Class 11

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.

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.

(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.

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:
Ψ_w = Ψ_s + Ψ_p

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 Ψ_s.

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.

(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 (Ψ_p). Because of the rigidity of the cell wall, the cell does not rupture. This turgor pressure is ultimately responsible for enlargement of cells.

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.

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.

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.

Punjab State Board PSEB 11th Class Biology Book Solutions Chapter 10 Cell Cycle and Cell Division Textbook Exercise Questions and Answers.

PSEB Solutions for Class 11 Biology Chapter 10 Cell Cycle and Cell Division

PSEB 11th Class Biology Guide Cell Cycle and Cell Division Textbook Questions and Answers

Question 1.
What is the average cell cycle span for a mammalian cell ?
Answer:
The average cell cycle span for a mammalian cell is 24 hours.

Question 2.
Distinguish cytokinesis from karyokinesis?
Answer:
Cytokinesis is the division of cytoplasm, whereas karyokinesis is the division of nucleus of the cell.

Question 3.
Describe the events taking place during interphase.
Answer:
The interphase, though called the resting phase, is the time during which the cell is preparing for division by undergoing cell growth and DNA replication. The interphase is divided into three further phases
(i) G₁ – phase (Gap 1)
(ii) S – phase (Synthesis)
(iii) G₂ – phase (Gap 2)
G₁ – phase corresponds to the interval between mitosis and initiation of DNA replication. During Ga-phase the cell is metabolically active and continuously grows but does not replicate its DNA. S or synthesis phase marks the period during, which, DNA synthesis or replication takes place. During this time, the amount of DNA per cell doubles.

During the G₂ – phase, proteins are synthesised in preparation for mitosis, while cell growth continues.
Cells in this stage remain metabolically active but no longer proliferate unless called on to do so depending on the requirement of the organism.

Question 4.
What is G₀ (quiescent phase) of cell cycle?
Answer:
G₀ is the quiescent stage of the cell cycle. It is also known as inactive stage of the cell cycle. Cells in this stage remain metabolically remain active, but no longer proliferate unless called on to do so depending on the requirement of the organisms.

Question 5.
Why is mitosis called equational division?
Answer:
Mitosis is the kind of cell division in which daughter cells have the same number and kind of chromosomes as that of parent cell. Mitosis is therefore, also known as equational division.

Question 6.
Name the stage of cell cycle at which one of the following events occur:
(i) Chromosomes are moved to spindle equator.
(ii) Centromere splits and chromatids separate.
(iii) Pairing between homologous chromosomes takes place.
(iv) Crossing over between homologous chromosomes takes place.
Answer:
(i) Prophase,
(ii) Anaphase,
(iii) Zygotene stage of meiosis-I,
(iv) Pachytene stage.

Question 7.
Describe the following:
(i) Synapsis
(ii) Bivalent
(iii) Chiasmata
Draw a diagram to illustrate your answer.
Answer:
(i) Synapsis: During meiosis-I, the process of pairing of two homologous chromosomes is known as synapsis. It is so exact that pairing is not merely* between corresponding chromosomes but between corresponding individual units.

(ii) Bivalent: The complex formed by a pair of synapsed homologous chromosomes is called a bivalent or a tetrad. It consists of four chromatids.

(iii) Chiasmata: The chiasmata formation is the indication of completion of crossing over and beginning of separation of chromosomes. The chiasma is formed when the chromosomal parts begin to repel each other except in the region where these are in contact. Thus, chiasmata formation is necessary for the separation of homologous chromosomes.

Question 8.
How does cytokinesis in plant cells differ from that in animal cells?
Answer:
Cytokinesis in an Animal Cell: It occurs by a process called cleavage. A cleavage furrow appears and at the site of the cleavage furrow the cytoplasm has a ring of microfilaments made of actin associated with molecules of the protein myosin. This ring of proteins will contract causing the furrow to deepen, which will then in turn pinch the cell into two separate cells.

Cytokinesis in a Plant Cell: In a plant cell, vesicles containing cell wall material collect at the middle of the parent cell. They will fuse and form a membranous cell plate. This plate will grow outward and it accumulates more cell wall material. Eventually, the plate will fuse with the plasma membrane and the cell plate contents will join the parental cell wall. This results in two different daughter cells.

Question 9.
Find examples where the four daughter cells from meiosis are equal in size and where they are found unequal in size.
Answer:
The four daughter haploid cells may or may not be equal in size at the end of meiosis-II with telophase-II

Question 10.
Distinguish anaphase of mitosis from anaphase-1 of meiosis. Ans. Differences between anaphase of mitosis and anaphase-I of meiosis.
Answer:

Anaphase of Mitosis	Anaphase-I of Meiosis
1. Each chromosome arranged at the metaphase plate is split simultaneously and the two daughter chromatids, begin to migrate towards the two opposite poles.	The spindle fibres contract and pull the centromeres of homologous chromosomes towards the opposite poles.
2. The centromere of each chromosome is towards the pole with arms of chromosome trailing behind.	The centromere is not divided, so half of the chromosomes of parent nucleus go to one pole and the remaining half in the opposite pole.
3. In this stage (a) centromeres split and chromatids separate. (b) chromatids move to opposite poles.	In this stage (a) homologous chromosomes separate. (b) sister chromatids remain associated at their centromere.

Question 11.
List the main differences between mitosis and meiosis.
Answer:

Mitosis	Meiosis
1. The division occurs in somatic cells.	It occurs in reproductive cells.
2. It is a single division.	It is a double division.
3. The daughter cells resemble each other as well as their mother cell.	The daughter cells neither resemble one another nor their mother cell.
4. Replication of chromosomes occurs before every mitotic division.	Replication of chromosomes occurs only once though meiosis is a double division.
5. Mitosis does not introduce variations.	Meiosis introduces variations.
6. Mitosis is required for growth, repair and healing.	Meiosis is involved in sexual reproduction.

Question 12.
What is the significance of meiosis?
Answer:
Meiosis is the mechanism by which conservation of specific chromosome number of each species is achieved across generations in sexually reproducing organisms.
It also increases the genetic variability in the population of organisms from one generation to the next. Variations are very important for the process of evolution.

Question 13.
Discuss with your teacher about:
(i) haploid insects and lower plants where cell-division occurs, and
(ii) some haploid cells in higher plants where cell-division does
not occur.
Answer:
(i) In some insects and lower plants fertilization is immediately followed by zygotic meiosis, which leads to the production of haploid organisms. This type of life cycle is known as haplontic life cycle.

(ii) The phenomenon of polyploidy can be observed in some haploid cells in higher plants in which cell division does not occur. Polyploidy is a state in which cells contain multiple pairs of chromosomes than the basic set. Polyploidy can be artificially induced in plants by applying colichine to cell culture.

Question 14.
Can there be mitosis without DNA replication in S-phase?
Answer:
No, DNA replication is required for doubling the chromosome number.

Question 15.
Can there be DNA replication without cell division?
Answer:
Yes, DNA replication takes place during the synthesis stage of interphase in the cell cycle. It is totally independent of cell division. After G₂ -phase a cell may or may not enter into the M-phase.

Question 16.
Analyse the events during every stage of cell cycle and notice how the following two parameters change?
(i) Number of chromosomes (n) per cell
(ii) Amount of DNA content (C) per cell.
Answer:
(i) The number of chromosomes (n) is double in the prophase, metaphase and anaphase stage of an organism. In the telophase stage, during daughter cell formation, the number of chromosomes get reduced to half.

(ii) Amount of DNA content (C) per cell also changes during different phases of cell division. It is double in the prophase, metaphase and anaphase stage of an organism. In the telophase stage, during daughter cell formation, the number of chromosomes get reduced to half than that of their parents. ’

Punjab State Board PSEB 11th Class Biology Book Solutions Chapter 14 Respiration in Plants Textbook Exercise Questions and Answers.

PSEB Solutions for Class 11 Biology Chapter 14 Respiration in Plants

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

Question 1.
Differentiate between:
(a) Respiration and Combustion
(b) Glycolysis and Krebs’ cycle
(c) Aerobic respiration and Fermentation
Answer:
(a) Differences between Respiration and Combustion

Respiration	Combustion
1. It is the breakdown of complex compounds through oxidation within the cells, leading to release of considerable amount of energy.	Combustion is the complete burning of glucose, which produces CO2 and H2O and yields energy which is given out as heat.
2. It is a controlled biochemical process.	It is an uncontrolled physicochemical process.
3. Many chemical bonds break simultaneously by releasing large amounts of energy.	Chemical bonds break one after another to release energy.
4. Enzymes are involved.	Enzymes are not involved.

(b) Differences between Glycolysis and Krebs’ Cycle

Glycolysis	Krebs’ Cycle
1. It is also called as Embden Meyerhof-Parnas (EMP) pathway.	It is also known as citric acid cycle or tricarboxylic acid cycle.
2. It occurs in the cytosol of prokaryotes as well as eukaryotes.	In eukaryotes, it occurs in matrix of mitochondria, and in prokaryotes, it occurs in cytoplasm.
3. It starts with the oxidation of glucose.	It starts with the oxidation of pyruvic acid.
4. It is an enzyme-controlled 10 steps process by which glucose, fructose or sucrose is reduced to form pyruvic acid with the production of ATP and NADH2.	It involves 8 steps to oxidize two molecules of acetyl CO-A

(c) Differences between Aerobic Respiration and Fermentation

Aerobic Respiration	Fermentation
1. The presence of O2 is required for complete oxidation of organic substances.	Incomplete oxidation of glucose occurs in the absence of oxygen.
2. It releases CO2, water, and a large amount of energy present in the substrate.	Pyruvic acid is converted to CO2 and ethanol and less amount of energy is released.
3. Only two molecules of ATP are produced.	Many molecules of ATP are produced.
4. NADH is oxidized to NAD+ slowly.	This reaction is very vigorous under aerobic respiration.

Question 2.
What are respiratory substrates? Name the most common respiratory substrate.
Answer:
The compounds oxidized during the process of respiration are called respiratory substrates. Carbohydrates, especially glucose, act as respiratory substrates. Fats, proteins, and organic acids also act as respiratory substrates.

Question 3.
Give the schematic representation of glycolysis.
Answer:
Glycolysis

• The scheme of reactions was given by Embden, Meycrhof and Pumas, hence it is also called EMP Pathway; it occurs in the cytoplasm of cells.
• In this process, glucose is partially oxidized/ converted into two molecules of pyruvic acid.
• Glucose is phosphorylated with the help of ATP into Glucose-6-Phosphate, catalyzed by the enzyme hexokinase.
• Glucose-6-Phosphate is converted into its isomer, Fructose-6-Phosphate, catalyzed by isomerase.
• Fructose-6-Phosphate is then phosphorylated with the use of ATP into Fructose 1,6-bisphosphate, catalyzed by
  phosphofructokinase.
• Fructose 1,6-bisphosphate is split into one molecule of 3-pho sphoglyceraldchyd and one molecule of
  dihydroxy-acetone phosphate; these two products are interconvertible.
• 3-phosphoglyceraldehyde is oxidized to 1, 3 bi phosphoglycerate, where NAD is reduced to NADH.
• 1,3-bisphosphoglycerate is split into 3-phosphoglycerate along with the formation of ATP, catalyzed by phosphoglycerate kinase.

• 3-Phosphoglycerate is subsequently converted into 2-phosphoglycerate and then into Phosphoenol Pyruvate (PEP).
• PEP is converted into pyruvate along with the formation of ATP, catalyzed by the enzyme pyruvate kinase.

In this pathway, ATP molecules are formed in two ways :

1. Direct/substrate phosphorylation of ADP to ATP.
2. Oxidation of NADH to NAD ’ and ATP formation through electron transport (oxidative phosphorylation).

Four molecules of ATP are formed directly in the following steps:

• When 1, 3 bi phosphoglyceric acids is converted into 3-phosphoglyceric acid and
• When phosphoenolpyruvate is converted into pyruvic acid.

Two molecules of ATP are consumed, one in each of the following steps:

1. Phosphorylation of glucose to glucose-6- phosphate and
2. Conversion of fructose-6-phosphate into fructose 1, 6 bisphosphates.

When 3-phosphoglyceraldehyde (PGAL) is converted into 1, 3 bi phosphoglycerates, two redox equivalents are removed in the form of two hydrogen atoms from PGAL and transferred to NAD⁺, which is reduced to NADH ⁺ H⁺.
Oxidation of two molecules of NADH to NAD yields six molecules of ATP during electron transport. The end products of glycolysis are two molecules each of pyruvic acid, NADH, and ATP.

Question 4.
What are the main steps in aerobic respiration? Where does it take place?
Answer:
The main steps in aerobic respiration are as follows :

• Glycolytic breakdown of glucose into pyruvic acid.
• Oxidative decarboxylation of pyruvic acid to acetyl Co-A (acetyl coenzyme-A).
• Krebs’ cycle.
• Terminal oxidation and phosphorylation in respiratory chain.
  It occurs inside the mitochondrial matrix.

Question 5.
Give the schematic representation of an overall view of Kreb’s cycle.
Answer:
Tricarboxylic Acid Cycle or Kreb’s Cycle:

• It is known as tricarboxylic acid cycle, since the first product, citric acid (hence citric acid cycle) is a tricarboxylic acid (has three COOH groups).
• The sequence of reactions in citric acid cycle was first followed by Sir Hans Kreb; hence it is also known as Kreb’s cycle.
• During this cycle of reactions, 3 molecules of NAD and one molecule of FAD are reduced to NADH and FADH respectively.
• Respiration in Plants 115:
• One ATP molecule is produced directly from GTP formed by substrate-level phosphorylation.

The reactions of aerobic oxidation of Pyruvic acid can be summed up as :
Pyruvic acid + 4NAD+ + FAD+ + 2H₂O + ADP + Pi → 3CO₂ + 4NADH + 4H⁺ + FADH₂ + ATP

Question 6.
Explain ETS.
Answer:
Electron Transport System (ETS):

• ETS occurs in the electron transport particles (ETP) on the inner surface of the inner membrane of mitochondria.
• NADH formed in glycolysis and citric acid cycle are oxidized by NADH-dehydrogenase (complex I) and the electrons are transferred to ubiquinone, via FMN.
• Ubiquinone also receives reducing equivalents via FADH generated during the oxidation of succinate by succinate dehydrogenase (complex II).
• The reduced ubiquinone called ubiquinol is then oxidized by transfer of electrons to cytochrome c, via cytochrome be bc₁ -complex (complex III).
• Cytochrome c acts as a mobile carrier between complex III and complex IV.
• Complex IV refers to cytochrome c oxidase complex containing cytochromes a and a3 and two copper centres.
• When the electrons are shunted over the carriers via complex I to IV in the electron transport chain, they are coupled to ATP synthetase (complex V) for the formation of ATP from ADP and IP.
• Oxygen functions as the terminal acceptor of electrons and is reduced to water along with the hydrogen atoms.

Question 7.
Distinguish between the following:
(a) Aerobic respiration and anaerobic respiration.
(b) Glycolysis and fermentation.
(c) Glycolysis and citric acid cycle.
Answer:
(a) Differences between Aerobic and Anaerobic Respiration

Aerobic Respiration	Anaerobic Respiration
1. Presence of oxygen is essential.	Occurs without oxygen.
2. Complete oxidation of substrates occurs into CO2 and H2O.	Incomplete degradations of substrates.
3. End products are non-toxic.	End products are toxic when accumulated in large amount.
4. Involves glycolysis, Krebs’ cycle, and terminal oxidation.	Involves only glycolysis followed by incomplete breakdown of pyruvic acid.
5. 36 ATP molecules are produced from each glucose molecule.	Only two molecules of ATP are produced from each glucose molecule.

(b) Differences between Glycolysis and Fermentation:

Glycolysis	Fermentation
1. it occurs in cytoplasm of the cell and in all living things.	it occurs in yeast and muscle cells in animals when oxygen is not sufficient for cellular respiration.
2. Glucose undergoes partial oxidation to form two molecules of pyruvic acid.	The enzyme-like pyruvic acid decarboxylase and alcohol dehydrogenase catalyze these reactions.
3. The end products are CO2, H2O and energy.	The end products are CO2 and ethanol and in animal cells lactic acid is released.

(c) Differences between Glycolysis and Krebs’ Cycle

Glycolysis	Krebs’ Cycle
1. It is also called as Embden Meyerhof-Parnas (EMP) pathway.	It is also known as citric acid cycle or tricarboxylic acid cycle.
2. It occurs in the cytosol of prokaryotes as well as eukaryotes.	In eukaryotes, it occurs in matrix of mitochondria and in prokaryotes, it occurs in cytoplasm.
3. It starts with the oxidation of glucose.	It starts with the oxidation of pyruvic acid.
4. It is an enzyme-controlled 10 steps process by which glucose, fructose or sucrose is reduced to form pyruvic acid with the production of ATP and NADH2.	It involves 8 steps to oxidize two molecules of acetyl CO-A

Question 8.
What are the assumptions made during the calculation of net gain of ATP?
Answer:
The calculations of net gain of ATP can be made only on certain assumptions :

• There is a sequential, orderly pathway functioning, with one substrate forming the next and with glycolysis, TCA cycle and ETS pathway following one after another.
• The NADH synthesized in glycolysis is transferred into the mitochondria and undergoes oxidative phosphorylation.
• None of the intermediates in the pathway are utilised to synthesize any other compound.
• Only glucose is being respired-no other alternative substrates are entering in the pathway at any of the intermediary stages.

But this kind of assumptions are not really valid in a living system. All pathways work simultaneously and do not take place one after another. Substrates enter the pathways and are withdrawn from it as end when necessary; ATP is utilized as and when needed; enzymatic rates are controlled by multiple means. In overall steps, there is a net gain of 36 ATP molecules during aerobic respiration of one molecule of glucose.

Question 9.
Discuss ‘the respiratory pathway is an amphibolic pathway.’
Answer:
Aniphibolic Pathway: Glucose is the favored substrate for respiration. All carbohydrates are usually first converted into glucose before they are used for respiration. Other substrates can also be respired but then they do not enter the respiratory pathway at the first step.

Since respiration involves breakdown as well as synthesis of substrates, the respiratory process involves both catabolism and anabolism. That is why respiratory pathway is considered to be an amphibolic pathway rather than as a catabolic one.

Question 10.
Define RQ. What is its value for fats?
Answer:
Respiratory quotient (RQ) or respiratory ratio can be defined as the ratio of the volume of CO₂ evolved to the volume of O₂ consumed during respiration. The value of respiratory quotient depeñds on the type of respiratory substrate. Its value is one for carbohydrates. However, it is always less than one for fats as fats consume more oxygen for respiration than carbohydrates.

It can be illustrated through the example of tripalmitin fatty acid, which consumes 145 molecules of O₂ for respiration while 102 molecules of CO₂ are evolved.The RQ value for tripalmitin is 0.7.

Question 11.
What is oxidative phosphorylation?
Answer:
It refers to the formation of ATP in the mitochondria, utilizing the energy obtained by the oxidation of organic molecules.

Question 12.
What is the significance of step-wise release of energy in respiration?
Answer:
During oxidation within a cell, all the energy contained in respiratory substrates is not released free in a single step. It is released in a series of slow step-wise reactions controlled by enzymes and it is trapped as chemical energy in the form of ATP. The significance of this step-wise release of energy is that some energy is used to synthesize ATP. This ATP is stored for the later utilization wherever required. Hence, ATP acts as the energy currency of cell. The energy

stored in ATP can be utilized for following:

• In various energy-requiring processes of organisms.
• The carbon skeleton produced during respiration is used as precursors for the synthesis of other molecules.

Punjab State Board PSEB 11th Class Biology Book Solutions Chapter 9 Biomolecules Textbook Exercise Questions and Answers.

PSEB Solutions for Class 11 Biology Chapter 9 Biomolecules

PSEB 11th Class Biology Guide Biomolecules Textbook Questions and Answers

Question 1.
What are macromolecules? Give examples.
Answer:
Chemical compounds, which are found in the acid insoluble fraction are called macromolecules or biomacromolecules. For example, proteins, lipids and carbohydrate, etc.

Question 2.
Illustrate a glycosidic, peptide and a phosphodiester bond.
Answer:
Glycosidic Bond: A glycosidic bond is a type of functional group that joins a carbohydrate (sugar) molecule to another group, which may or may not be another carbohydrate.

Peptide Bond: A peptide bond (amide bond) is a chemical bond formed between two molecules when the carboxyl group of one molecule reacts with the amine group of the other molecule, thereby releasing a molecule of water (H20).
H₂O.

Phosphodiester Bond: A phosphodiester bond is a group of strong covalent bonds between a phosphate group and two other molecules over two ester bonds. In DNA and RNA, the phosphodiester bond is the linkage between the 3′ carbon atom of one sugar molecule and the 5’carbon of another, deoxyribose in DNA and ribose in RNA.

Question 3.
What is meant by tertiary structure of proteins?
Answer:
Tertiary Structure of Protein; The overall shape of a single protein molecule; the spatial relationship of the secondary structures to one another. Tertiary structure is generally stabilized by non-local interactions, most commonly the formation of a hydrophobic core, but also through salt bridges, hydrogen bonds, disulfide bonds,1 and even post-translational modifications. The term “tertiary structure” is often used as synonymous with the term fold. The tertiary structure is what controls the basic function of the protein.

Question 4.
Find and write down structures of 10 interesting small molecular weight biomolecules. Find if there is any industry which manufactures the compounds by isolation. Find out who are the buyers?
Answer:

Question 5.
Proteins have primary structure. If you are given a method to know which amino acid is at either of the two termini (ends) of a protein, can you connect this information to purity or homogeneity of a protein?
Answer:
The sequence of amino acids, i.e., the positional information in a protein which is the first amino acid, which is second and so on is called the primary structure of a protein. The first amino acid is also called as N-terminal amino acid. The last amino acid is called the C-terminal amino acid. Yes, we can connect this information to purity or homogeneity of a protein. Based on number of amino and carboxyl groups, there are acidic (e.g., glutamic acid), basic (lysine) and neutral (valine) amino acids, proteins may be acidic, basic and neutral.

Question 6.
Find out and make a list of proteins used as therapeutic agents. Find other applications of proteins (e. g, cosmetics etc.)
Answer:
Some proteins and their functions are as follows:

Proteins	Functions
1. Collagen	Intercellular ground substance
2. Trypsin	Enzyme
3. Insulin	Hormone
4. Antibody	Fights against infections
5. Receptors	Sensory reception (example-taste)
6. Glut-4	Enables glucose transport in cells
7. Keratolytic protein	Used to soften hard skin
8. Egg protein	Used for skin tightening

Question 7.
Explain the composition of triglyceride.
Answer:
Triglycerides are composed of two types of molecules, i.e., glycerol i (3 carbon molecules) and fatty acids which attach to the glycerol at the alcohol unit. The following is a structural representation of a triglyceride at the molecular level.

Fatty acids are chains of hydrocarbons 4-22 (or more) carbons a long with a carboxyl group at one end. If each carbon has two hydrogen atoms, the fatty acid is saturated. If two carbon atoms are double-bonded, so that there is less hydrogen in the fatty acid, it is unsaturated (monounsaturated). If more than two carbon atoms are unsaturated, the fatty acid is polyunsaturated.

Question 8.
Can you describe what happens when milk is converted into curd or yoghurt, from your understanding of proteins.
Answer:
Milk contains a protein called casein. This protein gives milk its characteristic white colour. It is of high nutritional value because it contains all the essential amino acids required by man’s body. The curd forms because of the chemical reaction between lactic acid bacteria and casein. When curd is added to milk, the lactic acid bacteria present in it cause coagulation of casein and thus, convert it into curd.

Question 9.
Can you attempt building models of biomolecules using commercially available atomic models (Ball and Stick models).
Answer:
Yes, we can make models of biomolecules using commercially available atomic models.

Question 10.
Attempt titrating an amino acid against a weak base and discover the number of dissociating (ionisable) functional groups in the amino acid.
Answer:
When an amino acid is titrated against a weak base, it dissociates and gives two functional groups:
(i) -COOH group (carboxylic group)
(ii) Amino group (NH_2/sub>)

Question 11.
Draw the structure of the amino acid, alanine.
Answer:

Question 12.
What are gums made of? Is fevicol different?
Answer:
Gums are made of carbohydrates, i. e., L-rhamnose, D-galactose and D-galacturonic acid, etc. Fevicol is different from natural gums. It is a synthetic product.

Question 13.
Find out a qualitative test for proteins, fats and oils, amino acids and test any Fruit juice, saliva, sweat and urine for them.
Answer:

• Test for Proteins: Biuret test if Biuret’s reagent added to protein, then the colour of the reagent changes light blue to purple.
• Test for Fats and Oils: Grease or test.
• Test for Amino Acids: Ninhydrin test. If Ninhydrin reagent is added to the solution, then the colourless solution changes to pink, blue or purple depending upon the amino acid.

Question 14.
Find out how much cellulose is made by all the plants in the biosphere and compare it with how much of paper is manufactured by man and hence what is the consumption of plant material by man annually. What a loss of vegetation!
Answer:
Approximately, 100 billion tonnes of cellulose are made per year by all the plants in the biosphere and it takes 17 full grown trees to make one ton of paper. Trees are also used to fulfil the other requirements of man such as for timber, food, medicines, etc. Hence, it is difficult to calculate the annual consumption of plant material by man.

Question 15.
Describe the important properties of enzymes.
Answer:
Important properties of enzymes are given below :

• Enzymes are proteins which catalyse biochemical reactions in the cells.
• They are denatured at high temperatures.
• Enzymes generally function in a narrow range of temperature and pH. Each enzyme shows its highest activity at a particular temperature and pH called the optimum temperature and optimum pH.
• With the increase in substrate concentration, the velocity of the eyzymatic reaction rises at first. The reaction ultimately reaches a maximum velocity (V_max) which is not exceeded by any further rise in concentration of the
  substrate.
• The activity of an enzyme is also sensitive to the presence of specific chemicals that bind to the enzyme.
• Enzymes are substrate specific in their action.

Punjab State Board PSEB 11th Class Biology Book Solutions Chapter 12 Mineral Nutrition Textbook Exercise Questions and Answers.

PSEB Solutions for Class 11 Biology Chapter 12 Mineral Nutrition

PSEB 11th Class Biology Guide Mineral Nutrition Textbook Questions and Answers

Question 1.
‘All elements that are present in a plant need not be essential to its survival’. Comment.
Answer:
All elements that are present in a plant need not be essential to its survival because they do not directly involved in the composition of their body. However, if the concentration of micronutrients such as Fe, Mn, Cu, Zn, Cl, etc., rise above their critical values, they appear to be toxic for the plant.

Question 2.
Why is purification of water and nutrient salts so important in studies involving mineral nutrition using hydroponics?
Answer:
It is to know the essentiality of a mineral element in the life cycle of a plant. Further, it helps in improving the deficiency symptoms of the plants. The nutrient solution must be adequetly aerated to obtain the optimal growth.

Question 3.
Explain with examples : macronutrients, micronutrients, beneficial nutrients, toxic elements and essential elements.
Answer:
(i) Macronutrients: These are generally present in plant tissues in large amount (in excess 10 m mole kg’1 of dry matter). The macronutrients include carbon, hydrogen, oxygen, nitrogen, phosphorus, sulphur, potassium, calcium and magnesium.

(ii) Micronutrients: Micronutrients or trace elements, are needed in very small amount (less than 10m mole kg~: of dry matter). These include iron, manganese, copper, molybdenum, zinc, boron, chlorine and nickel.

(iii) Beneficial Nutrients: The elements which are not essential for plants, but their presence are beneficial for the growth and development. Such, elements are called beneficial elements.

(iv) Toxic Elements: Any mineral ion concentration in tissues, that reduces the dry weight of tissues by about 10 % is considered toxic. For example, Mn inhibit the absorption of other elements.

(v) Essential Elements: The macronutrients including carbon, hydrogen, oxygen, nitrogen, phosphorus, sulphur, potassium, calcium and magnesium, which are require directly for the growth and metabolism of the plants and whose deficiency produces certain symptoms in the plants are known as essential elements.

Question 4.
Name at least five different deficiency symptoms in plants. Describe them and correlate them with the concerned mineral deficiency.
Answer:
The kind of deficiency symptoms shown in plants include chlorosis, necrosis, stunted plant growth, premature fall of leaves and buds, and inhibition of cell division.

• Chlorosis is the loss of chlorophyll leading to yellowing in leaves. This symptom is caused by the deficiency of elements N, K, Mg, S, Fe, Mn, Zn and Mo.
• Necrosis or death of tissue, particularly leaf tissue, is due to the deficiency of Ca, Mg, Cu, K.
• Lack or low level of N, K, S, Mo causes an inhibition of cell division.
• Some elements like N, S, Mo delay flowering if their concentration in plants is low.

Question 5.
If a plant shows a symptom which could develop due to deficiency of more than one nutrient, how would you find out experimentally, the real deficient mineral element?
Answer:
Every element shows certain characteristic deficiency symptoms in the plants. The deficiency of any one element cannot be met by supplying some other element. So, by absorbing the type of deficiency symptom, we can determine the real deficient mineral element.

Question 6.
Why is that in certain plants deficiency symptoms appear first in younger parts of the plant, while in others they do so in mature organs?
Answer:
For elements that are actively mobilised within the plants and exported to young developing tissues, the deficiency symptoms tend to appear first in the older tissues. For example, the deficiency symptoms of nitrogen, potassium and magnesium are visible first in the senescent leaves. In the older leaves, biomolecules containing these elements are broken down, making these elements available for mobilising to younger leaves.

The deficiency symptoms tend to appear first in the young tissues, whenever the elements are relatively immobile and are not transported out of the mature organs. For example, elements like sulphur and calcium are a part of the structural component of the cell and hence are not easily released.

Question 7.
How are the minerals absorbed by the plants?
Answer:
Mechanism of Absorption of Minerals: The process of absorption can occur into following two main phases :
(i) In the first phase, an initial rapid uptake of ions into the ‘free space’ or ‘outer space’ of cells the apoplast is passive.

(ii) In the second phase of uptake, the ions are taken in slowly into the ‘inner space’ the symplast of the cells. The passive movement of ions into the apoplast usually occurs through ion-channels, the trans-membrane proteins that function as selective pores. On the other hand, the entry or exit of ions to and from the symplast requires the expenditure of metabolic energies. The movement of ions is usually called the inward movement into the cells is influx and the outward movement, efflux.

Question 8.
What are the conditions necessary for fixation of atmospheric nitrogen by Rhizobium ? What is their role in Nitrogen-fixation?
Answer:
The first essential condition for nitrogen fixation is legume-bacteria relationship. Rhizobium bacteria cause nodule formation for this association. The enzyme nitrogenase is highly sensitive to the molecular oxygen. The nodules protect these enzymes by an oxygen scavenger called leghaerrloglobin.
Rhizobium bacteria are free living in soil. They are symbionts, which can fix atmospheric nitrogen for plants.

Question 9.
What are the steps involved in formation of a root nodule?
Answer:
Steps in Nodule Formation: Nodule formation involves a sequence of multiple interactions between Rhizobium and roots of the host plant. Principal stages in the nodule formation are given below:

• Rhizobia multiply and colonise the surroundings of roots and get attached to epidermal and root hair cells.
• The root-hairs curl and the bacteria invade the root-hair.
• An infection thread is produced carrying the bacteria into the cortex of the root, where they initiate the nodule formation in the cortex of the root. Then the bacteria are released from the thread into the cells which leads to the differentiation of specialised nitrogen fixing cells.
• The nodule thus formed, establishes a direct vascular connection with the host for exchange of nutrients.

Question 10.
Which of the following statements are true? If false, correct them:
(a) Boron deficiency leads to stout axis.
(b) Every mineral element that is present in a cell is needed by the cell.
(c) Nitrogen as a nutrient element, is highly immobile in the plants.
(d) It is very easy to establish the essentiality of micronutrients because they are required only in trace quantities.
Answer:
(a) True
(b) All the mineral elements present in a cell are not needed by the cell. For example, plants growing near radioactive mining sites tend to accumulate large amounts of radioactive compounds. These compounds are not essential for the plants.

(c) Nitrogen as a nutrient element is highly mobile in plants. It can be mobilised from the old and mature parts of a plant to its younger parts.
(d) True

Punjab State Board PSEB 11th Class Biology Book Solutions Chapter 13 Photosynthesis in Higher Plants Textbook Exercise Questions and Answers.

PSEB Solutions for Class 11 Biology Chapter 13 Photosynthesis in Higher Plants

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

Question 1.
By looking at a plant externally can you tell whether a plant is C₃ or C₄? Why and how?
Answer:
Usually plants growing in dry conditions use C₄-pathways. It cannot be said conclusively, if the plant is a C₃ or C₄ by looking at external appearance.

Question 2.
By looking at which internal structure of a plant can you tell whether a plant is C₃ or C₄? Explain.
Answer:
The particularly large cells around the vascular bundles of the C₄– pathway plants are called bundle sheath cells and the leaves, which have such anatomy are said to have ‘Kranz’ anatomy.
‘Kranz’ means ‘wreath’ and is a reflection of the arrangement of cells.
The bundle sheath cells may form several layers around the vascular bundles they are characterised by having a large number of chloroplasts, thick walls impervious to gaseous exchange and no intercellular spaces.

Question 3.
Even though a very few cells in a C₄ plant carry out the biosynthetic-Calvin pathway, yet they are highly productive. Can you discuss why?
Answer:
The productivity of a plant is measured by the rate at which it photosynthesizes. The amount of carbon dioxide present in a plant is directly proportional to the rate of photosynthesis. C₄ plants have a mechanism for increasing the concentration of carbon dioxide. In C₄ plants, the Calvin cycle occurs in the bundle-sheath cells.

The C₄ compound (malic acid) from the mesophyll cells is broken down in the bundle sheath cells. As a result, CO₂ is released. The increase in CO₂ ensures that the enzyme RuBisCo does not act as an oxygenase, but as a carboxylase.
This prevents photorespiration and increases the rate of photosynthesis. Thus, C₄ plants are highly productive.

Question 4.
RuBisCO is an enzyme that acts both as a carboxylase and oxygenase. Why do you think RuBisCO carries out more carboxylation in C₄ -plants?
Answer:
RuBisCO has a much greater affinity for CO₂ than for O₂. It is the relative concentration of O₂ and CO₂ that determines which of the two will bind to the enzyme. In C₄ -plants some O₂ does bind to RuBisCO and hence, CO₂ fixation is decreased. Here the RuBP instead of being converted to two molecules of PGA binds with O₂ to form one molecule and phosphoglycolate in a pathway called photorespiration.

In the photorespiratory pathway, there is neither synthesis of sugars, nor of ATP. Rather it results in the release of CO₂ with the utilization of ATP. in the photorespiratory pathway, there is no synthesis of ATP or NADPH. Therefore, photorespiration is a wasteful process.

In C₄-plants, photorespiration does not occur. This is because they have a mechanism that increases the concentration of CO₂ at the enzyme site. This takes place when the C₄ acid from the mesophyll is broken down in the bundle cells to release CO₂, this results in increasing the intracellular concentration of CO₂. In turn, this ensures that the RuBisCO functions as a carboxylase minimizing the oxygenase activity.

Question 5.
Suppose there were plants that had a high concentration of chlorophyll-b but lacked chlorophyll a, would it carry out
photosynthesis? Then why do plants have chlorophyll-b and other accessory pigments?
Answer:
‘Though chlorophyll is the major pigment responsible for trapping light, other thylakoid pigments like chlorophyll-b, xanthophylls and carotenoids, which are called accessory pigments, also absorb light and transfer the energy to chlorophyll-a. Indeed, they not only enable a wider range of wavelengths of incoming light to be utilized for photosynthesis but also protect chlorophyll-a from photo-oxidation.

Question 6.
Why is the colour ola leaf kept n the dark frequently yellow or pale green? Which pigment do you think is more stable?
Answer:
This is due to the interconversion of pigments, i.e., change of green chlorophyll pigment into yellow-colored carotenoids. The carotene pigment is more stable.

Question 7.
Look at leaves of the same plant on the shady side and compare it with the leaves on the sunny side. Or, compare the potted plants kept in the sunlight with those in the shade. Which of them has leaves that are darker green? Why?
Answer:
The leaves of the same plant on the sunny side are dark green as compare it with the leaves on the sunny side due to more chlorophyll pigment.

Question 8.
Figure shows the effect of light on the rate of photosynthesis. Based on the graph, answer the following questions:
(a) At which point/s (A, B or C) in the curve is light a limiting factor?
(b) What could be the limiting factor/s in region A?
(c) What do C and D represent on the curve?

Answer:
(a) Points K-C of the curve, the rate did not increase with an increase in its concentration because under these conditions, light becomes limiting factor.

(b) The rate of photosynthesis shows proportionate increase upto a certain CO₂ concentration (In region A of the curve), beyond which the rate again hcomes constant, not showing any increase by increasing CO₂ concentration.

(c) lithe light inrensiry is doubled, i.e., the plants are exposed to 2 units of light, CO₂ concentration again becomes limiting factor beyond this concentration (Points C and D represent on the curve.)

Question 9.
Give comparison between the following:
(a) C₃ and C₄ pathways
(b) Cyclic and non-cyclic photophosphorylation
(c) Anatomy of leaf in C₃ and C₄ plants
Answer:
(a) Comparison between C₃ and C₄ pathways

C3 Pathway	C4 Pathway
The primary acceptor of CO2 is RuBP, a 5 carbon compound.	The primary acceptor of CO2 is PEP, a 3 carbon compound.
It operated under low concentration of CO2 in mesophyll cells.	It can operate under very low CO2 concentration in mesophyll cells.
CO2 once fixed is not released back.	CO2 once fixed is released back in bundle sheath cells.
Fixation of one molecule of CO2 needs 3ATP and 2 NADPH2 molecules. It requires 18ATP for the synthesis of one molecule of glucose.	C4 pathway requires 30 ATP for the synthesis of one molecule of glucose.
C3 -cycle operates in all categories of plants.	It operates in only C4-plants.

(b) Comparison between cyclic and non-cyclic photophosphorylation

Cyclic Photophosphorylation	Non-cyclic Photophosphorylation
It occurs in photosystem-I in stromal or intergranal thylakoids.	It is carried out by both PS-I and PS-II in the granal thylakoids.
It is not connected to photolysis of water so no oxygen is evolved.	It is connected with photolysis of water, so oxygen is evolved in it.
It is activated by light of 700 nm wavelength.	It occurs in 680 nm as well as 700 nm wavelength.
It generates ATP only there is no formation of NADPH2.	It produces both ATP as well as NADPH2.
Chlorophyll does not receive any electrons from donor.	The source of electrons is photolysis of water.
This system does not take part in photosynthesis except in bacteria.	This system is connected with CO2 fixation and is dominant in green plants.

(C) Comparison between C₃ and C₄ leaves.

C3 Leaves	C4 Leaves
Bundle-sheath cells are absent.	Bundle-sheath cells are present.
RuBisCo is present in the mesophyll cells.	RuBisCo is present in the bundle sheath cells.
The first stable compound produced is 3-phosphoglycerate, a three-carbon compound.	The first stable compound produced is oxaloacetic acid; a four-carbon compound.
‘ Photorespiration occurs.	‘ Photorespirarion does not occur.

Punjab State Board PSEB 11th Class Biology Book Solutions Chapter 19 Excretory Products and their Elimination Textbook Exercise Questions and Answers.

PSEB Solutions for Class 11 Biology Chapter 19 Excretory Products and their Elimination

PSEB 11th Class Biology Guide Excretory Products and their Elimination Textbook Questions and Answers

Question 1.
Define Glomerular Filtration Rate (GFR).
Answer:
Glomerular Filtration Rate (GFR) is the amount of the filtrate formed by the kidneys per minute. GFR in a healthy individual is approximately 125 mt/mm, i.e., 180 L day.

Question 2.
Explain the autoregulatory mechanism of GFR.
Answer:
Regulation of GFR: The kidneys have built-in mechanisms for the regulation o glomerular filtration rate. One such efficient mechanism is carried out by the juxtaglomerular apparatus (JGA). JGA is a special sensitive region formed by cellular modifications in the distal convoluted tubule and the afferent arteriole at the location of their contact. A fall in GFR can activate the JG cells to release renin which can stimulate the glomerular blood flow and thereby the GFR back to normal.

Question 3.
Indicate whether the following statements are true or false:
(a) Micturition is carried out by a reflex.
(b) Al)H helps In water elimination, making the urine hypotonia.
(c) Protein-free fluid is filtered from blood plasma into the Bowman’s capsule.
(d) Henle’s loop plays an important role in concentrating the urine.
(e) Glucose is actively reabsorbed in the proximal convoluted tubule.
Answer:
(a) True,
(b) False,
(C) True,
(d) True,
(e) True.

Question 4.
Give a brief account of the counter-current mechanism.
Answer:
Counter-current Mechanism

• The counter-current mechanism takes place in Henle’s loop and vasa recta.
• The flow of filtrate in the two limbs of Henle’s loop is in opposite directions and thus, forms a counter-current.
• The flow of blood through the two limbs of vasa recta is also in a counter-current pattern.
• NaCl is transported by the ascending limb of Henle’s loop which is exchanged with the descending limb of vasa recta.
• NaCl is returned to the interstitium by the ascending portion of vasa recta.
• Similarly small amount of urea enters the thin segment of the ascending limb of Henle’ loop, which is transported back to the interstitium by the collecting tubule.
• This transport of substances facilitated by the special arrangement of Henle’s loop and vasa recta is called the counter-current mechanism,
• It helps to maintain a concentration gradient in the medullary interstitium, which facilitates an easy passage of water from the collecting tubule thereby concentrating the filtrate (urine).

Question 5.
Describe the foie of liver, lungs and skin in excretion.
Answer:
Liver, lungs and skin also play an important role in the process of excretion. Role of the Liver: Liver is the largest gland in vertebrates. It helps in the excretion of cholesterol, steroid hormones, vitamins, drugs, and other waste materials through bile. Urea is formed in the liver by the ornithine cycle. Ammonia-a toxic substance is quickly changed into urea in the liver and thence eliminated from the body. Liver also changes the decomposed haemoglobin pigment into bile pigments called bilirubin and biliverdin.

Role of the Lungs: Lungs help in the removing waste materials such as carbon dioxide from the body.
Role of the Skin: Skin has many glands which help in excreting waste products through pores. It has two types of glands-sweat and sebaceous glands.

Sweat glands are highly vascular and tubular glands that separate the waste products from the blood and excrete them in the form of sweat. Sweat excretes excess salt and water from the body.
Sebaceous glands are branched glands that secrete an oily secretion called sebum.

Question 6.
Explain micturition.
Answer:
Micturition: The process of release of urine is called micturition and the neural mechanisms causing it is called the micturition reflex. An adult human excretes on an average 1 to 1.5 L of urine per day. The urine formed is a light yellow coloured watery fluid which is slightly acidic (pH 6.0) and has a characteristic odour.

Question 7.
Match the items of Column-I with those of Column-II:

Column-I	Column-II
(a) Ammonotelism	(i) Birds
(b) Bowman’s capsule	(ii) Water reabsorption
(c) Micturition	(iii) Bony fish
(d) Uricotelism	(iv) Urinary bladder
(e) ADH	(v) Renal tubule

Answer:

Column-I	Column-II
(a) Ammonotelism	(iii) Bony fish
(b) Bowman’s capsule	(v) Renal tubule
(c) Micturition	(iv) Urinary bladder
(d) Uricotelism	(i) Birds
(e) ADH	(ii) Water reabsorption

Question 8.
What is meant by the term osmoregulation?
Answer:
It is the phenomenon of regulation of change in the concentration of body fluids according to the concentration of external environment. Le., most marine invertebrates, some freshwater invertebrates, hagfish (a vertebrate).

Question 9.
Terrestrial animals are generally either republic or uricotelic, not ammonotelic, why?
Answer:
Terrestrial adaptation requires the production of less toxic nitrogenous wastes like urea and uric acid for the conservation of water. Aquatic animals excret atnmonia which requires large amounts of water to dissolve. This huge quantity of water is easily available to such animals from surroundings.

However, for terrestrial animals, such a huge quantity of water is not available, hence, they modify their original water product NH3 to comparatively less toxic products like urea and uric acid which requires less amount of water for their excretion.

Question 10.
What is the significance of juxtaglomerular apparatus (JGA) in Sidney function?
Answer:
The juxtaglomerular apparatus (JGA) plays an important role in monitoring and regulation of kidney functioning by hormonal feedback, mechanism, involving the hypothalamus and to a certain extent, the heart.

Question 11.
Name the following:
(a) A chordate animal having flame cells as excretory structures.
(b) Cortical portions projecting between the medullary pyramids in the human kidney.
(c) A loop of capillary running parallel to the Henle’s loop.
Answer:
(a) Flarworms,
(b) Columns of Bertini,
(c) Vasa recta

Question 12.
Fill in the gaps:
(a) Ascending limb of Henle’s loop is ………………………… to water whereas the descending limb is …………………….. to it.
(b) Reabsorption of water from distal parts of the tubules is facilitated by hormone ……………………………. .
(c) Dialysis fluid contains all the constituents as in plasma except …………………………… .
(d) A healthy adult human excretes (on an average) ………………………… gm of urea/day.
Answer:
(a) impermeable, permeable;
(b) vasopressin or ADH.;
(c) the nitrogenous wastes;
(d) 25-30.

Punjab State Board PSEB 11th Class Biology Book Solutions Chapter 4 Animal Kingdom Textbook Exercise Questions and Answers.

PSEB Solutions for Class 11 Biology Chapter 4 Animal Kingdom

PSEB 11th Class Biology Guide Animal Kingdom Textbook Questions and Answers

Question 1.
What are the difficulties that you would face in classification of animals, if common fundamental features are not taken into account?
Answer:
The major difficulties in the classification of animals are on the following lines :

• Some show cellular level of organisation, some have tissue level and even some have organ system level of organisation.
• Regarding symmetry, some are radially symmetrical, while some have bilateral symmetry.
• Some have open circulatory system, while others have closed type.
• Regarding digestion, some animals have extracellular digestion, while others have intracellular digestion.
• In case of body cavity, some have true coelom while others are pseudocoelomates.
• Regarding reproduction, some have only asexual reproduction, while others reproduce both by sexual and asexual means.
  So, these are difficulties that zoologists face in the classification of animals.

Question 2.
If you are given a specimen, what are the steps that you would follow to classify it?
Answer:
If I am given an animal specimen, then I will classify it on the basis of fundamental features which are common to all animal types inspite of the presence of some major differences in the structure and form of animals. The features taken into consideration during classification of animal are as follows:

• The type of arrangement of cells.
• Body symmetry.
• Nature of coelom.
• Pattern of digestive system.
• Type of circulatory system.
• Type of methods of reproduction.

Question 3.
How useful is the study of the nature of body cavity and coelom in the classification of animals?
Answer:
Presence or absence of a cavity between the body wall and the gut wall is very important in classification. The body cavity which is lined by, mesoderm, is called coelom.

• Animals possessing coelom are called coelomates, e.g., annelids, molluscs, arthropods, echinoderms, hemichordates and chordates.
• In some animals, the body cavity is not mesoderm, instead the mesoderm is present as scattered pouches in between the ectoderm and endoderm. Such a body cavity is called pseudocoelom and the animals possessing them are called pseudocoelomates, e.g., Aschelminthes.
• The animals in which the body cavity is absent are known as acoelomates, e.g. , Platyhelminthes.

Question 4.
Distinguish between intracellular and extracellular digestion?
Answer:
Differences between intracellular and extracellular digestion are as follows:
(i) Intracellular Digestion: It occurs inside the living cells with the help of lysosomal enzymes. Food particle is taken in through endocytosis. It forms a phagosome which fuses with a lysosome. The digested material pass into the cytoplasm. The undigested matter is throw out by exocytosis. It occurs in Amoeba, Paramecium, etc.,

(ii) Extracellular Digestion: In case of coelentrates digestion occurs in gastrovascular cavity. This cavity has gland cells which secret digestive enzymes over the food. The partially digested fragmented food particles are ingested by nutritive cells. It occurs in Hydra, Aurelia, etc.

Question 5.
What is the difference between direct and indirect development?
Answer:
In direct development the embryo directly develops into an adult, while in indirect development there is an intermediate larval stage. Certain members of arthropods show larval stage of development.

Question 6.
What are the peculiar features that you find in parasitic platyhelminthes?
Answer:
Hooks and suckers are present in the parasitic forms. They are parasitic flatworms commonly called flukes. The body is unsegmented leaf like, which is covered by a thick living tegument. There is no epidermis. The mouth is anterior and is armed with suckers for attachment in the host. Life history includes larval stage and involves more than one hosts.
Examples : Fasciola (the liver fluke), Schistosoma (the blood fluke.)

Question 7.
What are the reasons that you can think of for the arthropods to constitute the largest group of the animal kingdom?
Answer:
Arthropods constitute the largest group of the animal kingdom. It is estimated that the Arthropoda population of the world is approximately a billion (10¹⁸) individuals, in terms of species diversity, number of individuals and geographical distribution. It is the most successful phylum on the Earth that have ever existed. Arthropods are equipped with jointed appendages, which are variously adapted for walking, swimming, feeding, sensory reception and defence. The appendages of abdomen are associated with locomotion, reproduction and in some cases with defence as well.

The appendages of head are related to defence, whereas those of thorax are mainly associated with locomotion. These features are responsible for its large diversity.

Question 8.
Water vascular system is the characteristic of which group of the following?
(a) Porifera
(b) Ctenophora
(c) Echinodermata
(d) Chordata
Answer:
Echinodermata have the water vascular system.

Question 9.
‘All vertebrates are chordates but all chordates are not vertebrates’.’Justify the statement.
Answer:
Notochord is a characteristic feature of all chordates. The members of sub-phylum – Vertebrata possess notochord during the embryonic stage. But in adults the notochord is replaced by a cartilaginous or bony vertebral column. Whereas in member of other Sub-phyla of Chordata the notochord remain as such. The urochordate and cephalochordates retain the notochord during their entire life cycle. Thus, the absence of notochord in adult vertebrates suggest that all vertebrates are chordates but all chordates are not vertebrates.

Question 10.
How important is the presence of air bladder in Pisces?
Answer:
In fishes, air bladder regulates buoyancy and helps in floating in water. If it is absent, animals need to swim constantly to avoid sinking.

Question 11.
What are the modifications that are observed in birds that help them fly?
Answer:
Flight adaptations in birds are as follows :

• Boat-shaped body helps to propel through the air easily.
• Feathery covering of body to reduce the friction of air.
• Holding the twigs automatically by hindlimbs.
• Extremely powerful muscles that enables the wings to work during flight.
• Bones are light, hollow and provide more space for muscle attachment. Presence of pneumatic bones which reduce the weight of body and help in flight.
• The first four thoracic vertebrae are fused to form a furculum for walking of the wings.
• Lungs are solid and elastic and have associated air sacs.
• The power of accomodation of eyes is well developed due to the presence of comb-like structure pecten.
• A single left ovary and oviduct to reduce the body weight.

Question 12.
Could the number of eggs or young ones produced by an oviparous and viviparous mother be equal? Why?
Answer:
The number of eggs or young ones produced by an oviparous or viviparous mother cannot be equal. An oviparous mother gives rise to more number of eggs as some of them die during hatching and as they have to pass through a large number of developmental stages before becoming an adult. On the other hand, a viviparous mother gives rise to fewer number of young ones because there are less chances of their death. Moreover, they did not have to pass through any larval stage.

Question 13.
Segmentation in the body is first observed in which of the following:
(a) Platyhelminthes
(b) Annelida
(c) Aschelminthes
(d) Arthropoda
Answer:
Segmentation in the body is first observed in Annelida. This phenomenon is known as metamerism.

Question 14.
Match the following:

A. Operculum	1. Ctenophora
B. Parapodia	2. Mollusca
C. Scales	3. Porifera
D. Comb plates	4. Reptilia
E. Radula	5. Annelida
F. Hairs	6. Cyclostomata and Chondrichthyes
G. Choanocytes	7. Mammalia
H. Gill slits	8. Osteichthyes

Answer:

A. Operculum	8. Osteichthyes
B. Parapodia	5. Annelida
C. Scales	4. Reptilia
D. Comb plates	1. Ctenophora
E. Radula	2. Mollusca
F. Hairs	7. Mammalia
G. Choanocytes	3. Porifera
H. Gill slits	6. Cyclostomata and Chondrichthyes

Question 15.
Prepare a list of some animals that are found parasitic on human heings.
Answer:
A list of parasitic animals on human beings:

Parasite	In Part of Human Body
Leishmania donovani	Blood
Trichomonas vaginalis	Vagina of human female
Plasmodium vivax	Blood
Taenia solium	Intestine
Ascaris lumbricoides	Small intestine
Wuchereria bancrofti	Lymphatic and muscular system
Loa loa	Eyes
Fasciola hepatica	Liver and bile ducts
Entamoeba histolytica	Intestine
Trypanosoma gambiense	Blood

Punjab State Board PSEB 11th Class Biology Book Solutions Chapter 3 Plant Kingdom Textbook Exercise Questions and Answers.

PSEB Solutions for Class 11 Biology Chapter 3 Plant Kingdom

PSEB 11th Class Biology Guide Plant Kingdom Textbook Questions and Answers

Question 1.
What is the basis of classification of algae?
Answer:
Basis of classification of algae are as follows:

• Kinds of pigments.
• Nature of reserve food.
• Kinds, number and points of insertion of flagella of motile cells.
• Presence or absence of organised nucleus in the cell.

Question 2.
When and where does reduction division take place in the life cycle of a liverwort, a moss, a fern, a gymnosperm and an angiosperm?
Answer:
Reduction division in the life cycle of a liverwort, a moss, a fern and a gymnosperm take place during the production of spores from spore mother cells. In case of an angiosperm, the reduction division occurs during pollen grain formation from anthers and during production of embryo sac from ovule.

Question 3.
Name three groups of plants that bear archegonia. Briefly describe the life cycle of any one of them.
Answer:
Three groups of plants that bear archegonia are bryophytes, pteridophytes and gymnosperms.
Life Cycle of a Pteridophyte: The life cycle of a pteridophyte consists of two morphologically distinct phases:
1. The gametophytic phase
2. The sporophytic phase.
These two phases come one after another in the life cycle of a pteridophyte. This phenomenon is called alternation of generation. The gametophyte is haploid with single set of chromosomes. It produces male sex organs antheridia and female sex organs archegonia.

• The antheridia may be embedded or projecting type. Each antheridium has single layered sterile jacket enclosing a mass of androcytes.
• The androcytes are flask-shaped, sessile or shortly stalked and differentiated into globular venter and tubular neck.
• The archegonium contains large egg, which is non-mo tile.
• The antherozoids after liberation from antheridium, reaches up to the archegonium fuses with the egg and forms a diploid structure known as zygotes.
• The diploid zygote is the first cell of sporophytic generation. It is retained inside the archegonium and forms the embryo.
• The embryo grows and develop to form sporophyte which is differentiated into roots, stem and leaves.
• At maturity the plant bears sporangia, which encloses spore mother cells.
• Each spore mother cell gives rise to four haploid spores which are usually arranged in tetrads.
• The sporophytic generation ends with the production of spores.
• Each spore is the first cell of gametophytic generation. It germinates to produce gametophyte and completes its life cycle.

Question 4.
Mention the ploidy of the following:
Protonemal cell of a moss, primary endosperm nucleus in dicot, leaf cell of a moss, prothallus cell of a fern, gemma cell in Marchantia, meristem cell of monocot, ovum of a liverwort and zygote of a fern.
Answer:
Protonemal cell of a moss – haploid
Primary endosperm nucleus in dicot – triploid
Leaf cell of a moss – haploid
Prothallus cell of a fern – haploid
Gemma cell in Marchantia – haploid
Meristem cell of monocot – diploid
Ovum of a liverwort – haploid
Zygote of a fern – diploid

Question 5.
Write a note on economic importance of algae and gymnosperms.
Answer:
Economic Importance of Algae

• Red algae provides food, fodder and commercial products. Porphyra tenera is rich in protein, carbohydrates and vitamin-A, B, E and C.
• Corallina has vermifuge properties.
• Agar-agar a gelatin substance used as solidifying agent in culture media is obtained from Gelidium and Gracilaria algae. Funori is a glue used as adhesive and in sizing textiles, papers, etc. Chondrus is most widely used in sea weed in Europe.
• Mucilage extracted from Chondrus is used in sampoos, shoe polish and creams.
• Carrageenin is a sulphated polysaccharide obtained from cell wall of Chondrus crispus and Gigartina and is used in confectionary, bakery, jelly, creams, etc.

Economic Importance of Gymnosperms

• Gymnosperms hold soil particles and thus check soil erosion.
• Many gymnosperms are grown in gardens as ornamental plants, i.e., Cycas, Thiya, Araucaria, Taxus, Agathis, Maiden hair tree, etc.
• Sago is a kind of starch obtained from cortex and pith of stem and seeds of Cycas. Roasted seeds of Pinus geradiana (chilgoza) are used as dry fruit.
• Paper pulp is obtained from wood of Picea (spruce), Gnetum, Pinus (pine) and Larix (larck).
• The wood of Juniperus virginiana (red cedar) is used to make pencils, holders and cigar boxes. Wood of Taxus is heaviest amongst soft woods and is used for making bows for archery.
• Dry leaves of Cycas are used to make baskets and brooms. Needles of Pinus in making fibre board. Electric and telephone poles are made of stem of conifers.
• Essential oils are obtained from Juniperus, Tsugo, Picea, Abies, Cedrus, etc. Resins are obtained from many species of Pinus.

Question 6.
Both gymnosperms and angiosperms bear seeds, then why are they classified separately?
Answer:
Both gymnosperms and angiosperms bear seeds, but they are yet classified separately. Because, in case of gymnosperms the seeds are naked, i.e., the seeds are not produced inside the fruit but in case of angiosperms the seeds are enclosed inside the fruit.

Question 7.
What is heterospory? Briefly comment on its significance. Give two examples.
Answer:
Heterospory is the phenomenon in which a plant produces two types of spores, namely microspores and megaspores.
Heterospory is significant in the following ways:

• Microspores give rise to male gametophyte and megaspores give rise to female gametophyte.
• Female gametophyte is retained on the parent plant. The development of zygote takes place within the female gametophyte.
• This leads to formation of seeds.
  Examples: All gymnosperms and all angiosperms, Pinus, Gnetum, neem, peepal, etc.

Question 8.
Explain briefly the following terms with suitable examples:
(i) Protonema
(ii) Antheridium
(iii) Archegonium
(iv) Diplontic
(v) Sporophyll
(vi) Isogamy.
Answer:
(i) Protonema: It is the juvenile stage of a moss. It results from the germinating meiospore. When fully grown, it consists of a slender green, branching system of filaments called the protonema.

(ii) Antheridium: The male sex organ of bryophyte and pteridophyte is known as antheridium. It has a single-layered sterile jacket enclosing in a large number of androcytes. The androcytes’ metamorphose into flagellated motile antherozoids.

(iii) Archegonium: The female sex organ of bryophytes, which is multicellular and differentiated into neck and venter. The neck consists of neck canal cells and venter contains the venter canal cells and egg.

(iv) Diplontic: A kind of life cycle in which the sporophyte is the dominant, photosynthetic, independent phase of the plant and alternate with haploid gametophytic phase is known as diplontic life cycle.

(v) Sporophyll: The sporangium bearing structure in case of Selaginella is known as sporophyll.

(vi) Isogamy: It is the process of fusion between two similar gametes, i.e., Chlamydomonas.

Question 9.
Differentiate between the following:
(i) red algae and brown algae,
(ii) liverworts and moss,
(iii) homosporous and heterosporous pteridophytes.
(iv) syngamy and triple fusion.
Answer:
(i) Differences between Red Algae and Brown Algae

Red Algae	Brown Algae
1. It belongs to the It belongs to the class-Rhodophyceae	It belongs to the It belongs to the class Phaeophyceae.
2. It is red in colour due to the presence of pigments chlorophyll-a, c and phycoerythrin. Example: Stylolema, Rhodela.	It is brown in colour due to the presence Example: Sargassum, Microcystis.

(ii) Differences between Liverworts and Moss

Liverwort	Moss
1. These are the member of class-Hepaticopsida of bryophyta.	These belongs to class-Bryop’sida of bryophyta.
2. Thallus is dorsoventrally flattened and lobed liver like	Thallus is leafy and radially symmetrical.
3. Rhizoids are unicellular.	Rhizoids are multicellular
4. Elaters are present in capsule to assist dispersal of spores.	Elaters are absent, but peristome teeth are present in the capsule to assist dispersal of spores.

(iii) Differences between Homosporous and Heterosporous Pteridophytes

Homosporous Pteridophyte	Heterosporous Pteridophyte
Pteridophytes, which produce only one kind of spores. Example: Lycopodium	These produce two kinds of spores, i.e., large megaspore and smaller microspore. Example: Selaginella

(iv) Differences between Syngamy and Triple Fusion

Syngamy	Triple Fusion
It is the act of fusion of one male gamete with the egg cell to form zygote.	The act of fusion of second male gamete with secondary nucleus to form triploid enddsperm is called triple endosperm is called triple fusion.

Question 10.
How would you distinguish monocots from dicots?
Answer:

Dicotyledons (Dicots)	Monocotyledons (Monocots)
•» Tap root system	Fibrous root system
•» Two cotyledons	One cotyledon
•» Reticulate Venation	Parallel venation
•» Tetramerous or Pentamerous flowers	Trimerous flowers

Question 11.
Match the followings (column I with column II)

Column I	Column II
(a) Chlamydomonas	(i) Moss
(b) Cycas	(ii) Pteridophyte
(c) Selaginella	(iii) Algae
(d) Sphagnum	(iv) Gymnosperm

Answer:

Column I	Column II
(a) Chlamydomonas	(iii) Algae
(b) Cycas	(iv) Gymnosperm
(c) Selaginella	(ii) Pteridophyte
(d) Sphagnum	(i) Moss

Question 12.
Describe the important characteristics of gymnosperms.
Answer:
Characteristics of gymnosperms are as follows :

• Naked-seeded plants, i.e., their ovules are exposed and not enclosed in ovaries. Hence, the seeds are naked without fruits.
• Tap root system is present. They show symbiotic as speciation with fungi
  to form mycorrhizae or with N₂-fixing cyanobacteria to form colloidal roots as in Cycas.
• Leaves are large and needle-shaped.
• Vascular tissues are well developed.
• Gymnosperms are heterosporous.
• Pollination by wind and deposited in ovules.
• Fertilisation occurs in archegonia.
• Retention of female gametophyte inside the ovule and the ovules on the sporophytic plant for complete development is responsible for the development of seed habit.

Punjab State Board PSEB 11th Class Biology Book Solutions Chapter 20 Locomotion and Movement Textbook Exercise Questions and Answers.

PSEB Solutions for Class 11 Biology Chapter 20 Locomotion and Movement

PSEB 11th Class Biology Guide Locomotion and Movement Textbook Questions and Answers

Question 1.
Draw the diagram of a sarcomere of skeletal muscle showing different regions.
Answer:

Question 2.
Define sliding-filament theory of muscle contraction.
Answer:
The sliding-filament, theory states that the contraction of a muscle fibre takes place by the sliding of the thin filaments over the thick filaments.

Question 3.
Describe the important steps in muscle contraction.
Answer:
Mechanism of Muscle Contraction:

• The mechanism of muscle contraction is explained by the sliding filament theory.
• This theory states that contraction of a muscle fibre is due to the sliding of the thin (actin) filaments over the thick (myosin) filaments.
• Muscle contraction is initiated by a neural signal from the central nervous system through a motor neuron.
• When the neural signal reaches the neuromuscular junction, it releases a neurotransmitter, i.e., acetylcholine, which generates an action potential in the sarcolemma.
• This spreads through the muscle fibre and causes the release of Ca⁺⁺ ions from the sarcoplasmic reticulum into the sarcoplasm.
• The Ca⁺⁺ ions bind to the subunit of troponin and brings about conformational changes; this removes the masking of the active site for myosin.
• The myosin head binds to the active site on actin to form a cross-bridge; this utilises energy from the hydrolysis of ATP.
• This pulls the actin filaments towards the centre of A-band.
• As a result, the Z-lines limiting the sarcomere are pulled closer together, causing a shortening of the sarcomere or contraction of muscle.
• Thus, during muscle contraction, the length of A band remains unchanged, while that of I-band decreases.
• The myosin goes back to its relaxed state.
• A new ATP binds and the cross-bridge is broken and the actin filaments slide out of A-band.
• The cycle of cross bridge-formation and cross bridge breakage continues till the Ca⁺⁺ ions are pumped back to the sarcoplasmic reticulum which leads to the masking of the active site on F-actin.

Question 4.
Write true or false. If false change the statement so that it is true.
(a) Actin is present in thin filament.
(b) H-zone of striated muscle fibre represents both thick and thin filaments.
(c) Human skeleton has 206 bones.
(d) There are 11 pairs of ribs in man.
(e) Sternum is present on the ventral side of the body.
Answer:
(a) True
(b) False, H-zone represents thick filaments
(c) True
(d) False, there are 12 pairs of ribs in man.
(e) True

Question 5.
Write the difference between:
(a) Actin and Myosin
(b) Red and White Muscles
(c) Pectoral and Pelvic Girdle
Answer:
(a) Differences between Actin and Myosin Filament

Actin Filaments	Myosin Filaments
1. These are found in I-band.	These are found in A-band.
2. These are thin.	These are thick.
3. Cross bridges (heads) are absent.	Cross bridges (heads) are present.
4. It is a globular protein with low molecular weight.	It is a heavy molecular weight polymerised protein.

(b) Differences between Red and White Muscles

Red Muscles	White Muscles
1. In some muscles, myoglobin content is high, which gives a reddish colour to them, such muscles are called red muscles.	Some muscles possess very less quantity of myoglobin, so they appear whitish called as white muscles.
2. These contain plenty of mitochondria.	These have less number of mitochondria but amount of sarcoplasmic reticulum is high.
3. These are called aerobic muscles.	They depend on anaerobic process of energy.

(c) Differences between Pectoral and Pelvic Girdle

Pectoral Girdle	Pelvic Girdle
1. It helps in the articulation of upper limbs.	It helps in the articulation of lower limbs.
2. It is situated in the pectoral region of the body.	It is situated in the pelvic region of the body.
3. Each half of pectoral girdle is formed of a clavicle and a scapula.	Pelvic girdle consists of two coxal bones.
4. Scapula is a large triangular flat bone and clavicle is a long slender bone.	Each coxal bone is formed of three bones, ilium, ischium and pubis.
5. An expanded process, acromion from scapula forms a depression called glenoid cavity, which articulates with the head of humerus to form shoulder joint.	Ilium, ischium and pubis fuse at a point to form a cavity called acetabulum to which the thigh bone articulates.

Question 6.
Match Column-I with Column-II

Column-I	Column-II
(a) Smooth muscle	(i) Myoglobin
(b) Tropomyosin	(ii) Thin filament
(c) Red muscle	(iii) Sutures
(d) Skull	(iv) Involuntary

Answer:

Column-I	Column-II
(a) Smooth muscle	(iv) Involuntary
(b) Tropomyosin	(ii) Thin filament
(c) Red muscle	(i) Myoglobin
(d) Skull	(iii) Sutures

Question 7.
What are the different types of movements exhibited by the cells of human body?
Answer:
Cells of the human body exhibit three main types of movements-amoeboid, ciliary and muscular.
(i) Amoeboid Movement: Some specialised cells in our body like macrophages and leucocytes in blood exhibit amoeboid movement. It is effected by pseudopodia formed by the streaming of protoplasm (as in Amoeba). Cytoskeletal elements like microfilaments are also involved in amoeboid movement.

(ii) Ciliary Movement: Ciliary movement occurs in most of our internal tubular organs which are lined by ciliated epithelium. The coordinated movements of cilia in the trachea help us in removing dust particles and some of the foreign substances inhaled along with the atmospheric air. Passage of ova through the female reproductive tract is also facilitated by the ciliary movement.

(iii) Muscular Movement: Movement of our limbs, jaws, tongue, etc., require muscular movement. Locomotion requires a perfect coordinated activity of muscular, skeletal and neural systems.

Question 8.
How do you distinguish between a skeletal muscle and a cardiac muscle?
Answer:

Skeletal Muscle	Cardiac Muscle
1. The, cells of skeletal muscles are unbranched.	1. The cells of cardiac muscles are branched.
2. Intercalated disks are absent.	2. The cells are joined with one another by intercalated disks that help in coordination or synchronization of the heartbeat.
3. Alternate light and dark bands are present.	3. Faint bands are present.
4. They are voluntary muscles.	4. They are involuntary muscles.
5. They contract rapidly and get fatigued in a short span of time.	5. They contract rapidly but do not get fatigued easily.
6. They are present in body parts such as the legs, tongue, hands, etc.	6. These muscles are present in the heart and control the contraction and relaxation of the heart.

Question 9.
Name the type of joint between the following:
(i) Atlas/Axis
(ii) Carpal/Metacarpal of thumb
(iii) Between phalanges
(iv) Femur/Acetabulum
(v) Between cranial bones
(vi) Between pubic bones in the pelvic girdle.
Answer:
(i) Pivot joint
(ii) Saddle joint
(iii) Hinge joint
(iv) Ball and socket joint
(v) Fibrous joint
(vi) Cartilagenous joint

Question 10.
Fill in the blank spaces.
(a) All mammals (except a few) have ………………………………. cervical vertebra.
(b) The number of phalanges in each limb of human is ……………………………………
(c) Thin filament of myofibril contains 2 ‘F’ actins and two other proteins namely ………………………. and ………………..
(d) In a muscle fibre Ca²⁺ is stored in ………………………..
(e) ………………….. and ……………………………….. pairs of ribs are called floating ribs.
(f) The human cranium is made of …………………………. bones.
Answer:
(a) seven
(b) fourteen.
(c) troponin and tropomyosin
(d) sarcoplasm
(e) 11 th; 12th
(f) eight