2025 AQA A LEVEL BIOLOGY PAPER 2
ULTIMATE BUNDLE 200 PRACTICE
QUESTIONS WITH DETAILED
RATIONALES ,GENETICS,
ECOSYSTEMS, CONTROL & GENE
EXPRESSION | A STUDY GUIDE
Section 1: Energy Transfers (Photosynthesis & Respiration) (Questions
1–10)
1. Which molecule is the final electron acceptor in the non-cyclic
photophosphorylation pathway of photosynthesis?
A. Oxygen
B. NADP⁺
C. Cytochrome complex
D. Plastocyanin
Rationale: In non-cyclic photophosphorylation, electrons from
Photosystem I are passed to NADP⁺ (along with H⁺ ions) to form
reduced NADP (NADPH).
Answer:B
2. During the Calvin cycle, how many molecules of Triose Phosphate
(TP) are required to regenerate 5 molecules of Ribulose Bisphosphate
(RuBP)?
A. 3
B. 5
C. 6
D. 10
,***Rationale: It takes 6 TP to make 1 glucose (2 exit the cycle). The
remaining 10 TP molecules are required to regenerate 6 RuBP molecules.
Wait, let's re-calculate stoichiometry. 6 CO₂ + 6 RuBP → 12 TP. 2 TP
leave. 10 TP remain. 10 TP are rearranged to form 6 RuBP. So, to
regenerate 6 RuBP, you need 10 TP. To regenerate 5 RuBP? The ratio is
10 TP : 6 RuBP = 5:3. So for 5 RuBP, you need (5 * 10/6) = 8.33? No, the
question asks about the standard cycle turn. Let's rephrase for clarity
based on standard AQA knowledge.
Correction: Standard question: How many TP are needed to regenerate 3
RuBP? Answer:5 TP.
Let's stick to the standard stoichiometry: 6 Turns produce 12 TP. 2 used
for hexose, 10 used to regenerate 6 RuBP.
Ratio TP:RuBP = 10:6 = 5:3.
If the question asks "How many TP to regenerate 3 RuBP?", answer is 5.
Let's adjust the question to a standard integer value.
Revised Q2: In one turn of the Calvin cycle producing one hexose sugar,
how many molecules of Triose Phosphate are used specifically to
regenerate Ribulose Bisphosphate?
A. 2
B. 6
C. 10
D. 12
Rationale: For every 6 CO₂ fixed, 12 TP are made. 2 TP are used to make
hexose sugar. The remaining 10 TP are used to regenerate 6 RuBP.
Answer:C
3. Which of the following correctly describes the role of oxygen in aerobic
respiration?
A. It acts as a catalyst in the Krebs cycle.
B. It combines with carbon to form CO₂.
C. It is the final electron acceptor in the electron transport chain,
forming water.
D. It is required for glycolysis to occur.
,Rationale: Oxygen accepts low-energy electrons and protons at the end
of the oxidative phosphorylation chain to form water. Without it, the
chain stops.
Answer:C
4. Where does the link reaction take place in a eukaryotic cell?
A. Cytoplasm
B. Outer mitochondrial membrane
C. Mitochondrial matrix
D. Inner mitochondrial membrane (cristae)
Rationale: The link reaction (conversion of pyruvate to acetyl CoA)
occurs in the mitochondrial matrix.
Answer:C
5. What is the net gain of ATP molecules per molecule of glucose during
anaerobic respiration in yeast?
A. 2
B. 4
C. 32
D. 38
Rationale: Anaerobic respiration only utilizes glycolysis, which produces
a net gain of 2 ATP. The subsequent fermentation steps regenerate NAD⁺
but produce no additional ATP.
Answer:A
6. Which pigment is found at the reaction center of Photosystem II?
A. Chlorophyll b
B. Carotene
C. P680
D. P700
Rationale: P680 is the specific chlorophyll a molecule at the reaction
center of PSII that absorbs light best at 680nm. P700 is for PSI.
Answer:C
, 7. In the electron transport chain of respiration, energy released by
electrons is used to:
A. Synthesize glucose directly.
B. Pump protons (H⁺) from the matrix into the intermembrane space.
C. Reduce NAD⁺ to NADH.
D. Break down pyruvate.
Rationale: The energy creates an electrochemical gradient by pumping
H⁺ ions across the inner mitochondrial membrane.
Answer:B
8. Limiting factors for photosynthesis include all of the following
EXCEPT:
A. Light intensity
B. Carbon dioxide concentration
C. Oxygen concentration
D. Temperature
Rationale: While high oxygen can inhibit photosynthesis via
photorespiration, it is not typically listed as a primary limiting factor in
the same way as light, CO₂, and temperature in standard A-Level
contexts.
Answer:C
9. Which coenzyme is reduced during glycolysis and the Krebs cycle?
A. FAD only
B. NAD⁺ only
C. Both NAD⁺ and FAD
D. ATP
Rationale: NAD⁺ is reduced in glycolysis, the link reaction, and the
Krebs cycle. FAD is reduced only in the Krebs cycle. Therefore, both are
reduced across the whole process.
Answer:C
10. Chemiosmosis refers to:
ULTIMATE BUNDLE 200 PRACTICE
QUESTIONS WITH DETAILED
RATIONALES ,GENETICS,
ECOSYSTEMS, CONTROL & GENE
EXPRESSION | A STUDY GUIDE
Section 1: Energy Transfers (Photosynthesis & Respiration) (Questions
1–10)
1. Which molecule is the final electron acceptor in the non-cyclic
photophosphorylation pathway of photosynthesis?
A. Oxygen
B. NADP⁺
C. Cytochrome complex
D. Plastocyanin
Rationale: In non-cyclic photophosphorylation, electrons from
Photosystem I are passed to NADP⁺ (along with H⁺ ions) to form
reduced NADP (NADPH).
Answer:B
2. During the Calvin cycle, how many molecules of Triose Phosphate
(TP) are required to regenerate 5 molecules of Ribulose Bisphosphate
(RuBP)?
A. 3
B. 5
C. 6
D. 10
,***Rationale: It takes 6 TP to make 1 glucose (2 exit the cycle). The
remaining 10 TP molecules are required to regenerate 6 RuBP molecules.
Wait, let's re-calculate stoichiometry. 6 CO₂ + 6 RuBP → 12 TP. 2 TP
leave. 10 TP remain. 10 TP are rearranged to form 6 RuBP. So, to
regenerate 6 RuBP, you need 10 TP. To regenerate 5 RuBP? The ratio is
10 TP : 6 RuBP = 5:3. So for 5 RuBP, you need (5 * 10/6) = 8.33? No, the
question asks about the standard cycle turn. Let's rephrase for clarity
based on standard AQA knowledge.
Correction: Standard question: How many TP are needed to regenerate 3
RuBP? Answer:5 TP.
Let's stick to the standard stoichiometry: 6 Turns produce 12 TP. 2 used
for hexose, 10 used to regenerate 6 RuBP.
Ratio TP:RuBP = 10:6 = 5:3.
If the question asks "How many TP to regenerate 3 RuBP?", answer is 5.
Let's adjust the question to a standard integer value.
Revised Q2: In one turn of the Calvin cycle producing one hexose sugar,
how many molecules of Triose Phosphate are used specifically to
regenerate Ribulose Bisphosphate?
A. 2
B. 6
C. 10
D. 12
Rationale: For every 6 CO₂ fixed, 12 TP are made. 2 TP are used to make
hexose sugar. The remaining 10 TP are used to regenerate 6 RuBP.
Answer:C
3. Which of the following correctly describes the role of oxygen in aerobic
respiration?
A. It acts as a catalyst in the Krebs cycle.
B. It combines with carbon to form CO₂.
C. It is the final electron acceptor in the electron transport chain,
forming water.
D. It is required for glycolysis to occur.
,Rationale: Oxygen accepts low-energy electrons and protons at the end
of the oxidative phosphorylation chain to form water. Without it, the
chain stops.
Answer:C
4. Where does the link reaction take place in a eukaryotic cell?
A. Cytoplasm
B. Outer mitochondrial membrane
C. Mitochondrial matrix
D. Inner mitochondrial membrane (cristae)
Rationale: The link reaction (conversion of pyruvate to acetyl CoA)
occurs in the mitochondrial matrix.
Answer:C
5. What is the net gain of ATP molecules per molecule of glucose during
anaerobic respiration in yeast?
A. 2
B. 4
C. 32
D. 38
Rationale: Anaerobic respiration only utilizes glycolysis, which produces
a net gain of 2 ATP. The subsequent fermentation steps regenerate NAD⁺
but produce no additional ATP.
Answer:A
6. Which pigment is found at the reaction center of Photosystem II?
A. Chlorophyll b
B. Carotene
C. P680
D. P700
Rationale: P680 is the specific chlorophyll a molecule at the reaction
center of PSII that absorbs light best at 680nm. P700 is for PSI.
Answer:C
, 7. In the electron transport chain of respiration, energy released by
electrons is used to:
A. Synthesize glucose directly.
B. Pump protons (H⁺) from the matrix into the intermembrane space.
C. Reduce NAD⁺ to NADH.
D. Break down pyruvate.
Rationale: The energy creates an electrochemical gradient by pumping
H⁺ ions across the inner mitochondrial membrane.
Answer:B
8. Limiting factors for photosynthesis include all of the following
EXCEPT:
A. Light intensity
B. Carbon dioxide concentration
C. Oxygen concentration
D. Temperature
Rationale: While high oxygen can inhibit photosynthesis via
photorespiration, it is not typically listed as a primary limiting factor in
the same way as light, CO₂, and temperature in standard A-Level
contexts.
Answer:C
9. Which coenzyme is reduced during glycolysis and the Krebs cycle?
A. FAD only
B. NAD⁺ only
C. Both NAD⁺ and FAD
D. ATP
Rationale: NAD⁺ is reduced in glycolysis, the link reaction, and the
Krebs cycle. FAD is reduced only in the Krebs cycle. Therefore, both are
reduced across the whole process.
Answer:C
10. Chemiosmosis refers to:
