CLEP Biology Cellular Respiration and Photosynthesis 5 — Questions and Answers
Question 1: What is the primary reason why the theoretical ATP yield from glucose oxidation is never achieved in living cells?
- Cells cannot synthesize enough ADP to accept the phosphate groups
- Proton leakage, variable P/O ratios, and energy costs of membrane transport reduce efficiency (Correct answer)
- Oxygen availability limits the rate of the ETC
- The citric acid cycle can only run half the number of turns predicted
Correct answer: Proton leakage, variable P/O ratios, and energy costs of membrane transport reduce efficiency
Real ATP yields are lower than theoretical maxima because protons leak across the inner membrane, P/O ratios vary, and ATP/ADP transport across the mitochondrial membrane consumes some proton-motive force.
Question 2: In the context of the Calvin cycle, what does the term 'regeneration of RuBP' refer to?
- The synthesis of new RuBP from CO2 and G3P using photon energy
- The ATP-dependent conversion of Calvin cycle intermediates back into ribulose-1,5-bisphosphate (Correct answer)
- The release of RuBP from Rubisco after carboxylation
- The reduction of 3-PGA to form RuBP using NADPH
Correct answer: The ATP-dependent conversion of Calvin cycle intermediates back into ribulose-1,5-bisphosphate
Regeneration of RuBP is the third stage of the Calvin cycle, where ATP is used to convert 5-carbon intermediates (ribulose-5-phosphate) back into RuBP so CO2 fixation can continue.
Question 3: Which regulatory molecule activates phosphofructokinase-1 (PFK-1), signaling that the cell needs more energy?
- High ATP levels
- High citrate levels
- AMP (adenosine monophosphate) (Correct answer)
- High NADH levels
Correct answer: AMP (adenosine monophosphate)
AMP activates PFK-1, the key regulatory enzyme of glycolysis, signaling low energy charge and stimulating glucose breakdown to generate more ATP.
Question 4: What structural feature of the inner mitochondrial membrane dramatically increases its surface area for ATP synthesis?
- Outer membrane pores (porins)
- Cristae (infoldings of the inner membrane) (Correct answer)
- The intermembrane space
- Ribosomes attached to the outer surface
Correct answer: Cristae (infoldings of the inner membrane)
Cristae are extensive infoldings of the inner mitochondrial membrane that greatly increase its surface area, accommodating more ETC complexes and ATP synthase molecules.
Question 5: What is the fate of the G3P (glyceraldehyde-3-phosphate) molecules produced in the Calvin cycle?
- All G3P exits the chloroplast to form glucose in the cytoplasm
- About 5/6 of G3P is used to regenerate RuBP, and 1/6 exits to form glucose or other organic molecules (Correct answer)
- G3P is immediately converted back to CO2 in a feedback loop
- G3P is stored within the thylakoid lumen as starch
Correct answer: About 5/6 of G3P is used to regenerate RuBP, and 1/6 exits to form glucose or other organic molecules
For every 6 G3P produced (from 3 turns of the Calvin cycle), 5 are used to regenerate RuBP and only 1 net G3P exits to be used for glucose synthesis or other biosynthesis.
Question 6: Brown adipose tissue generates heat instead of ATP during cold exposure. Which protein is responsible for this thermogenesis?
- ATP synthase (Complex V)
- Thermogenin (uncoupling protein 1, UCP1) (Correct answer)
- Cytochrome c oxidase (Complex IV)
- NADH dehydrogenase (Complex I)
Correct answer: Thermogenin (uncoupling protein 1, UCP1)
Thermogenin (UCP1) creates a proton channel that allows H+ to leak back across the inner mitochondrial membrane without going through ATP synthase, dissipating the proton gradient as heat.
Question 7: Which statement correctly describes the difference between P680 and P700 in the light reactions?
- P680 is in PSI and P700 is in PSII; both absorb the same wavelength
- P680 is the reaction center of PSII with a stronger oxidizing power; P700 is the reaction center of PSI with a weaker oxidizing power (Correct answer)
- P680 produces NADPH directly; P700 splits water to release O2
- P700 absorbs shorter wavelengths than P680 and is more easily excited
Correct answer: P680 is the reaction center of PSII with a stronger oxidizing power; P700 is the reaction center of PSI with a weaker oxidizing power
P680 in PSII has a stronger oxidizing power (more positive redox potential) allowing it to oxidize water, while P700 in PSI has a weaker oxidizing power but produces a powerful reductant to ultimately reduce NADP+.
CLEP Biology Cellular Respiration and Photosynthesis 5
What is the primary reason why the theoretical ATP yield from glucose oxidation is never achieved in living cells?