PSEB 12th Class Biology Solutions Chapter 14 Ecosystem

Punjab State Board PSEB 12th Class Biology Book Solutions Chapter 14 Ecosystem Textbook Exercise Questions and Answers.

PSEB Solutions for Class 12 Biology Chapter 14 Ecosystem

PSEB 12th Class Biology Guide Ecosystem Textbook Questions and Answers

Question 1.
Fill in the blanks.
(a) Plants are called as ……………………………. because they fix carbon dioxide.
(b) In an ecosystem dominated by trees, the pyramid (of numbers) is ………………….. type.
(c) In aquatic ecosystems, the limiting factor for the productivity is ………………………. .
(d) Common detritivores in our ecosystem are …………………………. .
(e) The major reservoir of carbon on earth is …………………………….. .
Answer:
(a) autotrophs
(b) inverted
(c) light
(d) earthworms
(e) oceans.

Question 2.
Which one of the following has the largest population in a food chain?
(a) Producers
(b) Primary consumers
(c) Secondary consumers
(d) Decomposers
Answer:
(a) Producers (decomposers can be maximum but they are excluded from the food chain ).

Question 3.
The second trophic level in a lake is
(a) Phytoplankton
(b) Zooplankton
(c) Benthos
(d) Fishes
Answer:
(b) Zooplankton
Zooplankton are primary consumers in aquatic food chains that feed upon phytoplankton. Therefore, they are present at the second trophic level in a lake.

PSEB 12th Class Biology Solutions Chapter 14 Ecosystem

Question 4.
Secondary producers are
(a) Herbivores
(b) Producers
(c) Carnivores
(d) None of the above
Answer:
(d) None of the above
Plants are the only producers. Thus, they are called primary producers. There are no other producers in a food chain.

Question 5.
What is the percentage of photosynthetically active radiation (PAR) in the incident solar radiation?
(a) 100%
(b) 50 %
(c) 1-5%
Answer:
(b) 50%
Out of total incident solar radiation, about fifty percent of it forms photosynthetically active radiation or PAR.

Question 6.
Distinguish between
(a) Grazing food chain and detritus food chain
(b) Production and decomposition
(c) Upright and inverted pyramid
(d) Food chain and Food web
(e) Litter and detritus
(f) Primary and secondary productivity
Answer:
(a) Grazing food chain and detritus food chain

Grazing food chain Detritus food chain
1. In this food chain, energy is derived from the Sun. In this food chain, energy comes from organic matter (or detritus) generated in trophic levels of the grazing food chain.
2. It begins with producers, present at the first trophic level. The plant biomass is then eaten by herbivores, which in turn are consumed by a variety of carnivores. begins with detritus such as dead bodies of animals or fallen leaves, which are then eaten by decomposers or detrivores. These detritivores are in turn consumed
by their predators.
3. This food chain is usually large. It is usually smaller as compared to the grazing food chain.

(b) Production and decomposition

Production Decomposition
1. It is the rare of producing organic matter (food) by producers. It is the process of breaking down of complex organic matter or biomass from the body of dead plants and animals with the help of decomposers into organic raw material such as CO2, H2O and other nutrients.
2. It depends on the photosynthetic cápacity of the producers. It occurs with the help of decomposers.
3. Sunlight is required by plants for primary production. Sunlight is not required for decomposition by clecomposers.

PSEB 12th Class Biology Solutions Chapter 14 Ecosystem

(c) upright and inverted pyramid

Upright pyramid Inverted pyramid
1. The pyramid of energy is upright. always The pyramid of biomass and the pyramid of, numbers can be inverted.
2. In the upright pyramid, the number and biomass of organisms in the producer level of an ecosystem is the highest, which keeps on decreasing at each trophic level in a food chain. In an inverted pyramid, the number and biomass of organisms in the producer level of an ecosystem is the lowest, which keeps on increasing at each tropic level.

(d) Food chain and food web

Food chain Food web
1. The transfer of energy from producers to top consumers through a series of organisms is called food chain. A number of food chain inter-connected with each other forming a web-like pattern is called food web.
2. One organism holds only one position. One organism can hold more than one position.
3. The flow of energy can be easily calculated. The flow of energy is very difficult to calculate.
4. It is always straight and proceed in a progressive straight line. Instead of straight line, it is a series of branching lines.
5. Competition is limited to members of same trophic level. Competition is amongst members of same and different trophic levels.

(e) Litter and detritus

Litter Detritus
l. It is made of dried fallen plant matter. It is freshly deposited organic matter, i. e. remains of plants and animals.
2. It is found above the ground. It is found both above and below the ground.

(f) Primary and secondary productivity

Primary productivity Secondary_productivity
1. Primary productivity is the amount of energy accumulation or amount of biomass produced per unit area over a time period. Secondary productivity is the rate of formation of new organic matter by consumer.
2.  It is of two types, gross primary productivity (GPP) and net primary productivity (NPP). They are
related as: GPP – R = NPP, where R is respiratory losses.
It is also of two types gross secondary productivity (GSP) and net secondary productivity (NSP). They are related as: NSP = GSP – R
Where R is respiratory losses.

Question 7.
Describe the components of an ecosystem.
Answer:
An ecosystem is defined as an interacting unit that includes both the biological community as well as the non-living components of an area. The living and the non-living components of an ecosystem interact amongst themselves and function as a unit, which gets evident during the processes of nutrient cycling, energy flow, decomposition, and productivity. There are many ecosystems such as ponds, forests, grasslands, etc.

The two components of an ecosystem are as follows :
(a) Biotic Component: It is the living component of an ecosystem that includes biotic factors such as producers, consumers, decomposers, etc. Producers include plants and algae. They contain chlorophyll pigment, which helps them carry out the process of photosynthesis in the presence of light.

Thus, they are also called converters or transducers. Consumers or heterotrophs are organisms that are directly (primary consumers) or indirectly (secondary and tertiary consumers) dependent on producers for their food. Decomposers include micro-organisms such as bacteria and fungi. They obtain nutrients by breaking down the remains of dead plants and animals.

(b) Abiotic Component: They are the non-living component of an ecosystem such as light, temperature, water, soil, air, inorganic nutrients, etc.

PSEB 12th Class Biology Solutions Chapter 14 Ecosystem

Question 8.
Define ecological pyramids and describe with examples, pyramids of number and biomass.
Answer:
An ecological pyramid is a graphical representation of various ecological parameters such as the number of individuals present at each trophic level, the amount of energy, or the biomass present at each trophic level.
Ecological pyramids represent producers at the base, while the apex represents the top-level consumers present in the ecosystem.
There are three types of pyramids:

  1. Pyramid of numbers
  2. Pyramid of energy
  3. Pyramid of biomass

1. Pyramid of Numbers: It is a graphical representation of the number of individuals present at each trophic level in a food chain of an ecosystem. The pyramid of numbers can be upright or inverted depending on the number of producers. For example, in a grassland ecosystem, the pyramid of numbers is upright.

In this type of a food chain, the number of producers (plants) is followed by the number of herbivores (mice), which in turn is followed by the number of secondary consumers (snakes) and tertiary carnivores (eagles). Hence, the number of individuals at the producer level will be the maximum, while the number of individuals present at top carnivores will be least.
PSEB 12th Class Biology Solutions Chapter 14 Ecosystem 1
On the other hand, in a parasitic food chain, the pyramid of numbers is inverted. In this type of a food chain, a single tree (producer) provides food to several fruit-eating birds, which in turn support several insect species.

2. Pyramid of Biomass: A pyramid of biomass is a graphical representation of the total amount of living matter present at each trophic level of an ecosystem. It can be upright or inverted. It is upright in grasslands and forest ecosystems as the amount of biomass present at the producer level is higher than at the top carnivore level.
The pyramid of biomass is inverted in a pond ecosystem as the biomass of fishes far exceeds the biomass of zooplankton (upon which they feed).
PSEB 12th Class Biology Solutions Chapter 14 Ecosystem 2

Question 9.
What is primary productivity? Give brief description of factors that affect primary productivity.
Answer:
It is defined as the amount of organic matter or biomass produced by producers per unit area over a period of time. Primary productivity of an ecosystem depends on the variety of environmental factors such as light, temperature, water, precipitation, etc. It also depends on the availability of nutrients and the availability of plants to carry out photosynthesis.

Question 10.
Define decomposition and describe the processes and products of decomposition.
Answer:
Decomposition is the process that involves the breakdown of complex organic matter or biomass from the body of dead plants and animals with the help of decomposers into inorganic raw materials such as carbon dioxide, water, and other nutrients. The various processes involved in decomposition are as follows :
1. Fragmentation: It is the first step in the process of decomposition. It involves the breakdown of detritus into smaller pieces by the action of detritivores such as earthworms.

2. Leaching: It is a process where the water-soluble nutrients go down into the soil layers and get locked as unavailable salts.

3. Catabolism: It is a process in which bacteria and fungi degrade detritus through various enzymes into smaller pieces.

4. Humification: The next step is humification which leads to the formation of a dark-colored colloidal substance called humus, which acts as reservoir of nutrients for plants.

5. Mineralisation: The humus is further degraded by the action of microbes, which finally leads to the release of inorganic nutrients into the soil. This process of releasing inorganic nutrients from the humus is known as mineralization.

Decomposition produces a dark-colored, nutrient-rich substance called humus. Humus finally degrades and releases inorganic raw materials such as C02, water, and other nutrients in the soil.

PSEB 12th Class Biology Solutions Chapter 14 Ecosystem

Question 11.
Give an account of energy flow in an ecosystem.
Answer:
Energy enters an ecosystem from the Sun. Solar radiations pass through the atmosphere and are absorbed by the Earth’s surface. These radiations help plants in carrying out the process of photosynthesis.
Also, they help maintain the Earth’s temperature for the survival of living organisms. Some solar radiations are reflected by the Earth’s surface.

Only 2-10% of solar energy is captured by green plants (producers) during photosynthesis to be converted into food.
The rate at which the biomass is produced by plants during photosynthesis is termed as ‘gross primary productivity.
When these green plants are consumed by herbivores, only 10% of the stored energy from producers is transferred to herbivores. The remaining 90 % of this energy is used by plants for various processes such as respiration, growth, and reproduction. Similarly, only 10% of the energy of herbivores is transferred to carnivores. This is known as ten percent law of energy flow.
PSEB 12th Class Biology Solutions Chapter 14 Ecosystem 3

Question 12.
Write important features of a sedimentary cycle in an ecosystem.
Answer:
Sedimentary cycles have their reservoirs in the Earth’s crust or rocks. Nutrient elements are found in the sediments of the Earth. Elements such as sulphur, phosphorus, potassium, and calcium have sedimentary cycles.
Sedimentary cycles are very slow. They take a long time to complete their circulation and are considered as less perfect cycles. This is because during recycling, nutrient elements may get locked in the reservoir pool, thereby taking a very long time to come out and continue circulation. Thus, it usually goes out of circulation for a long time.

Question 13.
Outline salient features of carbon cycling in an ecosystem.
Answer:
The carbon cycle is an important gaseous cycle which has its reservoir pool in the atmosphere.
All living organisms contain carbon as a major body constituent. Carbon is a fundamental element found in all living forms. All biomolecules such as carbohydrates, lipids, and proteins required for life processes are made of carbon.
Carbon is incorporated into living forms through a fundamental process called ‘photosynthesis’.

Photosynthesis uses sunlight and atmospheric carbon dioxide to produce a carbon compound called ‘glucose’.
This glucose molecule is utilised by other living organisms. Thus, atmospheric carbon is incorporated in living forms.
Now, it is necessary to recycle this absorbed carbon dioxide back into the atmosphere to complete the cycle.

There are various processes by which carbon is recycled back into the atmosphere in the form of carbon dioxide gas.
The process of respiration breaks down glucose molecules to produce carbon dioxide gas. The process of decomposition also releases carbon dioxide from dead bodies of plants and animals into the atmosphere. Combustion of fuels, industrialization, deforestation, volcanic eruptions and forest fires act as other major sources of carbon dioxide.
PSEB 12th Class Biology Solutions Chapter 14 Ecosystem 4

PSEB 12th Class Physics Important Questions Chapter 2 Electrostatic Potential and Capacitance

Punjab State Board PSEB 12th Class Physics Important Questions Chapter 2 Electrostatic Potential and Capacitance Important Questions and Answers.

PSEB 12th Class Physics Important Questions Chapter 2 Electrostatic Potential and Capacitance

Very short answer type questions

Question 1.
A charge ‘q’ is moved from a point A above a dipole of dipole moment ‘p’ to a point B below the dipole in equatorial plane without acceleration. Find the work done in the process.
PSEB 12th Class Physics Important Questions Chapter 2 Electrostatic Potential and Capacitance 1
Answer:
Work done in the process is zero. Because, equatorial plane of a dipole is equipotential surface and work done in moving charge oh equipotential surface is zero.
W = qVAB = q × 0 = 0

Question 2.
A point charge Q is placed at point O as shown in the figure. Is the potential difference (VA – VB) positive, negative or zero if Q is
(i) positive
(ii) negative
PSEB 12th Class Physics Important Questions Chapter 2 Electrostatic Potential and Capacitance 2
Answer:
Let the distance of points A and B from charge Q be rA and rB, respectively.
∴ Potential difference between points A and B
VA – VB = \(\frac{q}{4 \pi \varepsilon_{0}}\left[\frac{1}{r_{A}}-\frac{1}{r_{B}}\right]\)
As rA = OA, rB = rB = OB
and rA < rB, \(\frac{1}{r_{A}}>\frac{1}{r_{B}}\)
There,[latex]\frac{1}{r_{A}}-\frac{1}{r_{B}}[/latex] has positive value.
(VA – VB) depends on the nature of charge q.
(VA – VB) is positive when Q > 0
(VA – VB ) is negative when Q < 0

PSEB 12th Class Physics Important Questions Chapter 2 Electrostatic Potential and Capacitance

Question 3.
A hollow metal sphere of radius 5 cm is charged such that potential on its surface is 10 V. What is the potential at the centre of the sphere?
Answer:
The electric potential at every point inside the charged spherical shell is same and equal to the electric potential on its surface.
The electric potential at the centre of sphere is 10 V.

Question 4.
Can two equipotential surfaces intersect each other? Justify your answer.
No, two equipotential surfaces cannot intersect each other because

  1. Two normals can be drawn at intersecting point on two surfaces which gives two directions of E at the same point which is impossible.
  2. Also two values of potential at the same point is not possible.

Question 5.
Why electrostatic potential is constant throughout the volume of the conductor and has the same value as on its surface?
Since, electric field intensity inside the conductor is zero. So, electrostatic potential is a constant.
But, E = –\(\frac{\Delta V}{\Delta r}\)
∴ E = 0, ΔV = 0
or V2 – V1 =0, V2 – V1
The potential at every point inside the conductor remains same.

Question 6.
In a certain 0.5 cm3 of space, electric potential is found to be 7 V throughout. What is the electric field in this region?
Answer:
Zero, because electric potential is same throughout as
E = \(\frac{d V}{d r}\)

PSEB 12th Class Physics Important Questions Chapter 2 Electrostatic Potential and Capacitance

Question 7.
Distinguish between a dielectric and a conductor.
Answer:
Dielectrics are non-conductors and do not have free electrons at all. While conductor has free electrons in its any volume which makes it able to pass the electricity through it.

Question 8.
The given graph shows the variation of charge q versus potential difference V for two capacitors C1 and C2. Both the capacitors have same plate separation but plate area of C2 is greater than that C1. Which line (A or B) corresponds to C1 and why?
PSEB 12th Class Physics Important Questions Chapter 2 Electrostatic Potential and Capacitance 3
Answer:
Line B corresponds to Q because slope (q/V) of B is less than slope of A.

Question 9.
Do free electrons travel to region of higher potential or lower potential? (NCERT Exemplar)
Answer:
Free electrons would travel to regions of higher potentials as they are negatively charged.

Question 10.
Can there be a potential difference between two adjacent conductors carrying the same charge? (NCERT Exemplar)
Answer:
Yes, if the sizes are different.

PSEB 12th Class Physics Important Questions Chapter 2 Electrostatic Potential and Capacitance

Question 11.
Prove that, if an insulated, uncharged, conductor is placed near a charged conductor and no other conductors are present, the uncharged body must intermediate in potential between that of the charged body and that of infinity. (NCERT Exemplar)
Answer:
Let us take any path from the charged conductor to the uncharged conductor along the direction of electric field. Therefore, the electric potential decrease along this path.
Now, another path from the uncharged conductor to infinity will again continually lower the potential further. This ensures that the uncharged body must be intermediate in potential between that of the charged body and that of infinity.

Question 12.
A test charge q is made to move in the electric field of a point charge Q along two different closed paths [figure]. First path has sections along and perpendicular to lines of electric field.
Second path is a rectangular loop of the same area as the first loop. How does the work done compare in the two cases? (NCERT Exemplar)
PSEB 12th Class Physics Important Questions Chapter 2 Electrostatic Potential and Capacitance 4
Answer:
As electric field is conservative, work done will be zero in both the cases.

Short answer type questions

Question 1.
What is electrostatic shielding? How is this property used in actual practice? Is the potential in the cavity of a charged conductor zero?
Answer:
Whatever be the charge and field configuration outside, any cavity in a conductor remains shielded from outside electric influence. The field inside a conductor is zero. This is known as electrostatic shielding.

  • Sensitive instruments are shielded from outside electrical influences by enclosing them in a hollow conductor.
  • During lightning it is safest to sit inside a car, rather than near a tree. The metallic body of a car becomes an electrostatic shielding from lightening.

Potential inside the cavity is not zero. Potential is constant.

Question 2.
Two capacitors of unknown capacitances C1 and C2 are connected first in series and then in parallel across a battery of 100 V. If the energy stored in the two combinations is 0.045 J and 0.25 J respectively, determine the value of C1 and C2. Also calculate the charge on each capacitor in parallel combination.
Answer:
Energy stored in a capacitor,
PSEB 12th Class Physics Important Questions Chapter 2 Electrostatic Potential and Capacitance 5
Q1 = C1V = 38.2 × 10-6 × 100 = 38.2 × 10-4 C
Q2 = C2V = 11.8 × 10-6 × 100 = 11.8 × 10-4C

PSEB 12th Class Physics Important Questions Chapter 2 Electrostatic Potential and Capacitance

Question 3.
Calculate the equivalent capacitance between points A and B in the circuit below. If a battery of 10 V is connected across A and B, calculate the charge drawn from the battery by the circuit.
PSEB 12th Class Physics Important Questions Chapter 2 Electrostatic Potential and Capacitance 6
Answer:
PSEB 12th Class Physics Important Questions Chapter 2 Electrostatic Potential and Capacitance 7
PSEB 12th Class Physics Important Questions Chapter 2 Electrostatic Potential and Capacitance 8

Question 4.
A parallel plate capacitor is charged by a battery to a potential. The battery is disconnected and a dielectric slab is inserted to completely fill the space between the plates.
How will
(a) its capacitance
(b) electric field between the plates and
(c) energy stored in the capacitor be affected? Justify your answer giving necessary mathematical expression for each case.
Answer:
On introduction of dielectric slab in a isolated charged capacitor.
(a) The capacitance (C’) becomes K times of original capacitor as
C = \(\frac{\varepsilon_{0} A}{d}\)
and C ‘ = \(\frac{K \varepsilon_{0} A}{d}\)

(b) The total charge on the capacitor remains conserved on introduction of dielectric slab. Also, the capacitance of capacitor increases to K times of original values.
∴ CV = C’V’
CV = (KC)V’
⇒ V’ = \(\frac{V}{K}\)
∴ New electrical field,
E’ = \(\frac{V^{\prime}}{d}=\left(\frac{V / K}{d}\right)=\left(\frac{V}{d}\right) \frac{1}{K}=\frac{E}{K}\)
∴ On introduction of dielectric medium new electric field E’ becomes \(\) times of its original value.
(c) Energy stored initially,
U = \(\frac{q^{2}}{2 C}\)
Energy stored later,
U’ = \(\frac{q^{2}}{2(K C)}\) [< C’ = KG]
where, K = dielectric constant of medium
⇒ U’ = \(\frac{1}{K}\left(\frac{q^{2}}{2 C}\right)\)
⇒ U’ = \(\frac{1}{K}\) × U
The energy stored in the capacitor decreases and becomes \(\frac{1}{K}\) times of original energy.

PSEB 12th Class Physics Important Questions Chapter 2 Electrostatic Potential and Capacitance

Question 5.
Two identical parallel plate capacitors A and B are connected to a battery of V volts with the switch S closed. The switch is now opened and the free space between the plates of the capacitors is filled with a dielectric of dielectric constant K. Find the ratio of the total electrostatic energy stored in both capacitors before and after the introduction of the dielectric.
PSEB 12th Class Physics Important Questions Chapter 2 Electrostatic Potential and Capacitance 9
Answer:
Let C be the capacitance of each capacitor.
With switch S closed, the two capacitor are in parallel.
∴ Equivalent capacitance is 2 C.
∴ Energy stored = \(\frac{1}{2}\)(2C)V2
U1 = CV2 …………… (1)
PSEB 12th Class Physics Important Questions Chapter 2 Electrostatic Potential and Capacitance 10
Now, when switch is opened and then free space of capacitors are filled with dielectric, the capacitance of each capacitor will be KC. For capacitor B, the charge will remain as before and for A, the potential difference will remain same.

Charge on each capacitor in the previous case will be CV.
∴ Energy stored in capacitor A in circuit case is
PSEB 12th Class Physics Important Questions Chapter 2 Electrostatic Potential and Capacitance 11
PSEB 12th Class Physics Important Questions Chapter 2 Electrostatic Potential and Capacitance 12

Question 6.
Consider two conducting spheres of radii R1 and R2 with R1 > R2. If the two are at the same potential, the larger sphere has more charge than the smaller sphere. State whether the charge density of the smaller sphere is more or less than that of the larger one. (NCERT Exemplar)
Answer:
PSEB 12th Class Physics Important Questions Chapter 2 Electrostatic Potential and Capacitance 13

PSEB 12th Class Physics Important Questions Chapter 2 Electrostatic Potential and Capacitance

Question 7.
Calculate potential on the axis of a ring due to charge Q uniformly distributed along the ring of radius R.
(NCERT Exemplar)
Answer:
Let us take point P to be at a distance x from the centre of the ring, as shown in figure. The charge element dq is at a distance x from point P. Therefore, V can be written as
PSEB 12th Class Physics Important Questions Chapter 2 Electrostatic Potential and Capacitance 14
PSEB 12th Class Physics Important Questions Chapter 2 Electrostatic Potential and Capacitance 15

Long answer type questions

Question 1.
(i) Explain using suitable diagrams, the difference in the behaviour of a
(a) conductor and
(b) dielectric in the presence of external electric field. Define the terms polarisation of a dielectric and write its relation with susceptibility.

(ii) A thin metallic spherical shell of radius R carries a charge Q on its surface. A point charge \(\frac{Q}{2}\) is placed at its centre C and an other charge +2Q is placed outside the shell at a distance x from the centre as shown in the figure. Find (a) the force on the charge at the centre of shell and at the point A, (b) the electric flux through the shell.
PSEB 12th Class Physics Important Questions Chapter 2 Electrostatic Potential and Capacitance 16
Answer:
(i) When a capacitor is placed in an external electric field, the free charges present inside the conductor redistribute themselves in such a manner that the electric field due to induced charges opposes the external field within the conductor. This happens until a static- situation is achieved, i.e., when the two fields cancels each other and the net electrostatic field in the conductor becomes zero.

In contrast to conductors, dielectrics are non- conducting substances, i.e., they have no charge carriers. Thus, in a dielectric, free movement of charges is not possible. It turns out that the external field induces dipole moment by stretching molecules of the dielectric. The collective effect of all the molecular dipole moments is the net charge on the surface of the dielectric which produces a field that opposes the external field. However, the opposing field is so induced does not exactly cancel the external field, ft only reduces it. The extent of the effect depends on the nature of dielectric.

Both polar and non-polar dielectric develop net dipole moment in the presence of an external field. The dipole moment per unit volume is called polarisation and is denoted by P for linear isotropic dielectrics.
P = %E

(a) At point C, inside the shell.
The electric field inside a spherical shell is zero. Thus, the force experienced by charge at the centre C will also be zero.
PSEB 12th Class Physics Important Questions Chapter 2 Electrostatic Potential and Capacitance 17

PSEB 12th Class Physics Important Questions Chapter 2 Electrostatic Potential and Capacitance

Question 2.
(a) Find the ratio of the potential differences that must he applied across the parallel and series combination of two capacitors C 1and C2 with their capacitances in the ratio 1:2 so that the energy stored in the two cases becomes the same.

(b) Show that the potential energy of a dipole making angle θ with the direction of the field is given by u(θ) = – p \(\). Hence find out the amount of work done in rotating it from the position of unstable equilibrium to the stable equilibrium.
PSEB 12th Class Physics Important Questions Chapter 2 Electrostatic Potential and Capacitance 18
PSEB 12th Class Physics Important Questions Chapter 2 Electrostatic Potential and Capacitance 19

(b) As charges +q and -q traverse equal distance under equal and opposite forces; therefore, net work done in bringing the dipole in the region of electric field perpendicular to field-direction will be zero, i.e.,W1 = 0.
PSEB 12th Class Physics Important Questions Chapter 2 Electrostatic Potential and Capacitance 20
Now, the dipole is rotated and brought to orientation making an angle θ with the field direction (i.e., θ0 = 90° and θ1 = θ°), therefore, work done
W2 = pE (cosθ0 – cosθ 1)
= pE (cos 90° – cos θ) = -pE cos θ

∴Total work done in bringing the electric dipole from infinity, i.e., electric potential energy of electric dipole. Thus, work done by external torque in rotating a dipole in uniform electric field is stored as the potential energy of the system.

U = W1 + W2 = 0 – pEcosθ
= -pE cosθ
In vector form
U = – \(\vec{p} \cdot \vec{E}\)
For rotating dipole from position of unstable equilibrium (θ0 =180°) to the stable equilibrium (θ = 0°)
∴W = pE (cos 180° – cos0°)
= pE (-1 -1) = -2 pE

PSEB 11th Class Biology Solutions Chapter 11 Transport in Plants

Punjab State Board PSEB 11th Class Biology Book Solutions Chapter 11 Transport in Plants Textbook Exercise Questions and Answers.

PSEB Solutions for Class 11 Biology Chapter 11 Transport in Plants

PSEB 11th Class Biology Guide Transport in Plants Textbook Questions and Answers

Question 1.
What are the factors affecting the rate of diffusion?
Answer:
Factors affecting the rate of diffusion are as follows:

  • Gradient of concentration
  • Permeability of membrane
  • Temperature
  • Pressure

Question 2.
What are porins? What role do they play in diffusion?
Answer:
The porins are proteins that form huge pores in the outer membranes of the plastids, mitochondria and some bacteria allowing molecules up to the size of small proteins to pass through. Porins facilitate diffusion.

PSEB 11th Class Biology Solutions Chapter 11 Transport in Plants

Question 3.
Describe the role played by protein pumps during active transport in plants.
Answer:
Protein pumps use energy to carry substances across the cell membrane. These pumps can transport substances from a low concentration to a high concentration (‘uphill’ transport). Transport rate reaches a maximum when all the protein transporters are being used or are saturated. Like enzymes, the carrier protein is very specific in what it carries across the membrane. These proteins are sensitive to inhibitors that react with protein side chains.

Question 4.
Explain why pure water has the maximum water potential.
Answer:
Water molecules possess kinetic energy. In liquid and gaseous form, they are in random motion that is both rapid and constant. The greater the concentration of water in a system, the greater is its kinetic energy or ‘water potential’. Hence, pure water will have the greatest water potential. Water potential is denoted by the Greek symbol psi or \p and is, expressed in pressure units such as pascals (Pa).

Question 5.
Differentiate between the following:
(a) Diffusion and Osmosis
(b) Transpiration and Evaporation
(c) Osmotic pressure and Osmotic Potential
(d) Imbibition and Diffusion
(e) Apoplast and Symplast Pathways of Movement of Water in Plants
(f) Guttation and Transpiration

(a) Differences between Diffusion and Osmosis

Diffusion Osmosis
1. It is a movement of molecules from high concentration to low concentration. It is a movement of molecules from high concentration to low concentration
2. It does not require any driving force. It occurs in response to a driving force.

(b) Differences between Transpiration and Evaporation

Transpiration Evaporation
1. It is the loss of water through the aerial parts of plants. It is the loss of water from free surface of water.
2. It occurs in living tissues. It occurs in non-living surfaces.
3. It is both physical and physiological process. It is only a physical process, controlled by environmental factors.

PSEB 11th Class Biology Solutions Chapter 11 Transport in Plants

(c) Differences between Osmotic Pressure and Osmotic Potential

Osmotic Pressure Osmotic Potential
1. It is the pressure required to stop the movement of water molecules through a semipermeable

membrane.

It is the amount by which water potential is reduced by the presence of solute.
2. Osmotic pressure is the positive pressure. Osmotic potential is negative.

(d) Differences between Imbibition and Diffusion

Imbibition Diffusion
It is a special type of diffusion, where water is absorbed by solids-colloids causing them to increase in volume. For example, absorption of water by dry seeds and dry wood. In diffusion, molecules move in a random fashion. It is not dependent on a living system.

(e) Differences between Apoplast and Symplast Pathways of Movement of Water in Plants

Apoplast Symplast
1. It is the system of adjacent cell walls that is continuous throughout the plant except casparian strips of the endodermis of the roots. It is the system of interconnected protoplast.
2. Water moves through the intercellular spaces and the walls of cells. Water travels through the cytoplasm
3. Movement does not involve crossing the cell membrane. Water has to move in cells through the cell membrane.

(f) Differences between Guttation and Transpiration

Guttation Transpiration
1. It occurs through hydathodes, present at the vein ends. It occurs through general surface stomata and lenticles.
2. It occurs in leaves only. It can occur through all aerial parts.
3. It does not occur in deficient water conditions and never leads to wilting. It can occur in water deficient conditions leading to wilting.
4. It is regulated by humidity, temperature and presence of water in soil. It is regulated by a number of external and internal factors such as relative humidity, temperature, opening and closing of stomata, etc.

PSEB 11th Class Biology Solutions Chapter 11 Transport in Plants

Question 6.
Briefly describe water potential. What are the factors affecting it?
Answer:
Water potential is the potential energy of water relative to pure free water (e.g., deionised water). It quantifies the tendency of water to move from one area to another due to osmosis, gravity, mecanical pressure or matrix effects including surface tension. Water potential is measured in units of pressure and is commonly represented by the Greek letter (psi). This concept has proved especially useful in understanding water movement within plants, animals and soil.

Water potential of a cell is affected by both solute and pressure potential. The relationship between them is as follows:
Ψw = Ψs + Ψp

Question 7.
What happens when a pressure greater than the atmospheric pressure is applied to pure water or a solution?
Answer:
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 be build up in a plant system when water enters a plant cell due to diffusion causing a pressure build up against the cell wall. It makes the cell turgid, this increases the pressure potential. Pressure potential is usually positive. It is denoted by Ψs.

Question 8.
(a) With the help of well-labelled diagrams, describe the process of plasmolysis in plants, giving appropriate examples.
(b) Explain what will happen to a plant cell if it is kept in a solution having higher water potential.
Answer:
(a) Plasmolysis occurs when water moves out of the cell and the cell membrane of a plant cell shrinks away from its cell wall. This occurs when the cell is kept in a solution that is hypertonic (has more solutes) to the protoplasm. Water moves out from the cell through diffusion and causes the protoplasm to shrink away from the walls. In such situation, cell becomes plasmolysed.

When the cell is placed in an isotonic solution. There is not flow of water towards the inside or outside. If the external solution balances the osmotic pressure of the cytoplasm, it is said to be isotonic. When the water flow into the cell and out of the cells are in equilibrium the cell is called flaccid.
PSEB 11th Class Biology Solutions Chapter 11 Transport in Plants 1
(b) When the plant cell is kept in a solution having high water potential (hypotonic solution or dilute solution as compared to cytoplasm), water diffuses into the cell causing the cytoplasm to build up a pressure against the wall, called turgor pressure. The pressure exerted by the protoplasts due to entry of water against the rigid walls is called pressure potential (Ψp). Because of the rigidity of the cell wall, the cell does not rupture. This turgor pressure is ultimately responsible for enlargement of cells.

PSEB 11th Class Biology Solutions Chapter 11 Transport in Plants

Question 9.
How is the mycorrhizal association helpful in absorption of water and minerals in plants?
Answer:
A mycorrhiza is a symbiotic association of a fungus with a root system. The fungal filaments form a network around the young root or they penetrate the root cells. The hyphae have a very large surface area that absorb mineral ions and water from the soil from a much larger volume of soil that perhaps a root cannot do. The fungus provides minerals and water to the roots, in turn the roots provide sugars and N-containing compounds to the mycorrhizae.

Question 10.
What role does root pressure play in water movement in plants?
Answer:
Root pressure can provide only modest push during water transport in plants. The main role of root pressure is in re-establishing the continuous chain of water molecules in the xylem. The continuous chain often breaks due to enormous tension created by transpiration pull.

Question 11.
Describe transpiration pull model of water transport in plants. What are the factors influencing transpiration? How is it useful to plants?
Answer:
Transpiration occurs mainly through the stomata in the leaves. As water evaporates through the stomata, since the thin film of water over the cells is continuous, it results in pulling of water, molecule by molecule, into the leaf from the xylem. Also, because of lower concentration of water vapour in the atmosphere as compared to the substomatal cavity and intercellular spaces, water diffuses into the surrounding air. This creates a transpiration pull.

Factors Affecting Transpiration: Temperature, light, humidity and wind speed.
Importance of Transpiration: Transport of liquids and minerals is facilitated because of transpiration.

Question 12.
Discuss the factors responsible for ascent of xylem sap in plants.
Answer:
The transpiration driven ascent of xylem sap depends mainly on the following physical properties of water:

  • Cohesion: Mutual attraction between water molecules.
  • Adhesion: Attraction of water molecules to polar surfaces (such as the surface of tracheary elements).
  • Surface Tension: Water molecules are attracted to each other in the liquid phase more than to water in the gas phase.

These properties give water high tensile strength, i.e., an ability to resist a pulling force and high capillarity, i.e., the ability to rise in thin tubes. In plants, capillarity is aided by the small diameter of the tracheary elements, the tracheids and vessel elements.

PSEB 11th Class Biology Solutions Chapter 11 Transport in Plants

Question 13.
What essential role does the root endodermis play during mineral absorption in plants?
Answer:
The endodermis of roots have many transport proteins embedded in their plasma membrane. They let some solutes cross the membrane but not all. Transport proteins in endodermis cells enable plant cells lo adjust the quantity and types of solutes to be absorbed from the soil. It regulates the quantity and type of minerals and ions, that reach the xylem tissue of plants.

Question 14.
Explain why xylem transport is unidirectional and phloem transport bi-directional?
Answer:
The source sink (food making tissue-tissue which stores food) relationship is variable in plants so, the direction of movement in the phloem can be upwards downwards, i.e., bi-directional. It is opposite to xylem, where the movement is always unidirectional. Hence, unlike one-way flow of water in transpiration, food in phloem sap can be transported in any required direction so long there is a source of sugar and a sink is able to use, store or remove the sugar. Here, in case of unidirectional flow in xylem tissue, it is important to note that root endodermis because of the layer of suberin has the ability to actively transport ions in one direction only.

Question 15.
Explain pressure flow hypothesis of translocation of sugars in plants.
Answer:
The Pressure Flow or Mass Flow Hypothesis: The accepted mechanism used for the translocation of sugars from source to sink is called the pressure flow hypothesis. As glucose is prepared at the source (by photosynthesis) it is converted to sucrose (a disaccharide). The sugar is then moved in the form of sucrose into the companion cells and then into the living phloem sieve tube cells by active transport.

As osmotic pressure builds up the phloem sap will move to areas of lower pressure. At the sink, osmotic pressure must be reduced. Again active transport is necessary to move the sucrose out of the phloem sap and into the cells which will use the sugar converting it into energy, starch or cellulose. As sugars are removed, the osmotic pressure decreases and water moves out of the phloem.

Hydrostatic pressure in the phloem sieve tube increases, pressure flow begins and the sap moves through the phloem. Meanwhile, at the sink, incoming sugars are actively transported out of the phloem and removed as complex carbohydrates. The loss of solute produces a high water potential in the phloem and water passes out, returning eventually to xylem.

PSEB 11th Class Biology Solutions Chapter 11 Transport in Plants

Question 16.
What causes the opening and clog” T of guard cells of stomata during transpiration?
Answer:
The immediate cause of the opening or closing-of the stomata is a change in the turgidity of the guard cells. The inner wall of each guard cell, towards the pore or stomatal, aperture, is thick and elastic. When, turgidity increases within the two guard cells flanking each stomatal aperture or pore, the thin outer walls bulge out and force the inner walls into a crescent shape. The opening of the stoma is also aided due to the orientation of the microfibrils in the cell walls of the guard cells. Cellulose microfibrils are oriented radially rather than longitudinally making it easier for the stoma to open. When the guard cells lose turgor, due to water loss (or water stress) the elastic inner walls regain their original shape, the guard cells become flaccid and the stoma closes.

PSEB 12th Class Biology Important Questions Chapter 4 Reproductive Health

Punjab State Board PSEB 12th Class Biology Important Questions Chapter 4 Reproductive Health Important Questions and Answers.

PSEB 12th Class Biology Important Questions Chapter 4 Reproductive Health

Very short answer type questions

Question 1.
Reproductive health refers only to healthy reproductive functions. Comment. [NCERT Exemplar]
Answer:
Reproductive health refers to the total well-being in all aspects of reproduction, i.e., physical, behavioural, psychological and social.

Question 2.
The present population growth rate in India is alarming. Suggest ways to check it. [NCERT Exemplar]
Answer:

  • By increasing marriageable age.
  • By promoting use of birth control measure.
  • By educating people about consequences of un- controlled population growth.

Question 3.
Why do intensely lactating mothers not generally conceive? [NCERT Exemplar]
Answer:
Due to suppression of gonadotropins, ovulation and menstrual cycle do not take place.

PSEB 12th Class Biology Important Questions Chapter 4 Reproductive Health

Question 4.
Name an IUD that you would recommend to promote the cervix hostility to sperms.
Answer:
The hormone releasing IUD’s, e.g. progestasert, LNG-20 are , recommended to promote the cervix hostility to sperms.

Question 5.
Mention any tvgo events that are exhibited by the intake of oral contraceptive pills to prevent pregnancy in humans.
Answer:
Two events that are exhibited by the intake of oral contraceptive pills to prevent pregnancy in humans are ovulation and implantation.

Question 6.
Why is tubectomy considered a contraceptive method?
Answer:
Tubectomy involves cutting a piece of the fallopian tube and tying its ends. This way, the sperms are not able to reach the egg and it acts as a contraceptive method.

Question 7.
Mention the primary aim of the ‘Assisted Reproductive Technology’ (ART) programme. [NCERT Exemplar]
Answer:
‘Assisted Reproductive Technology’ (ART) is the collection of certain I1 special techniques. The primary aim of the ART programmes is to assist infertile couples to have children through certain special techniques (like ZIFT, IUT, GIFT, ICSI, AI, etc.), when corrective treatment for infertility problems is not possible.

Question 8.
Expand GIFT and ICSI.
Answer:
GIFT: Gamete Intra Fallopian Transfer.
ICSI: Intra Cytoplasmic Sperm Injection.

PSEB 12th Class Biology Important Questions Chapter 4 Reproductive Health

Short answer type questions

Question 1.
Comment on the RCH programme of the government to improve the reproductive health of the people. [NCERT Exemplar]
Answer:
The basic aims of the RCH programmes are creating public awareness f regarding reproduction-related aspects and providing facilities to build up a healthy society with added emphasis on the health of mother and child.

Question 2.
(a) List any four characteristics of an ideal contraceptive.
(b) Name two intrauterine contraceptive devices that effect the motility of sperms.
Answer:
(a) An ideal contraceptive should be:

  • user friendly,
  • easily available,
  • effective,
  • reversible with no or least side-effects,
  • non-interfering with the sexual drive/desire and/or the sexual act of the user, (any four)

(b) Intrauterine Devices (IUDs): Lippes loop, copper releasing IUDs (Cu-T, Cu-7, multiload 375).

Question 3.
Name two hormones that are constituents of contraceptive pills. Why do they have high and effective contraceptive value? Name a commonly prescribed non-steroidal oral pill.
Answer:
Hormonal preparations (progestogens or progesterons and estrogens) are highly effective contraceptive because they inhibit ovulation and implantation, e.g., Mala-D, Mala-L. Morning after pills are used as emergency contraceptives, to avoid pregnancy due to rape or casual unprotected intercourse.

“Saheli”, a new oral pill is used “once-a-week” a non-steroidal preparation With very less side effects and high contraceptive value developed by CDRI in Lucknow, India.

PSEB 12th Class Biology Important Questions Chapter 4 Reproductive Health

Question 4.
Name and explain the surgical method advised to human males and females as a means of birth control. Mention its one advantage and one disadvantage.
Answer:
In the males-Vasectomy. In this method, a small part of the vas deferens is removed or tied up through a small incision on the scrotum. In the females-Tubectomy. In this method, a small part of the fallopian tube is removed or tied up through a small incision in the abdomen or through vagina.
Advantage: Highly effective
Disadvantage: Reversibility is very poor

Question 5.
A childless couple has agreed for a test tube baby programme. List only the basic steps of the procedure would involve to conceive the baby.
Answer:
IVF is the technique used in the case of childless couple. In IVF or In Vitro fertilisation, fertilisation is carried out in a glass container outside the body of the mother. Purified semen is poured over the mature retrieved oocytes. The fertilised eggs are separated and allowed to remain in culture medium, maintained in incubator for 48-72 hours. During the period, the fertilised egg undergoes cleavage and reach 4-8 celled stage, 2-3 fertilised 4-8 celled embryos are transferred or implanted .into the uterus of the recipient surrogate mother, for further development up to delivery.
Note: Excess fertilised oocytes are cryopreserved for use in case of implantation failure.

Question 6.
Why is medical termination of pregnancy (MTP) carried out?
Answer:
MTP is carried out to get rid of unwanted pregnancies. It is also essential when the foetus is suffering from an incurable disease or when continuation of the pregnancy could be harmful or even fatal to the mother and/or foetus.

PSEB 12th Class Biology Important Questions Chapter 4 Reproductive Health

Long answer type questions

Question 1.
Your school has been selected by the Department of Education to organise and host an interschool seminar on “Reproductive Health-Problems and Practices.” However, many parents are reluctant to permit their wards to attend it. Their argument is that the topic is “too embarrassing.”
Put forth four arguments with appropriate reasons and explanation to justify the topic to be very essential and timely.
Answer:
The selection of your school, to host a seminar on “Reproductive Health-Problems and Practices” is a matter great pride for the prestigious institute. The students will have an opportunity to listen to the diverse ideas, suggested by the learned speakers.

It is sad that many parents are reluctant to permit their wards to attend the seminar assuming that the topic is too embarrassing. The following arguments will justify the relevance of the topic in the present time:
(i) Introduction of sex education and the proper information about reproductive organs, adolescence and related changes will protect the youth from social evils like sex-abuse and sex-related crimes.

(ii) Right information about safe, healthy and hygienic sexual practices, sexually transmitted diseases (STDs) would help the people to lead a reproductive healthy life.

(iii) Decline in sex-ratio is a matter of great concern. The Govt, has put a I statutory ban on female foeticide. Both girls and boys have equal rights and equal opportunities in all spheres of life.

(iv) India is facing another problem of population explosion. It is eating, almost all the benefits of overall development. The benefits of development are not trickling down to the poor at lower strata. There is need for family planning, socially conscious healthy families of desired size i.e., Hum Do Humare Do.

PSEB 12th Class Biology Solutions Chapter 4 Reproductive Health

Punjab State Board PSEB 12th Class Biology Book Solutions Chapter 4 Reproductive Health Textbook Exercise Questions and Answers.

PSEB Solutions for Class 12 Biology Chapter 4 Reproductive Healths

PSEB 12th Class Biology Guide Reproductive Health Textbook Questions and Answers

Question 1.
What do you think is the significance of reproductive health in a society?
Answer:
Reproductive health is the total well being in all aspects of reproduction. It includes physical, emotional, behavioural, and social well being. Sexually transmitted diseases such as AIDS, gonorrhoea, etc. are transferred from one individual to another through sexual contact. It can also lead to unwanted pregnancies.

Hence, it is necessary to create awareness among people, especially the youth, regarding various reproduction related aspects as the young individuals are the future of the country and they are most susceptible of acquiring sexually transmitted diseases. Creating awareness about the available birth control methods, sexually transmitted diseases and their preventive measures, and gender equality will help in bringing up a socially conscious healthy family. Spreading. awareness regarding uncontrolled population growth and social evils among young individuals will help in building up a reproductively healthy society.

Question 2.
Suggest the aspects of reproductive health which need to be given special attention in the present scenario.
Answer:
Reproductive health is the total well being in all aspects of reproduction. The aspects which have to be given special attention in the present scenario are as follows:
1. Counselling and creating awareness among people, especially the youth, about various aspects of reproductive health, such as sexually transmitted diseases, available contraceptive methods, case of pregnant mothers, adolescence, etc.

2. Providing support and facilities such as medical assistance to people during pregnancy, STDs, abortions, contraceptives, infertility, etc. for building a reproductively healthy society.

PSEB 12th Class Biology Solutions Chapter 4 Reproductive Health

Question 3.
Is sex education necessary in schools? Why?
Answer:
Yes, introduction of sex education in schools is necessary. It would provide right information to young individuals at the right time about various aspects of reproductive health such as reproductive organs, puberty, and adolescence related changes, safe sexual practices, sexually transmitted diseases, etc.

The young individual or adolescents are more susceptible in acquiring various sexually transmitted diseases. Hence, providing information to them at the right time would help them to lead a reproductively healthy life and also protect them from the myths and misconceptions about various sex related issues.

Question 4.
Do you think that reproductive health in our country has improved in the past 50 years? If yes, mention some such areas of improvement.
Answer:
Yes, the reproductive health has tremendously improved in India in the last 50 years. The areas of improvement are as follows:

  1. Massive child immunisation programme, which has lead to a decrease in the infant mortality rate.
  2. Maternal and infant mortality rate, which has been decreased drastically due to better post natal care.
  3. Family planning, which has motivated people to have smaller families.
  4. Use of contraceptive, which has resulted in a decrease in the rate of sexually transmitted diseases and unwanted pregnancies.

Question 5.
What are the suggested reasons for population explosion?
Answer:
The human population is increasing day by day, leading to population explosion. It is because of the following two major reasons:
(a) Decreased death rate
(b) Increased birth rate and longevity
The death rate has decreased in the past 50 years. The factor leading to decreased death rate and increased birth rate are control of diseases, awareness and spread of education, improvement in medical facilities, ensured food supply in emergency situation, etc. All this has resulted in an increase in the longevity of an individual.

Question 6.
Is the use of contraceptives justified? Give reasons.
Answer:
Yes, the use of contraceptives is absolutely justified. The human population is increasing tremendously. Therefore, to regulate the population growth by regulating reproduction has become a necessary demand in the present times. Various contraceptive devices have been devised to reduce unwanted pregnancies, which help in bringing down the increased birth rate and hence, in checking population explosion.

PSEB 12th Class Biology Solutions Chapter 4 Reproductive Health

Question 7.
Removal of gonads cannot be considered as a contraceptive option. Why?
Answer:
Contraceptive devices are used to prevent unwanted pregnancy and to prevent the spreading of STDs. There are many methods, such as natural, barrier, oral, and surgical methods, that prevent unwanted pregnancy. However, the complete removal of gonads cannot be a contraceptive option because it will lead to infertility and unavailability of certain hormones that are required for normal functioning of accessory reproductive parts. Therefore, only those contraceptive methods can be used that prevent the chances of fertilisation rather than making the person infertile forever.

Question 8.
Amniocentesis for sex determination is banned in our country. Is this ban necessary? Comment.
Answer:
Amniocentesis is a pre-natal diagnostic technique that is used to determine the sex and metabolic disorders of the developing foetus in the mother’s uterus through the observation of the chromosomal patterns. This method was developed so as to determine any kind of genetic disorder present in the foetus. However, unfortunately, this technique is being misused to detect the sex of the child before birth and the female foetus is then aborted. Thus, to prevent the increasing female foeticides, it is necessary to ban the usage of amniocentesis technique for determining the sex of a child.

PSEB 12th Class Biology Solutions Chapter 4 Reproductive Health

Question 9.
Suggest some methods to assist infertile couples to have children.
Answer:
Infertility is the inability of a couple to produce a baby even after unprotected intercourse. It might be due to abnormalities present in either male or female, or might be even both the partners. The techniques used to assist infertile couples to have children are as follows:
(a) Test Tube Babies: This involves in-vitro fertilisation where the sperms meet the egg outside the body of a female. The zygote, hence produced, is then transferred in the uterus or fallopian tube of a normal female. The babies produced from this method are known as test tube babies.

(b) Gamete Intra Fallopian Transfer (GIFT): It is a technique that involves the transfer of gamete (ovum) from a donor into the fallopian tube of the recipient female who is unable to produce eggs, but has the ability to conceive and can provide right conditions for the development of an embryo.

(c) Intra Cytoplasmic Sperm Injection (ICSI): It is a method of injecting sperm directly into the ovum to form an embryo in laboratory.

(d) Artificial Insemination: Artificial insemination is a method of transferring semen (sperm) from a healthy male donor into the vagina or uterus of the recipient female. It is employed when the male partner is not able to inseminate the female or has low sperm counts.

Question 10.
What are the measures one has to take to prevent from contracting STDs?
Answer:
Sexually transmitted diseases (STDs) get transferred from one individual to the other through sexual contact. Adolescents and young adults are at the greatest risk of acquiring these sexually transmitted diseases. Hence, creating awareness among the adolescents regarding its after-effects can prevent them from contracting STDs. The use of contraceptives, such as condoms, etc. while intercourse, can prevent the transfer of these diseases. Also, sex with unknown partners or multiple partners should be avoided as they may have such diseases. Specialists should be consulted immediately in case of doubt so as to assure early detection and cure of the disease.

Question 11.
State True/False with explanation.
(a) Abortions could happen spontaneously too. (True/False)
(b) Infertility is defined as the inability to produce a viable offspring and is always due to abnormalities/defects in the female partner. (True/False)
(c) Complete lactation could help as a natural method of contraception. (True/False)
(d) Creating awareness about sex related aspects is an effective method to improve reproductive health of the people. (True/False)
Answer:
(a) False
Abortion is term given for medical termination of pregnancy.

(b) False
Infertility is defined as the inability of the couple to produce baby even after unprotected coitus. It might occur due to abnormalities/defects in either male or female or both.

(c) False
Complete lactation or lactational amenorrhea is a natural method of contraception. Flowever, it is limited till lactation period, which continues till six months after parturition.

(d) True.

PSEB 12th Class Biology Solutions Chapter 4 Reproductive Health

Question 12.
Correct the following statements:
(a) Surgical methods of contraception prevent gamete formation.
(b) All sexually transmitted diseases are completely curable.
(c) Oral pills are very popular contraceptives among the rural women.
(d) In E. T. techniques, embryos are always transferred into the uterus.
Answer:
(a) Surgical methods of contraception prevent the flow of gamete during intercourse.
(b) Some of the sexually transmitted diseases are curable if they are detected early and treated properly. AIDS is still an incurable disease.
(c) Oral pills are very popular contraceptives among urban women.
(d) In embryo transfer technique, 8 celled embryos are transferred into the fallopian tube while more than 8 celled embryos are transferred into the uterus.

PSEB 12th Class Biology Important Questions Chapter 3 Human Reproduction

Punjab State Board PSEB 12th Class Biology Important Questions Chapter 3 Human Reproduction Important Questions and Answers.

PSEB 12th Class Biology Important Questions Chapter 3 Human Reproduction

Very short answer type questions

Question 1.
Why are the human testes located outside the abdominal cavity? Name the pouch in which they are present.
Answer:
The human testes need lower temperature, 2-2.5°C less than the body temperature, for the formation of sperms which is provided outside the body. Testes are present in scrotal sac or scrotum.

Question 2.
Write the location and function of the sertoli cells in humans.
Answer:
Sertoli cells are present in seminiferous tubules. They provide nutrition to the germ cells or sperms.

Question 3.
Mention the location and the function of leydig cells in humans.
Answer:
Leydig cells are present in seminiferous tubules. They synthesise and secrete androgens.

PSEB 12th Class Biology Important Questions Chapter 3 Human Reproduction

Question 4.
The path of sperm transport is given below. Provide the missing steps in blank boxes. [NCERT Exemplar]
PSEB 12th Class Biology Important Questions Chapter 3 Human Reproduction 1
Answer:
Vasa efferentia, Vas deferens.

Question 5.
Female reproductive organs and associated functions are given below in column A and B. Fill in the blank boxes. [NCERT Exemplar]

Column A Column B
Ovaries Ovulation
Oviduct A
B Pregnancy
Vagina Birth

Answer:
A – Fertilisation
B – Uterus

Question 6.
What is the role of cervix of the human female system in reproduction? [NCERT Exemplar]
Answer:
Cervix helps in regulating the passage of sperms into the uterus and forms the birth canal to facilitate parturition.

Question 7.
When do the oogenesis and the spermatogenesis initiate in human females and males respectively?
Answer:
In females; oogenesis initiates during foetal life. In males, spermatogenesis begins at the time of puberty.

PSEB 12th Class Biology Important Questions Chapter 3 Human Reproduction

Question 8.
Mention the importance of LH surge during menstrual cycle. [NCERT Exemplar]
Answer:
LH surge is essential for the events leading to ovulation.

Question 9.
Menstrual cycles are absent during pregnancy. Why? [NCERT Exemplar]
Answer:
The high levels of progesterone and estrogens during pregnancy suppress the release of gonadotropins required for the development of new follicles. Therefore, new cycle cannot be initiated.

Question 10.
How does the sperm penetrate through the zona pellucida in [ human ovum?
Answer:
The sperm penetrates through zona pellucida with the help of secretions from acrosome.

Question 11.
Mention the function of trophoblast in human embryo.
Answer:
Trophoblast is the outer layer of blastocyst which helps in the attachment of blastocyst to the endometrium of the uterus.

Question 12.
Explain the function of umbilical cord.
Answer:
It transports nutrients and respiratory gases and metabolic wastes to and from mother and foetus.

PSEB 12th Class Biology Important Questions Chapter 3 Human Reproduction

Question 13.
Given below are the stages in human reproduction. Write them ‘ in correct sequential order.
Insemination, Gametogenesis, Fertilisation, Parturition, Gestation, Implantation. [NCERT Exemplar]
Answer:
Gametogenesis, Insemination, Fertilisation, Implantation, Gestation, Parturition.

Question 14.
What stimulates pituitary to release the hormone responsible for parturition? Name the hormone.
Answer:
The signal from the fully developed foetus and placenta or the foetal , ejection reflex induces mild uterine contraction. The hormone released is oxytocin.

Question 15.
Name the important mammary gland secretions that help in resistance of the new bom baby. [NCERT Exemplar]
Answer:
Colostrum

Question 16.
Mention the function of mitochondria in sperm.
Answer:
It provide energy for the movement of sperm tail.

Short answer type questions

Question 1.
Write the function of each of the following:
(a) Middle piece in human sperm.
(b) Luteinising hormone in human males.
Answer:
(a) Provides energy for movement.
(b) Stimulates synthesis and secretion of androgens or male hormones for spermatogenesis.

PSEB 12th Class Biology Important Questions Chapter 3 Human Reproduction

Question 2.
Differentiate between mayor structural changes in the human ovary during the follicular and luteal phase of the menstrual cycle.
Answer:

Follicular phase Luteal phase
1. During this, primary follicles grow to become fully mature Graafian follicle. During this, remaining part of Graafian follicle transforms into corpus luteum.
2. Endometrium regenerates through proliferation. Endometrium further thickens secreting progesterone for implantation after fertilisation. If fertilisation does not occur, corpus luteum degenerates.

Question 3.
Explain the events in a normal woman during her menstrual cycle on the following days :
(a) Pituitary hormone levels from 8 to 12 days.
(b) Uterine events from 13 to 15 days.
(c) Ovarian events from 16 to 23 days.
Answer:
(a) The level of LH and FSH secreted by anterior lobe of pituitary, stimulated by GnRH, increases.

(b) The endometrium of the uterus regenerates through proliferation. It grows and become thickened. There is repair of ruptured blood vessels and new blood capillaries develop. Uterine glands elongate.

(c) The remnant of Graafian follicle forms corpus luteum which secretes large amount of progesterone essential for maintenance of endometrium for implantation and for pregnancy.

Question 4.
What happens to corpus luteum in human female if the ovum is (a) fertilised, (b) not fertilised?
Answer:
(a) In case the ovum is fertilised, the corpus luteum persists and secretes a large amount of progesterone. The progesterone is essential for maintenance of endometrium, a necessity for implantation and for pregnancy.

(b) In the absence of fertilisation, corpus luteum degenerates. The level of LH and progesterone decreases very low. The absence of hormonal support, leads to the disintegration of endometrium and results in menstrual flow.

PSEB 12th Class Biology Important Questions Chapter 3 Human Reproduction

Question 5.
A large number of married couples the world over are childless. It is shocking to know that in India the female partner is often blamed for the couple being childless.
(a) Why in your opinion the female partner is often blamed for such situations in India? Mention any two values that you as a biology student can promote to check this Social evil.
(b) State any two reasons responsible for the cause of infertility.
(c) Suggest a technique that can help the couple to have a child where the problem is with the male partner.
Answer:
(a) The female partner is wrongly blamed for not bearing the child. It is due to the lack of proper information and knowledge about the reproduction and reproductive organs. Being a biology student I will advise and explain the couple as well as their other family members. Both male and female are equally responsible for bearing child. The two value, promoted are (i) concern for others (ii) scientific
temperament.

(b) Infertility is the inability to produce children inspite of unprotected sex and sexual co-habitation. It may be due to (i) physical/congenital disease (ii) immuno- logical or even physiological reason.

(c) In case, the problem is with male partner, artificial insemination (AI) is adopted. Semen collected either from the husband or a healthy donor is artificially introduced in the vagina or into the uterus of the female.

Question 6.
Name and explain the role of inner and middle walls of the human uterus.
Answer:
The inner wall of the uterus is called endometrium. It supports foetal growth and helps in placenta formation after implantation.
The middle wall of the uterus is called myometrium, exhibits strong contraction during delivery of baby.

Question 7.
Write a brief account of the structure and functions of placenta. [NCERT Exemplar]
Answer:
Placenta connects the foetus to the uterus through an umbilical chord. Both the foetal and the maternal tissues contribute to its formation. The foetal part is the chorionic villi and the maternal part is the uterine mucosa.

PSEB 12th Class Biology Important Questions Chapter 3 Human Reproduction

Long answer type questions

Question 1.
(a) Briefly explain the events of fertilisation and implantation in an adult human female.
(b) Comment on the role of placenta as an endocrine gland.
Or
Name the stage of human embryo of which it gets implanted.
Explain the process of implantation.
Or Draw a labelled diagram of a human blastocyst. How does it get implanted in the uterus?
Answer:
(a) (i) Fertilisation : Fertilisation occurs, if the ovum and sperms are transported simultaneously to the ampullary-isthmic junction and involve fusion of sperm with an ovum.
Secretions of acrosome of sperm help it to enter into the cytoplasm of ovum through zona pellucida and the plasma membrane. It induces meiotic division-II to form haploid ovum (ootid) and secondary polar body. The fusion of sperm with ovum to form diploid zygote is called fertilisation.
PSEB 12th Class Biology Important Questions Chapter 3 Human Reproduction 2
(ii) Implantation: Zygote undergoes cleavage to form a solid mass of 16 cells-morula, with daughter cells called blastomeres. Morula develops into a embryo with about 64 cells and with a cavity called blastocoel and the embryo is termed as blastocyst. It consists of outer layer of cells-trophoblast and inner cell mass.

The trophoblast gets attached to the endometrium- uterine wall of mother, after 7 days of fertilisation by a process called implantation leading to pregnancy. The uterine cells divide rapidly and cover blastocyst. The blastocyst gets embedded in the endometrium. Inner cell mass forms embryo.
PSEB 12th Class Biology Important Questions Chapter 3 Human Reproduction 3

(b) Placenta beside providing nutrients to the foetus also act as an endocrine gland. Placenta secretes human chorionic gonadotrophin (hCG), human placental lactogen (hPL), estrogen, progesterone etc., and later relaxin is secreted by ovary which facilitates parturition. Increased level of hormones like cortisol, prolactin and thyroxine etc., help in foetal growth, metabolic changes in the mother and maintenance of pregnancy.

PSEB 12th Class Biology Important Questions Chapter 15 Biodiversity and Conservation

Punjab State Board PSEB 12th Class Biology Important Questions Chapter 15 Biodiversity and Conservation Important Questions and Answers.

PSEB 12th Class Biology Important Questions Chapter 15 Biodiversity and Conservation

Very short answer type questions

Question 1.
Why is genetic variation important in the plant Rauwolfia vomitoria? [NCERT Exemplar]
Answer:
Genetic variation affects the variation in potency and concentration of me drug reserpine in the medicinal plant Rauwolfia vomitoria.

Question 2.
Name the type of biodiversity represented by the following:
(i) 50,000 different strains of rice in India
(ii) Estuaries and alpine meadows in India.
Answer:
(i) Genetic diversity
(ii) Ecological diversity

Question 3.
Identify ‘a’ and ‘b’ in the figure given below representing proportionate number of major vertebrate taxa.
PSEB 12th Class Biology Important Questions Chapter 15 Biodiversity and Conservation 1
Answer:
(a) Mammals
(b) Amphibians

Question 4.
Suggest two practices giving one example by each, that help protect rare or threatened species.
Answer:

  1. By using cryopreservation (preservation at -196°C) technique, sperms, eggs, tissues, and embryos can be stored for long period in gene banks, seed banks etc.
  2. Plants are propagated in vitro using tissue culture methods.

Question 5.
According to David Tilman, greater the diversity greater is the primary productivity. Can you think of a very low diversity man-made ecosystem that has high productivity? [NCERT Exemplar]
Answer:
Agricultural fields like wheat field or paddy field which are also examples of monoculture practices.

PSEB 12th Class Biology Important Questions Chapter 15 Biodiversity and Conservation

Question 6.
What is Red Data Book? [NCERT Exemplar]
Answer:
The Red Data Book is a compilation of data on species threatened with extinction and is maintained by IUCN. ‘

Question 7.
What is the expanded form of IUCN? [NCERT Exemplar]
Answer:
International Union for Conservation of Nature.

Question 8.
Why Western Ghats in India have been declared as biological hotspots?
Answer:
Western Ghats are biological hotspots because they have species richness and species evenness.

Question 9.
What is a national park?
Answer:
It is a protected area reserved for wildlife where human activities are not permitted.

Question 10.
What is the difference between endemic and exotic species? [NCERT Exemplar]
Answer:
Endemic species are native species restricted to a particular geographical region. Exotic species are species which are introduced from other geographical regions into an area.

Question 11.
How is the presently occurring species extinction different from the earlier mass extinctions? [NCERT Exemplar]
Answer:
Species extinction occurring at present is due to anthropogenic or man-made causes whereas the earlier extinction was due to natural causes.

Question 12.
Differentiate between in situ and ex situ approaches of conservation of biodiversity.
Answer:

In situ approach Ex-situ approach
1. It involves protection of endangered species of plants and animals. It involves protection of endangered species by removing them from the natural habitat.
2. This is done by protecting the natural habitat or ecosystem. This is done by placing the species under special care.

Short answer type questions

Question 1.
What is biodiversity? Why is it a matter of concern now?
Answer:
Biodiversity is the occurrence of different types of genes, gene pools, species, habitats and ecosystems at a particular place and various parts of earth. It is a matter of concern because species are continuously lost, limiting the diversity and this will affect our survival and well-being on earth due to the changes in environment.

PSEB 12th Class Biology Important Questions Chapter 15 Biodiversity and Conservation

Question 2.
Where would you expect more species biodiversity-in tropics or in polar regions? Give reasons in support of your answer.
Answer:
More biodiversity is found in the tropics. This is because tropical regions remain undisturbed from frequent glaciations as in polar regions. Also, the tropics are less seasonal/more constant.

Question 3.
Is it true that there is more solar energy available in the tropics? Explain briefly. [NCERT Exemplar]
Answer:
As one moves from the equator to the polar regions, the length of the day decreases and the length of the night increases. The length of day and night are same at the equator. Therefore, it is true that there is more solar energy available in the tropics.

Question 4.
The given graph alongside shows species-area relationship. Write the equation of the curve ‘a’ and explain.
Answer:
PSEB 12th Class Biology Important Questions Chapter 15 Biodiversity and Conservation 2
The equation of the curve ‘a’ is S = CAZ.
(i) Within a region, species richness increases with increasing explored area but only up to a limit.
(ii) Relationship between species richness and area for a wide variety of taxa turns out to be a rectangular hyperbola.

Question 5.
Explain ‘rivet popper hypothesis. Name the ecologist who proposed it.
Answer:
Paul Ehrlich proposed the rivet popper hypothesis. This hypothesis states that in an airplane (ecosystem) all parts are joined together using thousands of rivets (species). If every passenger traveling in it starts popping a rivet to take home (causing a species to become extinct), it may not affect flight safety (proper functioning of the ecosystem) initially but as more and more rivets are removed, the plane becomes dangerously weak over a period of time. Also, which rivet is removed may also be critical like loss of rivets on the wings (key species) is more serious threat to flight safety than loss of few rivets on the seats or windows inside the plane.

Question 6.
How do human activities cause desertification?
Answer:
Human activities like over-cultivation, unrestricted grazing, deforestation, and poor irrigation practices result in arid patches of land. The fertile topsoil that may take centuries to develop is eroded due to these activities. When large barren patches extend and meet over time, a desert is created. Increased urbanization is also one of the causes of desertification.

Question 7.
Why are conventional methods not suitable for the assessment of biodiversity of bacteria? [NCERT Exemplar]
Answer:
Many bacteria are not culturable under normal conditions in the laboratory. This becomes a problem in studying their morphological, biochemical, and other characterizations which are useful for their assessment.

Question 8.
List any four techniques where the principle of ex-situ conservation of biodiversity has been employed.
Answer:
Ex-situ Conservation (off-site conservation)

  1. Zoological parks and botanical gardens.
  2. Wildlife safari parks, aquaria.
  3. Preservation of germplasm-seed gene banks, tissue culture, cryopreservation.
  4. Sacred plants grown in homes, villages, and religious places.

Long answer type questions

Question 1.
(a) Why should we conserve biodiversity? How can we do it?
(b) Explain the importance of biodiversity hotspots and sacred groves.
Or
Why should biodiversity be conserved? List any two ethical arguments in its support. ‘
Answer:
(a) Need for Conservation of Biodiversity: Reasons for conservation of biodiversity can be grouped into three categories :

  1. narrowly utilitarian
  2. broadly utilitarian and
  3. ethical.

(i) Narrowly Utilitarian: The reasons for conserving biodiversity are obvious because of their :
(a) direct economic benefits such as

  • food (cereals, pulses, fruits)
  • firewood
  • fiber
  • construction material
  • products of medicinal importance
  • industrial products (tannins, gums, lubricants, dyes, resins, perfumes).

(b) More than 25% of the drugs are derived from plants.

PSEB 12th Class Biology Important Questions Chapter 15 Biodiversity and Conservation

(c) 25,000 species of plants are used as traditional medicines by native people.

(ii) Broadly Utilitarian: Biodiversity plays a major role in providing ecosystem services that nature provides and which cannot be given a price tag are :

  • Production of Oxygen
  • Pollination of flowers by bees, bumblebees, birds, and bats, etc.
  • Resulting in the formation of fruits and seeds.
  • Aesthetic pleasures like bird watching, walking through the thick forests, waking up to bulbul’s song, etc.

(iii) Ethical :
(a) We share this planet with millions of plants, animals, and microbe species. Every species has an intrinsic value even if it is not of current or any economic value to us.

(b) We have an essential duty to care for their well-being and pass on the biological legacy in a proper form to our future generations. We can conserve biodiversity by two major approaches

  • In situ conservation (on site/ conservation)
  • Ex situ conservation (off-site conservation).

(c) Hotspots are the areas identified by conservationists for the very high level of species richness and high degree of endemism (species confined to a particular area and not found anywhere else). Hotspots help in protection of certain biodiversity-rich regions.

Sacred groves are the tracts of forest set aside where all the trees and wildlife within are given religious sanctity and total protection. Such sacred groves are found in Khasi and Jaintia hills in Meghalaya, Aravalli Hills of Rajasthan etc.

PSEB 12th Class Biology Solutions Chapter 3 Human Reproduction

Punjab State Board PSEB 12th Class Biology Book Solutions Chapter 3 Human Reproduction Textbook Exercise Questions and Answers.

PSEB Solutions for Class 12 Biology Chapter 3 Human Reproduction

PSEB 12th Class Biology Guide Human Reproduction Textbook Questions and Answers

Question 1.
Fill in the blanks:
(a) Humans reproduce …………………. .(asexually/sexually)
Answer:
sexually.

(b) Humans are …………………… .(oviparous/viviparous/ovoviviparous)
Answer:
viviparous.

(c) Fertilisation is ………………………. in humans, (external/internal)
Answer:
internal

(d) Male and female gametes are …………………. .(diploid/haploid)
Answer:
haploid.

(e) Zygote is ………………….. .(diploid/haploid)
Answer:
diploid.

(f) The process of release of ovum from a mature follicle is called ……………………. .
Answer:
ovulation.

(g) Ovulation is induced by a hormone called …………………… .
Answer:
luteinising hormone.

(h) The fusion of male and female gametes is called ……………………… .
Answer:
fertilisation.

(i) Fertilisation takes place in ………………… .
Answer:
fallopian tube (ampullary-isthmic junction).

(j) Zygote divides to form ……………….., which is implanted in uterus.
Answer:
blastocyst

(k) The structure which provides vascular connection between foetus and uterus is called …………………… .
Answer:
placenta.

PSEB 12th Class Biology Solutions Chapter 3 Human Reproduction

Question 2.
Draw a labelled diagram of male reproductive system.
Answer:
PSEB 12th Class Biology Solutions Chapter 3 Human Reproduction 1
PSEB 12th Class Biology Solutions Chapter 3 Human Reproduction 2

Question 3.
Draw a labelled diagram of female reproductive system.
Answer:
PSEB 12th Class Biology Solutions Chapter 3 Human Reproduction 3
PSEB 12th Class Biology Solutions Chapter 3 Human Reproduction 4

PSEB 12th Class Biology Solutions Chapter 3 Human Reproduction

Question 4.
Write two major functions each of testis and ovary.
Answer:
Functions of the testis
(a) They produce male gametes called spermatozoa by the process of spermatogenesis.
(b) The leydig cells of the seminiferous tubules secrete the male sex hormone called testosterone. Testosterone aids the development of secondary sex characteristics in males.

Functions of the ovary
(a) They produce female gametes called ova by the process of oogenesis.
(b) The growing Graafian follicles secrete the female sex hormone called estrogen. Estrogen aids the development of secondary sex characteristics in females.

Question 5.
Describe the structure of a seminiferous tubule.
Answer:
The production of sperms in the testes takes place in a highly coiled structure called the seminiferous tubules. These tubules are located in the testicular lobules. Each seminiferous tubule is lined by germinal epithelium. It is lined on its inner side by two types of cells namely spermatogonia aid sertoli cells respectively. Spermatogonia are male germ cells which produce primary spermatocytes by meiotic divisions. Primary spermatocytes undergo further meiotic division to form secondary spermatocytes and finally, spermatids. Spermatids later metamorphoses into male gametes called spermatozoa. Sertoli cells are known as nurse cells of the testes as they provide nourishment to the germ cells. There are large polygonal cells known as interstitial cells or leydig cells just adjacent to seminiferous tubules. These cells secrete the male hormone called testosterone.
PSEB 12th Class Biology Solutions Chapter 3 Human Reproduction 5

Question 6.
What is spermatogenesis? Briefly describe the process of spermatogenesis.
Answer:
Spermatogenesis is the process of the production of sperms from the immature germ cells in males. It takes place in seminiferous tubules present inside the testes. During spermatogenesis, a diploid spermatogonium (male germ cell) increases its size to form a diploid primary spermatocyte. This diploid primary spermatocyte undergoes first meiotic division (meiosis I), which is a reductional division to form two equal haploid secondary spermatocytes. Each secondary spermatocyte then undergoes second meiotic division (meiosis II) to form two equal haploid spermatids. Hence, a diploid spermatogonium produces four haploid spermatids. These spermatids are transformed into spermatozoa (sperm) by the process called spermiogenesis.
PSEB 12th Class Biology Solutions Chapter 3 Human Reproduction 6

PSEB 12th Class Biology Solutions Chapter 3 Human Reproduction

Question 7.
Name the hormones involved in regulation of spermatogenesis.
Answer:
Follicle-stimulating hormones (FSH) and luteinising hormones (LH) are secreted by gonadotropin releasing hormones from the hypothalamus. These hormones are involved in the regulation of the process of spermatogenesis. FSH acts on sertoli cells, whereas LH acts on leydig cells of the testis and stimulates the process of spermatogenesis.

Question 8.
Define spermiogenesis and spermiation.
Answer:
Spermiogenesis : It is the process of transforming spermatids into matured spermatozoa or sperms.
Spermiation : It is ‘the process when mature spermatozoa are released from the sertoli cells into the lumen of seminiferous tubules.

Question 9.
Draw a labelled diagram of sperm.
Answer:
PSEB 12th Class Biology Solutions Chapter 3 Human Reproduction 7

Question 10.
What are the major components of seminal plasma?
Answer:
Semen (produced in males) is composed of sperms and seminal plasma. The major components of the seminal plasma in the male reproductive system are mucus, spermatozoa, and various secretions of accessory glands. The seminal plasma is rich in fructose, calcium, ascorbic acid, and certain enzymes. It provides nourishment and protection to sperms.

PSEB 12th Class Biology Solutions Chapter 3 Human Reproduction

Question 11.
What are the major functions of male accessory ducts and glands?
Answer:
The male accessory ducts are vasa efferentia, epididymis, vas deferens, and rete testis. They play an important role in the transport and temporary storage of sperms. On the contrary, male accessory glands are seminal vesicles, prostate glands, and bulbourethral glands. These glands secrete fluids that lubricate the reproductive system and sperms. The sperms get dispersed in the fluid which makes their transportation into the female body easier. The fluid is rich in fructose, ascorbic acid, and certain enzymes. They also provide nutrients and activate the sperm.

Question 12.
What is oogenesis? Give a brief account of oogenesis.
Answer:
Oogenesis is the process of the formation of a mature ovum from the oogonia in females. It takes place in the ovaries. During oogenesis, a diploid oogonium or egg mother cell increases in size and gets transformed into a diploid primary oocyte. This diploid primary oocyte undergoes first meiotic division i.e., meiosis I or reductional division to form two unequal haploid cells. The smaller cell is known as the first polar body, while the larger cell is known as the secondary oocyte. This secondary oocyte undergoes second meiotic division i.e., meiosis II or equational division and gives rise to a second polar body and an ovum. Hence, in the process of oogenesis, a diploid oogonium produces a single haploid ovum while two or three polar bodies are produced.
PSEB 12th Class Biology Solutions Chapter 3 Human Reproduction 8

Question 13.
Draw a labelled diagram of a section through ovary.
Answer:
PSEB 12th Class Biology Solutions Chapter 3 Human Reproduction 9

PSEB 12th Class Biology Solutions Chapter 3 Human Reproduction

Question 14.
Draw a labelled diagram of a Graafian follicle.
Answer:
PSEB 12th Class Biology Solutions Chapter 3 Human Reproduction 10

Question 15.
Name the functions of the following:
(a) Corpus luteum
(b) Endometrium
(c) Acrosome
(d) Sperm tail
(e) Fimbriae
Answer:
(a) Corpus Luteum: It is formed from the ruptured Graafian follicle. It secretes progesterone hormone during the luteal phase of the menstrual cycle. A high level of progesterone inhibits the secretions of FSH and LH, thereby preventing ovulation. It also allows the endometrium of the uterus to proliferate and to prepare itself for implantation.

(b) Endometrium: It is the innermost lining of the uterus. It is rich in glands and undergoes cyclic changes during various phases of the menstrual cycle to prepare itself for the implantation of the embryo.

(c) Acrosome: It is a cap-like structure present in the anterior part of the head of the sperm. It contains hyaluronidase enzyme, which hydrolyses the outer membrane of the egg, thereby helping the sperm to penetrate the egg at the time of fertilisation.

(d) Sperm Tail: It is the longest region of the sperm that facilitates the movement of the sperm inside the female reproductive tract.

(e) Fimbriae: They are finger-like projections at the ovarian end of the fallopian tube. They help in the collection of the ovum (after ovulation), which is facilitated by the beating of the cilia.

Question 16.
Identify True/False statements. Correct each false statement to make it true.
(a) Androgens are produced by Sertoli cells. (True/False)
(b) Spermatozoa get nutrition from Sertoli cells. (True/False)
(c) Leydig cells are found in ovary. (True/False)
(d) Leydig cells synthesise androgens. (True/False)
(e) Oogenesis takes place in corpus luteum. (True/False)
(f) Menstrual cycle ceases during pregnancy. (True/False)
(g) Presence or absence of hymen is not a reliable indicator of virginity or sexual experience. (True/False)
Answer:
(a) Androgens are produced by Sertoli cells.
False Correct : Leydig cells.

(b) Spermatozoa get nutrition from Sertoli cells.
True

(c) Leydig cells are found in ovary.
False Correct : spermatogonia.

(d) Leydig cells synthesise androgens.
True

(e) Oogenesis takes place in corpus luteum.
False Correct : ovaries

(f) Menstrual cycle ceases during pregnancy.
True

(g) Presence or absence of hymen is not a reliable indicator of virginity or sexual experience.
True

PSEB 12th Class Biology Solutions Chapter 3 Human Reproduction

Question 17.
What is menstrual cycle? Which hormones regulate menstrual cycle? ,
Answer:
The menstrual cycle is a series of cyclic physiologic changes that take place inside the female reproductive tract in primates. The whole cycle takes around 28 days to complete. The end of the cycle is accompanied by the breakdown of uterine endothelium, which gets released in the form of blood and mucus through the vagina. This is known as menses.

The follicle stimulating hormone (FSH), luteinising hormone (LH), L estrogen, and progesterone are the various hormones that regulate the menstrual cycle. The level of FSH and LH secreted from the anterior pituitary gland increases during the follicular phase. FSH secreted under the influence of RH (releasing hormone) from the hypothalamus , stimulates the conversion of a primary follicle into a graafian follicle.

The level of LH increases gradually leading to the growth of follicle and f secretion of estrogen. Estrogen inhibits the secretion of FSH and stimulates the secretion of luteinising hormone. It also causes the thickening of the uterine endometrium. The increased level of LH causes the rupturing of the graafian follicle and release the ovum into the fallopian tube. The ruptured graafian follicle changes to corpus luteum and starts secreting progesterone hormone during the luteal phase.

Progesterone hormone helps in the maintenance and preparation of endometrium for the implantation of the embryo. High levels of progesterone hormone in the blood decrease the secretion of LH and FSH, therefore inhibiting further ovulation.

Question 18.
What is parturition? Which hormones are involved in induction of parturition?
Answer:
Parturition is the process of giving birth to a baby as the development of the foetus gets completed in the mother’s womb. The hormones involved in this process are oxytocin and relaxin. Oxytocin leads to the contraction of smooth muscles of myometrium of the uterus, which directs the full term foetus towards the birth canal. On the other hand, relaxin hormone causes relaxation of the pelvic ligaments and prepares the uterus for child birth.

Question 19.
In our society the women are often blamed for giving birth to [ daughters. Can you explain why this is not correct?
Answer:
All human beings have 23 pairs of chromosomes. Human males have 22 pairs of autosomes and contain one or two types of sex chromosome. They are either X or Y. On the contrary, human females have 22 pairs of autosomes and contain only the X sex chromosome. The sex of an individual is determined by the type of the male gamete (X or Y), which fuses with the X chromosome of the female. If the fertilising sperm is X, then the baby will be a girl and if it is Y, then the baby will be a boy.
Hence, it is incorrect to blame a woman for the gender of the child.

PSEB 12th Class Biology Solutions Chapter 3 Human Reproduction

Question 20.
How many eggs are released by a human ovary in a month? How many eggs do you think would have been released if the mother gave birth to identical twins? Would your answer change if the twins bom were fraternal?
Answer:
An ovary releases an egg every month. When two babies are produced in succession, they are called twins. Generally, twins are produced from a single egg by the separation of early blastomeres resulting from the first zygotic cleavage. As a result, the young ones formed will have the same genetic make-up and are thus, called identical twins.

If the twins born are fraternal, then they would have developed from two separate eggs. This happens when two eggs’ (one from each ovary) are released at the same time and get fertilised by two separate sperms. Hence, the young ones developed will have separate genes and are therefore, called non-identical or fraternal twins.

Question 21.
How many eggs do you think were released by the ovary of a female dog which gave birth to 6 puppies?
Answer:
Dogs and rodents are polyovulatory species. In these species, more than one ovum is released from the ovary at the time of ovulation. Hence, six eggs were released by the ovary of a female dog to produce six puppies.

PSEB 11th Class Biology Important Questions Chapter 10 Cell Cycle and Cell Division

Punjab State Board PSEB 11th Class Biology Important Questions Chapter 10 Cell Cycle and Cell Division Important Questions and Answers.

PSEB 11th Class Biology Important Questions Chapter 10 Cell Cycle and Cell Division

Very short answer type questions

Question 1.
During which phase of mitotic cell division, chromosomes gets separated?
Answer:
During anaphase.

Question 2.
Does mitosis occurs before or after the interphase?
Answer:
Yes, mitosis occurs before or after the interphase, as dividing phase (meiosis or mitosis) and interphase are considered only as the major phases of a cell cycle.

PSEB 11th Class Biology Important Questions Chapter 10 Cell Cycle and Cell Division

Question 3.
Mitosis cell division helps in regeneration process. How?
Answer:
Mitosis helps in regeneration by keeping all the somatic cells of an organism genetically similar, so that they are able to regenerate part or whole of the organism.

Question 4.
Given that average duplication time of E. coli is 20 minutes. How much time will two E. coli cells take to become 32 cells?
Answer:
2 hours (2n = 25 = 2 × 2 × 2 × 2 × 2 = 32 generations).

Question 5.
If a tissue has 1024 cells at a given time, how many cycles of mitosis had the original parental single cell undergone?
[NCERT Exemplar]
Answer:
10 (2n, where n =10 generations).

Question 6.
Two key events take place during S-phase in animal cells, i.e., DNA replication and duplication of centriole. In which parts of the cell do these events occur? [NCERT Exemplar]
Answer:
DNA replication in the nucleus. Centriole duplication in the cytoplasm.

Question 7.
At what stage of meiosis, formation of tetrads occurs? Name it.
Answer:
Tetrads are formed during pachytene of prophase-I (meiosis-I).

PSEB 11th Class Biology Important Questions Chapter 10 Cell Cycle and Cell Division

Question 8.
Meiosis is essential in sexually reproducing organisms. How?
Answer:
Meiosis is essential in sexually reproducing organisms because it keeps the chromosome number constant.

Question 9.
Which cells of our body do not divide?
Answer:
Neuron cells stops dividing soon after the birth of a child.

Question 10.
The diagram shows a bivalent at prophase-I of meiosis. Which of the four chromatids can cross over? [NCERT Exemplar]
Answer:
The non-sister chromatids of homologous pair of chromosome undergo meiosis.
PSEB 11th Class Biology Important Questions Chapter 10 Cell Cycle and Cell Division 1

Short answer type questions

Question 1.
Imagine a situation if there was no meiosis. Then what would have happened to the next generation?
Answer:
In the absence of meiosis the next generation would have double the number of chromosomes after fusion of gametes. This would have resulted in the birth of an altogether new species. The maintenance of characters set would have been possible only through asexual reproduction.

Question 2.
Give a description of metaphase I of meiosis.
Answer:
Metaphase I: The bivalent chromosomes align on the equatorial plate. The microtubules from the opposite poles of the spindle attach to the pair of homologous chromosomes.

PSEB 11th Class Biology Important Questions Chapter 10 Cell Cycle and Cell Division

Question 3.
Describe telophase I of meiosis.
Answer:
Telophase I

  • The nuclear membrane and nucleolus reappear, cytokinesis follows and this is called as diad of cells.
  • Although in many cases the chromosomes do undergo some dispersion, they do not reach the extremely extended state of the interphase nucleus.
  • The stage between the two meiotic divisions is called interkinesis and is generally short lived. Interkinesis is followed by prophase II, a much simpler prophase than prophase I.

Question 4.
What is the process of cell division in prokaryotes?
Answer:
Prokaryotes do not have nucleus. So, there is no elaborate karyokinesis, as seen in eukaryotes. In prokaryotes the replication of DNA starts the process of cell division. Once genetic material is replicated, it is followed by division of cytoplasm. The process is known as binary fission.

Question 5.
How does meiosis facilitate creation of offsprings, with distinct characters?
Answer:
Meiosis happens during gametogenesis and as a result gametes have half the number of chromosomes. During fertilization, when gametes fuse together two different sets of chromosomes make a new set. This results in an offspring, who has distinct characters, compared to parents.

Question 6.
What is the significance of mitosis?
Answer:
Significance of Mitosis

  • In multicellular organisms, body growth is by mitotic divisions of the cells.
  • Replacement of worn out tissues/cells (e.g., blood cells, skin cells) and repair of the injured tissues is by mitosis.
  • In unicellular organisms, mitosis are involved in asexual reproduction
    (multiplication of cells).
  • In plants, vegetative propagation involves only mitotic divisions and genetically identical individuals are produced.
  • Uncontrolled cell divisions in certain tissues/organ (cancer) result in tumours.

PSEB 11th Class Biology Important Questions Chapter 10 Cell Cycle and Cell Division

Long answer type questions

Question 1.
Briefly describe the significance of cell division.
Answer:
Cell division is significant in the following ways :

  • Cell Multiplication: Cell division is a means of cell multiplication or formation of new cells from pre-existing cells.
  • Continuity: It maintains continuity of living matter generation after generation.
  • Multicellular Organisms: The body of a multicellular organism is formed of innumerable cells. They are formed by repeated divisions of a single cell or zygote. As the number of cells increases, many of them begin to differentiate, form tissues and organisms.
  • Cell Size: Cell division helps in maintenance of a particular cell size which is essential for efficiency and control of cell activities.
  • Genetic Similarity: The common type of cell division or mitosis maintains genetic similarity of all the cells in an individual despite being different, i.e., structurally and functionally.

Question 2.
Explain meiosis-II in an animal cell.
Answer:
All these happen in the two haploid nuclei simultaneously.

  • Prophase-II: It takes short time. Spindle formation begins and the chromosomes become short. Two chromatids, are joined to a single centromere. Nuclear membrane and nucleolus disintegrate.
  • Metaphase-II: At the equator, the chromosomes lie and spindle is formed. The centromere of every chromosomes is joined to the spindle fibre and centromere also divides.
  • Anaphase-II: The daughter chromosomes are formed. Chromatids move towards their poles with the spindle fibres.
  • Telophase-II: Reaching at the poles, chromosomes form nuclei which are haploid (n) daughter nuclei. Again nuclear membrane is constructed. Nucleolus now becomes clearly visible.
  • Cytokinesis: It occurs and four daughter cells are formed which are haploid (n). It may occur once or twice (i.e., in meiosis-I and II) or only after the meiosis-II cell division.

PSEB 11th Class Biology Important Questions Chapter 10 Cell Cycle and Cell Division

Question 3.
Describe briefly the phases of meiotic division.
Answer:
Meiotic division takes place in germ cells. The number of chromosomes is reduced to half in daughter cells.

Meiotic cell division is divided into two phases, i.e., meiotic-I and II.
In the meiotic-I division, the homologous chromosomes pair to form bivalent. Exchange of genetic material takes place. The chromosomes now separate and get distributed into daughter cells.

Long prophase-I is divided into five sub-stages, i.e., leptotene, zygotene, pachytene, diplotene and diakinesis. During metaphase-I, the bivalents arrange themselves on equatorial plate with their arms on the plate but the centromere is directed towards opposite pole. It is followed by anaphase-I. Now, the homologous chromosomes repel each other, move to the opposite poles with both their chromatids. In this way each pole gets half the chromosomes number of the parent cell.

In telophase-I, the nuclear envelope and nucleolus again appear. The centromere of each chromosome breaks, separating the chromatids, one each to a daughter cell. The meiotic cell division maintains chromosome number of a species.

As a result of meiotic division, the four daughter cells are formed with half the chromosome number (haploid) in each cell.
PSEB 11th Class Biology Important Questions Chapter 10 Cell Cycle and Cell Division 2

PSEB 12th Class Biology Solutions Chapter 15 Biodiversity and Conservation

Punjab State Board PSEB 12th Class Biology Book Solutions Chapter 15 Biodiversity and Conservation Textbook Exercise Questions and Answers.

PSEB Solutions for Class 12 Biology Chapter 15 Biodiversity and Conservation

PSEB 12th Class Biology Guide Biodiversity and Conservation Textbook Questions and Answers

Question 1.
Name the three important components of biodiversity.
Answer:
Biodiversity is the variety of living forms present in various ecosystems. It includes variability among life forms from all sources including land, air, and water. Three important components of biodiversity are as follows :

  • Genetic diversity
  • Species diversity
  • Ecological diversity

Question 2.
How do ecologists estimate the total number of species present in the world?
Answer:
The diversity of living organisms present on the Earth is very vast. According to an estimate by researchers, it is about seven million. the total number of species present in the world is calculated by statistical comparison between species richness of a well-studied group of insects of temperate and tropical regions. Then, these ratios are extrapolated with other groups of plants and animals to calculate the total species richness present on the Earth.

Question 3.
Give three hypotheses for explaining why tropics show greatest levels of species richness.
Answer:
There are three different hypotheses proposed by scientists for explaining species richness in the tropics.

  1. Tropical latitudes receive more solar energy than temperate regions, which leads to high productivity and high species diversity.
  2. Tropical regions have less seasonal variations and have a more or less constant environment. This promotes the niche specialization and thus, high species richness.
  3. Temperate regions were subjected to glaciations during the ice age, while tropical regions remained undisturbed which led to an increase in the species diversity in this region.

Question 4.
What is the significance of the slope of regression in a species-area relationship?
Answer:
The slope of regression (z) has a great significance in order to find a species-area relationship. It has been found that in smaller areas (where the species-area relationship is analyzed), the value of slopes of regression is similar regardless of the taxonomic- group or the region. However, when a similar analysis is done in larger areas, then the slope of regression is much steeper.

PSEB 12th Class Biology Solutions Chapter 15 Biodiversity and Conservation

Question 5.
What are the major causes of species losses in a geographical region?
Answer:
Biodiversity is the variety of living forms present in various ecosystems. It includes variability among life forms from all sources including land, air, and water. Biodiversity around the world is declining at a very fast pace. The following are the major causes for the loss of biodiversity around the world :

(i) Habitat Loss and Fragmentation: Habitats of various organisms are altered or destroyed by uncontrolled and unsustainable human activities such as deforestation, slash and burn agriculture, mining, and urbanisation. This results in the breaking up of the habitat into small pieces, which effects the movement of migratory animals and also, decreases the genetic exchange between populations leading to a declination of species.

(ii) Over-exploitation: Due to over-hunting and over-exploitation of various plants and animals by humans, many species have become endangered or extinct (such as the tiger and the passenger pigeon).

(iii) Alien Species Invasions: Accidental or intentional introduction of non-native species into a habitat has also led to the declination or extinction of indigenous species. For example, the Nile perch introduced in Lake Victoria in Kenya led to the extinction of more than two hundred species of native fish in the lake.

(iv) Co-extinction: In a native habitat, one species is connected to the other in an intricate network. The extinction of one species causes the extinction of other species, which is associated with it in an obligatory way. For example, the extinction of the host will cause the extinction of its parasites.

Question 6.
How is biodiversity important for ecosystem functioning?
Answer:
An ecosystem with high species diversity is much more stable than an ecosystem with low species diversity. Also, high biodiversity makes the ecosystem more stable in productivity and more resistant towards disturbances such as alien species invasions and floods.

If an ecosystem is rich in biodiversity, then the ecological balance would not get affected. As we all know, various trophic levels are connected through food chains. If anyone organism or all organisms of any one trophic level is killed, then it will disrupt the entire food chain. For example, in a food chain, if all plants are killed, then all deer’s will die due to the lack of food.

If all deer’s are dead, soon the tigers will also die. Therefore, it can be concluded that if an ecosystem is rich in species, then there will be other food alternatives at each trophic level which would not allow any organism to die due to the absence of their food resource. Hence, biodiversity plays an important role in maintaining the health and ecological balance of an ecosystem.

Question 7.
What are sacred groves? What is their role in conservation?
Answer:
Sacred groves are tracts of forest which are regenerated around places of worship. Sacred groves are found in Rajasthan, Western Ghats of Karnataka, and Maharashtra, Meghalaya, and Madhya Pradesh. Sacred grows help in the protection of many rare, threatened, and endemic species of plants and animals found in an area. The process of deforestation is strictly prohibited in this region by tribals. Hence, the sacred grove biodiversity is a rich area.

Question 8.
Among the ecosystem services are control of floods and soil erosion. How is this achieved by the biotic components of the ecosystem?
Answer:
The biotic components of an ecosystem include the living organisms such as plants and animals. Plants play a very important role in controlling floods and soil erosion. The roots of plants hold the soil particles together, thereby preventing the top layer of the soil to get eroded by wind or running water. The roots also make the soil porous, thereby allowing groundwater infiltration and preventing floods. Hence, plants are able to prevent soil erosion and natural calamities Fucii as floods and droughts. They also increase the fertility of soil and biodiversity.

PSEB 12th Class Biology Solutions Chapter 15 Biodiversity and Conservation

Question 9.
The species diversity of plants (22 percent) is much less than that of animals (72 percent). What could be the explanations to how animals achieved greater diversification?
Answer:
72 percent of species recorded on the Earth are animals and only 22 percent species are plants. There is quite a large difference in their percentage This is because animals have adapted themselves to ensure their survival in changing environments in comparison to plants. For example, insects and other animals have developed a complex nervous system to control and coordinate their body structure. Also, repeated body/ segments with paired appendages and external cuticles have made insects versatile and have given them the ability to survive in vain JUS habitats as compared to other life forms.

Question 10.
Can you think of a situation where we deliberately want to make a species extinct? How would you justify it?
Answer:
Yes, there are various kinds of parasites and disease-causing microbes that we deliberately want to eradicate from the Earth. Since these micro-organisms are harmful to human beings, scientists are working hard to fight against them.

Scientists have been able to eliminate smallpox virus from the world through the use of vaccinations. This shows than humans deliberately want to make these species extinct. Several other eradication programs such as polio and Hepatitis B vaccinations are aimed to eliminate these disease-causing microbes.