Rajasthan Board RBSE Class 12 Biology Chapter 10 Photosynthesis
RBSE Class 12 Biology Chapter 10 Multiple Choice Questions
Question 1.
Which metal element is found in the centre of the chlorophyll molecule?
(a) Fe
(b) Mg
(c) Ne
(d) Cu
Answer:
(b) Mg
Question 2.
PS-II is related to –
(a) Photolysis of water
(b) Reduction of \({ CO }_{ 2 }\)
(c) Flowering
(d) All the above
Answer:
(a) Photolysis of water
Question 3.
In PS-I and PS-II the reaction centres respectively are –
(a) \({ P }_{ 700 }\) and \({ P }_{ 680 }\)
(b) \({ P }_{ 600 }\) and \({ P }_{ 700 }\)
(c) \({ P }_{ 580 }\) and \({ P }_{ 700 }\)
(d) \({ P }_{ 700 }\) and \({ P }_{ 580 }\)
Answer:
(a) \({ P }_{ 700 }\) and \({ P }_{ 680 }\)
Question 4.
\({ O }_{ 2 }\) evolution is related with which of the following?
(a) PS – I
(b) PS – II
(c) Phytochrome
(d) All the above
Answer:
(b) PS – II
Question 5.
Which of the following process occurs in \({ C }_{ 4 }\) plants and not in \({ C }_{ 3 }\) plants.
(a) Glycolysis
(b) Photorespiration
(c) Transpiration
(d) Photosynthesis
Answer:
(b) Photorespiration
Question 6.
Unit of photosynthesis is –
(a) Quantasome
(b) Microsome
(c) Peroxisome
(d) Spherosome
Answer:
(a) Quantasome
Question 7.
Photolysis of water necessarily requires –
(a) Mn
(b) Mg
(c) Zn
(d) Fe
Answer:
(a) Mn
Question 8.
Select correct statement for photosynthesis –
(a) \(CO\) and \({ H }_{ 2 }{ O }\) are oxidised
(b) \({ CO }_{ 2 }\) and \({ H }_{ 2 }{ O }\) are reduced
(c) \({ H }_{ 2 }{ O }\) is reduced and \({ CO }_{ 2 }\) oxidized
(d) \({ H }_{ 2 }{ O }\) is oxidized and \({ CO }_{ 2 }\) reduced
Answer:
(d) \({ H }_{ 2 }{ O }\) is oxidized and \({ CO }_{ 2 }\) reduced
Question 9.
What is the source of oxygen, liberated in photosynthesis?
(a) \({ H }_{ 2 }{ O }\)
(b) \({ CO }_{ 2 }\)
(c) \({ H }_{ 2 }{ O }\) and \({ CO }_{ 2 }\) both
(d) Neither \({ O }_{ 2 }\) nor \({ CO }_{ 2 }\)
Answer:
(a) \({ H }_{ 2 }{ O }\)
Question 10.
What is the site of the dark reaction of photosynthesis?
(a) Grana of chloroplast
(b) Stroma of chloroplast
(c) The outer membrane of the chloroplast
(d) The inner membrane of the chloroplast
Answer:
(b) Stroma of chloroplast
Question 11.
Which of the following show Kranz anatomy?
(a) \({ C }_{ 3 }\) plants
(b) \({ C }_{ 4 }\) plants
(c) Succulents
(d) All the above
Answer:
(b) \({ C }_{ 4 }\) plants
Question 12.
The first stable product of the \({ C }_{ 4 }\) cycle is –
(a) Pyruvic acid
(b) Oxaloacetic acid
(c) Malic acid
(d) None of the above
Answer:
(b) Oxaloacetic acid
Question 13.
For reduction of 6 molecules of \({ CO }_{ 2 }\), reducing the power generated during non-cyclic photophosphorylation is depicted as –
(a) 24\({ { H }^{ + } }\)
(b) 36\({ { H }^{ + } }\)
(c) 18\({ { H }^{ + } }\)
(d) 12\({ { H }^{ + } }\)
Answer:
(a) 24\({ { H }^{ + } }\)
Question 14.
The wavelength of photosynthetically active radiation is –
(a) 340 – 450 nm
(b) 400 – 700 nm
(c) 500 – 600 nm
(d) 450 – 950 nm
Answer:
(b) 400 – 700 nm
RBSE Class 12 Biology Chapter 10 Very Short Answer Questions
Question 1.
Define photosynthesis.
Answer:
The process, by which green plants convert \({ CO }_{ 2 }\) absorbed from the atmosphere and \({ H }_{ 2 }{ O }\) absorbed from the soil into carbohydrate in the presence of sunlight, is called photosynthesis.
In other words synthesis of carbohydrates by green plants from \({ CO }_{ 2 }\) and \({ H }_{ 2 }{ O }\) in the presence of sunlight is called photosynthesis.
Question 2.
What is the first stable product of the reaction of photosynthesis?
Answer:
In \({ C }_{ 3 }\) plants the first stable product of \({ CO }_{ 2 }\) fixation is phosphoglyceric acid (PGA) whereas, in \({ C }_{ 4 }\) plants, the first stable product is oxalic acetic acid (OAA).
Question 3.
Write the formula of Chl. ‘a’ and Chl ‘b’. What is the main difference between chlorophyll ‘a’ and chlorophyll ‘b’.
Answer:
CHL. ‘a’ \({ C }_{ 55 }\)\({ H }_{ 72 }\)\({ O }_{ 5 }\) \({ N }_{ 4 }\) Mg
CHL. ‘b’ \({ C }_{ 55 }\)\({ H }_{ 70 }\)\({ O }_{ 6 }\) \({ N }_{ 4 }\) Mg
In chlorophyll ‘a’ \({ CH }_{ 3 }\) group is found on 3rd carbon of second pyrrole ring whereas in chlorophyll ‘b’ in place of \({ – }{ CH }_{ 3 }\), aldehyde (\({ – }{ CHO }\)) group is found.
Question 4.
Write a full form of NADP.
Answer:
Nicotinamide Adenine Dinucleotide Phosphate.
Question 5.
Name the cell organelles which participate in photorespiration.
Answer:
Chloroplasts Peroxisomes and Mitochondria.
Question 6.
Who is considered as father/founder of plant physiology?
Answer:
Stephen Hales is considered as the father of plant physiology.
Question 7.
What are the site of light reaction and the dark reaction of the process of photosynthesis?
Answer:
The light reaction takes place in the grana region of chloroplasts and dark reaction is completed in the stroma region of chloroplasts.
Question 8.
What is the law of limiting factor?
Answer:
According to Blackman, when some process is affected by more than one factors, then the rate of the process is determined by the factor present in a minimum amount. This is called Blackman’s law of limiting factor.
Question 9.
Which protein is found in the greatest amount in the biosphere?
Answer:
RuBISCO – Ribulose biphosphate carboxylase.
Question 10.
In which part of visible spectrum”red drop” occurs.
Answer:
The red drop occurs when the light of wavelength greater (longer) than 680 nm is supplied to plant.
Question 11.
In photosynthesis which pigments serve as accessory pigments.
Answer:
Carotenoids and phycobilins act as accessory pigments.
Question 12.
Why photorespiration is considered as a harmful or destructive activity?
Answer:
In the process of photorespiration, food is oxidised in the presence of \({ O }_{ 2 }\) and energy is not released in the form of ATP. Hence it is considered as a harmful or destructive activity.
RBSE Class 12 Biology Chapter 10 Short Answer Questions
Question 1.
Which pigments are involved in the process of photosynthesis?
Answer:
The main pigments involved in photosynthesis in plants are chlorophyll ‘a’ and chlorophyll ‘b’ whereas carotenoids and phycobilins (when present) act as accessory pigments.
Question 2.
Write a brief account of the structure of chloroplast.
Answer:
Chloroplasts are found in the mesophyll cells of leaves and other green parts of a plant.
- Each chloroplast is a discoid or lens-shaped structure bounded by a double membrane envelope.
The inner space is differentiated into two regions. - The matrix part of this organelles is called stroma and is filled with proteinaceous fluid containing 70s ribosomes, small circular DNA, osmophilic droplets dissolved salts and several enzymes.
- In the stroma, the region is found lamellar structures which are arranged one over the other like a stack of coins. The lamellar structures are thylakoids and this region comprising of lamellar structures is called granum.
- Each chloroplast may contain several grana which remain interconnected by stroma lamellae or frets.
- Photosynthetic pigments, chlorophyll, carotenoids are found in the grana region of a chloroplast. Hence this acts as a site of light reaction of photosynthesis.
- Dark reaction is completed in the stroma region of the chloroplast.
Question 3.
What is the contribution of Blackman in plant physiology?
Answer:
Blackman (1905) first proposed that the process of photosynthesis is completed in two steps.
- Light reaction
- Dark reaction.
Blackman also proposed the law of limiting factor. According to this principle, if some process is affected by several factors, then it’s rate is determined by the factor which is present in the minimum amount.
Question 4.
Explain in brief, the structure of chlorophyll.
Answer:
Chlorophyll is a bipolar macromolecule. The head part of this is made up of porphyrin ring and the tail is made up of phytol chain.
- The top part is formed by four pyrrole molecules arranged in a cycle and form the porphyrin ring.
- In the centre of this ring is found magnesium atom.
- On the 7th carbon of fourth pyrrole ring, is attached a phytol chain which forms the phytol tail.
- The head part of chlorophyll is hydrophilic and the tail part is hydrophobic.
Question 5.
Explain the difference between \({ C }_{ 3 }\) and \({ C }_{ 4 }\) cycle.
Answer:
Question 6.
\({ CO }_{ 2 }\) fixation through crassulacean acid metabolism is a physiological adaptation in xerophytic and succulent plants. Explain.
Answer:
Crassulacean acid metabolism is normally found in succulent plants and plants growing under xeric environment. In these plants, leaves are thick and fleshy and the stomata in the leaves open during the night and remain closed during day time.
- These plants fix \({ CO }_{ 2 }\) during the night when stomata are open and store it in the form of organic acid.
- During day time \({ CO }_{ 2 }\) is released by decarboxylation and use the \({ CO }_{ 2 }\) to synthesize carbohydrate by Calvin cycle.
- Hence these plants can use atmospheric \({ CO }_{ 2 }\) without losing much water by transpiration because, during day time when the factors promoting transpiration are at the high magnitude, the stomata remain closed and open during the night only when factors relating to transpiration are at low magnitude.
Question 7.
What do you understand by photophosphorylation?
Answer:
Formation of ATP molecule from ADP and inorganic phosphate (Pi) in the presence of sunlight is called photophosphorylation. This process takes place in the grana region of chloroplast and solar energy is stored in the form of chemical energy in ATP molecule. Photophosphorylation is of two types:
- Cyclic photophosphorylation
- Non-cyclic photophosphorylation
RBSE Class 12 Biology Chapter 10 Essay Type Questions
Question 1.
Describe the process of light reaction of photosynthesis.
Answer:
The process of light reaction is completed in the grana region of chloroplast in the presence of sunlight. In this process, the electromagnetic energy of sunlight is converted into chemical energy and stored in the form of ATP and NADPH + \({ { H }^{ + } }\).
The process of light reaction is complete in the following steps:
- The excitement of the chlorophyll molecule and as a result, the electrons are emitted out.
- Photolysis of water.
- NADPH + \({ { H }^{ + } }\) formation.
- Cyclic and non-cyclic photophosphorylation.
Light Reaction:
- This phase is also called photochemical reaction or Hill’s reaction.
- This process involves several reactions leading to the conversion of light energy into chemical energy in the form of ATP and NADPH + \({ { H }^{ + } }\) and evolution of \({ O }_{ 2 }\).
- This process completed in the following steps:
1. Absorption of light energy and excitement of chlorophyll molecule.
- Chlorophyll pigment absorbs quantum energy of electromagnetic rays of light and the chlorophyll molecule become excited.
- The energised electrons are emitted out of the excited chlorophyll molecules and are accepted by an electron acceptor.
2. Photolysis of Water and Evolution of Oxygen:
- The energy absorbed by chlorophyll is used in break down of water into its component ions.
- Break down of \({ H }_{ 2 }{ O }\) and formation of \({ { H }^{ + } }\) and \({ { OH }^{ – } }\) ions in the presence of light are called photolysis of water
- This reaction results into liberation of \({ O }_{ 2 }\). Some part of this is used in respiration and remaining \({ O }_{ 2 }\) is released in the atmosphere.
3. Formation of NADPH + \({ { H }^{ + } }\):
- Hydrogen ions liberated in the photolysis of water reduce NADP and form NADPH + \({ { H }^{ + } }\).
- This process takes place on the surface of the thylakoid membrane in contact with stroma.
- For oxidation of 2 molecules of NADP + 4\({ { H }^{ + } }\) ions are required. These 4\({ { H }^{ + } }\) ions are formed by photolysis of 2 molecules of water.
- This process also requires four electrons (\({ 4e }^{ – }\)), which are obtained from PS-I.
- These 2 molecules of NADPH + \({ { H }^{ + } }\) are used in the reduction of one molecule of \({ CO }_{ 2 }\).
4. Cyclic photophosphorylation:
- In cyclic photophosphorylation electrons ejected out from \({ P }_{ 700 }\) return back to \({ P }_{ 700 }\) and during their transport through various electron acceptors bring about ATP synthesis. Hence it is called cyclic photophosphorylation.
- In this process, only PS-I is involved.
- Electrons ejected out from reaction centre \({ P }_{ 700 }\) travel through different electron acceptors such as
A(FeS) → Fd → Cyt. \({ b }_{ 6 }\) → Cyt. f → Plastocyanin and are finally recycled to \({ P }_{ 700 }\). - During this transport sufficient energy is released when electrons jump from Cyt. \({ b }_{ 6 }\) to Cyt. f. and the energy is used in the synthesis of ATP.
Non-Cyclic Photophosphorylation:
- In this process electrons ejected out from \({ P }_{ 680 }\) do not return back to Pan and travel to different electron acceptors and during this flow brings about ATP synthesis. Hence it is called as non-cyclic photophosphorylation.
- In this process, both PS-I and PS-II are involved.
- Electrons ejected out from reaction centre \({ P }_{ 680 }\) flow through a series of electron acceptors such as Pheophytin
→ Plastoquinone (PQ) → Cyt. \({ b }_{ 6 }\) → Cyt. f → Plastocyanin and finally lost to \({ P }_{ 700 }\). - From \({ P }_{ 700 }\) on electrons flow to ferredoxin where they bring about a reduction of NADP to NADPH + \({ { H }^{ + } }\).
- During this flow sufficient energy is released when electrons jump from plastoquinone to cytochrome \({ b }_{ 6 }\), and the energy is used in the synthesis of ATP.
Note: The outcome of non-cycle photophosphorylation is the formation of ATP, generation of NADPH + \({ { H }^{ + } }\) and release of oxygen (\({ O }_{ 2 }\)). Non-cyclic photophosphorylation is also called the Z scheme. A comparative account of cyclic and non-cyclic photophosphorylation is given in the following table:
- The net result of the reactions of PS-I and PS-II is the liberation of \({ O }_{ 2 }\) by photolysis of water, synthesis of ATP molecules and generation of NADPH + \({ { H }^{ + } }\).
- The ATP and NADPH + \({ { H }^{ + } }\) are used during the dark reaction in the reduction of \({ CO }_{ 2 }\) and formation of carbohydrate.
Question 2.
Explain \({ CO }_{ 2 }\) fixation by Calvin-Benson cycle.
Answer:
In the Calvin-Benson cycle, the first stable product formed during \({ CO }_{ 2 }\) fixation is a three-carbon comoundPhosphoglyceric acid (PGA). Hence it is also called a \({ C }_{ 3 }\) cycle. Formation of carbohydrate from \({ CO }_{ 2 }\) and \({ H }_{ 2 }{ O }\) in the process of photosynthesis is completed by the following steps:
- Phosphorylation of Ribulose monophosphate
- Carboxylation of Ribulose Biphosphate
- Reductive phase-Reduction of PGA to PGAL
- Formation of Hexose sugar
- Regeneration of Ribulose monophosphate
- The path of carbon i.e. fixation of \({ CO }_{ 2 }\) in carbohydrate was studied by Calvin-Benson and co-workers (1946-1953).
- This scientist used \({ C }^{ 14 }\) carbon-containing \({ CO }_{ 2 }\) in the experiment carried out on unicellular alga Chlorella and Scenedesmos.
- In this experiment, it was observed that during \({ CO }_{ 2 }\) fixation the first stable compound formed is a three-carbon (3-C) containing compound Phosphoglyceric Acid. (PGA). Hence this process is called the \({ C }_{ 3 }\) cycle.
- Calvin and Benson were awarded Noble Prize in 1961 for this work.
Note: The process of \({ CO }_{ 2 }\) fixation in the form of carbohydrate is completed in the following steps:
1. Phosphorylation of Ribulose monophosphate:
- During the dark reaction, a five-carbon compound-Ribulose biphosphate acts as \({ CO }_{ 2 }\) acceptor.
- This compound is formed by phosphorylation of Ribulose monophosphate in the presence of phosphorylation and ATP.
- For combining with 6 molecules of \({ CO }_{ 2 }\), 6 molecules of Ribulose biphosphate are formed as per the equation is given below.
2. Carboxylation of 1-5-Biphosphate:
- RuBP acts as an acceptor of \({ CO }_{ 2 }\).
- 6 molecules of RuBP react with 6 molecules of \({ CO }_{ 2 }\) and 6 molecules of water in the presence of carboxylase enzyme.
- This reaction results in the formation of 12 molecules of phosphoglyceric acid (PGA). Because PGA is a three-carbon compound and is the first stable product of the process, hence dark reaction is also called the \({ C }_{ 3 }\) cycle.
- This reaction is also described as carboxylate phase of dark reaction.
3. Reduction of 1-3 diphosphate glyceric acid:
- 12 molecules of 1-3 diphosphate glyceric acid are reduced to 12 molecules of 3 phosphoglyceraldehydes in the presence of enzyme triose phosphate dehydrogenase.
- The reduction of PGA into PGAL is brought about by the reducing power (NADPH + \({ { H }^{ + } }\)) generated during light reaction.
Note: Out of 12 molecules of 3-phosphoglyceraldehyde so formed, only 2 molecules react to form one molecule of hexose sugar. Remaining 10 molecules enter in a series of complex biochemical reactions and regenerate 6 molecules of Ribulose monophosphate, which again enter into Calvin’s cycle.
4. Formation of Hexose Sugar (Synthesis phase):
2 molecules of 3 phosphoglyceraldehydes formed in the reductive phase react to form one molecule of Glucose by following reactions:
The above reactions are also called Glycolytic Reversion because these take place in the opposite direction of Glycolysis.
5. Regenerative Phase or Regeneration of Ribulose 5-Monophosphate:
- We know that 6 molecules of \({ CO }_{ 2 }\) and 6 molecules of Ribulose biphosphate react to form 12 molecules of phosphoglyceric acid at the beginning of the Calvin-Benson cycle.
- We also know that 6 molecules of Ribulose biphosphate are formed by phosphorylation of 6 molecules of Ribulose 5-monophosphate.
- Hence for continuity of the cycle, regeneration of 6 molecules of Ribulose 5-monophosphate is necessary at the end of the cycle.
- This regeneration of 6 molecules of Ribulose 5-monophosphate takes place through the following reactions involving 10 molecules of PGAL formed during the reductive phase.
(1) Formation of dihydroxyacetone phosphate: 4 molecules of 3-PGAL give rise to 4 molecules of DHAP in the presence of triosephosphate isomerase.
(2) Formation of Fructose 1-6 diphosphate: 2 molecules each of PGAL and DHAP react in the presence of Aldolase and form 2 molecules of Fructose 1-6 diphosphate.
(3) Formation of Fructose 6-Phosphate: Fructose 1-6 diphosphate reacts with \({ H }_{ 2 }{ O }\) in the presence of enzyme phosphatase and forms fructose 6-phosphate.
(4) Formation of Xylulose and Erythrose sugars: 2 molecules of Fructose 6-phosphate react with 2 molecules of PGAL in the presence of enzyme transketolase and produce 2 molecules of Xylulose-5 phosphate along with 2 molecules of Erythrose-4 phosphate.
(5) Formation of Sedoheptulose 1-7 diphosphate: 2 molecules of erythrose-4 phosphate react with 2 molecules of DHAP in the presence of transaldolase to form 2 molecules of sedoheptulose 1-7 diphosphate. Sedoheptulose 1-7, diphosphate formed in this reaction combines with water in presence of phosphatase and converts in sedoheptulose 7-phosphate and \({ H }_{ 3 }{ PO }_{ 4 }\).
(6) Formation of Ribose-5 phosphate and xylulose 5-phosphate:
(7) An overview of the above reactions [(i)-(vi)] shows that 10 molecules of PGAL are involved in these and the outcome of the reactions is the formation of 4 molecules of Xylulose 5-phosphate and 2 molecules of Ribose-5-Phosphate. These molecules finally react to form 6 molecules of Ribulose monophosphate as shown below:
Thus the reactions of regenerative phase result in the formation of 6 molecules of Ribulose-5-phosphate which change to 6 molecules of Ribulose 1-5 biphosphate (RuBP) by the following reaction.
Question 3.
What do you understand by photophosphorylation? Explain the process in detail.
Answer:
Formation of ATP molecule from ADP and inorganic phosphate (Pi) in the presence of sunlight is called photophosphorylation. This process takes place in the grana region of chloroplast and solar energy is stored in the form of chemical energy in ATP molecule.
- The formation of ATP from ADP in chloroplast by use of light energy is called photophosphorylation.
- This was discovered by Arnon et. al. (1954).
Photophosphorylation is of two types:
- Cyclic photophosphorylation
- non-cyclic photophosphorylation
1. Cyclic photophosphorylation:
- In cyclic photophosphorylation electrons ejected out from \({ P }_{ 700 }\) return back to \({ P }_{ 700 }\) and during their transport through various electron acceptors bring about ATP synthesis. Hence it is called cyclic photophosphorylation.
- In this process, only PS-I is involved.
- Electrons ejected out from reaction centre \({ P }_{ 700 }\) travel through different electron acceptors such as
A(FeS) → Fd → Cyt. \({ b }_{ 6 }\) → Cyt. f → Plastocyanin and are finally recycled to \({ P }_{ 700 }\). - During this transport sufficient energy is released when electrons jump from Cyt. \({ b }_{ 6 }\) to Cyt. f. and the energy is used in the synthesis of ATP.
2. Non-Cyclic Photophosphorylation:
- In this process electrons ejected out from \({ P }_{ 680 }\) do not return back to Pan and travel to different electron acceptors and during this flow bring about ATP synthesis. Hence it is called as non-cyclic photophosphorylation.
- In this process, both PS-I and PS-II are involved.
- Electrons ejected out from reaction centre P680 flow through a series of electron acceptors such as Pheophytin.
→ Plastoquinone (PQ) → Cyt. \({ b }_{ 6 }\) → Cyt. f → Plastocyanin and finally lost to \({ P }_{ 700 }\). - From \({ P }_{ 700 }\) on electrons flow to ferredoxin where they bring about the reduction of NADP to NADPH + \({ { H }^{ + } }\).
- During this flow sufficient energy is released when electrons jump from plastoquinone to cytochrome \({ b }_{ 6 }\), and the energy is used in the synthesis of ATP.
Note: The outcome of non-cycle photophosphorylation is the formation of ATP, generation of NADPH + \({ { H }^{ + } }\) and release of oxygen (\({ O }_{ 2 }\)). Non-cyclic photophosphorylation is also called the Z scheme. A comparative account of cyclic and non-cyclic photophosphorylation is given in the following table:
- The net result of the reactions of PS-I and PS-II is the liberation of \({ O }_{ 2 }\) by photolysis of water, synthesis of ATP molecules and generation of NADPH + \({ { H }^{ + } }\).
- The ATP and NADPH + \({ { H }^{ + } }\) are used during the dark reaction in the reduction of \({ CO }_{ 2 }\) and formation of carbohydrate.
Question 4.
Describe the Hatch-Slack cycle and write its significance.
Answer:
Plants in which \({ CO }_{ 2 }\) fixation occurs through a four-carbon compound are called \({ C }_{ 4 }\) plants. In these plants, the first stable compound of dark reaction is a four-carbon compound-oxaloacetic acid. Hatch and Slack (1966) worked out the complete path of carbon in these plants (\({ C }_{ 4 }\) plants). Hence this cycle is called HatchSlack cycle. This cycle is found in many monocots such as Maize, Sugarcane, Pearl millet, Grasses etc. and also some dicots such as Amaranthus, Euphorbia etc.
- Kaprilav (1960) while working on maize and Kortschak et.al (1965) working on sugarcane plant observed that the first stable carbon compound of \({ CO }_{ 2 }\) fixation in photosynthesis is a four-carbon compound-oxaloacetic acid. (OAA).
- M.D. Hatch and C.R. Slack(1966) confirmed the finding of Kaprilav and Kortschak et.al. and worked out the complete path of the carbon cycle in these plants. Hence this cycle is called the Hatch-Slack cycle.
- This cycle has been found to occur in monocot plants (Maize, Sugarcane, Pearl millet etc.) and also in some dicot. plants (Amaranthus, Euphorbia etc.).
- As the first stable carbon compound is not a \({ C }_{ 3 }\) compound but a four carbon-containing compounds-oxaloacetic acid, it is called a \({ C }_{ 4 }\) Cycle or dicarboxylic acid cycle.
(1) The structural peculiarity of \({ C }_{ 4 }\) plants:
- The \({ C }_{ 4 }\) plants have an interesting type of leaf anatomy.
- In these plants, two types of cells participate in photosynthesis.
- The mesophyll cells
- bundle sheath cells.
- The mesophyll cells have small-sized chloroplasts having chloroplasts well-developed grana whereas the bundle sheath cells have large-sized chloroplasts without grana.
- Hence in Caplants light reaction take place in the mesophyll cells and the dark reaction is completed in bundle sheath parenchyma.
- The bundle sheath parenchyma is arranged around the vascular bundle of leaves in one or two layers or rings in the form of Wreath.
- In German language, Wreath is called Kranz and hence leaf anatomy of this type has been called as Kranz anatomy.
(2) Mechanism of \({ C }_{ 4 }\) Cycle:
- Atmospheric \({ CO }_{ 2 }\) enters the leaf through stomata and is absorbed by the mesophyll cells of \({ C }_{ 4 }\) plants.
- Here a three-carbon compound Phosphoenolpyruvate (PEP) acts as an acceptor of \({ CO }_{ 2 }\) Phosphoenolpyruvate reacts with \({ CO }_{ 2 }\) in the presence of carboxylase enzyme and forms a four-carbon compound-Oxalic Acetic Acid (OAA).
- Oxaio acetic acid is reduced by NADPH + \({ { H }^{ + } }\) and is converted into Malic acid.
- From mesophyll cells, malic acid is passed to bundle sheath parenchyma
- In bundle sheath parenchyma, decarboxylation of malic acid takes place and \({ CO }_{ 2 }\) is released and pyruvic acid is formed.
- Pyruvic acid passes back to mesophyll cells, where it is converted into phosphoenolpyruvate by the use of ATP molecules, and thus the cycle continues.
- \({ CO }_{ 2 }\) released in bundle sheath parenchyma enters into \({ C }_{ 3 }\) cycle and forms hexose molecule.
- Thus fixation of atmospheric \({ CO }_{ 2 }\) takes place in the mesophyll cells through \({ C }_{ 4 }\) cycle and conversion of \({ CO }_{ 2 }\) into sugar takes place in bundle sheath parenchyma by \({ C }_{ 3 }\) cycle.
(3) Importance of \({ C }_{ 4 }\) plants and \({ C }_{ 4 }\) cycle:
- \({ C }_{ 4 }\) plants are capable of photosynthesis in relatively low concentration of \({ CO }_{ 2 }\) and hence biologically important.
- Photorespiration does not occur in \({ C }_{ 4 }\) plants and so the productivity of \({ C }_{ 4 }\) plants is more than \({ C }_{ 3 }\) plants.
- PEP carboxylase enzyme is active even at low concentration of \({ CO }_{ 2 }\).
- \({ C }_{ 4 }\) plants can successfully survive at a place having a shortage of water and high temperature (30°C – 45°C) and hence are more successful in tropics.
Significance of the Hatch-Slack Cycle:
- \({ C }_{ 4 }\) plants are capable of photosynthesis in relatively low concentration of \({ CO }_{ 2 }\).
- The productivity of \({ C }_{ 4 }\) plants is high because photorespiration does not occur in \({ C }_{ 4 }\) plants.
- \({ C }_{ 4 }\) plants can survive easily at places having a shortage of water and under high temperature (30°C – 45°C) conditions.
Question 5.
Write notes on the following:
- Photosynthetic pigments
- Photosystem I and Photosystem-II
- Photolysis of water
- Importance of the \({ C }_{ 4 }\) cycle.
- Photo-respiration and Photosynthesis
Answer:
1. Photosynthetic pigments are as follows:
- Chlorophyll
- Carotenoids
- Phycobilins
Chlorophyll ‘a’ is universal pigment and is the main pigment. A special type of chlorophyll ‘a’ acts as a reaction centre. Carotenoids and phycobilins act as accessory pigments.
2. Photosystem I and Photosystem II:
Photosystem I:
- Reaction centre is \({ P }_{ 700 }\).
- Participates in both cyclic and non-cyclic photophosphorylation.
- It is situated in the thylakoid membrane found in grana as well as stroma region.
Photosystem II:
- Reaction centre is \({ P }_{ 680 }\).
- Participates only in non-cyclic photophosphorylation.
- It is located in the thylakoid membrane found in the grana region only.
3. Photolysis of Water:
The break down of \({ H }_{ 2 }{ O }\) into its component ions by chlorophyll molecule using solar energy is called photolysis of water. This results in the evolution of \({ O }_{ 2 }\).
4. Importance of the \({ C }_{ 4 }\) cycle:
- Plants showing \({ C }_{ 4 }\) cycle can successfully photosynthesize in relatively low concentration of \({ CO }_{ 2 }\) because in \({ C }_{ 4 }\) plants Phospho enol pyruvate carboxylase enzyme is found which is active even at low concentration of \({ CO }_{ 2 }\).
- \({ C }_{ 4 }\) cycle enables plants to grow successfully at relatively dry places and having a high atmospheric temperature.
5. Photo-respiration and Photosynthesis:
- Photorespiration: Respiration taking place in addition to normal respiration in photosynthetic parts of the plant in the presence of sunlight is called photorespiration.
- Photosynthesis: Synthesis of carbohydrates using \({ CO }_{ 2 }\) and \({ H }_{ 2 }{ O }\) as raw material and sunlight as a source of energy by chlorophyllous parts of plants is called photosynthesis.
Question 6.
Discuss in detail the factors affecting the process of photosynthesis.
Answer:
The process of photosynthesis is affected by several factors. These can be divided into two categories:
(1) External Factors or Atmospheric factors:
- Light
- Temperature
- \({ CO }_{ 2 }\)
- \({ H }_{ 2 }{ O }\)
- \({ O }_{ 2 }\)
(2) Internal Factors:
- Chlorophyll
- Amount of stored food
- Structure of leaf
(1) External Factors:
1. Light:
Light is the most important factor because sunlight is a source of energy for the synthesis of carbohydrates. Light may affect the process of photosynthesis in three forms.
(a) Type of Sunlight:
The only visible spectrum of sunlight with wavelength 400nm – 780nm is used in photosynthesis. Rate of photosynthesis is highest when both blue and red part of light are supplied together to plant as compared to when these lights are supplied separately. Chlorophyll absorbs more energy from these two parts of the visible spectrum.
(b) Light intensity:
- The amount of light absorbed by plants depends upon the form of the leaves and their arrangement (phyllotaxy).
- On average, only 3.0% of the total light is absorbed by chlorophyll pigment and remaining is reflected or refracted.
- An increase in the intensity of the light may result in an increase in photosynthesis if temperature and \({ CO }_{ 2 }\) is not a limiting factor.
- At a certain stage, a further increase in light intensity fails to increase the rate of photosynthesis because some other factor may become limiting factors.
- Increase in light intensity beyond a certain limit may damage chloroplast and other cell organelles by photooxidation and photosynthetic apparatus are destroyed.
- This is called Solarization.
Note: At a certain light intensity, the amount of \({ CO }_{ 2 }\) used in photosynthesis and the amount of \({ CO }_{ 2 }\) produced in respiration become equal. This point is known as the Compensation point.
(c) Duration of Light:
- It has been observed that plants may carry out photosynthesis when continuous light is supplied for a relatively long period of time.
- Normally, 10 to 12 hours duration of light availability is required to help plants to carry out sufficient photosynthesis.
2. Temperature:
Photosynthesis takes place in a wide range of temperature. Some gymnosperms such as Juniperus may perform photosynthesis at -35°C and in some xerophytes photosynthesis may take place at 55°C. Algae found in hot waters may perform photosynthesis at a high temperature of 75°C. In the majority of plants rate of photosynthesis gradually increases between 10°C – 35°C. At high temperature, photosynthetic enzymes begin to get denatured. Binding of RuBISCO with \({ CO }_{ 2 }\) decreases resulting in to decrease in the rate of photosynthesis.
3. Carbon dioxide (\({ CO }_{ 2 }\)):
The amount of carbon dioxide in the atmosphere usually is 0.03% (300 ppm). As the amount of \({ CO }_{ 2 }\) increases, the rate of photosynthesis keeps on increasing up to some time until some other factor becomes limiting factors. Normally the rate of photosynthesis increases by an increase in \({ CO }_{ 2 }\) concentration up to 1.0%, but an increase in \({ CO }_{ 2 }\) concentration beyond this has a toxic effect on plants. Gaseous exchange stops due to the closing of stomatal pore.
4. Water:
Water is an important reactant in the process of photosynthesis. It acts as a hydrogen donor in this process. Only 1.0% of total water absorbed by plants is used in photosynthesis. Hence normally water does not become a limiting factor in photosynthesis.
It may affect photosynthesis indirectly. Greater shortage of soil water may indirectly affect photosynthesis because the shortage of water may cause the closure of stomata. As a result, gaseous exchange stops, the water potential of the leaf becomes low and enzymatic activity may be adversely affected.
5. Oxygen:
Increase in the concentration of \({ O }_{ 2 }\) affects the rate of photosynthesis. Oxygen acts as a competitive inhibitor for enzyme RuBISCO. In \({ C }_{ 3 }\) plants RuBISCO begins to act as an oxygenase enzyme in place of carboxylase under the increased concentration of \({ O }_{ 2 }\). At this stage, photorespiration begins causing a reduction in \({ CO }_{ 2 }\) fixation.
Internal Factors:
Rate of photosynthesis is affected by several internal factors i.e. plant-related factors. Some of these are as follows:
1. Chlorophyll:
Chlorophyll is the primary pigment related to photosynthesis and converts light energy into chemical energy. Normally the rate of photosynthesis increases with an increase in the amount of chlorophyll if other factors remain optimum or do not become a limiting factor.
2. Amount of stored food:
The end product of photosynthesis is stored in plant cells. The continuous accumulation of the end products in the cells decreases the rate of photosynthesis. The translocation of end products to other parts increases the rate again.
3. The internal structure of leaf:
Rate of photosynthesis also depends on the number of stomata per unit area, their structure and distribution on leaves. A higher number of stomata and a longer period of their opening promotes absorption of \({ CO }_{ 2 }\) and hence the rate of photosynthesis increases.
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