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PSEB 11th Class Biology Important Questions Chapter 2 Biological Classification

July 30, 2024July 29, 2024 by Prasanna

Punjab State Board PSEB 11th Class Biology Important Questions Chapter 2 Biological Classification Important Questions and Answers.

PSEB 11th Class Biology Important Questions Chapter 2 Biological Classification

Very short answer type questions

Question 1.
Name the organisms living in salty areas. [NCERT Exemplar]
Answer:
Halophiles live in habitats having high salinity and high light intensity.

Question 2.
Name the kingdom under which cyanobacteria have been classified.
Answer:
Cyanobacteria are prokaryotic organisms belong to kingdom – Monera.

Question 3.
What is the principle underlying the use of cyanobacteria in agricultural fields for crop improvement? [NCERT Exemplar]
Answer:
Some cyanobacteria live in mutually beneficial relationship with plants, obtaining food and shelter and fixing nitrogen for the plant. They also reduce soil alkalinity and improve soil texture.

Question 4.
Are chemosynthetic bacteria autotrophic or heterotrophic? [IMCERT Exemplar]
Answer:
Chemosynthetic bacteria are- autotrophs because they are able to synthesize their food from inorganic raw material with the help of energy obtained from chemical reaction.

Question 5.
Fusion of two gametes which are dissimilar in size is termed as. [NCERT Exemplar]
Answer:
Anisogamy, it occurs in Ceratium a dinoflagellate.

Question 6.
Why are cysts formed in protistans?
Answer:
Cysts formation helps to over come unfavourable condition.

Question 7.
What kind of nutrition occurs in a parasite?
Answer:
Parasites have phagotrophic and absorptive type of nutrition.

Question 8.
An association between roots of higher plants and fungi is called.
Answer:
Mycorrhiza is a symbiotic association between a fungus and the root of a plant. Mycorrhiza does not penetrate deep but remains in the superficial layers of the soil.

Question 9.
How the saprotrophic Basidiomycetes are able to decompose plant matter?
Answer:
The saprotrophic Basidiomycetes can decompose plant matter because they have enzymes for metabolising both cellulose and lignin.

Question 10.
Which are the most advanced group of fungi?
Answer:
Basidiomycetes.

Question 11.
What is capsid and how it is useful for viruses?
Answer:
Capsid is a proteinaceous covering around the virus. It protects the nucleoid from damage from physical and chemical agents.

Question 12.
Which enzyme is present in bacteriophages?
Answer:
Lysozyme is present in the region that comes in contact with host cell.

Short answer type questions

Question 1.
Write five beneficial usage or effects of bacteria.
Answer:
(a) Curdling of milk
(b) Lactobacillus is an important commensal in the gut flora of humans.
(c) Penicillin antibiotics are prepared by bacteria.
(d) Bacteria is a good decomposer, so it assists in completing the energy cycle.
(e) Rhizobium bacteria helps in nitrogen fixation.

Question 2.
What is the role of methanogens?
Answer:
Methanogens are a type of bacteria which live in the gut of ruminating animals. They assist those animals in digestion and the byproduct of that digestive process is methane. More number of livestock population results in increased methane level in the environment leading to global warming. So, indirectly methanogens can be responsible for global warming.

Question 3.
Cyanobacteria plays a major role in our ecology. Discuss.
Answer:
Cyanobacteria, also known as ‘blue-green algae’ help in carbon fixation in a major way on the ocean surface. They are helpful in nitrogen fixation in paddy fields leading to a better harvest. About 80% of photosynthesis on ocean surface is done by cyanobacteria. So, it can be said that they play a major role in our ecology.

Question 4.
Give a diagrammatic representation of classification of Protista.
Answer:
PSEB 11th Class Biology Important Questions Chapter 2 Biological Classification 1

Long answer type questions

Question 1.
Answer the following:
(i) A poisonous mushroom having white spores
(ii) Edible Basidiomycetes
(iii) Used in brewing industry
(iv) It is searched by trained pets
(v) Root like, cards like hyphal masses having a distinct growing point.
(vi) Non-motile meiospores develop in Basidiomycetes
(vii) Compact groups of hyphal produced to overcome unfavourable conditions.
Answer:
(i) toadstools
(ii) mushrooms and young pufballs
(iii) brewer’s yeast Saccharomyces cerevisiae
(iv) Truffles (tuber like underground fungus)
(v) Rhizomorphs perennate during periods of scarcity of food and water.
(vi) Basidiospores (develop exogenously)
(vii) Sclerotia under favourable conditions each one forms a new mycelium.

PSEB 11th Class Physics Solutions Chapter 12 Thermodynamics

July 30, 2024July 29, 2024 by Prasanna

Punjab State Board PSEB 11th Class Physics Book Solutions Chapter 12 Thermodynamics Textbook Exercise Questions and Answers.

PSEB Solutions for Class 11 Physics Chapter 12 Thermodynamics

PSEB 11th Class Physics Guide Thermodynamics Textbook Questions and Answers

Question 1.
A geyser heats water flowing at the rate of 3.0 liters per minute from 27°C to 77°C. If the geyser operates on a gas burner, what is the rate of consumption of the fuel if its heat of combustion is 4.0 x 10⁴ J/g?
Solution:
Water is flowing at a rate of 3.0 liter/min.
The geyser heats the water, raising the temperature from 27°C to 77°C.
Initial temperature, T₁ = 27°C
Final temperature, T₂ = 77°C
Rise in temperature, ΔT = T₂ -T₁
= 77-27 = 50°C

Heat of combustion = 4 x 10⁴ J/g
Specific heat of water, C = 4.2 J g^-1 °C^-1
Mass of flowing water, m = 3.0 litre/min = 3000 g/min
Total heat used, ΔQ = mC ΔT
= 3000 x 4.2 x 50
= 6.3 x 10⁵ J/min
∴ Rate of consumption = \(\frac{6.3 \times 10^{5}}{4 \times 10^{4}}\) = 15.75 g/min

Question 2.
What amount of heat must be supplied to 2.0 x 10^-2 kg of nitrogen (at room temperature) to raise its temperature by 45°C at constant pressure? (Molecular mass of N2 = 28; R = 8.3 J mol^-1K^-1).
Solution:
Mass of nitrogen, m = 2.0 x 10^-2 kg = 20g
Rise in temperature, ΔT = 45°C
Molecular mass of N₂,M =28
Universal gas constant, R = 8.3 J mol^-1K^-1
Number of moles, n = \(\frac{m}{M}\)
= \(\frac{2.0 \times 10^{-2} \times 10^{3}}{28}\) = 0.714
Molar specific heat at constant pressure for nitrogen,
C_p = \(\frac{7}{2}\) R = \(\frac{7}{2}\) x 8.3
= 29.05J mol^-1 K^-1

The total amount of heat to be supplied is given by the relation
ΔQ = nC_pΔT
= 0.714 x 29.05 x 45 = 933.38 J
Therefore, the amount of heat to be supplied is 933.38 J

Question 3.
Explain why
(a) Two bodies at different temperatures T₁ and T₂ if brought in thermal contact do not necessarily settle to the mean temperature (T₁ + T₂) / 2.
(b) The coolant in a chemical or a nuclear plant (i.e., the liquid used to prevent the different parts of a plant from getting too hot) should have high specific heat.
(c) Air pressure in a car tyre increases during driving.
(d) The climate of a harbour town is more temperate than that of a town in a desert at the same latitude.
Answer:
(a) When two bodies at different temperatures T₁ and T₂ are brought in thermal contact, heat flows from the body at the higher temperature to the body at the lower temperature till equilibrium is achieved, i.e., the temperatures of both the bodies become equal. The equilibrium temperature is equal to the mean temperature (T₁ + T₂) / 2 only when the thermal capacities of both the bodies are equal.

(b) The coolant in a chemical or nuclear plant should have a high specific heat. This is because higher the specific heat of the coolant, higher is its heat-absorbing capacity and vice versa. Hence, a liquid having a high specific heat is the best coolant to be used in a nuclear or chemical plant. This would prevent different parts of the plant from getting too hot.

(c) When a car is in motion, the air temperature inside the car increases because of the motion of the air molecules. According to Charles’ law, temperature is directly proportional to pressure. Hence, if the temperature inside a tyre increases, then the air pressure in it will also increase.

(d) A harbour town has a more temperate climate (i.e., without the extremes of heat or cold) than a town located in a desert at the same latitude. This is because the relative humidity in a harbour town is more than it is in a desert town.

Question 4.
A cylinder with a movable piston contains 3 moles of hydrogen at standard temperature and pressure. The walls of the cylinder are made of a heat insulator, and the piston is insulated by having a pile of sand on it. By what factor does the pressure of the gas increase if the gas is compressed to half its original volume?
Solution:
The cylinder is completely insulated from its surroundings. As a result, no heat is exchanged between the system (cylinder) and its surroundings. Thus, the process is adiabatic.
Initial pressure inside the cylinder = P₁
Final pressure inside the cylinder = P₂
Initial volume inside the cylinder = V₁
Final volume inside the cylinder = V₂
Ratio of specific heats, γ = 1.4 For an adiabatic process,
we have
P₁V₁^γ = P₂V₂^γ
The final volume is compressed to half of its initial volume.
PSEB 11th Class Physics Solutions Chapter 12 1
Hence, the pressure increases by a factor of 2.639.

Question 5.
In changing the state of a gas adiabatically from an equilibrium state A to another equilibrium state B, an amount of work equal to 22.3 J is done on the system. If the gas is taken from state A to B via a process in which the net heat absorbed by the system is 9.35 cal, how much is the net work done by the system in the latter case? (Take 1 cal = 4.19 J)
Solution:
The work done (W) on the system while the gas changes from state A to state Bis 22.3 J.
This is an adiabatic process. Hence, change in heat is zero.
ΔQ = 0
ΔW = -22.3 J (since the work is done on the system)
From the first law of thermodynamics, we have:
ΔQ = ΔU + ΔW

where, ΔU = Change in the internal energy of the gas .
ΔU = ΔQ – ΔW = -(-22.3 J)
ΔU = +22.3 J
When the gas goes from state A to state B via a process, the net heat absorbed by the system is
ΔQ =9.35cal = 9.35 x 4.19 = 39.1765J
Heat absorbed, ΔQ = ΔU + ΔW
ΔW = ΔQ – ΔU = 39.1765 – 22.3 – = 16.8765J
Therefore, 16.88 J of work is done by the system.

Question 6.
Two cylinders A and B of equal capacity are connected to each other via a stopcock.
A contains a gas at standard temperature and pressure.
B is completely evacuated. The entire system is thermally insulated.
The stopcock is suddenly opened. Answer the following:
(a) What is the final pressure of the gas in A and B?
(b) What is the change in internal energy of the gas?
(c) What is the change in the temperature of the gas?
(d) Do the intermediate states of the system (before settling to the final equilibrium state) lie on its P -V – T surface?
Solution:
(a) 0.5 atm
The volume available to the gas is doused as soon as the stopcock between cylinders A and B is opened? Since volume is inversely proportional to pressure, the pressure will decrease to one-half of the original value. Since the initial pressure of the gas is 1 atm, the pressure in each cylinder will be 0.5atm.

(b) Zero
The internal energy of the gas can change only when work is done by or on the gas. Since in this case no work is done by or on the gas, the internal energy of the gas will not change.

(c) Zero
Since no work is being done by the gas during the expansion of the gas, the temperature of the gas will not change at all.

(d) No
The given process is a case of free expansion. It is rapid and cannot be controlled. The intermediate states do not satisfy the gas equation and since they are in non-equilibrium states, they do not lie on the P-V-T surface of the system.

Question 7.
A steam engine delivers 5.4 x 10⁸ J of work per minute and services 3.6 x 10⁹ J of heat per minute from its boiler. What is the efficiency of the engine? How much heat is wasted per minute?
Solution:
Work done by the steam engine per minute, W = 5.4 x 10⁸ J
Heat supplied from the boiler, H = 3.6 x 10⁹ J
Efficiency of the engine = \(\frac{\text { Output energy }}{\text { Input energy }}\)
∴ η = \(\frac{W}{H}=\frac{5.4 \times 10^{8}}{3.6 \times 10^{9}}\) = 0.15
Hence, the percentage efficiency of the engine is 15%.
Amount of heat wasted= 3.6 x 10⁹ – 5.4 x 10⁸
= 30.6 x 10⁸ = 3.06 x 10⁹ J
Therefore, the amount of heat wasted per minute is 3.06 x 10⁹J.

Question 8.
An electric heater supplies heat to a system at a rate of 100 W. If system performs work at a rate of 75 joules per second. At what rate is the internal energy increasing?
Solution:
Heat is supplied to the system at a rate of 100 W.
∴ Heat supplied, ΔQ = 100 J/s
The system performs at a rate of 75 J/s.
∴ Work done, ΔW = 75 J/s

From the first law of thermodynamics, we have
ΔQ = ΔU + ΔW
where ΔU = Rate of change in internal energy
ΔU = ΔQ – ΔW = 100 – 75 = 25 J/s = 25W
Therefore, the internal energy of the given electric heater increases at a rate of 25 W.

Question 9.
A thermodynamic system is taken from an original state to an intermediate state by the linear process shown in figure given below.
PSEB 11th Class Physics Solutions Chapter 12 2
Its volume is then reduced to the original value from E to F by an isobaric process. Calculate the total work done by the gas from D to E to F.
Solution:
Total work done by the gas from D to E to F = Area of ΔDEF
Area of ΔDEF = \(\frac{1}{2}\) DF x EF
where, DF = Change in pressure
=600 N/m²
= 300N/m² = 300N/m²
FE = Change in volume
5.0 m³ – 2.0 m³ = 3.0m³
Area of ADEF = \(\frac{1}{2} \) x 300 x 3 = 450 J
Therefore, the total work done by the gas from D to E to F is 450 J.

Question 10.
A refrigerator is to maintain eatables kept inside at 9°C. If the room temperature is 36°C, Calculate the coefficient of performance.
Solution:
Temperature inside the refrigerator, T₁ = 9°C = 273 + 9 = 282 K
Room temperature, T₂ = 36°C = 273+36
Coefficient of performance = \(\frac{T_{1}}{T_{2}-T_{1}}\)
= \(\frac{282}{309-282}=\frac{282}{27}\)
309-282 = 10.44
Therefore, the coefficient of performance of the given refrigerator is 10.44.

PSEB 11th Class Biology Important Questions Chapter 15 Plant Growth and Development

July 30, 2024July 29, 2024 by Prasanna

Punjab State Board PSEB 11th Class Biology Important Questions Chapter 15 Plant Growth and Development Important Questions and Answers.

PSEB 11th Class Biology Important Questions Chapter 15 Plant Growth and Development

Very short answer type questions

Question 1.
Identify the actively dividing cells in plants.
Answer:
Meristems are the actively dividing cells present in the plants.

Question 2.
What happens if the meristematic cells ever cease to divide?
Answer:
If meristematic cells cease to divide, the growth of the plant will be hindered and will undergo a period of dormancy depending upon the seasonal changes in the climate. „

Question 3.
Growth is one of the characteristics of all living organisms? Do unicellular organisms also grow? If so, what are the parameters?
Answer:
Yes, unicellular organisms also grow. Their cell size increases up to a certain fixed dimension only.

Question 4.
Mention the name of the internal factors that control development in plants.
Answer:
Internal factors that control development in plants are as follows:

Genetic factors (intracellular)
Plant growth regulators (intercellular).

Question 5.
Identify the plant hormone-related with intermodal elongation.
Answer:
Gibberellin is the plant hormone-related with internodal elongation.

Question 6.
Mention the name the growth regulator, which was first isolated from endosperm of maize. Give its main biological activity.
Answer:
Zeatin is the growth regulator isolated from endosperm of maize. It controls cell division (cytokinesis) even in non-meristematic tissues.

Question 7.
In most plants, the terminal bud suppresses the development of lateral buds into branches. What is this phenomenon called? Name one phytohormone that can promote this phenomenon.
Answer:
The phenomenon is called apical dominance. Auxin is the phytohormone involved in prompting this phenomenon.

Question 8.
Which air pollutant also acts as a plant hormone?
Answer:
Ethylene.

Question 9.
How do gibberellin help in promoting seed germination?
Answer:
The gibberellin mobilize storage reserves by amylases during germination of seeds.

Question 10.
What are the plant organs responsible for the perception of light variation? What is the pigment responsible for this perception?
Answer:

Leaves are mainly responsible for perception of light intensity in plants.
The pigment that performs this perception is called phytochrome.

Question 11.
Name the hormones involved in photoperiodism.
Answer:
Florigen is the hormone involved in photoperiodism.

Question 12.
Beetroot is often known as a long-day plant. Explain why?
Answer:
Beetroot is known as long-day plant because flowering takes place when the plants are exposed to day length longer than a critical period.

Short answer type questions

Question 1.
An owner of an apple orchard wants to get better yield and wants to wait for good market conditions to sell his apples. Which PGR should he use to achieve his goals?
Answer:
He should use Gibberellins. Gibberellins help increase the size of apples. Moreover, they also delay senescence so apples can be left on branches for a longer duration. This will give the orchard owner enough time to wait for good market conditions.

Question 2.
What are plasticity and heterophylly?
Answer:
In some plants, certain structures show different forms, in response to environment or to phases of life. This ability is known as plasticity.
PSEB 11th Class Biology Important Questions Chapter 15 Plant Growth and Development 1
For example, in cotton, coriander, and larkspur, leaves of juvenile plants are different in shape compared to leaves in mature plants. This is called heterophylly. Juvenile In buttercup shape of leaves produced in air is different from that produced in water.

Question 3.
What are the favorable conditions for seed germination?
Answer:
Favorable conditions for seed germination are given below:

Proper temperature
Moisture
Sunlight
Oxygen.

Question 4.
What are the various man-made meant of overcoming seed dormancy?
Answer:
Man-made means of overcoming seed dormancy are given below :
The seed-coat barrier in some seeds can be broken by mechanical abrasions using knives, sandpaper, etc. or by vigorous shaking.

Effect of inhibitory substances can be removed by subjecting the seeds to chilling conditions or by application of certain chemicals like gibberellic acid and nitrates. Changing environment conditions: like light and temperature.

Question 5.
What do you understand by apical dominance?
Answer:
In most of the higher plants, growing apical bud inhibits the growth of lateral (axillary buds). This phenomenon is called apical dominance. Removal of short tips (decapitation) normally results in the growth of lateral buds. Decapitation is used in tea plantations to get more leaves from a plant.

Long answer type questions

Question 1.
Mention the phenomenon of growth in plants. Explain the phases of growth in detail.
Answer:
Growth is defined as a permanent or irreversible increase in dry weight, mass or volume of cell, organ or organism.
Plant growth takes place in three steps or phases cell, division (meristematic), cell elongation and cell maturation.
(i) Cell Division (Meristematic) Phase:

It is also called formative phase.
New cells are produced by mitotic divisions of the pre-existing cells. The meristematic cells have thin cellulose walls with abundant plasmodesma connections, dense protoplasm and conspicuous nuclei.
In higher plants cell division occurs in meristems or growing points.
As the formation of new cells requires intense biosynthetic activity, the rate of respiration in the cells of formation phase is very high.

(ii) Cell Elongation Phase:

It is also called phase of cell enlargement.
This phase lies just behind the growing points and is mainly responsible for growth of plant parts.
The newly formed cells, produced informative phase undergo enlargement.
The cell walls of the enlarging of cell show plastic extension through enzymatic loosening of microfibrils and deposition of new materials.
The enlarging cell also develops a central vacuole, rate of respiration is high but less than that of the cells in the formative phase.
Thus, this phase is characterized by cell enlargement, new cell wall deposition and increased vacuolation.

(iii) Cell Maturation Phase

This phase occurs just behind the phase of elongation.
The enlarged cells develop into particular type of cells by undergoing structural and physiological differentiation.
Hence, at this phase, all the diverse tissue types observed in root or stem.

PSEB 11th Class Biology Important Questions Chapter 8 Cell: The Unit of Life

July 30, 2024July 29, 2024 by Prasanna

Punjab State Board PSEB 11th Class Biology Important Questions Chapter 8 Cell: The Unit of Life Important Questions and Answers.

PSEB 11th Class Biology Important Questions Chapter 8 Cell: The Unit of Life

Very short answer type questions

Question 1.
Give one difference between the characteristic feature of Gram positive and Gram negative bacteria.
Answer:
Gram positive bacteria retain crystal violet dye and stain blue in colour, while Gram negative bacteria loose dye colour on washing.

Question 2.
What is the name given to the infoldings of plasmalemma in fungal cell below the wall?
Answer:
Mesosomes.

Question 3.
What does ‘S’ refers in a 70S and 80S ribosome? [NCERT Exemplar]
Answer:
The ‘S’ refers to Svedberg units of sedimentation coefficient. The sedimentation coefficient is a measure of the speed of the sedimentation for a particular cell organelle in ultracentrifuge.

Question 4.
What are inclusion bodies?
Answer:
Reserve material in prokaryotic cells are stored in the cytoplasm in the form of inclusion bodies.

Question 5.
In which organelle the proteins required for functioning of nucleus are formed?
Answer:
Proteins required are formed in cytoplasm.

Question 6.
Which is considered to be the main arena of cellular activities in plant and animal cells?
Answer:
Cytoplasm is considered to be the main arena of cellular activities.

Question 7.
Mention a single membrane bound organelle, which is rich in hydrolytic enzymes. [NCERT Exemplar]
Answer:
Lysosome.

Question 8.
What is the significance of vacuole in a plant cell? [NCERT Exemplar]
Answer:
The vacuole in a plant cell helps to maintain osmotic pressure for turgidity and osmosis. It also stores useful as well as waste substances.

Question 9.
Why are the mitochondria and plastids called semi-autonomous particles?
Answer:
These are called so, because they are not dependent upon nuclear DNA and cytoplasmic ribosomes for the synthesis of proteins, while other organelles are dependent.

Question 10.
What is referred to as satellite chromosome? [NCERT Exemplar]
Answer:
Sometimes, few chromosomes have non-staining secondary constrictions at a constant location. This gives appearance of a small fragment called satellite. The chromosome having satellite are known as satellite chromosomes.

Question 11.
What do you meant by 9+2 pattern of organisation?
Answer:
It means that 9 microtubule doublets surround two single microtubules.

Question 12.
Define the microbodies.
Answer:
Membrane bound minute vesicles that contain various enzymes are called microbodies. These are present in both plant and animal cells.

Short answer type questions

Question 1.
Differentiate between prokaryotic and eukaryotic cells.

Prokaryotic Cell	Eukaryotic Cell
•» Nuclear membrane absent	Nuclear membrane present
•» Cell organelles absent (except ribosome)	Cell organelles present
•» Endomembrane system absent	Endomembrane system present
•» Example: bacteria	Example: RBC, neuron

Question 2.
Describe passive transport, osmosis and active transport in plasma membrane.
Answer:
Passive Transport: The plasma membrane is selectively permeable to some molecules present on either side of it. Many molecules can move briefly across the membrane without any requirement of energy and this is called the passive transport.

Osmosis: Neutral solutes may move across the membrane by the process of simple diffusion along the concentration gradient, i.e., from higher concentration to the lower. Water may also move across this membrane from higher to lower concentration. Movement of water by diffusion is called osmosis.

Active Transport: As the polar molecules cannot pass through the nonpolar lipid bilayer, they require a carrier protein of the membrane to facilitate their transport across the membrane. A few ions or molecules are transported across the membrane against their concentration gradient, i.e., from lower to the higher concentration. Such a transport is an energy dependent process, in which ATP is utilised and is called active transport, e.g., Na⁺/K⁺ Pump.

Question 3.
What is endomembrane system in a cell?
Answer:
Certain cell organelles function in coordinated manner. Their v coordination makes the endomembrane system of cell. Following structures comprise the endomembrane system:

Endoplasmic Reticulum
Golgi Complex
Lysosome, and
Vacuoles.

Question 4.
Cells in the epithelial tissue are held together with very little intercellular material (matrix). Specialised junctions provide both structural and functional links between individual cells. Three types of cell junctions have been recognised.
(a) Name the three types of cell junctions and write their functions.
(b) What value is shown by such an arrangement?
Answer:
(a) The three types of cell junctions are:

Tight junctions,
Gap junctions and
Adhering junctions

Functions of cell junctions are ‘

Tight junctions prevent the leaking of substances across a tissue.
Gap junctions facilitate the cells to communicate with each other by connecting their cytoplasm.
Adhering junctions perform cementing of the adjacent cells to keep them together.

(b) No individual can function alone in a society. We need some form of interaction and help from others; so lend a helping hand to the needy.

Question 5.
Give a brief description of ribosomes.
Answer:
Ribosomes are the granular structures first observed under the electron microscope as dense particles by George Palade (1953). They are composed of ribonucleic acid (RNA) and proteins and are not surrounded by any membrane. The eukaryotic ribosomes are 80S while the prokaryotic ribosomes are 70S. Here ‘S’ stands for the sedimentation coefficient; it indirectly is a measure of density and size. Both 70S and 80S ribosomes are composed of two subunits.

Question 6.
Write a short note on-
(i) Cytoskeleton,
(ii) Cilia & Flagella
Answer:
(i) Cytoskeleton: An elaborate network of filamentous proteinaceous structures present in the cytoplasm is collectively referred to as the cytoskeleton. The cytoskeleton in a cell are involved in many functions such as mechanical support, motility, maintenance of the shape of the cell.

(ii) Cilia and Flagella: Cilia (sing.: cilium) and flagella (sing.: flagellum) are hair-like outgrowths of the cell membrane. Cilia are small structures which work like oars, causing the movement of either the cell or the surrounding fluid. Flagella are comparatively longer and responsible for cell movement. The prokaryotic bacteria also possess flagella but these are structurally different from that of the eukaryotic flagella.

Long answer type questions

Question 1.
Write the functions of the following: [NCERT Exemplar]
(i) Centromere
(ii) Cell wall
(iii) Smooth ER
(iv) Golgi apparatus
(v) Centrioles
Answer:
(i) Functions of Centromere
(a) It is a narrow non-stainable area which join chromatids together to form a chromosome.
(b) The centromere thus, keep the two chromatids of a chromosome in an intact stage.
(c) This is an essential step for chromosomes of a cell during cell division whether it may be mitosis or meiosis.

(ii) Functions of Cell Wall
(a) It helps in providing a definite shape to the cell and also protects protoplasm against any mechanical injury, i.e., damage and infection.
(b) It also helps in cell-to-cell interaction.
(c) It provides barrier to undesirable macromolecules and attack of pathogens.

(iii) Functions of Smooth Endoplasmic Reticulum
(a) It provides mechanical support to colloidal complex of cytoplasmic matrix.
(b) It holds various cell organelles in position.
(c) It conducts information from outside to inside of cell.

(iv) Functions of Golgi Apparatus
(a) It performs the function of packaging material.
(b) It acts as an important site for the formation of glycoproteins and glycolipids.
(c) It helps in the production of complex carbohydrates other than glycogen and starch.
(d) It helps in the formation of cell wall.

(v) Functions of Centrioles
(a) Formation of new centrioles from pre-existing one’s during cell division.
(b) They form basal bodies, which in turn form cilia and flagella.
(c) They form spindle fibres that give rise to spindle apparatus during cell division .in animal cells.

Question 2.
Is there a species or region specific type of plastids? How does one distinguish one from the other? [NCERT Exemplar]
Answer:
Plastids are both region or species specific. These are as follows :
(i) Proplastids: These are colourless, rounded but amoeboid plastid precursors, found in meristematic and newly formed cells of plants. It has a double membrane envelope that surrounds a colourless matrix, containing DNA, ribosomes and reserve food. A few vesicles and lamellae also occur in the matrix.

(ii) Leucoplasts: These are colourless plastids that occur in non-green plant cells commonly near the nucleus. They are as follows:
(a) Amyloplasts: These leucoplasts store starch, e.g., tuber of potato, grain of rice and wheat.
(b) Elaioplasts: These store fats, e.g., Rose.
(c) Aleuroplasts: They are protein storing plastids, e.g., Castor endosperm.

(iii) Chromoplasts: These are non-photosynthetic coloured plastids which synthesise and store carotenoid pigments. They appear, orange, red or yellow in colour. These mostly occur in ripe fruits (tomato and chillies) carrot roots, etc.

(iv) Chloroplasts: These are photosynthetic plastids, which are green in colour and found in the leaves of all green plants. They have lamellae organised in the form of grana.

PSEB 11th Class Biology Solutions Chapter 15 Plant Growth and Development

July 30, 2024July 29, 2024 by Prasanna

Punjab State Board PSEB 11th Class Biology Book Solutions Chapter 15 Plant Growth and Development Textbook Exercise Questions and Answers.

PSEB Solutions for Class 11 Biology Chapter 15 Plant Growth and Development

PSEB 11th Class Biology Guide Plant Growth and Development Textbook Questions and Answers

Question 1.
Define growth, differentiation, development, dedifferentiation, redifferentiation, determinate growth, meristem and growth rate.
Answer:
Growth: It is an irreversible permanent increase in size of an organ or its parts or even of an individual cell.
Determinate Growth: Although growth in most of the plant parts is unlimited, certain parts grow up to a certain level and then stop growing. This kind of growth is known as determinate growth.

Development: It is the process of whole series of changes which an organism goes through during its life cycle.
Meristem: The cells of which the capacity to divide and self-perpetuate.

Growth Rate: The increased growth per unit time is termed as growth rate. Thus, rate of growth can be expressed mathematically. An organism or a part of the organism can produce more cells in a variety of ways. The growth rate shows an increase that may be arithmetic or geometrical.

Differentiation: The cells derived from root apical and shoot-apical meristems and cambium differentiate and mature to perform specific functions. This act leading to maturation is termed as differentiation.

Dedifferentiation: The living differentiated cells, that by now have lost the capacity to divide can regain the capacity of division under certain conditions. This phenomenon is termed as dedifferentiation. For example, formation of meristems; interfascicular cambium, and cork cambium from fully differentiated parenchyma cells. Redifferentiation: While undergoing dedifferentiation plant cells once again lose their capacity to divide and become mature. This process is called redifferentiation.

Question 2.
Why is not anyone parameter good enough to demonstrate growth throughout the life of a flowering plant?
Answer:
Anyone parameter is not good enough to demonstrate growth throughout the life of a flowering plant because the plants exhibit different types of growth during different stages of their life cycle. In the seedling stage, they are in state of active cell division (i.e., mitotic divisions), then they undergo active cell enlargement stage during growing stage. In the reproductive or flowering stage of their life cycle, they exhibit reductional divisions. Finally, after the formation of various organs, they undergo cell differentiation or get matured.

Question 3.
Describe briefly:
(a) Arithmetic growth
(b) Geometric growth
(c) Sigmoid growth curve
(d) Absolute and relative growth rates
Answer:
(a) Arithmetic Growth: In arithmetic growth, following mitotic cell division only One daughter cell continues to divide while the other differentiates and matures. Example is a root elongating at a constant rate.

(b) Geometric Growth: In geometrical growth, in most systems, the initial growth is Size of slow (lag phase), and it organ increases rapidly thereafter at an exponential rate (log or exponential phase). Here, both the progeny cells following mitotic cell division retain the ability to divide and continue to do so.
PSEB 11th Class Biology Solutions Chapter 15 Plant Growth and Development 1
(c) Sigmoid Growth Curve: If we plot the parameter of growth against time, we get a typical sigmoid or S-curve. A sigmoid curve is a characteristic of living organism growing in a natural environment. It is typical for all cells, tissues and organs of a plant.

(d) Absolute and Relative Growth Rates: The measurement and the comparison of total growth per unit time is called the absolute growth rate. And the growth of the given system per unit time expressed on a common basis, e. g., per unit initial parameter is called the relative growth rate.

Question 4.
List five main groups of natural plant growth regulators. Write a note on discovery, physiological functions, and agricultural/ horticultural applications of any one of them.
Answer:
Five main groups of natural plant growth regulators are auxins, gibberellins, ethylene, cytokinins and abscisic acid.
Auxins
(i) Discovery: Auxins was first isolated from human urine. They are generally produced by the growing apices of the stems and roots. Auxins like IAA (indole acetic acid) and indole butyric acid (IBA) have been isolated from plants. NAA (naphthalene acetic acid) and 2, 4-D (2, 4-dichloro phenoxy acetic) are synthetic auxins.

(ii) Physiological Function: They help to initiate rooting in stem cuttings, an application widely used for plant propagation. Auxins promote flowering, i. e., in pineapples. They help to prevent fruit and leaf drop at early stages but promote the abscission of older mature leaves and fruits.

(iii) Agricultural/Horticultural Applications: Auxins also induce parthenocarpy, e. g., in tomatoes. They are widely used as herbicides 2, 4-D, widely used to kill dicotyledonous seeds, does not affect mature monocotyledonous plants. It is used to prepare seed-free lawns by gardeners. Auxin also controls xylem differentiation and helps in cell division.

Question 5.
What do you understand by photoperiodism and vernalization? Describe their significance.
Answer:
1. Photoperiodism: The response of plants to periods of day/night is termed as photoperiodism. The site of perception of light/dark duration are the leaves.

Significance: The significance of photoperiodism is in regulating flowering in plants. Flowering is an important step towards seed formation and seeds are responsible for continuing the generation of a plant.

2. Vernalisation: There are plants in which flowering is either quantitatively or qualitatively dependent on exposure to low temperatures. This phenomenon is termed as vernalization. Vernalisation refers especially to the promotion of flowering by a period of low temperature.

Significance: Vernalisation prevents precocious reproductive development late in the growing season. This enables the plant to have sufficient time to reach maturity.

Question 6.
Why is abscisic acid also known as stress hormone?
Answer:
Abscisic acid acts as a general plant growth inhibitor and an inhibitor of plant metabolism. ABA inhibits seed germination. ABA stimulates the closure of stomata in the epidermis and increases the tolerance of plants to various kinds of stresses. Therefore, it is also called the stress hormone.

Question 7.
‘Both growth and differentiation in higher plants are open Comments.
Answer:
Plant growth is unique because plants retain the capacity for unlimited growth throughout their life. This ability of the plants is due to the presence of meristems at certain locations in their body. The cells of such meristems have the capacity to divide and self-perpetuate. The product, however, soon loses the capacity to divide and such cells make up the plant body. This form of growth wherein new cells are always being added to the plant body by the activity of the meristem is called the open form of growth.

Question 8.
‘Both a short day plant and a long day plant can produce flower simultaneously in a given place’. Explain.
Answer:
Petkus winter rye (Secale cereal) gives responses of low temperature at very young seedlings or even at seed stage. If winter rye is shown in the spring, the seeds germinate and produce vegetative plants in the following summer. In this case, the period of vegetative growth is extended and flowering occurs only in the next summer when the cold requirements is fulfilled during winters. The same variety, if grown in early autumn produces flowers in the following summer.

Question 9.
Which one of the plant growth regulators would you use if you are asked to:
(a) induce rooting in a twig
(b) quickly ripen a fruit.
(c) delay leaf senescence
(d) induce growth in axillary buds
(e) ‘bolt’ a rosette plant
(f) induce immediate stomatal closure in leaves.
Answer:
(a) Cytokinins
(b) Ethylene
(c) Cytokinins
(d) Auxin
(e) Gibberellins
(f) Abscisic acid.

Question 10.
Would a defoliated plant respond to the photoperiodic cycle? Why?
Answer:
No, a defoliated plant do not respond to the photoperiodic cycle. Because leaves of a plant are the sites of light perception for the induction of flowering.

Question 11.
What would be expected to happen if:
(a) GA₃ is applied to rice seedlings.
(b) dividing cells stop differentiating.
(c) a rotten fruit gets mixed with unripe fruits.
(d) you forget to add cytokinin to the culture medium.
Answer:
(a) The rice seedlings show extraordinary elongation of stem and leaf sheaths.
(b) Tissue and organ differentiation will not take place.
(c) Unripe fruits will also get rotten due to the ethylene hormone secreted by rotten fruit.
(d) No root and shoot formation will take place.

PSEB 11th Class Biology Important Questions Chapter 7 Structural Organisation in Animals

July 30, 2024July 29, 2024 by Prasanna

Punjab State Board PSEB 11th Class Biology Important Questions Chapter 7 Structural Organisation in Animals Important Questions and Answers.

PSEB 11th Class Biology Important Questions Chapter 7 Structural Organisation in Animals

Very short answer type questions

Question 1.
A muscular fibre tapers at both the ends and does not show striations. Name the muscle fibre. [NCERT Exemplar]
Answer:
Smooth or non-striated muscle fibre.

Question 2.
Name few called the specialised connective tissues?
Answer:
Bones, cartilage and blood are the special types of connective tissues.

Question 3.
How does a gap junction facilitate the intercellular communication? [NCERT Exemplar]
Answer:
A gap junction facilitates the cells to communicate with each other by connecting the cvtoplasm of the adjacent cells.

Question 4.
Which tissue forms the ligaments? What is their function?
Answer:
Ligaments are formed of yellow elastic connective tissue. The ligaments join the bones together.

Question 5.
On which segment of the body, male and female genital apertures are present in earthworm?
Answer:
Male – 18th segment; Female – 14th segment.

Question 6.
Why the body segmentation in earthworm is called metameric segmentation ?
Answer:
In metameric segmentation, the external segmentation corresponds to the internal segmentation of the body.

Question 7.
In earthworm, from which segment intestine starts and where it ends?
Answer:
In earthworm, intestine starts from the 15th segment onwards and continues till the last segment.

Question 8.
Where the sclerites are present in cockroach? [NCERT Exemplar]
Answer:
In the exoskeleton all over the body.

Question 9.
Which month part of cockroach is comparable to our tongue? [NCERT Exemplar]
Ans.
Hypopharynx.

Question 10.
Name the upper lip and lower lip of the cockroach.
Answer:
Upper lip – Labrum
Lower lip – Labium

Question 11.
Why the blood of cockroach is not responsible for transporting respiratory gases?
Answer:
Because the respiratory pigment is absent in their blood.

Question 12.
List the parts of blood vascular system of cockroach.
Answer:
Haemocoel, heart and blood.

Short answer type questions

Question 1.
Answer:

Smooth Muscle	Skeletal Muscle	Cardiac Muscle
1. Spindle shaped cells	Striated unbranched cells	Striated branched cells
2. Found in muscles of internal organs	Found	Found in the heart
3. Control involuntary actions	Control voluntary actions	Control heart’s pumping which is involuntary

Question 2.
What do you understand by special junctions between cells? Which type of special junction are found in epithelial tissues?
Answer:
All cells are held together with intercellular material. These materials form junctions between cells.
There are three types of cell junctions found in epithelial tissue.
(a) Tight Junctions: These junctions help in stopping leakage of substances across a tissue.
(b) Adhering Junctions: These junctions keep neighbouring cells together.
(c) Gap Junctions: These junctions connect cytoplasm of adhering cells, and facilitate exchange of materials.

Question 3.
Explain fertilization and development in Pheretima.
Answer:
Fertilization and Development in Pheretima

A mutual exchange of sperm occurs between two worms during mating. One worm has to find another worm and they mate juxtaposing opposite gonadal openings exchanging packets of sperms called spermatophores.
Mature sperm and egg cells and nutritive fluid are deposited in cocoons produced by the gland cells of clitellum. Fertilisation and development occur within the cocoons which are deposited in soil.
The cocoon holds the worm embryos. After about 3 weeks each cocoon produces two to twenty baby wopns with an average of four. Earthworms development is direct, i.e., there is no larva formed.

Question 4.
Describe mouth parts of the cockroach. ,
Answer:
Mouth Parts of Cockroach: Anterior end of the head bears appendages forming biting and chewing type of mouth parts. The mouth ‘ parts consist of following structures:
(a) Labrum (upper lip),
(b) A pair of mandibles,
(c) A pair of maxillae and
(d) Labium (lower lip).
Apart from these a median flexible lobe, acting as tongue (hypopharynx), lies within the cavity enclosed by the mouth parts.

Question 5.
Explain in brief the nervous system of cockroach.
Answer:
Nervous System of Cockroach

The nervous system of cockroach consists of a series of fused, segmentally arranged ganglia joined by paired longitudinal connectives on the ventral side.
Three ganglia lie in the thorax, and six in the abdomen. The nervous system of cockroach is spread throughout the body.
The head holds a bit of a nervous system while the rest is situated along the ventral (belly-side) part of its body.

Question 6.
Explain digestion in frogs.
Answer:
Digestion in Frog

In the stomach digestion of food takes place by the action of HCl and gastric juices secreted from the walls of the stomach.
Partially digested food called chyme is passed from stomach to the first part of the intestine, the duodenum. The duodenum receives bile from gall bladder and pancreatic juices from the pancreas through a common bile duct.
Bile emulsifies fat and pancreatic juices digest carbohydrates and proteins. Final digestion takes place in the intestine.
Digested food is absorbed by the numerous finger-like folds in the inner wall of intestine called villi and microvilli.

Question 7.
Explain in brief the central nervous system of frog.
Answer:
Central Nervous System of Frog ,

Brain is enclosed in a bony structure called brain box or skull (cranium).
The brain is divided into fore-brain, mid-brain and hind-brain.
Forebrain includes olfactory lobes, paired cerebral hemispheres and unpaired diencephalon.
The midbrain is characterised by a pair of optic lobes.
Hind-brain consists of cerebellum and medulla oblongata.
The medulla oblongata passes out through the foramen magnum and continues into spinal cord, which is enclosed in the vertebral column.
There are ten pairs of cranial nerves arising from the brain.

Question 8.
How do different senses work in frog? Explain in brief. ,
Answer:
Sense Organs in Frog : Frog has different types of sense organs which are as follows:
(a) Sensory papillae or organs of touch,
(b) Taste buds.
(c) Nasal epithelium for the sense of smell,
(d) Eyes for vision and
(e) Tympanum ‘with internal ears for hearing.

Out of these, eyes and internal ears are well-organised structures and the rest are cellular aggregations around nerve endings. Eyes in a frog are a pair of spherical structures situated in the orbit in skull. These are simple eyes. External ear is absent in frogs and only tympanum can be seen externally. The ear is an organ of hearing as well as balancing.

Long answer type questions

Question 1.
Which features distinguish blood from lymph?
Answer:
Differences between blood and lymph are given below :

Blood	Lymph
1. It contains plasma, erythrocytes, leucocytes and platelets.	It contains plasma and leucocytes.
2. The presence of haemoglobin imparts red colour to it.	It is colourless as haemoglobin is absent.
3. Its plasma contains more protein, calcium and phosphrous as compared to lymph.	Its plasma has fewer protein and less calcium and phosphorus than blood.
4. Contains moderate amount of CO₂and other metabolic waste.	Contains excessive amount of CO₂and other metabolic waste.

Question 2.
In which segment does the following structures lies in the earthworm’s body?
(i) Spermathecae
(ii) Pharynx
(iii) Gizzard
(iv) Intestine
(v) Septal nephridia
(vi) Ovary
(vii) Testes
(viii) Typhlosole
(ix) Lateral heart
(x) Pharyngeal nephridia
Answer:
(i) Spermathecae – 6th, 7th, 8th, 9th
(ii) Pharynx – 4th
(iii) Gizzard – 8th
(iv) Intestine – 15th to last
(v) Septal nephridia – 15th to last
(vi) Ovary – 13th
(vii) Testes – 10th, 11th
(viii) Typhlosole – 26th-95th
(ix) Lateral heart – 7th, 9th
(x) Pharyngeal nephridia – 4th-6th

PSEB 11th Class Biology Solutions Chapter 7 Structural Organisation in Animals

July 30, 2024July 29, 2024 by Prasanna

Punjab State Board PSEB 11th Class Biology Book Solutions Chapter 7 Structural Organisation in Animals Textbook Exercise Questions and Answers.

PSEB Solutions for Class 11 Biology Chapter 7 Structural Organisation in Animals

PSEB 11th Class Biology Guide Structural Organisation in Animals Textbook Questions and Answers

Question 1.
Answer in one word or one line:
(i) Give the common name of Periplanata americana.
(ii) How many spermathecae are found in earthworm?
(iii) What is the position of ovaries in cockroach?
(iv) How many segments are present in the abdomen of cockroach?
(v) Where do you find Malpighian tubules?
Answer:
(i) American cockroach
(ii) 4 pairs
(iii) Under the 4th, 6th abdominal terga
(iv) 10 segments in adults
(v) At the junction of midgut and hindgut in cockroach.

Question 2.
Answer the following:
(a) What is the function of nephridia?
(b) How many types of nephridia are found in earthworm based on their location?
Answer:
(a) The Function of Nephridia: The nephridia regulate the volume and composition of the body fluids. The nephridium starts out as a funnel that collects excess fluid from coelomic chamber. The funnel connects with a tubular parts of the nephridium, which delivers the wastes through a pore to the surface in the body wall in the digestive tube.

(b) Based on their location, there are following three types of nephridia in earthworm:

Septal nephridia.
Pharyngeal nephridia.
Integumentary nephridia.

Question 3.
Draw a labelled diagram of the reproductive organs of an earthworm.
Answer:
PSEB 11th Class Biology Solutions Chapter 7 Structural Organisation in Animals 1

Question 4.
Draw a labelled diagram of alimentary canal of a cockroach.
Answer:
PSEB 11th Class Biology Solutions Chapter 7 Structural Organisation in Animals 2

Question 5.
Distinguish between the following:
(a) Prostomium and peristomium.
(b) Septal nephridium and pharyngeal nephridium.
Answer:
(a) Prostomium and Peristomium: The first segment of earthworm with a ventral mouth is known as peristomium. Prostomium is a dorsal, lobe which is present on the ventral mouth.

(b) Septal Nephridia and Pharyngeal Nephridia: Septal nephridia are present on both the sides of intersegmental septa of segment 15 to the ‘ last that open into intestine.
The pharyngeal nephridia are closed (no nephrostome) nephridia present as three paired groups (of about 100) in 4th, 5th and 6th segments.

Question 6.
What are the cellular components of blood?
Answer:
The cellular components of blood are red blood cells, white blood cells and platelets.

Question 7.
What are the following and where do you find them in animal body (a) Chondrocytes (b) Axons (c) Ciliated epithelium.
Answer:
(a) Chondrocytes: These are the matrix secreting cells of the cartilage. These are found in the cartilage of connecting tissue.

(b) Axon: It is a long fibre, the distal end of which is branched. Each branch terminates as a bulb like structure called synaptic knob. The axon transmit nerve impulses away from the cell body.

(c) Ciliated Epithelium: If the columnar or cuboidal cells of columnar and cuboidal epithelium bear cilia on their free surface they are called ciliated epithelium.

Question 8.
Describe various types of epithelial tissues with the help of labelled diagrams.
Answer:
Epithelial Tissues: Epithelial tissues provide covering to the inner and outer lining of various organs. There are following two types of epithelial tissues :
1. Simple epithelium: Simple epithelium is composed of a single layer of cells. It functions as a lining for body cavities, ducts and tubes.

2. Compound epithelium: The compound epithelium consists of two or more cell layers. It has protective function as it does in our skin. It covers the dry surface of the skin, the moist surface of buccal cavity, pharynx, inner lining of ducts of salivary glands and of pancreatic ducts.
On the basis of structural modifications of the cells, simple epithelium is further divided into three types. These are :
(a) Squamous epithelium: The squamous epithelium is made of a single thin layer of flattened cells with irregular boundaries. They are found in the walls of blood vessels and air sacs of lungs and are involved in functions like forming a diffusion boundary.

(b) Cuboidal epithelium: The cuboidal epithelium is composed of a single layer of cube-like cells. This is commonly found in the ducts of glands and tubular parts of nephrons in kidneys. Its main functions are secretion and absorption.

(c) Columnar epithelium: The columnar epithelium is composed of a single layer of tall and slender cells. They are found in the lining of stomach and intestine and help in absorption and secretion.
(i) When the columnar or cuboidal cells bear cilia on their free surface they are called ciliated epithelium. Their function is to move particles or mucus in a specific direction over the epithelium organs like bronchioles and fallopian tubules.
(ii) Some of the columnar or cuboidal cells get specialized for secretion are called glandular epithelium. They are unicellular and multicellular.
PSEB 11th Class Biology Solutions Chapter 7 Structural Organisation in Animals 3

Question 9.
Distinguish between:
(a) Simple epithelium and compound epithelium
(b) Cardiac muscle and striated muscle
(c) Dense regular and dense irregular connective tissue
(d) Adipose tissue and blood tissue
(e) Simple gland and compound gland
Answer:
(a) Differences between Simple and Compound Epithelium

Simple Epithelium	Compound Epithelium
1. It is composed of a single layer of	It consists of two or more cell layers.
2. It functions as a lining for body cavities, ducts and tubes.	It is protective in function like our skin.

(b) Differences between Cardiac and Striated Muscle

Cardiac Muscle	Striated Muscle
1. It occurs only in the wall of heart.	It occurs in the body wall, limb, tongue, pharynx, etc.
2. They are short and cylindrical with truncate ends.	They are long and cylindrical with blunt ends.
3. They have nerve supply from brain and autonomous nerve system	They have nerve supply from central nervous system.

Dense Regular Connective Tissue	Dense Irregular Connective Tissue
Collagen fibres are present in rows between many parallel bundle of fibres. Examble: Tendons	Fibroblasts and many fibre are present that are oriented differently. Examble: Cartilage, bones and blood.

(d) Differences between Adipose Tissue and Blood Tissue

Adipose Tissue	Blood Tissue
1. It is a soft gel like connective tissue.	It is a fluid connective tissue.
2. It is partitioned into lobules by septa.	There are no partitions.
3. It is a storage tissue.	It is a transport tissue.
4. Matrix is secreted by the cells.	Matrix is not secreted by the cells.
5. It contain fibres.	Fibres are not conspicuous.
6. Adipocytes contain fat droplets.	No cell of the tissue contains fat droplets.

(e) Differences between Simple and Compound Gland

Simple Gland	Compound Gland
1. These glands have single unbranched duct.	These glands have branched system of ducts.
2. These may be simple tubular glands, simple coiled tubular glands and simple alveolar glands.	These may be compound tubular glands, compound alveolar glands

Question 10.
Mark the odd one in each series:
(a) Areolar tissue, blood, neuron, tendon
(b) RBC, WBC, platelets, cartilage
(c) Exocrine, endocrine, salivary gland; ligament
(d) Maxilla, mandible, labrum, antennae
(e) Protonema, mesothorax, metathorax, coxa
Answer:
(a) Neuron,
(b) Cartilage,
(c) Ligament,
(d) Antennae,
(e) Protonema.

Question 11.
Match the terms in column I with those in column II.

Column I	Column II
A. Compound epithelium	1. Alimentary canal
B. Compound eye	2. Cockroach
C. Septal nephridia	3. Skin
D. Open circulatory system	4. Mosaic vision
E. Typhlosole	5. Earthworm
F. Osteocytes	6. Phallomere
G. Genitalia	7. Bone

Answer:

Column I	Column II
A. Compound epithelium	3. Skin
B. Compound eye	4. Mosaic vision
C. Septal nephridia	5. Earthworm
D. Open circulatory system	2. Cockroach
E. Typhlosole	1. Alimentary canal
F. Osteocytes	7. Bone
G. Genitalia	6. Phallomere

Question 12.
Mention briefly about the circulatory system of earthworm.
Answer:
Pheretima exhibits a closed type of blood vascular system, consisting of blood vessels, capillaries and heart. Due to closed circulatory system, blood is confined to the heart and blood vessels. Contractions keep blood circulating in one direction. Smaller blood vessels supply the gut, nerve cord, and the body wall. Blood glands are present on the 4th, 5th and 6th segments. They produce blood cells and haemoglobin which is dissolved in blood plasma. Blood cells are phagocytic in nature. Earthworms lack specialised breathing devices. Respiratory exchange occurs through moist body surface into their blood stream.

Question 13.
Draw a neat diagram of digestive system of frog.
Answer:
PSEB 11th Class Biology Solutions Chapter 7 Structural Organisation in Animals 5

Question 14.
Mention the function of the following:
(a) Ureters in frog
(b) Malpighian tubules
(c) Body wall in earthworm
Answer:
(a) Functions of Ureters in Frog: In male frog, two ureters emerge from the kidneys. The ureters act as urinogenital duct which opens into the cloaca. Thus, the ureters carry both sperms and excretory wastes to the cloaca. In female frog, the ureters and oviduct open separately in the cloaca. The ureters in frog, thus acts as carrier of sperms and ova.

(b) Functions of Malpighian Tubules of Cockroach: Excretion is carried out by Malpighian tubules. Each tubule is lined by glandular cells. They absorb excretory waste products and converts them into uric acid which is excreted out through the hindgut.

(c) Functions of Body Wall of Earthworm: The body wall of earthworm has five layers – cuticle, epidermis, circular muscle layer, longitudinal muscle layer, peritoneum.

Cuticle is a non-cellular elastic layer.
The columnar cells of body wall provide support and therefore, are also known as supporting cells.
Epidermis also has gland cells, receptor cells and basal cells.
The glandular cell secrete mucus and thus, keep the skin moist, this
help in cutaneous respiration.
The last layer of the body wall is the outer membrane of the coelom called coelomic epithelium. The various muscle layers of the body wall provide strength and rigidity.

PSEB 11th Class Biology Important Questions Chapter 16 Digestion and Absorption

July 30, 2024July 29, 2024 by Prasanna

Punjab State Board PSEB 11th Class Biology Important Questions Chapter 16 Digestion and Absorption Important Questions and Answers.

PSEB 11th Class Biology Important Questions Chapter 16 Digestion and Absorption

Very short answer type questions

Question 1.
What do we call the type of teeth attachment in which each tooth, is embedded in a socket of jaws bones? [NCERT Exemplar]
Answer:
Thecodont dentition.

Question 2.
Out of the three types of salivary glands, which is situated in the upper palate on either side of cheek.
Answer:
Parotid gland (type of salivary gland) is situated in the upper palate on the either side of the cheek.

Question 3.
If a person is suffering from reflux oesophagitis, which part of his alimentary canal is not functioning properly?
Answer:
In reflux oesophagitis, the oesophagus does not function properly.

Question 4.
HCl is secreted in stomach. Give the name of the cells that secrete it.
Answer:
Oxyntic cells (parietal cells) are the stomach epithelial cells, that secretes gastric acid, i.e., HCl.

Question 5.
Crypts of Lieberkuhn are found in which part of the alimentary canal.
Answer:
These are found in small intestine portion of alimentary canal.

Question 6.
Give the name of the enzymes involved in the breakdown of nucleotides into sugars and bases? [NCERT Exemplar]
Answer:
Nucleosidases.

Question 7.
Trypsinogen is an inactive enzyme of pancreatic juice. An enzyme, enterokinase activates it. Which tissue/cells secrete this enzyme? How is it activated? [NCERT Exemplar]
Answer:
The cells of duodenum secrete enzyme enterokinase. It is activated by food in the duodenum.

Question 8.
Mention the function of saliva other than digestion.
Answer:
Apart from taking part in digestion, saliva also helps to lubricate the food for swallowing.

Question 9.
State the source of trypsin and the food constituent which this enzyme hydrolyses.
Answer:
Pancreatic juice is the source of trypsin which hydrolyses proteins into peptides.

Question 10.
If the bile duct is completely blocked. How would it affect the digestion of food?
Answer:
If the bile duct is blocked completely, the bile will fail to reach the small intestine and the digestion of fats gets affected.

Question 11.
By which type of mechanism, amino acids are absorbed in our body?
Answer:
Small amounts of amino acids are absorbed by active transport and some are absorbed by the facilitated transport.

Question 12.
As fatty acids and glycerol are not absorbed directly. Name the form in which fatty acids are converted to get absorbed?
Answer:
Fatty acids and glycerols are not absorbed directly into the bloodstream. Thus, they are absorbed in the form of small droplets called micelles.

Question 13.
Which type of absorption takes place in large intestine?
Answer:
Absorption of water, some minerals and drugs takes place in the large intestine.

Question 14.
What happens in the condition when bile from the liver crystalises?.
Answer:
When bile from the liver get crystallised, person form stones in the body.

Short answer type questions

Question 1.
What is the role of tongue indigestion?
Answer:
Tongue helps in mixing the food properly with salivary enzymes. Moreover, tongue has tastebuds which give the sense of different tastes. Eating is a complex process which needs involvement of olfactory and visual senses as well. Alongwith tongue all these senses help in picking the right food.

Question 2.
What is the function of large intestine?
Answer:
Functions of large intestine are as follows:

Absorption of some water, minerals and certain drugs.
Secretion of mucus. Mucus helps in adhering the waste particles together and lubricates it for easy passage.

Question 3.
What is digestive waste and how is it removed from the body?
Answer:
After digestion and absorption of food is over the residue left makes the digestive waste. The digestive wastes, solidified into coherent faeces in the rectum initiate a neural reflex causing an urge or desire for its removal. The egestion of faeces to the outside through the anal opening (defaecation) is a voluntary process and is carried out by a mass peristaltic movement.

Question 4.
How does the nervous system control the activities of gastrointestinal tract?
Answer:
The sight, smell and/or presence of food in the oral cavity can stimulate the secretion of saliva. Gastric and intestinal secretions are also stimulated by similar neural signals. Muscular activities of alimentary canal is coordinated by both local and CNS neural mechanisms. Hormonal control of secretion of digestive enzymes is carried out by local hormones.

Question 5.
Write a short note on-Disorders of digestive system.
Answer:
Disorders of Digestive System
(i) Inflammation of intestinal tract
This is the most common disorder of the digestive system.
It is caused by infections by bacteria or viruses and also by parasites like roundworm, hookworm, pinworm, etc.

(ii) Jaundice
It is the infection and inflammation of the liver.
Bile pigments are present in the blood and cause yellowing Of eyes, skin, etc.

(iii) Vomiting
It is the egestion of contents of the stomach through the mouth.

(iv) Diarrhoea
The abnormal frequency of bowel movement and increased liquidity of the faecal discharge is known as diarrhoea.
It reduces the absorption of food.

(v) Constipation
In constipation, the faeces are retained within the rectum as the bowel movements occur irregularly.

(vi) Indigestion
In indigestion, the food is not properly digested leading to a feeling of fullness.
The causes of indigestion are inadequate enzyme secretion, anxiety, food poisoning, overeating, spicy food, etc.

Long answer type questions

Question 1.
As a result of intestinal disease, parts of the alimentary canal are sometimes surgically removed. Discuss the effect of this removal on lifestyle and digestive function:
(i) the stomach
(ii) the colon.
Answer:
(i) Removal of the stomach, which is the main organ for digestion of protein, leads to a change in the patient’s diet. It is the stomach where proteins are first broken down into polypeptides by pepsin in gastric juice before they can be acted on by pancreatic enzymes and intestinal enzymes. The patient must lower down the amount of protein in his diet. Animal meat especially red meat, which is rich in proteins must be avoided.

(ii) Removal of the colon, which is responsible for absorption of water from undigested food, result in loss of water (dehydration) inpatient. The patient needs to drink plenty of fluids to replace water loss. In most cases, where the whole colon is removed, a surgical procedure.is done to attach the small intestine to the rectum to allow for recta elimination of liquid stools. A small pouch is created in the lower abdomen to collect the stool. The patient has to learn to regulate his bowel movements.

Question 2.
A person had roti and dal for his lunch. Trace the changes that will take place during its complete passage through the alimentary canal. [NCERT Exemplar]
Answer:
Changes that will take place in food (roti and dal) through the passage of alimentary canal are given below:

The food substances are first masticated by the teeth in the mouth, where carbohydrate part of the food is digested by the action of salivary amylase enzyme secreted by the salivary glands.
This partially digested food reaches the stomach, where it receives acidic HCl and mainly the protein part of the food is digested by the action of proteolytic enzymes.
The lipid part of the food is digested by the bile secreted by the gall bladder.
In the small intestine, particularly in the duodenum, this semi-digested food is finally digested by the digestive enzymes present in the intestinal and pancreatic juices.
After digestion, the broken down products of food, i.e., amino acids, glycerol, starch, etc., are observed mainly in the small intestine.
The undigested remains of food will finally pass through the anus.

PSEB 11th Class Biology Solutions Chapter 16 Digestion and Absorption

July 30, 2024July 29, 2024 by Prasanna

Punjab State Board PSEB 11th Class Biology Book Solutions Chapter 16 Digestion and Absorption Textbook Exercise Questions and Answers.

PSEB Solutions for Class 11 Biology Chapter 16 Digestion and Absorption

PSEB 11th Class Biology Guide Digestion and Absorption Textbook Questions and Answers

Question 1.
Choose the correct answer among the following:
(i) Gastric juice contains
(a) pepsin, lipase and rennin
(b) trypsin, lipase and rennin
(c) trypsin, pepsin and lipase
(d) trypsin, pepsin and rennin

(ii) Succus entericus is the name given to
(a) a junction between ileum and large intestine
(b) intestinal juice
(c) swelling in the gut
(d) appendix
Answer:
(i) (a) Pepsin, lipase, and rennin
(ii) (b) Intestinal juice.

Question 2.
Match column I with column II.

Column I	Column II
A. Bilirubin and biliverdin	1. Parotid
B. Hydrolysis of starch	2. Bile
C. Digestion of fat	3. Lipases
D. Salivary gland	4. Amylases

Answer:

Column I	Column II
A. Bilirubin and biliverdin	2. Bile
B. Hydrolysis of starch	4. Amylases
C. Digestion of fat	3. Lipases
D. Salivary gland	1. Parotid

Question 3.
Answer briefly:
(a) Why are villi present in the intestine and not in the stomach?
(b) How does pepsinogen change into its active form?
(c) What are the basic layers of the wall of alimentary canal?
(d) How does bile help in the digestion of fats?
Answer:
(a) The mucosa layer of alimentary canal forms small finger-like foldings called villi in the small intestine. The cells lining the villi produce numerous microscopic projections called microvilli giving a brush border appearance. These modifications increase the surface area enormously.
Villi are supplied with the network of capillaries and large lymph vessel called the lacteal mucosal.

(b) The inactive form of enzyme pepsinogen is activated by Rd.

(c) The wall of alimentary canal from esophagus to rectum possesses four layers namely serosa, muscularis, sub-mucosa and mucosa. Serosa is the outermost layer, followed by muscularis, sub-mucosa and mucosa.

(d) Bile salts help in emulsification of lipids and activate the lipases.

Question 4.
State the role of pancreatic juice in digestion of proteins.
Answer:
The pancreatic juice contains inactive enzymes trypsinogen, chymotrypsinogen, procarboxypeptidases, amylases, lipases, and nucleases. Trypsinogen is aëtivated by an enzyme, enterokinase, secreted by the intestinal mucosa into active trypsin, which in turn activates the other enzymes in the pancreatic juice. Proteins, proteases and peptones (partially hydrolyzed proteins) in the chyme reaching the intestine are acted upon by the proteolytic enzymes of pancreatic juice as given below:
PSEB 11th Class Biology Solutions Chapter 16 Digestion and Absorption 1

Question 5.
Describe the process of digestion of protein in stomach.
Answer:
The food mixes thoroughly with the acidic gastric juice of the stomach by the churning movements of its muscular wall and is called the chyme. The pepsinogen, on exposure to hydrochloric acid gets converted into the active enzyme pepsin, the proteolytic enzyme of the stomach. Pepsin converts proteins into proteoses and peptones (peptides).

HCl provides the acidic pH (pH 1.8) optimal for pepsins. Rennin is a proteolytic enzyme found in gastric juice of infants which helps in the digestion of milk proteins.

Question 6.
Give the dental formula of human beings.
Answer:
The dental formula of human beings is
\(\frac{2123}{2123} \times 2\).

Question 7.
Bile juice contains no digestive enzymes, yet it is important for digestion. Why?
Answer:
Bile is yellowish-green alkaline solution with 89-98% water, having no digestive enzymes. The bile released into the duodenum contains bile pigments (bilirubin and biliverdin), bile salts, cholesterol, and phospholipids but no enzymes. Bile helps in emulsification of fats, i.e., breaking down of the fats into very small micelles. Bile also activates lipases.

Question 8.
Describe the digestive role of chymotrypsin. Which two other digestive enzymes of the same category are secreted by its source gland?
Answer:
Chymotrypsin is the active form of chymotrypsinogen. It is activated by trypsin. It curdles milk. Nucleases like DNA ase and RNAase and pancreatic lipase are other enzymes secreted by the pancreas.

Question 9.
How are polysaccharides and disaccharides digested?
Answer:
The chemical process of digestion of carbohydrates is initiated in the oral cavity by the hydrolytic action of the carbohydrate splitting enzyme, the salivary amylase. About 30 percent of starch is hydrolyzed here by this enzyme (optimum pH 6.8) into a disaccharide-maltose. Further, carbohydrates in the chyme are hydrolyzed by pancreatic amylase into disaccharides.

Question 10.
What would happen if HCl were not secreted in the stomach?
Answer:
The mucus and bicarbonates present in the gastric juice play an important role in lubrication and protection of the mucosal epithelium from excoriation by the highly concentrated hydrochloric acid.

HCl provides the acidic pH (pH 1.8) optimal for pepsins. Rennin is a proteolytic enzyme found in gastric juice of infants which helps in the digestion of milk proteins. Small amount of lipases are also secreted by gastric glands.

Question 11.
How does butter in your food get digested and absorbed in the body?
Answer:
Bile helps in emulsification of fats, i. e., breaking down of the fats into very small micelles. Bile also activates lipases.

Question 12.
Discuss the main steps in the digestion of proteins as the food passes through different parts of the alimentary canal.
Answer:
Digestion of Protein in Stomach: The proenzyme pepsinogen, on exposure to HCl, gets converted into active enzyme pepsin.

Pepsin always outs in acidic medium (pH 1.8). In infants, main proteins are digested by rennin.

Digestion of Protein in Small Intestine: Pancreatic juice contains proenzyme trypsinogen. It is activated by enterokinase, secreted by intestinal mucosa, into active trypsin. Trypsin acts in alkaline medium.

The dipeptides are changed into amino acids by the enzyme succus enterics (intestinal juice).

Question 13.
Explain the term ‘the codont’ and ‘diphyodont’.
Answer:
Each tooth is embedded in a socket of jaw bone. This type of attachment is called thecodont. The majority of mammals including human beings forms two sets of teeth during their life, a set of temporary milk or deciduous teeth replaced by a set of permanent or adult teeth. This type of dentition is called diphyodont.

Question 14.
Name different types of teeth and their number in an adult human.
Answer:
An adult human has 32 permanent teeth, which are of four different types (heterodont dentition), i.e., incisors (I), canine (C), premolars (Pm), and molars (M), and their number are 4, 2, 4, 6 respectively.

Question 15.
What are the functions of liver?
Answer:
Liver is the largest gland in human body which is mainly responsible for the digestion of food.
Role of liver in digestion of food :

Its hepatic cells secrete bile juice which passes through the hepatic duct into the gall bladder.
It has its major role in digestion and processing of proteins.
Bile secreted by it is mainly responsible for digestion of fats for easy absorption in the body.
It also responsible for the removal of toxins from blood.

PSEB 11th Class Biology Important Questions Chapter 6 Anatomy of Flowering Plants

July 30, 2024July 29, 2024 by Prasanna

Punjab State Board PSEB 11th Class Biology Important Questions Chapter 6 Anatomy of Flowering Plants Important Questions and Answers.

PSEB 11th Class Biology Important Questions Chapter 6 Anatomy of Flowering Plants

Very short answer type questions

Question 1.
The product of photosynthesis is transported from the leaves to various parts of the plant and stored in some cells before being utilised. What are the cells/tissues that store them?
Answer:
Parenchyma.

Question 2.
What is the function of phloem parenchyma? [NCERT Exemplar]
Answer:
It takes part in lateral conduction of food and supply of water from xylem.

Question 3.
What is the epidermal cell modification in plants which prevent [NCERT Exemplar]
Answer:
Cuticularised trichoblasts or epidermal hair which produce a stationary layer of air over the surface that reduces isolation and rate of transpiration.

Question 4.
The cells of this tissue are living and show angular wall thickenings. They also provide mechanical support. The tissue is: [NCERT Exemplar]
(a) xylem
(b) sclerenchyma
(c) collenchyma
(d) epidermis
Answer:
(c) The collenchyma is a simple permanent tissue, which provide mechanical support.

Question 5.
In which vascular bundles, phloem lies on the outer side and xylem towards the inner side or central axis?
Answer:
Collateral vascular bundles.

Question 6.
What is present on the surface of the leaves which helps the plant prevent loss of water but is absent in roots? [NCERT Exemplar]
Answer:
Cuticle layer and wax.

Question 7.
What constitutes the cambial ring? [NCERT Exemplar]
Answer:
Fusion of interfascicular and intrafascicular cambium strips.

Question 8.
The cross-section of a plant material showed the following features when viewed under the microscope (a) Vascular bundles were radially arranged (b) Four xylem strands with exarch condition of protoxylem. To which organ should it be assigned? [ NCERT Exemplar]
Answer:
Dicot root.

Question 9.
What do hardwood and softwood stand for? [NCERT Exemplar]
Answer:
Hardwood contains xylem vessels, i.e., dicot wood. Softwood contains tracheids only, i. e., gymnospermous wood.

Question 10.
Give one basic functional difference between phellogen and phelloderm.
Answer:
Phellogen is meristematic and divides to produce new cells. Phelloderm stores food materials.

Short answer type questions

Question 1.
Differentiate among parenchyma, collenchyma and sclerenchyma.

Parenchyma	Collenchyma	Sclerenchyma
1. Living cells.	1. Living cells.	1. Dead cells.
2. Forms	2. Found below epidermis.	2. Usually found in epidermis.
3. Chloroplast present.	3. Chloroplast present sometimes.	3. Chloroplast absent.
4. Performs many vital functions.	4. Provides mechanical support to growing parts.	4. Provides mechanical support to organs.

Question 2.
Mention key differences between xylem and phloem.
Answer:
Differences between xylem and phloem

Xylem	Phloem
1. Composed of tracheids, vessels and xylem parenchyma.	1. Composed of sieve tubes, companion cells and phloem parenchyma.
2. Facilitate conduction of water and minerals from roots.	2. Facilitate conduction of food from leaves.

Question 3.
What is the difference between simple tissue and complex tissue in plants?
Answer:
Simple tissues are composed of similar cells. Complex tissues are composed of dissimilar cells. Simple tissues provide bulk and mechanical support to plants. Complex tissues are meant for transportation of substances and they also provide mechanical support.

Question 4.
What is the difference between monocot and dicot leaves?
Answer:
Stomata are found on both surfaces in monocot leaves, while they are found on ventral surface only, in dicot leaves. In monocot leaves venation is parallel, which is evident by similar size of vascular bundles. In dicot leaves venation is reticulate, so vascular bundles are of various sizes.

Question 5.
What are the differences between meristematic tissues and permanent tissues?
Answer:
Differences between meristematic tissues and permanent tissues

Meristematic Tissues	Permanent Tissues
1. Cells keep on dividing.	1. Cells stop dividing.
2. Growth is the basic function.	2. Protection is the basic function.
3. Found in tips of roots and stem.	3. Found in girth and periphery.

Long answer type questions

Question 1.
Distinguish between following:
(i) Exarch and endarch condition of protoxylem
(ii) Stele and vascular bundles
(iii) Protoxylem and metaxylem
(iv) Interfascicular cambium and intrafascicular cambium
(v) Open and closed vascular bundles
(vi) Stem hair and root hair [NCERT Exemplar]
Answer:
(i) In exarch condition protoxylem towards the periphery. In endarch
condition protoxylem towards the centre.
(ii) Stele is the arrangement of vascular tissues and vascular bundle is a group of xylem and phloem.
(iii) Protoxylem is an early formed xylem and metaxylem is late formed xylem.
(iv) Interfascicular cambium is formed by permanent tissues. Intrafascicular cambium is present in between the primary xylem and primary phloem of a vascular bundle.
(v) Open vascular bundles have intrafascicular cambium and show secondary growth. Closed vascular bundles do not have intrafascicular cambium.
(vi) Stem hair are multicellular, whereas root hair are unicellular.