Samacheer Kalvi 11th Bio Botany Solutions Chapter 13 Photosynthesis

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Tamilnadu Samacheer Kalvi 11th Bio Botany Solutions Chapter 13 Photosynthesis

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Samacheer Kalvi 11th Bio Botany Photosynthesis Text Book Back Questions and Answers

Question 1.
Assertion (A): Increase in proton gradient inside lumen responsible for ATP synthesis
Reason (R): Oxygen evolving complex of PSI located on thylakoid membrane facing Stroma, releases H⁺ ions
a) Both Assertion and Reason are True
b) Assertion is Tine and Reason is False
c) Reason is True Assertion is False
d) Both Assertion and Reason are False
Answer:
(b) Assertion is Tine and Reason is False

Question 2.
Which chlorophyll molecule does not have a phytol tail?
(a) Chl – a
(b) Chl – b
(c) Chl – c
(d) Chl – d
Answer:
(c) Chl – c

Question 3.
The correct sequence of flow of electrons in the light reaction is
a) PS II, Plastoquinone, cytochrome PS I, ferredoxin
b) PS I, plastoquinone, cytochrome, PS II, ferredoxin.
c) PS II, ferredoxin, plastoquinone, cytochrome, PS I.
d) PS I, plastoquinone, cytochrome, PS II, ferredoxin
Answer:
(a) PS II, Plastoquinone, cytochrome PS I, ferredoxin

Question 4.
For every CO₂ molecule entering the C₃ cycle, the number of ATP & NADPH required:
(a) 2 ATP + 2 NADPH
(b) 2 ATP + 3 NADPH
(c) 3 ATP + 2 NADPH
(d) 3 ATP + 3 NADPH
Answer:
(c) 3 ATP + 2 NADPH.

Question 5.
Identify true statement regarding light reaction of photosynthesis?
a) Splitting of water molecule is associate with PS I.
b) PS I and PS II involved in the formation of NADPH + H⁺
c) The reaction center of PSI is Chlorophyll a with absorption peak at 680nm.
d) The reaction center of PS II is Chlorophyll a with absorption peak at 700 nm.
Answer:
(b) PS I and PS II involved in the formation of NADPH + H⁺

Question 6.
Two groups (A & B) of bean plants of similar size and same leaf area were placed in identical conditions. Group A was exposed to light of wavelength 400 – 450 nm and Group B to light of wavelength of 500 – 550 nm. Compare the photosynthetic rate of the 2 groups giving reasons.
Answer:
The rate of photosynthesis in group A bean plants is more than what is found in Group B plants because the rate of absorption of light is more at the wavelength is less beyond the wavelength of 500 – 550 nm.

Question 7.
A tree is believed to be releasing oxygen during nighttime. Do you believe the truthfulness of this statement? Justify your answer by giving reasons?
Answer:
It is a belief that some trees such as the Aloe vera Peepal tree, some palm varieties grown as indoor plants release oxygen during night time. There is no scientific evidence for this So it is not true. This is because oxygen is evolved during the light reaction of photosynthesis only. This reaction occurs only in the presence of light. Therefore oxygen cannot be released during nighttime.

Question 8.
Grasses have an adaptive mechanism to compensate. photorespiratory losses – Name and describe the mechanism.
Answer:
Rate of respiration is more in light than in dark. Photorespiration is the excess respiration taking place in photosynthetic cells due to absence of CO₂ and increase of O₂. This condition changes the carboxylase role of RUBISCO into oxygenase. C₂ Cycle takes place in chloroplast, peroxisome and mitochondria. RUBP is converted into PGA and a 2C – compound phosphoglycolate by Rubisco enzyme in chloroplast. Since the first product is a 2C – compound, this cycle is known as C₂ Cycle. Phosphoglycolate by loss of phosphate becomes glycolate.

Glycolate formed in chloroplast enters into peroxisome to form glyoxylate and hydrogen peroxide. Glyoxylate is converted into glycine and transferred into mitochondria. In mitochondria, two molecules of glycine combine to form serine. Serine enters into peroxisome to form hydroxy pyruvate. Hydroxy pyruvate with help of NADH + H⁺ becomes glyceric acid. Glyceric acid is cycled back to chloroplast utilising ATP and becomes Phosphoglyceric acid (PGA) and v enters into the Calvin cycle (PCR cycle). Photorespiration does not yield any free energy in the form of ATP. Under certain conditions 50% of the photosynthetic potential is lost because of Photorespiration

Question 9.
In Botany class, teacher explains, Synthesis of one glucose requires 30 ATPs in C₄ plants and only 18 ATPs in C₃ plants. The same teacher explains C₄ plants are more advantageous than C₃ plants. Can you identify the reason for this contradiction?
Answer:
C₄ plants requires 30 ATPs and 12 NADPH + H⁺ to synthesize one glucose, but C₃ plants require only 18 ATPs and 12 NADPH + H⁺ to synthesize one glucose molecule. If then, how can you say C₄ plants are more advantageous? C₄ plants are more advantageous than C₃ plants because C₄ photosynthesis is advantages over C₃ plant, because C₄ photosynthesis avoids photorespiration and is thus potentially more efficient than C₃ plants. Due to the absense of photorespiration, carbon di oxide compensation point for C₄ is lower than that of C₃ plants.

Question 10.
When there is plenty of light and higher concentration of O₂, what kind of pathway does the plant undergo? Analyse the reasons.
Answer:
Photorespiration is the excess respiration taking place in photosynthetic cells due to absence of CO₂ and increase of O₂. This condition changes the carboxylase role of the RUBISCO (RUBP carboxylase oxygenase) enzyme into oxygenase. C₂ cycle or photorespiration begins and operates in the chloroplast, Peroxisome, and Mitochondria.
It protects, photosynthetic cells from photooxidative damage.

Samacheer Kalvi 11th Bio Botany Photosynthesis Additional Questions & Answers

I. Choose the correct answer (1 Mark)

Question 1.
The radiation that reaches the earth & the visible spectrum …….. are
a) 300 to 2600 nm & 390 to 793 nm
b) 390 to763 nm & 560 to987nm
c) 440 to l600 nm & 770 to689nm
d) 480 to 1800 nm & 660 to 978 nm
Answer:
a) 300 to 2600 nm & 390 to 763 n

Question 2.
Who has first explained the importance of chlorophyll in photosynthesis:
(a) Joseph Priestly
(b) Dutrochet
(c) Stephen Hales
(d) Lavoisier
Answer:
(b) Dutrochet

Question 3.
The number of chlorophyll molecules required to fix one molecules of CO₂ are
a) 1500
b) 2500
c) 3500
d) 4400
Answer:
b) 2500

Question 4.
Who received 1988 Nobel prize for his work on photosynthesis in Rhodobacter:
(a) Emerson and Arnold
(b) Ruben and Kamem
(c) Arnon, Allen and Whatley
(d) Huber, Michael and Dissenhofer
Answer:
(d) Huber, Michael and Dissenhofer

Question 5.
Carotenoids are mostly
a) Methyl group of chemicals
b) Ferric compounds
c) Tetra terpenes
d) Triterpenoids
Answer:
c) Tertra terpenes

Question 6.
Indicate the correct statement:
(a) Grana lamellae have only PS I
(b) Stroma lamellae have only PS II
(c) Grana lamellae have both PS I and PS II
(d) Stroma lamellae have both PS I and PS II
Answer:
(c) Grana lamellae have both PS I and PS II

Question 7.
Match the following:

A. Cyanobacteria	(i) Chlorophyll D
B. Green algae	(ii) Chlorophyll C
C. Brown algae	(iii) Chlorophyll A
D. Red algae	(iv) Chlorophyll B

(a) A – (iii); B – (i); C – (iv); D – (ii)
(b) A – (ii); B – (iii); C – (iv); D – (i)
(c) A – (iii); B – (iv); C – (i); D – (ii)
(d) A – (iii); B – (iv); C – (ii); D – (i)
Answer:
(d) A – (iii); B – (iv); C – (ii); D – (i)

Question 8.
ATP required to release electron is
a) 48 ATP
b) 22 ATP
c) 12 ATP
d) 18 ATP
Answer:
d) 18 ATP

Question 9.
Biosynthesis of chlorophyll ‘a’ requires:
(a) Mg, Fe, Cu, Zn, Mn, K and nitrogen
(b) Mg, Fe, Cu, Mo, Mn, K and nitrogen
(c) Mg, Cu, Zn, Mo, Mn, K and nitrogen
(d) Mg, Fe, Cu, Zn, Mo, K and nitrogen
Answer:
(a) Mg, Fe, Cu, Zn, Mn, K and nitrogen

Question 10.
One complete light reaction involves ……………….. quanta of light
a) 50
b) 45
c) 48
d) 90
Answer:
c) 48

Question 11.
Almost all carotenoid pigments have:
(a) 50 carbon atoms
(b) 40 carbon atoms
(c) 30 carbon atoms
(d) 60 carbon atoms
Answer:
(b) 40 carbon atoms

Question 12.
The 5C – primary acceptor of CO₂ in the Calvin cycle is
a) OAA
b) RUBP
c) PEA
d) RUBISCO
Answer:
b) RUBP

Question 13.
The visible spectrum of light ranges between:
(a) 200 to 2800 nm
(b) 300 to 2600 nm
(c) 200 to 800 nm
(d) 300 to 2400 nm
Answer:
(b) 300 to 2600 nm

Question 14.
Water splitting is concerned with photosystem
a) P.S I
b) P.S II
c) Both (a) & (b)
d) P.S II
Answer:
b) P.S II

Question 15.
Indicate the correct statement in respect to Hill’s reaction:
(i) During photosynthesis oxygen is evolved from water
(ii) Electrons for the reduction of CO₂ are obtained from H₂S.
(iii) During photosynthesis oxygen is evolved from CO₂
(iv) Electrons for the reduction of CO₂ are obtained from water

(a) (i) and (ii)
(b) (ii) and (iii)
(c) (i) and (iv)
(d) (ii) and (iv)
Answer:
(c) (i) and (iv)

Question 16.
Photosystem usually involved in both cyclic and non – cyclic phosphorylation is
a) PS-I
b) PS II
c) Both (a) & (b)
d) P. 890
Answer:
b) PS II

Question 17.
Find out the odd one:
(a) Ferredoxin
(b) Succinate
(c) Cytochrome b₆ – f
(d) Plastocyanin
Answer:
(b) Succinate

Question 18.
C3 cyclic is not known as
a) Dark reaction
b) Biosynthetic phase
c) P C.R. Cycle
d) CAM Cycle
Answer:
d) CAM Cycle

Question 19.
The chemiosmotic theory was proposed by:
(a) S. Michael
(b) R. Hill
(c) P. Mitchell
(d) G. Root
Answer:
(c) P. Mitchell

Question 20.
Maximum O₂ evolution occurs from
a) forests
b) Marine phytoplankton
c) Crops
d) Landmass
Answer:
b) Marine phytoplankton

Question 21.
The key enzyme in the carboxylation reaction is:
(a) Ribulose dehydrogenase
(b) Carboxylase
(c) Carboxylase oxygenase
(d) Carboxyl anhydrase
Answer:
(c) Carboxylase oxygenase

Question 22.
Photo reaction centres in higher plants are
a) P700
b) P 680
c) Both a & b
d) Chlorophyll a
Answer:
c) Both a & b

Question 23.
In bundle sheath cells, malic acid undergoes decarboxylation and produces 3 carbon compound:
(a) Glyceric acid and CO₂
(b) Glyceraldehyde and CO₂
(c) Pyruvic acid and CO₂
(d) None of the above
Answer:
(c) Pyruvic acid and CO₂

Question 24.
Water-soluble pigment is
a) Chlorophyll
b) Carotene
c) Phycobillin
d) Xanthophyll
Answer:
c) Phycobillin

Question 25.
Crassulacean acid metabolism or CAM cycle was first observed in:
(a) sugarcane
(b) bryophyllum
(c) mango
(d) banana
Answer:
(b) bryophyllum

Question 26.
Glycolate protects plant cells from:
(a) Photophosphorylation
(b) Photo reduction
(c) Photo oxidation
(d) Photolysis
Answer:
(c) Photo oxidation

Question 27.
The site of photophosphorylation in Chloroplast is
a) granum
b) matrix
c) chloroplast surface
d) None of these
Answer:
a) granum

Question 28.
Hormones like gibberellin:
(a) increases the rate of photosynthesis
(b) increase respiration
(c) decrease the rate of photosynthesis
(d) decrease the rate of transpiration
Answer:
(a) increases the rate of photosynthesis

Question 29.
Bacterial photosynthesis differs from higher plant photosynthesis in:
(a) utilizing water as electron donar
(b) releasing O₂
(c) releasing sulphur instead of oxygen
(d) utilizing SO₂ as electron donar
Answer:
(c) releasing sulphur instead of oxygen

Question 30.
Agranal chloroplasts are characteristics of
a) Mesophyll of pea leaves
b) Bundle sheaths of Mango leaves
c) Mesophyll of Maize leaves
d) Bundle sheath of sugarcane leaves
Answer:
d) Bundle sheath of sugarcane leaves.

II. Answer the following (2 Marks)

Question 1.
What is Bioluminescence? or Chemiluminescence?
Answer:
Bioluminescence is the production and emission of light by a living organism. Bioluminescence is rare in true plants – Eg. Fire flies, glow worms etc. – Also observed in some fungi.

Question 2.
Define photosynthesis.
Answer:
Photosynthesis is referred to as photochemical oxidation and reduction reactions carried out with help of light, converting solar energy into Chemical energy.

Question 3.
Name the Photosynthetic pigments of Bacteria
Answer:
Bacterio chlorophyll a, Bacteria chlorophyll b, Chlorobium chlorophyll 650, Chlorobium chlorophyll 666.

Question 4.
What is thylakoid? Explain how they are arranged?
Answer:
A sac like membranous system called thylakoid or lamellae is present in stroma and they are arranged one above the other forming a stack of coin like structure called granum (plural grana).

Question 5.
What is Emerson’s Enhancement Effect (EEE)?
Answer:
Emerson found when monochromatic light of longer wavelength (Farred) When supplemental with shorter wavelength (red light) enhanced the photosynthetic yield and recovered red drag.

This enhancement of photosynthetic yield is referred to as EEE.
I) P.S – yield of farred light (710 nm) =10
II) P.S- yield of red light (650 nm) = 43.5
(I) + (II) = 72.5

EEE = 72.5

Question 6.
Define photosynthetic pigment.
Answer:
A photosynthetic pigment is a pigment that is present in chloroplasts or photosynthetic bacteria which captures the light energy necessary for photosynthesis.

Question 7.
Match the following:

A. Xanthophyll	(i) Lycopene
B. Phycocyanin	(ii) Red algae
C. Carotene	(iii) Brown algae
D. Phycoerythin	(iv) Cyanobacteria

Answer:
A – (iii), B – (iv), C – (i), D – (ii).

Question 8.
What is carbon di oxide compensation point?
Answer:

• Carbon di oxide compensation point is that point at which the rate of Photosynthesis equals the rate of Respiration.
• At this point there is no exchange of oxygen and carbon di oxide.
• It occur when light is not the limiting factor and the concentration of CO₂ is 50 -100 ppm.

Question 9.
Write down any two properties of light.
Answer:
Two properties of light:

1. Light is a transverse electromagnetic wave.
2. It consists of oscillating electric and magnetic fields that are perpendicular to each other and perpendicular to the direction of propagation of the light.

Question 10.
What is meant by non – oxygenic photosynthesis?
Answer:
In some organism, oxygen is not evolved during photosynthesis and is called Non-oxygenic and Anaerobic Photosynthesis (NOAP).
Eg. Green sulphur, Purple sulphur, and green filamentous bacteria.

Question 11.
Define the term fluorescence.
Answer:
The electron from first singlet state (SI) returns to ground state (SO) by releasing energy in the form of radiation energy (light) in the red region and this is known as fluorescence.

Question 12.
Notes on xanthophylls.
Answer:

• Yellow pigments like carotene but contain oxygen.
• Lutein – responsible for colour change of leaves during autumn season.
• Eg. Lutein, violaxanthin, & Fuco xanthin.

Question 13.
Define photophosphorylation.
Answer:
Phosphorylation is the process of synthesis of ATP by the addition of inorganic phosphate to ADP. The addition of phosphate here takes place with the help of light generated electron and so it is called as photophosphorylation.

Question 14.
What are the phases of dark reaction?
Answer:
Dark reaction consists of three phases:

• Carboxylation (fixation)
• Reduction (Glycolytic Reversal)
• Regeneration

Question 15.
What is meant by photolysis of water or photo-oxidation of water?
Answer:

• The splitting up of water molecules into OH^– ions and H+ ions.
• The process of photolysis is associated with oxygen Evolving complex (OEC or water splitting complex in pigment system II.
• It is catalysed by Mn⁺⁺and Cl^– ions.
• At the end of Photolysis 4H⁺, 4e^– and O² are evolved from water.

Question 16.
What is meant by carbon dioxide compensation point?
Answer:
When the rate of photosynthesis equals the rate of respiration, there is no exchange of oxygen and carbon dioxide and this is called as carbon dioxide compensation point.

Question 17.
What, are the internal factors, that affect photosynthesis?
Answer:
Pigments, protoplasmic factor, accumulation of carbohydrates, anatomy of leaf and hormones.

Question 18.
What is meant by ‘Z’ scheme?
Answer:
During Non cyclic photophosphorylation in the light reaction of Photosynthesis, the electron flow looks like the appearance of letter ‘Z’ and so known as Z – scheme.

Question 19.
How does water affect the rate of photosynthesis?
Answer:
Photolysis of water provides electrons and protons for the reduction of NADP, directly. Indirect roles are stomatal movement and hydration of protoplasm. During water stress, the supply of NADPH + H⁺ is affected.

Question 20.
Why do we consider Cyclic electron transport or PS I – considered to be formed earlier in evolution?
Answer:

• P.S. I – need the light of longer wavelength (P – 700nm)
• P. S. I – acts under low light intensity.
• P.S.I – acts under low CO₂ concentration.
• P.S.I – function even under anaerobic conditions.
• These are the aspects that help to consider PSI as an earlier formed one in evolution.

III. Answer the following. (3 Marks)

Question 1.
Hydrogen energy is the future hope of the world – justify.
Answer:

• Yes – Hydrogen stands apart as a promising alternative energy source.
• Hydrogen is a component of water. There is more than enough water on earth to supply a global hydrogen fuel system.
• Hydrogen is the highest energy content of any common fuel by weight.
• It is clean energy as there is no carbon emission produced directly during this cycle.

Uses: To run vehicles, to heat homes office, to produce electricity, fuel aircraft, etc. Thus it is a part of the world’s energy future.

Question 2.
How is the chlorophyll synthesized?
Answer:
Chlorophyll is synthesized from intermediates of respiration and photosynthesis. Succinic acid an intermediate of Krebs cycle is activated by the addition of coenzyme A and it reacts with a simple amino acid glycine and the reaction goes on to produce chlorophyll ‘a’. Biosynthesis of chlorophyll ‘a’ requires Mg, Fe, Cu, Zn, Mn, K and nitrogen. The absence of any one of these minerals leads to chlorosis.

Question 3.
What are phycobilins?
Answer:
They are proteinaceous pigments, soluble in water, and do not contain Mg and Phytol tails. They exist in two forms such as:

1. Phycocyanin found in cyanobacteria.
2. Phycoerythrin found in Rhodophyceae algae (Red algae).

Question 4.
What are the conclusions of Hill’s reaction?
Answer:
The conclusions of Hill’s reaction:

1. During photosynthesis, oxygen is evolved from water.
2. Electrons for the reduction of CO₂ are obtained from water.
3. A reduced substance produced, later helps to reduce CO₂.

Question 5.
What are the 3 excited states of Chlorophyll?
Answer:
When a photon of light collides with the chlorophyll molecule, an electron from the outermost orbit is moved to a higher energy orbit causing excitation of chlorophyll. ‘

1. First singlet state (S₁)
2. Second singlet state (S₂)
3.  First triplet state (T₁)

Question 6.
Explain the term phosphorescence.
Answer:
Electron from Second Singlet State (S₂) may return to next higher energy level (S₁) by losing some of its extra energy in the form of heat. From first singlet state (S₁) electron further drops to first triplet state (T₁). Triplet State is unstable having half life time of 10^-3 seconds and electrons returns to ground state with emission of light in red region called as phosphorescence. Phosphorescence is the delayed emission of absorbed radiations. Pathway of electron during Phosphorescence:
S₂ → S₁ → T₁ → S₀

Question 7.
Describe the method of carboxylation.
Answer:
The acceptor molecule Ribulose 1,5 Bisphosphate (RUBP) a 5 carbon compound with the help of RUBP carboxylase oxygenase (RUBISCO) enzyme accepts one molecule of carbon dioxide to form an unstable 6 carbon compound. This 6C compound is broken down into two molecules of 3 – carbon compound phosphoglyceric acid (PGA).

Question 8.
Explain phase-3 of dark reaction.
Answer:
Regeneration of RUBP involves the formation of several intermediate compounds of 6 – carbon, 5 – carbon, 4 – carbon and 7 – carbon skeleton. Fixation of one carbon dioxide requires 3 ATPs and 2 NADPH + H⁺, and the fixation of 6 CO₂ requires 18 ATPs and 12 NADPH + H⁺ during C₃ cycle. One 6 carbon compound is the net gain to form hexose sugar.

Overall equation for dark reaction:
6CO₂ + 18 ATP + 12 NADPH + H⁺ → C₆H₁₂O₆ + 6H₂O + 18 ADP + 18 Pi + 12 NADP⁺

Question 9.
Notes on RUBISCO.
Answer:

• RUBISCO is known as RUBP carboxylase oxygenase enzyme.
• This is the most abundant protein found on earth.
• It constitutes 16% of the chioroplast protein.
• It act as carboxvlase in the presence of CO₂
• It act as oxygenase in the absence of CO₂

Question 10.
Mention the significances of C₄ cycle.
Answer:
The significances of C₄ cycle:

1. Plants having C₄ cycle are mainly of tropical and subtropical regions and are able to survive in environments with low CO₂ concentration.
2. C₄ plants are partially adapted to drought conditions.
3. Oxygen has no inhibitory effect on C₄ cycle since PEP carboxylase is insensitive to O₂.
4. Due to absence of photorespiration, CO₂ Compensation Point for C₄ is lower than that of C₃ plants.

Question 11.
What is the type of carbon pathway in xerophytic plants?
Answer:
Crassulacean Acid Metabolism or CAM cycle is one of the carbon pathways identified in succulent plants growing in semi-arid or xerophytic condition. This was first observed in crassulaceae family plants like Bryophyllum, Sedum, Kalanchoe and is the reason behind the name of this cycle. It is also noticed in plants from other families eg: Agave, Opuntia, Pineapple and Orchids.

Question 12.
Write clown the significance of CAM cycle.
Answer:

• It is advantageous for succulent plants to obtain CO₂, from malic acid when stomata are closed.
• During day time stomata are closed and CO₂ is not taken in but continue their photosynthesis.
• Stomata are closed during day & help plants to avoid transpiration and water loss.

IV. Answer the following (5 Marks)

Question 1.
Explain in detail about absorption spectrum and action spectrum of light.
Answer:
1. Absorption spectrum: The term absorption refers to complete retention of light, without reflection or transmission. Pigments absorb different Wavelengths of light. A curve obtained by plotting the amount of absorption of different wavelengths of light by a pigment is called its absorption spectrum.

• Chlorophyll ‘a’ and chlorophyll ‘b’ absorb quanta from blue and red region.
• Maximum absorption peak for different forms of chlorophyll ‘a’ is 670 to 673, 680 to 683 and 695 to 705 nm.
• Chlorophyll ‘a’ 680 (P680) and Chlorophyll ‘a’ 700 (P700) function as trap centre for PS II and PS I respectively.

2. Action Spectrum: The effectiveness of different wavelength of light on photosynthesis is measured by plotting against quantum yield. The curve showing the rate of photosynthesis at different wavelengths of light is called action spectrum. From the graph showing action spectrum, it can be concluded that maximum photosynthesis takes place in blue and red region of the spectrum. This wavelength of the spectrum is the absorption maxima for Chlorophyll (a) and Chlorophyll (b). The Action Spectrum is instrumental in the discovery of the existence of two photosystems in O₂ evolving photosynthesis.

Question 1.
Describe the structure of the chloroplast.
Answer:
A chloroplast is the site of photosynthesis.
Structure:

• Double membrane-bound organelle.
• Discoid or lens-shaped.
• 4-10 um in diameter, 1.33 um thickness.
• The space between the membranes is 100 – 200 A.

Stroma:

• Proteinaceous matrix –
• A sac-like membranous system called thylakoid or lamellae is present in the stroma.
• Thylakoids are stacked like coin – known as granum. (grana – plural).
• In each chloropiast 40 – 80 grana and each grana consists of 5 to 30 thylakoids.
• The thylakoids are known as grana lamellae and the lamellae in the stroma is known as stroma lamellae.
• A thinner lamella known as fret membrane connects grana.
• Disc size of thylakoid is 0.25 to 0.8 micron in diameter.
• Pigment system I is located on outer thylakoid membrane.
• Pigement system II is located on inner membrane facing thylakoid lumen.
• Grana lamellae have both PS I & PS II.
• Stroma lamellae have only PS I
• Chloropiast contains 30 – 35% proteins, 20 – 30% phopholipids, 5 -10% chlorophylls, 4 – 5% carotenoids.
• It also contain 70s Ribosome, Circular DNA, Starch grain.
• Inner membrane consists of spherical structures called Quantasomes.
• Stroma – contain photosynthetic enzyme.

Question 3.
Explain the process of photolysis water with a suitable diagram.
Answer:
The process of Photolysis is associated with Oxygen Evolving Complex (OEC) or water splitting complex in pigment system II and is catalyzed by the presence of Mn⁺⁺ and Cl^–. When the pigment system II is active it receives light and the water molecule splits into OH^– ions and H⁺ ions. The OH^– ions unite to form water molecules again and release O₂ and electrons. Photolysis of water is due to strong oxidant which is yet unknown and designated as Z or Yz.

Widely accepted theory proposed by Kok et al., (1970) explaining photo – oxidation of water is water oxidizing clock (or) S’ State Mechanism. It consists of a series of 5 states called as S₀, S₁, S₂, S₃ and S₄. Each state acquires positive charge by a photon (hv) and after the S₄ state it acquires 4 positive charges, four electrons and evolution of oxygen. Two molecules of water go back to the S₀. At the end of photolysis 4H⁺, 4e^– and O₂ are evolved from water.

Question 4.
Describe the process of non – cyclic photophosphorylation.
Answer:
When photons are activated reaction centre of pigment system II (P680), electrons are moved to the high energy level. Electrons from high energy state passes through series of electron carriers like pheophytin, plastoquinone, cytochrome complex, plastocyanin and finally accepted by PS I (P700). During this movement of electrons from PS II to PS I ATP is generated. PS I (P700) is activated by light, electrons are moved to high energy state and accepted by electron acceptor molecule ferredoxin reducing Substance (FRS). During the downhill movement through ferredoxin, electrons are transferred to NADP⁺ and reduced into NADPH + H⁺ (H⁺ formed from splitting of water by light).

Electrons released from the photosystem II are not cycled back. It is used for the reduction of NADP⁺ in to NADPH + H⁺. During the electron transport it generates ATP and hence this type of photophosphorylation is called non – cyclic photophosphorylation. The electron flow looks like the appearance of letter ‘Z’ and so known as Z scheme.

When there is availability of NADP⁺ for reduction and when there is splitting of water molecules both PS I and PS II are activated. Non-cyclic electron transport PS I and PS II both are involved cooperatively to transport electrons from water to MADP⁺. In oxygenic species non – cyclic electron transport takes place in three stages.

1. Electron transport from water to P680: Splitting of water molecule produce electrons, protons and oxygen. Electrons lost by the PS II (P680) are replaced by electrons from splitting of water molecules.
2. Electron transport from P680 to P700: Electron flow starts from P680 through a series of electron carrier molecules like pheophytin, plastoquinone (PQ), cytochrome b₆ – f complex, plastocyanin (PC) and finally reaches P700 (PS I).
3. Electron transport from P700 to NADP: PS I (P700) is excited now and the electrons pass to high energy level. When electron travels downhill through ferredoxin, NADP⁺ is reduced to NADPH + H⁺.

Question 5.
Explain chemiosmotic theory with suitable I diagram.
Answer:
The chemiosmotic theory was proposed by P. Mitchell (1966). According to this theory, electrons are transported along the membrane through PS I and PS II and connected by Cytochrome b₆ – f complex. The flow of electrical current is due to difference in electrochemical potential of protons across the membrane. Splitting of water molecule takes place inside the membrane. Protons or H+ ions accumulate within the lumen of the thylakoid (H⁺ increase 1000 to 2000 times). As a result, proton concentration is increased inside the thylakoid lumen.

These protons move across the membrane because the primary acceptor of electrons is located outside the membrane. Protons in stroma less in number and creates a proton gradient. This gradient is broken down due to the movement of a proton across the membrane to the stroma through CFO of the ATP synthase enzyme. The proton motive force created inside the lumen of thylakoid or chemical gradient of H⁺ ion across the membrane stimulates ATP generation.

The evolution of one oxygen molecule (4 electrons required) requires 8 quanta of light. C₃ plants utilise 3 ATPs and 2 NADPH + H⁺ to evolve one Oxygen molecule. To evolve 6 molecules of Oxygen 18 ATPs and 12 NADPH + H⁺ are utilised. C₄ plants utilise 5 ATPs and 2 NADPH + H⁺ to evolve one oxygen molecule. To evolve 6 molecules of Oxygen 30 ATPs and 12 NADPH + H⁺ are utilised.

Question 6.
Explain chemiosmotic theory.
Answer:

• Proposed by Mitchell in 1966.
• It explains the formation of ATP molecules in photosynthesis & Respiration.
• ATP synthesis in photosynthesis is linked to the development of proton gradient across the thylakoid membrane

Development of Proton gradient:

• Splitting of water molecules takes place on the inner side of thylakoid membrane. The protons produced are accumulated in the thylakoid lumen.
• With the transport of electrons through photosystems, protons are also transported across membranes.
• The number of protons decreases in the stroma and increases in the lumen. Thus creates a proton gradient.
• The proton gradient is broken due to the movement of proton across the membrane to the stroma through CFO of the ATP synthetase enzyme.
• The proton motive force or chemical gradient of  H⁺ ion across the membrane catalyses ATP generation.
• This membrane of ATP formation known as chemiosmotic mechanism of ATP formation.

Question 7.
Give the schematic diagram of photorespiration.
Answer:
The schematic diagram of photorespiration:

Question 8.
Distinguish between photorespiration and dark respiration.
Answer:
Photo respiration:

• It takes place in photosynthetic green cells.
• It takes place only in the presence of light.
• It involves chloroplast, peroxisome and mitochondria.
• It does not involve Glycolysis, Kreb’s Cycle, and ETS.
• Substrate is glycolic acid.
• It is not essential for survival.
• No phosphorylation and yield of ATP.
• NADH₂ is oxidised to NAD⁺.
• Hydrogen peroxide is produced.
• End products are CO₂ and PGA.

Dark respiration:

• It takes place in all living cells.
• It takes place all the time.
• It involves only mitochondria.
• It involves glycolysis, Kreb’s Cycle and ETS.
• Substrate is carbohydrates, protein or fats.
• Essential for survival.
• Phosphorylation produces ATP energy.
• NAD⁺ is reduced to NADH₂.
• Hydrogen peroxide is not produced.
• End products are CO₂ and water.

CHECK YOUR GRASP
Textbook Page No: 123

Question 1.
(i) Name the products produced from Non – Cyclic photophosphorylation?
(ii) Why does PS II require electrons from water?
(iii) Can you find the difference in the Pathway of electrons during PS I and PS II?
Answer:
(i) The products of non-cyclic phosphorylation are NADPH + H⁺ and ATP.
(ii) The electrons received from water are responsible for the production of ATP and NADPH + H⁺ through electron transport system in PS I and PS II.
(iii) Yes. Electron flow starts from P680 through a series of electron carrier molecules and finally reaches P700 (PSI). From PS I the electrons travel downhill through ferredoxin, NADP⁺ is recorded to NADPH + H⁺.

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