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Q.1 (A) Choose the correct option


Balbharathi solution, chapter 6, plant water relation, science, biology, maharashtra board, hsc, full solution, latest edition,

Multiple Choice Question:

1.In soil, water available for absorption by root is ______.

OPTIONS

  • gravitational water
  • capillary water
  • hygroscopic water
  • combined water

2.The most widely accepted theory for ascent of sap is ______.

OPTIONS

  • capillarity theory
  • root pressure theory
  • diffusion
  • transpiration pull theory

3.Water movement between the cells is due to ______.

OPTIONS

  • T.P.
  • W.P.
  • DPD
  • incipient plasmolysiS

4.In guard cells, when sugar is converted into starch, the stomatal pore ______.

OPTIONS

  • closes almost completely
  • opens partially
  • opens fully
  • remains unchanged

5.Surface tension is due to ____________.

OPTIONS

  • diffusion
  • osmosis
  • gravitational force
  • cohesion

6.Which of the following type of solution has a lower level of solutes than the solution?

OPTIONS

  • Isotonic
  • Hypotonic
  • Hypertonic
  • Anisotonic

7.During rainy season wooden doors warp and become difficult to open or to close because of ______

OPTIONS

  • plasmolysis
  • imbibition
  • osmosis
  • diffusion

8.Water absorption takes place through ______.

OPTIONS

  • lateral roots
  • root cap
  • root hair
  • primary root

9.Due to low atmospheric pressure the rate of transpiration will ____________.

OPTIONS

  • increase
  • decrease rapidly
  • decrease slowly
  • remain unaffected

10.Osmosis is a property of ______.

OPTIONS

  • solute
  • solvent
  • solution
  • membrane

Very short answer question.

1.What is osmotic pressure?

Explain the term osmosis.

SOLUTION

i. The pressure exerted due to osmosis is called osmotic pressure.

ii. Osmotic pressure is a pressure of the solution, which is required in opposite direction, so as to stop the entry of solvent molecules into the cell.

OR

Osmotic pressure of a solution is equivalent to the pressure which must be exerted upon it to prevent flow of solvent across a semipermeable membrane.

2.Name the condition in which protoplast of the plant cell shrinks.

SOLUTION

Plasmolysis

3.What happens when a pressure greater than the atmospheric pressure is applied to pure water or a solution?

SOLUTION 1

The water potential of pure water or a solution increases on the application of pressure values more than atmospheric pressure. For example: when water diffuses into a plant cell, it causes pressure to build up against the cell wall. This makes the cell wall turgid. This pressure is termed as pressure potential and has a positive value.

SOLUTION 2

If a pressure greater than atmospheric pressure is applied to pure water or a solution, its water potential increases. It is equivalent to pumping water from one place to another. Pressure can build up in a plant system when water enters a plant cell due to diffusion causing a pressure built up against the cell wall, it makes the cell turgid.

4.Which type of solution will bring about deplasmolysis?

SOLUTION

Hypotonic solution can bring about deplasmolysis.

5.Which type of plants have negative root pressure?

SOLUTION

The plants in which transpiration occurs rapidly especially during midsummer shows negative root pressure.

6.In which conditions transpiration pull will be affected?

SOLUTION

For transpiration pull to operate, the water column should be unbroken and continuous. However, due to temperature fluctuations during day and night, gas bubbles may enter in water column breaking the continuity.

7.Mention the shape of guard cells in Cyperus.

SOLUTION

In Cyperus, both kidney-shaped and dumbbell-shaped guard cells are present.

8.Why do diurnal changes occur in osmotic potential of guard cells?

SOLUTION

1. According to Steward, diurnal changes occur in the osmotic potential of guard cells due to starch-sugar inter-conversion.

2. Whereas according to Levitt active transport of potassium ions into the guard cells and out of them causes diurnal changes in the osmotic potential of guard cells.

3. Endo-osmosis and exo-osmosis occur due to diurnal changes in osmotic potential of guard cells.

9.What is the symplast pathway?

SOLUTION

When water passes across from one living cell to another living cell through plasmodesmata, then it is called the symplast pathway. It is also called the trans-membrane pathway.


Answer the following question.

1.Describe the mechanism for absorption of water.

SOLUTION

A mechanism for absorption of water:

1. In plants, water is absorbed mainly by two processes: Passive absorption and Active absorption

2. Passive absorption:

a. About 98% of the total water absorbed in plants occurs passively.

b. In passive absorption, living cells of the root do not play an important role in water absorption.

c. The driving force is transpiration pull and it thus proceeds through the DPD gradient.

d. There is no expenditure of energy (ATP) as water moves in accordance with the concentration gradient. Hence, it is passive absorption.

e. Passive absorption occurs during day time when transpiration is in progress. It stops at night when transpiration stops.

f. Rapid transpiration creates tension in the xylem vessel due to negative water potential. This tension is transmitted to xylem in the roots. Consequently, water is pulled upwards passively.

g. During passive absorption, no ATP is utilized. Thus, the rate of respiration is not affected.

3. Active absorption:

a. In this water is absorbed due to the activity of roots.

b. Root cells play an active role in the absorption of water.

c. The driving force is the root pressure developed, in the living cells of the root.

d. Active absorption occurs usually at night when transpiration stops due to closure of stomata.

e. As water absorption is against the DPD gradient, there is an expenditure of ATP (energy) generated through the respiratory activity of cells.

2.Discuss theories of water translocation.

SOLUTION

Theories of water translocation:

i. Various theories have been put forth to explain the mechanism of translocation of water. These theories include Vital force theory, Relay pump theory, Physical force theory, Root pressure theory, etc.

ii. Root Pressure Theory (Vital Theory): This theory was proposed by J. Priestley. According to this theory, the activity of living cells of the root is responsible for the translocation of water.

iii. Capillarity theory (physical force theory): This theory was put forth by Boehm in (1863). According to this theory, physical forces and dead cells are responsible for the ascent of sap.

iv. Cohesion- tension theory (Transpiration pull theory): This theory was put forth by Dixon and Jolly (1894). This is presently a widely accepted theory explaining the ascent of sap in plants. This theory is based on two principles i.e. Cohesion and adhesion, and transpiration pull.

3.What is transpiration?

SOLUTION

Transpiration:

The loss of water in the form of vapor is called transpiration that occurs through leaves, stem, flowers, and fruits. 

4.Describe the mechanism of opening and closing of stomata.

SOLUTION

Mechanism of opening and closing of stomata:

1. The opening and closing of stoma is controlled by turgor of guard cells.

2. During day time, guard cells become turgid due to endo-osmosis.

3. Thus turgor pressure is exerted on the thin walls of guard cells.

4. Being elastic and thin, lateral walls are stretched out.

5. Due to kidney or dumb-bell like shape, inner thick walls are pulled apart to open (widen) the stoma.

6. During night time, guard cells become flaccid due to exo-osmosis.

7. Flaccidity closes the stoma almost completely.

8. Endo-osmosis and exo-osmosis occur due to diurnal changes in the osmotic potential of guard cells.

9. According to starch-sugar inter-conversion theory (Steward 1964), during day time, enzyme phosphorylase converts starch to sugar, thus increasing the osmotic potential of guard cells causing entry of water, thereby guard cells are stretched and stoma widens. The reverse reaction occurs at night bringing about the closure of the stoma.

Starch(Stomata open)⇌(Night)Phosphorylase (Day)Sugar(Stomata close)

10. According to the theory of proton transport (Levitt-1974), stomatal movement occurs due to the transport of protons H+ and K+ ions. During the daytime, starch is converted into malic acid. Malic acid dissociates to form malate ions and protons. Protons are transported to subsidiary cells and K+ ions are imported from them.

Potassium Malate is formed that increases osmolarity and causes endoosmosis. Uptake of K+ ions is always accompanied by Cl ions. At night, uptake of K+ and Cl ions is prevented by abscisic acid, changing the permeability of guard cells. Due to this guard cells become hypotonic and thereby become flaccid.

5.What is transpiration?

SOLUTION

Transpiration:

The loss of water in the form of vapor is called transpiration that occurs through leaves, stem, flowers, and fruits. 

6.Explain the role of transpiration.

SOLUTION

Role of transpiration:

i. It removes excess of water.

ii. It helps in the passive absorption of water and minerals from the soil.

iii. It helps in the ascent of sap.

iv. As stomata are open, gaseous exchange required for photosynthesis and respiration is facilitated.

v. It maintains the turgor of the cells.

vi. Transpiration helps in reducing the temperature of leaf and in imparting a cooling effect.

7.What is the significance of transpiration?

SOLUTION

Significance of transpiration:

1. It removes excess of water.

2. It helps in the passive absorption of water and minerals from the soil.

3. It helps in the ascent of sap.

4. As stomata are open, gaseous exchange required for photosynthesis and respiration is facilitated.

5. It maintains the turgor of the cells.

6. Transpiration helps in reducing the temperature of leaf and in imparting a cooling effect.

8.Explain the root pressure theory and its limitations.

SOLUTION

Root pressure theory (Vital theory):

1. This theory was proposed by J. Priestley.

2. According to this theory, the activity of living cells of root is responsible for the translocation of water.

3. When a stem of a potted plant is cut few inches above the soil by a sharp knife, xylem sap is seen flowing out/ oozing out through the cut end.

4. This exudation at the cut end of the stem is a good proof for the existence of root pressure.

5. As water absorption by roots is a constant and continuous process, hydrostatic pressure is developed in the living cells of cortex of the root. This is termed as root pressure (coined by S. Hales).

6. It is due to root pressure water along with dissolved minerals is not only forced into xylem but it is also conducted upwards against the gravity.

7. Root pressure seems to be largely an osmotic phenomenon and its development is an active process.

8. The value of root pressure is +1 to +2 bars which is enough to pump water to a height of 10 to 20 meters.

9. The factors like oxygen, moisture, the temperature of the soil, salt contents, etc. influence the root pressure.

Limitations of root pressure theory:

Although ascent of sap takes place due to root pressure, there are certain objections raised, such as;

1. It is not applicable to plants taller than 20 meters.

2. Ascent of sap can also occur even in the absence of a root system.

3. Root pressure value is almost nearly zero in taller gymnosperm trees.

4. In actively transpiring plants, no root pressure is developed.

5. Xylem sap under normal condition is under tension i.e. it shows negative hydrostatic pressure or high osmotic pressure.

Thus, root pressure is not the sole mechanism explaining the ascent of sap in all plants of varying heights.

9.Explain capillarity theory of water translocation.

SOLUTION

Capillarity theory of water translocation:

1. This theory was put forth by Boehm in (1863).

2. According to this theory, physical forces and dead cells are responsible for the ascent of sap. For e.g. Wick dipped in an oil lamp, shows capillarity due to which oil is raised upwards. The conduction of water in a straw dipped in water is raised to a certain height because of capillarity. The height to which water is raised depends on the diameter of the straw.

3. Capillarity is because of surface tension, and forces of cohesion (attraction between like molecules) and adhesion (attraction between unlike molecules).

4. Xylem vessel/ tracheid with its lumen can be compared with straw.

5. Water column exists because of combined cohesive and adhesive forces of water and xylem wall, due to capillarity.

6. Due to capillarity, water is raised or conducted upwards against gravity, to few centimeters only.

10.Why is transpiration is called ‘a necessary evil’?

SOLUTION

Curtis (1926) regarded transpiration as ‘a necessary evil’, because;

1. For stomatal transpiration to occur, stoma must remain open, during day time.

2. When stomata are open then only the gaseous exchange needed for respiration and photosynthesis will take place.

3. If stomatal transpiration stops, it will directly affect the productivity of the plant through the loss of photosynthetic and respiratory activity.

4. Hence for productivity, stomata must remain open.

5. Consequently transpiration cannot be avoided.

11.Explain the movement of water in the root.

SOLUTION

Journey of water from soil to xylem in roots (from epiblema upto xylem in the stelar region):

1. Water is absorbed by root hair cells through processes like imbibition, diffusion, osmosis which occur sequentially.

2. Water passes through the epidermal cell (epiblema), cortex, endodermis, Casparian strip, pericycle, and then to protoxylem.

3. When the root hair cell absorbs water it becomes turgid. Its turgor pressure increases, but its DPD value decreases.

4. However, the immediately adjacent cortical cell inner to it, has more DPD value, because its O. P. is more.

5. Therefore, cortical cells will absorb water from the turgid root hair cell. It then becomes turgid.

6. The flaccid root hair cell now absorbs water from the soil.

7. Water from the turgid cortical cell is absorbed by the inner cortical cell and the process goes on.

8. Thus, a gradient of suction pressure (DPD) is developed from cells of epiblema to the cortex of the root.

9. Consequently water moves rapidly across the root through loosely arranged living cells of cortex, followed by passage cells of endodermis and finally into the cell of pericycle.

10. Protoxylem is in close proximity to the pericycle.

11. It is due to root pressure, water from pericycle is forced into the xylem.

12. Pathway of water across the root occurs in two types: Apoplast pathway and Symplast pathway

13. Apoplast pathway: When some amount of water passes across the root through the cell wall and the intercellular spaces of cortical cells of the root, it is then called the apoplast pathway. This pathway occurs up to endodermis.

14. Symplast pathway: When water passes across from one living cell to another living cell through plasmodesmata, then it is called the symplast pathway. It is also called the trans-membrane pathway.


Very short answer question.

12.What is osmotic pressure?

Explain the term osmosis.

SOLUTION

i. The pressure exerted due to osmosis is called osmotic pressure.

ii. Osmotic pressure is a pressure of the solution, which is required in opposite direction, so as to stop the entry of solvent molecules into the cell.

OR

Osmotic pressure of a solution is equivalent to the pressure which must be exerted upon it to prevent flow of solvent across a semipermeable membrane.

13.Define and or explain the term:

Diffusion

SOLUTION

  1. Diffusion means to disperse.
  2. Diffusion can be defined as the movement of ions/ atoms/ molecules of a substance from the region of their higher concentration to the region of their lower concentration till equilibrium is reached.
  3. The movement is due to the kinetic energy of the molecules.
  4. Water passes into the cell by diffusion through a freely permeable cell wall.

Define and or explain the term:

14.Plasmolysis

SOLUTION 1

Plasmolysis – The shrinkage of cytoplasm of a living cell as a result of exosmosis is known as plasmolysis.

SOLUTION 2

  1. When a living cell is placed in a hypertonic solution, exo-osmosis occurs. This is called plasmolysis.
  2. During plasmolysis, the protoplast of the cell shrinks and recedes from the cell wall due to which cell becomes flaccid. Such a cell is called a plasmolysed cell.
  3. In a plasmolyzed cell, a gap is developed between the cell wall and the protoplast. This gap is filled up by the outer solution.

15.imbibition

SOLUTION

1. Imbibition is swelling up of hydrophilic colloids due to the adsorption of water.

OR

The adsorption of water by hydrophilic compounds is called imbibition.

2. Substance that adsorbs water/liquid is called imbibant and water/ liquid that gets imbibed is called imbibate.

3. The root hair cell wall is made up of pectic compounds and cellulose which are hydrophilic colloids.

4. During imbibition, water molecules get tightly adsorbed without the formation of a solution.

5. Imbibition continues until the equilibrium is reached. In other words, water moves along the concentration gradient.

6. Imbibition is significant in soaking of seeds, swelling up of dried raisins, kneading of flour etc.

16.Guttation

SOLUTION

  1. The loss of water in the form of liquid is called guttation.
  2. It occurs through special structures called water stomata or hydathodes.

17.Transpiration

SOLUTION

i. The loss of water in the form of vapour is called transpiration that occurs through leaves, stem, flowers and fruits.

ii. Transpiration occurs through three main sites – cuticle, stomata, and lenticels.

Define and or explain the term:

18.Ascent of sap

SOLUTION

The transport of water with dissolved minerals from the root to other aerial parts like stem and leaves, against the gravity, is called translocation or ascent of sap.

19.Active absorption

SOLUTION

  1. In this water is absorbed due to the activity of roots.
  2. Root cells play an active role in the absorption of water.
  3. The driving force is the root pressure developed, in the living cells of the root.
  4. Active absorption occurs usually at night when transpiration stops due to the closure of stomata.
  5. As water absorption is against the DPD gradient, there is an expenditure of ATP (energy) generated through the respiratory activity of cells.

Answer the following question.

1.Define and or explain the term:

DPD

SOLUTION

  1. Diffusion pressure of pure solvent (pure water) is always more than the diffusion pressure of the solvent in a solution. The difference in the diffusion pressures of pure solvent and the solvent in a solution is called Diffusion Pressure Deficit (DPD) or Suction Pressure (SP).
  2. The term DPD was coined by B.S. Meyer (1938). Nowadays, term water potential is used for DPD.
  3. In colloquial language, the term DPD is actually the thirst of a cell with which it absorbs water from the surroundings.
  4. The water around the cell wall has more diffusion pressure than cell sap. Due to this, water moves in the cell by diffusion. 

2.Define and or explain the term:

Turgor pressure

SOLUTION

Turgor pressure (T.P) is the pressure exerted by turgid cell sap on to the cell membrane and cell wall.

3.Define and or explain the term:

Water potential

SOLUTION

i. Chemical potential of water is called water potential.

ii. It is represented by Greek letter psi (ψ).

iii. The unit of measurement of water potential is bars/ pascals/ atmospheres.

iv. Water potential of protoplasm is equal but opposite in sign to DPD. It has a negative value.

v. Water potential of pure water is always zero. The addition of any solute in it decreases its psi (ψ) value. Therefore, it has a negative value.

vi. Difference between water potential of the adjacent cells decides the movement of water through plasmodesmata across the cells.

vii. Water always flows from less negative potential to more negative water potential (i.e. from high water potential area to low water potential area).

4.Define and or explain the term:

Wall pressure

SOLUTION

The cell wall is thick and rigid, exerts a counter pressure on the cell sap. This is called Wall pressure (W. P).

5.Define and or explain the term:

Root pressure

SOLUTION

  1. During the absorption of water, the continuous flow of water develops hydrostatic pressure in living cells of the root. This is called root pressure.
  2. Root pressure causes water to flow from pericycle into the xylem. It also causes upward conduction of water against gravity.
  3. A manometer is used to measure the root pressure.

6.Distinguish between Osmotic pressure and Turgor pressure.

SOLUTION

Osmotic pressureTurgor pressure
Osmotic pressure is a pressure of the solution, which is required in opposite direction, so as to stop the entry of solvent molecules into the cell.Turgor pressure is the pressure exerted by turgid cell sap on to the cell membrane and cell wall.

7.Distinguish between Diffusion and Osmosis

SOLUTION

DiffusionOsmosis
1. It takes place in solid, gas, or liquid medium.1. It takes place only in a liquid medium.
2. It does not require the presence of a semi-permeable membrane.2. It requires the presence of a semipermeable membrane.
3. In diffusion, the movement of ions/atoms/molecules from a region of higher concentration to the region of lower concentration takes place.3. In osmosis, diffusion of the only solvent from a lower concentration of solution to a higher concentration of solution occurs
4. It is influenced by the diffusion pressure4. It is only influenced by the turgor pressure.

8.Enlist macronutrients and micronutrients required for plant growth.

Long answer question.

Write on macro- and micro nutrients required for plant growth.

SOLUTION

  1. Macronutrients:
    Some minerals like C, H, O, P, N, S, Mg, K, Ca required in large quantity for normal growth of the plant, are called macro elements. Macronutrients are required in large quantities. They mainly play nutritive and structural roles.
  2. Micronutrients:
    Some minerals like Cu, Mo, Mn, Cl, Bo, Zn required in small quantities for the growth of a plant, are called microelements.
    Micronutrients are required in traces because they function in the catalytic role as co-factors.

9.How are the minerals absorbed by the plants?

SOLUTION

i. The analysis of plant ash demonstrates that minerals are absorbed by plants from soil and surroundings.

ii. Minerals are absorbed by plants in the ionic (dissolved) form, mainly through roots and then transported.

iii. Mineral ion absorption is independent of water absorption.

iv. It can occur in two ways i.e. active and passive absorption.

v. In passive absorption, the movement of mineral ions into root cells occurs as a result of diffusion. Mineral ions diffuse from a region of their higher concentration to a region of their lower concentration without the expenditure of energy.

vi. Most minerals in the soil are charged particles hence, they cannot pass across the cell membranes. Hence most of the minerals are absorbed actively with the expenditure energy.

vii. Inactive absorption, minerals are absorbed against the concentration gradient with the expenditure energy.

viii. Absorbed mineral ions are pulled in an upward direction along with xylem sap because of transpiration pull.

ix. Hence, mineral ions are pulled from the source (root) and are transported ascendingly through the sap to the needed areas like apical, lateral, young leaves, developing flowers, fruits, seeds, and storage organs.

x. Mineral ions get unloaded by fine veins through the process of diffusion in the vicinity of cells. Cells uptake them actively.

Long answer question.

1.Describe structure of root hair.

SOLUTION

1. Root hair is a cytoplasmic extension (prolongation) of epiblema cell.

2. Each root hair may be approximately 1 to 10 mm long and tube-like structure.

3. It is colourless, unbranched, short-lived (ephemeral), and very delicate.

4. It has a large central vacuole surrounded by a thin film of cytoplasm, plasma membrane and thin cell wall, which is two-layered.

5. Outer layer is composed of pectin and the inner layer is made up of cellulose.

6. Cell wall of a root hair is freely permeable but the plasma membrane is selectively permeable.

2.Write on journey of water from soil to xylem in roots.

SOLUTION

1. Water is absorbed by root hair cells through processes like imbibition, diffusion, osmosis which occur sequentially.

2. Water passes through the epidermal cell (epiblema), cortex, endodermis, casparian strip, pericycle and then to protoxylem.

3. When root hair cell absorbs water it becomes turgid. Its turgor pressure increases, but its DPD value decreases.

4. However, the immediately adjacent cortical cell inner to it, has more DPD value because its O. P. is more.

5. Therefore, the cortical cells will absorb water from the turgid root hair cell. It then becomes turgid.

6. The flaccid root hair cell now absorbs water from the soil.

7. Water from the turgid cortical cell is absorbed by the inner cortical cell and the process goes on.

8. Thus, a gradient of suction pressure (DPD) is developed from cells of epiblema to the cortex of the root.

9. Consequently water moves rapidly across the root through loosely arranged living cells of cortex, followed by passage cells of endodermis and finally into the cell of pericycle.

10. Protoxylem is in close proximity with pericycle.

11. It is due to root pressure, water from pericycle is forced into the xylem.

12. Pathway of water across the root occurs in two types: Apoplast pathway and Symplast pathway

13. Apoplast pathway: When some amount of water passes across the root through the cell wall and the intercellular spaces of cortical cells of the root, it is then called the apoplast pathway. This pathway occurs up to endodermis.

14. Symplast pathway: When water passes across from one living cell to another living cell through plasmodesmata, then it is called the symplast pathway. It is also called the trans-membrane pathway.

3.Explain cohesion theory for translocation of water.

SOLUTION

1. This theory was put forth by Dixon and Jolly (1894).

2. This is presently a widely accepted theory explaining the ascent of sap in plants.

3. This theory is based on two principles i.e. Cohesion and adhesion, and transpiration pull.

4. Cohesion and adhesion:

a. A strong force of attraction between water molecules is called cohesive force.

b. While a strong force of attraction between water molecules and the lignified wall of the lumen of the xylem vessel, is called adhesive force.

c. Due to combined cohesive and adhesive forces a continuous water column is developed (formed) in the xylem right from root up to the tip of the topmost leaf in the plant.

5. Transpiration pull:

a. The transpiration pull developed in the leaf vessel is transmitted down to the root and thus accounts for the ascent of sap.

b. Excess water is lost in the form of vapour, mainly through the stomata found on a leaf.

c. This water loss increases the D.P.D. of mesophyll cells. These cells withdraw water ultimately from the xylem in the leaf.

d. In other words, due to continuous transpiration, a gradient of suction pressure (i.e. D.P.D.) is developed right from guard cells up to the xylem in the leaf. This will create a tension (called a negative pull or transpiration pull) in the xylem.

e. Consequently, the water column is pulled out of xylem. Thus, water is pulled upwards passively against the gravity leading to the ascent of sap.

4.Write mechanism of opening and closing of stoma.

SOLUTION

1. Opening and closing of stoma is controlled by the turgor of guard cells.

2. During day time, guard cells become turgid due to endo-osmosis.

3. Thus turgor pressure is exerted on the thin walls of guard cells.

4. Being elastic and thin, lateral walls are stretched out.

5. Due to kidney or dumb-bell like shape, inner thick walls are pulled apart to open (widen) the stoma.

6. During night time, guard cells become flaccid due to exo-osmosis.

7. Flaccidity closes the stoma almost completely.

8. Endo-osmosis and exo-osmosis occur due to diurnal changes in the osmotic potential of guard cells.

9. According to starch-sugar inter-conversion theory (Steward 1964), during day time, enzyme phosphorylase converts starch to sugar, thus increasing the osmotic potential of guard cells causing entry of water, thereby guard cells are stretched and stoma widens. The reverse reaction occurs at night bringing about the closure of the stoma.

Starch(Stomata open)⇌(Night)Phosphorylase (Day)Sugar(Stomata close)

10. According to the theory of proton transport (Levitt-1974), stomatal movement occurs due to the transport of protons H+ and K+ ions. During the daytime, starch is converted into malic acid. Malic acid dissociates to form malate ions and protons. Protons are transported to subsidiary cells and K+ ions are imported from them. Potassium Malate is formed that increases osmolarity and causes endosmosis. The uptake of K+ ions is always accompanied by Cl ions. At night, uptake of K+ and Cl– ions is prevented by abscisic acid, changing the permeability of guard cells. Due to this guard cells become hypotonic and thereby become flaccid.

5.What is hydroponics? How is it useful in identifying the role of nutrients?

SOLUTION

  1. Hydroponics is a technique in which plants are grown in nutrient solutions in absence of soil. Roots are immersed in an adequately aerated, dilute, and defined solution of nutrients. Purified water and mineral salts are used in the nutrient medium.
  2. In hydroponics, the concentration of a particular mineral in a solution of nutrients in which roots are immersed can be increased or decreased. By this method, essential elements can be identified and their deficiency symptoms can be discovered.
    Thus it helps to identify the role of nutrients in plant growth.

6.Explain the active absorption of minerals.

SOLUTION

1. Uptake of mineral ions against the concentration gradient is called active absorption.

2. Such movement requires an expenditure of energy by the absorbing cell. This energy is derived from respiration and is supplied through ATP.

3. The rate of active absorption of minerals depends upon respiration.

4. When the roots are deprived of oxygen, they show a sudden drop in the active absorption of minerals. The mineral ions accumulated in the root hair pass into the cortex and finally reach the xylem.

5. The minerals in the xylem are then carried along with water to other parts of the plant along the transpiration stream and are subsequently assimilated into organic molecules and then redistributed to other parts of the plant through the phloem.

7.Enlist macronutrients and micronutrients required for plant growth.

Long answer question.

Write on macro- and micro nutrients required for plant growth.

SOLUTION

  1. Macronutrients:
    Some minerals like C, H, O, P, N, S, Mg, K, Ca required in large quantity for normal growth of the plant, are called macro elements. Macronutrients are required in large quantities. They mainly play nutritive and structural roles.
  2. Micronutrients:
    Some minerals like Cu, Mo, Mn, Cl, Bo, Zn required in small quantities for the growth of a plant, are called microelements.
    Micronutrients are required in traces because they function in the catalytic role as co-factors.

                  COMPLETED


Chapter 5, Origin and Evolution of Life, hsc, biology, science, maharashtra bore, latest edition, full solution,

Chapter 5: Origin and Evolution of Life

Multiple choice question.

1.Who proposed that the first form of life could have come from per- existing nonliving organic molecules?

OPTIONS

  • Alfred Wallace
  • Oparin and Haldane
  • Charles Darwin
  • Louis Pasteur

2.The sequence of origin of life may be-

OPTIONS

  • Organic materials- inorganic materials – Eobiont- colloidal aggregates- cell.
  • Inorganic materials – organic materials – colloidal aggregates – Eobiont- cell
  • Organic materials- inorganic materials – colloidal aggregates – cell
  • Inorganic materials- organic materials – Eobiont- colloidal aggregates – cell

3.In Hardy – Weinberg equation, the frequency of homozygous recessive individual is represented by:

OPTIONS

  • p2
  • pq
  • q2
  • 2pq

4.Select the analogous organs.

OPTIONS

  • Forelimbs of whale and bat
  • Flippers of dolphins and penguin
  • Thorn and tendrils of Bougainvillea and Cucurbita.
  • Vertebrates hearts or brains.

5.Archaeopteryx is known as missing link because it is a fossil and share characters of both

OPTIONS

  • Fishes and amphibians
  • Annelida and arthropoda
  • Birds and reptiles
  • Chordates and nonchordates

6.Identify the WRONG statement regarding evolution.

OPTIONS

  • Darwin’s variations are small and directional
  • Mutations are random and nondirectional
  • Adaptive radiations leads to divergent evolution
  • Mutations are non – random and directional

7.Gene frequency in a population remain constant due to –

OPTIONS

  • Mutation
  • Migration
  • Random mating
  • Non- random mating

8.Which of the following characteristic is not shown by the ape?

OPTIONS

  • Prognathous face
  • Tail is present
  • Chin is absent
  • Forelimbs are longer than hind limbs

9.______ can be considered as a connecting link between ape and man.

OPTIONS

  • Australopithecus
  • Homo habilis
  • Homo erectus
  • Neanderthal man

10.The cranial capacity of Neanderthal man was

OPTIONS

  • 600 cc
  • 940 cc
  • 1400 cc
  • 1600 cc

Very short answer question.

1.Define the Gene pool

SOLUTION

The total genetic information encoded in the sum total of genes in a Mendelian population is called gene pool.

2.Define the Gene frequency

SOLUTION

The proportion of an allele in the gene pool, to the total number of alleles at a given locus, is called gene frequency.

3.Define Organic evolution.

SOLUTION

Organic evolution can be defined as slow, gradual, continuous and irreversible changes through which the present-day complex forms of the life developed (or evolved) from their simple pre-existing forms.

4.Define Population.

SOLUTION

According to this theory all individuals of the same species constitute a population.

5.Define Speciation.

SOLUTION

The process of formation of a new species from the pre-existing species is called speciation.

6.What is adaptive radiation?

SOLUTION

The process of evolution which results in the transformation of original species to many different varieties is called adaptive radiation.

7.If variation occurs in a population by chance alone and not by natural selection and brings a change in frequencies of an allele. What is it called? 

SOLUTION

If the variation in a population occurs by chance alone and not by natural selection and brings about a change in frequencies of an allele, it is called genetic drift.

8.State the Hardy – Weinberg equilibrium.

SOLUTION

The Hardy-Weinberg equilibrium law states that ‘at the equilibrium point, both the gene (allele) frequency and genotypic frequency remain constant from generation to generation’.

9.What are homologous organs?

SOLUTION

Homologous organs are those organs, which are structurally similar but perform different functions.

10.What is vestigeal organ?

SOLUTION

Vestigeal organs or rudimentary organs are imperfectly developed and non-functional, degenerate structures that were functional in some related and other animals or in ancestors.

11.What is the scientific name of the modern man?

SOLUTION

The scientific name of the modern man is Homo sapiens.

12.What is coacervate?

SOLUTION

Coacervates are colloidal aggregations of hydrophobic proteins and lipids (lipoid bubbles).

13.Which period is known as “age of Reptilia”?

SOLUTION

Jurassic period is known as age of Reptilia.

14.Name the ancestor of human which is described as a man with an ape brain.

SOLUTION

Australopithecus is the ancestor of humans which is described as a man with an ape brain.


Short answer question.

1.Write a note on Genetic drift.

SOLUTION

1. Any alternation in allete frequency in the natural population by chance, is called genetic drift. e.g. Elimination of a particular allele from a population due to events like accidental death prior to mating of an organism.

2. The concept of genetic drift was first given Sewall Wright, and is hence also called as the Sewall Wright effect.

3. Genetic drifts are random or directionless.

4. The effect of genetic drift is more significant in small population than in large population.

5. Due to genetic drift, some alleles of a population are lost or reduced by chance and some others may be increased.

6. Sometimes, a few individuals become isolated from the large population and they produce a new population in a new geographical areas. The allele frequency of the new population becomes different. The original drifted population (i.e. colonizing ancestor/ pioneer) becomes ‘founders’ and the effect is called the founder effect.

7. A bottleneck effect is seen when much of a population is killed due to a natural disaster and only a few remaining individuals are left to begin a new population.

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2.Enlist the different factors that are responsible for changing gene frequency.

SOLUTION

The four major factors that are responsible for changing gene frequency are as follows:

1. Gene flow (Migration):

Gene flow is the movement of genes into or out of a population. Gene movement may be in the form of migration of organism, or gametes (dispersal of pollens) or segments of DNA (transformation). Thus, gene flow alters gene frequency causing evolutionary changes.

2. Genetic drift:

Any random fluctuation (alteration) in allele frequency, occurring in the natural population by pure chance, is called genetic drift. For example, when the size of a population is severely reduced due to natural disasters like earthquakes, floods, fires, etc. cause the elimination of particular alleles from a population. Smaller populations have greater chances for genetic drift. Thus, genetic drift will result in the change in the gene frequency and has the potential to bring about evolutionary change.

3. Natural selection:

Natural selection is the process by which better adapted organisms grow and produce more offsprings in the population. It brings about evolutionary changes by favouring differential reproduction of genes that bring about changes in gene frequency from one generation to the next generation.

iv. Mutations: www.asterclasses.com

Sudden permanent heritable changes are called mutations. Mutation can occur in the gene, in the chromosome and in the chromosome number. Mutation that occurs within the single gene, is called point mutation or gene mutation or in a larger segment of genes by chromosomal aberrations. Both point mutations and chromosomal aberrations can alter gene frequency. Mutation leads to the change in the phenotype of the organism, causing variation.

3.Draw a graph to show that natural selection leads to disruptive change.

SOLUTION

4.Give the significance of fossils.

SOLUTION

1. Fossils are the dead remains of plants and animals that lived in past in various geological layers.

2. The study of fossils provides the most convincing and direct evidence of evolution.

3. Study of fossils tells us that life forms were not the same millions of years ago (mya).

4. The geological time’s scale is based on fossil records.

5. From the fossil records we can trace the complete evolutionary history of animals.

6. Study of fossils is an important aspect of evolution since it can be used in paleontology and anthropology for determining age of the fossils and deducing information about their ancestors.

5.Write the objections to Mutation theory of Hugo de Vries.

SOLUTION

Objections to Mutation Theory are as follows:

1. The large and discontinuous variations observed by Hugo de Vries were actually due to chromosomal aberrations. Gene mutations usually bring about only minor changes.

2. Rate of mutation is very slow as compared to the requirement of evolution.

3. Chromosomal aberrations have little significance in evolution as they are quite unstable.

6.What is disrruptive selection?

SOLUTION

Disruptive Natural selection is a selection in which more number of individuals acquire peripheral character value at both ends of the distribution curve.

7.Give an example of disrruptive selection

SOLUTION

1. In this selection, nature selects extreme phenotypes and eliminate intermediates.

2. This results in the formation of two peaks in the distribution of traits.

3. This kind of selection is rare.

4. It ensures the effect on the entire gene pool of a population, considering all mating types or systems.

e.g. Disruptive selection was observed in the different beak sizes of African seed cracker finches. The birds have different sizes of the beak and they feed on seeds. The available seeds were of two kinds i.e., small and large-sized seeds. Large beak sized birds feed on large seeds while small beak sized birds feed on small seeds and their number was increased. Intermediate beak sized birds were unable to feed on either type of seeds so their population decreased gradually and then were eliminated by natural selection.


1.Match the following:

Column- IColumn- II
1. August Weismanna. Mutation theory
2. Hugo de vriesb. Germplasm theory
3. Charl Darwinc. Theory of acquired characters
4. Lamarkd. Theory of natural selection

SOLUTIONS

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Column- IColumn- II
1. August Weismannb. Germplasm theory
2. Hugo de vriesa. Mutation theory
3. Charl Darwind. Theory of natural selection
4. Lamarkc. Theory of acquired characters
Asterclasses

2.What is adaptive radiation? Explain with suitable example.

SOLUTION

1. The process of evolution which results in the transformation of original species to many different varieties is called adaptive radiation.

2. Darwin’s Finches is one of the best examples of adaptive radiation. During his visit to Galapagos Islands, Charles Darwin also noticed a variety of small birds. These birds are now called Darwin’s finches. Darwin concluded that the American mainland species of the bird was the original one from which they migrated to the different islands of Galapagos. These birds adapted to the different environmental conditions of these islands. From original seed-eating features, many other forms with altered beaks evolved into insectivorous features.

3. Another example of adaptive radiation is Australian Marsupials. In Australia, there are many marsupial mammals who evolved from a common ancestor.

3.By taking industrial melanism as one example. Explain the concept of natural selection.

SOLUTION

1. Natural selection encourages those genes or traits that assure the highest degree of adaptive efficiency between the population and its environment.

2. Industrial melanism is one of the best examples of natural selection.

3. In Great Britain, before industrialization (1845) grey white-winged moths (Biston betularia) were more in number than black-winged moth (Biston carbonara).

4. These moths are nocturnal and during the day time they rest on a tree trunk.

5. White-winged moths were camouflaged (hide in the background) well with the lichen-covered trees that helped them to escape from the predatory birds.

6. However, the black-winged moth resting on lichen-covered tree trunks were easy victims for the predatory birds and their number was reduced.

7. During the industrial revolution, large number of industries came up in Great Britain.

8. The industries released black sooty smoke that covered and killed the lichens growing on a tree and turn the tree black due to pollution.

9. This change became an advantage to the black-winged moths that camouflaged well with the black tree trunks and their number increased

10. The white-winged moths however became victims to predatory birds due to which their number reduced. Thus, natural selection has resulted in the establishment of a phenotypic trait in the changing environmental conditions.

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4.Describe the Urey and Miller experiment.

SOLUTION

Stanley L. Miller and his teacher Harold C. Urey provided the first experimental evidence in support of chemical evolution theory of Oparin.

1. Apparatus and Procedure:

They designed a glass-apparatus called spark-discharge apparatus. The apparatus was first sterilized and evacuated. Methane, ammonia, and hydrogen gases were pumped in the proportion of 1:2:2 into the glass chamber. A tube carrying water vapour was also connected to the chamber. The lightning effect was mimicked by the action of electric discharge in the chamber.

The process of evaporation and precipitation was also stimulated by the use of heating mantle and condenser respectively.

The mixture of CH4, NH3, H2 was exposed continuously to electric discharge for several days causing the gases to interact, after which these were condensed.

2. Observation:

It was observed that the liquid collected in the U-tube turned brown.

3. Results and Conclusions:

Chemical analysis of this liquid reported the presence of simple organic compounds. (urea, amino acids, lactic acid, etc.). This experiment strongly supports that the simple molecules present in the Earth’s early atmosphere combined to form the organic building blocks of life.

5.What is Isolation?

SOLUTION

It is the separation of the population of a particular species into smaller units which prevents interbreeding between them. A barrier that prevents gene flow or exchange of genes between isolated populations, is called isolating mechanism.

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5.Describe the different types of reproductive isolations.

SOLUTION

A number of isolating mechanisms are operated in nature and therefore divergence and speciation may occur.

The isolating mechanisms are of two types namely, geographical isolation and reproductive isolation.

1. Geographical Isolation:

It is also called as physical isolation. It occurs when an original population is divided into two or more groups by geographical barriers such as rivers, oceans, mountains, glaciers, etc. These barriers prevent interbreeding between isolated groups.

The separated groups are exposed to different kinds of environmental factors and they acquired new traits by mutations. The separated populations develop distinct gene pool and they do not interbreed. Thus, new species have been formed by geographical isolation. e.g. Darwin’s Finches.

2. Reproductive Isolation:

Reproductive isolation occurs due to change in genetic material, gene pool and structure of genital organs. It prevents interbreeding between populations. Types of Isolating Mechanisms:

a. Pre-mating or pre-zygotic isolating mechanism:

This mechanism prevents fertilization and zygote formation.

i. Habitat isolation or (Ecological isolation): Members of a population living in the same geographic region but occupying separate habitats in such a way that potential mate do not meet.

ii. Seasonal or temporal isolation: Members of a population living in the same geographic region but are sexually mature at different years or different times of the year.

iii. Ethological isolation: Due to specific mating behavior the members of the population do not mate.

iv. Mechanical Isolation: Members of two populations have a difference in the structure of reproductive organs.

2. Post-mating or Post-zygotic barriers:

i. Gamete mortality: Gametes have a limited life span. Due to one or the other reasons, if the union of the two gametes does not occur in the given time, it results in gamete mortality.

ii. Zygote mortality: Here, egg is fertilized but the zygote dies due to one or the other reasons.

iii. Hybrid sterility: Hybrids develop to maturity but become sterile due to the failure of proper gametogenesis (meiosis).

e.g. Mule is an inter-generic hybrid that is sterile.

6.What is Genetic variations?

SOLUTION

Genetic variations are caused due to various aspects of mutation, recombination, and migration. The change in gene and gene frequencies is known as genetic variation. 

7.Explain the different factors responsible for genetic variations.

SOLUTION

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Genetic variations are caused by the following factors:

a. Gene Mutation:

Sudden permanent heritable change is called a mutation. Mutation can occur in the gene, in the chromosome, and in the chromosome number. The mutation that occurs within the single gene is called point mutation or gene mutation.

Mutation leads to the change in the phenotype of the organism, causing variation.

b. Genetic recombination:

In sexually reproducing organisms, during gamete formation, the exchange of genetic material occurs between non-sister chromatids of homologous chromosomes. This is called crossing over. It produces new genetic combinations that result in variations. Fertilization between opposite mating gametes leads to various recombinations resulting in the phenotypic variations causing a change in the frequencies of alleles.

c. Gene flow:

Gene flow is the movement of genes into or out of a population. Gene movement may be in the form of migration of organism, or gametes (dispersal of pollens) or segments of DNA (transformation). Gene flow also alters gene frequency causing evolutionary changes.

d. Genetic drift:

Any random fluctuation (alteration) in allele frequency, occurring in the natural population by pure chance, is called genetic drift.

e.g. When the size of a population is severely reduced due to natural disasters like earthquakes, floods, fires, etc., it causes the elimination of particular alleles from a population. Smaller populations have greater chances for genetic drift. It will result in a change in the gene frequency. Genetic drift is also an important factor for evolutionary change.

e. Chromosomal aberrations:

The structural and morphological change in chromosome due to rearrangement is called chromosomal aberrations. It changes the arrangement of the genes (order or sequence) that results in the variation.

Chromosomal aberrations occur due to the following reasons:

1. Deletion: Loss of genes from the chromosome.

2. Duplication: Genes are repeated or doubled in number on the chromosome.

3. Inversion: A particular segment of the chromosome is broken and gets reattached to the same chromosome in an inverted position due to 180° twists. There is no loss or gain of the gene complement of the chromosome. www.asterclasses.com

4. Translocation: Transfer (transposition) of a part of a chromosome or a set of genes to a non-homologous chromosome is called translocation. It is affected naturally by the transposons present in the cell.

Chromosomal aberrations-


Long answer question.

1.Complete the chart.

EraDominating groupof animal
1. Cenozoic____________
2. _________Reptiles
3. Palaeozic____________
4. _________Invertebrates

SOLUTION

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EraDominating groupof animal
1. CenozoicMammals
2. MesozoicReptiles
3. PalaeozicAmphibians
4. PalaeozoicInvertebrates

          COMPLETED


Chapter 4, Molecular Basis of Inheritance, hsc, biology, science, maharashtra bore, latest edition, full solution,

Multiple Choice Question:

1.Griffith worked on ____________.

OPTIONS

  • Bacteriophage
  • Drosophila
  • Frog eggs
  • Streptococci

2.The molecular knives of DNA are _______.

OPTIONS

  • Ligases
  • Polymerases
  • Endonucleases
  • Transcriptase

3.Translation occurs in the _______.

OPTIONS

  • nucleus
  • cytoplasm
  • nucleolus
  • lysosomes

4.The enzyme required for transcription is _______.

OPTIONS

  • DNA polymerase
  • RNA polymerase
  • Restriction enzyme
  • RNAase

5.Transcription is the transfer of genetic information from _______.

OPTIONS

  • DNA to RNA
  • tRNA to mRNA
  • DNA to mRNA
  • mRNA to tRNA

6.Which of the following is NOT part of protein synthesis?

OPTIONS

  • Replication
  • Translation
  • Transcription
  • All of these

7.In the RNA molecule, which nitrogen base is found in place of thymine?

OPTIONS

  • Guanine
  • Cytosine
  • Thymine
  • Uracil

8.How many codons are needed to specify three amino acids?

OPTIONS

  • 3
  • 6
  • 9
  • 12

9.Which out of the following is not an example of an inducible operon?

OPTIONS

  • Lactose operon
  • Histidine operon
  • Arabinose operon
  • Tryptophan operon

10.Place the following event of translation in the correct sequence

i. Binding of met-tRNA to the start codon.

ii. Covalent bonding between two amino acids.

iii. Binding of second tRNA.

iv. Joining of small and large ribosome subunits.

OPTIONS

  • iii, iv, i, ii
  • i, iv, iii, ii
  • iv, iii, ii, i
  • ii, iii, iv, i

Very Short Answer Question:

1.What is the function of an RNA primer during DNA synthesis?

SOLUTION

RNA primers provide the starting point for DNA polymerase to initiate synthesizing a new DNA strand.

2.Why the genetic code is considered as commaless?

SOLUTION

Genetic code is commaless: There is no gap or punctuation mark between successive/ consecutive codons.

3.What is genome?

SOLUTION

The term genome refers to the total genetic constitution of an organism.

OR

It is a complete copy of genetic information (DNA) or one complete set of chromosomes (monoploid or haploid) of an organism.

4.Which enzyme does remove supercoils from replicating DNA?

SOLUTION

Super helix relaxing enzyme removes supercoils from replicating DNA.

5.Why are Okazaki fragments formed on lagging strand only?

SOLUTION

  1. The two strands in DNA are antiparallel i.e. one strand runs in 5’ → 3’ direction whereas the other runs in 3’ → 5’.
  2. The DNA polymerase synthesizes a new DNA strand in 5’ → 3’ direction only.
  3. Leading template is synthesized continuously and lagging template is synthesized discontinuously.
  4. Due to 5’ → 3’ polymerizing activity of DNA polymerase Okazaki fragments are formed only on lagging strand only.

6.When does DNA replication takes place?

SOLUTION

DNA replication occurs in the S-phase of the interphase of the cell cycle, prior to cell division.

7.Define term- codon and codogen.

SOLUTION

Codon:

A sequence of three adjacent nucleotides in mRNA that codes for one amino acid are known as a codon.

Codogen:

It is the smallest possible sequence (triplet) of nucleotides present on the DNA strand which can specify one particular amino acid.

8.What is degeneracy of genetic code?

SOLUTION

Usually, the single amino acid is encoded by a single codon. However, some amino acids are encoded by more than one codon. e.g. Cysteine has two codons, while isoleucine has three codons. This is called the degeneracy of the code. Degeneracy of the code is explained by the Wobble hypothesis. Here, the first two bases in different codons are identical but the third one varies.

9.Which are the nucleosomal ‘core’ histones?

SOLUTION

The nucleosome core is made up of two molecules of each of four types of histone proteins viz. H2A, H2B, H3 and H4.

Short Answer Question:

1.Write a short note on DNA packaging in the eukaryotic cell.

SOLUTION

  1. The organization of DNA is much more complex in eukaryotes.
  2. Histone proteins are rich in lysine and arginine residues which are basic in nature and are positively charged.
  3. These histones organize themselves to make a unit of 8 molecules known as histone octamer.
  4. The negatively charged helical DNA is wrapped around the positively charged histone octamer, forming a structure known as a nucleosome.
  5. The nucleosome core is made up of two molecules of each of four types of histone proteins viz. H2A, H2B, H3 and H4. H1 protein binds the DNA thread where it enters (arrives) and leaves the nucleosome.
  6. One nucleosome approximately contains 200 base pair long DNA helix wound around it.
  7. About 146 base pair long segment of DNA remains present in each nucleosome.
  8. Nucleosomes are the repeating units of chromatin, which are threadlike, stained (coloured) bodies present in nucleus. These look like ‘beads-on-string’, when observed under an electron microscope.
  9. DNA helix of 200 bp wraps around the histone octamer by 1¾ turns.
  10. Six such nucleosomes get coiled and then form solenoid that looks like coiled telephone wire.
  11. The chromatin is packed to form a solenoid structure of 30 nm diameter (300Å) and further supercoiling tends to form a looped structure called chromatin fiber, which further coils and condenses at the metaphase stage to form the chromosomes.
  12. The packaging of chromatin at higher levels, needs an additional set of proteins that are called Non-Histone Chromosomal proteins (NHC).

2.Enlist the characteristics of genetic code.

SOLUTION

Genetic code of DNA has certain following characteristics:

1. Genetic code is a triplet code:

The sequence of three consecutive bases constitutes a codon, which specifies one particular amino acid. The base sequence in a codon is always in 5’ → 3’ direction. In every living organism, genetic code is a triplet code.

2. Genetic code has distinct polarity:

Genetic code shows definite polarity i.e. direction. It is always read in 5’ → 3’ direction and not in 3’ → 5’ direction. Otherwise the message will change e.g. 5’ AUG 3’

3. Genetic code is non-overlapping:

Code is non-overlapping i.e. each single base is a part of only one codon. Adjacent codons do not overlap.

4. Genetic code is commaless:

There is no gap or punctuation mark between successive/ consecutive codons.

5. Genetic code has degeneracy:

Usually, the single amino acid is encoded by a single codon. However, some amino acids are encoded by more than one codon. e.g. Cysteine has two codons, while isoleucine has three codons. This is called the degeneracy of the code. Degeneracy of the code is explained by the Wobble hypothesis. Here, the first two bases in different codons are identical but the third one varies.

6. Genetic code is universal:

In most of the living organisms, the specific codon specifies the same amino acid. e.g. Codon AUG always specifies amino acid methionine.

7. Genetic code is non-ambiguous:

The specific amino acid is encoded by a particular codon. Alternatively, two different amino acids will never be encoded by the same codon.

8. Initiation codon and termination codon:

AUG is always an initiation codon in any and every mRNA. AUG codes for amino acid methionine. Out of 64 codons, three codons viz. UAA, UAG, and UGA are termination codons that terminate/ stop the process of elongation of a polypeptide chain, as they do not code for any amino acid.

9. Codon and anticodon:

A codon is a part of DNA e.g. AUG is codon. It is always represented as 5’ AUG 3’. Anticodon is a part of tRNA. It is always represented as 3’UAC 5’.

3.Write a note on applications of DNA fingerprinting.

SOLUTION

  1. In forensic science, DNA fingerprinting is used to solve problems of rape and some complicated murder cases.
  2. DNA fingerprinting is used to find out the biological father or mother or both, of the child, in case of disputed parentage.
  3. DNA fingerprinting is used in the pedigree analysis in cats, dogs, horses and humans.

4.Explain the role of lactose in ‘Lac Operon’.

SOLUTION

  1. A few molecules of lactose enter into the cell by an enzyme permease.
  2. A small amount of this enzyme is present even when the operon is switched off.
  3. A few molecules of lactose, act as inducer and bind to the repressor.
  4. This repressor – inducer complex fails to join with the operator gene, which is then turned on.
  5. Structural genes produce all enzymes. Thus, lactose acts as an inducer of its own breakdown.
  6. When the inducer level falls, the operator is blocked again by the repressor. So structural genes are repressed/inactivated again. This is negative feedback.

Short Answer Question:

1.Write a note on Human genome project (HGP).

SOLUTION

The human genome project was initiated in 1990 under the International administration of the Human Genome Organization (HUGO).

This project was coordinated by the US Department of Energy and the National Institute of health. Additional contributors included universities across the United States and international partners in the United Kingdom, France, Germany, Japan and China.

The Human Genome Project was completed in 2003.

Following are the main aims of the human genome project:

1. Mapping the entire human genome at the level of nucleotide sequences.

2. To store the information collected from the project in databases.

3. To develop tools and techniques for analysis of the data.

4. Transfer of the related technologies to the private sectors, such as industries.

5. Taking care of the legal, ethical and social issues which may arise from the project

2.Describe the structure of ‘Operon’.

SOLUTION

The concept of the operon was first proposed by Jacob and Monod. A unit of genetic material that functions in a coordinated manner by means of a regulator, an operator, a promoter, and one or more structural genes that are transcribed together is called an operon. The clusters of genes with related functions are called operons.

Components of operon:

1. Regulator gene:

i. This gene controls the operator gene in cooperation with an inducer present in the cytoplasm.

ii. The regulator gene precedes the promoter gene. It may not be present immediately adjacent to the operator gene.

iii. The regulator gene produces a protein called repressor protein.

iv. The repressor binds with the operator gene and represses (stops) its action. Therefore, it is called regulator protein.

2. Promoter gene:

i. This gene precedes the operator gene. It is present adjacent to the operator gene.

ii. RNA polymerase enzyme binds to the promoter gene.

iii. The promoter gene base sequence determines which strand of DNA acts a template.

iv. When the operator gene is turned on, the enzyme moves over the operator gene and transcription of structural genes starts.

3. Operator gene:

i. This gene lies adjacent to the structural genes and controls their functioning.

ii. When the operator gene is turned on by an inducer, the structural genes produce mRNA.

iii. The operator gene is turned off by a product of the repressor gene.

4. Structural gene:

i. When lactose is added to the E. coli culture, the structural genes produce mRNA which in turn produces polypeptides, on the ribosomes.

ii. The polypeptides formed, act as enzymes to metabolize lactose in the cell.

iii. There are 3 structural genes in the sequence lacZ, lacY and lacA.

iv. Enzymes produced by these genes are β-galactosidase, permease, and transacetylase respectively.

Short Answer Question:

1.In the figure below A, B and C are three types of ______.

SOLUTION

In the given figure A (Messenger RNA), B (Ribosomal RNA) and C (Transfer RNA) are three types of Ribonucleic acids (RNA).

2.Identify the labeled structures on the following diagram of translation.

  • Part A is the _______
  • Part B is the _______
  • Part A is the _______

SOLUTION

  • Part A is the Anticodon present on the anticodon loop of tRNA.
  • Part B is the Amino acid
  • Part A is the Large subunit of ribosome

3.Match the entries in column I with those of column II and choose the correct answer.

Column IColumn II
A. Alkali treatmenti. Separation of DNA fragments on gel slab
B. Southern blottingii. Split DNA fragments into single strands
C. Electrophoresisiii. DNA transferred to nitrocellulose sheet
D. PCRiv. X-ray photography
E. Autoradiographyv. Produce fragments of different sizes
F. DNA treated with RENvi. DNA amplification

SOLUTION

Column IColumn II
A. Alkali treatmentii. Split DNA fragments into single strands
B. Southern blottingiii. DNA transferred to nitrocellulose sheet
C. Electrophoresisi. Separation of DNA fragments on gel slab
D. PCRvi. DNA amplification
E. Autoradiographyiv. X-ray photography
F. DNA treated with RENv. Produce fragments of different sizes

Long Answer Question:

1.Explain the process of DNA replication.

SOLUTION

The process by which DNA duplicates to form identical copies is known as replication.

Semi-conservative method of replication:

1. After replication, each daughter DNA molecule has one old and other new strands.

2. As parental DNA is partly conserved in each daughter’s DNA, the process of replication is called semi-conservative.

3. The model of semi-conservative replication was proposed by Watson and Crick.

4. The semi-conservative model of DNA replication using the heavy isotope of nitrogen N15 and E. coli was experimentally proved by Meselson and Stahl (1958).

Mechanism of replication is as follows:

a. Activation of Nucleotides:

i. The four types of nucleotides of DNA i.e. dAMP, dGMP, dCMP and dTMP are present in the nucleoplasm.

ii. They are activated by ATP in presence of an enzyme phosphorylase.

iii. This results in the formation of deoxyribonucleotide triphosphates i.e. dATP, dGTP, dCTP and dTTP. This process is known as Phosphorylation.

b. Point of Origin or Initiation point:

i. Replication begins at a specific point ‘O’ origin and terminates at point ‘T’.

ii. Origin is flanked by ‘T’ sites. The unit of DNA in which replication occurs is called replicon.

iii. In prokaryotes, there is only one replicon however in eukaryotes, there are several replicons in tandem.

iv. At the point ‘O’, enzyme endonuclease nicks one of the strands of DNA, temporarily.

v. The nick occurs in the sugar-phosphate backbone or the phosphodiester bond.

c. Unwinding of DNA molecule:

i. Enzyme DNA helicase breaks weak hydrogen bonds in the vicinity of ‘O’.

ii. The strands of DNA separate and unwind. This unwinding is bidirectional and continues as ‘Y’ shaped replication fork.

iii. Each separated strand acts as a template.

iv. The two separated strands are prevented from recoiling (rejoining) by SSBP (Single-strand binding proteins).

v. SSB proteins remain attached to both the separated strands for facilitating the synthesis of new polynucleotide strands.

d. Replicating fork:

i. The point formed due to the unwinding and separation of two strands appears like a Y-shaped fork, called replicating/ replication fork.

ii. The unwinding of strands imposes strain which is relieved by the super-helix relaxing enzyme.

e. Synthesis of new strands:

i. Each separated strand acts as a mould or template for the synthesis of a new complementary strand.

ii. It requires a small RNA molecule, called RNA primer.

iii. RNA primer attaches to the 3’ end of the template strand and attracts complementary nucleotides from the surrounding nucleoplasm.

iv. These nucleotides bind to the complementary nucleotides on the template strand by forming hydrogen bonds (i.e. A=T or T=A; G = C or C = G).

v. The newly bound consecutive nucleotides get interconnected by phosphodiester bonds, forming a polynucleotide strand.

vi. The synthesis of a new complementary strand is catalyzed by enzyme DNA polymerase. 7. The new complementary strand is always formed in 5’→ 3’ direction.

f. Leading and Lagging strand:

i. The template strand with free 3’ end is called a leading template and with free 5’ end is called a lagging template.

ii. The process of replication always starts at the C-3 end of the template strand and proceeds towards C-5 end.

iii. As both the strands of the parental DNA are antiparallel, new strands are always formed in 5’ → 3’ direction.

iv. One of the newly synthesized strands which develop continuously towards the replicating fork is called the leading strand.

v. Another new strand develops discontinuously away from the replicating fork and is called the lagging strand.

vi. Maturation of Okazaki fragments: DNA synthesis on the lagging template takes place in the form of small fragments called as Okazaki fragments (named after scientist Okazaki).

vii. Okazaki fragments are joined by the enzyme DNA ligase.

viii. RNA primers are removed by DNA polymerase and replaced by DNA sequence with the help of DNA polymerase-I in prokaryotes and DNA polymerase-α in eukaryotes.

ix. Finally, DNA gyrase (topoisomerase) enzyme forms a double helix to form daughter DNA molecules.

g. Formation of daughter DNA molecules:

Asterclasses

i. At the end of the replication, two daughter DNA molecules are formed.

ii. In each daughter’s DNA, one strand is parental and the other one is totally newly synthesized.

iii. Thus, 50% is contributed by mother DNA. Hence, it is described as semiconservative replication.

2.Describe the process of transcription in protein synthesis.

SOLUTION

The process of copying of genetic information from one (template) strand of DNA into a single-stranded RNA transcript is called transcription.

The process of transcription is as follows:

  1. For transcription, promoter, structural gene, and terminator (together called transcription unit) are required.
  2. The DNA strand used for the synthesis of RNA is called antisense or template strand which is oriented in 3′ → 5′ direction, while the other strand not involved in RNA synthesis is called the coding strand. It is oriented in 5′ → 3′ direction.
  3. A small DNA sequence which provides a binding site for RNA polymerase is called promoter which is present towards 5′ end/upstream, while a small DNA sequence which terminates the transcription process called terminator is present towards 3′ end/downstream.
  4. The process of transcription, in both prokaryotes and eukaryotes, involves three stages viz. Initiation, Elongation, and Termination.
  5. During initiation, RNA polymerase binds to the promoter and moves along the DNA and causes local unwinding of DNA duplex into two chains in the region of the gene.
  6. Exposed ATCG bases project into the nucleoplasm.
  7. Only one strand functions as template (antisense strand) and the other strand is complementary which is actually a coding strand (sense strand).
  8. During elongation, the ribonucleoside triphosphates join bases of the DNA template chain.
  9. As transcription proceeds, the hybrid DNA-RNA molecule dissociates and makes mRNA molecules free.
  10. As transcription proceeds, the hybrid DNA-RNA molecule dissociates and makes mRNA molecules free.
  11.  

3.Describe the process of translation in protein synthesis.

SOLUTION

Definition:

The translation is the mechanism in which codons of mRNA are translated and specific amino acids in a sequence form a polypeptide on ribosomes.

The process of translation requires amino acids, mRNA, tRNA, ribosomes, ATP, Mg++ ions, enzymes, elongation, translocation and release factors.

  1. About 20 different types of amino acids available in the cytoplasm are known to form proteins.
  2. DNA controls the synthesis of proteins having amino acids in a specific sequence. This control is possible through the transcription of mRNA. Genetic code is specific for particular amino acids.
  3. RNAs serve as intermediate molecules between DNA and protein.
  4. Ribosomes serve as a site for protein synthesis. Each ribosome consists of large and small subunits. These subunits occur separately in the cytoplasm. Only during protein synthesis, in presence of Mg++ ions, these two subunits get associated together.

Mechanism of translation (Synthesis of polypeptide chain):

It involves three steps initiation, elongation and termination:

a. Initiation:

1. Activation of amino acids is essential before translation initiates.

2. The amino acid is activated by utilizing energy from ATP molecule. This amino acid binds with the amino acid binding site of tRNA and forms of tRNA- amino acid complex.

3. A small subunit of ribosome attaches to the mRNA at 5’ end.

4. The initiator codon, AUG is present on mRNA which initiates the process of protein synthesis.

5. Initiator charged tRNA (with activated amino acid methionine) binds with the initiation codon (AUG) by its anticodon (UAC) through hydrogen bonds.

6. It carries activated amino acid methionine (in eukaryotes) or formyl methionine (in prokaryotes).

7. It occupies the P site of the ribosome and the A- the site is vacant.

8. Now the large subunit of ribosome joins with the smaller subunit that requires Mg++ ions.

b. Elongations:

During this process, activated amino acids are added one by one to first amino acid (methionine). Addition of Amino acid occurs in 3 Step cycle –

1. Codon recognition- Amino acyl tRNA molecule enters the ribosome at A-site. Anticodon binds with the codon by hydrogen bonds.

2. Amino acid on the first initiator tRNA at P-site and amino acid on tRNA at A-site join by peptide bond. Here enzyme Ribozyme acts as a catalyst. At this time first tRNA at ‘P’ site is kicked off.

3. Translocation-

The tRNA at A-site carrying a dipeptide at A-site moves to the P site. This process is called translocation. In translocation, both the subunits of ribosome move along in relation to tRNA and mRNA. Hence, tRNA carrying dipeptide now gets positioned at ‘P’ site of the ribosome, making ‘A’ site vacant. At this site, then next charged tRNA molecule carrying amino acid will be received. During this process, the first uncharged tRNA is discharged from E-site. This process of arrival of tRNA- amino acid complex, the formation of the peptide bond, ribosomal translocation, and removal of the previous tRNA, are repeated.

c. Termination and release of polypeptide:

1. Towards the 3’ end of mRNA, there is a stop codon (UAA/ UAG/ UGA). It is exposed at the A-site.

2. It is not read and joined by the anticodon of any tRNA.

3. The release factor binds to the stop codon, thereby terminating the translation process.

4. The polypeptide is now released in the cytoplasm.

5. Two subunits of ribosome dissociate and last tRNA is set free in the cytoplasm.

6. mRNA also has some additional sequences that are not translated and are referred as untranslated regions (UTR).

7. The UTRs are present at both 5’-end (before start codon) and at 3’- end (after stop codon). They are required for an efficient translation process.

8. Finally, mRNA is also released in the cytoplasm. It gets denatured by nucleases immediately. Hence mRNA is short-lived.

4.Describe the ‘Lac-operon’.

SOLUTION

1. Lactose or lac operon of E. coli is an inducible operon. The operon is switched on when a chemical inducer- lactose is present in the medium.

2. Jacob and Monad proposed the classical model of Lac operon.

3. The Lac operon consists of the promoter site (P), regulatory site (i), and operator site (O).

4. It also has three structural genes, namely z, y and each producing an enzyme.

5. The following three enzymes are required for the metabolism of lactose in the cell.

Name of geneEnzyme producedFunction
lac zβ-galactosidaseLactose–β-galactosidase———–Glucose+GalactoseLactose→β-galactosidaseGlucose+Galactose
lac yPermeaseEntry of lactose in the cell
lac aTransacetylaseTransfers acetyl group from Acetyl CoA to β-galactosidase

6. If glucose is not available for cells, they will require another source of energy such as lactose.

7. If lactose is not available, the repressor protein produced by repressor gene will attach to the operator and block RNA polymerase.

8. Lactose acts as an inducer. If lactose is available, it will prevent the repressor from binding the operator, by forming an inducer-repressor complex and allow RNA polymerase to transcribe mRNA.

9. RNA polymerase will attach to the promoter and will begin transcribing mRNA.

10. RNA polymerase first transcribes the lac z gene which is responsible for synthesizing β-galactosidase.

11. RNA polymerase moves on to the next gene, lac y that synthesizes the enzyme permease.

12. RNA polymerase finally moves to the lac a gene that is responsible for synthesizing transacetylase.

13. β-galactosidase, permease and transacetylase are enzymes in the metabolic pathway used to get energy from lactose.

14. After lactose is used up and levels decrease, the repressor will attach to the operator blocking the production of β-galactosidase, permease and transacetylase, so that lactose levels increase.

Long Answer Question:

The DNA molecule is double-stranded and the RNA molecule is single-stranded.

OPTIONS

  • True
  • False

SOLUTION

True.

  1. The DNA is responsible for storing genetic information and also preserving it for the next generation of cells. Thus, it needs to be stable and resistant to enzymatic or oxidative alteration.
  2. The double-stranded DNA molecule is so designed that the part which stores the genetic information; the nitrogenous base pairs are stacked inward.
  3. Phosphate groups 
  4. (PO4-)
  5.  keep the base pairs safe inside form the backbone. DNA is further wrapped up around histone proteins into chromosomes which keep it condensed. This would not be possible if it were only single-stranded.
  6. RNA is not meant to last long and acts as a template carrying information which is copied from the DNA. Using protein-synthesizing machinery, it forms a protein. After which is disintegrates and digested.

Long Answer Question:

The process of translation occurs at the ribosome.

OPTIONS

  • True
  • False

SOLUTION

True.

Ribosomes are sites for protein synthesis. Ribosome is responsible for holding mRNA incorrect position. Therefore, the process of translation occurs at the ribosomes.

Long Answer Question:

The job of mRNA is to pick up amino acids and transport them to the ribosomes.

OPTIONS

  • True
  • False

SOLUTION

False.

The job of tRNA is to pick up amino acids and transport them to the ribosomes. mRNA carries message in the form of code from DNA.

Transcription must occur before translation may occur.

OPTIONS

  • True
  • False

SOLUTION

True.

Transcription is a process of formation of mRNA whereas translation is process of protein synthesis. For protein synthesis mRNA is required to act as a template. Therefore, transcription must occur before translation may occur.

6.Guess (i) the possible location of DNA on the collected evidence from a crime scene and (ii) the possible sources of DNA.

EvidencePossible location of DNA on the evidenceSources of DNA
e.g. Eyeglassese.g. Ear piecese.g. Sweat, Skin
Bottle, Can, GlassSides, mouthpiece__________
__________HandleSweat, skin, blood
Used cigaretteCigarette butt____________
Bite mark____________Saliva
__________Surface areaHair, semen, sweet, urine

SOLUTION

EvidencePossible location of DNA on the evidenceSources of DNA
e.g. Eyeglassese.g. Ear piecese.g. Sweat, Skin
Bottle, Can, GlassSides, mouthpieceSaliva, sweat
Door, baseball bat, a similar weaponHandleSweat, skin, blood
Used cigaretteCigarette buttSaliva
Bite markPersons skinor clothingSaliva
Blanket, pillow, bedsheet, dirty laundrySurface areaHair, semen, sweat, urine

COMPLETED


Chapter 3, Inheritance and Variation, hsc, biology, science, maharashtra bore, latest edition, full solution,

Multiple choice question.

1.Phenotypic ratio of incomplete dominance in Mirabilis jalapa.

OPTIONS

  • 2 : 1 : 1
  • 1 : 2 : 1
  • 3 : 1
  • 2 : 2

2.In a dihybrid cross, F2 generation offsprings show four different phenotypes while the genotypes are _______.

OPTIONS

  • six
  • nine
  • eight
  • sixteen

3.A cross between an individual with an unknown genotype for a trait with recessive plant for that trait is _______.

OPTIONS

  • back cross
  • reciprocal cross
  • monohybrid cross
  • test cross

4.When phenotypic and genotypic ratios are the same, then it is an example of _______.

OPTIONS

  • incomplete dominance
  • complete dominance
  • multiple alleles
  • cytoplasmic inheritance

5.If the centromere is situated near the end of the chromosome, the chromosome is called ______.

OPTIONS

  • metacentric
  • acrocentric
  • sub-metacentric
  • telocentric

6.Chromosomal theory of inheritance was proposed by _______.

OPTIONS

  • Sutton and Boveri
  • Watson and Crick
  • Miller and Urey
  • Oparin and Halden

7.If the genes are located in a chromosome as p-q-r-s-t, which of the following gene pairs will have the least probability of being inherited together?

OPTIONS

  • p and q
  • r and s
  • s and t
  • p and s

8.Find the mismatch pair:

OPTIONS

  • Down’s syndrome = 44 + XY
  • Turner’s syndrome = 44 + XO
  • Klinefelter syndrome = 44 + XXY
  • Super female = 44 + XXX

9.A colour-blind man marries a woman, who is homozygous for normal colour vision, the probability of their son being colour blind is –

OPTIONS

  • 0%
  • 25%
  • 50%
  • 100%

Very Short Answer Question.

1.Explain the statement of Test cross is back cross but back cross is not necessarily a test cross.

SOLUTION

  1. In back cross F1 generation can be crossed with either dominant or recessive parent.
  2. But in test cross, F1 generation is crossed with a recessive parent only.
  3. Thus, in the back cross, if F1 generation is crossed with a recessive parent it will be a test cross, but if F1 generation is crossed with a dominant parent it will not be a test cross. Therefore, the test cross is a back cross but the back cross is not necessarily a test cross.

2.Explain the statement of Law of dominance is not universal.

SOLUTION

  1. According to law of dominance, when two homozygous individuals with one or more sets of contrasting characters are crossed, the alleles (characters) that appear in F1 are dominant and those which do not appear in F1 are recessive.
  2. In many cases, the dominance is not complete or absent. This can be explained by two deviations of Mendel’s law of dominance: Incomplete dominance and codominance. Thus, law of dominance is significant and true, but it is not universally applicable.

3.Define dihybrid cross

SOLUTION

A cross between parents differing in two heritable traits is called a dihybrid cross.

OR

A cross between two pure (homozygous) parents in which the inheritance pattern of two pairs of contrasting characters is considered simultaneously is called a dihybrid cross.

1.Define the following:

1.Homozygous

SOLUTION 1

Homozygous – Diploid condition where both the alleles are identical is called homozygous.

SOLUTION 2

Homozygous (pure): An individual having identical alleles for a particular character is homozygous for that character. It is pure or true breeding for that trait. e.g. TT, tt.

2.Heterozygous

SOLUTION 1

Heterozygous – Diploid condition where both the alleles are different is called heterozygous.

SOLUTION 2

Heterozygous: An individual possessing contrasting (dissimilar) alleles for a particular trait is called heterozygous. It is a hybrid and does not breed true for that trait. e.g. Tt

3.Test cross.

SOLUTION

Test cross: The cross between F1 hybrid and its homozygous recessive parent is called a test cross.


Very Short Answer Question.

1.What is allosome?

SOLUTION

The chromosomes which are responsible for the determination of sex are known as Allosomes (sex chromosomes).

2.What is crossing over?

SOLUTION

  1. Crossing over is a process that produces new combinations (recombinations) of genes by interchanging and exchanging of corresponding segments between non-sister chromatids of homologous chromosomes.
  2. It occurs during pachytene of prophase I of meiosis.
  3. The mechanism of crossing over consists of four sequential steps such as synapsis, tetrad formation, crossing over, and terminalisation.
  4. The phenomenon of crossing over is universal and it is necessary for the natural selection because it increases the chances of variation.

3.Give one example of the autosomal recessive disorder.

SOLUTION

Autosomal recessive traits Phenyl ketonuria (PKU), Cystic fibrosis, and Sickle cell anaemia.

4.What are X-linked genes?

SOLUTION

The genes which are present on the non-homologous region of X-chromosome are known as X-linked genes.

5.What are holandric traits?

SOLUTION

The traits that are controlled by genes present only on the Y chromosome are known as holandric traits.

6.Give an example of a chromosomal disorder caused due to nondisjunction of autosomes.

SOLUTION

Down syndrome is an example of a chromosomal disorder caused due to non-disjunction of autosomes.

7.Give one example of complete sex linkage?

SOLUTION

Examples of X-linked traits are hemophilia, red-green colour blindness, myopia (near sightedness), and for Y-linked are hypertrichosis, ichthyosis, etc.


Short Answer Question.

1.Enlist seven traits of pea plant selected/ studied by Mendel.

SOLUTION

Following are the seven traits of pea plant selected/studied by Mendel:

CharacterContrasting form / traits
 DominantRecessive
1. Height of stemTallDwarf
2. Colour of flowerPurpleWhite
3. Position of flowerAxialTerminal
4. Pod shapeInflatedConstricted
5. Pod colourGreenYellow
6. Seed shapeRoundWrinkled
7. Seed colour (cotyledon)YellowGreen

2.Why law of segregation is also called the law of purity of gametes?

SOLUTION

  1. A diploid organism contains two factors for each trait in its diploid cells and the factors segregate during the formation of gametes.
  2. The two alleles (contrasting characters) do not mix, alter or dilute each other and the gametes formed are ‘pure’ for the characters which they carry.
  3. A gamete may carry either a dominant or recessive factor but not both.

Hence, this law is also called the law of purity of gametes.

3.Write a note on pleiotropy.

SOLUTION

  1. When a single gene controls two (or more) different traits it is called pleiotropic gene and the phenomenon is called pleiotropy or pleiotropism.
  2. The phenotypic ratio is 1:2 instead of 3:1 because of the death of recessive homozygote. The disease, sickle-cell anaemia, is caused by a gene HbS.
  3. Normal or healthy gene HbA is dominant. The carriers (heterozygotes HbA /HbS) show signs of mild anaemia as their RBCs become sickle shaped i.e. half-moon-shaped only under abnormally low O2 concentration.
  4. The homozygotes with recessive gene HbS die of fatal anaemia.
  5. Thus, the gene for sickle-cell anaemia is lethal in homozygous condition and produces sickle cell trait in the heterozygous carrier. Two different expressions are produced by a single gene.

4.What are the reasons for Mendel’s success?

SOLUTION

Following are the reasons for Mendel’s success:

  1. His experiments were carefully planned and involved a large sample.
  2. He carefully recorded the number of plants of each type and expressed his results as ratios.
  3. In the pea plant, contrasting characters can be easily recognized.
  4. The seven different characters in the pea plant were controlled by a single factor each.
  5. The factors are located on separate chromosomes and these factors are transmitted from generation to generation.

5.“Father is responsible for determination of sex of child and not the mother”. Justify.

Asterclasses.com

SOLUTION

  1. A human male has 44 autosomes + XY sex chromosomes, whereas a female has 44 autosomes + XX sex chromosomes.
  2. During gamete formation in male, the diploid germ cells in testis undergo spermatogenesis to produce two types of haploid sperms, 50% sperms contain 22 autosomes and X chromosome while 50% sperms contain 22 autosomes and Y chromosome.
  3. In females, the diploid germ cells in ovaries undergo oogenesis to produce only one type of egg. All eggs contain 22 autosomes and the X chromosome.
  4. Thus human male is heterogametic and female is homogametic.
  5. If a sperm containing X chromosome fertilizes the egg (ovum), the diploid zygote formed after fertilization grows into a female child.
  6. If a sperm containing Y chromosome fertilizes the egg, then diploid zygote formed after fertilization grows into a male child.
  7. The sex of a child depends on the type of sperm fertilizing the egg and hence the father is responsible for the determination of sex of the child and not the mother

6.What is a linkage? How many linkage groups do occur in human being?

SOLUTION

The tendency of two or more genes presents on the same chromosomes to be inherited together is known as linkage. The haploid number of chromosomes in humans is 23 therefore there are 23 linkage groups in humans.

7.Write note on –PKU.

SOLUTION

  1. Phenylketonuria is an inborn metabolic disorder caused due to deficiency of phenylalanine hydroxylase enzyme.
  2. Phenylketonuria is caused due to recessive autosomal genes.
  3. When recessive genes are present in homozygous condition, phenylalanine hydroxylase enzyme is not produced.
  4. This enzyme is essential for the conversion of amino acid phenylalanine into tyrosine.
  5. Due to the absence of this enzyme, phenylalanine is not converted into tyrosine.
  6. Hence, phenylalanine and its derivatives are accumulated in blood and cerebrospinal fluid (CSF).
  7. It affects development of the brain and causes mental retardation.
  8. Excess phenylalanine is excreted in urine, hence this disease is called phenylketonuria.

8.Compare X chromosome and Y chromosome.

SOLUTION

X ChromosomeY Chromosome
1. It is metacentric, hence appears X shaped.1. It is acrocentric, hence appears Y shaped
2. It is longer than ‘Y’ chromosomes.2. It is shorter than ‘X’ chromosomes.
3. It contains a large amount of euchromatin and a small amount of heterochromatin.3. It contains large amount of heterochromatin and small amount of euchromatin.
4. It is found in both males and females.4. It is found only in males.
5. Non-homologous part of X chromosome shows more genes than Y chromosome.5. Non-homologous part of Y chromosome contains few genes as compared to X chromosome.
6. X – linked genes are present on the X chromosome.6. Y-linked genes (Holandric genes) are present on Y chromosome.
7. Genes present on X chromosome show crisscross inheritance.7. Genes present on Y chromosome show straight inheritance.

9.Explain the chromosomal theory of inheritance.

SOLUTION

The chromosomal theory of inheritance was proposed by Sutton and Boveri.

Following are the postulates of chromosomal theory of inheritance:

  1. Chromosomes are found in pairs in somatic or diploid cells.
  2. During gamete formation, homologous chromosomes pair, segregate and assort independently at meiosis. Due to this, each gamete contains only one chromosome of a pair.
  3. Hereditary characters are carried by chromosomes which are present in the nucleus of these gametes.
  4. Gametes (sperm and egg) contain all the hereditary characters. They form the link between parents and offsprings.
  5. The union of sperm and egg during fertilization restores the diploid number of chromosomes.

10.Observe the given pedigree chart and answer the following question.

Identify whether the trait is sex-linked or autosomal.

SOLUTION

The trait represented in given pedigree is sex linked trait.

11.Observe the given pedigree chart and answer the following question.

Give an example of a trait in human beings which shows such a pattern of inheritance.

SOLUTION

Haemophilia, colour blindness are examples of sex-linked traits in humans.


1.Match the column-I with column-II and re-write the matching pairs.

Column-IColumn-II
1. 21 trisomya. Turner’s syndrome
2. X-monosomyb. Klinefelter’s syndrome
3. Holandric traitsc. Down’s syndrome
4. Feminized maled. Hypertrichosis

SOLUTION

Column-IColumn-II
1. 21 trisomyc. Down’s syndrome
2. X-monosomya. Turner’s syndrome
3. Holandric traitsd. Hypertrichosis
4. Feminized maleb. Klinefelter’s syndrome

Very Short Answer Questions.

1.Define dihybrid cross

SOLUTION

A cross between parents differing in two heritable traits is called a dihybrid cross.

OR

A cross between two pure (homozygous) parents in which the inheritance pattern of two pairs of contrasting characters is considered simultaneously is called a dihybrid cross.

2.Explain a dihybrid cross with suitable example and checker board method.

SOLUTION

The phenotypic ratio of different types of offsprings (with different combinations) obtained in F2 generation of dihybrid cross is called the dihybrid ratio. It is 9 : 3 : 3 : 1.

For example, when we cross a yellow round seed pea plant with a green wrinkled seed pea plant, we get 9 yellow round, 3 yellow wrinkled, 3 green round and 1 green wrinkled plants in the F2 generation.

F2 generation → 

YRYryRyr
 
YRYYRR Yellow roundYYRr Yellow roundYyRR Yellow roundYyRr Yellow round
YrYYRr Yellow roundYYrr Yellow wrinkledYyRr Yellow roundYyrr Yellow wrinkled
yRYyRR Yellow roundYyRr Yellow roundyyRR Green roundyyRr Green round
yrYyRr Yellow roundYyrr Yellow wrinkledyyRr Green roundyyrr Green wrinkled

Phenotypic ratio: Yellow round = 9 ;Yellow wrinkled = 3; Green round = 3; Green wrinkled = 1

Dihybrid ratio → 9 : 3 : 3 : 1

Genotypic ratio → 

YYRR YYRr YyRR YyRr YYrr Yyrr yyRR yyRr yyrr

  1   :   2    :   2   :   4   :   1  :  2  :   1   :  2  :   1

3.Explain with suitable example an independent assotrment.

SOLUTION

Law of independent assortment:

The law states that, when a hybrid possessing two (or more) pairs of contrasting factors (alleles) forms gametes, the factors in each pair segregate independently of the other pair.

OR

The law of independent assortment states that, when two parents differing from each other in two or more pairs of contrasting characters are crossed, then the inheritance of one pair of character is independent of the other pair of character

  1. This law is based on a dihybrid cross.
  2. It describes how different genes or alleles present on separate chromosomes independently separate from each other, during the formation of gametes. These alleles are then randomly united in fertilization.
  3. In dihybrid cross, F2 phenotypic ratio 9:3:3:1 indicates that the two pairs of characters behave independent of each other. It can be concluded that the two characters under consideration are assorted independently giving rise to different combinations.
Asterclasses.com

F2 generation → 

TRTrtRtr
  ↓
TRTTRR Tall redTTRrTall redTtRRTall redTtRrTall red
TrTTRrTall redTTrrTall whiteTtRrTall redTtrrTall white
tRTtRRTall redTtRrTall redttRR Dwarf redttRr Dwarf red
trTtRrTall redTtrrTall whitettRr Dwarf redttrr Dwarf white

Result: Tall red = 9; Tall white = 3; Dwarf red = 3; Dwarf white = 1

Phenotypic ratio → 9 : 3 : 3 : 1

Genotypic ratio → 

1    :     2     :  2   :   4  :    1    :   2  :   1   :   2  :   1

TTRR  TTRr  TtRR  TtRr  ttRR  ttRr  TTrr  Ttrr   ttrr

From the above results, it is obvious that the inheritance of character of tallness is not linked with the red colour of the flower. Similarly, the character of dwarfness is not linked with the white colour of the flower. This is due to the fact that in the above cross, the two pairs of characters segregate independently. In other words, there is an independent assortment of characters during inheritance.

4.Define test cross and explain its significance.

SOLUTION

Definition:

The cross between F1 hybrid and its homozygous recessive parent is called a test cross.

Significance of test cross:

  1. It helps to determine whether individuals exhibiting dominant character are genotypically homozygous or heterozygous.
  2. It has wide application in plant breeding experiments.
  3. In rapid crop improvement programmes, test cross is used to introduce a useful recessive trait in the hybrids.

4.What is parthenogenesis?

SOLUTION

Parthenogenesis:

The process by which a female produces offspring from unfertilized eggs is known as parthenogenesis.

5.Explain the haplo-diploid method of sex determination in the honey bee.

SOLUTION

Haplo-diploid method of sex determination in the honey bee:

Asterclasses.com
  1. In honey bees, the chromosomal mechanism of sex determination is of haplo-diploid type.
  2. In this type, the sex of the individual is determined by the number of the set of chromosomes received.
  3. Females are diploid (2n=32) and males are haploid (n=16).
  4. The female produces haploid eggs (n=16) by meiosis and male produces haploid sperms (n=16) by mitosis.
  5. If the egg is fertilized by sperm, the zygote develops into a diploid female (2n=32) (queen and worker) and an unfertilized egg develops into a haploid male (n=16) (Drone) by way of parthenogenesis.
  6. The diploid female gets differentiated into either worker or queen bee depending on the food they consume during their development.
  7. Diploid larvae which get royal jelly as food develops into queen (fertile female) and other develops into workers (sterile females).

6.In the answer for inheritance of X-linked genes, Madhav had shown carrier male. His answer was marked incorrect. Madhav was wondering why his marks were cut. Explain the reason.

SOLUTION

A male has X and Y chromosomes. The X-linked genes do not have their alleles on Y chromosome. Therefore, for males to suffer from disease only one copy of a defective gene is sufficient. In the inheritance of X-linked genes, females may be carriers because they have two X chromosomes and may carry one normal and another defective gene. This is not possible in the case of males due to the presence of a single X chromosome.

7.With the help of a neat labelled diagram, describe the structure of chromosome.

SOLUTION

Structure of chromosome:

i. Chromosomes are highly condensed and therefore are clearly visible in metaphase stage of cell division.

ii. A typical chromosome consists of two chromatids joined together at centromere also known as primary constriction.

iii. Primary constriction consists of a disk shape plate called kinetochore. During cell division, spindle fibres get attached to the kinetochore.

iv. Apart from primary constriction, some few chromosomes possess additional one or two constrictions called secondary constriction.

v. At secondary constriction I (nucleolar organizer), the nucleolus becomes organized during interphase.

vi. A satellite body (SAT body) is attached at secondary constriction II, in very few chromosomes.

vii. Each chromatid in turn contains a long, unbranched, slender, highly coiled double-stranded DNA thread, called chromonema, extending through the length of the chromatid.

viii. The ends of the chromosome (i.e. chromatids) are known as telomeres.

Structure of Chromosome-

8.What is cris-cross inheritance?

SOLUTION

The inheritance of characters from the father to his grandson through his daughter is called criss-cross inheritance. 

Long answer type question.

9.Explain cris-cross inheritance with suitable example.

SOLUTION

Criss-cross inheritance can be explained with the help of two examples: colour blindness and haemophilia.

i. Colour blindness:

a. A person suffering from colour blindness cannot differentiate between red and green colours. Both these colours appear grey to the colour blind person.

b. It is caused due to recessive X-linked genes (Xc) which prevent the formation of colour sensitive cells in the retina that are necessary for distinguishing red and green colours.

c. Dominant X linked gene (XC) is necessary for the formation of colour sensitive cells in the retina of eye.

d. The homozygous recessive females (XcXc) and hemizygous recessive male (XcY) are unable to distinguish between red and green colours. The frequency of colour blind women is much less than colour blind men.

e. When a normal man marries a carrier woman, half of their sons may be colour blind, while the remaining half will have normal vision. All their daughters will have normal vision and half of them will be carriers for the disease.

f. If a colour blind male (XcY) marries a female with normal vision (XCXC), then all the offsprings will have normal vision. The sons will have normal vision but daughters will be carriers for the disease. The carriers have normal vision.

ii. Haemophilia (Bleeder’s disease):

a. Haemophilia is X-linked recessive disorder in which blood fails to clot or coagulates very slowly.

b. The genes for normal clotting are dominant over the recessive genes for haemophilia.

c. The person having the recessive gene for haemophilia is deficient in clotting factors (VIII or IX) in blood.

d. Even minor injuries cause continuous bleeding, hence haemophilia is also called as bleeder’s disease.

e. The recessive gene for haemophilia is located on a nonhomologous region of X chromosome.

f. As there is no corresponding allele on Y chromosome to suppress its expression, so men suffer from this disease.

g. Women suffer only when both X chromosomes have recessive genes (alleles).

1. If a haemophilic male (XhY) marries a female with the normal clotting of blood (XX), then all the offsprings will show normal clotting of blood. The sons will have normal clotting of blood, but daughters will be carriers for the disease. The carriers have normal clotting of blood.

2. When carrier woman (XHXh) marries a normal man (XHY), then all the daughters will have normal clotting of blood but half of them will be carriers for the disease. Half the sons will be haemophilic while the remaining will have normal clotting of blood.

asterclasses

Long answer type question.

10.Describe the different types of chromosomes.

SOLUTION

Following are the four types of chromosomes with respect to the position of the centromere: Acrocentric (j shaped), Telocentric (i shaped), Sub-metacentric (L shaped) and Metacentric (V-shaped).

Type of chromosomeName of chromosomePosition of centromere
MetacentricMiddle of the chromosome
Sub-metacentricSome distance away from the centre of chromosomes
AcrocentricNear one end of the chromosome
TelocentricAt one end

                              COMPLETED


Chapter 2, Reproduction in Lower and Higher Animals, hsc, biology, science, maharashtra bore, latest edition, full solution,

Multiple choice question.

1.The number of nuclei present in a zygote is _______.

OPTIONS

  • two
  • one
  • four
  • eight

2.Which of these is the male reproductive organ in humans?

OPTIONS

  • sperm
  • seminal fluid
  • testes
  • ovary

3.Attachment of embryo to the wall of the uterus is known as _______.

OPTIONS

  • fertilization
  • gestation
  • cleavage
  • implantation

4.Rupturing of follicles and discharge of ova is known as _______.

OPTIONS

  • capacitation
  • gestation
  • ovulation
  • copulation

5.In human female, the fertilized egg gets implanted in uterus _______.

OPTIONS

  • After about 7 days of fertilization
  • After about 30 days of fertilization
  • After about two months of fertilization
  • After about 3 weeks of fertilization

6.Test tube baby technique is called _______.

OPTIONS

  • in vivo fertilization
  • in situ fertilization
  • in vitro fertilization
  • artificial insemination

7.The given figure shows human sperm. Various parts of it are labelled as A, B, C, and D. Which labelled part represents acrosome?

OPTIONS

  • B
  • C
  • D
  • A

8.Presence of beard in boys is a _______.

OPTIONS

  • primary sex organ
  • secondary sexual character
  • secondary sex organ
  • primary sexual character

Answer in one sentence.

1.What is the difference between a foetus and an embryo?

SOLUTION

The embryo is the developing organism from fertilization to the end of the eighth week of development. It develops into the foetus.

The foetus is the developing organism from the beginning of the third month to birth.

2.Outline the path of sperm upto the urethra.

SOLUTION

Seminiferous tubules → Rete testis → Vasa efferentia → Epididymis → Vas deferens → Ejaculatory duct → Urethra

3.Which glands contribute fluids to the semen?

SOLUTION

The seminal vesicle, prostate gland and Cowper’s / Bulbourethral gland contribute fluids to the semen.

4.Name the endocrine glands involved in maintaining the sex characteristics of males.

SOLUTION

Testes (gonads)

5.Where does fertilization and implantation occur?

SOLUTION

Fertilization usually takes place in the ampulla of the fallopian/uterine tube, while implantation occurs in the endometrium of the uterus.

6.Enlist the external genital organs in the female.

SOLUTION

The external genital organs of female include parts external to the vagina, collectively called vulva (covering or wrapping), or pudendum. They include the following parts:

  1. Vestibule:
    It is a median vertical depression of the vulva enclosing the urethral and vaginal opening.
  2. Labia minora:
    These are another pair of thin folds inner to the labia majora with which they merge posteriorly to form the fourchette (frenulum), while towards the anterior end they converge into a hood-like covering around the clitoris.
  3. Clitoris:
    A small conical and sensitive projection lying at the anterior end of labia minora. It has a pair of erectile tissue i.e. corpora cavernosa which is homologous to the penis.
  4. Labia majora:
    These are a pair of fleshy folds of skin forming the boundary of the vulva. They are homologous to the scrotum. They surround and protect the other parts of the external genitalia and enclose the urethral and vaginal openings in the vestibule.
  5. Mons pubis:
    It is a fleshy elevation above the labia majora. The Mons pubis and outer part of labia majora show pubic hair.

7.Give two differences between blastula and gastrula.

SOLUTION

BlastulaGastrula
1. It is formed during the early mitotic division through the process of blastulation.1. It is formed at a later stage through the process of gastrulation.
2. Blastula is formed from morula.2. Gastrula is formed blastula.
3. It is formed through rapid mitotic divisions.3. It is formed by slower mitotic divisions.
4. The cells do not move during the formation of the blastula.4. The cells move through morphogenetic movement during the formation of the gastrula.
5. Germinal layers absent5. The formation of three germinal layers occurs.
6. Cells are undifferentiated.6. Cells undergo differentiation

8.What is the difference between embryo and zygote?

SOLUTION

EmbryoZygote
1. It is multicellular.1. It is unicellular.
2. It follows the stage of zygote formation. 2. It is the first stage of development after fertilization.
3. Embryogenesis results in the formation of the embryo.3. Fertilization results in zygote formation.
4. It is formed in the uterus.4. It is formed in the fallopian/ uterine tube.

Fill in the blank:

1.The primary sex organ in human male is _______.

SOLUTION

The primary sex organ in human male is the testis.

2.The _______ is also called the womb.

SOLUTION

The uterus is also called the womb.

3.Sperm fertilizes ovum in the _______of fallopian tube.

SOLUTION

Sperm fertilizes ovum in the ampulla of fallopian tube.

4.The disc-like structure which helps in the transfer of substances to and from the fetus’s body is called _______.

SOLUTION

The disc-like structure which helps in the transfer of substances to and from the fetus’s body is called placenta.

5.Gonorrhoea is caused by _______ bacteria.

SOLUTION

Gonorrhoea is caused by Neisseria gonorrhoeae bacteria.

6.The hormone produced by the testis is _______.

SOLUTION

The hormone produced by the testis is testosterone.


Short answer question.

1.Write a note on budding in Hydra.

SOLUTION

  1. Asexual reproduction in Hydra takes place through budding.
  2. Budding normally occurs in favorable conditions.
  3. In Hydra, a small outgrowth is produced towards the basal end of the body.
  4. It develops as a bud which grows and forms tentacles.
  5. This bud eventually develops (get transformed) into a new individual.
  6. The young Hydra gets detached from the parent and becomes an independent new organism.

Budding in Hydra-

2.Explain the different methods of reproduction occurring in sponges.

SOLUTION

  • Asexual reproduction in sponges:
  1. Sponges reproduce asexually via. gemmule formation. Gemmule is an internal bud formed only in sponges to overcome unfavorable conditions.
  2. The structure of the gemmule includes the micropyle, spicule, inner layer, archaeocytes, and outer layer.
  3. It possesses an asexually produced mass or aggregation of dormant cells known as archaeocytes. These cells are capable of developing into a new organism (totipotent).
  4. The archaeocytes get coated by a thick resistant layer of secretion by amoebocytes.
  5. Monaxon spicules (developed by growth along a single axis) are secreted by scleroblasts in between the inner and outer membrane.
  6. On return of favorable conditions of water and temperature, the gemmules hatch and develop into a new individual.
  7. There is a minute opening called micropyle through which the cells (new individuals) come out during favourable conditions.
  • Sexual reproduction in sponges:
  1. Poriferans that reproduce by the sexual method are hermaphrodites and produce sperms and eggs at different times.
  2. Sperms disseminated into the water column, are subsequently captured by female sponges of the same species.
  3. Inside the female, the sperm is transported to eggs by means of archaeocytes.
  4. Fertilization occurs and zygotes develop into ciliated larvae.
  5. Once these larvae are in the water column, they settle and develop into juvenile sponges.
  6.  

3.Write a note on IVF.

SOLUTION

IVF (In-vitro Fertilization):

It is a process of fertilization where an egg is combined with sperm outside the body in a test tube or glass plate to form a zygote under simulated conditions in the laboratory. The zygote or early embryos (with up to 8 blastomeres) could be then transferred into the fallopian tube for further development.

4.Comment on any two mechanical contraceptive methods.

SOLUTION

Mechanical means / Barrier methods:

In this method, the ovum and sperm are prevented from physically meeting with the help of barriers. These mechanical barriers are of three types.:

1. Condom:

It is a thin rubber sheath that is used to cover the penis of the male during copulation.

It prevents the entry of ejaculated semen into the female reproductive tract. It can thus prevent conception. It is a simple and effective method and has no side effects.

Condoms should be properly discarded after every use.

Condom is also a safeguard against STDs and AIDS.

e.g. “Nirodh” is the most widely used contraceptive by males. It is easily available and is given free by the government.

2. Diaphragm, cervical caps and vaults:

These devices used by the female are made up of rubber. They prevent conception by blocking the entry of sperms through the cervix. The device is inserted into the female reproductive tract to cover the cervix during copulation.

3. Intra-uterine devices (IUDs):

These clinical devices are plastic or metal objects. A doctor or trained nurse places the IUDs into the uterus. These devices include Lippes loop, copper releasing IUDs (Cu-T, Cu7, multiload 375), and hormone-releasing IUDs (LNG-20, progestasert).

i. Lippes loop:

It is a plastic double “s” loop. It attracts the macrophages stimulating them to accumulate in the uterine cavity. Macrophages increase phagocytosis of sperms within the uterus and act as a contraceptive.

ii. Copper releasing IUDs:

Suppress sperm motility and the fertilizing capacity of sperms.

iii. Hormone releasing IUDs:

Make the uterus unsuitable for implantation and the cervix hostile to the sperms. It delays pregnancy for a longer period.

Drawbacks: Spontaneous expulsion, occasional haemorrhage, and chances of infection are the drawbacks of IUDs.

5.Write a note on tubectomy.

SOLUTION

  1. Tubectomy is a permanent birth control method in women.
  2. It is performed by removing a small part of the fallopian tube or tying it up through a small incision in the abdomen or through the vagina.
  3. These techniques are highly effective but their reversibility is highly poor.
  4. Sometimes it becomes necessary to use these methods either to prevent pregnancy or to delay or space pregnancy due to personal reasons.
  5. This method blocks gamete transport and prevents pregnancy.

6.Give the name of causal organism of syphilis and write on its symptoms.

SOLUTION

  1. Causative agent:
    Treponema pallidum (Bacteria)
  2. Symptoms:
    Primary lesion called chancre at the site of infection. The chancre is formed on the external genitalia, skin rashes, and mild fever, inflamed joints, loss of hair. Paralysis, Degenerative changes occur in the heart and brain.

7.What is colostrum?

SOLUTION

  1. Colostrum is sticky and yellow fluid secreted by the mammary glands soon after childbirth.
  2. It contains proteins, lactose, and mother’s antibodies e.g. IgA.
  3. The fat content in colostrum is low.
  4. The antibodies present in it helps in developing resistance for the new born baby at a time when its own immune response is not fully developed.

Answer the following question.

1.Describe the phases of the menstrual cycle and their hormonal control.

SOLUTION

The menstrual cycle involves a series of cyclic changes in the ovary and the female reproductive tract, mainly in the uterus.

It is divided into four phases:

i. Menstrual phase

ii. Proliferative phase

iii. Ovulatory phase

iv. Secretory phase or luteal phase

i. Menstrual phase:

The beginning of each cycle is taken as the first day when menses or loss of blood takes place. During this phase, about 45-100ml of blood is lost.

This phase lasts for approximately five days (average 3-7 days).

The blood in the menstrual discharge does not clot due to the presence of fibrinolysin.

Menstrual phase occurs when an ovulated egg does not get fertilized and it is thereby shed out along with the menstruum. This process is also referred to as the ‘funeral of unfertilized egg’.

Changes in the uterus:

The endometrium of the uterus breaks down under the effect of prostaglandins released due to decreased levels of progesterone and estrogen.

During menses, the blood, tissue fluid, mucus, endometrial lining, and the unfertilized oocyte is discharged through the vagina. Also, the endometrial lining becomes very thin i.e. about 1 mm.

Changes in the ovary:

During these five days, many primordial follicles develop into primary and few of them into secondary follicles under the effect of FSH.

ii. Proliferative phase / Follicular phase / Post menstrual phase:

This phase is the duration between the end of menstruation and the release of ovum (ovulation). The duration of this phase is more variable than other phases. Generally, it extends from 5th to 13th day of the menstrual cycle.

Changes in the ovary:

Generally, out of 6 to 12 secondary follicles that proceed to develop, only one develops into a Graafian follicle (mature follicle). while the rest of the follicles degenerate (atresia). The stimulation for proliferation of new follicles is influenced by GnRH which stimulates release of FSH. The developing secondary follicles secrete the hormone estrogen.

Changes in the uterus:

Endometrium begins to regenerate under the effect of gradually increasing the number of estrogens. Regeneration also involves the formation of endothelial cells, endometrial or uterine glands, and network of blood vessels. The thickness of the endometrium reaches 3-5 mm.

iii. Ovulatory phase:

It is the shortest phase of menstrual cycle.

Changes in the ovary:

It involves rupturing of the mature Graafian follicle and release of an ovum (secondary oocyte) into the pelvic cavity; usually on 14th day of the menstrual cycle. Rapid secretion of LH by a positive feedback mechanism causes the mature follicle to rupture. Ovulation may be accompanied by mild or severe pains in lower abdomen.

iv. Secretory phase / Luteal phase:

It is the phase between ovulation and the beginning of the next menses. This phase is the longest phase. It lasts for 14 days i.e., from 15th to 28th day of the cycle.

Changes in the ovary:

After release of secondary oocyte, remaining tissue of Graafian follicle transforms into a corpus luteum under the effect of LH. The corpus luteum releases progesterone, small amount of estrogen, and inhibin.

The ovulated egg may get fertilized within 24 hours. In the absence of fertilization: the Corpus luteum can survive for only two weeks and then degenerates into a white scar called corpus albicans. In case of fertilization: The embryo is implanted, there is a secretion of human chorionic gonadotropin (hCG), which extends the life of corpus luteum and stimulates its secretory activity. The presence of hCG in maternal blood and urine is an indicator of pregnancy. In absence of fertilization, the next menstrual cycle begins.

Changes in the uterus:

Under the influence of progesterone and estrogen, the endometrial glands grow, become coiled, and start uterine secretions. Endometrium becomes more vascularized and thickens up to 8-10 mm. Inhibin stops secretion of FSH. These changes are necessary for fertilization and subsequent implantation.

2.Explain the steps of parturition.

SOLUTION

Parturition is the process of giving birth to a baby. The physical activities involved in parturition like uterine and abdominal contractions, dilation of the cervix, and passage of baby are collectively called labour. Labour is accompanied by a localised sensation of discomfort or agony called labor pains. Parturition involves the following three steps:

1. Dilation stage:

Uterine contractions begin from the top, forcing the baby towards the cervix. Contractions are accompanied by pain caused by compression of blood vessels. Oxytocin induces uterine contractions which become stronger and stronger due to stimulatory reflex. As the baby is pushed down in the uterus, its head comes to lie against the cervix. The cervix gets dilated and the vagina also shows similar dilation. This stage of labour can normally last up to few hours. It ends in the rupturing of amniotic sac of the foetus.

2. Expulsion stage:

During this stage, the uterine and abdominal contractions become stronger. In normal delivery, the fetus passes out through the cervix and vagina with head in the forward direction. It takes around 20 to 60 minutes. The umbilical cord is tied and cut off close to the baby’s navel.

3. After birth:

After the delivery of the baby, the placenta separates from the uterus and is expelled out as “after birth”, due to severe contractions of the uterus. This process happens within 10 to 45 minutes of delivery.

3.Explain the histological structure of testis.

SOLUTION

Histology of Testis:

1. Externally, the testis is covered by three layers. These are:

a. Tunica vaginalis: It is the outermost incomplete peritoneal covering made up of connective tissue and epithelium.

b. Tunica albuginea: It is the middle layer formed by collagenous connective tissue.

c. Tunica vasculosa/vascularis: It is the innermost layers. It is a thin and membranous layer.

2. Each testis is divided into about 200-300 testicular lobules by fibres from tunica albuginea. Each lobule has 1 to 4 highly coiled seminiferous tubules.

3. Each seminiferous tubule is internally lined by a single layer of cuboidal germinal epithelial cells (spermatogonia) and few large pyramidal cells called Sertoli or sustentacular cells.

4. The germinal epithelial cells undergo gametogenesis to form spermatozoa.

5. Sertoli cells provide nutrition to the developing sperms.

6. Various stages of spermatogenesis can be seen in the seminiferous tubules. The innermost spermatogonial cell (2n), primary spermatocyte (2n), secondary spermatocyte (n), spermatids (n) and sperms (n).

7. Between seminiferous tubules, few groups of interstitial cells (Cells of Leydig) are present

8. After puberty, interstitial cells produce a type of androgen i.e. testosterone.

4.Describe the structure of blastula.

SOLUTION

Blastulation is the process of formation of the hollow and multicellular blastocyst. The process of blastulation can be summarized as follows:

  1. The embryo (blastocyst) that enters the uterus remains floating in uterine cavity for 2-4 days after its entry i.e. till the end of 7th day after fertilization.
  2. The outer layer of cells seen in the morula now forms the layer called the trophoblast.
  3. Cells from the trophoblast begin to absorb the glycogen rich uterine milk
  4. The blastocyst doubles in size from 0.15 mm to 0.30 mm.
  5. With more fluid entering inside the blastocyst cavity is formed.
  6. These outer cells become flat and are called trophoblast cells (since they help only in absorbing nutrition for the developing embryo).
  7. The larger inner cells form inner cell mass or embryoblast (the embryo proper develops from the embryoblasts).
  8. These remain attached to the trophoblasts on only one side.
  9. The trophoblast cells in contact with the embryonal knob are called cells of Rauber.
  10. At this stage, the blastocyst shows polarity i.e. the side with inner cell mass is called the embryonal end and the side opposite to it is the abembryonic end.
  11. By the end of the 7th day the blastocyst is fully formed and ready for implantation and gastrulation.
  12. The function of zona pellucida is to prevent the implantation of the embryo at an abnormal site. It does not expose the sticky and phagocytic trophoblast cells till it reaches the implantation site i.e. within the uterus, after which the zona pellucida ruptures.

5.Explain the histological structure of ovary in human.

SOLUTION

Histological structure of ovary:

Each ovary is a compact structure differentiated into a central part called medulla and the outer part called the cortex. The cortex is covered externally by a layer of germinal epithelium. The stroma of loose connective tissue of the medulla has blood vessels, lymph vessels, and nerve fibers.

The outer cortex is more compact and granular.

It shows large number of tiny masses of cells called ovarian follicles. These are collectively formed from the immature ova originating from cells of the dorsal endoderm of the yolk sac. The cells migrate to the gonadal ridge during embryonic development and divide mitotically.

Now these cells are called oogonia.

As the oogonia continue to grow in size they are surrounded by a layer of granulosa cells. This assembly forms the rudiments of the ovarian follicles. The process of oogenesis starts much before the birth of the female baby and by the end of twelve weeks the ovary is fully formed. The ovary has more than two million primordial follicles in it. The cells of the germinal epithelium give rise to groups of oogonia projecting into the cortex in the form of cords called egg tubes of Pfluger. Each cord at its end has a round mass of oogonial cells called egg nests, from which the primordial ovarian follicles develop. Each primordial follicle has, at its center a large primary oocyte (2n) surrounded by a single layer of flat follicular cells. The primary oocyte starts with its meiotic division but gets arrested it at meiosis I.

Of the two million primordial follicles embedded in the fetal ovary only about one million remains at birth and only about 40,000 remain at the time of puberty.

T.S. of Ovary-

The histological structure of the ovary shows the different stages of development of the oocyte in the ovary. These changes are cyclic and occur during each menstrual cycle. This development involves maturation of the primordial follicles into primary, secondary and Graafian follicles.

Each primary follicle has multi-layered cuboidal follicular cells. The stroma cells add theca over the follicle, which then changes into a secondary follicle.

There is the growth of the oocyte and the granulosa cells increase in number. They start producing the hormone estrogen.

The secondary follicle grows into the Graafian follicle by the addition of more follicular cells.

As this process of maturation of follicles takes place, they begin to move towards the surface of the ovary. The Graafian follicle presses against the thin wall of the ovary giving it a blistered appearance.

The egg is released from the Graafian follicle during ovulation and the remaining part of the follicle changes into a temporary endocrine gland called corpus luteum.

If fertilization does not take place the corpus luteum degenerates into a white scar called corpus albicans.

6.Describe the various methods of birth control to avoid pregnancy.

SOLUTION

Contraceptive methods are of two main types i.e. temporary and permanent.

1. Temporary methods:

i. Natural method/ Safe period / Rhythm method:

In the natural method, the principle of avoiding chances of fertilization is used. A week before and a week after menstrual bleeding is considered a safe period for sexual intercourse.

This method is based on the fact that ovulation occurs on the 14th day of the menstrual cycle. Drawback: High rate of failure.

ii. Coitus Interruptus or withdrawal:

In this method, the male partner withdraws his penis from the vagina just before ejaculation, so as to avoid insemination.

Drawback: Pre-ejaculation fluid may contain sperms and this can cause fertilization.

iii. Lactational amenorrhea (absence of menstruation):

This method is based on the fact that ovulation does not occur during the period of intense lactation following parturition. Therefore, as long as the mother breastfeeds the child fully, chances of conception are almost negligible.

Drawbacks: High chances of failure.

iv. Chemical means (spermicides):

In this method, chemicals like foam, tablets, jellies, and creams are used by the female partner. Before sexual intercourse, if these chemicals are introduced into the vagina, they adhere to the mucous membrane, immobilize and kill the sperms.

Drawback: It may cause allergic reaction. This method also has chances of failure.

v. Mechanical means / Barrier methods:

In this method, the ovum and sperm are prevented from physically meeting with the help of barriers.

These mechanical barriers are of three types.:

1. Condom:

It is a thin rubber sheath that is used to cover the penis of the male during copulation. It prevents the entry of ejaculated semen into the female reproductive tract. It can thus prevent conception. It is a simple and effective method and has no side effects. Condoms should be properly discarded after every use. A condom is also a safeguard against STDs and AIDS.

e.g.“Nirodh” is the most widely used contraceptive by males. It is easily available and is given free by the government.

2. Diaphragm, cervical caps and vaults:

These devices used by the female are made up of rubber. They prevent conception by blocking the entry of sperms through the cervix. The device is inserted into the female reproductive tract to cover the cervix during copulation.

3. Intra-uterine devices (IUDs):

These clinical devices are plastic or metal objects. A doctor or trained nurse places the IUDs into the uterus. These devices include the Lippes loop, copper releasing IUDs (Cu-T, Cu7, multiload 375), and hormone-releasing IUDs (LNG-20, progestasert).

  • Lippes loop:
    It is a plastic double “s” loop. It attracts the macrophages stimulating them to accumulate in the uterine cavity. Macrophages increase phagocytosis of sperms within the uterus and act as a contraceptive.
  • Copper releasing IUDs:
    Suppress sperm motility and the fertilizing capacity of sperms.
  • Hormone releasing IUDs:
    Make the uterus unsuitable for implantation and cervix hostile to the sperms. It delays pregnancy for a longer period.
    Drawbacks: Spontaneous expulsion, occasional haemorrhage and chances of infection are the drawbacks of IUDs.

vi. Physiological (Oral) Devices:

Physiological devices are used in the form of tablets/ pills. It is an oral contraceptive, used by the female which contains progesterone and estrogen. These hormones inhibit ovulation; hence no eggs are released from the ovary of the female using this pill and thus conception cannot occur.

They also alter the quality of cervical mucus to prevent the entry of sperms. The pill “Saheli” is an oral contraceptive for females which is non-steroidal. Saheli is to be taken once in a week. These pills are sponsored by the Government. Saheli is now a part of the National Family Programme as an oral contraceptive pill in India.

Drawback: Oral contraceptive pills have side effects such as nausea, weight gain, tenderness of breast, and slight blood loss between menstrual periods.

vii. Other contraceptives:

The birth control implant is a contraceptive used by the female. e.g. implanon, explanon, etc. It is a tiny, thin rod about the size of a matchstick. It is implanted under the skin of the upper arm and contains progesterone and estrogen. Their mode of action is similar to that of pills. They prevent pregnancy for 3-4 years.

2. Permanent Methods:

The permanent birth control method in men is called vasectomy and in women it is called tubectomy. These are surgical methods, also called sterilization. In vasectomy a small part of the vas deferens is tied and cut. In tubectomy, a small part of the fallopian tube is tied and cut. This blocks gamete transport and prevent pregnancy.

7.What are the goals of RCH programme.

SOLUTION

The goals of the Reproductive and Child Healthcare (RCH) programme are as follows:

i. To create awareness among people about various aspects related to reproduction.

ii. To provide facilities to people in order to understand and build up reproductive health.

iii. To provide support for building up a reproductively healthy society.

iv. To bring about a change mainly in three critical health indicators i.e. reducing total infertility rate, infant mortality rate, and maternal mortality rate.

8.Which hormones are involved in parturition?

SOLUTION

Parturition is controlled by a complex neuroendocrine mechanism.

1. Signals arise from the fully formed foetus and placenta cause mild uterine contractions.

2. This is accompanied by rise in estrogen- progesterone ratio, increase in oxytocin receptors in uterine muscles.

3. Increase in hormone ratio causes vigorous contractions of myometrium of uterus at the end of pregnancy.

4. The fully developed foetus gives signals for the uterine contractions by secreting Adrenocorticotropic Hormone (ACTH) from pituitary and corticosteroid from adrenal gland.

5. This, in turn, triggers the release of oxytocin from mother’s pituitary gland, which acts on the uterine muscles of the mother and causes vigorous uterine contractions leading to the expulsion of the baby from the uterus.

9.Which is the function of male accessory glands?

SOLUTION

Male accessory glands secrete substances that protect the gametes and facilitate their movement.

1. Seminal vesicles:

These are a pair of small fibromuscular pouches present on the posterior side of the urinary bladder. They secrete a seminal fluid (alkaline) containing citric acid, fructose, fibrinogen and prostaglandins. About 60% of the total volume of semen is made up of seminal fluid.

Fructose provides energy to sperms for swimming, while fibrinogen helps in coagulation of semen after ejaculation for quick propulsion into the vagina. The prostaglandins stimulate reverse peristalsis in the vagina and uterus aiding the faster movement of sperms towards the egg in the female body.

2. Prostate gland:

The prostate gland consists of 20 to 30 lobes and is located under the urinary bladder. It surrounds the urethra. It releases a milky white, alkaline fluid called prostatic fluid into the urethra. Prostatic fluid forms about 30% of the total volume of semen. It contains citric acid, acid phosphatase, and various other enzymes.

The acid phosphatase protects the sperm from the acidic environment of the vagina.

3. Cowper’s glands:

Cowper’s glands are also known as bulbourethral glands. These are pea-sized and lie on either side of membranous urethra. They secrete a viscous, alkaline, mucous like fluid which acts as a lubricant during copulation.

Semen: It is the viscous, alkaline and milky fluid (pH 7.2 to 7.7) ejaculated by the male reproductive system.

Generally, 2.5 to 4.0 ml of semen is given out during a single ejaculation and it contains about 400 million sperms.

Semen contains secretion of the epididymis and the accessory glands for nourishing (fructose), neutralizing acidity (Ca++, bicarbonates), activation for movement (prostaglandins).

10.What is capacitation? Give its importance.

SOLUTION

Capacitation generally requires 5-6 hours during which the acrosome membrane becomes thin, Ca++ enters the sperm, and sperm tails begin to show rapid whiplash movements.

As a result of capacitation, sperms become extra active and begin to start moving upwards from the vagina to the uterus and to the oviducts. Prostaglandins activate the sperms. The vestibular secretions of the female also enhance sperm’s motility. The sperms swim at an average speed of 1.5 to 3.0 mm/min and reach the ampulla. The contraction of uterus and fallopian tubes stimulated by oxytocin of females also aids in the movement of sperm. After capacitation the sperms may reach the ampulla within 5 minutes.

Answer the following question.

1.Explain the following parts of male reproductive system along with labelled diagram showing these parts- Testis, vasa deferentia, epididymis, seminal vesicle, prostate gland and penis.

SOLUTION

asterclasses

The male reproductive system consists of the following:

i. Primary sex organs (Gonads): Testes

ii. Accessory organs: Accessory ducts and accessory glands

iii. External genitalia: Scrotum and Penis

i. Primary sex organs: Testes

Testes are a pair of primary sex organs which are mesodermal in origin. They are located outside the abdomen in a pouch called scrotum, i.e. extra-abdominal in position. Testes develop in the abdominal cavity (early foetal life) and later descend into the scrotal sac through a passage called inguinal canal. They are suspended in the scrotal sac by the spermatic cord. Testes are connected to the wall of scrotum by a short fibromuscular band called gubernaculum.

They are oval in shape, about 4 to 5 cm long, 2 to 3 cm wide, and 3 cm thick.

The outermost covering of the testis is formed of a dense fibrous membrane called tunica albuginea.

ii. Accessory sex organs:

It includes accessory ducts, accessory glands.

a. Accessory ducts:

1. Rete testis:

The seminiferous tubules of the testis at the posterior surface form a network of tubules called rete testis. The rete testis opens into vasa efferentia. 2. Vasa efferentia:

Vasa efferentia are 12-20 fine tubules arising from the rete testis and join to the epididymis. They carry sperms from the testis and open into the epididymis.

3. Epididymis:

It is a long and highly coiled tube which is differentiated into an upper caput-, middle corpus- and lower cauda epididymis. The sperms undergo maturation in the epididymis.

4. Vasa deferens:

The vas deferens travels up to the abdominal cavity and loops over the ureter to open into the urethra. Vas deferens join the seminal vesicle to form ejaculatory duct.

5. Ejaculatory ducts:

The ejaculatory duct passes through the prostate gland and opens into the urethra.

6. Urethra:

The urethra provides a common passage for the urine and semen and hence is also called urinogenital duct. In males the urethra is long and extends through the penis. It opens to the outside by an opening called the urethral meatus or urethral orifice.

b. Accessory glands:

1. Seminal vesicles:

These are a pair of small fibromuscular pouches present on the posterior side of the urinary bladder. They secrete a seminal fluid (alkaline) containing citric acid, fructose, fibrinogen, and prostaglandins. About 60% of the total volume of semen is made up of seminal fluid. Fructose provides energy to sperms for swimming, while fibrinogen helps in coagulation of semen after ejaculation for quick propulsion into the vagina. The prostaglandins stimulate reverse peristalsis in vagina and uterus aiding the faster movement of sperms towards the egg in the female body.

2. Prostate gland:

Prostate gland consists of 20 to 30 lobes and is located under the urinary bladder. It surrounds the urethra. It releases a milky white, alkaline fluid called prostatic fluid into the urethra. Prostatic fluid forms about 30% of the total volume of semen. It contains citric acid, acid phosphatase and various other enzymes.

The acid phosphatase protects the sperm from the acidic environment of the vagina.

3. Cowper’s glands:

Cowper’s glands are also known as bulbourethral glands. These are pea-sized and lie on either side of membranous urethra. They secrete a viscous, alkaline, mucous like fluid which acts as a lubricant during copulation.

Semen: It is the viscous, alkaline and milky fluid (pH 7.2 to 7.7) ejaculated by the male reproductive system. Generally, 2.5 to 4.0 ml of semen is given out during a single ejaculation and it contains about 400 million sperms. Semen contains secretion of the epididymis and the accessory glands for nourishing (fructose), neutralizing acidity (Ca++, bicarbonates), activation for movement (prostaglandins).

iii. External genitalia:

a. Penis:

The penis is the male copulatory organ. It is cylindrical and muscular with three bundles of erectile tissue: a pair of postero-lateral tissue called corpora cavernosa and a median corpus spongiosum. The swollen tip of the penis is called glans penis. It is covered by a loose fold of skin called foreskin or prepuce.

b. Scrotum:

It is a loose pouch of pigmented skin lying behind the penis and is divided into a right and left scrotal sac by a septum of tunica dartos made of smooth muscle fibres. The foetal testes are guided into and retained in the scrotum by a short fibro muscular band called gubernaculum. The testes remain suspended in scrotum by a spermatic chord.

The failure of the testis to descend into the scrotum is called cryptorchidism which also results in sterility.

The cremaster and dartos muscles of scrotum help in drawing testes close or away from the body. This helps in maintaining the temperature of the testis 2-3 0C lower than the normal body temperature, necessary for spermatogenesis.

Long answer question.

2.Describe female reproductive system of human.

SOLUTION

The human female reproductive system consists of:

i. Internal genitalia:

It includes ovaries, oviducts, uterus, vagina.

Asterclasses

a. Ovary:

It is the primary female sex organ. It is a solid, oval, or almond-shaped organ. It is 3 cm in length, 1.5 cm in breadth, and 1 cm thick. It is located in the upper lateral part of the pelvis near the kidneys. Each ovary is held in position by ligaments by attaching it to the uterus and the abdominal wall. The largest of these is the broad ligament formed by a fold of peritoneum. It holds the ovary, oviduct and the uterus to the dorsal body wall. The ovarian ligament attaches the ovary to the uterus.

Functions:

Its main function is production of egg or ovum and the female reproductive hormones.

The ovary produces five hormones viz. estrogen, progesterone, relaxin, activin and inhibin.

b. Oviduct / Fallopian tube / Uterine tube:

These are a pair of muscular ducts lying horizontally over the peritoneal cavity. The proximal part of the tube lies close to the ovary and distally it opens into the uterus. Each tube is 10 to 12 cm in length. It is internally lined by ciliated epithelium. It can be divided into three regions:

1. Infundibulum:

The proximal funnel like the part with an opening called ostium surrounded by many finger-like processes called fimbriae (of these at least one is long and connected to the ovary). The cilia and the movement of fimbriae help in transporting the ovulated egg to the ostium.

2. Ampulla:

It is the middle, long and straight part of the oviduct. Fertilization of the ovum takes place in this region.

3. Isthmus / Cornua:

The distal narrow part of the duct opening into the uterus.

Functions:

Fallopian tubes carry the released egg from the ovary to the uterus. Ampulla provides the site for fertilization of the ovum.

c. Uterus:

It is commonly also called as the womb.

It is a hollow, muscular, pear-shaped organ, located above and behind the urinary bladder. It is about 7.5 cm long, 5 cm broad and 2.5 cm thick. Internally the uterine wall can be distinguished into three layers: The outermost perimetrium, middle thick muscular myometrium, made up of thick layer of smooth muscles. Vigorous contractions of these muscles cause labour during parturition (childbirth). The innermost layer called endometrium or mucosal membrane is made up of stratified epithelium. The thickness of this layer regularly undergoes changes during the menstrual cycle. It is richly supplied with blood vessels and uterine glands. These provide nourishment to the developing foetus.

The uterus can be divided into three regions:

1. Fundus: It is the upper dome shaped part. Normally implantation of the embryo occurs in the fundus.

2. Body: It is the broad part of the uterus which gradually tapers downwards.

3. Cervix: It is the narrow neck about 2.5 cm in length. It extends into the vagina. Its passage has two openings: an internal os/ orifice towards the body, and an external os/ orifice towards the vagina.

Functions:

Uterus receives the ovum. It provides site for implantation, gestation and parturition. It forms placenta for the development of foetus.

d. Vagina:

It is a tubular, female copulatory organ, 7 to 9 cm in length.

It lies between the cervix and the vestibule.

The vaginal wall has an inner mucosal lining, the middle muscular layer and an outer adventitia layer.

The mucosal epithelium is stratified and non-keratinized and stores glycogen.

There are no glands but the cervical secretion of mucus is received in the vagina.

The opening of the vagina into the vestibule is called vaginal orifice. The vaginal orifice is partially covered by the hymen.

Functions:

The vagina acts as a passage for menstrual flow as well as a birth canal during parturition.

ii. External genitalia (Vulva):

The external genital organs of female include parts external to the vagina, collectively called vulva (covering or wrapping), or pudendum. They include the following parts:

a. Vestibule:

It is a median vertical depression of vulva enclosing the urethral and vaginal opening.

b. Labia minora:

These are another pair of thin folds inner to the labia majora with which they merge posteriorly to form the fourchette (frenulum), while towards anterior end they converge into a hood-like covering around the clitoris.

c. Clitoris:

A small conical and sensitive projection lying at the anterior end of labia minora. It has a pair of erectile tissue i.e. corpora cavernosa which is homologous to the penis.

d. Labia majora:

These are a pair of fleshy folds of skin forming the boundary of the vulva. They are homologous to the scrotum. They surround and protect the other parts of the external genitalia and enclose the urethral and vaginal openings in the vestibule.

e. Mons pubis:

It is a fleshy elevation above the labia majora. The Mons pubis and outer part of labia majora show pubic hair.

iii. Accessory glands:

a. Vestibular glands / Bartholin’s glands:

It is a pair of glands homologous to the Bulbourethral or Cowper’s glands of the male. They open into the vestibule and release a lubricating fluid.

b. Mammary glands:

These are accessory organs of the female reproductive system for production and release of milk after parturition. The development of the mammary glands occur at puberty under the influence of estrogen and progesterone. Lactotropic hormone (LTH) or prolactin helps in the development of lactiferous tubules during pregnancy. The mammary glands are a pair of rounded structures present in the subcutaneous tissue of the anterior thorax in the pectoral region (from 2nd to 6th rib). These are modified sweat glands. Each mammary gland contains fatty connective tissue and numerous lactiferous ducts. The glandular tissue of each breast is divided into 15-20 irregularly shaped mammary lobes, each with alveolar glands and lactiferous duct.

Alveolar glands secrete milk which is stored in the lumen of alveoli. The alveoli open into mammary tubules. The tubules of each lobe join to form a mammary duct.

Many mammary ducts join to form a wider mammary ampulla, which is connected to lactiferous duct. These converge towards the nipple located near the tip of the breast.

It is surrounded by a dark brown coloured and circular area of the skin called areola.

3.Describe the process of fertilization.

SOLUTION

Fertilization is the process which involves the fusion of the haploid male and female gametes resulting in the formation of a diploid zygote (2n). The process of fertilization is internal and it usually takes place in the ampulla of the fallopian/uterine tube. The fertilized egg or zygote further develops into an embryo within the uterus.

The mechanism of fertilization is as follows:

i. Movement of sperm towards egg:

The ejaculated semen is made up of sperms and some other secretions. This coagulated semen undergoes liquefication and sperms become active. Once the sperms reach the vagina around 50% sperms are demobilized/broken/destroyed and the remaining sperms undergo capacitation.

Capacitation:

Capacitation generally requires 5-6 hours during which the acrosome membrane becomes thin, Ca++ enters the sperm, and sperm tails begin to show rapid whiplash movements. As a result of capacitation, sperms become extra active and begin to start moving upwards from the vagina to the uterus and to the oviducts. Prostaglandins activate the sperms. The vestibular secretions of the female also enhance sperms’ motility. The sperms swim at an average speed of 1.5 to 3.0 mm/min and reach the ampulla. The contraction of the uterus and fallopian tubes stimulated by oxytocin of females also aids in the movement of sperm. After capacitation, the sperms may reach the ampulla within 5 minutes.

ii. Entry of sperm into the egg:

Out of 200 to 400 million sperms, only few hundred manage to reach the ampulla, out of which only a single sperm fertilizes the ovum. After the sperm reaches the egg/ovum, its acrosome releases lysins: hyaluronidase and corona penetrating enzymes. These enzymes separate and dissolve the cells of corona radiata, so that the sperm head can pass through the zona pellucida of the egg. The zona pellucida has fertilizin receptor proteins (ZP3, ZP2). The fertilizin binds to specific acid protein- anti-fertilizin of sperm and brings about the attraction of sperms to the egg to enhance fertilization.

Acrosome reaction:

As the sperm head touches the zona pellucida in the animal pole region, its acrosome covering ruptures to release lytic enzymes, acrosin or zona lysin which act on the zona pellucida at the point of contact. This causes egg reaction during which a small fertilization cone/cone of reception is formed on the egg membrane. The sperm head comes in contact with this cone. It results in the production of a weak wave of depolarization. The plasma membrane of both cells dissolves at the point of contact.

The sperm nucleus and the centrioles enter the egg, while other parts remain outside.

As soon as the sperm head touches the vitelline membrane, a cortical reaction gets activated changing the vitelline membrane into a fertilization membrane by deactivating the sperm receptors of zona pellucida.

A distinct perivitelline space is created around the fertilization membrane.

This prevents any further entry of other sperms into the egg i.e. polyspermy is avoided.

iii. Activation of ovum:

The ovum before fertilization was at metaphase II stage. After the contact of sperm head to the vitelline membrane of egg, it gets activated to resume and complete meiosis II. After meiosis II, the second polar body is formed. The germinal vesicle organizes into female pronucleus also known as the true ovum or egg.

The fusion of egg and sperm:

The coverings of male and female pronuclei degenerate, allowing the chromosomal pairing. This results in the formation of a synkaryon by the process called syngamy or karyogamy. The zygote is thus formed. The proximal centriole received from the sperm helps in the formation of the synkaryon spindle and cleavage of the cell into two blastomeres.

The zygote is thus formed.

The proximal centriole received from the sperm helps in the formation of the synkaryon spindle and cleavage of the cell into two blastomeres.

4.Explain the process by which zygote divides and re-divides to form the morula.

SOLUTION

Asterclasses

The zygote formed as a result of syngamy is activated to divide.

1. Cleavage:

Cleavage is the process of early mitotic division of the zygote into a hollow multicellular blastula. It does not involve the growth of the daughter cells. The cells formed by cleavage are called blastomeres. Since there is no growth phase between the cleavages, the size of blastomeres will be reduced with every successive cleavage.

As the size reduces, the metabolic rate increases. Subsequent cleavages are thus faster than earlier ones. This requires rapid replication of DNA and high consumption of oxygen.

2. Process of cleavage:

In human, cleavage is holoblastic i.e. the whole zygote gets divided. The cleavage planes may be longitudinal or meridional and equatorial or horizontal. It is radial and indeterminate i.e. fate of each blastomere is not predetermined.

The 1st cleavage in the zygote is meridional and occurs at about 30 hours after fertilization.

It divides longitudinally into two blastomeres, one slightly larger than the other.

The 2nd cleavage is also longitudinal but at the right angle to the 1st one and occurs after 30 hours of 1st cleavage.

The 3rd cleavage is horizontal. After 3rd cleavage, the embryo is in the 8-cell stage.

While the cleavages occur, the young embryo is gradually being pushed towards the uterus.

By the end of 4th day after fertilization, the embryo is a solid ball of 16-32 cells and externally looking like mulberry. This stage is thus called a morula.

3. Morula:

The morula shows cells of two types:

a. smaller, clearer cells towards the outer side

b. inner cell mass of larger cells.

Cells are compactly arranged. Till the formation of morula, the zona pellucida is retained around the embryo and thus, there is no change in the overall size from zygote to morula. The morula reaches the isthmus and gains entry into the uterus by the end of day 4.


COMPLETED

Chapter 1, Reproduction in Lower and High plants, hsc, science, biology, maharashtra board, new edition, balbharathi solution,

Multiple choice question.

1.Insect pollinated flowers usually posses ______

OPTIONS

  • Sticky pollens with rough surface
  • Large quantities of pollens
  • Dry pollens with smooth surface
  • Light coloured pollens
2.In ovule, meiosis occurs in ______

OPTIONS

  • integument
  • nucellus
  • megaspore
  • megaspore mother cell
3.The ploidy level is not the same in ______.

OPTIONS

  • Integuments and nucellus
  • Root tip and shoot tip
  • Secondary nucleus and endosperm
  • Antipodals and synergids

4.Which of the following types require pollinator but the result is genetically similar to autogamy?

OPTIONS

  • Geitonogamy
  • Xenogamy
  • Apogamy
  • Cleistogamy

5.If diploid chromosome number in a flowering plant is 12, then which one of the following will have 6 chromosomes?

OPTIONS

  • Endosperm
  • Leaf cells
  • Cotyledons
  • Synergids

6.In angiosperms, endosperm is formed by/ due to ______

OPTIONS

  • free nuclear divisions of megaspore
  • polar nuclei
  • polar nuclei and male gamete
  • synergids and male gamete

Point out the odd one.

OPTIONS

  • Nucellus
  • Embryo sac
  • Micropyle
  • Pollen grain

Very short answer type question:

1.Name the part of gynoecium that determines the compatible nature of pollen grain.

SOLUTION

Pistil determines the compatible nature of pollen grain.

2.How many haploid cells are present in a mature embryo sac?

SOLUTION

Total 6 haploid cells are present in a mature embryo sac. They are antipodal cells (3), synergids (2), and egg cell (1).

3.Even though each pollen grain has 2 male gametes, why at least 20 pollen grains are required to fertilize 20 ovules in a particular carpel?

SOLUTION

During double fertilization, one of the male gamete of pollen grain fuses with egg cell, while other male gamete fuses with secondary nucleus. Thus to fertilize 20 ovules in a particular carpel, 20 pollen grains are required.

4.Define megasporogenesis.

SOLUTION

It is the process of formation of haploid megaspores from diploid megaspore mother cell (MMC) by meiotic division.

5.What is hydrophily?

SOLUTION

Pollination carried out by water is called hydrophily.

6.Name the layer which supplies nourishment to the developing pollen grains.

SOLUTION

Tapetum supplies nourishment to the developing pollen grains.

7.Define Parthenocarpy.

SOLUTION

It is the condition in which fruit is developed without the process of fertilization is called parthenocarpy.

8.Are pollination and fertilization necessary in apomixis?

SOLUTION

In apomixis, the embryo is formed without the formation of gametes and fertilization.

Thus, pollination and fertilization are not necessary for apomixis.

9.Name the parts of pistil which develop into fruits and seeds.

SOLUTION

After fertilization, the ovary of the pistil develops into fruit and ovules into seeds.

10.What is the function of filiform apparatus?

SOLUTION

Filiform apparatus guide the entry of pollen tube towards the egg.


Short Answer Question:

1.How polyembryony can be commercially exploited?

SOLUTION

  1. Polyembryony increases the chances of survival of the new plants.
  2. Genetically uniform parental type seedlings are obtained from nucellar embryos, thus nucellar adventive polyembryony is of great significance in horticulture.
  3. Plantlets obtained from these embryos are disease-free.
  4. These embryos can be isolated and grown on embryo culture to produce clones.

2.Pollination and seeds formation is very crucial for fruit formation. Justify the statement.

SOLUTION

  1. Pollination is a very important part of the life cycle of a flowering plant.
  2. The flowers must be pollinated in order to bring about the process of fertilization.
  3. Pollination brings male and female gametes of a flower together during fertilization.
  4. As a result of fertilization, ovary develops into fruits and ovules into seeds.
  5. Seeds on germination give rise to a new plant that further grows and develops fruits and seeds. Thus pollination and seed formation are required to create offsprings for the next generation.

3.Incompatibility is a natural barrier in the fusion of gametes. How will you explain this statement?

SOLUTION

  1. Incompatibility refers to inability of certain gametes even from genetically similar plant species to fuse with each other.
  2. It is considered as the most prevalent and effective device to avoid inbreeding and outbreeding.
  3. Pollen pistil interaction is a dynamic process that involves pollen recognition followed by promotion or inhibition of the pollen. 
  4. Chemical substances released by the style act as a barrier.
  5. Typically the pollen belonging to the correct mating type germinates on stigma, develops a pollen tube, and brings about fertilization.
  6. The pollens belonging to the other mating type are discarded.

Thus, incompatibility is a natural barrier in the fusion of gametes.

4.Describe three devices by which cross-pollination is encouraged in angiosperms by avoiding self- pollination.

SOLUTION

Genetic diversity is an essential factor for evolution by natural selection. Continued self-pollination results in inbreeding depression. Thus, plants have developed many devices to encourage cross-pollination.

Examples of outbreeding devices are as follows:

  1. Unisexuality:
    In this, the plant bears either male or female flowers. It is also called as dioecism.
    As flowers are unisexual, self-pollination is not possible. Plants may be monoecious,
    e.g. Maize or dioecious, e.g. Mulberry, Papaya.
  2. Dichogamy:
    In this, anthers and stigmas mature at different times in a bisexual flower due to which self-pollination is prevented. It can be further divided into two types:
    a. Protandry:
    In this type, anthers mature first, but the stigma of the same flower is not receptive at that time.
    e.g. in the disc florets of sunflower.
    b. Protogyny:
    In this type, stigma of carpel matures earlier than anthers of the same flower. e.g. Gloriosa.
  3. Prepotency:
    In this, pollen grains of other flowers germinate rapidly over the stigma than the pollen grains from the same flower, e.g. Apple.
  4. Heterostyly (heteromorphy):
    Plants like Primula (Primrose) produce two or three types of flowers in which stigmas and anthers are placed at different levels (heterostyly and heteroanthy).
    This prevents the pollens from reaching the stigma and pollinating it. In heteromorphic flowers, pollen grains produced from anther pollinate stigmas produced at the same level.
    Thus self-pollination is not possible in such cases.
  5. Herkogamy:
    It is a mechanical device to prevent self-pollination in a bisexual flower. In plants, a natural physical barrier is present between two sex organs and avoid contact of pollen with the stigma of the same flower, in e.g. Calotropis, pentangular stigma is positioned above the level of anthers (pollinia).
  6. Self-incompatibility (self-sterility):
    This is a genetic mechanism due to which the germination of pollen on the stigma of the same flower is inhibited, e.g. Tobacco, Thea.

Long Answer Question:

1.Describe the process of double fertilization.

SOLUTION

Double fertilization:

Asterclasses.com
  1. The fusion of one male gamete with an egg and that of another male gamete with a secondary nucleus is called as double fertilization.
    It is the characteristic feature of angiosperms.
    It was discovered by Nawaschin in the liliaceous plants like Lilium and Fritillaria.
  2. When pollen grain reaches the surface of the stigma, it germinates and forms a pollen tube.
  3. Pollen tube penetrates the stigma, style, ovary chamber and then enters the ovule.
  4. The growth of the pollen tube is guided by the chemicals secreted by the synergids.
  5. Usually, when a pollen tube enters the ovule through the micropyle, it is termed as porogamy.
    But in some cases, it enters through chalaza which is known as chalazogamy. In some plants, it enters by piercing the integuments which are called mesogamy.
  6. A pollen tube penetrates the embryo sac of ovule through its micropylar end.
  7. The pollen tube carrying male gametes penetrates in one of the synergids.
  8. Watery contents of synergid are absorbed by the pollen tube, due to which it ruptures and releases the contents, including the two non-motile male gametes.
  9. As non-motile male gametes are carried through a hollow pollen tube, it is known as siphonogamy that ensures fertilization to take place.
  10. Fertilization mainly involves two processes: Syngamy and Triple fusion.
    a. Syngamy:
    It is the fusion of haploid male gamete with a haploid female gamete (egg). It results in the formation of a diploid zygote which develops to form an embryo. Syngamy is a type of generative fertilization.
    b. Triple fusion:
    It is the fusion of second haploid male gamete with diploid secondary nucleus. It results in the formation of Primary Endosperm Nucleus (PEN) which develops into triploid endosperm. Triple fusion is a type of vegetative fertilization.
  11. In this process, both the male gametes participate, due to which fertilization occurs twice in the same embryo sac, hence it is described as double fertilization.

2.Explain the stages involved in the maturation of microspore into a male gametophyte.

SOLUTION

Asterclasses

Development of male gametophyte

  1. Pollen grain/microspore marks the beginning of male gametophyte, thus it is the first cell of the male gametophyte.
  2. It undergoes the first mitotic division to produce bigger, naked vegetative cells and small, thin-walled generative cells.
  3. The vegetative cell is rich in food and having an irregularly shaped nucleus.
  4. The generative cell floats in the cytoplasm of the vegetative cells.
  5. The second mitotic division is concerned with generative cells only and gives rise to two non-motile male gametes.
  6. The mitotic division of the generative cells takes place either in the pollen grain or in the pollen tube.
  7. The pollen grains are shed from the anther, at this two-celled stage in most of the angiosperms.

3.Explain the development of dicot embryo.

SOLUTION

A: Oospore.

B: Two celled proembryo.

e: embryonal initial;

t: suspensor initial;

m: Embryo sac membrane.

B1 : 4-celled I-shaped proembryo;

e1, e2: embryonal initial; s1, s2 : suspensor initial.

C: Further development of embryo.

S: Suspensor, h: Hypophysis; E: Embryonal mass

D: L. S. of ovule

Endo: Endosperm in free nuclear stage.

Anti: Antipodal tissue.

Embryo: Developing embryo

E: Embryo showing further development of embryonic octants and hypophysis.

F: L. S. of ovule. Endosperm becoming cellular.

G: Embryo; Cot: Cotyledons; Hypo: Hypocotyl; Rad: Radicle; R.c.: Rootcap;

H: Mature seed; Pl: Plumule. Endosperm has been consumed almost completely

Development of dicot embryo:

  1. The zygote divides to form two-celled proembryo.
  2. The larger cell towards the micropyle is called basal or suspensor initial cell and smaller cell towards chalaza is called terminal or embryonal initial cell.
  3. The suspensor cell divides transversely in one plane to produce filamentous suspensor of 6-10 cells.
  4. The first cell of the suspensor towards the micropylar end becomes swollen and functions as a haustorium.
  5. The lowermost cell of the suspensor is known as hypophysis.
  6. The suspensor helps in pushing the embryo in the endosperm.
  7. The embryonal initial undergoes three successive mitotic divisions to form octant.
  8. The planes of divisions are at right angles to each other.
  9. The lower tier of four cells of octant gives rise to hypocotyl and radicle whereas four cells of the upper-tier form the plumule and the one or two cotyledons.
  10. The hypophysis by further division gives rise to the part of radicle and root cap.
  11. Subsequently, the cells in the upper tier of the octant divide into several planes so as to become heart-shaped which then forms two lateral cotyledons and a terminal plumule.
  12. Further enlargement of hypocotyl and cotyledons result in a curvature of the embryo and it appears horseshoe-shaped.

4.Draw a labeled diagram of the L.S. of anatropous ovule and list the components of the embryo sac and mention their fate after fertilization.

SOLUTION

1. Structure of anatropous ovule:

Asterclasses

2. List the components of embryo sac and mention their fate after fertilization:

Components of the embryo sacFate after fertilization
OvuleSeed
EggEmbryo
NucellusPerisperm
Secondary nucleusEndosperm
Outer integumentTesta (outer seed coat)
Inner integumentTegmen (inner seed coat)
MicropyleAn opening in the seed (i.e. micropyle)
SynergidsDegenerate
AntipodalsDegenerate

Fill in the blank:

1.The _________ collect the pollen grains.

SOLUTION

The stigma collect the pollen grains.

2.The male whorl, called the ________ produces ________.

SOLUTION

The male whorl, called the androecium produces pollen grains.

3.The pollen grains represent the ________.

SOLUTION

The pollen grains represent the male gametophyte.

4.The ________contains the egg or ovum.

SOLUTION

The embryo sac contains the egg or ovum.

5.________takes place when one male gamete and the egg fuse together. The fertilized egg grows into a seed from which the new plants can grow.

SOLUTION

Syngamy (fertilization) takes place when one male gamete and the egg fuse together. The fertilized egg grows into a seed from which the new plants can grow.

6.The ______ is the base of the flower to which other floral parts are attached.

SOLUTION

The thalamus is the base of the flower to which other floral parts are attached.

7.________is the transfer of pollen grains from the anther of the flower to the stigma of the same or a different flower

SOLUTION

Pollination is the transfer of pollen grains from the anther of the flower to the stigma of the same or a different flower.

8.Once the pollen reaches the stigma, the pollen tube traverses down the ________to the ovary where fertilization occurs.

SOLUTION

Once the pollen reaches the stigma, the pollen tube traverses down the style to the ovary where fertilization occurs.

9.The ______ are coloured to attract the insects that carry the pollen. Some flowers also produce ______ or ______ that attracts insects.

SOLUTION

The petals are coloured to attract the insects that carry the pollen. Some flowers also produce sweet odour or nectar that attracts insects.

10The whorl ________is green that protects the flower until it opens.

SOLUTION

The whorl calyx is green that protects the flower until it opens.


1.Label the parts of seed.

SOLUTION

Match the column.

OPTIONS

  • A – V, B – I, C – II, D – IV
  • A – III, B – IV, C – I, D – V
  • A – IV, B – I, C – V, D – II
  • A – IV, B – V, C – III, D – II

       COMPLETED

Balbharati, solutions, for, Social, Science, History, and, Civics, 10th, Standard, SSC, Maharashtra, State, Board, chapter 2, The Electoral Process, [Latest edition],

Exercise | Q 1.1 | Page 80

Choose the correct option from the given options and complete the sentence. 

The Election Commissioner is appointed by the _______

President 

Prime Minister 

 Speaker of Loksabha 

 Vice President

SOLUTION

President

_________ was appointed as the first Chief Election Commissioner of independent India. 

Dr. Rajendra Prasad 

T.N. Sheshan 

Sukumar Sen 

Neela Satyanarayan

SOLUTION

Sukumar Sen

Constituencies are created by _________ committee of the Election Commission. 

Selection 

Delimitation 

Voting 

Timetable

SOLUTION

Delimitation


State whether the following statement are true or false. Give reason for your answer

The Elections Commission lays down the code of conduct during elections.

SOLUTION

True.

Reason: The Code of conduct is measure adopted by the Election Commission to ensure free and fair elections in India. It explains the rules that are to be followed by the Government, political parties and voters, before elections and during elections. Violation of the Code of Conduct can lead to termination of candidature or even imprisonment.

Under special circumstances the Election Commission holds re-elections in a particular constituency for a second time.

SOLUTION

True.

Under special circumstances the

Election Commission holds re-elections in a particular constituency for a second time. It is done in case of any dispute arising regarding the election for e.g. complaints criminal malpractices like booth capturing and in special circumstances of high NOTA votes.

The state government decides as to when and in how many stages the elections would be held in a particular State.

SOLUTION

False.

The Election Commission decides as to when and in how many stages the elections would be held in a particular State. It is the sole authority on conducting elections in Indian.


Write short note.

Journey from the ballot box to EVM machine

SOLUTION

Since the first General Election in 1951-52, the election process has undergone various changes to improve voter experience. One such change includes introduction of Electronic Voting Machine (EVM) in 1990s. It was a step towards preservation of environment since the machine eliminated the use of paper, it was now easier to vote for disabled people and promised an early declaration of results. The most notable feature of the EVM, however, was the NOTA (None of the above) option which enabled the voters to not vote for any candidate if they were not satisfied with him/her.

Reorganising the constituencies

SOLUTION

Reorganisation of the constituencies is the responsibility of the Delimitation committee of the Election Commission. The constituencies are reorganized from time to time on the basis of population density of an area, as tabulated in the decennial census reports. Uttar Pradesh is the most densely populated state of India, and hence has the largest no. of seats to Loksabha in the general election, each seat representing a constituency. At present there are 543 constituencies which are due for a reorganisation in 2021 on the basis of the next Census report.

Complete the following picture.

SOLUTION

Role of Election Commission: Conduct free and fair elections in the country.

Role of the voters: Cast their votes and participate in the election process

Role of political parties & their candidates: Follow the Code of Conduct.

Answer in brief. 

Explain the functions of the Election Commission

SOLUTION

The functions of the Election Commission are as follows:

1. Prepare voters list

It is responsible to preparing a list of eligible voters and updating existing voter’s list. It has the sole authority to issue voter identity cards.

2. Formulate the timetable and programme of elections.

It is responsible for conducting free and fair elections and decides when to conduct elections and how to conduct elections in every state.

3. Scrutinize candidate applications.

Every candidate, affiliated to a party or standing independent, has to fill an application with the election commission giving information about oneself. The commission then scrutinizes all the applications and allows the eligible candidates to contest.

4. Recognize political parties.

All political parties are required to be recognized by the Election commission. The commission has the right to derecognize a party as well. It is also responsible for allotting election symbols to the political parties.

5. Resolve disputes relating to elections.

The commission is responsible for resolution of any dispute arising regarding elections. It can accordingly declare any candidate disqualified and call for re-election in a constituency.

Write some additional information about post of the Election Commissioner.

SOLUTION

The Election Commissioner is an important office in the Government of India.

1. He is appointed by the President and is responsible for the smooth conduct of the election process. The Election Commissioner are mostly retired IAS officers appointed on an extended term.

2. At first there was just one Election Commissioner, later in 1989 the Commission was enlarged with a Chief Election Commissioner and two Election commissioners.

3. Sukumar Sen was the first Chief Election Commissioner of India and Om Prakash Rawat is the current Chief Election Commissioner.

4. The term of the Election Commissioner is for 6 years, draws salary at par with those of the Judges the Supreme Court of India.

5. The Commissioner can only be removed from office with two-thirds majority in Loksabha and Rajsabha on the grounds of misconduct or incapacity.

Explain the meaning of Code of Conduct

SOLUTION

Some important points on the Code of Conduct are:

1. Code of Conduct is a measure adopted by the Election Commission of India to ensure free and fair elections. It is use to control incidences of malpractices during the elections.

2. It is a set of guidelines for the Government, political parties and candidates to be followed before and during the elections.

3. It concerns rules and regulations with respect to speeches, election manifestos, processions and general conduct.

4. Its objective is to check misuse of power during the elections and curb malpractices e.g. hate speeches, liquor distribution, use of muscle force etc. Violation of the code of conduct can lead to cancellation of candidature.

5. Due to the strict observance of the code in the last few elections, the people have become more confident and aware of their rights and duties as voters during the election process.

Organise a mock poll in the school to understand the process of voting .

SOLUTION

Our school witness election every academic year. This year, as the election process was a part of the study, we decided to strictly observe the election processes. The details of the observation are given below:

1. Formation of school squads (which can be related to Formation of Constituencies):

• There were 4 squads namely blue, red, green and pink. It resembled the constituencies in national and state-level elections.

2. Filling of Nominations:

• The nomination of candidates is an important part of the election process.

• The regulations require that the candidate or the person who suggests his name files the nomination papers with the principal (Returning Officer in general election)

• Criteria were set for the qualification of those who become candidates.

3. Scrutiny of Nominations:

• The principal (Returning Officer) scrutinizes the nomination papers very carefully.

• If the eligibility of the student candidate is dissatisfied, he/she is officially stopped from contesting in the election.

• The candidates could withdraw their nomination papers till the prescribed period.

• The security deposit from candidates done in the general elections was not possible in the case of a school election. This step was omitted here.

4. Election Campaign:

Techniques of the election campaign and the tools employed by the squads and the independent candidates are many:

• Election Manifesto

• Electioneering (Activities and Techniques to Persuade Voters)

• Canvassing

• Meet the candidate event

• Class-to-class campaigning

In general elections, the candidates are assigned the tasks like:

• to address public meetings

• street corner meetings

• door-to-door canvassing

• new slogans are coined to attract the masses

• advertisements are released to the press (the popular daily and weekly newspapers)

• Radio and the Television are used to broadcast the speeches and panel-discussions of leaders of various parties

• electronic media plays the most effective role in creating people’s awareness of programs of the political parties

• Attend many interviews to newspapers and television agencies.

• Wide coverage is being given to all these events at regular intervals

5. Polling Personnel and the Polling:

• The election campaign was stopped 48 hours before the time when the poll concludes on the polling day.

• The vice-principal (Presiding Officer) manages the whole of the polling process and guarantee that all persons working under him adhere to the electoral norms and practices.

• The voter records his vote by placing the seal-mark against the name of the candidate he wants to vote. (in general election it is by pressing the button of the voting machine).

6. Counting of Votes and Declaration of Results:

• After the polling has ended the ballot boxes or the voting machines are sealed and carried under custody to the counting stations (concerned classrooms).

• Then the process of counting the votes began.

• The representatives of all the squads were present at the counting point.

• The candidate who obtains the highest number of votes is declared elected.

7. Submission of Account Relating to Election Expenses:

• The school budget law fixes the maximum limit of the expenses to be incurred by various proceedings on their election.

• In general election in India, the limit of election expenses for an Assembly election in most States was raised from Rs. 1.50 lakh to Rs. 6 lakh.

• It was enhanced from Rs. 4.50 lakh to Rs. 15 lakh for a Parliamentary contest.

• The candidates are required to file an account of the election expenses.

• It is a dishonest practice for a candidate to expend more money than the prescribed amount on his election.

• These steps were not present or required in the school election.

8. Election Disputes:

• The concerned teachers and the administrative head formed the election dispute cell in school.

• The Indian Constitution originally provided for the appointment of Election Tribunals for deciding disputes happening in connection with elections.

• The Nineteenth Amendment Act (1966) eliminated this provision and insisted that the election disputes would be decided by the High Courts.

Thus the scrutiny and recording of school election gave us a better understanding of the National election in India.


Balbharati, solutions, for, Social, Science, History, and, Civics, 10th, Standard, SSC, Maharashtra, State, Board, chapter 1, Working of the Constitution, [Latest edition],

Exercise | Q 1.1 | Page 74

Choose the correct option from the given options and complete the sentence. 

In Maharashtra __________ seats are reserved for women in local  self-governing institutions. 

25%

30%

40%

50%

Solution

50%

Which of the following laws created a favorable environment for women to secure freedom and self-development? 

Options

Right to Information Act 

Dowry Prohibition Act 

Food Security Act 

 None of the above

Solution

The Dowry Prohibition Act

The essence of democracy is ________

universal adult franchise. 

decentralisation of power. 

policy of reservation of seats. 

judicial decisions.

Solution

decentralisation of power.

Exercise | Q 1 | Page 74

State whether following statement is true or false. Give reason for your answer. 

Indian democracy is considered the largest democracy in the world.

Solution

True

Reason : Indian democracy is considered the largest democracy in the world because of political maturity. It’s the world’s second largest country in terms of population, people have direct representation in the legislative processes through regular elections and the right to adult suffrage.

Secrecy in the working of Government has increased due to the Right to Information.

Solution

False

Reason: Secrecy in the working of Government has decreased due to the Right to Information (Act 2005). Since 2000, the democratic reforms has been approached as ‘rights’ of citizen which has made governance more transparent and the government more accountable.

The nature of Constitution is seen as a living document.

Solution

True.

Reason: The Constitution is seen as a living document owing to its dynamic nature. It has to change according to changing conditions of the society and the Parliament has been entrusted with that responsibility.

Exercise | Q 1 | Page 74

Write short note.

Provisions regarding minorities.

Solution

The constitution has made several provisions for the protection of minorities in the country. It prohibits discrimination on the basis of caste, religion, race, language and region. Comprehensive provisions in the Fundamental Rights protect their right to equality, right against exploitation, and their cultural and educational rights.

Policy of reservation of seats.

Solution

Policy of reservation of seats is meant for the weaker section of the society who have been deprived of equal opportunity to education and employment for years. The Constitution accordingly mandates reservation of seats for scheduled caste, scheduled tribes and other backward classes in educational institutions and government services.

Women representation in the Loksabha.

Solution

Since independence, steps have been to address the issue of inadequate representation of women in political institutions. Under the 73rd and 74th amendment to the Constitution, 33% of seats in the local self- governing institutions are reserved for women. The representation of women in Lok Sabha has also seen a significant improvement with appointment of 12.15% women MPs in 2014 General elections.

Exercise | Q 1 | Page 74

Explain the following concept.

Right based approach

Solution

Since 2000, the approach towards democratic reform has evolved to be right based. Democratic reforms are necessitated as the ‘rights’ of the citizen, instead of approached as a part of the Directive principle of State policy. Citizens can take the government to the court in case their ‘rights’ are not met and thus, this change in approach has strengthened our democracy and made the govt. more accountable to the public. People now have the Right to Information, Right to Food, Right to Education and thus are more participative in the democratic processes.

Right to information

Solution

Right to information was enacted in 2005 to empower the citizen with vital information about the working and decision making process of the state. It makes the govt. more accountable and transparent, with increasing communication between the state and the citizen, hence building trust. It has reduced the element of secrecy in the working of government. The movement for RTI was first started in the state of Maharashtra in 2000 under the leadership of Anna Hazare. After the success of the state level efforts, the movement acquired national importance later in 2005.

Exercise | Q 1 | Page 74

Answer the following question.

What are the effects of reducing the voting age from 21 years to 18 years ?

Solution

The effects of reducing the voting age from 21 years to 18 years are:

1. Increased participation in the political processes. India has the largest number of voters compared to any other democratic country.

2. It encourages youth participation in public life, giving the younger generation an opportunity to be the change makers.

3. The change is not just quantitative but qualitative as well since new parties are emerging with the active support of the young voters.

4. It reflects the political maturity of India that empowers and actively working towards helping its citizens exercise their Right to vote.

5. Such provisions have made India the largest democracy in the world.

What is meant by establishment of social justice?

Solution

Social justice and equality are important objectives of our Constitution. It means establishing and ensuring a fair environment for the growth of every individual without any form of discrimination. Various provisions have been made to establish social justice in our society, some of these are:

Policy of Reservation: Seats are reserved for the weaker section of the society to enable them equal and fair opportunities for education and government services.

Scheduled Castes and Tribes (Prevention of Atrocities) Act: It prevents any act of injustice and atrocities committed against the people belonging to Scheduled Castes and Scheduled Tribes.

Provisions for Minority: There is a comprehensive provision in the Constitution that protects the fundamental rights of the minority to equality, freedom, education, right against exploitation, and the right to preserve their culture.

Laws for Women: Taking into account the problems of women, various policies have been formulated to remove illiteracy among women and avail them of equal opportunity for their development. The right to have an equal share in the property of their father and husband, the Dowry Prohibition Act, the Domestic Violence Prohibition Act, etc. have been conceptualized to create a favourable environment for women.

Which decision of the Court has resulted in protection of honour and dignity of women ?

Solution

The decision of the court in the cases of domestic violence, sexual harassment in work place etc. has resulted in protection of honour and dignity of women.

The most recent case in news had been the matter of Triple Talaq, where the practice was declared as unconstitutional as per the decision of the Supreme Court. It was an unfair practice subjected against Muslim women where divorce was pronounced with mere utterance of the word ‘Talaq’ trice on most trivial of issues with no formal legal proceedings, thus undermining various fundamental rights of women. It violated their dignity and fundamental right against discrimination on the basis of religion and gender.

Supreme Court as the apex judicial body is responsible for ensuring the preservation of the fundamental rights. Important judgments by the judiciary on various subjects have made the fundamental rights in the constitution more meaningful.

Project | Q 1 | Page 74

Which information can be secured with the help of right to information? Find out with the help of your teachers.

Solution

Under the provisions of the Act, Right to Information (RTI) is an act of the Parliament of India any citizen of India may ask for information from a “public authority” (a body of Government or “instrumentality of State”) which is necessary to reply immediately or within thirty days.

The Right to information in India is governed by two major bodies:

• Central Information Commission (CIC)

• State Information Commissions

The information which can be collected are:

1) Information from a public authority

2) Inspection of work, records etc of public authority

3) Details related to members of parliament and members of legislative assembly.

The important feature of RTI is that it is applicable to Indian citizens, but not associations or companies.

Make a list of concessions given by the Government for the students of minority community ?

Solution

AREA    SCHEMES

Educational   

1.Scholarship Schemes

2.Maulana Azad National Fellowship (MANF)

3.Padho Pardesh – Scheme of Internet Subsidy on Educational Loans for Overseas Studies for the Students Belonging to the Minority Communities

4.Naya Savera – Free Coaching and Allied Scheme

5.Nai Udaan – Support for Students for preparation of main Examination who clear Prelims conducted by UPSC/SSC , state public service commission (PSC) etc . 

Economic   

1.Seekho Aur Kamao (Learn & Earn)

2.USTTAD (Upgrading the skills and Training in Traditional Arts/Crafts for Development)

3.Nai Manzil

4.Concessional credit through National Minorities Dvelopment and Finance Corporation (NMDFC)

Infrastructural assistance    1.Pradhan Mantri Jan Vikas Karyakram (PMJVK)

Visit the official website of National election commission and collect more information about it.

Solution

(INFORMATION FROM THE INTERNET)

The Election Commission of India is an autonomous constitutional authority responsible for administering Union and State election processes in India. The body administers elections to the Lok Sabha, Rajya Sabha, State Legislative Assemblies in India, and the offices of the President and Vice President in the country.

A Constitutional Body

Election Commission of India is a permanent Constitutional Body. The Election Commission was established in accordance with the Constitution on 25th January 1950. The Commission celebrated its Golden Jubilee in 2001.

Originally the commission had only a Chief Election Commissioner. It currently consists of Chief Election Commissioner and two Election Commissioners.

For the first time, two additional Commissioners were appointed on 16th October 1989 but they had a very short tenure till 1st January 1990. Later, on 1st October 1993 two additional Election Commissioners were appointed. The concept of multi-member Commission has been in operation since then, with decision making power by majority vote.

Appointment & Tenure of Commissioners

The President appoints Chief Election Commissioner and Election Commissioners. They have a tenure of six years, or up to the age of 65 years, whichever is earlier. They enjoy the same status and receive salary and perks as available to Judges of the Supreme Court of India. The Chief Election Commissioner can be removed from office only through impeachment by Parliament.

Transaction of Business

The Commission transacts its business by holding regular meetings and also by circulation of papers. All Election Commissioners have equal say in the decision making of the Commission. The Commission, from time to time, delegates some of its executive functions to its officers in its Secretariat.

The Setup

The Commission has a separate Secretariat at New Delhi, consisting of about 300 officials, in a hierarchical set up.

Budget & Expenditure

The Secretariat of the Commission has an independent budget, which is finalized directly in consultation between the Commission and the Finance Ministry of the Union Government. The latter generally accepts the recommendations of the Commission for its budgets. The major expenditure on the actual conduct of elections is, however, reflected in the budgets of the concerned constituent units of the Union – States and Union Territories.

Political Parties & the Commission

Political parties are registered with the Election Commission under the law. The Commission ensures inner-party democracy in their functioning by insisting upon them to hold their organizational elections at periodic intervals.

Advisory Jurisdiction & Quasi-Judicial Functions

Under the Constitution, the Commission also has advisory jurisdiction in the matter of post-election disqualification of sitting members of Parliament and State Legislatures

Judicial Review

The decisions of the Commission can be challenged in the High Court and the Supreme Court of India by appropriate petitions. By long-standing convention and several judicial pronouncements, once the actual process of elections has started, the judiciary does not intervene in the actual conduct of the polls.

Media Policy

The Commission has a comprehensive policy for the media. It holds regular briefings for the mass media-print and electronic, on a regular basis, at close intervals during the election period and on specific occasions as necessary on other occasions.

Voter Education

Voters’ Participation in the democratic and electoral processes is integral to the successful running of any democracy and the very basis of wholesome democratic elections. Recognizing this, Election Commission of India, in 2009, formally adopted Voter Education and Electoral participation as an integral part of its election management.

International Co-operation

India is a founding member of the International Institute for Democracy and Electoral Assistance (IDEA), Stockholm, Swed

Take an interview of women representations from local self-governing institutions from your area.

Solution

The MLA of our district is a woman. She has been in the socio-cultural activities in our region for the last fifteen years. She is vibrant and active. An interview with her was very insightful and motivating.

Q: Do you see women reservation as part of a greater change in the social system?

Ans. I judge the excellence and the ambiance of politics will improve if more women come in. After bearing in mind the structure work for 45 years, it is understandable that we are the weaker sex, whether we like it or not. We are barred from jobs, the economic mainstream, decision-making processes, heritage rights… Whether one is fighting for the Scheduled Castes, the Backward Classes, or the minorities…. the main group that is affected are women. And things have not changed since then. Unless we are assured an access point by law we will never be allowed to participate. I really consider that. Some women say that we should come into decision-making positions on our own stream, but how many have come so far? To all those who say that they will do it on their own, I would like to ask why is the number of seats occupied by women in the legislatures deteriorating?

Q: You have completed some individual efforts at getting many numbers of women party tickets. What have you found to be the barriers?

Ans. We have tried but we have been unproductive because we are neither in the collection panels nor are we represented in decision-making bodies. We give lists, we struggle, we counsel, but they don’t listen to us. The election commission will reserve 33 percent of the seats by ballot for women. These are reserved for women for two terms and are then rotated. In the interim, women can establish themselves and if they are popular they will be able to contest on their own worth as general candidates by the next elections. So each ward has a chance to throw up women. We can launch a deadline of 25 years, or five general elections, by which time the reservations for women could be remoted. If women do not have an entry tip, how will they ever get in?

Q: What is your individual stand on the subject of 33 percent reservation for women?

Ans. As a result of reservation, one million women have been elected to local bodies. In many states they have even surpassed the quota for example, in Karnataka, women constitute 47 percent of the elected panchayat members. Bengal has also been a triumph story, as have Maharashtra, Kerala and Tamil Nadu. We need reforms not just for women the whole system of selection of candidates needs to be renovated.

Q: How did this issue get included in your party’s manifesto?

Ans. We had argued for its early introduction so that negotiations and debate could be allowed. Finally, when it was introduced, those who had never spoken about reservation for Backward Classes suddenly become the champions of Backward Class reservations. Between 1991, when the Panchayath bill was introduced, until 1996 when this bill is on the edge of being introduced, these people did not coherent any of their concerns for the Backward Class men! The issue gains importance only now because women are about to gain seats.

Q: Do you think this is just party games or something else?

Ans. No, no, even women in our party stood up to oppose the bill. There is a lobby of self-interest amongst women as well. The matter is different among the patriarchal viewpoint. They ask; Once decisions are taken, why have a women’s meeting?

Q: What is the status now?

Ans. A Select Committee has been set up. I am not recommending anything. Based on the various Commission reports we made reservations on the services. Nobody had spoken about proviso for Backward Classes or others in Parliament or legislatures. Women cut across all sections and by giving reservations to them you will be serving women from all sections to come forward. There are presently no reserved constituencies for the Backward Classes and minorities, so why should this issue get tied up with a reservation for women?

The interview session was so encouraging, and the MLA winded up the conversation by laying high remarks about the future of the whole society.


Samacheer, Kalvi, 10th, sslc, Science, Solutions, Chapter 6, Nuclear Physics, tamilnadu board, Physics,

Question 1.
Identify A, B, C, and D from the following nuclear reactions.

Solution:


Samacheer Kalvi 10th Science Solutions Chapter 6 Nuclear Physics 2

A is alpha particle, B is neutron, C is proton and D is electron.

Question 2.
A radon specimen emits radiation of 3.7 × 103 GBq per second. Convert this disintegration in terms of a curie, (one curie = 3.7 × 1010 disintegration per second)
Solution:
1 Bq = one disintegration per second
one curie = 3.7 × 1010 Bq

Question 3.
92U235 experiences one α – decay and one β – decay. Find the number of neutrons in the final daughter nucleus that is formed.
Solution:
Let X and Y be the resulting nucleus after the emission of the alpha and beta particles respectively.

Number of neutrons = Mass number – Atomic number = 231 – 91 = 140.

Question 4.
Calculate, the amount of energy released when a radioactive substance undergoes fusion and results in a mass defect of 2 kg.
Solution:
Mass defect in the reaction (m) = 2 kg
Velocity of light (c) = 3 × 108 ms-1
By Einstein’s equation,
Energy released E = mc2
= 2 × (3 × 108)2
= 1.8 × 1017 J.

Samacheer Kalvi 10th Science Nuclear Physics Textual Evaluation

I. Choose the correct answer

Question 1.
Man – made radioactivity is also known as _____.
(a) Induced radioactivity
(b) Spontaneous radioactivity
(c) Artificial radioactivity
(d) (a) & (c).
Answer:
(d) (a) & (c).

Question 2.
Unit of radioactivity is:
(a) roentgen
(b) curie
(c) becquerel
(d) all the above
Answer:
(d) all the above

Question 3.
Artificial radioactivity was discovered by _____.
(a) Becquerel
(b) Irene Curie
(c) Roentgen
(d) Neils Bohr.
Answer:
(b) Irene Curie

Question 4.
In which of the following, no change in mass number of the daughter nuclei takes place:
(i) a decay;
(ii) P decay
(iii) y decay
(iv) neutron decay
(a) (i) is correct
(b) (ii) and (iii) are correct
(c) (i) and (iv) are correct
(d) (ii) and (iv) are correct
Answer:
(b) (ii) and (iii) are correct

Question 5.
_____ isotope is used for the treatment of cancer.
(a) Radio Iodine
(b) Radio Cobalt
(c) Radio Carbon
(d) Radio Nickel.
Answer:
(b) Radio Cobalt

Question 6.
Gamma radiations are dangerous because:
(a) it affects eyes and bones
(b) it affects tissues
(c) it produces genetic disorder
(d) it produces an enormous amount of heat
Answer:
(c) it produces genetic disorder

Question 7.
_____ aprons are used to protect us from gamma radiations.
(a) Lead oxide
(b) Iron
(c) Lead
(d) Aluminium.
Answer:
(c) Lead

Question 8.
Which of the following statements is / are correct?
(i) α particles are photons
(ii) Penetrating power of γ radiation is very low
(iii) Ionization power is maximum for α rays
(iv) Penetrating power of γ radiation is very high
(a) (i) & (ii) are correct
(b) (ii) & (iii) are correct
(c) (iv) only correct
(d) (iii) & (iv) are correct.
Answer:
(d) (iii) & (iv) are correct.

Question 9.
Proton-Proton chain reaction is an example of:
(a) Nuclear fission
(b) α – decay
(c) Nuclear fusion
(d) β – decay
Answer:
(c) Nuclear fusion

Question 10.
In the nuclear reaction X126⟶α decay zYA, the value of A & Z.
(a) 8, 6
(b) 8, 4
(c) 4, 8
(d) cannot be determined with the given data.
Answer:
(c) 4, 8

Question 11.
Kamini reactor is located at _____.
(a) Kalpakkam
(b) Koodankulam
(c) Mumbai
(d) Rajasthan.
Answer:
(a) Kalpakkam

Question 12.
Which of the following is/are correct?
(i) Chain reaction takes place in a nuclear reactor and an atomic bomb.
(ii) The chain reaction in a nuclear reactor is controlled.
(iii) The chain reaction in a nuclear reactor is not controlled.
(iv) No chain reaction takes place in an atom bomb.
(a) (i) only correct
(b) (i) & (ii) are correct
(c) (iv) only correct
(d) (iii) & (iv) are correct
Answer:
(b) (i) & (ii) are correct

II. Fill in the blanks

Question 1.
One roentgen is equal to ______ disintegrations per second?
Answer:
3.7 × 1010.

Question 2.
Positron is an _____.
Answer:
antiparticle of electron.

Question 3.
Anaemia can be cured by _____ isotope.
Answer:
Radio iron (Fe59).

Question 4.
Abbreviation of ICRP _____.
Answer:
International Commission on Radiological Protection.

Question 5.
_____ is used to measure the exposure rate of radiation in humans.
Answer:
Roentgen.

Question 6.
_____ has the greatest penetration power.
Answer:
Gamma ray.

Question 7.
zYA→Z+1YA+X; Then X is _____.
Answer:
−1e0 (β decay).

Question 8.
zXA→YAZ This reaction is possible in _____ decay.
Answer:
Gamma (γ).

Question 9.
The average energy released in each fusion reaction is about _____ J.
Answer:
3.84 × 10-12.

Question 10.
Nuclear fusion is possible only at an extremely high temperature of the order of _____ K.
Answer:
107 to 109.

Question 11.
The radioisotope of _____ helps to increase the productivity of crops.
Answer:
phosphorous (P – 32).

Question 12.
If radiation exposure is 100 R, it may cause _____.
Answer:
fatal disease.

III. State whether the following statements are true or false: If false, correct the statement

Question 1.
Plutonium -239 is a fissionable material.
Answer:
True.

Question 2.
Elements having an atomic number greater than 83 can undergo nuclear fusion.
Answer:
False.
Correct Statement: Elements having an atomic number greater than 83 can undergo nuclear fusion.

Question 3.
Nuclear fusion is more dangerous than nuclear fission.
Answer:
False.
Correct Statement: Nuclear fission is more dangerous than nuclear fusion. Because the average energy released in fission (3.2 × 10-11 J) process is more than the average energy released in fusion (3.84 × 10-12 J).

Question 4.
Natural uranium U-238 is the core fuel used in a nuclear reactor.
Answer:
False.
Correct Statement: U-238 is not a fissile material but are abundant in nature. But in a reactor, this can be converted into a fissile material Pu239 and U233. Only fissile materials are used in the fuel of a nuclear reactor.

Question 5.
If a moderator is not present, then a nuclear reactor will behave like an atom bomb.
Answer:
True.

Question 6.
During one nuclear fission on an average, 2 to 3 neutrons are produced.
Answer:
True.

Question 7.
Einstein’s theory of mass-energy equivalence is used in nuclear fission and fusion.
Answer:
True.

IV. Match the following

Question 1.

1. BARC(a) Kalpakkam
2. India’s first atomic power station(b) Apsara
3. IGCAR(c) Mumbai
4. The first nuclear reactor in India(d) Tarapur

Answer:
1. (c) Mumbai
2. (d) Tarapur
3. (a) Kalpakkam
4. (b) Apsara

Question 2.

1. Fuel(a) lead
2. Moderator(b) heavy water
3. Coolant(c) Graphite
4. Shield(d) Uranium

Answer:
1. (d) uranium
2. (c) Graphite
3. (b) heavy water
4. (a) lead

Question 3.

1. Soddy Fagan(a) Natural radioactivity
2. Irene Curie(b) Displacement law
3. Henry Becquerel(c) Mass energy equivalence
4. Albert Einstein(d) Artificial Radioactivity

Answer:
1. (b) Displacement law
2. (d) Artificial Radioactivity
3. (a) Natural radioactivity
4. (c) Mass energy equivalence

Question 4.

1. Uncontrolled fission Reaction(a) Hydrogen Bomb
2. Fertile material(b) Nuclear Reactor
3. Controlled fission Reaction(c) Breeder reactor
4. Fusion reaction(d) Atom bomb

Answer:
1. (d) Atom bomb
2. (c) Breeder reactor
3. (b) Nuclear Reactor
4. (a) Hydrogen Bomb

Question 5.

1. Co – 60(a) Age of fossil
2. I – 13(b) Function of Heart
3. Na – 24(c) Leukaemia
4. C – 14(d) Thyroid disease

Answer:
1. (c) Leukemia
2. (d) Thyroid disease
3. (b) Function of Heart
4. (a) Age of fossil

V. Arrange the following in the correct sequence

Question 1.
Arrange in descending order, on the basis of their penetration power.

  1. Alpha rays
  2. Beta rays
  3. Gamma rays
  4. Cosmic rays.

Answer:

  1. Gamma rays
  2. Beta rays
  3. Alpha rays
  4. Cosmic rays.

Question 2.
Arrange the following in the chronological order of discovery.

  1. A nuclear reactor
  2. Radioactivity
  3. Artificial radioactivity
  4. Discovery of radium.

Answer:

  1. Radioactivity (1896)
  2. Discovery of radium (1898)
  3. Artificial radioactivity (1934)
  4. Nuclear reactor (1942).

VI. Use the analogy to fill in the blank

Question 1.
Spontaneous process : Natural Radioactivity, Induced process: _____.
Answer:
Artificial radioactivity
(or)
Man – made activity.

Question 2.
Nuclear Fusion : Extreme temperature, Nuclear Fission: _____.
Answer:
Room temperature.

Question 3.
Increasing crops : Radio phosphorous, Effective functioning of heart: _____.
Answer:
Radio sodium (Na24).

Question 4.
Deflected by electric field : α ray, Null Deflection: _____.
Answer:
γ ray (Gamma – ray).

VII. Numerical Problems

Question 1.
88Ra226 experiences three α-decay. Find the number of neutrons in the daughter element.
Solution:
88Ra226 consider as a parent element that is 88X226 and their daughter element is zYA
According to α decay process,
88X26⟶3α decay 82214+3α decay
During the 3α decay, the atomic number decreases by 6 and mass number decreases by 12.
So the number of neutrons in the daughter element
N = A – Z
N = 214 – 88 = 126
Number of neutrons in the daughter element N = 126.

Question 2.
A cobalt specimen emits induced radiation of 75.6 millicurie per second. Convert this disintegration in to becquerel (one curie = 3.7 × 1010 Bq).
Solution:
Cobalt specimen emits induced radiation = 75.6 millicurie per second
(1 curie = 3.7 × 1010 Bq)
So 75.6 millicurie = 75.6 × 103 × 1 curie
= 75.6 × 10-3 × 3.7 × 1010 Bq
= 279.72 × 107
= 2.7972 × 109 Bq
75.6 millicurie per second is equivalent to 2.7972 × 109 Bq.

VIII. Assertion and Reason Type Questions

Mark the correct choice as
(a) If both the assertion and the reason are true and the reason is the correct explanation of the assertion.
(b) If both the assertion and the reason are true, but the reason is not the correct explanation of the assertion.
(c) Assertion is true, but the reason is false.
(d) Assertion is false, but the reason is true.

Question 1.
Assertion: A neutron impinging on U235, splits it to produce Barium and Krypton.
Reason: U-235 is a fissile material.
Answer:
(a) If both the assertion and the reason are true and the reason is the correct explanation of the assertion.

Question 2.
Assertion: In a β – decay, the neutron number decreases by one.
Reason: In β – decay atomic number increases by one.
Answer:
(d) The assertion is false, but the reason is true.
Explanation: In β – decay there is no change in the mass number of the daughter nucleus but the atomic number increases by one.

Question 3.
Assertion: Extreme temperature is necessary to execute nuclear fusion.
Reason: In nuclear fusion, the nuclei of the reactants combine releasing high energy.
Answer:
(a) If both the assertion and the reason are true and the reason is the correct explanation of the assertion.

Question 4.
Assertion: Control rods are known as ‘Neutron seeking rods’
Reason: Control rods are used to perform a sustained nuclear fission reaction.
Answer:
(a) If both the assertion and the reason are true and the reason is the correct explanation of the assertion.
Explanation: Control rods are used to control the number of neutrons in order to have a sustained the chain reaction. They absorb the neutrons, (they seeking the neutrons)

IX. Answer in one or two words (VSA)

Question 1.
Who discovered natural radioactivity?
Answer:
Henri Becquerel was discovered natural radioactivity.

Question 2.
Which radioactive material is present in the ore of pitchblende?
Answer:
Uranium

Question 3.
Write any two elements which are used for inducing radioactivity?
Answer:

  1. Boron and Aluminium.
  2. Alpha particle and neutron.

Question 4.
Write the name of the electromagnetic radiation which is emitted during a natural radioactivity.
Answer:
Gamma rays

Question 5.
If A is a radioactive element which emits an α-particle and produces 104Rf259. Write the atomic number and mass number of the element A.
Answer:
In α decay
zXAα decay 263×z−2YA−4+2He4(α decay )106X263⟶α decay 104Rf259+2He4
In element A having atomic number is 106 and mass number is 263.

Question 6.
What is the average energy released from a single fission process?
Answer:
The average energy released from a single fission process is about 3.2 × 10-11 J.

Question 7.
Which hazardous radiation is the cause for the genetic disorders (or) effect?
Answer:
Radioactive radiations

Question 8.
What is the amount of radiation that may cause the death of a person when exposed to it?
Answer:
When the body is exposed to about 600 R, it leads to death.

Question 9.
When and where was the first nuclear reactor built?
Answer:
The first nuclear reactor was built in 1942 in Chicago, USA.

Question 10.
Give the SI unit of radioactivity.
Answer:
Becquerel

Question 11.
Which material protects us from radiation?
Answer:
Lead coated aprons and lead gloves should be used while working with the hazardous area. These materials are used to protects us from radiation.

X. Answer the following questions in a few sentences.

Question 1.
Write any three features of natural and artificial radioactivity.
Answer:

Natural radioactivityArtificial radioactivity
1. Emission of radiation due to the self-disintegration of a nucleus.1. Emission of radiation due to the disintegration of a nucleus through the induced process.
2. Alpha, Beta and Gamma radiations are emitted.2. Mostly elementary particles such as neutron, positron, etc. are emitted.
3. It is a spontaneous process.3. It is an induced process.

Question 2.
Define critical mass.
Answer:
The minimum mass of fissile material necessary to sustain the chain reaction is called ‘critical mass (mc). It depends on the nature, density and the size of the fissile material.

Question 3.
Define One roentgen.
Answer:
One roentgen is defined as the quantity of radioactive substance which produces a charge of 2.58 × 10-4 coulomb in 1 kg of air under standard conditions of pressure, temperature and humidity.

Question 4.
State Soddy and Fagan’s displacement law.
Answer:
During a radioactive disintegration, the nucleus which undergoes disintegration is called a parent nucleus and that which remains after the disintegration is called the daughter nucleus.

Question 5.
Give the function of control rods in a nuclear reactor.
Answer:
Control rods are used to control the number of neutrons in order to have sustained chain reaction. Mostly boron or cadmium rods are used as control rods. They absorb the neutrons.

Question 6.
In Japan, some of the newborn children are having congenital diseases. Why?
Answer:
During the Second World War American, a bomber dropped the nuclear weapons over the Japanese cities of Hiroshima and Nagasaki. In the explosion of the atomic bomb to release the high energy dangerous radiation. In the explosion period, Japanese peoples are affected by radiation. This is the reason in Japan, some of the newborn children are having congenital diseases.

Question 7.
Mr Ramu is working as an X – ray technician in a hospital. But, he does not Wear the lead aprons. What suggestion will you give to Mr Ramu?
Answer:
X – rays have a destructive effect on living tissue. When the human body is exposed to X – rays, it causes redness of the skin, sores and serious injuries to the tissues and glands. They destroy the white corpuscles of the blood. If you don’t wear the lead aprons these kinds of diseases formed in your body. In my suggestion, you must wear lead aprons.

Question 8.
What is stellar energy?
Answer:
Fusion reaction that takes place in the cores of the Sun and other stars results in an enormous amount of energy, which is called as stellar energy.

Question 9.
Give any two uses of radioisotopes in the field of agriculture?
Answer:

  • The radioisotope of phosphorus (P – 32) helps to increase the productivity of crops.
  • The radiations from the radioisotopes can be used to kill the insects and parasites and prevent the wastage of agricultural products.

XI. Answer the following questions in detail.

Question 1.
Explain the process of controlled and uncontrolled chain reactions.
Answer:
(a) Controlled chain reaction

  • In the controlled chain reaction, the number of neutrons released is maintained to be one. This is achieved by absorbing the extra neutrons with a neutron absorber leaving only one neutron to produce further fission.
  • Thus, the reaction is sustained in a controlled manner. The energy released due to a controlled chain reaction can be utilized for constructive purposes.
  • The controlled chain reaction is used in a nuclear reactor to produce energy in a sustained and controlled manner.

(b) Uncontrolled chain reaction:

Question 2.
Compare the properties of Alpha, Beta and Gamma radiations.
Answer:

Propertiesα raysβ raysγ rays
What are they?Helium nucleus (2He4)consisting of two protons and two neutrons.They are electrons (−1e0), basic elementary particle in all atoms.They are electromagnetic waves consisting of photons.
ChargePositively charged particles. Charge of each alpha particle = +2eNegatively charged particles. Charge of each beta particle = -eNeutral particles. Charge of each gamma particle = zero
Ionising Power100 time greater than β rays and 10,000 times greater than γ raysComparatively lowVery less ionization power
Penetrating powerLow penetrating power (even stopped by a thick paper)Penetrating power is greater than that of α rays. They can penetrate through a thin metal foil.They have a very high penetrating power greater than that of β rays. They can penetrate through thick metal blocks.
Effect of an electric and magnetic fieldDeflected by both the fields. (in accordance with Fleming’s left-hand rule)Deflected by both the fields, but the direction of deflection is opposite to that for alpha rays. (in accordance with Fleming’s left-hand rule)They are not deflected by both the fields.
SpeedTheir speed ranges from 1/10 to 1/20 times the speed of light.Their speed can go up to 9/10 times the speed of light.They travel with the speed of light.

Question 3.
What is a nuclear reactor? Explain its essential parts with their functions.
Answer:
Nuclear reactor: A Nuclear reactor is a device in which the nuclear fission reaction takes place in a self – sustained and controlled manner to produce electricity.

Components of a Nuclear Reactor:
The essential components of a nuclear reactor are

  • Fuel: A fissile material is used as the fuel. The commonly used fuel material is uranium.
  • Moderator: A moderator is used to slow down the high energy neutrons to provide slow neutrons. Graphite and heavy water are commonly used moderators.
  • Control rod: Control rods are used to control the number of neutrons in order to have a sustained a chain reaction. Mostly boron or cadmium rods are used as control rods. They absorb the neutrons.
  • Coolant: A coolant is used to remove the heat produced in the reactor core, to produce steam. This steam is used to run a turbine in order to produce electricity. Water, air and helium are some of the coolants.
  • Protection wall: A thick concrete lead wall is built around the nuclear reactor in order to prevent the harmful radiations from escaping into the environment.

XII. HOT Questions

Question 1.
Mass number of a radioactive element is 232 and its atomic number is 90. When this element undergoes certain nuclear reactions, it transforms into an isotope of lead with a mass number 208 and an atomic number 82. Determine the number of alpha and beta decay that can occur.
Answer:
Mass number A = 232
Atomic number Z = 90
Daughter element:
Mass number A = 208
Atomic number Z = 82
Difference in mass number = 232 – 208 = 24
Difference in atomic number
= 90 – 82 = 8
Atomic number of α = 2
Atomic number of β = -1
Mass number of α = 4
Mass number of β = 0
Difference in mass number in transformations
= 24
Number of a decays = 244 = 6
Difference in atomic number = 8
ΔZ = 6α + 4β
= 6(2) + 4(-1)
= 12 – 4
= 8
∴ Number of β decays = 4
∴ Number of α decays = 6
∴ Number of β decays = 4

Question 2.
‘X – rays should not be taken often’. Give the reason.
Answer:

  • Radiation does involve in X – rays tests and isotope scans (in nuclear medicine) are too low to cause immediate hazardous effects.
  • If should be taken often, X – ray radiation from medical examinations though slightly increases one’s risk for cancer which can occur year or decades after X-ray exposure.

Question 3.
Cell phone towers should be placed far away from the residential area. why?
Answer:

  1. Living near a cell phone tower is not healthy. There is multiple health risks associated with living near a cell phone tower.
  2. Cell phone towers communicate by use pulsed microwave signals (radiofrequency radiation) with each other.
  3. That is the reason cell phone towers should be placed far away from the residential area.

Samacheer Kalvi 10th Science Nuclear Physics Additional Questions

I. Choose the best Answer.

Question 1.
Radium was discovered by _____.
(a) Marie curie
(b) Irene curie
(c) Henri Becquerel
(d) F. Joliot.
Answer:
(a) Marie Curie

Question 2.
How many radioactive substances discovered so far?
(a) 83
(b) 92
(c) 43
(d) 29
Answer:
(d) 29

Question 3.
The SI unit of Radioactivity is _____.
(a) Curie
(b) Rutherford
(c) Becquerel
(d) Roentgen (R).
Answer:
(c) Becquerel

Question 4.
Radioactivity is _____.
(a) increases with increase in temperature
(b) increases with increase in pressure
(c) depends on the number of electrons
(d) purely a nuclear phenomenon.
Answer:
(d) purely a nuclear phenomenon

Question 5.
Which of the following processes is a spontaneous process?
(a) Artifical radioactivity
(b) Natural radioactivity
(c) Photoelectric effect
(d) Collisions
Answer:
(b) Natural radioactivity

Question 6.
The charge of the β rays _____.
(a) 2e
(b) 0
(c) -e
(d) none of these.
Answer:
(c) -e

Question 7.
The charge of the γ rays _____.
(a) 2e
(b) 0
(c) -e
(d) none of these.
Answer:
(b) 0

Question 8.
The atomic number of the elements that exhibit artifical radioactivity is:
(a) more than 82
(b) more than 83
(c) less than 83
(d) less than 82
Answer:
(c) less than 83

Question 9.
Arrange α, β, γ rays in the increasing order of their ionizing power.
(a) α, β, γ
(b) β, α, γ
(c) γ, β, α
(d) γ, α, β.
Answer:
(c) γ, β, α

Question 10.
Which produces a charge of 2.58 × 10-4Coulomb in 1 Kg of air?
(a) Curie
(b) Becquerel
(c) Rutherford
(d) Roentgen
Answer:
(d) Roentgen

Question 11.
Ionising power of the γ rays _____.
(a) Comparatively very high ionization power
(b) 100 times greater than the α rays
(c) 100 times greater than the β rays
(d) Comparatively very less ionization power.
Answer:
(d) Comparatively very less ionization power.

Question 12.
Ionization power maximum for _____.
(a) neutrons
(b) α particles
(c) γ rays
(d) β particles.
Answer:
(b) α particles

Question 13.
Charge of gamma particle is:
(a) +2e
(b) -e
(c) Zero
(d) +1e
Answer:
(c) Zero

Question 14.
Which has low penetrating power?
(a) α rays
(b) γ rays
(c) β rays
(d) X rays.
Answer:
(a) α rays

Question 15.
In β – decay _____.
(a) atomic number decreases by one
(b) the mass number decreases by one
(c) proton number remains the same
(d) neutron number decreases by one.
Answer:
(d) neutron number decreases by one

Question 16.
In which decay the energy level of the nucleus changes:
(a) α – decay
(b) β – decay
(c) γ – decay
(d) neutron decay
Answer:
(c) γ – decay

Question 17.
In γ – decay _____.
(a) atomic number decreases by one
(b) there is no change in atomic and mass number
(c) energy only changes in the decay process
(d) both (b) and (c).
Answer:
(d) both (b) and (c).

Question 18.
The unit of decay constant is _____.
(a) no unit
(b) second
(c) second-1
(d) curie.
Answer:
(c) second-1

Question 19.
The range of temperature required for nuclear fusion is from:
(a) 107 to 109 K
(b) 10-9 to 10-7 K
(c) 105 to 109
(d) 105 to 107 K
Answer:
(a) 107 to 109 K

Question 20.
1 Rd is equal to _____.
(a) 106 decay / second
(b) 1 decay / second
(c) 3.7 × 1010 becquerel
(d) 1.6 × 1012 decay / second.
Answer:
(a) 106 decay / second

Question 21.
An element ZXA successively undergoes three α decays and four β decays and gets converted an element Y are respectively _____.
(a) Z−6YA−12
(b) Z+2YA−12
(c) Z−2YA−12
(d) Z−10YA−12.
Answer:
(c) Z−2YA−12

Question 22.
In the nuclear reaction 88Ra226 → X + 2He4X is:
(a) 90Th234
(b) 91Pa234
(c) 86Rn222
(d) 88Rn226
Answer:
(d) 88Rn226

Question 23.
Which one of the following is used in the treatment of skin diseases _____.
(a) Na24
(b) I31
(c) Fe59
(d) P32.
Answer:
(d) P32.

Question 24.
Anaemia can be diagnosed by _____.
(a) 15P31
(b) 15P32
(c) 26P59
(d) 11P24.
Answer:
(c) 26P59

Question 25.
Which is used as a coolant?
(a) Graphite
(b) Liquid sodium
(c) Boron
(d) Cadmium
Answer:
(b) Liquid sodium

Question 26.
The energy released per fission is _____.
(a) 220 MeV
(b) 300 MeV
(c) 250 MeV
(d) 200 MeV.
Answer:
(d) 200 MeV.

Question 27.
In the reaction 1N14 + 0n1 → X + 1H1 X is:
(a) 15P30
(b) 6C14
(c) 6C12
(d) 11Na23
Answer:
(c) 6C12

Question 28.
Natural uranium consists of _____.
(a) 99.72 % of U-238
(b) 0.28 % of U-238
(c) 0.72 % of U-238
(d) 99.28 % of U-238.
Answer:
(d) 99.28 % of U-238.

Question 29.
The number of power reactors in India is _____.
(a) 14
(b) 12
(c) 7
(d) 2.
Answer:
(a) 14

Question 30.
In the nucleus of 11Na23 the number of protons and neutrons are:
(a) 12, 11
(b) 10, 12
(c) 11, 12
(d) 11, 23
Answer:
(c) 11, 12

Question 31.
The moderator used in nuclear reactor is _____.
(a) cadmium
(b) boron carbide
(c) heavy water
(d) uranium (92U235).
Answer:
(c) heavy water

Question 32.
The first nuclear reactor was built at _____.
(a) Kalpakkam, India
(b) Hiroshima, Japan
(c) Chicago, USA
(d) Trombay, Bombay.
Answer:
(c) Chicago, USA

Question 33.
Which of the following is used in the treatment of skin cancer?
(a) Radio Cobalt
(b) Radio gold
(c) Radio Cobalt and radio gold
(d) none of the above
Answer:
(c) Radio Cobalt and radio gold

Question 34.
The explosion of an atom bomb is based on the principle of _____.
(a) uncontrolled fission reaction
(b) fusion reaction
(c) controlled fission reaction
(d) none of the above.
Answer:
(a) uncontrolled fission reaction

Question 35.
The reactor in which no moderator used is _____.
(a) fast breeder reactor
(b) pressurised water reactor
(c) pressurised heavy water reactor
(d) boiled water reactor.
Answer:
(a) fast breeder reactor

Question 36.
The number of neutrons present in 92U235is:
(a) 133
(b) 143
(c) 43
(d) 243
Answer:
(b) 143

Question 37.
In fast breeder, the coolant system used is _____.
(a) heavy water
(b) light water
(c) liquid sodium
(d) boiled water.
Answer:
(c) liquid sodium

Question 38.
The only reactor in the world which uses U-233 as fuel is _____.
(a) Zerlina
(b) Purnima
(c) Kamini
(d) Tires.
Answer:
(c) Kamini

Question 39.
The temperature of the interior of Sun is about _____.
(a) 1.4 × 107 K
(b) 108 K
(C) 14 × 107 K
(d) 600 K.
Answer:
(a) 1.4 × 107 K

Question 40.
Total energy radiated by Sun is about _____.
(a) 3.6 × 1028 Js-1
(b) 3.8 × 1028 Js-1
(c) 3.8 × 1026 Js-1
(d) 3.8 × 1023 Js-1.
Answer:
(c) 3.8 × 1026 Js-1


II. Fill in the blanks

Question 1.
Cathode rays are discovered by _____.
Answer:
J.J. Thomson.

Question 2.
Positive rays discovered by _____.
Answer:
Goldstein.

Question 3.
The chargeless particles are called neutron, it was discovered by _____.
Answer:
James Chadwick.

Question 4.
Ernest Rutherford explained that the mass of an atom is concentrated in its central part called _____.
Answer:
Nucleus.

Question 5.
The radioactive elements emit harmful radiations are ____, ____, ____ rays.
Answer:
alpha, beta, gamma.

Question 6.
_____ is an spontaneous process.
Answer:
Natural radioactivity.

Question 7.
The element whose atomic number is more than 83 undergoes _____.
Answer:
spontaneous process.

Question 8.
______ radioactive material is present in the ore of pitchblende.
Answer:
Uranium.

Question 9.
_____ are the example of artificial (or) man-made radioactive elements.
Answer:
Boron, Aluminium.

Question 10.
The element whose atomic number is less than 83 undergoes _____.
Answer:
induced radioactivity.

Question 11.
______ is an controlled manner.
Answer:
Artificial radioactivity.

Question 12.
Spontaneous radioactivity is also known as _____.
Answer:
Natural radioactivity.

Question 13.
One Curie is equal to _____ disintegrations per second.
Answer:
3.7 × 1010

Question 14.
One Rutherford (Rd) is equal to ______ disintegrations per second.
Answer:
106

Question 15.
The radioactive displacement law is framed by _____.
Answer:
Soddy and Fajan.

Question 16.
During the α decay process, the atomic number is ______ by 2 and the mass number is decreases by _____.
Answer:
decreases, 4.

Question 17.
In β-decay the atomic number increases by ____ unit and mass number _____.
Answer:
One, remains the same.

Question 18.
In α radiation, the charge of each alpha particle is _____.
Answer:
+2e.

Question 19.
In γ radiation, the charge of each gamma particle is _____.
Answer:
Zero.

Question 20.
In radioactive radiation, which one is travel with the speed of light _____.
Answer:
Gamma radiation.

Question 21.
zYA→z−2YA−4+X; Then X is _____.
Answer:
2He4 (α decay).

Question 22.
zYA→zYA+X; Then X is _____.
Answer:
γ decay.

Question 23.
The average energy released in each fission process in about _____.
Answer:
3.2 × 10-11 J.

Question 24.
Fissionable material is a radioactive element, which undergoes fission in a sustained manner when it absorbs a _____.
Answer:
Neutron.

Question 25.
_____ isotope is used to detect the presence of block in blood vessels and also used for the effective functioning of the heart.
Answer:
Na24 – Radio sodium.

Question 26.
_____ is used to cure goitre.
Answer:
Radio Iodine – I131

Question 27.
_____ is used to diagnose anaemia and also to provide treatment for the same.
Answer:
Radio – iron (Fe59).

Question 28.
Radio cobalt (Co60) and radio gold (Au198) are used in the treatment of _____.
Answer:
Skin cancer.

Question 29.
_____ are used to sterilize the surgical devices as they can kill the germs and microbes.
Answer:
Radiations.

Question 30.
The age of the earth, fossils, old paintings and monuments can be determined by _____. technique.
Answer:
Radiocarbon dating.

Question 31.
When the body is exposed to about 600 R, it leads to _____.
Answer:
Death.

Question 32.
Radioactive materials should be kept in a thick – walled container of _____.
Answer:
Lead.

Question 33.
_____ is used to remove the heat produced in the reactor core, to produce steam.
Answer:
Coolant.

Question 34.
The abbreviation of BARC is _____.
Answer:
Bhabha Atomic Research Centre.

Question 35.
India’s 1st nuclear power station is _____.
Answer:
Tarapur Atomic Power Station.

Question 36.
The first nuclear reactor built in India was _____.
Answer:
Apsara.

Question 37.
The total nuclear power operating sites in India is _____.
Answer:
7

Question 38.
The energy released in a nuclear fission process is about ______
Answer:
200 Mev.

Question 39.
The number of 0n1 released on an average per fission is _____.
Answer:
2.5.

Question 40.
A hydrogen bomb is based on the principle of _____.
Answer:
Nuclear fusion.

III. Match the following

Question 1.

1. Natural radioactivity(a) 3.7 × 1010decay/second
2. Artificial radioactivity(b) spontaneous process
3. 1 curie(c) 106decay/second
4. 1 Rd (Rutherford)(d) induced process

Answer:
1. (b) spontaneous process
2. (d) induced process
3. (a) 3.7 × 1010 decay / second
4. (c) 106 decay / second

Question 2.

1. Charge of each α particle(a) γ ray
2. Charge of each β particle(b) +2e
3. Penetration power is maximum(c) α ray
4. Ionisation power is maximum(d) zero

Answer:
1. (b) +2e
2. (d) zero
3. (a) γ ray
4. (e) α ray

Question 3.

1. Deuterium(a) −1e0
2. Protium(b) 1H3
3. Tritium(c) 2H4
4. α – decay(d) 1H1
5. β – decay(e) 1H2

Answer:
1. (e) 1H2
2. (d) 1H1
3. (b) 1H3
4. (c) 2H4
5. (a) −1e0

Question 4.

1. Uranium core bomb(a) fusion bomb
2. Plutonium core bomb(b) fission bomb
3. Hydrogen bomb(c) Nagasaki
4. Atom bomb(d) Hiroshima

Answer:
1. (d) Hiroshima
2. (c) Nagasaki
3. (a) fusion bomb
4. (b) fission bomb

Question 5.

1. Radio iron (Fe59)(a) treatment of skin diseases
2. Radio phosphorous (P32)(b) smoke detector
3. Radio gold (Au198)(c) diagnose anaemia
4. An isotope of Americium (Am241)(d) treatment of skin cancer

Answer:
1. (c) diagnose anaemia
2. (a) treatment of skin diseases
3. (d) treatment of skin cancer
4. (b) smoke detector

IV. Arrange the following in the correct sequence

Question 1.
Arrange α, β, γ rays in ascending order, on the basis of their penetrating power?
Answer:
Ascending order:

  • Alpha (α)
  • Beta (β)
  • Gamma (γ)

Question 2.
Arrange in ascending and descending order, on the basis of their Ionisation power.
Alpha (α), Beta (β), Gamma (γ)
Answer:

  1. Ascending order: Gamma (γ), Beta (β), Alpha (α)
  2. Descending order: Alpha (α), Beta (β), Gamma (γ)

Question 3.
Arrange in ascending and descending order, on the basis of their biological effect.
Alpha (α), Gamma (γ), Beta (β)
Answer:

  1. Ascending order: Alpha (α), Beta (β), Gamma (γ)
  2. Descending order: Gamma (γ), Beta (β), Alpha (α).

V. Numerical Problems

Question 1.
92U238 emits 8α particles and 6β particles. What is the neutron / proton ratio in the product nucleus?
Solution:

Question 2.
The element with atomic number 84 and mass number 218 change to another element with atomic number 84 and mass number 214. The number of α and β particles emitted are respectively?
Solution:
Samacheer Kalvi 10th Science Solutions Chapter 6 Nuclear Physics 8
Number of alpha decay, x = 1
Number of beta decay, y = 2.

Question 3.
The number of α and β particles emitted in the nuclear reaction 90Th228⟶83Bi12are respectively.
Solution:
Samacheer Kalvi 10th Science Solutions Chapter 6 Nuclear Physics 9
Number of α decay, x = 4
Number of β decay, y = 1.

VI. Assertion and Reason Type Questions

(a) If both the assertion and the reason are true and the reason is the correct explanation of the assertion.
(b) If both the assertion and the reason are true, but the reason is not the correct explanation of the assertion.
(c) If Assertion is true, but the reason is false.
(d) If Assertion is false, but the reason is true.
(e) If the Assertion and reason both are false.

Question 1.
Assertion: All the radioactive element are ultimately converted in lead.
Reason: All the elements above lead are unstable.
Answer:
(c) If Assertion is true, but the reason is false.
Explanation: When they are converted into a lead, the emission is stopped because the nucleus of lead is stable (or lead is most stable elements in radioactive series)

Question 2.
Assertion: Among the alpha, beta and gamma-ray a particle has maximum penetrating power.
Reason: The alpha particle is heavier than beta and gamma rays.
Answer:
(e) If the Assertion and reason both are false.
Explanation: The penetrating power is maximum in case of gamma rays because gamma rays are electromagnetic radiation of very small wavelength.

Question 3.
Assertion: The ionising power of β – particle is less compared to α – particles but their penetrating power is more.
Reason: The mass of β-particle is less than the mass of α-particle
Answer:
(b) If both the assertion and the reason are true, but the reason is not the correct explanation of the assertion.
Explanation: β – particle being emitted with very high speed compared to α – particle. Due to this reason, their loss of energy is very slow and they can penetrate the medium through a sufficient depth.

Question 4.
Assertion: Neutrons penetrate matter more readily as compared to protons.
Reason: Neutrons are slightly more massive than protons.
Answer:
(b) If both the assertion and the reason are true, but the reason is not the correct explanation of the assertion.
Explanation: Neutron is about 0.1 % more massive than a proton. But the unique thing about the neutron is that while it is heavy, it has no charge (it is neutral). This lack of charge gives it the ability to penetrate matter without interacting as quickly as the beta particles or alpha particles.

Question 5.
Assertion: zXA undergoes a decays and the daughter product is z−2YA−4
Reason: In α – decay, the mass number decreases by 4 and atomic number decreases by.
Answer:
(a) If both the assertion and the reason are true and the reason is the correct explanation of the assertion.
Explanation: zXA⟶z−2XA−4+2He4(α decay)

Question 6.
Assertion: Moderator is used to slowing down the high energy neutrons to provide slow neutrons.
Reason: Cadmium rods are used as control rods.
Answer:
(b) If both the assertion and the reason are true, but the reason is not the correct explanation of the assertion.
Explanation: Graphites and heavy water are commonly used moderators. This helps in moderator to slow down the fast neutrons.

Question 7.
Assertion: Alpha, beta and gamma radiations are emitted.
Reason: Nuclear fission process can be performed at room temperature.
Answer:
(a) If both the assertion and the reason are true and the reason is the correct explanation of the assertion.
Explanation: At room temperature, the nuclear fission process can perform breaking up of heavier nucleus into two smaller nuclei. In this process to emitted the alpha, beta and gamma radiations.

Question 8.
Assertion: An enormous amount of energy is released which is called stellar energy.
Reason: Fusion reaction that takes place in the cores of the Sun and other stars.
Answer:
(a) If both the assertion and the reason are true and the reason is the correct explanation of the assertion.
Explanation: The stars like our Sun emit a large amount of energy in the form of light and heat. This energy is termed as the stellar energy.

Question 9.
Assertion: Artificial radioactivity is a controlled process.
Reason: It is a spontaneous process – natural radioactivity.
Answer:
(b) If both the assertion and the reason are true, but the reason is not the correct explanation of the assertion.
Explanation: Artificial radioactivity is a controlled process. It is an induced process and man-made radioactivity.

Question 10.
Assertion: Gamma rays, penetrates through materials most effectively.
Reason: Gamma rays, which have the shortest wavelengths of all electromagnetic radiation.
Answer:
(a) If both the assertion and the reason are true and the reason is the correct explanation of the assertion.
Explanation: Gamma rays, which have the shortest wavelengths of all electromagnetic radiation. This is a reason they can penetrate through materials most effectively.

VII. Answer the following questions

Question 1.
Define ‘Radioactivity’.
Answer:
The phenomenon of nuclear decay of certain elements with the emission of radiations like alpha, beta, and gamma rays is called ‘radioactivity’.

Question 2.
By whom radioactivity is detected in pitchblende?
Answer:
Marie curie and Purie curie.

Question 3.
Define ‘Artificial Radioactivity’.
Answer:
The phenomenon by which even light elements are made radioactive, by artificial or induced methods, is called ‘Artificial radioactivity’ or ‘Man – made radioactivity’.

Question 4.
Define ‘One curie’.
Answer:
It is defined as the quantity of a radioactive substance which undergoes 3.7 × 1010 disintegrations in one second. This is actually close to the activity of 1 g of radium 226.
Curie = 3.7 × 1010 disintegrations per second.

Question 5.
In which elements artifical radioactivity is induced?
Answer:
Boron and aluminum

Question 6.
What is alpha decay (α decay)? give an example.
Answer:
A nuclear reaction in which an unstable parent nucleus emits an alpha particle and forms a stable daughter nucleus is called ‘alpha decay’.
E.g. Decay of uranium (U238) to thorium (Th234) with the emission of an alpha particle.
92U238→90Th234+2He4 (α – decay).

Question 7.
What is beta decay (β decay)? Give an example?
Answer:
A nuclear reaction, in which an unstable parent nucleus emits a beta particle and forms a stable daughter nucleus, is called ‘beta decay’.
E.g. Beta decay of phosphorous.
15P32→16S32+−1e0 (β – decay)

Question 8.
What is gamma decay (γ decay)?
Answer:
In a γ – decay, only the energy level of the nucleus changes. The atomic number and mass number of the radioactive nucleus remain the same.

Question 9.
State the value of Roentgen in terms of Coulomb.
Answer:
Roentgen = 2.58 × 10-4 Coulomb in / kg of air.

Question 10.
Define ‘nuclear fission’ Give an example.
Answer:
The process of breaking (splitting) up of a heavier nucleus into two smaller nuclei with the release of a large amount of energy and a few neutrons are called ‘nuclear fission’.
E.g. Nuclear fission of a uranium nucleus (U235)
92U235+0n1→56Ba141+36Kr92+30n1+Q( energy )

Question 11.
Define ‘Nuclear fusion’ Give an example.
Answer:
The process in which two light nuclei combine to form a heavier nucleus is termed as ‘Nuclear fusion’.
E.g. 1H2+1H2→2He4+Q( Energy )

Question 12.
Write down the types of the nuclear reactor.
Answer:
Breeder reactor, fast breeder reactor, pressurized water reactor, pressurized heavy water reactor, boiling water reactor, water – cooled reactor, gas – cooled reactor, fusion reactor and thermal reactor are some types of nuclear reactors, which are used in different places worldwide.

Question 13.
What is the safe limit of receiving radioactive radiations?
Answer:
100 m R per week

VIII. Answer in the details:

Question 1.
Explain the principle and working of an atom bomb?
Answer:
Atom bomb:
(i) The atom bomb is based on the principle of the uncontrolled chain reaction. In an uncontrolled chain reaction, the number of neutrons and the number of fission reactions multiply almost in a geometrical progression.

(ii) This releases a huge amount of energy in a very small time interval and leads to an explosion.

Structure:
(i) An atom bomb consists of a piece of fissile material whose mass is subcritical. This piece has a cylindrical void.

(ii) It has a cylindrical fissile material which can fit into this void and its mass is also subcritical. When the bomb has to be exploded, this cylinder is injected into the void using a conventional explosive.

(iii) The two pieces of fissile material join to form the supercritical mass, which leads to an explosion. During this explosion, a tremendous amount of energy in the form of heat, light and radiation is released.

(iv) A region of very high temperature and pressure is formed in a fraction of a second along with the emission of hazardous radiation like y rays, which adversely affect the living creatures. This type of atom bombs was exploded in 1945 at Hiroshima and Nagasaki in Japan during World War II.

Question 2.
State and define the units of radioactivity.
Answer:
Curie : It is the traditional unit of radioactivity. It is defined as the quantity of a radioactive substance which undergoes 3.7 × 1010 disintegrations in one second. This is actually close to the activity of lg of radium 226. 1 curie = 3.7 × 1010 disintegrations per second.

Rutherford (Rd) : It is another unit of radioactivity. It is defined as the quantity of a radioactive substance, which produces 106 disintegrations in one second.
1 Rd = 106 disintegrations per second.

Becquerel (Bq) : It is the SI unit of radioactivity is becquerel. It is defined as the quantity of one disintegration per second.

Roentgen (R) : It is the radiation exposure of γ and x-rays is measured by another unit called roentgen. One roentgen is defined as the quantity of radioactive substance which produces a charge of 2.58 × 10-4 coulomb in 1 kg of air under standard conditions of pressure, temperature and humidity.

Question 3.
Write down the features of nuclear fission and nuclear fusion.
Answer:

Nuclear FissionNuclear Fusion
1. The process of breaking up (splitting) of a heavy nucleus into two smaller nuclei is called ‘nuclear fission’.1. Nuclear fusion is the combination of two lighter nuclei to form a heavier nucleus.
2. Can be performed at room temperature.2. Extremely high temperature and pressure are needed.
3. Alpha, beta and gamma radiations are emitted.3. Alpha rays, positrons, and neutrinos are emitted.
4. Fission leads to emission of gamma radiation. This triggers the mutation in the human gene and causes genetic transform diseases.4. Only light and heat energy are emitted.

Question 4.
Write down the medical and industrial application of radioisotopes?
Answer:

  1. Radio sodium (Na24) is used for the effective functioning of the heart.
  2. Radio – Iodine (I131) is used to cure goitre.
  3. Radio – Iron is (Fe59) is used to diagnose anaemia and also to provide treatment for the same.
  4. Radio Phosphorous (P32) is used in the treatment of skin diseases.
  5. Radio Cobalt (Co60) and radio – gold (Au198) are used in the treatment of skin cancer.
  6. Radiations are used to sterilize the surgical devices as they can kill the germs and microbes.
  7. Radio cobalt (Co60) and radio – gold (Au198) are used in the treatment of skin cancer.
  8. Radiations are used to sterilize the surgical devices as they can kill the germs and microbes.

Question 5.
Write a note about stellar energy.
Answer:
The stars like our Sun emit a large amount of energy in the form of light and heat. This energy is termed as the stellar energy. Where does this high energy come from? All-stars contain a large amount of hydrogen. The surface temperature of the stars is very high which is sufficient to induce fusion of the hydrogen nuclei.

Fusion reaction that takes place in the cores of the Sun and other stars results in an enormous amount of energy, which is called as ‘stellar energy’. Thus, nuclear fusion or thermonuclear reaction is the source of light and heat energy in the Sun and other stars.

IX. Additional HOT Questions

Question 1.
Why is neutron so effective as bombarding particle?
Answer:
A neutron carries no charge. It easily penetrates even a heavy nucleus without being repelled or attracted by nucleus and electrons. So it serves as an ideal projectile for starting a nuclear reaction.

Question 2.
Is there any difference between electron and a beta particle.
Answer:
Basically, there is no difference between an electron and a beta particle. β particle is the name given to an electron emitted from the nucleus.

Question 3.
Why are the control rods made of cadmium?
Answer:
Cadmium has high cross – section for the absorption of neutrons.

Question 4.
Name two radioactive elements that are not found in observable quantities why is it so?
Answer:
Tritium and Plutonium are two radioactive elements that are not found in observable quantities in the universe.
It is because half-life period of each of two elements is very short compared to the age of the universe.


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