Previous year questions: Photosynthesis in Higher Plants

These ten questions come from the NCERT Exemplar chapter on photosynthesis in higher plants, the only official previous year source available for Class 11 Biology. The set moves from pigments and the electron path to the Calvin cycle, C4 anatomy and photorespiration. Marks follow the usual weightage for the question type in the Exemplar: one mark for a multiple choice, two marks for a very short answer, three marks for a short answer and five marks for a long answer. The chapter reference in each heading points to the item in the Exemplar.

Q1 (NCERT Exemplar, 1 mark; Chapter 13, Multiple Choice 1)

Which metal ion is a constituent of chlorophyll?

Answer: The porphyrin ring of chlorophyll holds a magnesium ion at its centre, so the correct choice is magnesium. Iron is part of the cytochrome carriers, copper is not the chlorophyll metal and zinc is a cofactor elsewhere. The magnesium point also explains why a magnesium deficiency makes older leaves yellow: without magnesium the chlorophyll molecule cannot be completed, and the older leaves lose their green colour first because the plant moves the scarce nutrient to younger leaves. The other choices are worth remembering as well. Iron is needed for the cytochrome and ferredoxin carriers, so iron deficiency also causes chlorosis, but iron is not the central metal of chlorophyll. Copper and zinc serve as cofactors in other reactions, which makes them plausible distractors in this favourite question.

Q2 (NCERT Exemplar, 1 mark; Chapter 13, Multiple Choice 2)

Which pigment acts directly to convert light energy into chemical energy?

Answer: Chlorophyll a is the primary pigment at the reaction centre and the only pigment that directly loses an electron to the electron transport chain. Chlorophyll b, xanthophylls and carotenoids are accessory pigments that pass absorbed energy to chlorophyll a, so the answer is chlorophyll a.

Q3 (NCERT Exemplar, 1 mark; Chapter 13, Multiple Choice 3)

Which range of wavelengths in nanometres is called photosynthetically active radiation?

Answer: Photosynthetically active radiation covers the visible range from about 400 to 700 nanometres, which is the light that the photosynthetic pigments can use. The shorter answers miss the visible span, and 100 to 390 is ultraviolet while 760 and above is infrared.

Q4 (NCERT Exemplar, 1 mark; Chapter 13, Multiple Choice 11)

The correct sequence of electron flow in the light reaction is which of these?

Answer: The electrons move from PS II through plastoquinone, the cytochromes and plastocyanin to PS I, and then to ferredoxin and NADP+ reductase. The choice that starts at PS II and ends with ferredoxin is correct. Any option that starts from PS I for the non-cyclic route or that places ferredoxin before PS I is wrong, and this pair of mistakes is what the question is testing.

Q5 (NCERT Exemplar, 2 marks; Chapter 13, Very Short Answer 6)

The reaction 3CO2 + 9ATP + 6NADPH plus water gives glyceraldehyde 3-phosphate plus 9ADP plus 6NADP plus 8Pi is analysed. How many ATP and NADPH are needed to fix one CO2, and where does the process occur?

Answer: Dividing each coefficient by three gives three ATP and two NADPH for one molecule of CO2, and the glyceraldehyde 3-phosphate produced is the three-carbon sugar phosphate of the Calvin cycle. The whole process runs in the stroma of the chloroplast, because the cycle enzymes, including RuBisCO, sit there. The reaction also shows the full accounting of the cycle: three ATP and two NADPH are spent for each CO2, one ATP of the three goes to the regeneration of RuBP, and the remaining two ATP with the two NADPH drive the reduction of 3-phosphoglycerate. Knowing this accounting makes the numerical questions on the Calvin cycle much easier, because every figure can be checked by multiplying or dividing these base figures.

Q6 (NCERT Exemplar, 2 marks; Chapter 13, Very Short Answer 9)

Where is NADP reductase located in the chloroplast, and what is its role in the proton gradient?

Answer: The enzyme sits on the outer, stroma-facing surface of the thylakoid membrane. Ferredoxin delivers electrons to it, and with protons taken from the stroma it converts NADP+ into NADPH. By taking protons out of the stroma, it helps keep the stroma less acidic than the lumen, and this difference is part of the gradient that drives ATP synthesis.

Q7 (NCERT Exemplar, 3 marks; Chapter 13, Short Answer 2)

Chlorophyll a is the primary pigment for the light reaction. What are accessory pigments, and what is their role?

Answer: Accessory pigments are chlorophyll b, the xanthophylls and the carotenoids, which cannot carry out the primary conversion themselves. They absorb wavelengths that chlorophyll a absorbs poorly, especially in the blue-green range, and transfer that energy to chlorophyll a. They also protect the reaction centre by dissipating excess light energy, which reduces photo-damage on very bright days.

Q8 (NCERT Exemplar, 3 marks; Chapter 13, Short Answer 8)

What conditions make RuBisCO act as an oxygenase, and what follows?

Answer: RuBisCO acts as an oxygenase when the oxygen concentration at its active site is high relative to carbon dioxide, which happens after stomata close on a hot, dry day when internal CO2 falls and oxygen from the light reaction accumulates. RuBP is then split into one molecule of 3-phosphoglycerate and one of phosphoglycolate, and the phosphoglycolate is recovered through the photorespiratory route, which consumes ATP and releases CO2 without making sugar.

Q9 (NCERT Exemplar, 3 marks; Chapter 13, Short Answer 12)

Six turns of the Calvin cycle are required to generate one mole of glucose. Explain.

Answer: Each turn fixes one CO2, so one turn adds one carbon atom to the cycle products. A glucose molecule has six carbon atoms, so six turns of carboxylation are needed to gather the six carbons. Six turns also mean 18 ATP and 12 NADPH are consumed, because each CO2 fixed costs three ATP and two NADPH. A useful check on the arithmetic: three turns give one molecule of glyceraldehyde 3-phosphate, which has three carbons, and two molecules of that sugar phosphate are needed for one glucose, so six turns and eighteen ATP follow directly. The cycle never produces glucose as a single direct step, and the stable sugar is drawn off only after the triose phosphate pool is large enough.

Q10 (NCERT Exemplar, 5 marks; Chapter 13, Long Answer 13)

Why does photorespiration not take place in C4 plants?

Answer: In C4 plants the first fixation of CO2 happens in the mesophyll cells and uses PEP carboxylase, an enzyme with a high affinity for CO2 and no affinity for oxygen. The four-carbon acid formed is transferred to the bundle sheath cells, which are wrapped tightly by thick walls and have almost no intercellular spaces. Inside the sheath the acid is decarboxylated, so the CO2 concentration around RuBisCO becomes much higher than the oxygen concentration. This high CO2 to oxygen ratio forces RuBisCO to work as a carboxylase, and the oxygenase reaction that starts photorespiration is suppressed. The physical separation of the two steps, with PEP carboxylase in the mesophyll and RuBisCO in the sheath, is the key difference from C3 plants, where both reactions happen in the same cell and compete directly. The C4 plant pays extra ATP to regenerate PEP, and in return it avoids the large carbon loss that photorespiration causes in C3 plants on hot, bright days. The anatomical arrangement and the enzyme separation work as a package, and an answer that names only Kranz anatomy without linking it to the CO2 concentration around RuBisCO misses the main point. The bundle sheath cells are also positioned so that the CO2 released from the four-carbon acid is trapped close to the Calvin cycle instead of leaking out through the leaf, and the thin mesophyll layer outside keeps the diffusion path short. A diagram earns marks in this question: sketch two concentric rings, label the inner ring as bundle sheath with RuBisCO and the outer ring as mesophyll with PEP carboxylase, and draw arrows for malate going in and pyruvate coming back. The same diagram covers the linked question on Kranz anatomy, so practise it once and reuse it for both in the final revision.