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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 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.

(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.