⁺ transport theory of stomatal opening.
Answer:

This theory was proposed by Levit (1974) and elaborated by Raschke (1975). According to this theory, the following steps are involved in the stomatal opening:
In light:

1. In guard cell, starch is converted into organic acid (malic acid).
2. Malic acid in guard cell dissociates to malate anion and proton (H⁺).
3. Protons are transported through the membrane into nearby subsidiary cells with the exchange of K⁺ (Potassium ions) from subsidiary cells to guard cells. This process involves an electrical gradient and is called ion exchange.
4. This ion exchange is an active process and consumes ATP for energy.
5. Increased K⁺ ions in the guard cell are balanced by Cl^– ions. Increase in solute concentration decreases the water potential in the guard cell.
6. Guard cell becomes hypertonic and favours the entry of water from surrounding cells.
7. Increased turgor pressure due to the entry of water opens the stomatal pore.

In Dark:

1. In dark photosynthesis stops and respiration continues with accumulation of CO₂ in the sub-stomatal cavity.
2. Accumulation of CO₂ in cell lowers the pH level.
3. Low pH and a shortage of water in the guard cell activate the stress hormone Abscisic acid (ABA).
4. ABA stops further entry of K⁺ ions and also induce K⁺ ions to leak out to subsidiary cells from guard cell.
5. Loss of water from guard cell reduces turgor pressure and causes closure of stomata.

Question 5.
Give an account of external factors, which affect the rate of transpiration.
Answer:
External or Environmental factors:
1. Atmospheric humidity: The rate of transpiration is greatly reduced when the atmosphere is very humid. As the air becomes dry, the rate of transpiration is also increased proportionately.

2. Temperature: With the increase in atmospheric temperature, the rate of transpiration also increases. However, at very high – temperatures stomata closes because of flaccidity and transpiration stop.

3. Light: Light intensity increases the temperature. As in temperature, transpiration is increased in high light intensity and is decreased in low light intensity. Light also increases the permeability of the cell membrane, making it easy for water molecules to move out of the cell.

4. Wind velocity: In still air, the surface above the stomata get saturated with water vapours and there is no need for more water vapour to come out. If the wind is breezy, water vapour gets carried away near leaf surface and DPD is created to draw more vapour from the leaf cells enhancing transpiration. However, high wind velocity creates an extreme increase in water loss and leads to a reduced rate of transpiration and stomata remain closed.

5. Atmospheric pressure: In low atmospheric pressure, the rate of transpiration increases. Hills favour high transpiration rate due to low atmospheric pressure. However, it is neutralized by low temperature prevailing in the hills.

6. Water: Adequate amount of water in the soil is a pre – requisite for optimum plant growth. Excessive loss of water through transpiration leads to wilting.

Question 6.
Describe the method of Ganongs potometer to measure the rate of transpiration.
Answer:
Ganongs potometer is used to measure the rate of transpiration indirectly. In this, the amount of water absorbed is measured and assumed that this amount is equal to the amount of water transpired. Apparatus consists of a horizontal graduated tube which is bent in opposite directions at the ends. One bent end is wide and the other is narrow. A reservoir is fixed to the horizontal tube near the wider end. The reservoir has a stopcock to regulate water flow. The apparatus is filled with water from reservoir.

A twig or a small plant is fixed to the wider arm through a split cock. The other bent end of the horizontal tube is dipped into a beaker containing coloured water. An air bubble is introduced into the graduated tube at the narrow end. Keep this apparatus in bright sunlight and observe. As transpiration takes place, the air bubble will move towards the twig. The loss is compensated by water absorption through the xylem portion of the twig. Thus, the rate of water absorption is equal to the rate of transpiration.

Question 7.
Explain Munch Mass Flow Hypothesis with its merits and objections.
Answer:
Mass flow theory was first proposed by Munch (1930) and elaborated by Crafts (1938). According to this hypothesis, organic substances or solutes move from the region of high osmotic pressure (from mesophyll) to the region of low osmotic pressure along the turgor pressure gradient. The principle involved in this hypothesis can be explained by a simple physical system as shown in figure.

Two chambers “A” and “B” made up of semipermeable membranes are connected by tube “T” immersed in a reservoir of water. Chamber “A” contains highly concentrated sugar solution while chamber “B” contains dilute sugar solution. The following changes were observed in the system.

1. The high concentration sugar solution of chamber “A” is in a hypertonic state which draws water from the reservoir by endosmosis.
2. Due to the continuous entry of water into chamber “A”, turgor pressure is increased.
3. Increase in turgor pressure in chamber “A” force, the mass flow of sugar solution to chamber “B” through the tube “T” along turgor pressure gradient.
4. The movement of solute will continue till the solution in both the chambers attains the state of isotonic condition and the system becomes inactive.
5. However, if new sugar solution is added in chamber “A”, the system will start to run again.

A similar analogous system as given in the experiment exists in plants:
Chamber “A” is analogous to mesophyll cells of the leaves which contain a higher concentration of food material in soluble form. In short “A” is the production point called “source”. Chamber “B” is analogous to cells of stem and roots where the food material is utilized. In short “B” is consumption end called “sink”. Tube “T” is analogous to the sieve tube of phloem.

Mesophyll cells draw water from the xylem (reservoir of the experiment) of the leaf by endosmosis leading to increase in the turgor pressure of mesophyll cell. The turgor pressure in the cells of stem and the roots are comparatively low and hence, the soluble organic solutes begin to flow en masse from mesophyll through the phloem to the cells of stem and roots along the gradient turgor pressure.

In the cells of stem and roots, the organic solutes are either consumed or converted into insoluble form and the excess water is released into xylem (by turgor pressure gradient) through cambium.

Merits:

1. When a woody or herbaceous plant is girdled, the sap contains high sugar containing exudates from cut end.
2. Positive concentration gradient disappears when plants are defoliated.

Objections:

1. This hypothesis explains the unidirectional movement of solute only. However, bidirectional movement of solute is commonly observed in plants.
2. Osmotic pressure of mesophyll cells and that of root hair do not confirm the requirements.
3. This theory gives passive role to sieve tube and protoplasm, while some workers demonstrated the involvement of ATP.

Question 8.
Write an essay on Lunde – gardh’s cytochrome pump theory of mineral transport.
Answer:
Lundegardh and Burstrom (1933) observed a correlation between respiration and anion absorption. When a plant is transferred from water to a salt solution the rate of respiration increases which is called,as anion respiration or salt respiration. Based on this observation Lundegardh (1950 and 1954) proposed cytochrome pump theory which is based on the following assumptions:

1. The mechanism of anion and cation absorption are different.
2. Anions are absorbed through cytochrome chain by an active process, cations are absorbed passively.
3. An oxygen gradient responsible for oxidation at the outer surface of the membrane and reduction at the inner surface.

According to this theory, the enzyme dehydrogenase on inner surface is responsible for the formation of protons (H⁺) and electrons (e^–). As electrons pass outward through electron transport chain there is a corresponding inward passage of anions.

Anions are picked up by oxidized cytochrome oxidase and are transferred to other members of chain as they transfer the electron to the next component. The theory assumes that cations (C⁺) move passively along the electrical gradient created by the accumulation of anions (A^–) at the inner surface of the membrane.

Main defects of the above theory are:

1. Cations also induce respiration.
2. Fails to explain the selective uptake of ions.
3. It explains absorption of anions only.

Solution To Activity
Textbook Page No: 63

Question 1.
Imbibition experiment: Collect 5 gm of gum from Drumstick tree or Babool tree or Almond tree. Immerse in 100 ml of water. After 24 hours observe the changes and discuss the results with your teacher.
Answer:

The gum will absorb large amount of water and swells. The phenomenon is called imbibition.

Textbook Page No: 65

Question 1.
Find the role of turgor pressure in sudden closing of leaves when we touch the ‘touch me not’ plant.
Answer:
When touched, this sensitive leaf reacts to stimulus as there is a higher pressure at that point and water in the vacuoles of the cells of the leaf lose water to the adjacent cell. This causes the leaves to close. If the leaves are left undisturbed for a few seconds, they slowly open up again and regain turgidity.

Textbook Page No: 75

Question 1.
Select a leafy twig of fully grown plant in your school campus. Cover the twig with a transparent polythene bag and tie the mouth of the bag at the base of the twig. Observe the changes after two hours and discuss with your teacher.
Answer:
Two hours and discuss with your teacher:

1. Select a leafy twig of a fully grown plant.
2. Cover the twig in a transparent polythene bag.
3. Tie the mouth of the bag.
4. Observe the bag after two hours.
5. Observation: Moisture will be observed inside the plastic bag because of transpiration of water from the plant twig.

Textbook Page No: 79

Question 1.
What will happen if an indoor plant is placed under fan and AC?
Answer:
When an indoor plant is placed under fan and AC, the transpiration of water from the plant may increase, because the wind from fan and the humidity from AC will increase transpiration of water from the plant.

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Tamilnadu Samacheer Kalvi 11th Bio Botany Solutions Chapter 10 Secondary Growth

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Samacheer Kalvi 11th Bio Botany Secondary Growth Text Book Back Questions and Answers

Question 1.
Consider the following statements In spring season vascular cambium:
(i) is less active
(ii) produces a large number of xylary elements
(iii) forms vessels with wide cavities of these

(a) (i) is correct but (ii) and (iii) are not correct
(b) (i) is not correct but (ii) and (iii) are correct
(c) (i) and (ii) are correct but (iii) is not correct
(d) (i) and (ii) are not correct but (iii) is correct
Answer:
(b) (i) is not correct but (ii) and (iii) are correct

Question 2.
Usually, the monocotyledons do not increase their girth, because:
(a) They possess actively dividing cambium
(b) They do not possess actively dividing cambium
(c) Ceases activity of cambium
(d) All are correct
Answer:
(b) They do not possess actively dividing cambium

Question 3.
In the diagram of lenticel identify the parts marked as A,B,C,D.

(a) A. Phellem, B. Complementary tissue, C. Phelloderm, D. Phellogen.
(b) A. Complementary tissue, B. Phellem, C. Phellogen, D. Phelloderm.
(c) A. Phellogen, B. Phellem, C. Pheiloderm, D. complementary tissue
(d) A. Phelloderm, B. Phellem, C. Complementary tissue, D. Phellogen
Answer:
(a) A. Phellem, B. Complementary tissue, C. Phelloderm, D. Phellogen.

Question 4.
The common bottle cork is a product of:
(a) Dermatogen
(b) Phellogen
(c) Xylem
(d) Vascular cambium
Answer:
(b) Phellogen

Question 5.
What is the fate of primary xylem in a dicot root showing extensive secondary growth?
(a) It is retained in the center of the axis
(b) It gets crushed
(c) May or may not get crushed
(d) It gets surrounded by primary phloem
Answer:
(b) It gets crushed

Question 6.
In a forest, if the bark of a tree is damaged by the horn of a deer, How will the plant overcome the damage?
Answer:
When the bark is damaged, the phellogem forms a complete cylinder around the stem and it gives rise to ring barks.

Question 7.
In which season the vessels of angiosperms are larger in size, why?
Answer:
In spring season the vessels are larger in size, because the cambium cells are very active during spring season.

Question 8.
Continuous state of dividing tissue is called meristem. In connection to this, what is the role of lateral meristem?
Answer:
Apical meristems produce the primary plant body. In some plants, the lateral meristem increase the girth of a plant. This type of growth is secondary because the lateral meristem are not directly produced by apical meristems. Woody plants have two types of lateral meristems: a vascular cambium that produces xylem, phloem tissues and cork cambium that produces the bark of a tree.

Question 9.
A timber merchant bought 2 logs of wood from a forest & named them A & B, The log A was 50 year old & B was 20 years old. Which log of wood will last longer for the merchant? Why?
Answer:
The wood of 50 years old will last longer than 20 years old wood, because timber from hard wood is more durable and more resistant to the attack of micro organisms and insect than the timber from sap wood.

Question 10.
A transverse section of the trunk of a tree shows concentric rings which are known as growth rings. How are these rings formed? What are the significance of these rings?
Answer:
The annual ring denotes the combination of early wood and late wood and the ring becomes evident to our eye due to the high density of late wood. Sometimes annual rings are called growth rings but it should be remembered all the growth rings are not annual. In some trees more than one growth ring is formed with in a year due to climatic changes. Additional growth rings are developed within a year due to adverse natural calamities like drought, frost, defoliation, flood, mechanical injury and biotic factors during the middle of a growing season, which results in the formation of more than one annual ring.

Such rings are called pseudo – or false – annual rings. Each annual ring corresponds to one year’s growth and on the basis of these rings, the age of a particular plant can easily be calculated. The determination of the age of a tree by counting the annual rings is called dendrochronology.

Samacheer Kalvi 11th Bio Botany Secondary Growth Other Important Questions & Answers

I. Choose the correct answer. (I Marks)
Question 1.
The roots and stems grow in length with the help of:
(a) cambium
(b) secondary growth
(c) apical meristem
(d) vascular parenchyma
Answer:
(c) apical meristem

Question 2.
The increase in the girth of plant is called:
(a) primary growth
(b) tertiary growth
(c) longitudinal growth
(d) secondary growth
Answer:
(d) secondary growth

Question 3.
The secondary vascular tissues include:
(a) secondary xylem and secondary phloem
(b) secondary xylem, cambium strip and secondary phloem
(c) secondary phloem and fascicular cambium
(d) secondary xylem and primary phloem
Answer:
(a) secondary xylem and secondary phloem

Question 4.
Choose the correct statements.
(i) A strip of vascular cambium is present between xylem and phloem of the vascular bundle.
(ii) Vascular cambium is believed originate from fusiform initials.
(iii) The vascular cambium is originated from procambium of vascular bundle
(iv) Vascular cambium is present between fusiform initials and ray initials

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

Question 5.
Match the following:

A. Xylem	(i) Treachery elements
B. Secondary xylem	(ii) Water transport
C. Phloem	(iii) Sieve elements
D. Secondary phloem	(iv) Food transport

(a) B – (i); A – (ii); C – (iii); D – (iv)
(b) B – (ii); A – (iii); C – (i); D – (iv)
(c) A – (ii); B – (i); C – (iv); D – (iii)
(d) A – (i); B – (ii); C – (iii); D – (iv)
Answer:
(c) A – (ii); B – (i); C – (iv); D – (iii)

Question 6.
The axial system of the secondary xylem includes:
(a) treachery elements, sieve elements, fibers and axial parenchyma
(b) treachery elements, fibers and axial parenchyma
(c) treachery elements and fibers
(d) sieve elements and axial parenchyma
Answer:
(b) treachery elements, fibers and axial parenchyma

Question 7.
The study of wood by preparing sections for microscopic observation is termed as:
(a) histology
(b) xylotomy
(c) phoemtomy
(d) anatomy
Answer:
(b) xylotomy

Question 8.
Ray cells are present between:
(a) primary xylem and phloem
(b) primary xylem and secondary xylem
(c) secondary xylem and phloem
(d) secondary phloem and cambium
Answer:
(c) secondary xylem and phloem

Question 9.
The axial system Consists of vertical files of:
(a) treachery elements and sieve elements
(b) treachery elements and apical parenchyma
(c) sieve elements are fibers
(d) treachery elements, fibers and wood parenchyma
Answer:
(d) treachery elements, fibers and wood parenchyma

Question 10.
Morus rubra has:
(a) porous wood
(b) soft wood
(c) spring wood
(d) sap wood
Answer:
(a) porous wood

Question 11.
Which of the statement is not correct?
(a) In temperate regions, the cambium is very active in winter season.
(b) In temperate regions, the cambium is very active in spring season.
(c) In temperate regions, cambium is less active in winter season.
(d) In temperate regions early wood is formed in spring season.
Answer:
(a) In temperate regions, the cambium is very active in winter season.

Question 12.
Usually more distinct annual rings are formed:
(a) in tropical plants
(b) in seashore plants
(c) in temperate plants
(d) in desert plants
Answer:
(c) in temperate plants

Question 13.
False annual rings are formed due to:
(a) rain
(b) adverse natural calamities
(c) severe cold
(d) none of the above
Answer:
(b) adverse natural calamities

Question 14.
determination of the age of a tree by counting the annual rings is called:
(a) chronology
(b) dendrochronology
(c) palaeology
(d) histology
Answer:
(c) palaeology

Question 15.
The age of American sequoiadendron tree is about:
(a) 350 years
(b) 3,000 years
(c) 3400 years
(d) 3500 years
Answer:
(d) 3500 years

Question 16.
The wood of Acer plant has:
(a) ring porous
(b) diffuse porous
(c) central porous
(d) none of the above
Answer:
(b) diffuse porous

Question 17.
In fully developed tyloses:
(a) only starchy crystals are present
(b) resin and gums only are present
(c) oil and tannins are present
(d) starchy crystals, resins, gums, oils, tannins or colored substances are present
Answer:
(d) starchy crystals, resins, gums, oils, tannins or colored substances are present

Question 18.
In bombax:
(a) the sieve tubes are blocked by tylose like outgrowths
(b) the resin ducts are blocked by tylose like outgrowths
(c) the phloem tube is blocked by tylose like out growths
(d) none of the above
Answer:
(a) the sieve tubes are blocked by tylose like outgrowths

Question 19.
Which of the statement is not correct?
(a) Sap wood and heart wood can be distinguished in the secondary xylem
(b) Sap wood is paler in colour
(c) Heart wood is darker in colour
(d) The sap wood conducts minerals, while the heart wood conduct water
Answer:
(d) The sap wood conducts minerals, while the heart wood conduct water

Question 20.
Timber from heart wood is:
(a) more fragile and resistant to the attack of insects
(b) more durable and more resistant to the attack of micro organism and insects
(c) more hard and less resistant to the attack of micro organism
(d) less durable and more resistant to the attack of micro organism and insects
Answer:
(b) more durable and more resistant to the attack of micro organism and insects

Question 21.
The dye, haematoxylin is obtained from:
(a) the heart wood of haematoxylum campechianum
(b) the sap wood of haematoxylum campechianum
(c) cambium cells of haematoxylum campechianum
(d) the seeds of haematoxylum campechianum
Answer:
(a) the heart wood of haematoxylum campechianum

Question 22.
Canada balsam is produced from:
(a) Pisum sativum
(b) resin of Arjuna plant
(c) Abies balsamea
(d) the root of Vinca rosea
Answer:
(c) Abies balsamea

Question 23.
Some commercially important phloem or bast fibres are obtained from:
(a) banana
(b) bamboo
(c) vinca rosea
(d) cannabis sativa
Answer:
(d) cannabis sativa

Question 24.
Phellogen comprises:
(a) homogeneous sclerenchyma cells
(b)homogeneous meristamatic cells
(c) homogeneous collenchyma cells
(d) none of the above cells
Answer:
(b)homogeneous meristamatic cells

Question 25.
Phelloderm is otherwise called as:
(a) primary cortex
(b) cork wood
(c) secondary cortex
(d) rhytidome
Answer:
(c) secondary cortex

Question 26.
Lenticel is helpful in:
(a) transportation of food
(b) photosynthesis
(c) exchanges of gases and transpiration
(d) transportation of water
Answer:
(c) exchanges of gases and transpiration

Question 27.
The antimalarial compound quinine is, extracted from:
(a) seeds of cinchona
(b) bark of cinchona
(c) leaves of cinchona
(d) flowers of cinchona
Answer:
(b) bark of cinchona

Question 28.
Gum Arabic is obtained from:
(a) Hevea brasiliensis
(b) Acacia Senegal
(c) Pinus
(d) Dilonix regia
Answer:
(b) Acacia Senegal

Question 29.
Turpentine used as thinner of paints is obtained from:
(a) Acacia Senegal
(b) Vinca rosea
(c) Hevea brasiliensis
(d) Pinus
Answer:
(d) Pinus

Question 30.
Rubber is obtained from:
(a) Bombax mori
(b) Hevea brasiliensis
(c) Quercus suber
(d) Morus rubra
Answer:
(b) Hevea brasiliensis

II. Answer the following. (2 Marks)

Question 1.
Define primary growth?
Answer:
The roots and stems grow in length with the help of apical meristems. This is called primary growth or longitudinal growth.

Question 2.
Mention the two lateral meristem responsible for secondary growth.
Answer:
The secondary growth in dicots and gymnosperms is brought about by two lateral meristems.

1. Vascular Cambium and
2. Cork Cambium

Question 3.
What is meant by vascular cambium?
Answer:
The vascular cambium is the lateral meristem that produces the secondary vascular tissues. i.e., secondary xylem and secondary phloem.

Question 4.
Define intrafascicular or fascicular cambium?
Answer:
A strip of vascular cambium that is believed to originate from the procambium is present between xylem and phloem of the vascular bundle. This cambial strip is known as intrafascicular or fascicular cambium.

Question 5.
Define interfascicular cambium?
Answer:
In between the vascular bundles, a few parenchymatous cells of the medullary rays that are in line with the fascicular cambium become meristematic and form strips of vascular cambium. It is called interfascicular cambium.

Question 6.
What is vascular cambial ring?
Answer:
This interfascicular cambium joins with the intrafascicular cambium on both sides to form a continuous ring. It is called a vascular cambial ring.

Question 7.
What is meant by stratified cambium?
Answer:
If the fusiform initials are arranged in horizontal tiers, with the end of the cells of one tier appearing at approximately the same level, as seen in Tangential Longitudinal Section (TLS), it is called storied (stratified) cambium.

Question 8.
Explain non – stratified cambium.
Answer:
In plants with long fusiform initials, they strongly overlap at the ends, and this type of cambium is called non – storied (non – startified) cambium.

Question 9.
Give a brief note on ray initials.
Answer:
These are horizontally elongated cells. They give rise to the ray cells and form the elements of the radial system of secondary xylem and phloem.

Question 10.
How does secondary xylem or wood form?
Answer:
The secondary xylem, also called wood, is formed by a relatively complex meristem, the vascular cambium, consisting of vertically (axial) elongated fusiform initials and horizontally (radially) elongated ray initials.

Question 11.
What is meant by spring wood?
Answer:
In the spring season, cambium is very active and produces a large number of xylary elements having vessels / tracheids with wide lumen. The wood formed during this season is called spring wood or early wood.

Question 12.
How does the autumn wood form?
Answer:
In winter, the cambium is less active and forms fewer xylary elements that have narrow vessels /  tracheids and this wood is called autumn wood or late wood.

Question 13.
Define growth rings?
Answer:
The annual ring denotes the combination of early wood and late wood and the ring becomes evident to our eye due to the high density of late wood. Sometimes annual rings are called growth rings.

Question 14.
Define dendroclimatology?
Answer:
It is a branch of dendrochronology concerned with constructing records of past climates and climatic events by analysis of tree growth characteristics, especially growth rings.

Question 15.
Explain diffuse porous woods with an example.
Answer:
Diffuse porous woods are woods in which the vessels or pores are rather uniform in size and distribution throughout an annual ring. eg: Acer

Question 16.
What is meant by ring porous woods?
Answer:
The pores of the early wood are distinctly larger than those of the late wood. Thus rings of wide and narrow vessels occur.

Question 17.
Define tyloses?
Answer:
In many dicot plants, the lumen of the xylem vessels is blocked by many balloon like ingrowths from the neighbouring parenchymatous cells. These balloons like structure are called tyloses.

Question 18.
Mention two plants from which bast fibres are obtained.
Answer:
Two plants from which bast fibres are obtained:

1. Flax – Linum ustitaissimum
2. Hemp – Cannabis sativa

Question 19.
Define Rhytidome?
Answer:
Rhytidome is a technical term used for the outer dead bark which consists of periderm and isolated cortical or phloem tissues ? formed during successive secondary growth, eg: Quercus.

Question 20.
What is polyderm? Explain briefly.
Answer:
Polyderm is found in the roots and underground stems. eg: Rosaceae. It refers to a special type of protective tissues consisting of uniseriate suberized layer alternating with multiseriate nonsuberized cells in periderm.

Question 21.
Define’bark’?
Answer:
The term ‘bark’ is commonly applied to all the tissues outside the vascular cambium of stem (i.e., periderm, cortex, primary phloem and secondary phloem).

Question 22.
What are the functions of lenticel?
Answer:
Lenticel is helpful in exchange of gases and transpiration called lenticular transpiration.

Question 23.
Explain briefly phelloderm.
Answer:
It is a tissue resembling cortical living parenchyma produced centripetally (inward) from the phellogen as a part of the periderm of stems and roots in seed plants.

Question 24.
What is the function of secondary phloem?
Answer:
Secondary phloem is a living tissue that transports soluble organic compounds made during photosynthesis to various parts of plant.

Question 25.
what is periderm?
Answer:
Whenever stems and roots increase in thickness by secondary growth, the periderm, a protective tissue of secondary origin replaces the epidermis and Often primary cortex. The periderm consists of phellem, phellogen, and phelloderm.

III. Answer the following. (3 Marks)

Question 1.
Distinguish between primary and secondary growth.
Answer:
1. Primary growth: The plant organs originating from the apical meristems pass through a period of expansion in length and width. The roots and stems grow in length with the help of apical meristems. This is tailed primary growth or longitudinal growth.

2. Secondary growth: The gymnosperms and most angiosperms, including some monocots, show an increase in thickness of stems and roots by means of secondary growth or latitudinal growth.

Question 2.
Explain fusiform initials.
Answer:
These are vertically elongated cells. They give rise to the longitudinal or axial system of the secondary xylem (treachery elements, fibers, and axial parenchyma) and phloem (sieve elements, fibers, and axial parenchyma).

Question 3.
Explain briefly about false annual rings.
Answer:
Additional growth rings are developed within a year due to adverse natural calamities like drought, frost, defoliation, flood, mechanical injury and biotic factors during the middle of a growing season, which results in the formation of more than one annual ring. Such rings are called pseudo – or false – annual rings.

Question 4.
Write down the differences between spring wood and autumn wood.
Answer:
The differences between spring wood and autumn wood:

Spring wood or Early wood	Autumn wood or Late wood
1. The activity of cambium is faster.	1. Activity of cambium is slower.
2. Produces large number of xylem elements.	2. Produces fewer xylem elements.
3. Xylem vessels /  trachieds have wider lumen.	3. Xylem vessels / trachieds have narrow lumen.
4. Wood is lighter in colour and has lower density.	4. Wood is darker in colour and has a higher density.

Question 5.
How do you distinguish between sap wood and heart wood?
Answer:

Sap wood (Alburnum)	Heart wood (Duramen)
1. Living part of the wood.	1. Dead part of the wood.
2. It is situated on the outer side of wood.	2.It is situated in the certre part of wood.
3. It is less in coloured.	3. It is dark in coloured.
4. Very soft in nature.	4. Hard in nature.
Tyloses are absent.	Tyloses are present.
5. It is not durable and not resistant to microorganisms.	5. It is more durable and resists microorganisms.

Question 6.
What are fossil resins? Explain with an example.
Answer:
Plants secrete resins for their protective benefits. Amber is a fossilized tree resinespecially from the wood, which has been appreciated for its colour and natural beauty since neolithic times. Much valued from antiquity to the present as a gemstone, amber is made into a variety of decorative objects. Amber is used in jewellery. It has also been used as a healing agent in folk medicine.

Question 7.
Write briefly about Cork cambium.
Answer:
It is a secondary lateral meristem. It comprises homogenous meristematic cells unlike vascular cambium. It arises from epidermis, cortex, phloem or pericycle (extrastelar in origin). Its cells divide periclinally and produce radially arranged files of cells. The cells towards the outer side differentiate into phellem (cork) and those towards the inside as phelloderm (secondary cortex).

Question 8.
Explain the term lenticel.
Answer:
Lenticel is raised opening or pore on the epidermis or bark of stems and roots. It is formed during secondary growth in stems. When phellogen is more active in the region of lenticels, a mass of loosely arranged thin – walled parenchyma cells are formed. It is called complementary tissue or filling tissue. Lenticel is helpful in exchange of gases and transpiration called lenticular transpiration.

Question 9.
Mention the benefits of bark in a tree.
Answer:
Bark protects the plant from parasitic fungi and insects, prevents water loss by evaporation and guards against variations of external temperature. It is an insect repellent, decay proof, fireproof and is used in obtaining drugs or spices. The phloem cells of the bark are involved in conduction of food while secondary cortical cells involved in storage.

Question 10.
Distinguish between Intrafascicular Interfascicular cambium.
Answer:
Between Intrafascicular Interfascicular cambium:

Intrafascicular cambium	Interfascicular cambium
1. Present inside the vascular bundles	1. Present in between the vascular bundles.
2. Originates from the procambium.	2. Originates from the medullary rays.
3. Initially it forms a part of the primary meristem.	3. From the beginning it forms a part of the secondary meristem.

IV. Answer In detail
Question 1.
Describe the activity of vascular with the help of diagram.
Answer:
Activity of Vascular Cambium:
The vascular cambial ring, when active, cuts off new cells both towards the inner and outer side. The cells which are produced outward form secondary phloem and inward secondary xylem. At places, cambium forms some narrow horizontal bands of parenchyma which passes through secondary phloem and xylem. These are the rays. Due to the continued formation of secondary xylem and phloem through vascular cambial activity, both the primary xylem and phloem get gradually crushed.

Question 2.
Describe the formation of sap wood and heart wood with suitabie diagram.
Answer:
Sap wood and heart wood can be distinguished in the secondary xylem. In any tree the outer part of the wood, which is paler in colour, is called sap wood are alburnum. The centre part of the wood, which is darker in colour is called heart wood or duramen. The sap wood conducts water while the heart wood stops conducting water. As vessels of the heart wood are blocked by tyloses, water is not conducted through them.

Due to the presence of tyloses and their contents the heart wood becomes coloured, dead and the hardest part of the wood. From the economic point of view, generally the heartwood is more useful than the sapwood. The timber form the heartwood is more durable and more resistant to the attack of microorganisms and insects than the timber from sapwood.

Question 3.
Draw and label the transverse section of dicot stem showing the secondary growth.

Answer:
The transverse section of dicot stem showing the secondary growth:

Question 4.
Distinguish between Phellem and Phelloderm.
Answer:
Phellem (Cork):

1. It is formed on the outer side of phellogen.
2. Cells are compactly arranged in regular tires and rows without intercellular spaces.
3. Protective in function.
4. Consists of nonliving cells with suberized walls.
5. Lenticels are present.

Phelloderm (Secondary cortex):

1. It is formed on the inner side of phellogen.
2. Cells are loosely arranged with intercellular spaces.
3. As it contains chloroplast, it synthesises and stores food.
4. Consists of living cells, parenchymatous in nature and does not have suberin.
5. Lenticels are absent.

Question 5.
Write down the economic importance of tree bark.
Answer:
The economic importance of tree bark:

Question 5.
Draw the different stages of secondary growth in a dicot root and label the parts.
Answer:
Stages of secondary growth in a dicot root and label the parts:

Solution To Activity
Textbook Page No: 38
Question 1.
Generally monocots do not have secondary growth, but palms and bamboos have woody stems. Find the reason.
Answer:
Some of the monocots like palm and bamboos show an increase in thickness of stems by means of secondary growth or latitudinal growth.

Textbook Page No: 48
Question 2.
Be friendly with your environment (Eco friendly) Why should not we use the natural products which are made by plant fibres like rope, fancy bags, mobile pouch, mat and gunny bags etc., instead of using plastics or nylon?
Answer:
We should not use the natural products, which are made by plants fibres, because, if we use more of plant products the greedy people will exploit the plant resources for making plant products and thereby depleting the tree cover, which in turn causes reduction in rain fall.

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Samacheer Kalvi 11th Bio Zoology Body Fluids and Circulation Text Book Back Questions and Answers

I. Multiple Choice Questions
Question 1.
What is the function of lymph?
(a) Transport of O₂ into brain
(b) Transport of CO₂ into lungs
(c) Bring interstitial fluid in blood
(d) Bring RBC and WBC in lymph node
Answer:
(c) Bring interstitial fluid in blood

Question 2.
Which one of the following plasma proteins is involved in the coagulation of blood?
(a) Globulin
(b) Fibrinogen
(c) Albumin
(d) Serum amylase
Answer:
(b) Fibrinogen

Question 3.
Which of the following WBCs are found in more numbers?
(a) Eosinophil
(b) Neutrophil
(c) Basophil
(d) Monocyte
Answer:
(b) Neutrophil

Question 4.
Which of the following is not involved in blood clotting?
(a) Fibrin
(b) Calcium
(c) Platelets
(d) Bilirubin
Answer:
(d) Bilirubin

Question 5.
Lymph is colourless because ………….
(a) WBC are absent
(b) WBC are present
(c) Haemoglobin is absent
(d) RBC are absent
Answer:
(c) Haemoglobin is absent

Question 6.
Blood group is due to the presence or absence of surface
(a) Antigens on the surface of WBC
(b) Antibodies on the surface of RBC
(c) Antigens on the surface of RBC
(d) Antibodies on the surface of WBC
Answer:
(c) Antigens on the surface of RBC

Question 7.
A person having both antigen A and antigen B on the surface of RBCs belongs to blood group
(a) A
(b) B
(c) AB
(d) O
Answer:
(c) AB

Question 8.
Erythroblastosis foetalis is due to the destruction of …………..
(a) Foetal RBCs
(b) Foetus suffers from atherosclerosis
(c) Foetal WBCs
(d) Foetus suffers from mianmata
Answer:
(a) Foetal RBCs

Question 9.
Dub sound of heart is caused by
(a) Closure of atrio-ventricular valves
(b) Opening of semi-lunar valves
(c) Closure of semi-lunar valves
(d) Opening of atrio-ventricular valves
Answer:
(c) Closure of semi-lunar values

Question 10.
Why is the velocity of blood flow the lowest in the capillaries?
(a) The systemic capillaries are supplied by the left ventricle, which has a lower cardiac output than the right ventricle.
(b) Capillaries are far from the heart, and blood flow slows as distance from the heart increases.
(c) The total surface area of the capillaries is larger than the total surface area of the arterioles.
(d) The capillary walls are not thin enough to- allow oxygen to exchange with the cells.
(e) The diastolic blood pressure is too low to deliver blood to the capillaries at a high flow rate.
Answer:
(c) The total surface area of the capillaries is larger than the total surface area of the arterioles.

Question 11.
An unconscious patient is rushed into the emergency room and needs a fast blood transfusion. Because there is no time to check her medical history or determine her blood type, which type of blood should you as her doctor, give her?
(a) A⁺
(b) AB
(c) O⁺
(d) O^–
Answer:
(c) O⁺

Question 12.
Which of these functions could or could not be carried out by a red blood cell?
(a) Protein synthesis
(b) Cell division
(c) Lipid synthesis
(d) Active transport
Answer:
(a) Protein synthesis: RBCs do not have ribosomes which are important for protein synthesis, They are concerned with transport of respiratory gases alone. Hence protein synthesis
could not take place in RBCs.

(b) Cells division: RBCs do not have numbers. They are produced in the bone marrow. They do not involve in cell division.

(c) Lipid Synthesis: Lipid synthesis occurs in endoplasmic reticulum (ER) and golgi complex. The ER is absent in RBCs. Hence lipid synthesis does not take place in RBCs.

(d) Active transport: Transport of respiratory gases between the alveoli to the blood vessels, blood vessel to the cells and vice versa take place due to difference in the partial pressure of O₂ and CO₂., Active transport of materials against concentration gradient does not take place in RBCs.

Question 13.
At the venous end of the capillary bed, the osmotic pressure is …………….
(a) Greater than the hydrostatic pressure
(b) Result in net outflow of fluids
(c) Results in net absorption of fluids
(d) No change occurs
Answer:
(a) Greater than the hydrostatic pressure

Question 14.
A patient’s chart reveals that he has a cardiac output of 7500mL per minute and a stroke volume of 50 mL. What is his pulse rate (in beats / min)
(a) 50
(b) 100
(c) 150
(d) 400
Answer:
(c) 150

Question 15.
At any given time there is more blood in the venous system than that of the arterial system. Which of the following features of the veins allows this?
(a) relative lack of smooth muscles
(b) presence of valves
(c) proximity of the veins to lymphatic’s
(d) thin endothelial lining
Answer:
(a) relative lack of smooth muscles

II. Short Answer Questions

Question 16.
Distinguish between arteries and veins?
Answer:

Arteries	Veins
1. Arteries are the blood vessels that carry blood away from the heart.	1. Veins are the blood vessels that carry blood to the heart.
2. Arteries carry oxygenated blood except pulmonary artery.	2. Veins carry deoxygenated blood except pulmonary veins.
3. Arteries usually lie deep inside the body.	3. Veins are usually located superficially.
4.  These are thick walled.	4. These are thin walled.
5. These do not have valves.	5. These have semilunar valves.
6. Blood pressure is high.	6.  Blood pressure is low.

Question 17.
Distinguish between open and closed circulation?
Answer:

Open circulation	Closed circulation
1. Open circulation, haemolymph is pumped by the heart which flows through blood vessels into the haemocoel.	1.In closed circulation, blood is pumped by the heart and flows through blood vessels
2. It is seen in arthropods and most molluscs.	2. It is seen in annelids, cephalopods and vertebrates

Question 18.
Distinguish between mitral valve and semi lunar valve?
Answer:

Mitral valve	Semilunar vales
1. The valve present between the left atrium left ventricle is called mitral valve.	1. The valves present at the openings of right and left ventricles into the pulmonary artery and aorta are semilunar valves.
2. It is made of two flaps.	2. These are of three half moon shaped cusps.

Question 19.
Right ventricular wall is thinner than the left ventricular wall. Why?
Answer:
The right ventricle pumps deoxygenated blood, to the lungs through pulmonary artery. The left ventricle pumps the oxygenated blood to all parts of the body through the aorta. Hence, left ventricle has to exert more pressure. Hence right ventricular wall is thinner but the left ventricular walls is thicker.

Question 20.
What might be the effect on a person whose diet has less iron content?
Answer:
A person whose diet has less iron content will become anaemic. The haemoglobin content of the blood will be less. The volume of oxygen carried by RBCs gets reduced. He/she may experience tiredness, weakness, fatigue etc. In order to overcome this deficiency one has to take iron rich diet.

Question 21.
Describe the mechanism by which the human heart beat is initiated and controlled?
Answer:
The rhythmic contraction and expansion of heart is called heart beat. The contraction of the heart is called systole and the relaxation of the heart is called diastole. The human heart is myogenic. The pacemaker cells are located in the right sinoatrial (SA) node.

On the left side of the right atrium, there is a mode called auriculo ventricular node (AV). Two special cardiac muscle fibres which originate from the AV node are called the bundle of His. It runs down into the interventricular spectrum and the fibres spread into the ventricle as the Purkinje fibres.

The pacemaker cells produce excitation through depolarization of their cell membrane. Early depolarization is slow and takes place by sodium influx and reduction in potassium efflux. Minimum potential is required to activate voltage gated calcium (Ca⁺) channels that cause rapid depolarization which results in action potential. The pace maker cells repolarise slowly via K⁺ efflux.

Question 22.
What is lymph? Write its function?
Answer:
About 90% of fluid that leaks from capillaries eventually seeps back into the capillaries and the remaining 10% is collected and returned to blood system by me of a series of tubules known as lymph vessels or lymphatics.

The fluid inside the lymphatics is called lymph. The lymphatic system consists of a complex network of thin walled ducts (lymphatic vessels), filtering bodies (lymph nodes) and a large number of lymphocytic cell concentrations in various lymphoid organism.

The lymphatic vessels have smooth walls that run parallel to the blood vessels, in the skin, along the respiratory and digestive tracts. These vessels serve as return ducts for the fluids that are continually diffusing out of the blood capillaries into the body tissues.

Lymph fluid must pass through the lymph nodes before it is returned to the blood. The lymph nodes that filter the fluid from the lymphatic vessels of the skin are highly concentrated in the neck, inguinal, axillaries, respiratory and digestive tracts.

The lymph fluid flowing out of the lymph nodes flow into large collecting duct which finally drains into larger veins that runs beneath the collar bone, the subclavian vein and is emptied into the blood stream. The narrow passages in the lymph nodes are the sinusoids that are lined with macrophages.

The lymph nodes successfully prevent the invading microorganisms from reaching the blood stream. Cells found in the lymphatics are the lymphocytes. Lymphocytes collected in the lymphatic fluid are carried via the arterial blood and are recycled back to the lymph. Fats are absorbed through lymph in the lacteals present in the villi of the intestinal wall.

Question 23.
What are the heart sounds? When and how are these sounds produced?
Answer:
Rhythmic contraction and expansion of heart is called heart beat. The contraction of the heart is called systole and the relaxation of the heart is called diastole. The heart normally beats 70-72 times per minute in a human adult. During each cardiac cycle two sounds are produced that can be heard through a stethoscope.

The first heart sound (lub) is associated with the closure of the tricuspid and bicuspid valves whereas Second heart sound (dub) is associated with the closure of the semilunar valves. These sounds are of clinical diagnostic significance. An increased heart rate is called tachycardia and decreased heart rate is called bradycardia.

Question 24.
Select the correct biological term. Lymphocytes, red cells, leucocytes, plasma, erythrocytes, white cells, haemoglobin, phagocyte, platelets, blood clot?

Question (a)
Disc shaped cells which are concave on both sides?
Answer:
Red blood cells

Question (b)
Most of these have a large, bilobed nucleus?
Answer:
Leucocytes

Question (c)
Enable red cells to transport blood?
Answer:
Haemoglobin

Question (d)
The liquid part of the blood?
Answer:
plasma

Question (e)
Most of them move and change shape like an amoeba?
Answer:
phagocyte

Question (f)
Consists of water and important dissolved substances?
Answer:
plasma

Question (g)
Destroyed in the liver and spleen after circulating in the blood for four months?
Answer:
RBCs

Question (h)
The substances which gives red colour to their cells?
Answer:
haemoglobin

Question (i)
Another name for red blood cells?
Answer:
Erythrocytes

Question (j)
Blood that has been changed to a jelly?
Answer:
Blood clot

Question (k)
A word that me cell eater?
Answer:
Phagocyte

Question (l)
Cells without nucleus?
Answer:
Red blood cells

Question (m)
White cells made in the lymphatic tissue?
Answer:
Lymphocytes

Question (n)
Blocks wound and prevent excessive bleeding?
Answer:
Platelets

Question (o)
Fragment of cells which are made in the bone marrow?
Answer:
Erythrocytes

Question (p)
Another name for white blood cells?
Answer:
Leucocytes

Question (q)
Slowly releases oxygen to blood cells?
Answer:
Red cells

Question (r)
Their function is to help blood clot in wounds?
Answer:
Platelets

Question 25.
Select the correct biological term?
Answer:
Cardiac muscle, atria, tricuspid valve, systole, auricles, arteries, diastole, ventricles, bicuspid valve, pulmonary artery, cardiac cycle, semi lunar valve, veins, pulmonary vein, capillaries, vena cava, aorta?

Question (a)
The main artery of the blood?
Answer:
Aorta

Question (b)
Valves between the left atrium and ventricle?
Answer:
Bicuspid valve

Question (c)
Technical name for relaxation of the heart?
Answer:
Diastole

Question (d)
Another name for atria?
Answer:
Arteries

Question (e)
The main vein?
Answer:
Vena cava

Question (f)
Vessels which carry blood away from the heart?
Answer:
Arteries

Question (g)
Two names for the upper chambers of the heart?
Answer:
Atria

Question (h)
Thick walled chambers of the heart?
Answer:
Atria

Question (i)
Carries blood from the heart to the lungs?
Answer:
Pulmonary Artery

Question (j)
Takes about 0.8 sec to complete?
Answer:
Cardiac cycle

Question (k)
Valves situated at the point where blood flows out of the heart?
Answer:
Semilunar values

Question (l)
Vessels which carry blood towards the heart?
Answer:
Veins

Question (m)
Carries blood from the lungs to the heart?
Answer:
Pulmonary veins

Question (n)
The two lower chambers of the heart?
Answer:
Ventricles

Question (o)
Prevent blood from re-entering the ventricles after entering the aorta?
Answer:
Semilunar valves

Question (p)
Technical name for one heart beat?
Answer:
Cardiac cycle

Question (q)
Valves between right atrium and ventricles?
Answer:
Tricuspid valve

Question (r)
Technical name for contraction of the heart?
Answer:
Systole

Question (s)
Very narrow blood vessels?
Answer:
Capillaries

Question 26.
Name and label the given diagram to show A, B, C, D, E, F, and G?
(A) Aorta
(B) Pulmonary trunk
(C) Left pulmonary veins
(D) Blocking the action of vasoconstrictor lowers the blood pressure. Give reasons.
(E) What is the role of ACH inhibitor in reducing blood pressure?
(F) What conditions one might expect if the blood pressure is not controlled?
Answer:
(A) Aortic arch
(B) Left pulmonary artery
(C) Left pulmonary veins
(D) Pulmonary trunk
(E) Left ventricle
(F) Right ventricle
(G) Inferior vena cava

Samacheer Kalvi 11th Bio Zoology Body Fluids and Circulation Additional Questions & Answers

I. Multiple Choice Questions
Choose The Correct Answer

Question 1.
Which of the following is not the function of circulatory system?
(a) Transport of respiratory gases
(b) Carrying of digested food materials
(c) Transport of hormones to target organism
(d) Removal of nitrogenous wastes from the body
Answer:
(d) Removal of nitrogenous wastes from the body

Question 2.
Which is known as liquid connective tissue?
(a) Plasma
(b) Blood
(c) Serum
(d) lymph
Answer:
(b) blood

Question 3.
What is the function of albumin?
(a) Transport of hormones
(b) Blood clothing
(c) Maintenance of osmotic pressure
(d) Immunity
Answer:
(c) Maintenance of osmotic pressure

Question 4.
Fibrinogen is concerned with ………….
(a) Transport of ions
(b) Tranport of liquids
(c) Transport of hormones
(d) Coagulaltion of blood
Answer:
(d) Coagulation of blood

Question 5.
The red colour of the RBC is due to the presence of a respiratory pigment ……………
(a) Haemoerythrin
(b) Haemoglobin
(c) Haemocyanin
(d) Chlorocronin
Answer:
(b) Haemoglobin

Question 6.
Which of the following are non-nucleated cells?
(a) WBCs
(b) Nerve cell
(c) RBCs
(d) Muscle cell
Answer:
(c) RBCs

Question 7.
What is haematocrit/packed cells volume?
(a) The ratio of WBCs to blood plasma
(b) The ratio of RBCs to blood plasma
(c) The ratio of platelets to blood plasma
(d) The ratio of plasma and blood cells
Answer:
(b) The ratio of RBCs to blood plasma

Question 8.
Which of the following is abundant in blood?
(a) Neutrophils
(b) Eosinophils
(c) Basophils
(d) Lymphocytes
Answer:
(a) Neutrophils

Question 9.
…………… are the blood cells which have two lobes which are joined by thin strands.
(a) Neutrophils
(b) Basophils
(c) Eosinophils
(d) Lymphocytes
Answer:
(c) Eosinophils

Question 10.
What is the percentage of lymphocytes among WBCs?
(a) 0.5 to 1.0%
(b) 1-3%
(c) 65%
(d) 28%
Answer:
(d) 28%

Question 11.
The macrophages in the sinusoids of the liver are called ………….
(a) Microglia
(b) Kupffer cells
(c) Alveolar macrophages
(d) Lymphocytes
Answer:
(b) Kupffer cells

Question 12.
‘A’ blood group has ………………….. antigen and …………………. antibody
(a) A, anti B
(b) AB, no antibodies
(c) No antigen, anti A, Anti B
(d) B, Anti A
Answer:
(d) B, Anti A

Question 13.
Erythroblastosis foetalis is a condition of incompatibility related to ……………..
(a) Rh antigen and Rh antibodies
(b) Anti A and antigen B
(c) Anti B and antigen A
(d) Antigens A and B
Answer:
(a) Rh antigen and Rh antibodies

Question 14.
The conversion of prothrombin into thrombin occurs in the presence of ……………..
(a) Potassium and vitamin D
(b) Sodium and vitamin B,2
(c) Calcium and vitamin K
(d) Iodine and vitamin E
Answer:
(c) Calcium and vitamin K

Question 15.
………………….. is the exceptional artery which carries deoxygenated blood.
(a) Pulmonary artery
(b) Corotid artery
(c) Coronary artery
(d) Femoral artery
Answer:
(a) Pulmonary artery

Question 16.
Pulmonary veins carry ………………. blood from lungs to ……………….
(a) Oxygenated, right auricle
(b) Deoxygenated, right auricle
(c) Deoxygenated, left auricle
(d) Oxygenated, left auricle
Answer:
(c) Deoxygenated, left auricle

Question 17.
The blood vessels that supply blood to the cardiac muscles with all nutrients are
(a) Coronary arteries
(b) Cerebral arteries
(c) Aorta
(d) Pulmonary veins
Answer:
(a) Coronary arteries

Question 18.
The opening between the left atrium and left ventricle is guarded by …………….
(a) Semilunar valves
(b) Mitral valve
(c) Tricuspid valve
(d) Flaps
Answer:
(b) Mitral valve

Question 19.
The heart normally beats times …………………. per minute in a human adult.
(a) 60-62
(b) 50-52
(c) 70-72
(d) 90-92
Answer:
(c) 70-72

Question 20.
Which wave shape occurs from the start of depolarisation of the atria to the beginning of ventricular depolarisation?
(a) P wave
(b) ST segment
(c) QRS complex
(d) PQ interval
Answer:
(d) PQ interval

Question 21.
In systemic circulation, blood from the …………………. ventricle is carried by a network of arteries, arterioles and capillaries to the tissues.
(a) Deoxygenated right
(b) Oxygenated left
(c) Oxygenated right
(d) Deoxygenated left
Answer:
(b) Oxygenated, left

Question 22.
Which hormone increases the heart beat?
(a) Acetylcholine
(b) Gastrin
(d) Epinephrine
(d) Oxytocin
Answer:
(c) Epinephrine

Question 23.
Thrombus in a coronary artery results in ………….
(a) Heart attack
(b) Stroke
(c) Hypertension
(d) Heart failure
Answer:
(a) Heart attack

Question 24.
Cerebral infarction is called ……….
(a) Heart attack
(b) Hypertension
(c) Heart failure
(d) Stroke
Answer:
(d) Stroke

Question 25.
The failure of the heart to pump out the normal stroke volume is a condition called ………….
(a) Cerebral thrombosis
(b) Hypertension
(c) Myocardial infarction
(d) Rheumatoid heart disease
Answer:
(c) Myocardial infarction

Question 26.
In which condition the heart muscles do not get oxygen supply?
(a) Stroke
(b) Ischemic heart disease
(c) Hypertension
(d) Heart attack
Answer:
(b) Ischemic heart disease

Question 27.
Which of the following is the autoimmune disease that damages the heart?
(a) Ischemic heart disease
(b) Myocardial infarction
(c) Cerebral thrombosis
(d) Rheumatic fever
Answer:
(d) Rheumatic fever

Question 28.
The life saving procedure, CPR was first used by ……………
(a) William Harvey
(b) Carl Landsteiner
(c) James Elam and Peter Safar
(d) Raymond de Viessens
Answer:
(c) James Elam and Peter Safar

II. Fill in the Blanks

Question 1.
The tissue fluid that surrounds the cell is ………..
Answer:
Interstitial fluid.

Question 2.
The fluid component of the blood is …………
Answer:
Plasma

Question 3.
The blood flowing into the capillary from an arteriole has a high …………. pressure.
Answer:
Hydrostatic

Question 4.
………….. is the plasma protein that facilitates the transport of ions, hormones, lipids and assists in immune function.
Answer:
Globulin

Question 5.
………….. is the respiratory pigment that facilitates the transport of gases.
Answer:
Haemoglobin

Question 6.
The RBCs are destroyed in the liver and ………….
Answer:
Spleen

Question 7.
………….. is the hormone that helps in differentiation of stem cells of bone marrow into erythrocytes.
Answer:
Erythropoietin

Question 8.
The ratio of red blood cells to blood plasma is expressed is …………
Answer:
Haematocrit

Question 9.
The granulocytes have in the cytoplasm.
Answer:
Granules

Question 10.
Neutrophils are also called …………
Answer:
Heterophils/polymorpho nuclear cells

Question 11.
…………… have distinctly dilobed nucleus and the lobes are joined by thin strands.
Answer:
Eosinophils

Question 12.
Basophils secrete substances such as ……………… serotonin and histones.
Answer:
Heparin

Question 13.
The macrophages of the central nervous system are the ………….
Answer:
Microglia

Question 14.
Platelets are produced from ………….
Answer:
Megakaryocytes

Question 15.
Surface antigens of RBCs are called …………..
Answer:
Agglutinogens

Question 16.
The ………………… acting on agglutinogen B is called anti B.
Answer:
Agglutinin

Question 17.
The condition called erythroblastosis foetalis can be avoided by administration of anti D antibodies called …………..
Answer:
Rhocum.

Question 18.
………………….. helps in the conversion of fibrinogen to fibrin threads.
Answer:
Thrombin

Question 19.
The plasma without fibrinogen is called …………
Answer:
Serum

Question 20.
The fluid inside lymphatics is called ………….
Answer:
Lymph

Question 21.
Lymphocytes collected in the lymphatic fluid are carried via the arterial blood and are recycled back to the ………….
Answer:
Lymph

Question 22.
Fats are absorbed through lymph in the present in the villi of the intestinal wall.
Answer:
Lacteals

Question 23.
The middle layer of the artery is composed of smooth muscles and an extra cellular matrix which contains a protein ……………
Answer:
Elastin

Question 24.
The tunica adventitia of the artery is composed of ………………………. fibres.
Answer:
Collagen

Question 25.
The blood vessels that carry blood away from the heart are called …………..
Answer:
Arteries

Question 26.
All arteries carry oxygenated blood except …………..
Answer:
Pulmonary artery

Question 27.
……………………….. are small, narrow, and thin walled which are connected to the capillaries.
Answer:
Arterioles.

Question 28.
The …………………….. are the site for exchange of materials between blood and tissues.
Answer:
Capillaries

Question 29.
The unidirectional flow of blood in veins is due to the presence of that prevents back flow of blood.
Answer:
Semi lunar valves

Question 30.
Blood vessels that supply blood to the cardiac muscles are …………..
Answer:
Coronary arteries

Question 31.
…………………….. circulatory system is seen in Arthropoda and most molluscs.
Answer:
Open

Question 32.
The right atrium receives blood.
Answer:
Oxygenated.

Question 33.
The left atrium receives ………………….. blood.
Answer:
Deoxygenated

Question 34.
The crocodile has a ……………………. chambered heart.
Answer:
Four

Question 35.
…………………….. guards the opening between the left atrium and left ventricle.
Answer:
Bicuspid valve/mitral valve

Question 36.
The myocardium of the ventricle is thrown into irregular muscular ridges called ……………
Answer:
Trabeculae cornea

Question 37.
The heart wall is made up of outer epicardium, middle myocardium and the inner …………..
Answer:
Endocardium

Question 38.
The heart is covered by a double membrane called ……………
Answer:
Pericardium

Question 39.
On the left side of the right atrium there is a node called ………….
Answer:
Auriculo ventricular node

Question 40.
The rhythmic contraction and expansion of heart is called ……………
Answer:
Heart beat

Question 41.
The contraction of the chambers of the heart is called ……………
Answer:
Systole

Question 42.
The relaxation of the chambers of the heart is called …………..
Answer:
Diastole

Question 43.
The ‘lub’ sound is associated with the closure of the ………………….. valves.
Answer:
Tricuspid and bicuspid

Question 44.
The ‘dub’ sound is associated with the closure of ……………………… valves.
Answer:
Semilunar

Question 45.
An increased heart beat is called ………….
Answer:
Tachycardia

Question 46.
The decreased heart beat is called ………….
Answer:
Bradycardia

Question 47.
The phase I of the cardiac cycle is called …………..
Answer:
Ventricular diastole

Question 48.
The ventricular systole is the …………………….. phase of the cardiac cycle.
Answer:
Third

Question 49.
The amount of blood pumped out by each ventricle per minute is called …………
Answer:
Cardiac output

Question 50.
……………………. is the number of beats of the heart per minute.
Answer:
Heart rate/pulse rate

Question 51.
………………………. is the volume of blood pumped out by one ventricle with each beat.
Answer:
Stroke volume

Question 52.
If the right side of the heart fails, it results in …………………… congestion.
Answer:
Peripheral

Question 53.
Frank-Startling effect protects the heart from abnormal increase in …………..
Answer:
Blood volume

Question 54.
………………………. is the pressure exerted on the surface of blood vessels by the blood.
Answer:
Blood pressure

Question 55.
……………………. is the pressure in the arteries as the chambers of the heart contracts.
Answer:
Systolic pressure

Question 56.
……………………… is the pressure in the arteries when the heart chambers relax.
Answer:
Diastolic pressure

Question 57.
Blood pressure is measured using a ………………………… and a stethoscope.
Answer:
Sphygmomanometer

Question 58.
The decrease in blood pressure upon standing is known as ……………………….. hypertension.
Answer:
Orthostatic

Question 59.
Orthostatic reflex triggers baroreceptor reflex and increases the mean …………
Answer:
Arterial pressure

Question 60.
Circulation of the blood was first described by …………..
Answer:
William Harvey

Question 61.
In …………………….. circulation, the blood from heart is taken to the lungs by pulmonary artery and the oxygenated blood from the lungs is emptied into the left auricle by the pulmonary vein.
Answer:
Pulmonary

Question 62.
Vasopressin and ………………………… are involved in the regulation of the kidneys results in vasoconstriction.
Answer:
Angiotensin II

Question 63.
Coronary heart disease occurs when the arteries are lined by …………..
Answer:
Atheroma

Question 64.
Uncontrolled hypertension may damage the heart, brain and ………….
Answer:
Kidneys

Question 65.
The cholesterol rich atheroma forms …………………… in the inner lining of the arteries making them less elastic.
Answer:
Plaques

Question 66.
………………………. in a caronary artery results in heart attack.
Answer:
Thrombus

Question 67.
Brain haemorrhage is a condition known as ………….
Answer:
Stroke.

Question 68.
The condition in which the part of the brain tissue that is supplied by damaged artery dies due to lack of oxygen is …………….
Answer:
Cerebral infarction

Question 69.
Ischemic pain in the heart muscles is called ………….
Answer:
Angina pectoris

Question 70.
Atheroma may partially block the ……………………….. and reduce the blood supply to the heart.
Answer:
Coronary artery

Question 71.
The common sites of varicose veins are legs, rectal-anal regions, oesophagus and the …………
Answer:
Spermatic cord

Question 72.
The prime defect in heart failure is a decrease in cardiac muscle …………..
Answer:
Contractility

Question 73.
Prolonged angina leads to death of the heart muscle resulting in …………
Answer:
Heart failure

Question 74.
The death of the muscle fibres of the heart due to reduced blood supply to the heart muscle is called …………..
Answer:
Myocardial infarction

Question 75.
……………………… is an autoimmune disease which occurs 2-4 weeks after streptococcal throat infection.
Answer:
Rheumatic heart disease

Question 76.
………………….. me a brief electric shock given to the heart to recover the function of the heart.
Answer:
Defibrillation

III. Answer The Following Questions

Question 1.
What are the two types of body fluids?
Answer:
The intra-cellular fluid present inside the cells and the extracellular fluid present outside the cells are the two types of body fluids.

Question 2.
What are the three types of extra-cellular fluids?
Answer:
The three types of extra-cellular fluids are the interstitial fluid, the plasma and lymph.

Question 3.
Explain the composition of blood?
Answer:
Blood is the most common body fluid that transports substances from one part of the body to the other. Blood is a connective tissue consisting of plasma (fluid matrix) and formed elements.

The plasma constitutes 55% of the total blood volume. The remaining 45% is the formed elements that consist of blood cells. The average blood volume is about 5000 ml (5L) in an adult weighing 70 Kg.

Plasma:
Plasma mainly consists of water (80 – 92%) in which the plasma proteins, inorganic constituents (0.9%), organic constituents (0.1%) and respiratory gases are dissolved.

The four main types of plasma proteins synthesized in the liver are albumin, globulin, prothrombin and fibrinogen. Albumin maintains the osmotic pressure of the blood. Globulin facilitates the transport of ions, hormones, lipids and assists in immune function.

Both Prothrombin and Fibrinogen are involved in blood clotting. Organic constituents include urea, amino acids, glucose, fats and vitamins; and the inorganic constituents include chlorides, carbonates and phosphates of potassium, sodium, calcium and magnesium.

The composition of plasma is always constant Immediately after a. meal, the blood in the hepatic portal vein has a very high concentration of glucose as it is transporting glucose from the intestine to the liver where it is stored.

The concentration of the glucose in the blood gradually falls after sometime as most of the glucose is absorbed. If too much of protein is consumed, the body cannot store the excess amino acids formed from the digestion of proteins.

The liver breaks down the excess amino acids and produces urea. Blood in the hepatic vein has a high concentration of urea than the blood in other vessels namely, hepatic portal vein and hepatic artery.

Formed elements:
Red blood cells/corpuscles (erythrocytes), white blood cells/corpuscles (Leucocytes) and platelets are collectively called formed elements.

Red blood cells:
Red blood cells are abundant than the other blood cells. There are about 5 million to 5.5 millions of RBC mnr3 of blood in a healthy man and 4.5-5.0 millions of RBC mm ° in healthy women.

The RBCs are very small with the diameter of about 7 pm (micrometer). The structure of RBC is shown in Figure. The red colour of the RBC is due to the presence of a respiratory pigment, haemoglobin dissolved in the cytoplasm.

Flaemoglobin plays an important role in the transport of respiratory gases and facilitates the exchange of gases with the fluid outside the cell (tissue fluid). The biconcave shaped RBCs increases the surface area to volume ratio, hence oxygen diffuses quickly in and out of the cell.

The RBCs are devoid of nucleus, mitochondria, ribosomes and endoplasmic reticulum. The absence of these organelles accommodates more haemoglobin thereby maximising the oxygen carrying capacity of the cell.

The average life span of RBCs in a healthy individual is about 120 days after which they are destroyed in the spleen (graveyard/cemetery of RBCs) and the iron component returns to the bone marrow for reuse.

Erythropoietin is a hormone secreted by the kidneys in response to low oxygen and helps in differentiation of stem cells of the bone marrow’ to erythrocytes (erythropoiesis) in adults. The ratio of red blood cells to blood plasma is expressed as Haematocrit (packed cell volume).

White blood cells:
(leucocytes) are colourless, amoeboid, nucleated cells devoid of haemoglobin and other pigments. Approximately 6000 to 8000 per cubic mm of WBCs are seen in the blood of an average healthy individual.

Depending on the presence or absence of granules, WBCs are divided into two types, granulocytes and agranulocytes. Granulocytes are characterised by the presence of granules in the cytoplasm and are differentiated in the bone marrow. The granulocytes include neutrophils, eosinophils and basophils.

Neutrophils are also called heterophils or polymorphonuclear (cells with 3-4 lobes of nucleus connected with delicate threads) cells which constitute about 60%-65% of the total WBCs. They are phagocytic in nature and appear in large numbers in and around the infected tissues.

Eosinophils have distinctly bilobed nucleus and the lobes are joined by thin strands. They are non-phagocytic and constitute about 2-3% of the total WBCs. Eosinophils increase during certain types of parasitic infections and allergic reactions.

Basophils are less numerous than any other type of WBCs constituting 0.5%-1.0% of the total number of leucocytes. The cytoplasmic granules are large sized, but fewer than eosinophils.

Nucleus is large sized and constricted into several lobes but not joined by delicate threads. Basophils secrete substances such as heparin, serotonin and histamines. They are also involved in inflammatory reactions.

Agranulocvtes are characterised by the absence of granules in the cytoplasm and are differentiated in the lymph glands and spleen. These are of two types, lymphocytes and monocytes. Lymphocytes constitute 28% of WBCs. These have large round nucleus and small amount of cytoplasm.

The two types of lymphocytes are B and T cells. Both B and T cells are responsible for the immune responses of the body. B cells produce antibodies to neutralize the harmful effects of foreign substances and T cells are involved in cell mediated immunity.

Monocytes (Macrophages) are phagocytic cells that are similar to mast cells and have kidney shaped nucleus. They constitute 1-3% of the total WBCs. The macrophages of the central nervous system are the ‘microglia’, in the sinusoids of the liver they are called ‘Kupffer cells’ and in the pulmonary region they are the ‘alveolar macrophages’.

Platelets are also called thrombocytes that are produced from megakaryocytes (special cells in bone marrow) and lack nuclei. Blood normally contains 1,50,000 – 3,50,000 platelets mm^-3 of blood. They secrete substances involved in coagulation or clotting of blood. The reduction in platelet number can lead to clotting disorders that result in excessive loss of blood from the body.

Question 4.
Explain the ABO blood groups?
Answer:
Depending on the presence or absence of surface antigens on the RBCs, blood group in individual belongs to four different types namely, A, B, AB and O. The plasma of A, B and O individuals have natural antibodies (agglutinins) in them.

Surface antigens are called agglutinogens. The antibodies (agglutinin) acting on agglutinogen A is called anti A and the agglutinin acting on agglutinogen B is called anti B.

Agglutinogens are absent in O blood group. Agglutinogens A and B are present in AB blood group and do not contain anti A and anti B in them. A, B and O are major allelic genes in ABO systems.

All agglutinogens contain sucrose, D-galactose, N-acetyl glucosamine and 11 terminal amino acids. The attachments of the terminal amino acids are dependent on the gene products of A and B. The reaction is catalysed by glycosyl transferanse.

Question 5.
Tabulate the distribution of antigens and antibodies is different blood groups?
Answer:
Distribution of antigens and antibodies in different blood groups:

Question 6.
Explain the role of Rh factor?
Answer:
Rh factor is a protein (D antigen) present on the surface of the red blood cells in majority (80%) of hum. This protein is similar to the protein present in Rhesus monkey, hence the term Rh. Individuals who carry the antigen D on the surface of the red blood cells are Rh⁺ (Rh positive) and the individuals who do not carry antigen D, are Rh^– (Rh negative). Rh factor compatibility is also checked before blood transfusion.

When a pregnant women is Rh⁺ and the foetus is Rh⁺ incompatibility (mismatch) is observed. During the first pregnancy, the Rh antigens of the foetus does not get exposed to the mother’s blood as both their blood are separated by placenta. However, small amount of the foetal antigen becomes exposed to the mother’s blood during the birth of the first child.

The mother’s blood starts to synthesize D antibodies. But during subsequent pregnancies the Rh antibodies from the mother (Rh^–) enters the foetal circulation and destroys the foetal RBCs. This becomes fatal to the foetus because the child suffers from anaemia and jaundice. This condition is called erythroblastosis foetalis. This condition can be avoided by administration of anti D antibodies (Rhocum) to the mother immediately after the first child birth.

Question 7.
Explain the process of coagulation of blood?
Answer:
If you cut your finger or when you get yourself hurt, your wound bleeds for some time after which it stops to bleed. This is because the blood clots or coagulates in response to trauma.

The mechanism by which excessive blood loss is prevented by the formation of clot is called blood coagulation or clotting of blood. The clotting process begins when the endothelium of the blood vessel is damaged and the connective tissue in its wall is exposed to the blood.

Platelets adhere to collagen fibres in the connective tissue and release substances that form the platelet plug which provides emergency protection against blood loss.

Clotting factors released from the clumped platelets or damaged cells mix with clotting factors in the plasma. The protein called prothrombin is converted to its active form called thrombin in the presence of calcium and vitamin K.

Thrombin helps in the conversion of fibrinogen to fibrin threads. The threads of fibrins become interlinked into a patch that traps blood cell and seals the injured vessel until the wound is healed.

After sometime fibrin fibrils contract, squeezing out a straw-coloured fluid through a meshwork called serum (Plasma without fibrinogen is called serum). Heparin is an anticoagulant produced in small quantities by mast cells of connective tissue which prevents coagulation in small blood vessels.

Question 8.
Explain the structure of blood vessels?
Answer:
The vessels carrying the blood are of three types; they are the arteries, veins and capillaries. These vessels are hollow structures and have complex walls surrounding the lumen. The blood vessels in hum are composed of three layers, tunica intima, tunica media and tunica externa.

The inner layer, tunica intima or tunica interna supports the vascular endothelium, the middle layer, tunica media is composed of smooth muscles and an extra cellular matrix which contains a protein, elastin. The contraction and relaxation of the smooth muscles results in vasoconstriction and vasodilation. The outer layer, tunica externa or tunica adventitia is composed of collagen fibres.

Arteries:
The blood vessels that carry blood away from the heart are called arteries. The arteries usually lie deep inside the body. The walls of the arteries are thick, non-collapsible to withstand high pressure. Valves are absent and have a narrow lumen. All arteries carry oxygenated blood, except the pulmonary artery.

The largest artery, the aorta (2.5 cm in diameter and 2 mm thick) branch into smaller arteries and culminates into the tissues as feed arteries. In the tissues the arteries branches into arterioles. As blood enters an arteriole it may have a pressure of 85 mm Hg (11.3 KPa) but as it leaves and flows into the capillary, the pressure drops to 35 mm Hg (4.7 KPa). (Note 1 mm Hg = 0.13 KPa.

SI unit of mm Hg is KiloPascal (KPa)). Arterioles are small, narrow, and thin walled which are connected to the capillaries. A small sphincter lies at the junction between the arterioles and capillaries to regulate the blood supply. Arteries do not always branch into arterioles, they can also form anastomoses.

Capillaries:
Capillary beds are made up of fine networks of capillaries. The capillaries are thin walled and consist of single layer of squamous epithelium. Tunica media and elastin fibres are absent. The capillary beds are the site for exchange of materials between blood and tissues.

The walls of the capillaries are guarded by semilunar valves. The blood volume in the capillaries is high but the flow of blood is slow. Mixed blood (oxygenated and deoxygenated) is present in the capillaries. The capillary bed may be flooded with blood or may be completely bypassed depending on the body conditions in a particular organ.

Veins:
Veins have thinner walls and a larger lumen and hence can be easily stretched. They carry deoxygenated blood except, the pulmonary vein. The blood pressure is low and the lumen has a wide wall which is collapsible.

Tunica media is thinner in veins than in arteries. Unidirectional flow of blood in veins is due to the presence of semilunar valves that prevents backflow of blood. Blood samples are usually taken from the veins rather than artery because of low pressure in the veins.

Question 9.
Write a short note on coronary blood vessels?
Answer:
Blood vessels that supply blood to the cardiac muscles with all nutrients and removes wastes are the coronary arteries and veins. Heart muscle is supplied by two arteries namely right and left coronary arteries.

These arteries are the first branch of the aorta. Arteries usually surround the heart in the manner of a crown, hence called coronary artery (L. Corona – crown). Right ventricle and posterior portion of left ventricle are supplied by the right coronary artery. Anterior and lateral part of the left ventricle is supplied by the left coronary arteries.

Question 10.
Compare the chambers of heart and the methods of circulation in fishes, amphibians, reptiles, crocodiles, birds and mammals?
Answer:
All vertebrates have muscular chambered heart. Fishes have two chambered heart. The heart in fishes consists of sinus venosus, an atrium, one ventricle and bulbus arteriosus or conus arteriosus.

Single circulation is seen in fishes. Amphibian have two auricles and one ventricle and no inter ventricular septum whereas reptiles except crocodiles have two auricles and one ventricle and an incomplete inter ventricular septum.

Thus mixing of oxygenated and deoxygenated blood takes place in the ventricles. This type of circulation is called incomplete double circulation. The left atrium receives oxygenated blood and the right atrium receives deoxygenated blood. Pulmonary and systemic circuits are seen in Amphibianand Reptiles.

The Crocodiles, Birds and Mammals have two auricles or atrial chambers and two ventricles, the auricles and ventricles are separated by inter auricular septum and inter ventricular septum. Hence there is complete separation of oxygenated blood from the deoxygenated blood. Pulmonary and systemic circuits are evident. This type of circulation is called complete double circulation.

Question 11.
Explain the structure of human heart?
Answer:
The structure of the heart was described by Raymond de Viessens, in 1706. Human heart is made of special type of muscle called the cardiac muscle. It is situated in the thoracic cavity and its apex portion is slightly tilted towards left. It weighs about 300g in an adult.

The size of our heart is roughly equal to a closed fist. Heart is divided into four chambers, upper two small auricles or atrium and lower two large ventricles.

The walls of the ventricles are thicker than the auricles due to the presence of papillary muscles. The heart wall is made up of three layers, the outer epicardium, middle myocardium and inner endocardium. The space present between the membranes is called pericardial space and is filled with pericardial fluid.

The two auricles are separated by inter auricular septum and the two ventricles are separated by inter ventricular septum. The separation of chambers avoids mixing of oxygenated and deoxygenated blood. The auricle communicates with the ventricle through an opening called auriculo ventricular aperture which is guarded by the auriculo ventricular valves.

The opening between the right atrium and the right ventricle is guarded by the tricuspid valve (three flaps or cusps), whereas a bicuspid (two flaps or cusps) or mitral valve guards the opening between the left atrium and left ventricle. The valves of the heart allows the blood to flow only in one direction, i.e., from the atria to the ventricles and from the ventricles to the pulmonary’ artery or the aorta. These valves prevent backward flow of blood.

The opening of right and left ventricles into the pulmonary artery and aorta are guarded by aortic and pulmonary valves and are called semilunar valves. Each semilunar valve is made of three half-moon shaped cusps. The myocardium of the ventricle is thrown into irregular muscular ridges called trabeculae comeae. The trabeculae comeae are modified into chordae tendinae. The opening and closing of the semilunar valves are achieved by the chordae tendinae.

The chordae tendinae are attached to the lower end of the heart by papillary muscles. Heart receives deoxygenated blood from various parts of the body through the inferior venacava and superior venacava which open into the right auricle. Oxygenated blood from lungs is drained into the left auricle through four pulmonary veins.

Question 12.
Explain the cardiac cycle?
Answer:
The events that occur at the beginning of heart beat and lasts until the beginning of next beat is called cardiac cycle. It lasts for 0.8 seconds. The series of events that takes place in a cardiac cycle.

PHASE 1:
Ventricular diastole- The pressure in the auricles increases than that of the ventricular pressure. AV valves are open while the semi lunar valves are closed. Blood flows from the auricles into the ventricles passively.

PHASE 2:
Atrial systole – The atria contracts while the ventricles are still relaxed. The contraction of the auricles pushes maximum volume of blood to the ventricles until they reach the end diastolic volume (EDV). EDV is related to the length of the cardiac muscle fibre. More the muscle is stretched, greater the EDV and the stroke volume.

PHASE 3:
Ventricular systole (isovolumetric contraction) – The ventricular contraction forces the AV valves to close and increases the pressure inside the ventricles. The blood is then pumped from the ventricles into the aorta without change in the size of the muscle fibre length and ventricular chamber volume (isovolumetric contraction).

PHASE 4:
Ventricular systole (ventricular ejection) – Increased ventricular pressure forces the semilunar valves to open and blood is ejected out of the ventricles without backflow of blood. This point is the end of systolic volume (ESV).

PHASE 5:
(Ventricular diastole) -The ventricles begins to relax, pressure in the arteries exceeds ventricular pressure, resulting in the closure of the semilunar valves. The heart returns to phase 1 of the cardiac cycle.

Question 13.
Explain cardiac output in man?
Answer:
The amount of blood pumped out by each ventricle per minute is called cardiac output (CO). It is a product of heart rate (HR) and stroke volume (SV). Heart rate or pulse is the number of beats per minute. Pulse pressure = systolic pressure – diastolic pressure. Stroke volume (SV) is the volume of blood pumped out by one ventricle with each beat. SV depends on ventricular contraction.

CO = HR x SV. SV represents the difference between EDV (amount of blood that collects in a ventricle during diastole) and ESV (volume of blood remaining in the ventricle after contraction). SV = EDV – ESV. According to Frank – Starling law of the heart, the critical factor controlling SV is the degree to which the cardiac muscle cells are stretched just before they contract.

The most important factor stretching cardiac muscle is the amount of blood returning to the heart and distending its ventricles, venous return.
During vigorous exercise, SV may double as a result of venous return.

Heart’s pumping action normally maintains a balance between cardiac output and venous return. Because the heart is a double pump, each side can fail independently of the other. If the left side of the heart fails, it results in pulmonary congestion and if the right side fails, it results in peripheral congestion. Frank – Starling effect protects the heart from abnormal increase in blood volume.

Question 14.
Explain the importance of blood pressure?
Answer:
Blood pressure is the pressure exerted on the surface of blood vessels by the blood. This pressure circulates the blood through arteries, veins and capillaries.

There are two types of pressure, the systolic pressure and the diastolic pressure. Systolic pressure is the pressure in the arteries as the chambers of the heart contracts. Diastolic pressure is the pressure in the arteries when the heart chambers relax.

Blood pressure is measured using a sphygmomanometer (BP apparatus). It is expressed as systolic pressure / diastolic pressure. Normal blood pressure in man is about 120/80mm Hg. Mean arterial pressure is a function of cardiac output and resistance in the arterioles. The primary reflex pathway for homeostatic control of mean arterial pressure is the baroreceptor reflex.

The baroreceptor reflex functions every morning when you get out of bed. When you are lying flat the gravitational force is evenly distributed. When you stand up, gravity causes blood to pool in the lower extremities.

The decrease in blood pressure upon standing is known as orthostatic hypotension. Orthostatic reflex normally triggers baroreceptor reflex. This results in increased cardiac output and increased peripheral resistance which together increase the mean arterial pressure.

Question 15.
Explain the recording of electrocardiogram?
Answer:
An electrocardiogram (ECG) records the electrical activity of the heart over a period of time using electrodes placed on the skin, arms, legs and chest. It records the changes in electrical potential across the heart during one cardiac cycle. The special flap of muscle which initiates the heart beat is called as sinu-auricular node or SA node in the right atrium.

It spreads as a wave of contraction in the heart. The waves of the ECG are due to depolarization and not due to contraction of the heart. This wave of depolarisation occurs before the beginning of contraction of the cardiac muscle. A normal ECG shows 3 waves designated as P wave, QRS complex and T wave.

P Wave (atrial depolarisation): It is a small upward wave and indicates the depolarisation of the atria. This is the time taken for the excitation to spread through atria from SA node. Contraction of both atria lasts for around 0.8-1.0 sec.

PQ Interval (AV node delay): It is the onset of P wave to the onset of QRS complex. This is from the start of depolarisation of the atria to the beginning of ventricular depolarisation. It is the time taken for the impulse to travel from the atria to the ventricles (0.12-0.21 sec). It is the measure of AV conduction time.

QRS Complex: (ventricular depolarisation) No separate wave for atrial depolarisation in the ECG is visible. Atrial depolarisation occurs simultaneously with the ventricular depolarisation.

The normal QRS complex lasts for 0.06-0.09 sec. QRS complex is shorter than the P wave, because depolarisation spreads through the Purkinjie fibres. Prolonged QRS wave indicates delayed conduction through the ventricle, often caused due to ventricular hypertrophy or due to a block in the branches of the bundle of His.

repolarisation. In the heart muscle, the prolonged depolarisation is due to retardation of K+ efflux and is responsible for the plateau.

The ST segment lasts for 0.09 sec. T wave (ventricular depolarisation): It represents ventricular depolarisation. The duration of the T wave is longer than QRS complex because repolarisation takes place simultaneously throughout the ventricular depolarisation.

Question 16.
Explain the regulation of cardiac activity?
Answer:
The type of heart in human is myogenic because the heart beat originates from the muscles of the heart. The nervous and endocrine systems work together with paracrine signals (metabolic activity) to influence the diameter of the arterioles and alter the blood flow.

The neuronal control is achieved through autonomic nervous system (sympathetic and parasympathetic). Sympathetic neurons release nor-epinephrine and adrenal medulla releases epinephrine.

The two hormones bind to p – adrenergic receptors and increase the heart rate. The parasympathetic neurons secrete acetylcholine that binds to muscarinic receptors and decreases the heart beat.

Vasopressin and angiotensin II, involved in the regulation of the kidneys, results in vasoconstriction while natriuretic peptide promotes vasodilation. Vagus nerve is a parasympathetic nerve that supplies the atrium especially the SA and the AV nodes.

Question 17.
What is hypertension?
Answer:
Hypertension is the most common circulatory disease. The normal blood pressure in man is 120/80 mmHg. In cases when the diastolic pressure exceeds 90 mm Hg and the systolic pressure exceeds 150 mm Hg persistently, the condition is called hypertension. Uncontrolled hypertension may damage the heart, brain and kidneys.

Question 18.
Explain the disorder of coronary heart disease?
Answer:
Coronary heart disease occurs when the arteries are lined by atheroma. The build-up of atheroma contains cholesterol, fibres, dead muscle and platelets and is termed Atherosclerosis.

The cholesterol rich atheroma forms plaques in the inner lining of the arteries making them less elastic and reduces the blood flow. Plaque grows within the artery and tends to form blood clots, forming coronary thrombus. Thrombus in a coronary artery results in heart attack.

Question 19.
Explain the disorder stroke?
Answer:
Stroke is a condition when the blood vessels in the brain bursts (Brain haemorrhage) or, when there is a block in the artery that supplies the brain, (atherosclerosis) or thrombus. The part of the brain tissue that is supplied by this damaged artery dies due to lack of oxygen (cerebral infarction).

Angina pectoris (ischemic pain in the heart muscles) is experienced during early stages of coronary heart disease. Atheroma may partially block the coronary artery and reduce the blood supply to the heart. As a result, there is tightness or choking with difficulty in breathing. This leads to angina or chest pain. Usually it lasts for a short duration of time.

Question 20.
Explain the disorder of myocardial infarction?
Answer:
The prime defect in heart failure is a decrease in cardiac muscle contractility. The Frank – Starling curve shifts downwards and towards the right such that for a given EDV, a failing heart pumps out a smaller stroke volume than a normal healthy heart. When the blood supply to the heart muscle or myocardium is remarkably reduced it leads to death of the muscle fibres.

This condition is called heart attack or myocardial infarction. The blood clot or thrombosis blocks the blood supply to the heart and weakens the muscle fibres. It is also called Ischemic heart disease due to lack of oxygen supply to the heart muscles. If this persists it leads to chest pain or angina. Prolonged angina leads to death of the heart muscle resulting in heart failure.

Question 21.
Explain the disorder of rheumatoid heart disease?
Answer:
Rheumatic fever is an autoimmune disease which occurs 2-4 weeks after throat infection usually a streptococcal infection. The antibodies developed to combat the infection cause damage to the heart. Effects include fibrous nodules on the mitral valve, fibrosis of the connective tissue and accumulation of fluid in the pericardial cavity.

Question 22.
Explain Cardio Pulmonary Resuscitation (CPR)?
Answer:
In 1956, James Elam and Peter Safar were the first to use mouth to mouth resuscitation. CPR is a life saving procedure that is done at the time of emergency conditions such as when a person’s breath or heart beat has stopped abruptly in case of drowning, electric shock or heart attack.

CPR includes rescue of breath, which is achieved by mouth to mouth breathing, to deliver oxygen to the victim’s lungs by external chest compressions which helps to circulate blood to the vital organiser.

CPR must be performed within 4 to 6 minutes after cessation of breath to prevent brain damage or death. Along with CPR, defibrillation is also done. Defibrillation me a brief electric shock is given to the heart to recover the function of the heart.

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Samacheer Kalvi 11th Bio Botany Tissue and Tissue System Text Book Back Questions and Answers

Question 1.
Refer to the given figure and select the correct statement:

(i) A, B, and C are histogen of shoot apex
(ii) A Gives rise to medullary rays
(iii) B Gives rise to cortex
(iv) C Gives rise to epidermis
(a) (i) and (ii) only
(b) (ii) and (in) only
(c) (i) and (iii) only
(d) (iii) and (iv) only
Answer:
(c) (i) and (iii) only

Question 2.
Read the following sentences and identify the correctly matched sentences.
(i) In exarch condition, the protoxylem lies outside of metaxylem.
(ii) In endarch condition, the protoxylem lie towards the centre.
(iii) In centarch condition, metaxylem lies in the middle of the protoxylem.
(iv) In mesarch condition, protoxylem lies in the middle of the metaxylem.
(a) (i), (ii) and (iii) only
(b) (ii), (iii) and (iv) only
(c) (i), (ii) and (iv) only
(d) All of these
Answer:
(c) (i), (ii) and (iv) only

Question 3.
In Gymnosperms, the activity of sieve tubes are controlled by:
(a) Nearby sieve tube members.
(b) Phloem parenchyma cells.
(c) Nucleus of companion cells.
(d) Nucleus of albuminous cells.
Answer:
(c) Nucleus of companion cells.

Question 4.
When a leaf trace extends from a vascular bundle in a dicot stem, what would be the arrangement of vascular tissues in the veins of the leaf?
(a) Xylem would be on top and the phloem on the bottom
(b) Phloem would be on top and the xylem on the bottom
(c) Xylem would encircle the phloem
(d) Phloem would encircle the xylem
Answer:
(a) Xylem would be on top and the phloem on the bottom

Question 5.
Grafting is successful in dicots but not in monocots because the dicots have:
(a) vascular bundles arranged in a ring
(b) cambium for secondary growth
(c) vessels with elements arranged end to end
(d) cork cambium
Answer:
(b) cambium for secondary growth

Question 6.
Why the cells of sclerenchyma and tracheids become dead?
Answer:
The cells of sclerenchyma and tracheids become dead because they lack protoplasm.

Question 7.
Explain sclereids with their types.
Answer:
Sclereids are dead cells, usually these are isodiametric but some are elongated too. The cell wall is very thick due to lignification. Lumen is very much reduced. The pits may simple or branched. Sclereids are mechanical in function. They give hard texture to the seed coats, endosperms etc., Sclereids are classified into the following types.

1. Branchysclereids or Stone cells: Isodiametric sclereids, with hard cell wall. It is found in bark, pith cortex, hard endosperm and fleshy portion of some fruits. eg: Pulp of Pyrus.
2. Macrosclereids: Elongated and rod shaped cells, found in the outer seed coat of leguminous plants. eg: Crotalaria and Pisum sativum.
3. Osteosclereids (Bone cells): Rod shaped with dilated ends. They occur in leaves and seed coats. eg: seed coat of Pisum and Hakea.
4. Astrosclereids: Star cells with lobes or arms diverging form a central body. They occur in petioles and leaves. eg: Tea, Nymphae and Trochodendron.
5. Trichosclereids: Hair like thin walled sclereids. Numerous small angular crystals are embedded in the wall of these sclereids, present in stems and leaves of hydrophytes. eg: Nymphaea leaf and Aerial roots of Monstera

Question 8.
What are sieve tubes? Explain.
Answer:
Sieve tubes are long tube like conducting elements in the phloem. These are formed from a series of cells called sieve tube elements. The sieve tube elements are arranged one above the other and form vertical sieve tube. The end wall contains a number of pores and it looks like a sieve. So it is called as sieve plate. The sieve elements show nacreous thickenings on their lateral walls. They may possess simple or compound sieve plates.

The function of sieve tubes are believed to be controlled by campanion cells In mature sieve tube, Nucleus is absent. It contains a lining layer of cytoplasm. A special protein (P. Protein = Phloem Protein) called slime body is seen in it. In mature sieve tubes, the pores in the sieve plate are blocked by a substance called callose (callose plug). The conduction of food material takes lace through cytoplasmic strands. Sieve tubes occur only in Angiosperms.

Question 9.
Distinguish the anatomy of dicot root from monocot root.
Answer:
The anatomy of dicot root from monocot root:

Question 10.
Distinguish the anatomy of dicot stem from monocot stem.
Answer:
The anatomy of dicot stem from monocot stem:

Samacheer Kalvi 11th Bio Botany Tissue and Tissue System Other Important Questions & Answers

I. Choose the correct answer. (1 Mark)
Question 1.
Who is the father of plant anatomy?
(a) David Muller
(b) Katherine Esau
(c) Nehemiah Grew
(d) Hofmeister
Answer:
(c) Nehemiah Grew

Question 2.
The study of internal structure and organisation of plant is called:
(a) plant taxonomy
(b) plant anatomy
(c) plant physiology
(d) plant ecology
Answer:
(b) plant anatomy

Question 3.
The book “Anatomy of seed plants” is written by:
(a) Hanstein
(b) Schmidt
(c) Nicholsen
(d) Katherine Esau
Answer:
(d) Katherine Esau

Question 4.
The term meristem is coined by:
(a) Nageli
(b) Robert
(c) Stevers
(d) Clowes
Answer:
(a) Nageli

Question 5.
Which of the statement is not correct?
(a) Meristematic cells are self perpetuating
(b) Meristematic cells are most actively dividing cells
(c) Meristematic cells have large vacuoles
(d) Meristematic cells have dense cytoplasm with prominent nucleus
Answer:
(c) Meristematic cells have large vacuoles

Question 6.
Apical cell theory is proposed by:
(a) David brown
(b) Hofmeister
(c) Land mark
(d) Clowes
Answer:
(b) Hofmeister

Question 7.
The tunica is:
(a) the peripheral zone of shoot apex, that forms cortex
(b) the inner zone of shoot apex, that forms stele
(c) the peripheral zone of shoot apex, that forms epidermis
(d) the inner zone of shoot apex, that forms cortex and stele
Answer:
(c) the peripheral zone of shoot apex, that forms epidermis

Question 8.
Which of the histogens gives rise to root cap?
(a) Plerome
(b) Periblem
(c) Dermatogen
(d) Calyptrogen
Answer:
(d) Calyptrogen

Question 9.
Quiescent centre concept was proposed by:
(a) Lindall
(b) Clowes
(c) Holstein
(d) Sanio
Answer:
(b) Clowes

Question 10.
Parenchyma cells which stores resin, tannins, calcium carbonate and calcium oxalate are termed as:
(a) critoblast
(b) chromoblasts
(c) idioblasts
(d) astroblasts
Answer:
(c) idioblasts

Question 11.
Petioles of banana is composed of:
(a) storage parenchyma
(b) stellate parenchyma
(c) angular collenchyma
(d) prosenchyma
Answer:
(b) stellate parenchyma

Question 12.
Which of the following statement is not correct?
(a) Sclerenchyma is a dead cell
(b) It lacks protoplasm
(c) The cell walls of these cells are uniformly thickened
(d) Sclerenchyma are actively dividing cells
Answer:
(d) Sclerenchyma are actively dividing cells

Question 13.
The seed coat of ground nut is made up of:
(a) stone cells
(b) osteosclereids
(c) macrosclereids
(d) parenchyma cells
Answer:
(b) osteosclereids

Question 14.
Plant fibers are modified:
(a) sclerenchyma cells
(b) collenchyma cells
(c) parenchyma cells
(d) none of the above
Answer:
(a) sclerenchyma cells

Question 15.
The term xylem was introduced by:
(a) Alexander
(b) Nageli
(c) Holstein
(d) Schemidt
Answer:
(b) Nageli

Question 16.
What type of xylem arrangement is seen in Selaginella sp?
(a) Endarch
(b) Exarch
(c) Centrarch
(d) Mesarch
Answer:
(c) Centrarch

Question 17.
In cross section, the tracheids are:
(a) hexagonal in shape
(b) rectangular in shape
(c) triangular in shape
(d) polygonal in shape
Answer:
(d) polygonal in shape

Question 18.
In grasses the guard cells in stoma are:
(a) bean shaped
(b) irregular shaped
(c) dumbbell shaped
(d) bell shaped
Answer:
(c) dumbbell shaped

Question 19.
Bulliform cells are present in:
(a) mango
(b) grasses
(c) ground nut
(d) potato
Answer:
(b) grasses

Question 20.
The sunken stomata:
(a) reduce water loss by transpiration
(b) increase water loss by transpiration
(c) increase heat loss by evaporation
(d) neither reduce nor increase water loss by transpiration
Answer:
(a) reduce water loss by transpiration

Question 21.
In Ocimum the trichomes are:
(a) non – glandular
(b) fibrous
(c) glandular
(d) none of these
Answer:
(c) glandular

Question 22.
In dicot stem, the hypodermis is generally:
(a) parenchymatous
(b) sclerenchymatous
(c) collenchymatous
(d) none of these
Answer:
(c) collenchymatous

Question 23.
Casparian strips contain thickenings of:
(a) calcium carbonate and calcium oxalate
(b) carbohydrate, protein and lignin
(c) crystal of calcium oxalate
(d) lignin, suberin and some other carbohydrates
Answer:
(d) lignin, suberin and some other carbohydrates

Question 24.
Indicate the correct statement:
(a) Albuminous cells in gymnosperms are a nucleated parenchyma cells.
(b) Albuminous cells in gymnosperms are nucleated collenchyma cells.
(c) Albuminous cells in gymnosperms are nucleated, thin walled parenchyma cells.
(d) Albuminous cells in gymnosperms are a nucleated sclerenchyma cells.
Answer:
(c) Albuminous cells in gymnosperms are nucleated, thin walled parenchyma cells.

Question 25.
Secondary phloem is derived from:
(a) apical meritesm
(b) vascular cambium
(c) primary phloem
(d) none of the above
Answer:
(b) vascular cambium

Question 26.
Which of the following statement is not correct?
(a) The outer most layer of the root is called piliferous layer.
(b) The chief function of piliferous layer is protection.
(c) Piliferous layer is made up of parenchyma cells with intracellular space.
(d) Piliferous layer is made up of parenchyma cells without intracellular space.
Answer:
(d) Piliferous layer is made up of parenchyma cells without intracellular space.

Question 27.
In beans, the metaxylem vessels are generally:
(a) polygonal in shape
(b) circular in shape
(d) rectangular in shape
(d) triangular in shape
Answer:
(a) polygonal in shape

Question 28.
Who discovered the Annular collenchyma?
(a) Clowes
(b) Sanio
(c) Nageli
(d) Duchaigne
Answer:
(d) Duchaigne

Question 29.
The main function of xylem is:
(a) to conduct the minerals to various parts of plants
(b) to conduct oxygen to various parts of plant body
(c) to conduct water and minerals from root to the other parts of the plant body
(d) to conduct stored food to various parts of plant body
Answer:
(c) to conduct water and minerals from root to the other parts of the plant body

Question 30.
In maize the vascular bundles are:
(a) scattered
(b) concentric
(c) excentric
(d) radial
Answer:
(a) scattered

Question 31.
stomata in leaves of a plant are used for:
(a) transpiration
(b) transpiration and gas exchange
(c) gas exchange
(d) none of the above
Answer:
(b) transpiration and gas exchange

Question 32.
Which of the statement is not correct?
(a) Palisade parenchyma cells are seen beneath the upper epidermis
(b) Palisade parenchyma cells contain more chloroplasts
(c) Palisade parenchyma cells are irregularly shaped
(d) The function of palisade parenchyma is photosynthesis
Answer:
(c) Palisade parenchyma cells are irregularly shaped

Question 33.
Spongy parenchyma cells are:
(a) irregularly shaped
(b) elongated cylindrical cells
(c) very lightly arranged cells
(d) with more number of chloroplasts than palisade parenchyma
Answer:
(a) irregularly shaped

Question 34.
The main function of spongy parenchyma is:
(a) photosynthesis
(b) exchange of gases
(c) exchange of minerals
(d) water transport
Answer:
(b) exchange of gases

Question 35.
All mesophyll cells in monocot leaf are nearly:
(a) isodiametric and thick walled
(b) irregular and thick walled
(c) isodiametric and thin walled
(d) irregular and thin willed
Answer:
(c) isodiametric and thin walled

Question 36.
Structurally, hydathodes are modified:
(a) cambium tissue
(b) parenchyma
(c) pith
(d) stomata
Answer:
(d) stomata

Question 37.
Hydathodes occurs in the leaves of:
(a) desert plants
(b) submerged aquatic plants
(c) floating aquatic weeds
(d) forest trees
Answer:
(b) submerged aquatic plants

Question 38.
The process of guttation is seen in:
(a) grasses
(b) dicot plants
(c) desert plants
(d) Nerium
Answer:
(a) grasses

Question 39.
Salt glands are present in:
(a) xerophytes
(b) hydrophytes
(c) halophytes
(d) merophytes
Answer:
(c) halophytes

Question 40.
The term sieve tubes is coined by:
(a) Schleiden
(b) Hanstein
(c) Tsehireh
(d) Hartig
Answer:
(d) Hartig

II. Answer the following. (2 Marks)

Question 1.
Define the tissue?
Answer:
A tissue is a group of cells that are alike in origin, structure and function. The study of tissue is called Histology.

Question 2. What are the different types of plant tissue?
Answer:
The two types of principal groups are:

1. Meristematic tissues
2. Permanent tissues.

Question 3.
Mention any, two characters of meriste – matic tissue.
Answer:
Two characters of meriste – matic tissue:

1. The meristematic cells are isodiametric and they may be, oval, spherical or polygonal in shape.
2. They have generally dense cytoplasm with prominent nucleus.

Question 4.
Mention the function of apical meristem.
Answer:
Present in apices of root and shoot. It is responsible for increase in the length of the plant, it is called as primary growth.

Question 5.
What is mean by carpus?
Answer:
It is the inner zone of shoot apex,that forms cortex and stele of shoot.

Question 6.
Explain apical cell theory?
Answer:
Apical cell theory is proposed by Nageli. The single apical cell or apical initial composes the root meristem. The apical initial is tetrahedral in shape and produces root cap from one side. The remaining three sides produce epidermis, cortex and vascular tissues. It is found in vascular cryptogams.

Question 7.
What is meant by angular collenchyma?
Answer:
It is the most common type of collenchyma with irregular arrangement and thickening at the angles where cells meet., eg: Hypodermis of Datum and Nicotiana.

Question 8.
Explain briefly Branchysciereids or Stone cells.
Answer:
Isodiametric sclereids, with hard cell wall. It is found in bark, pith cortex, hard endosperm and fleshy portion of some fruits. eg: Pulp of Pyrus.

Question 9.
What is Filiform Sclereids?
Answer:
The sclereids are present in the leaf lamina of Olea europaea. They are very much elongated fibre like and about 1mm length.

Question 10.
Distinguish between Libriform fibres and Fibre tracheids.
Answer:
Between Libriform fibres and Fibre tracheids:

Libriform fibres	Fibre tracheids
1. These fibres have slightly lignified secondary walls with simple pits. These fibres are long and narrow.	1. These are shorter than the libriform fibres with moderate secondary thickenings in the cell walls. Pits are simple or bordered.

Question 11.
What are bast fibres?
Answer:
These fibres are present in the phloem. Natural Bast fibres are strong and cellulosic. Fibres obtaining from the phloem or outer bark of jute, kenaf, flax and hemp plants. The so called pericyclic fibres are actually phloem fibres.

Question 12.
What is meant by endarch type of xylem arrangements?
Answer:
Protoxylem lies towards the .centre and meta xylem towards the periphery this condition is called Endarch. It is seen in stems.

Question 13.
What are the types of cells present in phloem?
Answer:
Phloem consists of four types of Cells:

1. Sieve elements
2. Companion cells
3. Phloem parenchyma
4. Phloem fibres.

Question 14.
Define epiblema?
Answer:
It is made up of single layer of parenchyma cells which are arranged compactly without intercellular spaces. It is devoid of epidermal pores and cuticle. Root hair is always single celled, it absorbs water and mineral salts from the soil. The another important function of piliferous layer is protection.

Question 15.
Explain bulliform cells in grasses.
Answer:
Some cells of upper epidermis (eg: Grasses) are larger and thin walled. They are called bulliform cells or motor cells. These cells are helpful for the rolling and unrolling of the leaf according to the weather change.

Question 16.
What is meant by Sunken Stomata?
Answer:
In some Xerophytic plants (eg: Cycas, Nerium), stomata is sunken beneath the abaxial leaf surface within stomatal crypts. The sunken stomata reduce water loss by transpiration.

Question 1 7.
Mention any two functions of epidermal tissue system in plants.
Answer:
Two functions of epidermal tissue system in plants:

1. This system in the shoot checks excessive loss of water due to the presence of cuticle.
2. Epidermis protects the underlying tissues.

Question 18.
Define chlorenchymo?
Answer:
Sometimes in young stem, chloroplasts develop in peripheral cortical cells, which is Called chlorenchyma.

Question 19.
What is meant by casparian strips?
Answer:
In true root endodermis, radial and inner tangential walls of endodermal cells possess thickenings of lignin, suberin and some other carbohydrates in the form of strips they are called casparian strips.

Question 20.
What are albuminous cells?
Answer:
The cytoplasmic nucleated parenchyma, is associated with the sieve cells of Gymnosperms. Albuminous cells in Conifers are analogous to companion cells of Angiosperms. It also called as strasburger cells.

Question 21.
Describe briefly radial types of vascular Bundles.
Answer:
Xylem and phloem are present on different radii alternating with each other. The bundles are separated by parenchymatous tissue. (Monocot and Dicot roots).

Question 22.
Define stele?
Answer:
All the tissues inside the endodermis comprise the stele. This includes pericycle, vascular system and pith.

Question 23.
What is meant by cambium?
Answer:
Cambium consists of brick shaped and thin walled meristematic cells. It is one to four layers in thickness. These cells are capable of forming new cells during secondary growth.

Question 24.
Define silica Cells?
Answer:
Some of the epidermal cells of the grass are filled with silica. They are called silica cells.

Question 25.
Define, hydathode?
Answer:
A hydathode is a type of epidermal pore, commonly found in higher plants. Structurally, hydathodes are modified stomata, usually located at leaf tips or margins, especially at the teeth. Hydathodes occur in the leaves of submerged aquatic plants such as ranunculus fluitans as well as in many herbaceous land plants.

III. Answer the following. (3 Marks)

Question 1.
Explain apical cell theory.
Answer:
Apical cell theory is proposed by Hofmeister (1852) and supported by Nageli (1859). A single apical cell is the structural and functional unit. This apical cell governs the growth and development of whole plant body. It is applicable in Algae, Bryophytes and in some Pteridophytes.

Question 2.
Explain histogen theory.
Answer:
Histogen theory is proposed by Hanstein (1868) and supported by Strassburgur. The shoot apex comprises three distinct zones.

1. Dermatogen: It is a outermost layer. It gives rise to epidermis.
2. Periblem: It is a middle layer. It gives rise to cortex.
3. Plerome: It is innermost layer. It gives rise to stele.

Question 3.
What is meant by quiescent centre concept?
Answer:
Quiescent centre concept was proposed by Clowes (1961) to explain root apical meristem activity. These centre is located between root cap and differentiating cells of the roots. The apparently inactive region of cells in root promeristem is called quiescent centre. It is the site of hormone synthesis and also the ultimate source of all meristematic cells of the meristem.

Question 4.
Explain the term “sclereids”.
Answer:
Sclereids are dead cells, usually these are isodiametric but some are elongated too. The cell wall is very thick due to lignification. Lumen is very much reduced. The pits may simple or branched. Sclereids are mechanical in function. They give hard texture to the seed coats, endosperms etc.

Question 5.
Explain briefly about plant fibres.
Answer:
Fibres are very much elongated sclerenchyma cells with pointed tips. Fibres are dead cells and have lignified walls with narrow lumen. They have simple pits. They provide mechanical strength and protect them from the strong wind. It is also called supporting tissues. Fibres have a great commercial value in cottage and textile industries.

Question 6.
Write briefly about xylem fibres.
Answer:
The fibres of sclerenchyma associated with the xylem are known as xylem fibres. Xylem fibres are dead cells and have lignified walls with narrow lumen. They cannot conduct water but being stronger provide mechanical strength. They are present in both primary and secondary xylem. Xylem fibres are also called libriform fibres.
The fibres are abundantly found in many plants. They occur in patches, in continuous bands and sometimes singly among other cells. Between fibres and normal tracheids, there are many transitional forms which are neither typical fibres nor typical tracheids. The transitional types are designated as fibre – tracheids. The pits of fibre – tracheids are smaller than those of vessels and typical tracheids.

Question 7.
Explain companion cells.
Answer:
The thin walled, elongated, specialized parenchyma cells, which are associated with die sieve elements, are called companion cells. These cells are living and they have cytoplasm and a prominent nucleus. They are connected to the sieve tubes through pits found in the lateral walls. Through these pits cytoplasmic connections are maintained between these elements. These cells are helpful in maintaining the pressure gradient in the sieve tubes. Usually the nuclei of the companion cells serve for the nuclei of sieve tubes as they lack them. The companion cells are present only in Angiosperms and absent in Gymnosperms and Pteridophytes. They assist the sieve tubes in the conduction of food materials.

Question 8.
Distinguish between meristematic tissue and permanent tissue.
Answer:
Between meristematic tissue and permanent tissue:

Meristematic tissue	Permanent tissue
1. Cells divide repeatedly	1. Do not divide
2. Cells are undifferentiated	2. Cells are fully differentiated
3. Cells are small and Isodiametric	3. Cells are variable in shape and size
4. Intercellular spaces are absent	4. Intercellular spaces are present
5. Vacuoles are absent	5. Vacuoles are present
6. Cell walls are thin	6. Cell walls maybe thick or thin
7. Inorganic inclusions are absent	7. Inorganic inclusions  are present

Question 9.
Write down the differences between tracheids and fibres.
Answer:
The differences between tracheids and fibres:

Tracheids	Fibres
1. Not much elongated	1. Very long cells
2. Posses oblique end walls	2. Posses tapering end walls
3. Cell walls are not as thick as fibtres	3. Cell wall are thick and lignified
4. Possess various types of thickenings	4. Possess only pitted thickenings
5. Responsible for the conduction and also mechanical support	5. Provide only mechanical support

Question 10.
Give a brief answer on subsidiary cells in plant leaves.
Answer:
Stomata are minute pores surrounded by two guard cells. The stomata occur mainly in the epidermis of leaves. In some plants addition to guard cells, specialised epidermal cells are present which are distinct from other epidermal cells. They are called Subsidiary cells. Based on the number and arrangement of subsidiary cells around the guard cells, the various types of stomata are recognised, The guard cells and subsidiary cells help in opening and closing of stomata during gaseous exchange and transpiration.

Question 11.
Explain the term trichomes.
Answer:
There are many types of epidermal outgrowths in stems. The unicellular or multicellular appendages that originate from the epidermal cells are called trichomes. Trichomes may be branched or unbranched and one or more one celled thick. They assume many shapes and sizes. They may also be glandular (eg: Rose, Ocimum) or non – glandular.

Question 12.
What do you understand about hypodermis in plant tissue system.
Answer:
One or two layers of continuous or discontinuous tissue present below the epidermis, is called hypodermis. It is protective in function. In dicot stem, hypodermis is generally collenchymatous, whereas in monocot stem, it is generally sclerenchymatous. In many plants collenchyma form the hypodermis.

Question 13.
What is meant by pith?’
Answer:
The central part of the ground tissue is known as pith or medulla. Generally this is made up of thin walled parenchyma cells with intercellular spaces. The cells in the pith generally stores starch, fatty substances, tannins, phenols, calcium oxalate crystals, etc.

Question 14.
Explain the piliferous layer as epiblema.
Answer:
The outermost layer of the root is known as piliferous layer. It consists of single row of thin – walled parenchymatous cells without any intercellular space. Epidermal pores and cuticle are absent in the piliferous layer. Root hairs that are found in the piliferous layer are always unicellular. They absorb waer and mineral salt from the soil. Root hairs are generally short lived. The main function of piliferous layer is protection of the inner tissues.

Question 15.
What is meant by stele in plant stem?
Answer:
The central part of the stem inner to the endodermis is known as stele. It consists of pericyle, vascular bundles and pith. In dicot stem, vascular bundles are arranged in a ring around the pith. This type of stele is called eustele.

Question 16.
Explain the nature of phloem in dicot stem.
Answer:
Primary phloem lies towards the periphery. It consists of protpphloem and metaphloem. Phloem consists of sieve tubes, companion cells and phloem parenchyma. Phloem fibres are absent in primary phloem. Phloem conduct organic foods material from the leaves to other parts of the plant body.

Question 17.
Explain the mesophyll layer of leaf.
Answer:
The ground tissue that is present between the upper and lower epidermis of the leaf is called mesophyll. Here, the mesophyll is not differentiated into palisade and spongy parenchyma. All the mesophyll cells are nearly isodiametric and thin walled. These cells are compactly arranged with limited intercellular spaces. They contain numerous chloroplasts.

Question 18.
Mention any three differences between stomata and hydathodes.
Answer:
Stomata:

1. Occur in epidermis of leaves, young stems.
2. Stomatal aperture is guarded by two guard cells.
3. The two guard cells are generally surrounded by subsidiary cell.

Hydathodes:

1. Occur at the tip or margin of leaves that are grown in moist shady place.
2. Aperture of hydathodes are surrounded by a ring of cuticularized cells.
3. Subsidiary cells are absent.,

Question 19.
What are halophiles? Explain briefly.
Answer:
Halophiles:

1. Plants that grow in salty environment are called Halophiles.
2. Plant growth in saline habitat developed numerous adaptations to salt stress. The secretion of ions by salt glands is the best known mechanism for regulating the salt content of plant shoots.
3. Salt glands typically are found in halophytes. (Plants that grow in saline environments)

IV. Answer in detail.

Question 1.
Explain Histogen theory, Korper Kappe Theory and Quiescent Centre Concept with diagrams.
Answer:
Histogen Theory: Histogen theory is proposed by Hanstein (1868) and supported by Strassburgur. The histogen theory as appilied to the root apical meristem speaks of four histogen in the meristem. They are respectively

1. Dermatogen: It is a outermost layer. It gives rise to root epidermis.
2. Periblem: it is a middle layer. It gives rise to cortex.
3. Plerome: It is innermost layer. It gives rise to stele.
4. Calyptrogen: it gives rise to root cap.
   

Korper Kappe Theory: Korper kappe theory is proposed by Schuepp. There are two zones in root apex – Korper and Kappe.

1. Korper zone forms the body.
2. Kappe zone forms The cap.

This theory is equivalent to tunica corpus theory of shoot apex.The two divisions are distinguished by the type of T (also called Y divisions). Korper is characterised by inverted T division and kappe by straight T divisions.

Quiescent Centre Concept: Quiescent centre concept was proposed by Clowes (1961) to explain root apical meristem activity. These centre is located between root cap and differentiating cells of the roots. The apparently inactive region of cells in root promeristem is called quiescent centre. It is the site of hormone synthesis and also the ultimate source of all meristematic cells of the meristem.

Question 2.
Describe the structure and function of different kinds of parenchyma tissues?
Answer:

Parenchyma is generally present in all organs of the plant. It forms the ground tissue in a plant. Parenchyma is a living tissue and made up of thin walled cells. The cell wall is made up of cellulose. Parenchyma cells may be oval, polyhedral, cylindrical, irregular, elongated or armed. Parenchyma tissue normally has prominent intercellular spaces. Parenchyma may store various types of materials like, water, air, ergastic substances. It is usually colourless. The turgid parenchyma cells help in giving rigidity to the plant body. Partial conduction of water is also maintained through parenchymatous cells. Occsionaliy Parenchyma cells which store resin, tannins, crystals of calcium carbonate, calcium oxalate are called idioblasts. Parenchyma is of different types and some of them are discussed as follows. Types of paranchyma:
(i) Aerenchyma: Parenchyma which contains air in its intercellular spaces. It helps in aeration and buoyancy. eg: Nymphae and Hydrilia.

(ii) Storage Parenchyma: parenchyma stores food materials. eg: Root and stem tubers.

(iii) Stellate Parenchyma: Star shaped parenchyma. eg: Petioles of Banana and Canna.

(iv) Chlorenchyma: Parenchyma cells with chlorophyll. Function is photosynthesis, eg: Mesophyll of leaves.

(v) Prosenchyma: parenchyma cells became elongated, pointed and slightly thick walled. It provides mechanical support.

Question 3.
Describe the types of tracheids with diagram.
Answer:

Types of secondary wall thickenings in tracheids and vessels:
Tracheids are dead, lignified and elongated cells with tapering ends. Its lumen is broader than that of fibres. In cross section, the tracheids are polygonal. There are different types of cell wall thickenings due to the deposition of secondary wall substances. They are annular (ring like), spiral (spring like), scalariform (ladder like) reticulate (net like) and pitted (uniformly thick except at pits). Tracheids are imperforated cells with bordered pits on their side walls. Only through this conduction takes place in Gymnosperms. They are arranged one above the other. Tracheids are chief water conducting elements in Gymnosperms and Pteridophytes. They also offer mechanical support to the plants.

Question 4.
Compare the different types of plant tissues.
Answer:
The different types of plant tissues:

Question 5.
Compare the vascular tissues of plant.
Answer:
The vascular tissues of plant:

Question 6.
Draw and label the various parts of T.S. of dicot root.
Answer:
Draw and label the various parts of T.S. of dicot root:

Question 7.
Explain in detail about the vascular bundles of monocot stem.
Answer:
1. Vascular bundles: Vascular bundles are scattered (atactostele) in the parenchyma ground tissue. Each vascular bundle is surrounded by a sheath of sclerenehymatous fibres called bundle sheath. The vascular bundles are conjoint, collateral, endarch and closed. Vascular bundles are numerous, small and closely arranged in the peripheral portion. Towards the centre, the bundles are comparatively large in size and loosely arranged. Vascular bundles are skull or oval shaped.

2. Phloem: The phloem in the monocot stem consists of sieve tubes and companion cells. Phloem parenchyma and phloem fibres are absent. It can be distinguished into an outer crushed protophloem and an inner metaphloem.

3. Xylem: Xylem vessels are arranged in the form of ‘Y’ the two metaxylem vessels at the base. In a mature bundle, the lowest protowylem disintegrates and forms a cavity known as protoxylem lacuna.

Question 8.
Draw and label the various parts of monocot stem.
Answer:
The various parts of monocot stem:

Question 9.
Explain the various parts of sunflower leaf with neat diagram.
Answer:
1. Anatomy of a Dicot Leaf – sunflower Leaf: Internal structure of dicotyledonous leaves reveal epidermis, mesophyll and vascular tissues.

2. Epidermis: This leaf is generally dorsiventral. It has upper and lower epidermis. The epidermis is usually made up of a single layer of cells that are closely packed. The cuticle on the upper epidermis is thicker than that of lower epidermis. The minute opening found on the epidermis are called stomata. Stomata are more in number on the lower epidermis than on the upper epidermis.

A stomata is surrounded by a pair of bean shaped cells are called guard cells. Each stoma internally opens into an air chamber. These guard cells contain chlotroplasts. The main function of epidermis is to give protection to the inner tissue called mesospyll. The cuticle helps to check transpiration. Stomata are used for transpiration and gas exchange.

3. Mesophyll: The entire tissue between the upper and lower epidermis is called mesophyll (GK meso = in the middle, phyllome = leaf). There are two regions in the mesophyll. They are palisade parenchyma and spongy parenchyma. Palisade parenchyma cells are seen beneath the upper epidermis. It consists of vertically elongated cylindrical cells in one or more layers. These are compactly arranged and are generally without intercellular spaces. Palisade parenchyma cells contain more chloroplasts than the spongy parenchyma cells.

The function of palisade parenchyma is photosynthesis. Spongy parenchyma lies below the palsied parenchyma. Spongy cells are irregularly shaped. These cells are very loosly arranged with numerous airspaces. As compared to palisade cells, the spongy cells contain number of chloroplasts. Spongy cells facilitate the exchange of gases with the help of air spaces. The air space that is found next to the stomata is called respiratory cavity or substomatal cavity. Å

4. Vascular tissue: Vascular tissue are present in the veins of leaf. Vascular bundles are conjoint collateral and closed. Xylem is present towards the upper epidermis, while the phloem towards the lower epidermis. Vascular bundles are surrounded by a compact layer by a parenchymatous cells called bundle sheath or border parenchyma.

Xylem consists of metaxylem and protoxylem elements. Protoxylem is present towards the upper epidermis, while the phloem consists of sieve tubes, companion cells and phloem parenchyma. Phloem fibres are absent. Xylem sonsists of vessels and xylem parenchyma. Tracheids and xylem fibres are absent.

Solution To Activity
Text Book Page No. 10
Question 1.
Cell lab: Students prepare the slide and identify the different types tissues.
Answer:
Preparing a slide of plant tissue.
Objective:

1. Using hand cutting method to make thin slice of dicot root.
2. To make slide and stain of plant sample.
3. To observe the plant sample under microscope.

Materials:

1. A young dicot root
2. Compound microscope
3. Slide
4. Cover slip
5. Eosin stain

Method:

1. Place 2 cm of young dicot root on a glass slide or plate.
2. Cut thin slices of the root through the region of maturation.
3. Stain it with Eosin.
4. Fix one or two of the sections in a slide and put a cover slip.
5. To observe the sample under a compound microscope and record the parts of the sample.

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Samacheer Kalvi 11th Bio Botany Biomolecules Text Book Back Questions and Answers

I. Multiple Choice Questions
Choose the correct answer
Question 1.
The most basic amino acid is …………… .
(a) Arginine
(b) Histidine
(c) Glycine
(d) Glutamine
Answer:
(a) Arginine

Question 2.
An example of feedback inhibition is  …………… .
(a) Cyanide action on cytochrome
(b) Sulpha drug on folic acid synthesiser bacteria
(c) Allosteric inhibition of hexokinase by glucose – 6 – phosphate
(d) The inhibition of succinic dehydrogenase by malonate
Answer:
(c) Allosteric inhibition of hexokinase by glucose – 6 – phosphate

Question 3.
Enzymes that catalyse interconversion of optical, geometrical or positional isomers are …………… .
(a) Ligases
(b) Lyases
(c) Hydrolases
(d) Isomerases
Answer:
(d) Isomerases

Question 4.
Proteins perform many physiological functions. For example some functions as enzymes. One of the following represents an additional function that some proteins discharge …………… .
(a) Antibiotics
(b) Pigment conferring colour to skin
(c) Pigments making colours of flowers
(d) Hormones
Answer:
(d) Hormones

Question 5.
Given below is the diagrammatic representation of one of the categories of small molecular weight organic compounds in the living tissues. Identify the category shown & one blank component “X” in it …………… .

Answer:
(a) Nucleoside
(b) Uracil.

Question 6.
Distinguish between nitrogenous base and a base found in inorganic chemistry.
Answer:
Nitrogenous Base:

1. Nitrogenous bases are organic molecules containing the element nitrogen & acts as a base in chemical reaction.
2. e.g. Adenine, Thymine

Base:

1. Bases are the substance that release hydroxide (OH^– ) ions in aqueous solution.
2. e.g. NaOH and Ca(OH)₂

Question 7.
What are the factors affecting the rate of enzyme reaction?
Answer:
(a) Temperature: Heating increases molecular motion. Thus the molecules of the substrate and enzyme move more quickly resulting in a greater probability of occurrence of the reaction. The temperature that promotes maximum activity is referred to as optimum temperature.

(b) pH: The optimum pH is that at which the maximum rate of reaction occurs. Thus the pH change leads to an alteration of enzyme shape, including the active site. If extremes of pH are encountered by an enzyme, then it will be denatured.

(c) Substrate Concentration: For a given enzyme concentration, the rate of an enzyme reaction increases with increasing substrate concentration.

(d) Enzyme Concentration: The rate of reaction is directly proportional to the enzyme concentration.

The Michaelis – Menton Constant (Km) and Its Significance:
When the initial rate of reaction of an enzyme is measured over a range of substrate concentrations (with a fixed amount of enzyme) and the results plotted on a graph. With increasing substrate concentration, the velocity increases – rapidly at lower substrate concentration. However the rate increases progressively, above a certain concentration of the substrate the curve flattened out. No further increase in rate occurs. This shows that the enzyme is working at maximum velocity at this point. On the graph, this point of maximum velocity is shown as V_Max.

Question 8.
Briefly outline the classification of enzymes.
Answer:
Enzymes are classified into six groups based on their mode of action.

Question 9.
Write the characteristic feature of DNA.
Answer:
The characteristic feature of DNA.

1. If one strand runs in the 5′ – 3′ direction, the other runs in 3′ – 5′ direction and thus are antiparallel (they run in opposite direction). The 5′ end has the phosphate group and 3’end has the OH group.
2. The angle at which the two sugars protrude from the base pairs is about 120°, for the narrow angle and 240° for the wide angle. The narrow angle between the sugars generates a minor groove and the large angle on the other edge generates major groove.
3. Each base is 0.34 nm apart and a complete turn of the helix comprises 3.4 nm or 10 base pairs per turn in the predominant B form of DNA.
4. DNA helical structure has a diameter of 20 Å and a pitch of about 3 Å. X – ray crystal study of DNA takes a stack of about 10 bp to go completely around the helix (360°).
5. Thermodynamic stability of the helix and specificity of base pairing includes
   • (a) The hydrogen bonds between the complementary bases of the double helix
   • (b) stacking interaction between bases tend to stack about each other perpendicular to the direction of helical axis. Electron cloud interactions (\({ \Pi -{ \Pi } }\)) between the bases in the helical stacks contribute to the stability of the double helix.
6. The phosphodiester linkages gives an inherent polarity to the DNA helix. They form strong covalent bonds, gives the strength and stability to the polynucleotide chain.
7. Plectonemic coiling – the two strands of the DNA are wrapped around each other in a helix, making it impossible to simply move them apart without breaking the entire structure. Whereas in paranemic coiling the two strands simply lie alongside one another, making them easier to pull apart.
8. Based on the helix and the distance between each turns, the DNA is of three forms – A DNA, B DNA and Z DNA.

Question 10.
Explain the structure and function of different types of RNA.
Answer:
1. mRNA (messenger RNA): Single stranded, carries a copy of instructions for assembling amino acids into proteins. It is very unstable and comprises 5% of total RNA polymer. Prokaryotic mRNA (Polycistronic) carry coding sequences for many polypeptides. Eukaryotic mRNA (Monocistronic) contains information for only one polypeptide.

2. tRNA (transfer RNA): Translates the code from mRNA and transfers amino acids to the ribosome to build proteins. It is highly folded into an elaborate 3D structure and comprises about 15% of total RNA. It is also called as soluble RNA.

3. rRNA (ribosomal RNA): Single stranded, metabolically stable, makeup the two subunits of ribosomes. It constitutes 80% of the total RNA. It is a polymer with varied length from 120 – 3000 nucleotides and gives ribosomes their shape. Genes for rRNA are highly conserved and employed for phylogenetic studies.

Entrance Examination Questions Solved
Choose the correct answer:
Question 1.
Who invented electron microscope? (2010 AIIMS, 2008 JIPMER)
(a) Janssen
(b) Edison
(c) Knoll and Ruska
(d) Landsteiner
Answer:
(c) Knoll and Ruska

Question 2.
Specific proteins responsible for the flow of materials and information into the cellare called …………… . (2009 AIIMS)
(a) Membrane receptors
(b) carrier proteins
(c) integral proteins
(d) none of these
Answer:
(b) carrier proteins

Question 3.
Omnis – cellula – e – cellula was given by …………… . (2007 AIIMS)
(a) Virchow
(b) Hooke
(c) Leeuwenhoek
(d) Robert Brown
Answer:
(a) Virchow

Question 4.
Which of the following is responsible for the mechanical support, protein synthesis and enzyme transport? (2007 AIIMS)
(a) cell membrane
(b) mitochondria
(c) dictyosomes
(d) endoplasmic reticulum
Answer:
(d) endoplasmic reticulum

Question 5.
Genes present in the cytoplasm of eukaryotic cells are found in …………… . (2006 AIIMS)
(a) mitochondria and inherited via egg cytoplasm
(b) lysosomes and peroxisomes
(c) Golgi bodies and smooth endoplasmic reticulum
(d) Plastids inherited via male gametes
Answer:
(a) mitochondria and inherited via egg cytoplasm

Question 6.
In which one the following would you expect to find glyoxysomes? (2005 AIIMS)
(a) Endosperm of wheat
(b) Endosperm of castor
(c) Palisade cells in leaf
(d) Root hairs
Answer:
(b) Endosperm of castor

Question 7.
A quantosome is present in …………… . (JIPMER 2012)
(a) Mitochondria
(b) Chloroplast
(c) Golgi bodies
(d) ER
Answer:
(b) Chloroplast

Question 8.
In mitochondria the enzyme cytochrome oxidase is present in …………… . (2012 JIPMER)
(a) Outer mitochondrial membrane
(b) inner mitochondrial membrane
(c) Stroma
(d) Grana
Answer:
(b) inner mitochondrial membrane

Question 9.
Which organelle is present in higher number in secretory cell? (2008 JIPMER)
(a) Mitochondria
(b) Chloroplast
(c) Nucleus
(d) Dictyosomes
Answer:
(d) Dictyosomes

Question 10.
Major site for the synthesis of lipids …………… . (2013 NEET)
(a) Rough ER
(b) smooth ER
(c) Centriole
(d) Lysosome
Answer:
(b) smooth ER

Question 11.
Golgi complex plays a major role in …………… . (2013 NEET)
(a) post translational modification of proteins and glycosidation of lipids
(b) translation of proteins
(c) Transcription of proteins
(d) Synthesis of lipid
Answer:
(a) post translational modification of proteins and glycosidation of lipids

Question 12.
Main arena of various types of activities of a cell is …………… . (2010 AIPMT)
(a) Nucleus
(b) Mitochondria
(c) Cytoplasm
(d) Chloroplast
Answer:
(c) Cytoplasm

Question 13.
The thylakoids in chloroplast are arranged in …………… . (2005 JIPMER)
(a) regular rings
(b) linear array
(c) diagonal direction
(d) stacked discs
Answer:
(d) stacked discs

Question 14.
Sequences of which of the following is used to know the phylogeny rRNA? (20022JIPMER)
(a) mRNA
(b) rRNA
(c) tRNA
(d) Hn RNA
Answer:
(b) rRNA

Question 15.
Structures between two adjacent cells which is an effective transport pathway? (2010 AIPMT)
(a) Plasmodesmata
(b) Middle lamella
(c) Secondary wall layer
(d) Primary wall layer
Answer:
(a) Plasmodesmata

Question 16.
In active transport carrier proteins are used, which use energy in the form of ATP to …………… .
(a) transport molecules against concentration gradient of cell wall
(b) transport molecules along concentration gradient of cell membrane
(c) transport molecules against concentration gradient of cell membrane
(d) transport molecules along concentration gradient of cell wall
Answer:
(c) transport molecules against concentration gradient of cell membrane

Question 17.
The main organelle involved in modification and routing of newly synthesised protein to their destinations is …………… . (AIPMT 2005)
(a) Mitochondria
(b) Glyoxysomes
(c) Spherosomes
(d) Endoplasmic reticulum
Answer:
(d) Endoplasmic reticulum

Question 18.
Algae have cell wall made up of …………… . (AIPMT 2010)
(a) Cellulose, galactans and mannans
(b) Cellulose, chitin and glucan
(c) Cellulose, Mannan and peptidoglycan
Answer:
(a) Cellulose, galactans and mannans

Samacheer Kalvi 11th Bio Botany Biomolecules Additional Questions and Answers

Question 1.
The percentage of water in the total cellular mass is …………… .
(a) 50%
(b) 60%
(c) 70%
(d) 80%
Answer:
(c) 70%

Question 2.
The metabolites which does not show any direct function in growth is called …………… metabolite.
(a) Primary
(b) Secondary
(c) Tertiary
(d) Quartemary
Answer:
(b) Secondary

Question 3.
Molecular formula for carbohydrates is …………… .
(a) (CH₂O)₂
(b) (CH₆O)
(C) (C₂H₂O)_n
(d) (CH₆O)_n
Answer:
(a) (CH₂O)₂

Question 4.
Number of carbon molecule in glucose is …………… .
(a) 4
(b) 6
(c) 8
(d) 12
Answer:
(b) 6

Question 5.
Number of sugar units in oligo saccharides are …………… .
(a) 6 to 10
(b) 1 to 10
(c) 2 to 8
(d) 2 to 10
Answer:
(d) 2 to 10

Question 6.
Which of the following is a trisaccharide?
(a) Maltose
(b) Stachyose
(c) Ramnose
(d) Aldose
Answer:
(c) Ramnose

Question 7.
…………… are also called as Glycan.
(a) Monosaccharides
(b) Disaccharides
(c) Polysaccharides
(d) Multisaccharides
Answer:
(c) Polysaccharides

Question 8.
Sucrose is a combination of …………… and fructose.
(a) α – glucose
(b) β – glucose
(c) Ketoses
(d) Maltose
Answer:
(a) α – glucose

Question 9.
…………… is also called as animal starch.
(a) Amylose
(b) Glycogen
(c) Glucose
(d) Glycerol
Answer:
(b) Glycogen

Question 10.
…………… reagent is used in starch test.
(a) Potassium permanganate
(b) Potassium iodide
(c) Calcium chloride
(d) Calcium iodide
Answer:
(b) Potassium iodide

Question 11.
Glycogen is not seen in …………… cells.
(a) liver
(b) skeletal
(c) muscle
(d) brain
Answer:
(d) Brain

Question 12.
Benedicts solution is nothing but …………… .
(a) Copper II sulphate
(b) Cuprous sulphate
(c) Cupric sulphate
(d) Copper I sulphate
Answer:
(a) Copper II sulphate

Question 13.
…………… is not a reducing sugar.
(a) Glucose
(b) Fructose
(c) Sucrose
(d) Ketose
Answer:
(c) Sucrose

Question 14.
…………… form the exoskeleton of insects & arthropods.
(a) N – acetyl glucosamine
(b) N – butyl glucosamine
(c) N – phenyl glucosamine
(d) N – methyl glucosamine
Answer:
(a) N – acetyl glucosamine

Question 15.
Number of fatty acids in triglyceride is …………… .
(a) 1
(b) 2
(c) 3
(d) 4
Answer:
(c) 3

Question 16.
The major structural component of cell membrane is …………… .
(a) glucolipids
(b) phospholipids
(c) proteolipids
(d) triglycerides
Answer:
(b) phospholipids

Question 17.
There are …………… different amino acids existing naturally.
(a) about 20
(b) about 10
(c) about 25
(d) about 22
Answer:
(a) about 20

Question 18.
A zwitterion also called as …………… ion.
(a) dipolar
(b) monopolar
(c) tripolar
(d) nonpolar
Answer:
(a) dipolar

Question 19.
…………… test is used as an indicator of the presence of protein.
(a) Biuret test
(b) Iodine test
(c) Benedict’s test
(d) Starch test
Answer:
(a) Biuret test

Question 20.
The competitive inhibitor is …………… for succinic dehydrogenase.
(a) malonate
(b) succinate
(c) oxalate
(d) citrate
Answer:
(a) malonate

Question 21.
…………… is the abundant protein in whole biosphere.
(a) RUBP
(b) NAD⁺
(c) NADPH
(d) RUBISCO
Answer:
(d) RUBISCO

Question 22.
…………… is an active enzyme with its non – protein component.
(a) Apoenzyme
(b) Holoenzyme
(c) Coenzymes
(d) Enzymes
Answer:
(b) Holoenzyme

Question 23.
Flavin adenine dinucleotide contains …………… which helps to accept hydrogen.
(a) ascolac acid
(b) cyanocobalamin
(c) riboflavin
(d) keratinine
Answer:
(c) riboflavin

Question 24.
…………… is a catalytic RNA.
(a) mRNA
(b) Ribozyme
(c) Ribonuclease
(d) rRNA
Answer:
(b) Ribozyme

Question 25.
…………… protects the end of the chromosomes from damage.
(a) Satellite
(b) Kinetochore
(c) Primary constriction
(d) Telomere
Answer:
(d) Telomere

Question 26.
Which is not a pyrimidine base?
(a) Cytosine
(b) Uracil
(c) Guanine
(d) Thymine
Answer:
(c) Guanine

Question 27.
Which type of DNA was described by Watson & Crick?
(a) Z – DNA
(b) α – DNA
(c) B – DNA
(d) A – DNA
Answer:
(c) B – DNA

Question 28.
According to Chargaff’s rule, the hydrogen bonding between Adenine and Thymine is …………… .
(a) 2
(b) 3
(c) 4
(d) Nil
Answer:
(a) 2

Question 29.
The first clear crystallographic evidence for helical structure of DNA was produced by …………… .
(a) Maurice Wilkins
(b) Rosalind Franklin
(c) Francis Crick
(d) Chargaff
Answer:
(b) Rosalind Franklin

Question 30.
According to Cargaff’s rule, A : T = G : C = …………… .
(a) 0
(b) 1
(c) >1
(d) <1
Answer:
(b) 1

Question 31.
A complete turn of the helix comprises …………… .
(a) 34 nm
(b) 3.4 nm
(c) 20 nm
(d) 2nm
Answer:
(b) 3.4 nm

Question 32.
Diameter of DNA helix is …………… .
(a) 34 Å
(b) 20 nm
(c) 34 nm
(d) 20 Å
Answer:
(d) 20 Å

Question 33.
RNA is …………… .
(a) Single stranded and stable
(b) Single stranded and unstable
(c) Double stranded and stable
(d) Double stranded and unstable
Answer:
(b) Single stranded and unstable

Question 34.
rRNA constitutes …………… of total RNA.
(a) 20%
(b) 70%
(c) 80%
(d) 15%
Answer:
(c) 80%

Question 35.
Shape to the ribosomes is provided by …………… .
(a) rRNA
(b) tRNA
(c) mRNA
(d) DNA
Answer:
(a) rRNA

Question 36.
Which RNA is also called as soluble RNA?
(a) rRNA
(b) tRNA
(c) mRNA
(d) ssRNA
Answer:
(b) tRNA

Question 37.
Which is the left – handed DNA?
(a) B – DNA
(b) A – DNA
(c) Z – DNA
(d) dsDNA
Answer:
(c) Z – DNA

Question 38.
Which of the following does not contain cell wall?
(a) Fungi
(b) Bacteria
(c) Mycoplasma
(d) Algae
Answer:
(c) Mycoplasma

Question 39.
The amino acid which is both an acid and a base is called …………… .
(a) Amphibolic
(b) Amphoteric
(c) Amphipathetic
(d) Anabolic
Answer:
(b) Amphoteric

Question 40.
…………… leads to the loss of 3D structure of protein.
(a) Annealing
(b) Extension
(c) Denaturation
(d) Polymerisation
Answer:
(c) Denaturation

Question 41.
Which of the following polysaccharides is used as solidifying agent in culture medium?
(a) Inulin
(b) Heparin
(c) Agar
(d) Keratan sulphate
Answer:
(c) Agar

Question 42.
Which is an anticoagulant?
(a) Inulin
(b) Heparin
(c) Agar
(d) Keratan sulphate
Answer:
(b) Heparin

Question 43.
Insulin is a polymer of …………… .
(a) sucrose
(b) fructose
(c) glucose
(d) maltose
Answer:
(b) fructose

II. Very Short Answer Type Questions (2 Marks)

Question 1.
Define cell pool and mention its constituents.
Answer:
The cell components are made of collection of molecules called as cellular pool, which consists of both inorganic and organic compounds.

Question 2.
Draw the molecular structure of water.
Answer:
the molecular structure of water:

Question 3.
Point out the percentage of water in human cell & a plant cell.
Answer:
Water makes upto 70% of human cell and upto 95% of mass of a plant cell.

Question 4.
What are metabolites?
Answer:
Metabolites are the organic compounds synthesized by plants, fungi and various microbes. They are the intermediates & products of metabolism.

Question 5.
Write the molecular formula for carbohydrates?
Answer:
(CH₂O)_n

Question 6.
Give an example for simple sugar with its formula.
Answer:
Glucose – C₆H₁₂O₆

Question 7.
Which type of sugar does sucrose belongs to? Write its monomer units.
Answer:
Sucrose is a disaccharides composed of α – glucose & fructose.

Question 8.
Classify polysaccharides based on function.
Answer:
Depending on the function, polysaccharides are of two types:

1. storage polysaccharide and
2. structural polysaccharide.

Question 9.
What are Glycans?
Answer:
Polysaccharides are also called as Glycans. They are made of hundreds of monosaccharide units.

Question 10.
Which is a common storage polysaccharide? Mention its monomer units.
Answer:
Starch is a storage polysaccharide made up of repeated units of amylose and amylopectin.

Question 11.
Which is an animal starch? Where can we see it in our body?
Answer:
Glycogen. It is found in liver cells & skeletal muscles.

Question 12.
Why oil does not get mixed if added with water?
Answer:
Oil is a lipid. Lipids are long hydrocarbon chains that are non-polar & thus hydrophobic, which avoids the oil to dissolve in water.

Question 13.
How saturated fatty acids differ from unsaturated fatty acids?
Answer:
Saturated factty acids have the hydrocarbon chain with single bond, whereas in unsaturated fatty acids the hydrocarbon chain will have double bonds.

Question 14.
How waxes are formed?
Answer:
Waxes are esters formed between a long chain alcohol and saturated fatty acids.

Question 15.
Why amino acids are amphoteric?
Answer:
The amino acid is both an acid and a base and is called amphoteric.

Question 16.
Name the various groups attached to the 4 valencies of carbon in an amino acid.
Answer:
The 4 valencies of carbon in an amino acid:

1. (NH₂)
   
2. an acidic carboxylic group (COOH) and
   
3. a hydrogen atom (H)
   
4. and side chain or variable R group.

Question 17.
Where the peptide bond is formed?
Answer:
A peptide bond is formed when the amino group of one amino acid reacts with carbonyl group of another amino acid.

Question 18.
Which was the first sequenced protein? Who had done it?
Answer:
First protein is insulin and it was sequenced by Fred Sanger.

Question 19.
Why proteins undergo conformational changes after its synthesis?
Answer:
After synthesis, the protein attains conformational change into a specific 3D form for proper functioning.

Question 20.
Mention the levels of protein organisation based on folding.
Answer:
According to the mode of folding, four levels of protein organisation have been recognised namely primary, secondary, tertiary and quaternary.

Question 21.
Define enzymes.
Answer:
Enzymes are globular proteins that catalyse the thousands of metabolic reactions taking place within cells and organism.

Question 22.
Name any four factors that affect enzyme reactions.
Answer:
Four factors that affect enzyme reactions:

1. pH
2. temperature
3. enzyme concentration and
4. substrate concentration.

Question 23.
What are inhibitors? Mention its types.
Answer:
Certain substances present in the cells may react with the enzyme and lower the rate of reaction. These substances are called inhibitors. It is of two types:

1. Competitive and
2. Non – competitive.

Question 24.
Differentiate Apoenzyme from Holoenzyme.
Answer:
Differ Apoenzyme from Holoenzyme:

Apoenzyme	Holoenzyme
1. Active enzyme with its non – protein component	1. Inactive enzyme without its non – protein component

Question 25.
What are Prosthetic groups? Give example.
Answer:
Prosthetic groups are organic molecules that assist in catalytic function of an enzyme. Example: Flavin adenine dinucleotide (FAD).

Question 26.
Draw a diagram showing the various components of enzymes.
Answer:
Catalytic site, Cofactor and Holoenzyme:

Question 27.
Write a note on Ribozyme.
Answer:
Ribozyme – Non – Protein Enzyme. A Ribozyme, also called as catalytic RNA; is a ribonucleic acid that acts as enzyme. It is found in ribosomes.

Question 28.
Give an example for following enzyme groups.
Answer:
An example for following enzyme groups:

1. Transferase – Ex: Transaminase
2. Isomerase – Ex: Isomerase
3. Oxidoreductase – Ex: Dehydrogenase
4. Lyase – Ex: Decarboxylase

Question 29.
Write the composition of DNA & RNA.
Answer:
Nitrogen base, pentose sugar and phosphate.

Question 30.
What is a nucleoside?
Answer:
A purine or a pyrimidine and a ribose or deoxyribose sugar is called nucleoside. A nitrogenous base is linked to pentose sugar through n-glycosidic linkage and forms a nucleoside.

Question 31.
What is a nucleotide?
Answer:
When a phosphate group is attached to a nucleoside it is called a nucleotide.

Question 32.
Name the two types of Purines and Pyrimidines.
Answer:
The two types of Purines and Pyrimidines:

1. Purines: Adenine and guanine
2. Pyrimidines: Cytosine and thymine (Uracil)

Question 33.
How DNA differs from RNA?
Answer:
DNA has thymine base, whereas RNA has uracil base. DNA has deoxyribose sugar, whereas RNA has ribose sugar.

Question 34.
Draw a simple diagram showing basic components of DNA.
Answer:
Deoxyribose sugar:

Question 35.
Which is the secondary structure of DNA? Who discovered it?
Answer:
B – DNA is the secondary structure of DNA. Watson & Crick discovered B – DNA.

Question 36.
State Chargaff’s rule.
Answer:
Chargaff’s Rule:
A = T; G = C
A + G = T + C,
A : T = G : C = 1.

Question 37.
Name the three forms of DNA.
Answer:
The three forms of DNA:

1. A – DNA
2. B – DNA and
3. Z – DNA.

Question 38.
Which is the soluble forms of RNA. Write its percentage composition of total RNA.
Answer:
tRNA is the soluble RNA which is about 15% of total RNA.

Question 39.
Name the types of RNA?
Answer:
The types of RNA:

1. mRNA
2. tRNA and
3. rRNA.

Question 40.
Draw the structure of transfer RNA.
Answer:
Transfer RNA (tRNA):

III. Short Answer Type Questions (3 Marks)

Question 1.
Distinguish between Macronutrients & Micronutrients.
Answer:
Macronutrients:

• Nutrients required in larger quantities for plant growth are called Macronutrients.
• e.g. Potassium and Calcium

Micronutrients:

• Nutrients required in trace amount for plant growth are called Micronutrients
• e.g. Zinc and Bora

Question 2.
Tabulate the various cellular components with their percentage.
Answer:
The various cellular components with their percentage:

Component	% of the total cellular mass
1. Water	1. 70
2. Proteins	2. 15
3. Carbohydrates	3. 3
4. Lipids	4. 2
5. Nucleic acids	5. 6
6. Ions	6. 4

Question 3.
List out the properties of Water.
Answer:
Properties of Water:

1. Adhesion and cohesion property
2. High latent heat of vaporisation
3. High melting and boiling point
4. Universal solvent
5. Specific heat capacity

Question 4.
How lattice formation occurs in water molecule?
Answer:
Two electro negative atoms of oxygen share a hydrogen bonds of two water molecule. Thus, they can stick together by cohesion and results in lattice formation.

Question 5.
Distinguish between Primary metabolite & Secondary metabolite.
Answer:
Between Primary metabolite & Secondary metabolite:

• Primary metabolites are those that are required for the basic metabolic processes like photosynthesis, respiration, etc Example: Lipase, a protein.
• Secondary metabolites does not show any direct function in growth and development of organisms. Example: Ricin, gums.

Question 6.
Define Polymerization.
Answer:
Polymerization, is a process in which repeating subunits termed monomers are bound into chains of different lengths. These chains of monomers are called polymers.

Question 7.
Explain the bond formation in sucrose molecule.
Answer:
Sucrose is formed from a molecule of α – glucose and a molecule of fructose. This is a condensation reaction releasing water. The bond formed between the glucose and fructose molecule by removal of water is called glycosidic bond. This is another example of strong, covalent bond.

Question 8.
How will you identify the presence of starch in a food sample.
Answer:
The presence of starch is identified by adding a solution of iodine in potassium iodide. Iodine molecules fit nearly into the starch helix, creating a blue – black colour.

Question 9.
Write a note on steroids.
Answer:
Steroids are complex compounds commonly found in cell membrane and animal hormones. e.g. Cholesterol which reinforces the structure of the cell membrane in animal cells and in an unusual group of cell wall deficient bacteria – Mycoplasma.

Question 10.
Draw the structure of basic amino acid.
Answer:
The structure of basic amino acid:

Question 11.
What is a Zwitterion? or What is an isoelectric point?
Answer:
A zwitterion also called as dipolar ion, is a molecule with two or more functional groups, of which at least one has a positive and other has a negative electrical charge and the net charge of the entire molecule is zero. The pH at which this happens is known as the isoelectric point.

Question 12.
Write briefly about protein denaturation.
Answer:
Denaturation is the loss of 3D structure of protein. Exposure to heat causes atoms to vibrate violently, and this disrupts the hydrogen and ionic bonds. Under these conditions, protein molecules become elongated, disorganised strands. Agents such as soap, detergents, acid, alcohol and some disinfectants disrupt the interchain bond and cause the molecule to be non – functional.

Question 13.
Why do some people have curly hair?
Answer:
Human hair is made of protein. The more the distance between the sulphur atoms, the more the proteins bend; the more the hair curls.

Question 14.
Write a note on RUBISCO.
Answer:
Ribulose biphosphate carboxylase oxygenase (RUBISCO) is an enzyme that catalyses the reaction between CO₂ and the CO₂ acceptor molecule in photosynthesis. It is the most abundant protein in the whole biosphere.

Question 15.
Differentiate Anabolic reaction and Catabolic reaction.
Answer:
Anabolic reaction:

• Anabolic reaction involves the building up of organic molecules.
• Ex: Synthesis of protein from amino acids.

Catabolic reaction:

• Catabolic reaction involves the breaking down of larger molecules.
• Ex: Breaking down of sugar in respiration.

Question 16.
What are Allosteric inhibitors?
Answer:
Compounds which modify enzyme activity by causing a reversible change in the structure of the enzyme active site. This in turn affects the ability of the substrate to bind to the enzyme. Such compounds are called allosteric inhibitors, e.g. The enzyme hexokinase which catalysis glucose to glucose – 6 phosphate in glycolysis is inhibited by glucose – 6 phosphate. This is an example for feedback allosteric inhibitor.

Question 17.
Explain in brief about End – product inhibitor. (Negative Feedback Inhibition)
Answer:
When the end product of a metabolic pathway begins to accumulate, it may act as an allosteric inhibitor of the enzyme controlling the first step of the pathway. Thus the product starts to switch off its own production as it builds up. The process is self – regulatory. As the product is used up, its production is switched on once again. This is called end – product inhibition.

Question 18.
Draw the structure of Purine & Pyrimidine.
Answer:
The structure of Pyrimidine & Purine:

Question 19.
Why the sugar in DNA is a deoxyribose?
Answer:
The sugar in DNA molecule is called 2’ – deoxyribose because there is no hydroxyl group at 2’ position.

Question 20.
How dinucleotide & polynucleotides are formed?
Answer:
Two nucleotides join to form dinucleotide that are linked through 3′ – 5′ phosphodiester linkage by condensation between phosphate groups of one with sugar of other. This is repeated many times to make polynucleotide.

Question 21.
Compare Plectonemic & Paranemic Coiling.
Answer:
Plectonemic coiling – the two strands of the DNA are wrapped around each other in a helix, making it impossible to simply move them apart without breaking the entire structure. Whereas in Paranemic coiling the two strands simply lie alongside one another, making them easier to pull apart.

Question 22.
Differentiate between Polycistronic & Monocistronic mRNA.
Answer:
Between Polycistronic & Monocistronic mRNA:

Polycistronic mRNA	Monocistronic mRNA
1. Polycistronic mRNA carry coding sequences for many Polypeptides	1. Monocistronic mRNA carry coding sequences for only one Polypeptide
2. Prokaryotic mRNA are polycistronic	2. Eukaryotic mRNA are monocistronic

Question 23.
What are Proteins?
Answer:
Proteins are polymers of 20 different amino acids, each of which has a distinct side chain with specific chemical properties. Each protein has a unique amino acid sequence which determines its 3D structure.

Question 24.
Herbivores can digest cellulose rich food, Why can’t human beings?
Answer:
Human cannot digest cellulose but herbivores can digest them with the help of bacteria present in the gut which produces enzymes cellulase. This is an example of mutualism.

Question 25.
How will you identify the presence of protein in food samples?
Answer:
The biuret test is used as an indicator of the presence of protein because it gives a purple colour in the presence of peptide bonds (-C- N-). To a protein solution an equal quantity of sodium hydroxide solution is added and mixed. Then a few drops of 0.5% copper (II) sulphate is added with gentle mixing. A distinct purple colour develops without heating.

Question 26.
Write a note on peptide bonds between amino acids.
Answer:
The amino group of one amino acid reacts with carboxyl group of other amino acid, forming a peptide bond. Two amino acids can react together with the loss of water to form a dipeptide. Long strings of amino acids linked by peptide bonds are called polypeptides. In 1953, Fred Sanger first sequenced the Insulin protein.

Question 27.
Which was the first alkaloid discovered? Mentions its uses.
Answer:
Morphine is the first alkaloid to be found. It comes from the plant Opium poppy (Papaver somniferum). It is used as a pain reliever in patients with severe pain levels and cough suppressant.

IV. Long Answer Type Questions (5 Marks)

Question 1.
How will you identify the presence of glucose in a given food sample?
Answer:
Aldoses and ketoses are reducing sugars. This means that, when heated with an alkaline solution of copper (II) sulphate (a blue solution called Benedict’s solution), the aldehyde or ketone group reduces Cu²⁺ ions to Cu⁺ ions forming brick red precipitate of copper (I) oxide. In the process, the aldehyde or ketone group is oxidised to a carboxyl group (-COOH).

This reaction is used as test for reducing sugar and is known as Benedict’s test. The results of Benedict’s test depends on concentration of the sugar. If there is no reducing sugar it remains blue. Sucrose is not a reducing sugar The greater the concentration of reducing sugar, the more is the precipitate formed and greater is the colour change.

Question 2.
Write a note on various levels of protein organisation.
Answer:
The primary structure is linear arrangement of amino acids in a polypeptide chain. Secondary structure arises when various functional groups are exposed on outer surface of the molecular interaction by forming hydrogen bonds. This causes the amino acid chain to twist into coiled configuration called α – helix or to fold into a flat β – pleated sheets.

Tertiary protein structure arises when the secondary level proteins fold into globular structure called domains. Quaternary protein structure may be assumed by some complex proteins in which more than one polypeptide forms a large multiunit protein. The individual polypeptide chains of the protein are called subunits and the active protein itself is called a multimer.

Question 3.
Enumerate the properties of Enzyme.
Answer:
The properties of Enzyme:

• Enzymes are globular proteins.
• They act as catalysts and effective even in small quantity.
• They remain unchanged at the end of the reaction.
• They are highly specific.
• They have an active site where the reaction takes place.
• Enzymes lower activation energy of the reaction they catalyse.

Question 4.
Draw a Flow Chart depicting the Carbohydrate Classification
Answer:
Flow Chart depicting the Carbohydrate Classification:

Question 5.
Explain the various types of chemical bonding in proteins.
Answer:
1. Hydrogen Bond: It is formed between some hydrogen atoms of oxygen and nitrogen in polypeptide chain. The hydrogen atoms have a small positive charge and oxygen and nitrogen have small negative charge. Opposite charges attract to form hydrogen bonds. Though these bonds are weak, large number of them maintains the molecule in 3D shape.

2. Ionic Bond: It is formed between any charged groups that are not joined together by peptide bond. It is stronger than hydrogen bond and can be broken by changes in pH and temperature.

3. Disulfide Bond: Some amino acids like cysteine and methionine have sulphur. These form disulphide bridge between sulphur atoms and amino acids.

4. Hydrophobic Bond: This bond helps some protein to maintain structure. When globular proteins are in solution, their hydrophobic groups point inwards away from water.

Question 6.
Explain Lock & Key Mechanism of Enzymatic reaction.
Answer:
Lock and Key Mechanism of Enzyme: In a enzyme catalysed reaction, the starting substance is the substrate. It is converted to the product. The substrate binds to the specially formed pocket in the enzyme – the active site, this is called lock and key mechanism of enzyme action. As the enzyme and substrate form a ES complex, the substrate is raised in energy to a transition state and then breaks down into products plus unchanged enzyme.

Question 7.
Describe the Competitive & Non – Competitive Inhibitors of enzyme.
Answer:
1. Competitive Inhibitor: Molecules that resemble the shape of the substrate and may compete to occupy the active site of enzyme are known as competitive inhibitors. For Example: the enzyme that catalyses the reaction between carbon dioxide and the CO₂ acceptor molecule in photosynthesis, known as ribulose biphosphate carboxylase oxygenase (RUBISCO) is competitively inhibited by oxygen / carbon – di – oxide in the chloroplast. The competitive inhibitor is malonate for succinic dehydrogenase.

2. Non – competitive Inhibitors: There are certain inhibitors which may be unlike the substrate molecule but still combines with the enzyme. This either blocks the attachment of the substrate to active site or change the shape so that it is unable to accept the substrate. For example the effect of the amino acids alanine on the enzyme pyruvate kinase in the final step of glycolysis.

Certain non – reversible / irreversible inhibitors bind tightly and permanently to an enzyme and destroy its catalytic properties entirely. These could also be termed as poisons. Example – cyanide ions which blocks cytochrome oxidase in terminal oxidation in cell aerobic respiration, the nerve gas sarin blocks a neurotransmitter in synapse transmission.

Question 8.
Give a detailed account on Enzyme Co – factors.
Answer:
Many enzymes require non – protein components called co – factors for their efficient activity. Co – factors may vary from simple inorganic ions to complex organic molecules.
They are of three types:

1. Inorganic ions, prosthetic groups and coenzymes.
2. Holoenzyme – active enzyme with its non – protein component.
3. Apoenzyme – the inactive enzyme without its non – protein component.

Inorganic ions help to increase the rate of reaction catalysed by enzymes. Example: Salivary amylase activity is increased in the presence of chloride ions. Prosthetic groups are organic molecules that assist in catalytic function of an enzyme. Flavin adenine dinucleotide (FAD) contains riboflavin(vit B2), the function of which is to accept hydrogen. ‘Haem’ is an iron – containing prosthetic group with an iron atom at its centre. Coenzymes are organic compounds which act as cofactors but do not remain attached to the enzyme. The essential chemical components of many coenzymes are vitamins. Eg. NAD, NADP, Coenzyme A, ATP.

Question 9.
Tabulate the various features of different forms of DNA.
Answer:
The various features of different forms of DNA:

Question 10.
Compare DNA with RNA?
Answer:
Compare DNA with RNA:

DNA	RNA
1. Deoxyribose sugar is present	1. Ribose sugar is present
2. Thymine is present	2. Uracil is present
3. More stable	3. Less stable
4. Double stranded	4. Single stranded
5. Types: A – DNA, B – DNA, Z – DNA	5. Type: mRNA, tRNA, rRNA
6. Genetic material for most of living organism except few viruses	6. Genetic material for few viruses only

V. Higher Order Thinking Skills (HOTs)

Question 1.
In which form does the glucose is stored in animal cells? Specify the cells?
Answer:
Glucose is stored in the form of glycogen. Glycogen is stored in liver cells and skeletal muscles, etc.

Question 2.
State the key differences between DNA & RNA.
Answer:
The key differences between DNA & RNA:

DNA	RNA
1. Double stranded	1. Single stranded
2. Thymine is the pyrimidine base	2. Uracil is the pyrimidine base

Question 3.
Aminoacids are the monomers of proteins. Similarly mention the monomers of nucleic acids along with its composition.
Answer:
The monomer unit of nucleic acids are nucleotides, which are composed of nitrogen base, pentose sugar and phosphoric acid.

Question 4.
Complete the equations.
(a) Nitrogen base + …………… . = Nucleoside.
(b) …………… + nucleoside = Nucleotide.
(c) Glucose + fructose = …………… .
Answer:
(a) sugar
(b) phosphoric acid and
(c) sucrose.

Question 5.
What happens if the sucrose is hydrolysed?
Answer:
On hydrolysis, the glycosidic bonds in sucrose gets splitted yielding glucose and fructose.

Question 6.
Name the types of bonds.
(a) Between amino acids of protein
(b) Between carboxyl group and glycerol of fatty acids and
(c) Between glucose units of cellulose.
Answer:
(a) Peptide bond
(b) Ester bond
(c) Glycosidic bond

Question 7.
Study the following equation and name the reaction A and B.

Answer:
Reaction A is glycogenolysis. Reaction B is glycogenesis.

Question 8.
Whether waxes are found in living organisms?
Answer:
Yes. Fur, feathers, fruits, leaves, skin, and exoskeleton of insects are naturally water – proofed with a coating of wax.

Question 9.
If dsDNA has 40% Guanine. Calculate the percentage of Adenine.
Answer:
According to Chargaff’s rule:
Guanine pairs with cytosine. It Guanine is 40%, then cytosine will also be 40%. Similarly, Adenine pairs with thymine, if guanine is 40%, the remaining 60% will be Adenine. So Thymine will also be 60%.
Thus, A : T = G : C = 1
and 60 : 60 = 40 : 40 = 1.

Question 10.
In an Eukaryotic cell, totally there are 10000 RNA molecules. Calculate the number of mRNA’s and tRNA’s if the count of rRNA is 8000.
Answer:
In a cell, rRNA contributes 80%, tRNA constitutes 15% and mRNA constitutes 5%. If rRNA is 8000 (80%), then tRNA count is 1500 (15%) and mRNA is 500 (5%).

Question 11.
Despite made of two different monomers amylose and amylopectin, starch is a homopolysaccharide – Comment.
Answer:
Starch is made up of amylose and amylopectin. Both are glucose polymers, hence starch is considered as homopolysaccharides.

Question 12.
How do you call a fatty acid as saturated or unsaturated?
Answer:
If the hydrocarbon chain is single bonded, then the fatty acid is said to be saturated. In unsaturated fatty acids, the hydrocarbon chain is double bonded.

Question 13.
Enzymes are biocatalysts – Justify.
Answer:
Enzymes are globular proteins that catalyze thousands of metabolic reactions taking place within cells and organisms. Hence enzymes are called as biological catalysts.

Question 14.
Starch, cellulose, glycogen and chitin are polysaccharides found among the following. Choose the one appropriate and write against each.
(a) Cotton fibre – …………… .
(b) Exoskeleton of ant – …………… .
(c) Liver –  …………… .
(d) Peeled potato – …………… .
Answer:
(a) Cellulose
(b) Chitin
(c) Glycogen and
(d) Starch.

Question 15.
Sucrose is not a reducing sugar. Why?
Answer:
Sucrose is a non – reducing sugar since it does not possess aldehyde or ketone group, which is responsible for reducing the alkaline solutions like copper (II) sulphate.

Question 16.
A DNA segment has a total of 1000 nucleotides, out of which 240 are adenine containing nucleotides. How many pyrimidine bases this DNA segment possess?
Answer:
Pyrimidine = 500.
According to Chargaff’s rule,
A = T,
A = 240, hence T = 240.
A + T = 240 + 240 = 480.
So, G + C = 1000 – 480 = 520.
G = C, Therefore, C = \(\frac {520}{2}\) = 260.
Thus, pyrimidine = C + T = 260 + 240 = 500.

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Samacheer Kalvi 11th Bio Zoology Respiration Text Book Back Questions and Answers

I. Multiple Choice Questions
Question 1.
Breathing is controlled by ………..
(a) cerebrum
(b) medulla oblongata
(c) cerebellum
(d) pons
Answer:
(b) medulla oblongata

Question 2.
Intercostal muscles are found between the ………..
(a) vertebral column
(b) sternum
(c) ribs
(d) glottis
Answer:
(c) ribs

Question 3.
The respiratory structures of insects are ………..
(a) tracheal tubes
(b) gills
(c) green glands
(d) lungs
Answer:
(a) tracheal tubes

Question 4.
Asthma is caused due to ………..
(a) inflammation of bronchus and bronchioles
(b) inflammation of branchiole
(c) damage of diaphragm
(d) infection of lungs
Answer:
(d) infection of lungs

Question 5.
The Oxygen Dissociation Curve is ………..
(a) sigmoid
(b) straight line
(c) curved
(d) rectangular hyperbola
Answer:
(a) sigmoid

Question 6.
The Tidal Volume of a normal person is ………..
(a) 800 mL
(b) 1200 mL
(c) 500 mL
(d) 1100- 1200 mL
Answer:
(c) 500 mL

Question 7.
During inspiration, the diaphragm ………..
(a) expands
(b) unchanged
(c) relaxes to become dome-shaped
(d) contracts and flattens
Answer:
(d) contracts and flattens

Question 8.
CO₂ is transported through blood to lungs as ………..
(a) carbonic acid
(b) oxyhaemoglobin
(c) carbamino haemoglobin
(d) carboxy haemoglobin
Answer:
(c) carbamino haemoglobin

Question 9.
When 1500 mL air is in the lungs, it is called ………..
(a) vital capacity
(b) tidal volume
(c) residual volume
(d) inspiratory reserve volume
Answer:
(b) tidal volume

Question 10.
Vital capacity is ………..
(a) TV + IRV
(b) TV + ERV
(c) RV + ERV
(d) TV + TRV + ERV
Answer:
(d) TV + TRV + ERV

Question 11.
After a long deep breath, we do not respire for some seconds due to ………..
(a) more CO₂ in the blood
(b) more O₂ in the blood
(c) less CO₂ in the blood
(d) less CL in the blood
Answer:
(b) more O₂ in the blood

Question 12.
Which of the following substances in tobacco smoke damage the gas exchange system?
(a) carbon monoxide and carcinogens
(b) carbon monoxide and nicotine
(c) carcinogens and tar
(d) nicotine and tar
Answer:
(d) nicotine and tar

Question 13.
Column I represents diseases and column II represents their symptoms. Choose the correctly paired option

Column I	Column II
(P) Asthma	(i) Recurring  of bronchitis
(Q) Emphysema	(ii) Accumulation of WBCs in alveolus
(R) Pneumonia	(iii) Allergy

(a) P = iii, Q = ii, R = i
(b) P = iii, Q = i, R = ii
(c) P = ii, Q = iii, R = i
(d) P = ii, Q = i, R = iii
Answer:
(a) P = iii, Q = ii, R = i

Question 14.
Which of the following best describes the process of gas exchange in the lungs?
(a) Air moves in and out of the alveoli during breathing.
(b) Carbon dioxide diffuses from deoxygenated blood in capillaries into the alveolar air.
(c) Oxygen and carbon dioxide diffuse down their concentration gradients between blood and alveolar air.
(d) Oxygen diffuses from alveolar air into deoxygenated blood.
Answer:
(c) Oxygen and carbon dioxide diffuse down their concentration gradients between blood and alveolar air.

Question 15.
Make the correct pairs.

Column I	Column II
(P) IC	i. maximum volume of air breathe in after forced.
(Q) EC	ii. Volume of air present after expiration in lungs.
(R) VC	iii. Volume of air inhaled after expiration.
(S) FRC	iv. Volume of air present after expiration in lungs.

(a) P – i, Q – ii, R – iii, S – iv
(b) P – ii, Q – iii, R – iv , S – i
(c) P – ii, Q – iii , R – i, S – iv
(d) P – iii, Q – iv , R – i, S – ii
Answer:
(d) P – iii, Q – iv , R – i, S – ii

Make the correct pairs.

Column I	Column II
(P) Tidal volume	i. 1000 to 1100 ml
(Q) Residual volume	ii. 500 ml
(R) Expiratory reserve volume	iii. 2500 to 3000 ml
(S) Inspiratory reserve volume	iv. 1100 to 1200 ml

(a) P – ii, Q – iv , R – i, S – iii
(b) P – iii. Q – ii , R – iv , S – i
(c) P – ii, Q – iv , R – iii, S – i
(d) P – iii , Q – iv , R – i, S – ii
Answer:
(a) P – ii, Q – iv , R – i, S – iii

II. Very Short Answer Questions

Question 17.
Name the respiratory organs of flatworm, earthworm, fish, prawn, cockroach and cat.
Answer:
Flatworm – Body surface Earthworm – Moist skin Fish – Gills Prawn – Gills Cockroach – Trachea Cat – Lungs

Question 18.
Name the enzyme that catalyses the bicarbonate formation in RBCs.
Answer:
Carbonic anhydrase.

Question 19.
Air moving from the nose to the trachea passes through a number of structures. List in order of the structures.
Answer:

Question 20.
Which structure seals the larynx when we swallow?
Answer:
Epiglottis.

III. Short Answer Questions

Question 21.
Resistance in the airways is typically low. Why? Give two reasons.
Answer:
The airway resistance is low because:

• The diameter of most airways is relatively large.
• For smaller airways there are many in parallel, making their combined diameter large.
• Air has a low viscosity.

Question 22.
How the body makes long-term adjustments when living in high altitude?
Answer:
When a person lives in higher altitude, the body makes respiratory and hematopoietic . adjustment. Kidneys accelerate the production of the hormone erythropoietin which stimulate the bone marrow to produce more RBCs. This improves the binding of 02 with haemoglobin.

Question 23.
Why is pneumonia considered a dangerous disease?
Answer:
Inflammation of the lungs due to infection caused by bacteria or virus is called pneumonia. The symptoms are sputum production, nasal congestion, shortness of breath, sore throat etc. The alveoli get filled with fluid or pus, making is difficult to breathe (lung abscesses).

Question 24.
Diffusion of gases occurs in the alveolar region only and not in any other part of the respiratory system. Discuss.
Answer:
The alveolar region is highly vascular. Each alveolus is made up of highly permeable and thin layers of squamous epithelial cells. The barrier between the alveoli and the capillaries is thin and diffusion of gases takes place from higher partial pressure to low er partial pressure. Hence, gaseous exchange takes place in the aboral region only but not in any other part of the respiratory system.

Question 25.
Sketch a flow chart to show the path way of air flow during respiration.
Answer:

Question 26.
Explain the conditions which creates problems in oxygen transport.
Answer:
When a person travels quickly from sea level to elevations above 8000 ft, where the atmospheric pressure and partial pressure of oxygen are lowered, the individual responds with symptoms of acute mountain sickness (AMS)- headache, shortness of breath, nausea and dizziness due to poor binding of O₂ with haemoglobin. When the person moves on a long-term basis to mountains from sea level his body begins to make respiratory and haematopoietic adjustments.

To overcome this situation kidneys accelerate production of the hormone erythropoietin, which stimulates the bone marrow to produce more RBCs. When a person descends deep into the sea, the pressure in the surrounding water increases which causes the lungs to decrease in volume.

This decrease in volume increases the partial pressure of the gases within the lungs. This effect can be beneficial, because it tends to drive additional oxygen into the circulation, but this benefit also has a risk, the increased pressure can also drive nitrogen gas into the circulation.

This increase in blood nitrogen content can lead to a condition called nitrogen narcosis. When the diver ascends to the surface too quickly a condition called ‘bends’ or decompression sickness occurs and nitrogen comes out of solution while still in the blood forming bubbles. Small bubbles in the blood are not harmful, but large bubbles can lodge in small capillaries, blocking blood flow or can press on nerve endings.

Decompression sickness is associated with pain in joints and muscles and neurological problems including stroke. The risk of nitrogen narcosis and bends is common in scuba divers. During carbon-dioxide poisoning, the demand for oxygen increases. As the 02 level in the blood decreases it leads to suffocation and the skin turns bluish black.

Entrance Examination Questions Solved
Choose the correct answer
Question 1.
The length of human trachea is about ……….. [Gujarat CETQB]
(a) 6 inches
(b) 12 cm
(c) 12 inches
(d) 18 cm
Answer:
(b) 12 cm

Question 2.
Hamburger’s phenomenon is also known as ……….. [CPMT1988,1991, AMU2001, JLPMER 2002]
(a) HCO₃ shift
(b) Na+ shift
(c) H+ shift
(d) Chloride shift
Answer:
(d) Chloride shift

Question 3.
Oxygen carrying capacity of blood is ……….. [CPMT1990]
(a) 20%
(b) 30%
(c) 40%
(d) 50%
Answer:
(a) 20%

Question 4.
Respiratory movements are controlled by ……….. [APMEE 1978, CPMT 1998]
(a) Cerebellum
(b) Cerebrum
(c) Medulla oblongata
(d) Crura cerebri
Answer:
(c) Medulla oblongata

Question 5.
At higher CO₂ concentration, oxygen dissociation curve of haemoglobin will ……….. [CPMTX 990]
(a) Move to left
(b) Move to right
(c) Become irregular
(d) Move upwardly
Answer:
(b) Move to right

Question 6.
Chloride shift is required for transport of ……….. [CPMT 1990]
(a) Nitrogen
(b) Oxygen
(c) Carbon dioxide
(d) Carbon dioxide and oxygen
Answer:
(c) Carbon dioxide

Question 7.
Volume of air inspired or expired with each normal breath is known ……….. [CMPTl 992, AMU 2000]
(a) Inspiratory capacity
(b) Total Lung capacity
(c) Tidal volume
(d) Residual volume
Answer:
(c) Tidal volume

Question 8.
Oxygen haemoglobin dissociation curve will shift to right on decrease of ……….. [AMU 1992]
(a) Acidity .
(b) Carbon dioxide concentration
(c) Temperature
(d) pH
Answer:
(d) pH

Question 9.
Double membrane pleural sac ……….. [JKCMEE 1992]
(a) Envelops the kidneys
(b) Envelops the brain
(c) Envelops the lungs
(d) Lines the nasal passage
Answer:
(c) Envelops the lungs

Question 10.
Volume of air remaining in lungs after maximum respiratory’ effort is ……….. [JKCMEE 1992, Har. PMT 2003]
(a) Vital capacity
(b) Residual volume
(c) Total lung capacity
(d) Tidal volume
Answer:
(b) Residual volume

Question 11.
In expiration, diaphragm becomes ………..
(a) Flattened
(b) Relaxed
(c) Straightened
(d) Arched
Answer:
(b) Relaxed

Question 12.
Carbon dioxide is transported from tissues to respiratory surface by only ………..
(a) Plasma and erythrocytes
(b) Plasma
(c) Erythrocytes
(d) Erythrocytes and leucocytes
Answer:
(a) Plasma and erythrocytes

Question 13.
Respiratory centre is situated in ……….. [CPMT1980,2002, BHU1995, MPPMT1998, RPAPT2006]
(a) Cerebellum
(b) Medulla oblongata
(c) Hypothalamus
(d) Cerebrum
Answer:
(b) Medulla oblongata

Question 14.
Air is breathed through ……….. [APMEE 1999]
(a) Trachea → lung → laiynx → phaiynx → alveoli
(b) Nose → larynx → pharynx → alveoli → bronchioles
(c) Nostrils → pharynx → larynx → trachea bronchi → bronchioles → alveoli
(d) Nose → mouth → lungs.
Answer:
(c) Nostrils → pharynx → larynx → trachea bronchi → bronchioles → alveoli

Question 15.
Which is false ?
(a) Blood from right side of heart is carried to lungs by pulmonary artery
(b) Pleura is double covering of kidney
(c) Pancreas is both exocrine & endocrine gland
(d) Scurvy is due to vitamin C deficiency
Answer:
(b) Pleura is double covering of kidney

Question 16.
Volume of air breathed in and out during effortless respiration is ………..
(a) residual volume
(b) vital volume
(c) tidal volume
(d) normal volume
Answer:
(c) Tidal volume

Question 17.
Body tissue obtain oxygen from haemoglobin due to its dissociation in tissues is caused by ……….. [MPPMT 1995]
(a) Low oxygen concentration and high carbon dioxide concentration
(b) Low oxygen concentration
(c) Low carbon dioxide concentration
(d) High carbon dioxide concentration
Answer:
(b) Low oxygen concentration

Question 18.
Lungs have a number of alveoli for ………..[MPPMT 1995]
(a) Having spongy texture and proper shape
(b) More surface area for diffusion of gases
(c) More space for increasing volume of inspired air
(d) More nerve supply
Answer:
(b) More surface area for diffusion of gases

Question 19.
Presence of large number of alveoli around alveolar ducts opening into bronchioles in mammalian lungs is ………..
(a) Inefficient system of ventilation with little of residual air
(b) Inefficient system of ventilation with high percentage of residual air
(c) An efficient system of ventilation with no residual air
(d) An efficient system of ventilation with little residual air
Answer:
(d) An efficient system of ventilation with little residual air

Question 20.
During transport of CO₂ blood does not become acidic due to ………..
(a) Neutralisation of H₂CO₂ by Na₂CO₃
(b) Absorption by leucocytes
(c) Blood buffers
(d) Non-accumulation
Answer:
(c) Blood buffers

Question 21.
At high altitude, RBCs of human blood will ………..
(a) Increase in number
(b) Decrease in number
(c) Decrease in size
(d) increase in size
Answer:
(a) Increase in number

Question 22.
CO₂ is transported ………..
(a) as dissolved in blood plasma
(c) as carbamino haemoglobin
Answer:
(d) as carbamino haemoglobin and carbonic acid

Question 23.
Maximum amount 70-75% of carbon dioxide transport occursa. [RPMT1996, 1998, MPPMT1998, CPMT 1998, BV 2002]
(a) Dissolved in plasma
(b) Carbaminohaemoglobin complex
(c) Bicarbonate
(d) None of the above
Answer:
(c) Bicarbonate

Question 24.
Trachea is lined with incomplete rings of ………..
(a) Fibrous cartilage
(b) Calcified cartilage
(c) Elastic cartilage
(d) Hyaline cartilage
Answer:
(d) Hyaline cartilage

Question 25.
Oxygen and carbon dioxide are transported in blood through ………..
(a) Platelets and corpuscles
(b) RBCs and WBCs
(c) WBCs and serum
(d) RBCs and plasma
Answer:
(d) RBCs and plasma

Question 26.
About 1500 ml of air left in lungs is called ………..
(a) Tidal volume
(b) Inspiratory reserve volume
(c) Residual volume
(d) Vital capacity
Answer:
(c) Residual volume

Question 27.
Which one protects the lungs? [BHU1990]
(a) Ribs
(b) Vertebral column
(c) Sternum
(d) All the above
Answer:
(d) All the above

Question 28.
Which one has the lowest value?
(a) Tidal volume
(b) Vital capacity
(c) Inspiratory reserve volume
(d) Expiratory reserve volume
Answer:
(b) Vital capacity

Question 29.
A child was killed through asphyxiation. Post morturm confirmed it because a piece of lung put in water ………..[MPPMT 1996]
(a) Settled down
(b) Kept floating
(c) Had blood spots
(d) None of the above
Answer:
(b) Kept floating

Question 30.
Amount of oxygen present in one gram of haemoglobin is ……….. [AII MS 1997, Har. PMT, 2000]
(a) 20 ml
(b) 1-34 ml
(c) 13-4 ml
(d) None of the above
Answer:
(b) 1-34 ml

Question 31.
A molecule of haemoglobin carries how many oxygen molecules ……….. [MPPMT 1997, CFMT 2002, JCMEE 2004]
(a) 1
(b) 2
(c) 3
(d) 4
Answer:
(d) 4

Question 32.
In carbon monoxide poisoning there is ……….. [AFMC 1997]
(a) Increase in carbon dioxide concentration
(b) Decrease in oxygen availability
(c) Decrease in free haemoglobin
(d) None of the above
Answer:
(c) Decrease in free haemoglobin

Question 33.
Exchange of gases in lung alveoli occurs through ……….. [AFMC 2002]
(a) Active transport
(b) Osmosis
(c) Simple diffusion
(d) Passive transport
Answer:
(c) Simple diffusion

Question 34.
Haemoglobin is …………
(a) Vitamin
(b) Skin pigment
(c) Blood carrier
(d) Respiratory pigmentc
Answer:
(d) Respiratory pigmentc

Question 35.
Vocal cords occur in ………..
(a) Pharynx
(b) Larynx
(c) Glottis
(d) Bronchial tube
Answer:
(b) larynx

Question 36.
The cells which do not respire ……….. [4FMC 2001]
(a) Epidermal cells
(b) Sieve cells
(c) Cortical cells
(d) Erythocytes
Answer:
(d) Erythocytes

Question 37.
Hiccough (hiccup) is due to activity of ………..
(a) Intercostal muscles
(b) Food in air tract
(e) Diaphragm
(d) Inadequate oxygen in environment
Answer:
(c) Diaphragm

Question 38.
Bicarbonate formed inside erythrocytes moves out to plasma while chloride of plasma pass into erythrocytes. The phenomenon is called ………..
(a) Bicarbonate shift
(b) Carbonation
(c) Hamburger phenomenon
(d) None of the above
Answer:
(c) Hamburger phenomenon

Question 39.
Respiratory centre of brain is stimulated by ……….. [AllMS 2000]
(a) Carbon dioxide content in venous blood
(b) Carbon dioxide content in arterial blood
(c) Oxygen content in venous blood
(d) Oxygen content in arterial blood
Answer:
(d) Oxygen content in arterial blood

Question 40.
A higher CO₂ concentration of blood causes ……….. [AMU 2001]
(a) Slow diffusion of CO₂ from blood
(b) Slow transport of CO₂ from blood
(c) Slow diffusion of O, from blood
(d) Both A and B
Answer:
(c) Slow diffusion of O₂ from blood

Question 41.
Gases diffuse over the respiratory surface because of ………..
(a) O, is more in alveoli than in blood
(b) 02 is more in blood than in tissues
(c) CO₂ is more in alveoli than in blood
(d) PCO₂ is more in blood than in tissues
Answer:
(a) 02 is more in alveoli than in blood

Question 42.
Dissociation curve of 02 (which is dissociation from Hb) shifts to the rights ………..
(a) 02 concentration decrease
(b) CO₂ concentration decreases
(c) CO₂ concentration increase
(d) Chloride concentration increases
Answer:
(c) CO₂ concentration increase

Question 43.
Thoracic cage of man is formed of ……….. [MPPMT 2002]
(a) Ribs and sternum
(b) Ribs, sternum and thoracic vertebrae
(c) Ribs,sternum and lumbar vertebrae
(d) Ribs and thoracic vertebrae
Answer:
(b) Ribs, sternum and thoracic vertebrae

Question 44.
Vital capacity of lung is equal to ……….
(a) IRV + ERV + TV
(b) IRV + ERV + TV – RV
(c) IRV + ERV + TV + RV
(d) IRV + ERV
Answer:
(a) IRV + ERV + TV

Question 45.
Dead space is ………….
(a) Upper respiratory tract
(b) Nasal chambers
(c) Alveolar space
(d) Lower respiratory tract
Answer:
(a) Upper respiratory tract

Question 46.
Carbon monoxide contained in Tobacco smoke ……….. [AIEEE 2003]
(a) Is carcinogenic
(b) Causes gastric ulcers
(c) Reduces oxygen carrying capacity of blood
(d) Raises blood pressure
Answer:
(c) Reduces oxygen carrying capacity of blood

Question 47 .
What is correct?
(a) Pulmonary ventilation is equal to alveolar ventilation
(b) Alveolar ventilation is less than pulmonary ventilation
(c) Alveolar ventilation is more than pulmonary ventilation
(d) Both are variable
Answer:
(b) Alveolar ventilation is less than pulmonary ventilation

Question 48.
Increase in CO₂ concentration shall cause ………..
(a) Slower and shallower breathing
(b) Slower and deeper breathing
(e) Faster and deeper breathing
(d) No effect on breathing
Answer:
(c) Faster and deeper breathing

Question 49.
Alveoli become enlarged and damaged with reduced surface area in heavy smokers. the condition is called ………..
(a) Silicosis
(b) Emphysema
(c) Asthma
(d) Bronchitis
Answer:
(b) Emphysema

Question 50.
SARS is caused by a variant of ……….. [AIIMS 2004]
(a) Pneumococcus pneumonia
(b) Common cold by Corona virus
(c) Asthma
(d) Bronchitis
Answer:
(b) Common cold by Corona virus

Question 51.
During inspiration ……….. [JJPMER 2004, RPMT 20051
(a) Diaphragm and external muscles relax
(b) Diaphragm and internal intercostal muscles relax
(c) Diaphragm and external intercostal muscles contract
(d) Diaphragm and internal intercostal muscles contract
Answer:
(c) Diaphragm and external intercostal muscles contract

Question 52.
Mountain sickness at high altitude is due to ……….. [CPMT 2005]
(a) Excess CO₂ in blood
(b) Decreased CO₂ in air
(c) Decreased partial pressure of oxygen
(d) Decreased efficiency of haemoglobin
Answer:
(c) Decreased partial pressure of oxygen

Question 53.
Capacity of human lungs for air in a healthy person is ………..
(a) 3000 ml
(b) 1500 ml
(c) 1000 ml
(d) 500 ml
Answer:
(a) 3000 ml

Question 54.
Rate of breathing is controlled by ………….
(a) Amount of freely available oxygen
(b) Carbon dioxide in blood
(c) Muscular functions of body
(d) All the above
Answer:
(b) Carbon dioxide in blood

Question 55.
During strenuous exercise,glucose is converted into ………..
(a) Glycogen
(b) Pyruvic acid
(c) Starch
(d) Lactic acid
Answer:
(d) Lactic acid

Question 56.
How much pulmonary air is expired normally? [ PMT 2005]
(a) 70°io
(b) 20%
(c) 25%
(d) 32%
Answer:
(d) 32%

Question 57.
Which is incorrect?
(a) Presence of non-respiratory air sacs increases efficiency of respiration in birds
(b) In insects, circulation body fluids serve to distribute oxygen to tissues
(c) Principle of counter – current flow facilitates efficient respiration in gills of fishes
(d) Residual air in lungs slightly decreases the efficiency of respiration in mammals
Answer:
(b) In insects, circulation body fluids serve to distribute oxygen to tissues

Question 58.
Percentage of oxygen being carried by blood plasma is ………..
(a) 6-9%
(b) 3-6%
(c) 2-3%
(d) 1-2%
Answer:
(c) 2-3%

Question 59.
Name of the pulmonary disease in which alveolar surface area involved in gas exchange is drastically reduced due to damage in the alveolar walls: [RE-NEET 2015]
(a) Asthma
(b) Pleurisy
(c) Emphysema
(d) Pneumonia
Answer:
(c) Emphysema

Question 60.
Asthma may be attributed to ……….. [AIPMT/NEET 2016]
(a) bacterial infection of the lungs
(b) allergic reaction of the mast cells in the lungs
(c) inflammation of the trachea
(d) accumulation of fluid in the lungs
Answer:
(b) allergic reaction of the mast cells in the lungs

Question 61.
Name the chronic respiratory disorder caused mainly by cigarette smoking: [RE-NEET 2016]
(a) Emphysema
(b) Asthma
(c) Respiratory acidosis
(d) Respiratory alkalosis
Answer:
(a) Emphysema

Question 62.
Lungs are made up of air-filled sacs, the alveoli. They do not collapse even after forceful expiration. [NEET 20171]
(a) Inspiratory Reserve Volume
(b) Tidal Volume
(c) Expiratory Reserve Volume
(d) Residual Volume
Answer:
(d) Residual Volume

Samacheer Kalvi 11th Bio Zoology Respiration Additional Questions & Answers

I. Multiple Choice Questions
Choose the correct answer

Question 1.
Insects respire through ………..
(a) body surface
(b) trachea
(c) gills
(d) book lungs
Answer:
(b) trachea

Question 2.
Which of the following does not belong to the conducting zone of the respiratory system?
(a) trachea
(b) bronchioles
(c) larynx
(d) alveoli
Answer:
(d) alveoli

Question 3.
The trachea is supported by the ………..
(a) long structure
(b) cartilage
(c) multiple cartilaginous rings
(d) ligaments
Answer:
(c) multiple cartilaginous rings

Question 4.
Which of the following is right regarding the bronchioles?
(a) trachea divides into bronchioles
(b) bronchioles have cartilaginous rings
(c) bronchioles collapse when the air pressure is more
(d) bronchioles are without cartilaginous rings
Answer:
(d) bronchioles are without cartilaginous rings

Question 5.
The lungs are protected ventrally by ………..
(a) vertebral column
(b) sternum
(c) ribs
(d) diaphragm
Answer:
(b) sternum

Question 6.
The function of pleural fluid is to ………..
(a) keep the lungs intact
(b) protect the lungs from getting firm
(c) maintain the texture of the lungs
(d) reduces friction when lungs expand and contract
Answer:
(d) reduce friction when lungs expand and contract

Question 7.
Which of the following happens during inspiration?
(a) Diaphragm becomes dome shaped
(b) internal intercostal muscles contract
(c) volume of the thoracic chamber increases
(d) ribs are pulled downward
Answer:
(c) volume of the thoracic chamber increases

Question 8.
The amount of air inspired or expired with each normal breath is called the ………..
(a) residual volume
(b) inspiratory reserve volume
(c) dead space
(d) tidal volume
Answer:
(d) tidal volume

Question 9.
The normal value of IRV is ………..
(a) 500 mL
(b) 2500-3000 mL
(c) 1100-1200 mL
(d) 150 mL
Answer:
(b) 2500-3000 mL

Question 10.
UV + RV =
(a) EC
(b) TLC
(c) IC
(d) VC
Answer:
(b) TLC

Question 11.
Which of the following is not involved in gaseous exchange?
(a) expiratory reserve volume
(b) inspiratory reserve volume
(c) residual volume
(d) dead space
Answer:
(d) dead space

Question 12.
Each haemoglobin can combine with molecules of oxygen.
(a) one
(b) four
(c) three
(d) two
Answer:
(b) four

Question 13.
The ferric state of haemoglobin ………..
(a) binds with oxygen
(b) binds with carbon dioxide
(c) does not bind with oxygen
(d) does not bind with carbon dioxide
Answer:
(c) does not bind with oxygen

Question 14.
Which of the following is incorrect with regard to the formation of oxyhaemoglobin in the alveoli?
(a) high pCO₂
(b) less H+ concentration
(c) low pCO₂
(d) low temperature
Answer:
(a) high pCO₂

Question 15.
Carbon dioxide is carried in the RBCs as ………..
(a) bicarbonate ions
(b) carbon dioxide
(c) carbonic acid
(d) carbaminohaemoglobin
Answer:
(d) carbaminohaemoglobin

Question 16.
The formation of is catalysed by the enzyme carbonic anhydrase in RBCs ………..
(a) carbonates
(b) carbaminohaemoglobin
(c) carbonic acid
(d) bicarbonates
Answer:
(c) carbonic acid

Question 17.
The chemosensitive area found close to the rhythm center is highly sensitive to ………..
(a) CO₂ alone
(b) H⁺ alone
(c) O₂
(d) CO₂ and H⁺
Answer:
(d) CO₂ and H⁺

Question 18.
Which is known as the respiratory rhythm centre?
(a) cerebellum
(b) pons varoli
(c) medulla oblongata
(d) infundibulium
Answer:
(c) medulla oblongata

Question 19.
When a person travels to higher elevations, he may develop the symptoms of acute mountain sickness. This is because of ………..
(a) nitrogen narcosis
(b) high level of carbon dioxide in blood
(c) low immunity
(d) poor binding of O₂ with haemoglobin
Answer:
(d) poor binding of O₂ with haemoglobin

Question 20.
Erythropoietin hormone stimulates the bone marrow to produce more ………..
(a) WBCs
(b) RBCs
(c) Platelets
(d) Electrolytes
Ansewer:
(b) RBCs

Question 21.
When a person goes deep into the sea, he/she develops complications due to ………..
(a) increased partial pressure of O₂
(b) increase in blood nitrogen content
(c) increased partial pressure of CO₂
(d) decrease in the pressure in the surrounding water
Answer:
(b) increase in blood nitrogen content

Question 22.
Scuba divers need to face the problem of ………..
(a) nitrogen narcosis
(b) low oxygen content
(c) high carbon dioxide content
(d) lower pressure
Answer:
(a) nitrogen narcosis

Question 23.
Narrowing and inflammation of bronchi and bronchioles is called as ………..
(a) emphysema
(b) bronchitis
(c) asthma
(d) pneumonia
Answer:
(c) asthma

Question 24.
Gradual break down of the thin walls of the alveoli decreasing the total surface area of the gaseous exchange is ………..
(a) bronchitis
(b) emphysema
(c) asthma
(d) tuberculosis
Answer:
(b) emphysema

Question 25.
What causes bronchitis?
(a) Decrease in the surface area of alveoli
(b) inflammation of bronchi and bronchioles
(c) inflammation of bronchi
(d) collection of fluid in the bronchi
Answer:
(c) inflammation of bronchi

Question 26.
Which of the following is caused by bacteria?
(a) pneumonia
(b) emphysema
(c) silicosis
(d) tuberculosis
Answer:
(d) tuberculosis

Question 27.
Which of the following is the occupational respiratory disease?
(a) tuberculosis
(b) emphysema
(c) silicosis
(d) pneumonia
Answer:
(c) silicosis

Question 28.
Why does smoking affect oxygen supply to the body?
(a) haemoglobin fails to bind with O₂
(b) haemoglobin binds with carbon monoxide
(c) Nicotine makes the heart heat faster
(d) Blood vessels are narrowed
Answer:
(b) haemoglobin binds with carbon monoxide

Question 29.
COPD is the ………..
(a) emphysema and chronic bronchitis
(b) asthma and emphysema
(c) emphysema
(d) asthma, chronic bronchitis and emphysema
Answer:
(d) asthma, chronic bronchitis and emphysema

II. Fill in the Blanks

Question 1.
Oxygen is necessary for breakdown of to release energy.
Answer:
Glucose.

Question 2
……………. produce sound in human beings.
Answer:
Larynx/vocal cords.

Question 3.
……………… are the respiratory organs in most of the aquatic arthropods and molasses.
Answer:
Gills.

Question 4.
The ciliated epithelial cells lining the trachea, bronchi and bronchioles secrete …………….
Answer:
Mucus.

Question 5
divides thoracic cavity and abdominal cavity.
Answer:
Diaphragm.

Question 6.
The movement of air between the atmosphere and the lungs is known as …………
Answer:
Ventilation/breathing.

Question 7
……………. is the movement of atmospheric air into the lungs.
Answer:
Inspiration.

Question 8.
……………… In a relaxed stage, the diaphragm is shaped.
Answer:
Dome.

Question 9.
The expulsion of air from the lungs to the atmosphere is known as ……………..
Answer:
Expiration.

Question 10
………….. in an instmment used to measure the volume of air involved in breathing movements.
Answer:
Spirometer.

Question 11.
The volume of air remaining in the lungs after a forceful expiration is ……………
Answer:
Residual volume. .

Question 12.
The maximum volume of air that can be moved out during a single breath following a maximal inspiration is …………..
Answer:
Vital capacity.

Question 13.
The total volume of air a person can inhale after normal inspiration is …………….
Answer:
Inspiratory capacity.

Question 14.
The total volume of air a person can exhale after normal inspiration is ……………..
Answer:
Expiratory capacity.

Question 15.
…………….. is the total volume of air which the lungs can accommodate after forced inspiration.
Answer:
Total lung capacity.

Question 16.
The amount of air that moves into the respiratory passage per minute is called ……………
Answer:
Minute respiratory volume.

Question 17
……………… is the air that is not involved in gaseous exchange.
Answer:
Dead space.

Question 18
………………. is the pressure contributed by an individual gas in a mixture of gases.
Answer:
Partial pressure.

Question 19.
Haemoglobin belongs to the class of …………..
Answer:
Conjugated protein.

Question 20.
Haemoglobin is made up of the iron and the rest colourless protein
Answer:
Haem, globin.

Question 21.
The ferric state of haemoglobin is called …………….
Answer:
Methemoglobin.

Question 22
……………. is the respiratory pigment present in the blood.
Answer:
Haemoglobin.

Question 23.
The blood of human being is red due to the presence of …………….
Answer:
Haemoglobin.

Question 24.
Every 100 mL of oxygenated blood can deliver about mL of O, to the tissues.
Answer:
5.

Question 25.
About 70% of CO₂ transport occurs as ions.
Answer:
Bicarbonate.

Question 26.
The reversible reaction of formation of carbonic acid from carbon dioxide and water is catalyzed by the enzyme called …………………
Answer:
Carbonic anhydrase.

Question 27
center present in pons Varoli region of the brain moderates the function of respiratory rhythm center.
Answer:
Pneumotaxic.

Question 28
……………… is the hormone secreted by kidneys to stimulate bone marrow to produce more RBCs.
Answer:
Erythropoietin.

Question 29.
The increase in the nitrogen content in blood heads to a condition called ………………..
Answer:
Nitrogen narcosis.

Question 30.
The skin turns bluish black during poisoning.
Answer:
Carbon dioxide.

Question 31.
Dust, drugs, pollen grains, fish, prawns are common for asthma.
Answer:
Allergens.

Question 32.
Cigarette smoking reduces the respiratory surface of the alveolar walls and it is known as ……………….
Answer:
Emphysema.

Question 33.
Cough, shortness of breath and sputum in the lungs are the symptoms of ………………
Answer:
Bronchitis.

Question 34.
Tuberculosis is caused by ……………….
Answer:
Mycobacterium tuberculae

Question 35.
Collection of fluid between the lungs and the chest wall is the complication of the disease …………..
Answer:
Tuberculosis.

Question 36.
Long exposure to dust gives rise to inflammation leading to ………………
Answer:
Fibrosis.

Question 37
……………… and asbestosis are occupational respiratory diseases resulting from inhalation of particles of silica from sand grinding and asbestos into the respiratory tract.
Answer:
Silicosis.

Question 38
……………… causes narrowing of blood vessels resulting in increase in blood pressure and coronary heart disease.
Answer:
Nicotine.

Question 39.
……………….. Smoking lowers count in men.
Answer:
Sperm.

Question 40
…………….. is the chemical present in the cigarette smoke that causes addiction.
Answer:
Nicotine.

Question 41.
Chronic obstructive pulmonary disease is due to
Answer:
Smoking.

Question 42.
A non-smoker who inhales cigarette smoke involves in smoking.
Answer:
Passive.

Question 43.
Emphysema, chronic bronchitis and asthma, COPD, cancer may be caused due to
Answer:
Smoking.

Question 44.
Cigarette smoke contains thousands of chemicals and even small quantities of
Answer:
Arsenic.

Question 45.
The present in the cigarette smoke damages the gaseous exchange.
Answer:
Tar.

III. Answer the following Questions

Question 1.
What is excretion?
Answer:
The exchange of oxygen and carbon dioxide between environment and cells of our body, where organic nutrients are broken down oxygenatically to release energy.

Question 2.
What are the functions of the respiratory system?
Answer:
The five primary functions of the respiratory system are –

• To exchange O₂ and CO₂ between the atmosphere and the blood.
• To maintain homeostatic regulation of body pH.
• To protect us from inhaled pathogens and pollutants.
• To maintain the vocal cords for normal communication (vocalization).
• To remove the heat produced during cellular respiration through breathing.

Question 3.
The rate of breathing in aquatic animals is faster than the terrestrial animals. Give reason.
Answer:
The amount of dissolved oxygen is very low in water compared to the amount of oxygen in the air. Hence the rate of breathing in aquatic animals is faster than the terrestrial animals.

Question 4.
What is the function of hairs and mucus in the nasal cavity?
Answer:
The hairs and mucus filter the dust particles present in the air passing through the nostrils.

Question 5.
What is the function of epiglottis?
Answer:
Epiglottis is a thin elastic flap at the junction of nasopharynx and larynx. It prevents the food from entering into the larynx and avoids choking on food.

Question 6.
What are the layers of the diffusion membrane of the alveolus?
Answer:
The diffusion membrane of the alveolus is made up of three layers. They are:

• The thin squamous epithelial cells of the alveoli.
• The endothelium of the alveolar capillaries and
• The basement substance found in between them.

Question 7.
What are the cells of the squamous epithelial cells of the alveoli?
Answer:
The thin squamous epithelial cells of the alveoli are composed of Type I and Type II cells. Type I cells are very thin so that gases can diffuse rigidly through them. Type II cells are thicker. They synthesize and secrete a substance called surfactant. It lowers the surface tension in the alveoli and prevents pulmonary edema.

Question 8.
How are lungs protected?
Answer:
The lungs are light spongy tissues enclosed in the thoracic cavity surrounded by an air tight space. It is bound dorsally by the vertebral column and ventrally by the sternum, laterally by the ribs and on the lower side by the dome shaped diaphragm.

Question 9.
What is pleura? What is its function?
Answer:
The lungs are covered by double layered pleural membrane containing several layers of elastic connective tissues and capillaries. It encloses the pleural fluid which reduces friction when the lungs expand and contract.

Question 10.
What are the characteristic features of respiratory surface?
Answer:
The surface area of respiratory surface is large and richly supplied with blood vessels.

• It is extremely thin and kept moist.
• It is in direct contact with the environment.
• It is permeable to respiratory gases.

Question 11.
Explain the human respiratory system.
Answer:
The human respiratory system constitutes the conducting zone and the respiratory zone. The conducting zone includes the external nostril, nasal cavity, pharynx, larynx, trachea, bronchi, bronchioles and the lungs which contain alveoli.

The parts from the external nostrils up to the bronchioles conduct the air and hence they are called the conducting zone. This zone humidifies and warms the incoming air. The nasal cavity has fine hairs and mucus which filter dust particles in the incoming air.

At the junction of the pharynx and larynx there is a flap called epiglottis. This closes the respiratory tract while swallowing thus preventing the entry of food particles into the trachea and choking. The ciliated epithelial cells of trachea, bronchi and bronchioles secrete mucus rich in glycoprotein.

Microorganism and dust particles attach in the mucus films and are carried upwards to pass down the gullet during normal swallowing. The trachea is a semi-flexible tube supported by multiple cartilaginous rings which expands upto the mid thoracic cavity.

And the level of the 5th theoretic vertebra, it divides into right and left bronchi which divide into secondary and tertiary bronchi and further divide into terminal bronchioles and respiratory bronchioles. Bronchi have ‘c’ shaped curved Cartilage plates to ensure that the air passage does not collapse as air pressure changes during breathing. The bronchioles do not have cartilaginous rings but they have rigidity to prevent them from collapsing but are surrounded by smooth muscle which contracts or relaxes to adjust the diameter of these airways.

The fine respiratory bronchioles terminate into highly vascularised thin-walled-pouch like air sacs called alveoli. It is made up of thin squamous epithelial cells, the endothelium of the alveolar capillaries and the basement substance are found in between them.

The thin squamous epithelial cells of the alveoli, the endothelium of the alveolar capillaries and the basement substance found in between them. The thin squamous epithelial cells of the alveoli are composed of Type I and Type II cells.

Type I cells are very thin so that gases can diffuse rapidly through them. Type II cells are thicker, synthesize and secrete a substance called Surfactant. The lungs are light spongy tissues endorsed in the thoracic cavity bound dorsally by the vertebral column and ventrally by the sternum, laterally by the ribs and on the lower side by the dome shaped diaphragm. The lungs are covered by double walled pleural membrane containing several layers of elastic connective tissues and capillaries. It encloses the pleural fluid. It reduces friction when the lungs expand and contract.

Question 12.
Explain the mechanism of breathing.
Answer:
The movement of air between the atmosphere and the lungs is known as ventilation or breathing. Inspiration and expiration are the two phases of breathing. Inspiration is the movement of atmospheric air into the lungs and expiration is the movement of alveolar air that diffuse out of the lungs.

Lungs do not contain muscle fibres but expands and contracts by the movement of the ribs and diaphragm. The diaphragm is a sheet of tissue which separates the thorax from the abdomen. In a relaxed state, the diaphragm is dome shaped.

Ribs are moved by the intercostal muscles. External and internal intercostal muscles found between the ribs and the diaphragm helps in creating pressure gradients. Inspiration occurs if the pressure inside the lungs (intrapulmonary pressure) is less than the atmospheric pressure; likewise expiration takes place when the pressure within the lungs is higher than the atmospheric pressure.

Inspiration is initiated by the contraction of the diaphragm muscles and external intercostal muscles, which pulls the ribs and sternum upwards and outwards and increases the volume of the thoracic chamber in the dorso-ventral axis, forcing the lungs to expand the pulmonary volume. The increase in pulmonary volume and decrease in the intrapulmonary’ pressure forces the fresh air from outside to enter the air passages into the lungs to equalize the pressure.

This process is called inspiration. Relaxation of the diaphragm allows the diaphragm and sternum to return to its dome shape and the internal intercostal muscles contract, pulling the ribs downward reducing the thoracic volume and pulmonary volume. This results in an increase in the intrapulmonary pressure slightly above the atmospheric pressure causing the expulsion of air from the lungs. This process is called expiration.

Question 13.
Write short note on respiratory volumes.
Answer:
Tidal Volume (TV) Tidal volume is the amount of air inspired or expired with each normal breath. It is approximately 500 mL., i.e. a normal human adult can inspire or expire approximately 6000 to 8000 mL of air per minute. During vigorous exercise, the tidal volume is about 4-10 times higher.

Inspiratory Reserve volume (IRV) Additional volume of air a person can inspire by forceful inspiration is called Inspiratory Reserve Volume. The normal value is 2500-3000 mL.

Expiratory Reserve volume (ERV) Additional volume of air a person can forcefully exhale by forceful expiration is called Expiratory Reserve Volume. The normal value is 1000-1100 mL. Residual Volume (RV) The volume of air remaining in the lungs after a forceful expiration. It is approximately 1100 – 1200 mL.

Question 14.
Write a short note on respiratory capacities.
Answer:
Vital capacity (VC) the maximum volume of air that can be moved out during a single breath following a maximal inspiration. A person first inspires maximally then expires maximally. VC = ERV + TV + IRV

1. Inspiratory capacity (IC) The total volume of air a person can inhale after normal expiration. It includes tidal volume and inspiratory reserve volume. IC = TV + IRV

2. Expiratory capacity (EC) The total volume of air a person can exhale after normal inspiration. It includes tidal volume and expiratory reserve volume. EC = TV + ERV

3. Total Lung Capacity (TLC) The total volume of air which the lungs can accommodate after forced inspiration is called Total Lung Capacity. This includes the vital capacity and the residual volume. It is approximately 6000 mL. TLC = VC+ RV

4. Minute Respiratory Volume The amount of air that moves into the respiratory passage per minute is called minute respiratory volume.

Question 15.
What is dead space?
Answer:
Some of the inspired air never reaches the gas exchange areas but fills the respiratory passages where exchange of gases does not occur. This air is called dead space. Dead space is not involved in gaseous exchange. It amounts to approximately 150mL.

Question 16.
How does gaseous exchange take place in the alveoli?
Answer:
The primary site for the exchange of gases is the alveoli. The uptake of 02 and the release of CO₂ occur between the blood and tissues by simple diffusion driven by partial pressure gradient of O₂ and CO₂. Partial pressure is the pressure contributed by an individual gas in a mixture of gases.

It is represented as pO₂ for oxygen and pCO₂ for carbon- dioxide. Due to pressure gradients, O₂ from the alveoli enters into the blood and reaches the tissues. CO₂ enters into the blood from the tissues and reaches alveoli for elimination. As the solubility of CO₂ is 20-25 times higher than that of O₂, the partial pressure of CO₂ is much higher than that of O₂.

Question 17.
Write the note on respiratory pigments.
Answer:
Haemoglobin Haemoglobin belongs to the class of conjugated protein. The iron containing pigment portion haem constitutes only 4% and the rest colourless protein of the histone class globin. Haemoglobin has a molecular weight of 68,000 and contains four atoms of iron, each of which can combine with a molecule of oxygen.

Methaemoglobin If the iron component of the haem moieties is in the ferric state, than the normal ferrous state, it is called methaemoglobin. Methaemoglobin does not bind O₂. Normally RBC contains less than 1% methaemoglobin.

Question 18.
Explain the transport of oxygen In blood.
Answer:
Molecular oxygen is carried in blood in two ways: bound to haemoglobin within the red blood cells and dissolved in plasma. Oxygen is poorly soluble in water, so only 3% of the oxygen is transported in the dissolved form. 97% of oxygen binds with haemoglobin in a reversible manner to form oxyhaemoglobin (Hb0₂). The rate at which haemoglobin binds with O₂ is regulated by the partial pressure of O₂.

Each haemoglobin carries maximum of four molecules of oxygen. In the alveoli high pO₂, low pCO₂, low temperature and less H+ concentration, favours the formation of oxyhaemoglobin, whereas in the tissues low p02, high pCO₂, high H+ and high temperature favours the dissociation of oxygen from oxyhaemoglobin.

A sigmoid curve (S-shaped) is obtained when percentage saturation of haemoglobin with oxygen is plotted against pO₂. This curve is called oxygen haemoglobin dissociation curve. This S-shaped curve has a steep slope for pO₂ values between 10 and 50 mm Hg and then flattens between 70 and 100 mm Hg. Under normal physiological conditions, every 100 mL of oxygenated blood can deliver about 5 mL of O₂ to the tissues.

Question 19.
Explain the transport of carbon dioxide.
Answer:
Blood transports CO₂ from the tissue cells to the lungs in three ways:
1. Dissolved in plasma :
About 7 – 10% of CO₂ is transported in a dissolved form in the plasma.

2. Bound to haemoglobin :
About 20 – 25% of dissolved CO₂ is bound and carried in the RBCs as carbaminohaemoglobin (Hb CO₂)
CO₂ + Hb → Hb CO₂ .

3. As bicarbonate ions in plasma about 70% of CO₂ is transported as bicarbonate ions. This is influenced by pC0₂ and the degree of haemoglobin oxygenation. RBCs contain a high concentration of the enzyme, carbonic anhydrase, whereas small amounts of carbonic anhydrase is present in the plasma.

→ At the tissues the pCO₂ is high due to catabolism and diffuses into the blood to form HCO₂ and H⁺ ions. When CO₂ diffuses into the RBCs, it combines with water forming carbonic acid (H₂CO₂) catalyzed by carbonic anhydrase. Carbonic acid is unstable and dissociates into hydrogen and bicarbonate ions.

Carbonic anhydrase facilitates the reaction in both directions.

The HCO₃^– moves quickly from the RBCs into the plasma, where it is carried to the lungs. At the alveolar site where pCO₂ is low, the reaction is reversed leading to the formation of CO₂ and water. Thus CO₂ trapped as HCO₃^– at the tissue level is transported to the alveoli and released out as CO₂. Every 100 mL of deoxygenated blood delivers 4 mL of CO₂ to the alveoli for elimination.

Question 20.
How is respiration regulated?
Answer:
A specialised respiratory center present in the medulla oblongata of the hind brain called respiratory rhythm center is responsible for this regulation. Pneumotaxic centre present in ponsvaroli region of the brain moderates the function of the respiratory rhythm centre to ensure normal breathing.

The chemosensitive area found close to the rhythm centre is highly sensitive to CO₂ and H⁺. And H⁺ are eliminated out by respiratory process. Receptors associated with the aortic arch and carotid artery send necessary signals to the rhythm center for remedial action. The role of O₂ is insignificant in the regulation of respiratory rhythm.

Question 21.
Write the flow chart of the events during inspiration and expiration.
Answer:
Events in inspiration and expiration

Question 22.
What is nitrogen narcosis? What is its effect?
Answer:
When a person descends deep into the sea, the pressure in the surrounding water increases which causes the lungs to decrease in volume. This decrease in volume increases the partial pressure of the gases within the lungs. This effect can be beneficial, because it tends to drive additional oxygen into the circulation, but this benefit also has a risk, the increased pressure can also drive nitrogen gas into the circulation. This increase in blood nitrogen content can lead to a condition called nitrogen narcosis.

When the diver ascends to the surface too quickly a condition called ‘bends’ or decompression sickness occurs and nitrogen comes out of solution while still in the blood forming bubbles. Small bubbles in the blood are not harmful, but large bubbles can lodge in small capillaries, blocking blood flow or can press on nerve endings. Decompression sickness is associated with pain in joints and muscles and neurological problems including stroke.

The risk of nitrogen narcosis and bends is common in scuba divers. During carbon-dioxide poisoning, the demand for oxygen increases. As the O₂ level in the blood decreases it leads to suffocation and the skin turns bluish black.

Question 23.
Explain the disorders of the respiratory system.
Answer:
Asthma – It is characterized by narrowing and inflammation of bronchi and bronchioles and difficulty in breathing. Common allergens for asthma are dust, drugs, pollen grains, certain food items like fish, prawn and certain fruits etc. Emphysema – Emphysema is chronic breathlessness caused by gradual breakdown of the thin walls of the alveoli decreasing the total surface area of a gaseous exchange, i.e., widening of the alveoli is called emphysema. The major cause for this disease is cigarette smoking, which reduces the respiratory surface of the alveolar walls.

Bronchitis- The bronchi when it gets inflated due to pollution smoke and cigarette smoking, causes bronchitis. The symptoms are cough, shortness of breath and sputum in the lungs. Pneumonia- Inflammation of the lungs due to infection caused by bacteria or virus is called pneumonia. The common symptoms are sputum production, nasal congestion, shortness of breath, sore throat, etc.

Tuberculosis- Tuberculosis is caused by Mycobacterium tuberculae. This infection mainly occurs in the lungs and bones. Collection of fluid between the lungs and the chest wall is the main complication of this disease. Occupational respiratory disorders- The disorders due to one’s occupation of working in industries like grinding or stone breaking, construction sites, cotton industries, etc.

Dust produced affects the respiratory tracts. Long exposure can give rise to inflammation leading to fibrosis. Silicosis and asbestosis are occupational respiratory diseases resulting from inhalation of particle of silica from sand grinding and asbestos into the respiratory tract. Workers, working in such industries must wear protective masks.

Question 24.
What are the effects of smoking?
Answer:
Smoking is inhaling the smoke from burning tobacco. There are thousands of known chemicals which includes nicotine, tar, carbon monoxide, ammonia, sulphur- dioxide and even small quantities of arsenic. Carbon monoxide and nicotine damage the cardiovascular system and tar damages the gaseous exchange system.

Nicotine is the chemical that causes addiction and is a stimulant which makes the heart beat faster and the narrowing of blood vessels results in raised blood pressure and coronary heart diseases. Presence of carbon monoxide reduces oxygen supply. Lung cancer, cancer of the mouth and larynx is more common in smokers than non-smokers.

Smoking also causes cancer of the stomach, pancreas and bladder and lowers sperm count in men. Smoking can cause lung diseases by damaging the airways and alveoli and results in emphysema and chronic bronchitis. These two diseases along with asthma are often referred as Chronic Obstructive Pulmonary Disease (COPD). When a person smokes, nearly 85% of the smoke released is inhaled by the smoker himself and others in the vicinity, called passive smokers, are also affected. Guidance or counselling should be done in such users to withdraw this habit.

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Samacheer Kalvi 11th Bio Botany Cell Cycle Text Book Back Questions and Answers

I. Multiple Choice Questions
Choose the correct answer
Question 1.
The correct sequence in cell cycle is …………… .
(a) S-M-G1-G2
(b) S-G1-G2-M
(c) G1-S-G2-M
(d) M-G-G2-S
Answer:
(c) G1-S-G2-M

Question 2.
If mitotic division is restricted in G1 phase of the cell cycle then the condition is known as …………… .
(a) S Phase
(b) G2 Phase
(c) M Phase
(d) G₀ Phase
Answer:
(d) G₀ Phase

Question 3.
Anaphase promoting complex APC is a protein degradation machinery necessary for proper mitosis of animal cells. If APC is defective in human cell, which of the following is expected to occur?
(a) Chromosomes will be fragmented
(b) Chromosomes will not condense
(c) Chromosomes will not segregate
(d) Recombination of chromosomes will occur
Answer:
(b) Chromosomes will not condense

Question 4.
In S phase of the cell cycle …………… .
(a) Amount of DNA doubles in each cell
(b) Amount of DNA remains same in each cell
(c) Chromosome number is increased
(d) Amount of DNA is reduced to half in each cell
Answer:
(a) Amount of DNA doubles in each cell

Question 5.
Centromere is required for …………… .
(a) Transcription
(b) Crossing over
(c) Cytoplasmic cleavage
(d) Movement of chromosome towards pole
Answer:
(d) Movement of chromosome towards pole

Question 6.
Synapsis occur between …………… .
(a) mRNA and ribosomes
(b) Spindle fibres and centromeres
(c) Two homologous chromosomes
(d) A male and a female gamete
Answer:
(c) Two homologous chromosomes

Question 7.
In meiosis crossing over is initiated at …………… .
(a) Diplotene
(b) Pachytene
(c) Leptotene
(d) Zygotene
Answer:
(b) Pachytene

Question 8.
Colchicine prevents the mitosis of the cells at which of the following stage …………… .
(a) Anaphase
(b) Metaphase
(c) Prophase
(d) Interphase
Answer:
(b) Metaphase

Question 9.
The paring of homologous chromosomes on meiosis is known as …………… .
(a) Bivalent
(b) Synapsis
(c) Disjunction
(d) Synergids
Answer:
(b) Synapsis

Question 10.
Anastral mitosis is the characteristic feature of …………… .
(a) Lower animals
(b) Higher animals
(c) Higher plants
(d) All living organisms
Answer:
(c) Higher plants

Question 11.
Write any three significance of mitosis.
Answer:
Exact copy of the parent cell is produced by mitosis (genetically identical).

1. Genetic stability – daughter cells are genetically identical to parent cells.
2. Repair of tissues – damaged cells must be replaced by identical new cells by mitosis.
3. Regeneration – Arms of star fish.

Question 12.
Differentiate between Mitosis and Meiosis.
Answer:
Difference Between Mitosis and Meiosis:

Difference Between Mitosis and Meiosis
Mitosis	Meiosis
1. One division	1. Two divisions
2. Number of chromosomes remains the same	2. Number of chromosomes is halved
3. Homologous chromosomes line up separately on the metaphase plate	3. Homologous chromosomes line up in pairs at the metaphase plate
4. Homologous chromosome do not pair up	4. Homologous chromosome pairup to form bivalent
5. Chiasmata do not form and crossing over never occurs	5. Chiasmata form and crossingover occurs
6. Daughter cells are genetically identical	6. Daughter cells are genetically different from the parent cells
7. Two daughter cells are formed	7. Four daughter cells are formed

Question 13.
Given an account of G₀ phase.
Answer:
Some cells exit G₁ and enters a quiescent stage called G₀, where the cells remain metabolically active without proliferation. Cells can exist for long periods in G₀ phase. In G₀ cells cease growth with reduced rate of RNA and protein synthesis. The G₀ phase is not permanent. Mature neuron and skeletal muscle cell remain permanently in G₀ .Many cells in animals remains in G₀ unless called onto proliferate by appropriate growth factors or other extracellular signals. G₀ cells are not dormant.

Question 14.
Differentiate Cytokinesis in plant cells and animal cells.
Answer:
1. Cytokinesis in Plant Cells:
Division of the cytoplasm often starts during telophase. In plants, cytokinesis cell plate grows from centre towards lateral walls centrifugal manner of cell plate formation. Phragmoplast contains microtubules, actin filaments and vesicles from golgi apparatus and ER. The golgi vesicles contains carbohydrates such as pectin, hemicellulose which move along the microtubule of the pharagmoplast to the equator fuse, forming a new plasma membrane and the materials which are placed their becomes new cell wall.

The first stage of cell wall construction is a line dividing the newly forming cells called a cell plate. The cell plate eventually stretches right across the cell forming the middle lamella. Cellulose builds up on each side of the middle lamella to form the cell walls of two new plant cells.

2. Cytokinesis in Animal Cells:
It is a contractile process. The contractile mechanism contained in contractile ring located inside the plasma membrane. The ring consists of a bundle of microfilaments assembled from actin and myosin. This fibril helps for the generation of a contractile force. This force draws the contractile ring inward forming a cleavage furrow in the cell surface dividing the cell into two.

Question 15.
Write about Pachytene and Diplotene of Prophase I.
Answer:
1. Pachytene: At this stage bivalent chromosomes are clearly visible as tetrads. Bivalent of meiosis I consists of 4 chromatids and 2 centromeres. Synapsis is completed and recombination nodules appear at a site where crossing over takes place between non – sister chromatids of homologous chromosome. Recombination of homologous chromosomes is completed by the end of the stage but the chromosomes are linked at the sites of crossing over. This is mediated by the enzyme recombinase.

2. Diplotene: Synaptonemal complex disassembled and dissolves. The homologous chromosomes remain attached at one or more points where crossing over has taken place. These points of attachment where ‘X’ shaped structures occur at the sites of crossing over is called Chiasmata. Chiasmata are chromatin structures at sites where recombination has been taken place. They are specialised chromosomal structures that hold the homologous chromosomes together.

Sister chromatids remain closely associated whereas the homologous chromosomes tend to separate from each other but are held together by chiasmata. This substage may last for days or years depending on the sex and organism. The chromosomes are very actively transcribed in females as the egg stores up materials for use during embryonic development. In animals, the chromosomes have prominent loops called lampbrush chromosome.

Samacheer Kalvi 11th Bio Botany Cell Cycle Additional Questions and Answers

I. Multiple Choice Questions
Choose the correct answer:
Question 1.
Most of the neurons in the brain are in …………… stage.
(a) G₁
(b) S
(c) G₂
(d) G₀
Answer:
(d) G₀

Question 2.
Short, constricted region in the chromosome is …………… .
(a) Kinetochore
(b) Centromere
(c) Satellite
(d) Telomere
Answer:
(b) Centromere

Question 3.
Robert Brown discovered the nucleus in the cells of …………… roots.
(a) Mirabilas
(b) Orchid
(c) Moringa
(d) Oryza
Answer:
(b) Orchid

Question 4.
Scientist who described chromosomes for the first time is …………… .
(a) Robert Brown
(b) Anton van Leeuwenhoek
(c) Boveri
(d) Anton Schneider
Answer:
(d) Anton Schneider

Question 5.
Number of chromosomes in onion cell is …………… .
(a) 8
(b) 16
(c) 32
(d) 64
Answer:
(a) 16

Question 6.
Longest part of the cell cycle is …………… .
(a) Prophase
(b) G₁ Phase
(c) Interphase
(d) Sphase
Answer:
(c) Interphase

Question 7.
Eukaryotic cells divides every …………… .
(a) 12
(b) 24
(c) 1
(d) 6
Answer:
(b) 24

Question 8.
Cell cycle was discovered by …………… .
(a) Singer & Nicolson
(b) Prevost & Dumans
(c) Schleider & Schwann
(d) Boveri
Answer:
(b) Prevost & Dumans

Question 9.
G₀ stage is called as …………… stage.
(a) Quiescent
(b) Metabolically active
(c) Synthesis of DNA
(d) Replication
Answer:
(a) Quiescent

Question 10.
…………… protein acts as major check point in phase.
(a) Porins
(b) Kinases
(c) Cyclins
(d) Ligases
Answer:
(c) Cyclins

Question 11.
Replication of DNA occurs at …………… phase.
(a) G₀
(b) G₁
(c) S
(d) G₂
Answer:
(c) S

Question 12.
Condensation of interphase chromosomes into mitotic forms is done by …………… proteins.
(a) MPF
(b) APF
(c) AMF
(d) MAF
Answer:
(a) MPF

Question 13.
Which of the following is also called as direct division?
(a) Amitosis
(b) Meiosis
(c) Mitosis
(d) Reduction division
Answer:
(a) Amitosis

Question 14.
Cells of mammalian cartilage undergoes …………… .
(a) Amitosis
(b) Meiosis
(c) Mitosis
(d) Equational division
Answer:
(a) Amitosis

Question 15.
Yeast cells undergo …………… .
(a) Open mitosis
(b) Closed mitosis
(c) Amitosis
(d) Meiosis
Answer:
(b) Closed mitosis

Question 16.
…………… is the longest phase in mitosis.
(a) Anaphase
(b) Telophase
(c) Prophase
(d) Interphase
Answer:
(c) Prophase

Question 17.
The DNA protein complex present in the centromere is …………… .
(a) Cyclin
(b) Kinesis
(c) MPF
(d) Kinetochore
Answer:
(d) Kinetochore

Question 18.
…………… protein induces the break down of cohesion proteins leading to chromatid separation during mitosis.
(a) APC
(b) MPF
(c) Cyclin
(d) Kinetochore
Answer:
(a) APC

Question 19.
Regeneration of arms of star fish is due to …………… .
(a) Meiosis
(b) Amitosis
(c) Mitosis
(d) Budding
Answer:
(c) Mitosis

Question 20
…………… is called as reduction division.
(a) Meiosis
(b) Mitosis
(c) Amitosis
(d) Budding
Answer:
(a) Meiosis

Question 21.
Bivalents occur at …………… stage.
(a) Zygotene
(b) Leptotene
(c) Pachytene
(d) Diplotene
Answer:
(a) Zygotene

Question 22.
Recombination of chromosomes occur at …………… .
(a) Zygotene
(b) Leptotene
(c) Pachytene
(d) Diplotene
Answer:
(c) Pachytene

Question 23.
Terminalisation of chiasmata occurs at …………… .
(a) Zygotene
(b) Leptotene
(c) Diakinesis
(d) Pachytene
Answer:
(c) Diakinesis

Question 24.
Number of daughter cells formed at the end of Meiosis I is …………… .
(a) 2
(b) 4
(c) 1
(d) 0
Answer:
(a) 2

Question 25.
…………… division leads to genetic variability.
(a) Mitotic
(b) Amitotic
(c) Meiotic
(d) Equational
Answer:
(c) Meiotic

Question 26.
Crossing over occurs at …………… stage.
(a) Leptotene
(b) Zygotene
(c) Pachytene
(d) Diplotene
Answer:
(c) Pachytene

Question 27.
Which of the following is not a mitogen?
(a) Giberellin
(b) Ethylene
(c) Kinetin
(d) Colchicine
Answer:
(d) Colchicine

Question 28.
In plants mitosis occurs at …………… cells.
(a) Sclerenchyma
(b) Meristem
(c) Xylem
(d) Parenchyma
Answer:
(b) Meristem

Question 29.
Which of the following alone is formed in the division of plant cells?
(a) Aster
(b) Centrioles
(c) Spindle
(d) Microtubules
Answer:
(c) Spindle

Question 30.
Amphiastral type cell division is seen in …………… cells.
(a) Fungal
(b) Algal
(c) Plant cells
(d) Animal
Answer:
(d) Animal

II. Very Short Answer Type Questions (2 Marks)

Question 1.
Name the two types of nuclear division.
Answer:
The two types of nuclear division:

1. Mitosis and
2. Meiosis.

Question 2.
Define Cell Cycle.
Answer:
A series of events leading to the formation of new cell is known as cell cycle.

Question 3.
Who discovered the Cell Cycle?
Answer:
Prevost & Dumans in 1824.

Question 4.
Draw a tabular column showing the duration of various phase in the cell cycle of human cell.
Answer:
A tabular column showing the duration of various phase in the cell cycle of human cell:

Cell cycle of a proliferating human cell
Phase	Time Duration (in hrs)
1. G2	1. 11
2. S	2. 8
3. G2	3. 4
4. M	4. 1

Question 5.
Define C – Value.
Answer:
C – Value is the amount in picograms of DNA contained within a haploid nucleus.

Question 6.
Which is the longest phase of cell cycle? What happens during that phase?
Answer:
Interphase is the longest phase. Cells are metabolically active and involved in protein synthesis and growth.

Question 7.
Name the phases which comprises the Interphase.
Answer:
The phases which comprises the Interphase:

1. G₁ Phase
2. S Phase and
3. G₂ Phase.

Question 8.
Name the proteins involved in the activation of genes & their proteins to perform cell division.
Answer:
Kinases & Cyclins.

Question 9.
What do you mean by G₀ stage?
Answer:
G₀ stage is called as quiescent stage, where the cells remain metabolically active without proliferation.

Question 10.
What is the role of MPF in Cell cycle?
Answer:
Maturation Promoting Factor (MPF) brings about condensation of interphase chromosomes into the mitotic form.

Question 11.
Distinguish between Karyokinesis & Cytokinesis.
Answer:
Between Karyokinesis & Cytokinesis:

• Karyokinesis: Karyokinesis refers to the nuclear division.
• Cytokinesis: Cytokinesis refers to the cytoplasmic division.

Question 12.
Point out any two cell – types which remain G₀ phase.
Answer:
Mature neurons and Skeletal muscle cells.

Question 13.
Why amitosis is called as incipient cell division?
Answer:
Amitosis is also called incipient cell division. Since there is no spindle formation and chromatin material does not condense.

Question 14.
List out the disadvantages of Amitosis.
Answer:
The disadvantages of Amitosis:

• Causes unequal distribution of chromosomes.
• Can lead to abnormalities in metabolism and reproduction.

Question 15.
Mitosis also called as equational division – Justify.
Answer:
At the end of mitosis the number of chromosomes in the parent and the daughter (Progeny) cells remain the same so it is also called as equational division.

Question 16.
Enumerate the stages of mitosis.
Answer:
Mitosis is divided into four stages prophase, metaphase, anaphase and telophase.

Question 17.
Define an aster.
Answer:
In animal cell the centrioles extend a radial array of microtubules towards the plasma membrane when they reach the poles of the cell. This arrangement of microtubules is called an aster. Plant cells do not form asters.

Question 18.
What is metaphase plate?
Answer:
The alignment of chromosome into compact group at the equator of the cell is known as metaphase plate.

Question 19.
What is Kinetochore?
Answer:
Kinetochore is a DNA – Protein complex present in the centromere DNA, where the microtubules are attached. It is a trilaminar disc like plate.

Question 20.
How will you calculate the length of the S period.
Answer:
Length of the S period = Fraction of cells in DNA replication × generation time.

Question 21.
Which type of cell division occurs in reproductive cells? What will be the result?
Answer:
Meiosis takes place in the reproductive organs. It results in the formation of gametes with half the normal chromosome number.

Question 22.
Define Synapsis.
Answer:
In Zygotene, pairing of homologous chromosomes takes place and it is known as synapsis.

Question 23.
What do you understand by independent assortment?
Answer:
The random distribution of homologous chromosomes in a cell in Metaphase I is called independent assortment.

Question 24.
Define Mitogen. Give an example.
Answer:
The factors which promote cell cycle proliferation is called mitogen.
Example: gibberellin. These increase mitotic rate.

Question 25.
What are mitotic poisons.
Answer:
Certain chemical components act as inhibitors of the mitotic cell division and they are called mitotic poisons.

Question 26.
Distinguish between Anastral & Amphiastral.
Answer:
Between Anastral & Amphiastral:
Anastral:

1. This is present only in plant cells.
2. No asters or centrioles are formed only spindle fibres are formed during cell division.

Amphiastral:

1. This is found in animal cells.
2. Aster and centrioles are formed at each pole of the spindle during cell division.

Question 27.
Draw a simple diagram to show the Amitosis.
Answer:
The Amitosis:

III. Short Answer Type Questions (3 Marks)

Question 1.
What is the role of nucleus in the cell?
Answer:
The role of nucleus in the cell:

• Control activities of the cell.
• Genetic information copied from cell to cell while the cell divides.
• Hereditary characters are passed onto new individuals when gametic cells fuse together in sexual reproduction.

Question 2.
What are restriction points? Mention its role in Cell cycle.
Answer:
The checkpoint called the restriction point at the end of G₁ it determines a cells fate whether it will continue in the cell cycle and divide or enter a stage called G₀ as a quiescent stage and probably as specified cell or die.

Question 3.
Point out the reasons responsible for the arresting of the cell in G₁ phase?
Answer:
Cells are arrested in G₁ due to:

• Nutrient deprivation
• Lack of growth factors or density dependant inhibition
• Undergo metabolic changes and enter into G₀ state.

Question 4.
Write a note on G₀ phase.
Answer:
Some cells exit G₁ and enters a quiescent stage called G₀, where the cells remain metabolically active without proliferation. Cells can exist for long periods in G₀ phase. In G₀ cells cease growth with reduced rate of RN A and protein synthesis. The G₀ phase is not permanent. Mature neuron and skeletal muscle cell remain permanently in G₀. Many cells in animals remains in G₀ unless called onto proliferate by appropriate growth factors or other extracellular signals. G₀ cells are not dormant.

Question 5.
List out the events taking place in S – Phase.
Answer:
S Phase – Synthesis phase – cells with intermediate amounts of DNA Growth of the cell continues as replication of DNA occur, protein molecules called histones are synthesised and attach to the DNA. The centrioles duplicate in the cytoplasm. DNA content increases from 2C to 4C.

Question 6.
Distinguish between Karyokinesis & Cytokinesis.
Answer:
Karyokinesis:

1. Involves division of nucleus.
2. Nucleus develops a constriction at the center and becomes dumbellshaped.
3. Constriction deepens and divides the nucleus into two.

Cytokinesis:

1. Involves division of cytoplasm.
2. Plasma membrane develops a constriction along nuclear constriction.
3. It deepens centripetally and finally divides the cell into two cells.

Question 7.
Explain the differences between closed and open mitosis.
Answer:
Between closed and open mitosis:

1. In closed mitosis, the nuclear envelope remains intact and chromosomes migrate to opposite poles of a spindle within the nucleus. Example: Yeast and slime molds.
2. In open mitosis, the nuclear envelope breaks down and then reforms around the 2 sets of separated chromosome. Example: Most plants and animals cells.

Question 8.
What happens to plant cells at the end of Telophase in Mitosis?
Answer:
In plants, phragmoplast are formed between the daughter cells. Cell plate is formed between the two daughter cells, reconstruction of cell wall takes place. Finally the cells are separated by the distribution of organelles, macromolecules into two newly formed daughter cells.

Question 9.
Bring out the significance of Meiosis.
Answer:
The significance of Meiosis:

• Meiosis maintains a definite constant number of chromosomes in organisms.
• Crossing over takes place and exchange of genetic material leads to variations among species. These variations are the raw materials to evolution. Meiosis leads to genetic variability by partitioning different combinations of genes into gametes through independent assortment.
• Adaptation of organisms to various environmental stress.

Question 10.
Differentiate between the mitosis of Plant Cell & Animal Cell.
Answer:
Plants:

1. Centrioles are absent
2. Asters are not formed
3. Cell division involves formation of a cell plate
4. Occurs mainly at meristem.

Animals:

1. Centrioles are present
2. Asters are formed
3. Cell division involves furrowing and cleavage of cytoplasm
4. Occurs in tissues throughout the body.

Question 11.
Explain briefly about Endomitosis.
Answer:
The replication of chromosomes in the absence of nuclear division and cytoplasmic division resulting in numerous copies within each cell is called endomitosis. Chromonema do not separate to form chromosomes, but remain closely associated with each other. Nuclear membrane does not rupture. So no spindle formation. It occurs notably in the salivary glands of Drosophila and other flies. Cells in these tissues contain giant chromosomes (polyteny), each consisting of over thousands of intimately associated, or synapsed, chromatids. Example: Polytene chromosome.

Question 12.
How G₀ cells help in Closing Technology?
Answer:
Since the DNA of cells in G₀, do not replicate. The researcher are able to fuse the donor cells from a sheep’s mammary glands into G₀, state by culturing in the nutrient free state. The G₀, donor nucleus synchronised with cytoplasm of the recipient egg, which developed into the clone Dolly.

IV. Long Answer Type Questions (5 Marks)

Question 1.
Draw and label the various stages of Prophase I.
Answer:
Label the various stages of Prophase I:

Question 2.
Explain in detail about the various stages of Prophase I.
Answer:
The various stages of Prophase I:
1. Prophase I – Prophase I is of longer duration and it is divided into 5 substages – Leptotene, Zygotene, Pachytene, Diplotene and Diakinesis.

2. Leptotene – Chromosomes are visible under light microscope. Condensation of chromosomes takes place. Paired sister chromatids begin to condense.

3. Zygotene – Pairing of homologous chromosomes takes place and it is known as synapsis. Chromosome synapsis is made by the formation of synaptonemal complex. The complex formed by the homologous chromosomes are called as bivalent (tetrads).

4. Pachytene – At this stage bivalent chromosomes are clearly visible as tetrads. Bivalent of meiosis I consists of 4 chromatids and 2 centromeres. Synapsis is completed and recombination nodules appear at a site where crossing over takes place between non – sister chromatids of homologous chromosome. Recombination of homologous chromosomes is completed by the end of the stage but the chromosomes are linked at the sites of crossing over. This is mediated by the enzyme recombinase.

5. Diplotene – Synaptonemal complex disassembled and dissolves. The homologous chromosomes remain attached at one or more points where crossing over has taken place. These points of attachment where ‘X’ shaped structures occur at the sites of crossing over is called.

6. Chiasmata: Chiasmata are chromatin structures at sites where recombination has been taken place. They are specialised chromosomal structures that hold the homologous chromosomes together. Sister chromatids remain closely associated whereas the homologous chromosomes tend to separate from each other but are held together by chiasmata. This substage may last for days or years depending on the sex and organism. The chromosomes are very actively transcribed in females as the egg stores up materials for use during embryonic development. In animals, the chromosomes have prominent loops called lampbrush chromosome.

7. Diakinesis – Terminalisation of chiasmata. Spindle fibres assemble. Nuclear envelope breaks down. Homologous chromosomes become short and condensed. Nucleolus disappears.

Question 3.
Describe the process of Cytokinesis in Plant cell & Animal Cell.
Answer:
1. Cytokinesis in Plant Cell: Division of the cytoplasm often starts during telophase. In plants, cytokinesis cell plate grows from centre towards lateral walls – centrifugal manner of cell plate formation. Phragmoplast contains microtubules, actin filaments and vesicles from golgi apparatus and ER. The golgi vesicles contains carbohydrates such as pectin, hemicellulose which move along the microtubule of the pharagmoplast to the equator fuse, forming a new plasma membrane and the materials which are placed there becomes new cell wall.

The first stage of cell wall construction is a line dividing the newly forming cells called a cell plate. The cell plate eventually stretches right across the cell forming the middle lamella. Cellulose builds up on each side of the middle lamella to form the cell walls of two new plant cells.

2. Cytokinesis in Animal Cells:
It is a contractile process. The contractile mechanism contained in contractile ring located inside the plasma membrane. The ring consists of a bundle of microfilaments assembled from actin and myosin. This fibril helps for the generation of a contractile force. This force draws the contractile ring inward forming a cleavage furrow in the cell surface dividing the cell into two.

Question 4.
What are the significances of Mitosis.
Answer:
Exact copy of the parent cell is produced by mitosis (genetically identical).

1. Genetic stability – Daughter cells are genetically identical to parent cells.
2. Growth – As multicellular organisms grow, the number of cells making up their tissue increases. The new cells must be identical to the existing ones.
3. Repair of tissues – Damaged cells must be replaced by identical new cells by mitosis.
4. Asexual reproduction – Asexual reproduction results in offspring that are identical to the parent. Example Yeast and Amoeba.
5. In flowering plants, structure such as bulbs, corms, tubers, rhizomes and runners are produced by mitotic division. When they separate from the parent, they form a new individual. The production of large numbers of offsprings in a short period of time, is possible only by mitosis. In genetic engineering and biotechnology, tissues are grown by mitosis (i.e. in tissue culture).
6. Regeneration – Arms of star fish

Question 5.
Explain the various phases in Cell Cycle.
Answer:
The different phases of cell cycle are as follows:
1. Interphase: Longest part of the cell cycle, but it is of extremely variable length. At first glance the nucleus appears to be resting but this is not the case at all. The chromosomes previously visible as thread like structure, have dispersed. Now they are actively involved in protein synthesis, at least for most of the interphase. C – Value is the amount in picograms of DNA contained within a haploid nucleus.

2. G₁ Phase: The first gap phase – 2C amount of DNA in cells of G₁ The cells become metabolically active and grows by producing proteins, lipids, carbohydrates and cell organelles including mitochondria and endoplasmic reticulum. Many checkpoints control the cell cycle. The checkpoint called the restriction point at the end of G₁ it determines a cells fate whether it will continue in the cell cycle and divide or enter a stage called G₀ as a quiescent stage and probably as specified cell or die. Cells are arrested in G₁ due to:

3. Nutrient deprivation: Lack of growth factors or density dependant inhibition. Undergo metabolic changes and enter into G₀ state. Biochemicals inside cells activates the cell division. The proteins called kinases and cyclins activate genes and their proteins to perform cell division. Cyclins act as major checkpoint which operates in G₁ to determine whether or not a cell divides.

4. G₀ Phase: Some cells exit G₁ and enters a quiescent stage called G₀, where the cells remain metabolically active without proliferation. Cells can exist for long periods in G₀ phase. In G₀ cells cease growth with reduced rate of RNA and protein synthesis. The G₀ phase is not permanent. Mature neuron and skeletal muscle cell remain permanently in G₀. Many cells in animals remains in G₀ unless called on to proliferate by appropriate growth factors or other extracellular signals. G₀ cells are not dormant.

5. S phase – Synthesis phase – cells with intermediate amounts of DNA. Growth of the cell continues as replication of DNA occur, protein molecules called histones are synthesised and attached to the DNA. The centrioles duplicate in the cytoplasm. DNA content increases from 2C to 4C.

6. G₂ – The second Gap phase – 4C amount of DNA in cells of G₂ and mitosis. Cell growth continues by protein and cell organelle synthesis, mitochondria and chloroplasts divide. DNA content remains as 4C. Tubulin is synthesised and microtubules are formed. Microtubles organise to form spindle fibre. The spindle begins to form and nuclear division follows.

One of the proteins synthesized only in the G₂ period is known as Maturation Promoting Factor (MPF). It brings about condensation of interphase chromosomes into the mitotic form. DNA damage checkpoints operates in G₁ S and G₂ phases of the cell cycle.

Question 6.
List out the important features of Chromosomes.
Answer:
The four important features of the chromosome are:
1. The shape of the chromosome is specific: The long, thin, lengthy structured chromosome contains a short, constricted region called centromere. A centromere may occur any where along the chromosome, but it is always in the same position on any given chromosome. The number of chromosomes per species is fixed: For example the mouse has 40 chromosomes, the onion has 16 and humans have 46.

2. Chromosomes occur in pairs: The chromosomes of a cell occur in pairs, called homologous pairs. One of each pair come originally from each parent. Example, human has 46 chromosomes, 23 coming originally from each parent in the process of sexual reproduction. Chromosomes are copied: Between nuclear divisions, whilst the chromosomes are uncoiled and cannot be seen, each chromosome is copied. The two identical structures formed are called chromatids.

V. Higher Order Thinking Skills (HOTs)

Question 1.
Given that the average duplication time of E.coli is 20 minutes. How much time will two E.coli cells takes to become 32 cells?
Answer:

One cells takes 80 minutes to form 16 cells. If 2 cells undergoes division simultaneously, it take 160 minutes (2 hours 40 minutes) to form 32 cells.

Question 2.
Complete the cell cycle by filling the gaps with respective phases.
Answer:
X= S phase or Synthesis phase
Y= M phase or Mitosis phase
Z= G₀ phase

Question 3.
Telophase is reverse of prophase – Comment
Answer:
Events in Prophase:

1. Nuclear membrane disappears
2. Nucleolus disappear
3. Spindle fibre begins to form
4. Chromosomes threads condeme to form chromosomes

Events in Telophase:

1. Nuclear membrane reappears
2. Nucleolus reappears
3. Spindle fibre disappears
4. Chromosomes decondeme to form chromosomes

Question 4.
Name the pathological condition when uncontrolled cell division occurs.
Answer:
Uncontrolled cell division & abnormal growth of cells leads to the pathological condition called tumor or cancer.

Question 5.
Microspores are produced in the multiples of four, why?
Answer:
Microspores are haploid spores produced from diploid microspores mother cells. Each microspores mother cell (2n) undergoes meiosis producing four Microspores (n). Because a complete meiotic division yields 4 cells. Thus microspores are produced in multiples of four.

Question 6.
Between Prokaryotes & Eukaryotes, which cell has a shorter cell division time.
Answer:
Prokaryotes like bacteria undergo simple form of cell division called binary fission which will get completed with in a hour, whereas Eukaryotic cell division (mitosis) takes nearly 24 hours to get completed. Hence Prokaryotes have shorter cell division time.

Question 7.
Though Prokaryotic cell division differs from Eukaryotic cell division, both show certain common aspects during cell division. Explain.
Answer:
Whether a cell is Prokaryote or Eukaryote, while undergoing division, the following events must occur in common.

1. Replication of DNA.
2. Cytokinesis at the end of cell division.

Question 8.
An anther has 1204 pollen grains. How many Pollen mother cells must have been there to produce them?Explain.
Answer:
301 – Pollen mother cells: 301 Pollen mother cells undergo meiosis producing 1204 pollen grains. Because at the end of meiosis, each pollen mother cells produces 4 pollen grains.

Question 9.
A cell has 32 chromosomes. It undergoes mitosis. What will be the chromosome number during metaphase?
Answer:
During S phase of interphase, the genetic material of the cell is duplicated. So during metaphase, the chromosome number(chromatid number) will be doubled thus 64 chromosomes (chromatids) will be present.

Question 10.
Why sibilings show disimilarities?
Answer:
Though born to same parents, siblings show dissimilarities and variation due to the crossing over and recombination of chromosomes during meiosis.

Question 11.
Ramu’s met with an accident while riding cycle and got wounded in his leg. After few days, the wound was healed and the skin becomes normal. How?
Answer:
Ramu’s wound was healed because of the mitotic division. As a result of mitosis, new cells are produced and damaged tissues were repaired resulting the damaged skin to become normal.

Question 12.
A flower of tomato plant following the process of sexual reproduction produces 240 viable seeds. What is the minimum number of microspore mother cells involved in this process?
Answer:
60 microspore mother cells are involved in providing 240 pollen grains. Because each microspore mother cell undergoes meiosis producing four pollen grains (i.e. 60 × 4 = 240). Each pollen grain produces two male gametes of which one undergoes true fertilization of ovule producing seeds. Other male gamete participate in double fertilization.

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Samacheer Kalvi 11th Bio Botany Cell: The Unit of Life Text Book Back Questions and Answers

I. Multiple Choice Questions
Choose the correct answer
Question 1.
The two subunits of ribosomes remain united at critical ion level of …………… .
(a) Magnesium
(b) Calcium
(c) Sodium
(d) Ferrous
Answer:
(a) Magnesium

Question 2.
Sequences of which of the following is used to know the phylogeny?
(a) mRNA
(b) rRNA
(c) tRNA
(d) Hn RNA
Answer:
(b) rRNA

Question 3.
Many cells function properly and divide mitotically even though they do not have …………… .
(a) Plasma membrane
(b) Cytoskeleton
(c) Mitochondria
(d) Plastids
Answer:
(d) Plastids

Question 4.
Keeping in view the fluid mosaic model for the structure of cell membrane, which one of the following statements is correct with respect to the movement of lipids and proteins from one lipid monolayer to the other…………… .
(a) Neither lipid nor proteins can flip – flop
(b) Both lipid and proteins can flip – flop
(c) While lipids can rarely flip – flop proteins cannot
(d) While proteins can flip – flop lipids cannot
Answer:
(c) While lipids can rarely flip – flop proteins cannot

Question 5.
Match the columns and identify the correct option:

Answer:
(c) iii, iv, i, ii.

Question 6.
Bring out the significance of phase contrast microscopy.
Answer:
Phase contrast microscope is used to observe living cells, tissues and the cells cultured invitro during mitosis.

Question 7.
State the protoplasm theory.
Answer:
Protoplasm theory was proposed by Max Schultze which states that the protoplasm is the living content of cell and is a complex colloidal system.

Question 8.
Distinguish between prokaryotes and eukaryotes.
Answer:
Between prokaryotes and eukaryotes:

Question 9.
Difference between plant and animal ceil.
Answer:

Plant Cell	Animal Cell
1. Usually they are larger than animal cells	1. Usually smaller than plant cells
2. Cell wall present in addition to plasma membrane and consists of middle lamellae, primary and secondary walls	2. Cell wall absent
3. Plasmodesmata present	3. Plasmodesmata absent
4. Chloroplast present	4. Chloroplast absent
5. Vacuole large and permanent	5. Vacuole small and temporary
6. Tonoplast present around vacuole	6. Tonoplast absent
7. Centrioles absent except motile cells of lower plants	7. Centrioles present
8. Nucleus present along the periphery of the cell	8. Nucleus at the centre of the cell
9. Lysosomes are rare	9. Lysosomes present
10. Synthesizes amino acids, coenzymes and vitamins required by them	10. Cannot synthesize aminoacids, coenzymes and vitamins required by them
11. Storage material is starch grains	11. Storage material is a glycogen granules

Question 10.
Draw the ultra structure of Plant Cell.
Answer:
The ultra structure of Plant Cell:

Samacheer Kalvi 11th Bio Botany Cell: The Unit of Life Additional Questions and Answers

I. Multiple Choice Questions
Choose the correct answer:
Question 1.
Scientist who named the unicellular particles as ‘animalcules’ …………… .
(a) Aristotle
(b) Robert Brown
(c) Antonie von Leeuwenhoek
(d) Robert Hooke
Answer:
(c) Antonie van Leeuwenhoek

Question 2.
Cell theory was modified by …………… .
(a) Schwann
(b) Schleiden
(c) Virchow
(d) Dutrochet
Answer:
(c) Virchow

Question 3.
Which of the following microscope produce 3D – image?
(a) Phase contrast
(b) TEM
(c) SEM
(d) Dark field
Answer:
(c) SEM

Question 4.
Which of the following electron opaque chemical is used in Electron microscope?
(a) Strontium
(b) Deuterium
(c) Palladium
(d) Uranium
Answer:
(c) Palladium

Question 5.
Medium for electron movement in TEM is …………… .
(a) Air
(b) Oil
(c) Water
(d) Vacuum
Answer:
(d) Vacuum

Question 6.
Resolving power of SEM is …………… .
(a) 5 – 10 Å
(b) 2 – 10 Å
(c) 5 – 20 nm
(d) 5 – 20 m
Answer:
(c) 5 – 20 nm

Question 7.
Which among the following is NOT an exception to cell theory?
(a) Viruses
(b) Viroids
(c) Prions
(d) Fungi
Answer:
(d) Fungi

Question 8.
Scientist who named the cytoplasm as “Sarcode” is …………… .
(a) Dujardin
(b) Corti
(c) Purkinje
(d) Hugo Van Mohl
Answer:
(a) Dujardin

Question 9.
The pH of protoplasm is around …………… .
(a) 6.6
(b) 6.7
(c) 6.8
(d) 6.9
Answer:
(c) 6.8

Question 10.
The refractive index of protoplasm is …………… .
(a) 1.4
(b) 2.4
(c) 3.4
(d) 0.4
Answer:
(a) 1.4

Question 11.
Histone proteins are seen in the DNA of …………… .
(a) Pseudokaryotes
(b) Prokaryotes
(c) Mesokaryotes
(d) Eukaryotes
Answer:
(d) Eukaryotes

Question 12.
Which of the following organelle is believed to be an endosymbiont?
(a) Ribosomes
(b) Mitochondrion
(c) Golgi bodies
(d) Nucleus
Answer:
(b) Mitochondrion

Question 13.
is the living content of the cell …………… .
(a) Cytoplasm
(b) Protoplasm
(c) Nucleoplasm
(d) Nucleus
Answer:
(b) Protoplasm

Question 14.
Fungal cell wall is made of …………… .
(a) Cutin
(b) Chitin
(c) Hemicellulose
(d) Pectin
Answer:
(b) Chitin

Question 15.
…………… acts as a channel between the protoplasm of adjacent cells.
(a) Middle lamellae
(b) Pits
(c) Plasmodesmata
(d) Primary cell wall
Answer:
(c) Plasmodesmata

Question 16.
Fluid mosaic model was proposed by …………… .
(a) Schleiden and Schwann
(b) Singer and Nicolson
(c) Binning and Roher
(d) G. Palade
Answer:
(b) Singer and Nicolson

Question 17.
Which is the largest of the internal membranes?
(a) Golgi bodies
(b) Endoplasmic reticulum
(c) Tonoplast
(d) Nuclear membrane
Answer:
(b) Endoplasmic reticulum

Question 18.
In plant cells, golgi bodies are found as small vesicles called …………… .
(a) Polysomes
(b) Cytosomes
(c) Cytosol
(d) Dictyosomes
Answer:
(d) Dictyosomes

Question 19.
organelle plays a major role in post translation process of protein …………… .
(a) Golgi bodies
(b) Nucleolus
(c) Ribosomes
(d) ER
Answer:
(a) Golgi bodies

Question 20.
Zymogen granules are synthesized in …………… .
(a) Lysosomes
(b) Golgi bodies
(c) Mitochondria
(d) Chloroplast
Answer:
(b) Golgi bodies

Question 21.
Altmann named mitochondrion as …………… .
(a) Apoplast
(b) Elaioplast
(c) Symplast
(d) Bioplast
Answer:
(d) Bioplast

Question 22.
DNA of mitochondrion is …………… .
(a) Helical
(b) Dumb – bell
(c) Circular
(d) Spiral
Answer:
(c) Circular

Question 23.
Mitochondria are inherited from parent …………… .
(a) Male
(b) Female
(c) Both
(d) None
Answer:
(b) Female

Question 24.
F₁ particles are also called as …………… .
(a) Polysomes
(b) Glyoxysomes
(c) Peroxisomes
(d) Oxysomes
Answer:
(d) Oxysomes

Question 25.
Elaioplasts store …………… .
(a) Starch
(b) Lipid
(c) Protein
(d) Chlorophyll
Answer:
(b) Lipid

Question 26.
The photosynthetic units are called as …………… .
(a) Oxysomes
(b) Quantosomes
(c) Thylakoids
(d) Chloroplasts
Answer:
(b) Quantosomes

Question 27.
Which organelle is not membrane bound?
(a) Mitochondrion
(b) Golgi bodies
(c) Chloroplast
(d) Ribosomes
Answer:
(d) Ribosomes

Question 28.
Ribosomes of mitochondrion are …………… .
(a) 16 S
(b) 80 S
(c) 70 S
(d) 50 S
Answer:
(c) 70 S

Question 29.
…………… mineral is required for structural cohesion of ribosomes.
(a) Ca²⁺
(b) H⁺
(c) Mg²⁺
(d) Cl^–
Answer:
(c) Mg²⁺

Question 30.
Lysosomes originate from …………… .
(a) Mitochondrion
(b) Nucleus
(c) ER
(d) Golgi bodies
Answer:
(d) Golgi bodies

Question 31.
In mammals, peroxisomes are seen in …………… cells.
(a) Brain
(b) Lung
(c) Liver
(d) Heart
Answer:
(c) Liver

Question 32.
Which organelle has a single unit membrane?
(a) Ribosomes
(b) Glyoxysomes
(c) Polysomes
(d) Nucleus
Answer:
(b) Glyoxysomes

Question 33.
The single unit membrane of vacuoles is called as …………… .
(a) Tonoplast
(b) Symplast
(c) Apoplast
(d) Amyloplast
Answer:
(a) Tonoplast

Question 34.
Vacuoles of Apple cells store …………… .
(a) Sucrose
(b) Malic acid
(c) Citrate
(d) Flavanoid
Answer:
(b) Malic acid

Question 35.
Ribosomal biogenesis occur at …………… .
(a) Mitochondrion
(b) Polysomes
(c) Nucleolus
(d) Chromosomes
Answer:
(c) Nucleolus

Question 36.
The term chromosome was introduced by …………… .
(a) Bridges
(b) Strasburger
(c) Waldeyer
(d) Poster
Answer:
(c) Waldeyer

Question 37.
Stability to chromosome is offered by …………… .
(a) Satellite
(b) Telomere
(c) Kinetochore
(d) Nucleolus
Answer:
(b) Telomere

Question 38.
Life span of the cells is determined by …………… .
(a) Kinetochore
(b) Satellite
(c) Chromatin
(d) Telomere
Answer:
(d) Telomere

Question 39.
The metacentric chromosomes are …………… shaped.
(a) L
(b) V
(c) J
(d) I
Answer:
(b) V

Question 40.
Polytene chromosomes are observed in …………… of Drosophila.
(a) Endocrine gland
(b) Gall bladder
(c) Salivary gland
(d) Exocrine gland
Answer:
(c) Salivary gland

Question 41.
Lampbrush chromosomes occur at …………… stage of meiotic Prophase I.
(a) Leptotene
(b) Diplotene
(c) Zygotene
(d) Pachytene
Answer:
(b) Diplotene

Question 42.
Number of basal rings in gram positive bacteria …………… .
(a) 2
(b) 4
(c) 6
(d) 8
Answer:
(a) 2

Question 43.
Microtubules are made of …………… .
(a) Dyenin
(b) Tubin
(c) Tubulin
(d) Nexin
Answer:
(c) Tubulin

Question 44.
Cytoplasm is stained …………… by eosin.
(a) Pink
(b) Blue
(c) Greenish blue
(d) Green
Answer:
(a) Pink

Question 45.
Key difference between plant cell & animal cell is …………… .
(a) Ribosomes
(b) Vacuoles
(c) Cell wall
(d) Centrioles
Answer:
(c) Cell wall

II. Very Short Answer Type Questions (2 Marks)

Question 1.
Name the scientist who proposed the cell theory.
Answer:
Matthias Schleiden and Theodor Schwann.

Question 2.
Define resolving power of a microscope.
Answer:
Resolving power or resolution refers to the ability of the lenses to show the details of object lying between two points. It is the finest detail available from an object. It can be calculated using the following formula.
⇒ Resolution = \(\frac {0.61λ}{NA}\),
where λ, = wavelength of the light and
NA is the numerical aperture.

Question 3.
Define magnification. How will you calculate it?
Answer:
The optical increase in the size of an image is called magnification. It is calculated by the following formula

Question 4.
List out the types of objective lenses used in Bright field microscope.
Answer:
The types of objective lenses used in Bright field microscope:

1. 5x
2. 10x
3. 45x
4. 100x

Question 5.
In Bright field microscope, where does the primary & secondary magnification occurs?
Answer:
Primary magnification is obtained through, objective lens and secondary magnification is obtained through eye piece lens.

Question 6.
List out the components of Electron microscope.
Answer:
The components of the microscope are as follows:

1. Electron Generating System
2. Electron Condensor
3. Specimen Objective
4. Tube Lens
5. Projector

Question 7.
Name the organisms that are exceptions to cell theory.
Answer:
Viruses, Virions and Prions.

Question 8.
Name the types of cells based on nuclear characteristics.
Answer:
The types of cells based on nuclear characteristics:

1. Prokaryotes
2. Mesokaryotes and
3. Eukaryotes.

Question 9.
Define nucleoid.
Answer:
Nucleoid refers to the primitive nucleus which is not bound by nuclear membrane and the DNA is without histone protein.

Question 10.
Name the organelles which are believed to be endosymbiont in Eukaryote cell.
Answer:
Chloroplast and Mitochondrion.

Question 11.
Write a note on endosymbiont theory.
Answer:
Two eukaryotic organelles believed to be the descendants of the endosymbiotic prokaryotes. The ancestors of the eukaryotic cell engulfed a bacterium and the bacteria continued to function inside the host cell.

Question 12.
Point out any four prokaryotes.
Answer:Four prokaryotes:

1. Bacteria
2. Blue Green algae
3. Mycoplasma and
4. Rickettsiae.

Question 13.
Why spirochaetae is said to be a prokaryote?
Answer:
Spirochaetae is a prokaryote since its DNA lies in nucleoid and not bound by nuclear membrane and also devoid of histone protein.

Question 14.
Name any two unique structures / organelles of a plant cell.
Answer:
Cell wall and Chloroplast.

Question 15.
What are the components of protoplasm?
Answer:
Protoplasm is composed of a mixture of small particles, such as ions, amino acids, monosaccharides, water, macro-molecules like nucleic acids, proteins, lipids and polysaccharides.

Question 16.
What is the cell wall composition of the following organism?
(a) Fungi
(b) Bacteria
(c) Algae
Answer:
(a) Fungi – Chitin and fungal cellulose.
(b) Bacteria – Peptidoglycan
(c) Algae – Cellulose, mannan and galactan.

Question 17.
Which cell wall layer of plant cell is laid during maturation? Mention its role.
Answer:
Secondary wall is laid during maturation. It plays a key role in determining the shape of a cell.

Question 18.
How pits are formed?
Answer:
In plant cell, at few regions the secondary wall layer is laid unevenly, whereas the primary wall and middle lamellae are laid continuously such regions are called pits.

Question 19.
What is the role of plasmodesmata in a plant cell?
Answer:
Plasmodesmata act as a channel between the protoplasm of adjacent cells through which many substances pass through.

Question 20.
Name the two types of Pits.
Answer:
The two types of Pits:

1. Simple pit and
2. Bordered pit.

Question 21.
Define flip – flop movement.
Answer:
The movement of membrane lipids from one side of the membrane to the other side by vertical movement is called flip – flopping or flip – flop movement.

Question 22.
Name the two types of protein seen in cell membrane.
Answer:
Two types of protein seen in cell membrane:

1. Integral proteins and
2. peripheral proteins.

Question 23.
Define cytoplasmic streaming.
Answer:
Cytoplasmic streaming refers to the movement of the cytoplasm along with the cellular materials inside the cell.

Question 24.
Name any four endomembrane structures.
Answer:
Four endomembrane structures:

1. Nuclear membrane
2. endoplasmic reticulum
3. vascular membrane
4. golgi bodies.

Question 25.
What are dictyosomes?
Answer:
In plant cells, the golgi bodies are found as small vesicles which are called as dictyosomes.

Question 26.
What are porins?
Answer:
Porins are the proteins in the outer membrane of mitochondrion which forms the channel that allows the free diffusion of molecules smaller than about 1000 daltons.

Question 27.
What are oxysomes?
Answer:
The inner membrane of mitochondrion consists of stalked particles called elementary particles or Fernandez Moran particles. F₁ particles or oxysomes.

Question 28.
How mitochondria helps in evolutionary studies?
Answer:
Mitochondrial DNA is used to track and date recent evolutionary time because it mutates 5 to 10 time faster than DNA in the nucleus.

Question 29.
Why mitochondrion is called as semi – autonomous organelle?
Answer:
Since mitochondria has it own DNA, it is called as Organelle.

Question 30.
Why mitochondria are called “Power houses of cell?
Answer:
Mitochondria are called Power house of a cell, as they produce energy rich (Adenosine Triphosphate) ATP.

Question 31.
Classify plastids based on colour.
Answer:
Chloroplast, Phaeoplast and Rhodoplast.

Question 32.
Classify plastids based on storage & mention their storage component.
Answer:

Plastids	Storage Content
1. Amyloplast	1. Starch
2. Elaioplast	2. Lipids / Oils
3. Aleuroplast or Proteoplast	3. Protein

Question 33.
What are Quantosomes?
Answer:
Quantosomes are small, rounded photosynthetic units present in thylakoids.

Question 34.
List out the functions of chloroplast.
Answer:
Photosynthesis, photo respiration.

Question 35.
What is Svedberg unit?
Answer:
Svedberg unit is a measure of a particle size dependent on the speed with which particle sediment in the ultracentrifuge.

Question 36.
Where the biogenesis of ribosomes occur?
Answer:
Biogenesis of ribosome are denova formation, auto replication and nucleolar origin.

Question 37.
What are Polysomes? State its function.
Answer:
During protein synthesis many ribosomes are attached to the single mRNA called polysomes or polyribosomes. The function of polysomes is the formation of several copies of a particular polypeptide during protein synthesis.

Question 38.
Define Autolysis.
Answer:
Autolysis is the process, where the lysosome causes self destruction of cell on insight of disease they destroy the cells.

Question 39.
List out the enzymes of lysosomes.
Answer:
Acid Hydrolases, Nuclease, Proteases, Glycosidases, Lipases, Phosphatases, Sulphatases and Phospholipidases.

Question 40.
Name any two organelles involved in photorespiration.
Answer:
Peroxisomes and Chloroplast.

Question 41.
Name few single unit membrane bound organelles.
Answer:
Peroxisomes, Glyoxysomes and Sphaerosomes.

Question 42.
What are Tonoplast?
Answer:
In plant cells vacuoles are large, bounded by a single unit membrane called Tonoplast.

Question 43.
How vacuoles helps to maintain the structure of a plant cell?
Answer:
The major function of plant vacuole is to maintain water pressure known as turgor pressure, which maintains the plant structure.

Question 44.
What are metachromatic granules?
Answer:
Inorganic inclusions in bacteria are polyphosphate granules (volutin granules) and sulphur granules. These granules are also known as metachromatic granules.

Question 45.
Which is the largest organelle in a cell? State its function.
Answer:
Nucleus. It controls all the cellular activities.

Question 46.
What is a pore complex?
Answer:
The pores are enclosed by circular structures called annuli. The pore and annuli forms the pore complex.

Question 47.
What are nucleolar organizers region?
Answer:
Secondary constrictions contains the genes for ribosomal RNA which induce the formation of nucleoli and are called nucleolar organizer regions.

Question 48.
Draw the types of chromosomes based on centromere position.
Answer:
The types of chromosomes based on centromere position:

Question 49.
What is a telomere?
Answer:
Telomere is the terminal part of chromosome. It offers stability to the chromosome. DNA of the telomere has specific sequence of nucleotides.

Question 50.
Classify chromosomes based on function.
Answer:
Autosomes and sex chromosomes.

Question 51.
Name any two giant chromosomes.
Answer:
Polytene chromosomes and lamp brush chromosomes.

Question 52.
Define proton motive force.
Answer:
The proton motive force is the force derived from the electrical potential and the hydrogen ion gradient across the cytoplasmic membrane.

Question 53.
Name the basal rings found in flagella of gram negative bacteria.
Answer:
The basal rings found in flagella of gram negative bacteria:

• L – Lipopolysacchride ring
• P – Peptidoglycan ring
• S – Super membrane ring and
• M – membrane ring.

Question 54.
(a) Define histochemistry.
(b) What is Somatic pairing?
Answer:
(a) The technique of staining the cells and tissue is called histochemical staining or histo chemistry.
(b) Maternal and paternal homologues remain associated side by side is called somatic pairing.

III. Short Answer Type Questions (3 Marks)

Question 1.
Explain the principle involved in Scanning Electron Microscope (SEM).
Answer:
In Scanning Electron Microscope (SEM) electrons are focused by means of lenses into a very fine point. The interaction of electrons with the specimen results in the release of different forms of radiation (such as auger electrons, secondary electrons, back scattered electrons) from the surface of the specimen. These radiations are then captured by an appropriate detector, amplified and then imaged on fluorescent screen.

Question 2.
Write a note on solation gelation property of protoplasm.
Answer:
The protoplasm exist either in semisolid (jelly – like) state called ‘gel’ due to suspended particles and various chemical bonds or may be liquid state called ‘sol’. The colloidal protoplasm which is in gel form can change into sol form by solation and the sol can change into gel by gelation. These gel – sol conditions of colloidal system are prime basis for mechanical behaviour of cytoplasm.

Question 3.
Explain the nuclear characters of Mesokaryotes.
Answer:
Mesokaryotes contains well organized nucleus with nuclear membrane and the DNA is organized into chromosomes but without histone protein components divides through amitosis similar with prokaryotes.

Question 4.
List out the unique characters of a Eukaryotic cell.
Answer:
Eukaryotes have true nucleus. The DNA is associated with protein bound histones forming the chromosomes. Membrane bound organelles are present.

Question 5.
Name the chemicals seen in the cell wall of plant cells.
Answer:
Cellulose, hemicellulose, pectin, lignin, cutin, suberin and silica.

Question 6.
Name the 3 distinct regions of plant cell wall.
Answer:
In plant, cell wall shows three distinct regions

1. Primary wall
2. Secondary wall and
3. Middle lamellae.

Question 7.
Explain the role of hemicellulose, pectin & glycoprotein in primary cell wall.
Answer:
Hemicellulose binds the microfibrils with matrix and glycoproteins control the orientation of microfibrils while pectin serves as filling material of the matrix.

Question 8.
In cell membrane, phospholipids undergo flip – flop movement but not the protein. Why?
Answer:
The phospholipids can have flip – flop movement because the phospholipids have smaller polar regions, whereas the proteins cannot flip – flop because the polar region is extensive.

Question 9.
Define signal transduction.
Answer:
The process by which the cells receive information from outside and respond is called signal transduction.

Question 10.
Distinguish between rough endoplasmic reticulum and smooth endoplasmic reticulum.
Answer:
Between rough endoplasmic reticulum and smooth endoplasmic reticulum:

RER	SER
1. Ribosomes are present in the outer surface of membrane	1. Ribosomes are absent on the membrane
2. It is involved in protein synthesis	2. It is involved in lipid synthesis

Question 11.
Write the major roles of Golgi bodies.
Answer:
Golgi complex plays a major role in post translational modification of proteins and glycosidation of lipids.

Question 12.
Which is the most abundant protein on Earth? Where it is encoded?
Answer:
Rubisco is the abundant protein on Earth. It is encoded by the chloroplast DNA.

Question 13.
Classify ribosomes with an example.
Answer:
Types of Ribosomes
1. 70S Ribosomes (subunit 30S and 50S): 3 RNA molecule

• 16SrRNA in 30S subunit
• 23S and 5S in 50S large subunit
  (Prokaryotic cells of Blue green algae, Bacteria, Mitochondria and Chloroplast of many Algae and higher plants)

2. 80S Ribosomes (subunits 40S and 60S): 4 RNA molecule

• 18Sr RNA in 40S small sub unit
• 28S, 5.8S and 5S in larger 60S subunit.
  (Eukaryotic cells of plants and animals).

Question 14.
Write a note on Glyoxysomes.
Answer:
Glyoxysome is a single membrane bound organelle. It is a sub cellular organelle and contains enzymes of glyoxylate pathway – oxidation of fatty acid occurs in glyoxysomes of germinating seeds.

Question 15.
What are cell inclusions? Give example.
Answer:
The cell inclusions are the non – living materials present in the cytoplasm. They are organic and inorganic compounds.  Example: Phosphate granules.

Question 16.
Draw and label the structure of Nucleus.
Answer:
The structure of Nucleus:

Question 17.
Distinguish between Euchromatin & Heterochromatin.
Answer:
The portion of Eukaryotic chromosome which is transcribed into mRNA contains active genes that are not tightly condensed during interphase is called Euchromatin. The portion of a Eukaryotic chromosome that is not transcribed into mRNA which remains condensed during interphase and stains intensely is called Heterochromatin.

Question 18.
Define Kinetochore.
Answer:
The centromere contains a complex system of protein fibres called kinetochore. Kinetochore is the region of chromosome which is attached to the spindle fibre during mitosis.

Question 19.
Write a note on SAT – chromosome.
Answer:
A satellite or SAT – chromosome are short chromosomal segment or rounded body separated from main chromosome by a relatively elongated secondary constriction. It is a morphological entity in certain chromosomes.

Question 20.
How telomeres helps in cancer studies?
Answer:
Maintenance of telomeres appears to be an important factor in determining the life span and reproductive capacity of cells so studies of telomeres and telomerase have the promise of providing new insights into conditions such as ageing and cancer.

Question 21.
Name the three types of centromere in eukaryotes.
Answer:
The three types of centromere in eukaryotes:

1. Point centromere
2. Regional centromere and
3. Holocentromere.

Question 22.
What are giant chromosomes?
Answer:
These chromosomes are larger in size and are called giant chromosomes. In certain plants they are found in the suspensors of the embryo. The polytene chromosome and lamp brush chromosome occur in animals and are also called as giant chromosomes.

Question 23.
How polyteny condition is achieved?
Answer:
Polyteny is achieved by repeated replication of chromosomal DNA several times without nuclear division and the daughter chromatids aligned side by side and do not separate (endomitosis).

Question 24.
What is Microphotography?
Answer:
Images of structures observed through microscopes can be further magnified, projected and saved by attaching a camera to the microscope by a microscope coupler or eyepiece adaptor. Picture taken using a inbuilt camera in a microscope is called microphotography or microphotograph.

Question 25.
Draw the Structure of Peroxisomes.
Answer:
The Structure of Peroxisomes:

IV. Long Answer Type Questions (5 Marks)

Question 1.
Write a detailed account on Dark field microscope.
Answer:
The dark field microscope was discovered by Z. Sigmondy (1905). Here the field will be dark but object will be glistening so the appearance will be bright. A special effect in an ordinary microscope is brought about by means of a special component called “Patch Stop Carrier”. It is fixed in metal ring of the condenser component.

Patch stop is a small glass device which has a dark patch at centre of the disc leaving a small area along the margin through which the light passes. The light passing through the margin will travel oblique like a hollow cone and strikes the object in the periphery, therefore the specimen appears glistening in a dark background.

Question 2.
Compare Transmission Electron Microscope with Scanning Electron Microscope.
Answer:
Transmission Electron Microscope with Scanning Electron Microscope:

Question 3.
List out the features of Cell Doctrine.
Answer:
The features of cell doctrine are as follows:

1. All organisms are made up of cells.
2. New cells are produced from the pre – existing cells.
3. Cell is a structural and functional unit of all living organisms.
4. A cell contains hereditary information which is passed on from cell to cell during cell division.
5. All the cells are basically the same in chemical composition and metabolic activities.
6. The structure and function of cell is controlled by DNA.
7. Sometimes the dead cells may remain functional as tracheids and vessels in plants and horny cells in animals.

Question 4.
Enumerate the properties of protoplasm.
Answer:
The properties of protoplasm:

1. Protoplasm is translucent, odourless and polyphasic fluid.
2. It is a crystal colloid solution which is a mixture of chemical substances forming crystalloid and colloidal solution.
3. Protoplasm exhibits three Brownian movement, amoeboid movement and cytoplasmic streaming or cyclosis. Viscosity of protoplasm is 2 – 20 centipoises. The Refractive index of the protoplasm is 1.4.
4. The pH of the protoplasm is around 6.8, contain 90% water (10% in dormant seeds).
5. Approximately 34 elements are present in protoplasm but only 13 elements are main or universal elements i.e. C, H, O, N, Cl, Ca, P, Na, K, S, Mg, I and Fe. Carbon, Hydrogen, Oxygen and Nitrogen form the 96% of protoplasm.
6. Protoplasm is neither a good nor a bad conductor of electricity.
7. Cohesiveness: Particles or molecules of protoplasm are adhered with each other by forces, such as Van der Waal’s bonds, that hold long chains of molecules together. This property varies with the strength of these forces.
8. Contractility: The contractility of protoplasm is important for the absorption and removal of water especially stomatal operations.
9. Surface Tension: The proteins and lipids of the protoplasm have less surface tension, hence they are found at the surface forming the membrane. On the other hand the chemical substances (NaCl) have high surface tension, so they occur in deeper parts of the cell protoplasm.

Question 5.
List out the functions of the Cell Wall.
Answer:
The cell wall plays a vital role in holding several important functions given below.

1. Offers definite shape and rigidity to the cell.
2. Serves as barrier for several molecules to enter the cells.
3. Provides protection to the internal protoplasm against mechanical injury.
4. Prevents the bursting of cells by maintaining the osmotic pressure.
5. Plays a major role by acting as a mechanism of defense for the cells.

Question 6.
Explain in detail about Fluid mosaic model.
Answer:
Jonathan Singer and Garth Nicolson (1972) proposed fluid model: It is made up of lipids and proteins together with a little amount of carbohydrate. The lipid membrane is made up of phospholipid. The phospholipid molecule has a hydrophobic tail and hydrophilic head. The hydrophobic tail repels water and water loving polar molecule are called hydrophilic molecule. They have polar phosphate group responsible for attracting water. Water hating non – polar molecule are called as hydrophobic molecule. They have fatty acid which is non – polar which cannot attract water.

Hydrophilic head attracts water. The proteins of the membrane are globular proteins which are found intermingled between the lipid bilayer most of which are projecting beyond the lipid bilayer. These proteins are called as integral proteins. Few are superficially attached on either surface of the lipid bilayer which are called as peripheral proteins. The proteins are involved in transport of molecules across the membranes and also act as enzymes, receptors or antigens.

Question 7.
Describe Endocytosis and Exocytosis.
Answer:
Cell surface membrane are able to transport individual molecules and ions. There are processes in which a cell can transport a large quantity of solids and liquids into cell endocytosis or out of cell exocytosis.

1. Endocytosis: During endocytosis the cell wraps the cell surface membrane around the material and brings it into cytoplasm inside a vesicle. There are two types of endocytosis:

• Phagocytosis – Particle is engulfed by membrane, which folds around it and forms a vesicle. The enzymes digest the material and products are absorbed by cytoplasm.
• Pinocytosis – Fluid droplets are engulfed by membrane, which forms vesicles around them.

2. Exocytosis: Vesicles fuse with plasma membrane and eject contents. This passage of material out the cell is known as exocytosis. This material may be a secretion in the case of digestive enzymes, hormones or mucus.

Question 8.
List out the functions of Golgi bodies.
Answer:
Functions of Golgi bodies:

1. Glycoproteins and glycolipids are produced.
2. Transporting and storing lipids.
3. Formation of lysosomes.
4. Production of digestive enzymes.
5. Cell plate and cell wall formation
6. Secretion of carbohydrates for the formation of plant cell walls and insect cuticles.
7. Zymogen granules (proenzyme / pre – cursor of all enzyme) are synthesized.

Question 9.
Draw & describe the structure of mitochondrion.
Answer:
Mitochondria are ovoid, rounded, rod shape and pleomorphic structures. Mitochondrion consists of double membrane, the outer and inner membrane. The outer membrane is smooth, highly permeable to small molecules and it contains proteins called Porins, which form channels that allows free diffusion of molecules smaller than about 1000 daltons and the inner membrane divides the mitochondrion into two compartments, outer chamber between two membranes and the inner chamber filled with matrix.

The inner membrane is convoluted (infoldings), called crista (plural: cristae). Cristae contain most of the enzymes for electron transport system. Inner chamber of the mitochondrion is filled with proteinaceous material called mitochondrial matrix. The inner membrane consists of stalked particles called elementary particles or Fernandez Moran particles. F₁ particles or Oxysomes. Each particle consists of a base, stem and a round head. In the head ATP synthase is present for oxidative phosphorylation. Inner membrane is impermeable to most ions, small molecules and maintains the proton gradient that drives oxidative phosphorylation.

Question 10.
Draw & describe the structure of Nucleus.
Answer:
Nucleus is an important unit of cell which control all activities of the cell. Nucleus holds the hereditary information. It is the largest among all cell organelles. It may be spherical, cuboidal, ellipsoidal or discoidal. It is surrounded by a double membrane structure called nuclear envelope, which has the inner and outer membrane. The inner membrane is smooth without ribosomes and the outer membrane is rough by the presence of ribosomes and is continues with irregular and infrequent intervals with the endoplasmic reticulum.

The membrane is perforated by pores known as nuclear pores which allows materials such as mRNA, ribosomal units, proteins and other macromolecules to pass in and out of the nucleus. The pores enclosed by circular structures called annuli. The pore and annuli forms the pore complex. The space between two membranes is called perinuclear space. Nuclear space is filled with nucleoplasm, a gelatinous matrix has uncondensed chromatin network and a conspicuous nucleoli. The chromatin network is the uncoiled, indistinct and remain thread like during the interphase. It has little amount of RNA and DNA bound to histone proteins in eukaryotic cells.

Question 11.
Write in detail about the 3 types of centromere in eukaryotes.
Answer:
There are three types of centromere in Eukaryotes. They are as follows:

1. Point Centromere: The type of centromere in which the kinetochore is assembled as a result of protein recognition of specific DNA sequences. Kinetochores assembled on point centromere bind a single microtubule. It is also called as localized centromere. It occurs in budding yeasts.
2. Regional Centromere: In regional centromere where the kinetochore is assembled on a variable array of repeated DNA sequences. Kinetochore assembled on regional centromeres bind multiple microtubules. It occurs in fission yeast cell, humans and so on.
3. Holocentromere: The microtubules bind all the along the mitotic chromosome. Example: Caenorbabditis elegans (transparent nematode) and many insects.

Question 12.
List the functions of Nucleus.
Answer:
Functions of the Nucleus:

1. Controlling all the cellular activities
2. Storing the genetic or hereditary information
3. Coding the information in the DNA for the production of enzymes and proteins.
4. DNA duplication and transcription takes place in the nucleus.
5. In nucleolus ribosomal biogenesis takes place.

Question 13.
Explain the structure and movement of Eukaryotic flagella.
Answer:
Structure: Eukaryotic Flagella are enclosed by unit membrane and it arises from a basal body. Flagella is composed of outer nine pairs of microtubules with two microtubules in its centre (9 + 2 arrangement). Flagella are microtubule projection of the plasma membrane. Flagellum is longer than cilium (as long as 200 m). The structure of flagellum has an axoneme made up microtubules and protein tubulin.

Movement: Outer microtubule doublet is associated with axonemal dynein which generates force for movement. The movement is ATP driven. The interaction between tubulin and dynein is the mechanism for the contraction of cilia and flagella. Dynein molecules uses energy from ATP to shift the adjacent microtubules. This movement bends the cilium or flagellum.

Question 14.
Describe the steps involved in cytologieal techniques.
Answer:
There are different types of mounting based on the portion of a specimen to be observed.

1. Whole mount: The whole organism or smaller structure is mounted over a slide and observed.
2. Squash: Is a preparation where the material to be observed is crushed/squashed onto a slide so as to reveal their contents. Example: Pollen grains, mitosis and meiosis in root tips and flower buds to observe chromosomes.
3. Smears: Here the specimen is in the fluid (blood and microbial cultures etc) are scraped, brushed or aspirated from surface of organ. Example: Epithelial cells.
4. Sections: Free hand sections from a specimen and thin sections are selected, stained and mounted on a slide. Example: Leaf and stem of plants.

Question 15.
Name any 5 common stains their colour & their affinity used in cytologieal studies.
Answer:
Common Stains used in Histochemistry

V. Higher Order Thinking Skills (HOTs)

Question 1.
What makes the plant cell more rigid than animal cells?
Answer:
Plants cells posses cell wall which provides sufficient rigidity and proper shape to them whereas in case of animal cells, cell wall is totally absent.

Question 2.
Cleaning organelle in the cell – Explain.
Answer:
Lysosomes contains a variety of hydrolytic enzymes, which can digest the materials within the cell. Thus lysosomes act as cleaning organelle of the cell.

Question 3.
Ribosomes are single membrane organelles present in both prokaryotes & eukaryotes. List out the sites where ribosomes are present in plant cell.
Answer:
The sites where ribosomes are present in plant cell:

1. Cytoplasm
2. On rough endoplasmic reticulum
3. Inside mitochondira & chloroplast.

Question 4.
What does ‘S’ refer in a 70 S and an 80 S ribosomes?
Answer:
‘S’ refers to Svedberg units named after Theoder Svedberg. The size of ribosome and their subunits are represented by Svedberg unit.

Question 5.
Briefly give the contributions of the following scientists in the field of cytology.
(a) Schleiden and Schwann
(b) Singer and Nicolson
Answer:
(a) Schleiden and Schwann proposed the cell theory.
(b) Singer & Nicolson proposed the fluid mosaic model of plasmomembrane.

Question 6.
Is extra genomic DNA is present in prokaryotes & Eukaryotes? If yes, locate them in both the types of organisms.
Answer:
Locate them in both the types of organisms:

1. In prokaryotes like bacteria, plasmids are the extra genomic DNA present in cytoplasm.
2. In Eukaryotes, the circular DNA present in matrix of mitochondria & chloroplast are extragenomic DNA.

Question 7.
Give possible reasons to call mitochondria & chloroplast as semi – autonomous organelles.
Answer:
Mitochondria & chloroplasts are considered as semi – autonomoius organelle due to following
facts.

1. Both mitochondria & chloroplasts have their own DNA, which can replicate independently.
2. They have their own ribosomes by which they self-synthesize same of their proteins without depending on cellular DNA.

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