Explanation
During regeneration 12 mol. of 3- phosphoglycerate are phosphorylated by 12 mol. of ATP to form 12 mol. of 1,3 - bisphosphoglycerate in presence of enzyme phosphoglycerol kinase. The molecules of ATP get converted to ADP.
12 mol. of 1,3 - bisphosphoglycerate are reduced to 12 mol. of 3- phosphoglyceraldehyde in presence of coenzyme NADPH. The reaction is catalyzed by the enzyme triosephosphate dehydrogenase. Inorganic phosphate is released.
Plants such as sugarcane and sorghum fix atmospheric carbon dioxide with the carboxylation of phosphoenol pyruvic acid to produce oxaloacetic acid (a C$$_4$$ acid) in the presence of phosphoenolpyruvate carboxylase (PEP carboxylase). Such plants are called C$$_4$$ plants.
C$$_4$$ plants have Kranz leaf anatomy. The vascular bundles, in these leaves, are surrounded by a layer of bundle sheath cells that contain a large number of chloroplasts.
Malic acid (Product of C$$_4$$ cycle) transport from mesophyll cell to bundle sheath where C$$_3$$ cycle starts which require RuBisco enzyme.
So, the correct answer is 'All a,b, and c'.
C$$_4$$ plants have Kranz anatomy of a leaf. The vascular bundles in these leaves are surrounded by a layer of bundle sheath cells that contain a large number of chloroplasts.
The chloroplasts in these leaves are dimorphic:
Bundle sheath- Chloroplast larger in size and arranged centripetally. It has unstacked thylakoids called agranal chloroplast. The cells perform C$$_3$$ cycle.
Mesophyll cells- Normal type of chloroplast. It has stacked thylakoids called granal chloroplast. The cells perform C$$_4$$ cycle.
Hence, the correct answer is 'Agranal bundle sheath chloroplasts and granal mesophyll chloroplasts'.
A. Aerobic respiration is the oxidative breakdown of respiratory substrates with the help of atmospheric O$$_2$ $.
B. Anaerobic respiration is the oxidative breakdown of respiratory substrates in absence of O$$_2$$.
C. Fermentation -The chemical breakdown of a substance by bacteria, yeasts, or other microorganisms, typically involving effervescence and the giving off of heat.
D. Photorespiration: It is inhibition of photosynthesis by high O$$_2$$ level. It was termed as Warburg effect. However, the inhibition was observed mostly in C$$_3$$ plants. The C$$_4$$ plants were hardly affected by varying O$$_2$$ concentrations. The process occurs in chloroplast, peroxisome and mitochondria.
Hence, the correct answer is 'Photorespiration'.
Mitochondria and chloroplast both have:
A. Chlorophyll a occurs with peak absorption at 700 nm in PS I and 680 nm in PS II.
B. Mn and Cl$$^-$$ are associated with photolysis of water.
C. No O$$_2$$ is evolved during cyclic photophosphorylation
D. In non electron transport process the excited electron from PS II is accepted by primary acceptor quinone, which sends them to an electron transport system consisting of plastoquinone, cytochrome complex and plastocyanin.The excited electron absorbed by PS I reaction center, then transfer to the ferredoxin(Fe), then the reduced Ferredoxin donates the electron to NADP$$^+$$.
So, the correct answer is 'PS I and PS II are both involved in non- cyclic photophosphorylation'.
C$$_4$$ plants can absorb CO$$_2$$ even in a much low CO$$_2$$ concentration while the C$$_3$$ plants fail to avail it. C$$_4$$ cycle is much more efficient photosynthetically. Primary Carboxylation of phosphoenolpyruvate acid (PEP) to form oxaloacetic acid (a 4- carbon compound) in presence of enzyme phosphoenolpyruvate carboxylase (PEP carboxylase) is an additional step in the C$$_4$$ plants that allow an efficient CO$$_2$$ utilization. Secondary carboxylation to synthesize carbohydrates occurs in bundle sheath.
Hence, the correct answer is 'Primary fixation of CO$$_2$$ is mediated via PEP carboxylase'.
In photorespiration, the glycine is transported out of peroxisomes into mitochondria, where two molecules of glycine interact to form one molecule each of serine, CO$$_2$$, NH$$_3$$.
2 Glycine + H$$_2$$O + NAD$$^+\rightarrow$$ Serine + CO$$_2$$+ NH$$_3$$ + NADH
When glyceric acid finally enters the chloroplast where it is phosphorylated to 3 PGA which enters in C$$_3$$ cycle, where during the reduction process, NAD$$^+$$ is reduced.
Hence, the correct answer is 'Mitochondria and chloroplasts'.
The photosystem II complex replaced its lost electrons from an external source; however, the two other electrons are not returned to photosystem II as they would in the analogous cyclic pathway. Instead, the still-excited electrons are transferred to a photosystem I complex, which boosts their energy level to a higher level using a second solar photon. The highly excited electrons are transferred to the acceptor molecule, but this time is passed on to an enzyme called Ferredoxin-NADP^+ reductase which uses them to catalyse the reaction (as shown):
This consumes the H+ ions produced by the splitting of water, leading to a net production of 1/2O2, ATP, and NADPH+H+ with the consumption of solar photons and water.
So, the correct option is 'Option A'.
Photorespiration: It is inhibition of photosynthesis by high O$$_2$$ level. It was termed as Warburg effect. The process occurs in chloroplast, peroxisome and mitochondria. However, the inhibition was observed mostly in C$$_3$$ plants. The C$$_4$$ plants were hardly affected by varying O$$_2$$ concentrations.’ This is because of the spatial separation of initial carbon dioxide fixation by the phosphoenolpyruvate carboxylase in the mesophyll cells and then the proceeding of the C$$_3$$ cycle in the bundle sheath cells with the involvement of the RuBiSCo enzyme. In this way, the RuBiSCo does not get to interact with the atmospheric oxygen.
Hence, the correct answer is 'C$$_4$$ plants'.
$$6PEP + 6RuBP + 6C{O_2} + 30ATP + 12NADPH \to 6PEP + 6RuBP + {C_6}{H_{12}}{O_6} + 30ADP + 30{H_3}P{O_4} + 12NAD{P^ + }$$
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