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Cell Respiration · Paper 2 — Free Biology HL Practice Questions

1MasterySAQ-SRole of mitochondria in energy production5 marksPaper 2~8 min
The below shows a simplified overview of aerobic respiration in a eukaryotic cell. Use the following ATP yields for parts (b) and (c): glycolysis =2 ATP= 2\,\text{ATP}; link reaction =0 ATP= 0\,\text{ATP}; Krebs cycle =2 ATP= 2\,\text{ATP}; oxidative phosphorylation =28 ATP= 28\,\text{ATP}.
(a)
State the location of the Krebs cycle within the mitochondrion. [1 mark]
(b)
Calculate the percentage of the total ATP yield, from one glucose molecule, that is produced inside the mitochondrion (Krebs cycle and oxidative phosphorylation combined). Show your working. [2 marks]
(c)
Explain why the link reaction and Krebs cycle are essential for the high ATP yield of aerobic respiration, even though they produce little ATP directly. [2 marks]
diagram

Solutions

2MasterySAQ-SRole of mitochondria in energy production7 marksPaper 2~11 min
A researcher investigates aerobic cell respiration using isolated rat liver mitochondria. The mitochondria are supplied with pyruvate, ADP, and inorganic phosphate (Pi\text{P}_\text{i}). Oxygen consumption is measured over 10 minutes; 0.5 μmol0.5\,\mu\text{mol} of O2\text{O}_2 is consumed. Assume: complete oxidation of one pyruvate via the link reaction and Krebs cycle yields 4 NADH4\,\text{NADH}, 1 FADH21\,\text{FADH}_2, and 1 ATP1\,\text{ATP} (substrate-level phosphorylation). Each NADH yields 2.5 ATP2.5\,\text{ATP}; each FADH₂ yields 1.5 ATP1.5\,\text{ATP}. Each O2\text{O}_2 molecule accepts 44 electrons; each NADH or FADH₂ donates 22 electrons to the electron transport chain.
(a)
State the role of the inner mitochondrial membrane in the reaction that directly consumes oxygen. [1 mark]
(b)
Determine the number of moles of pyruvate oxidised from the oxygen consumption data. Show all working. [2 marks]
(c)
Calculate the maximum amount of ATP, in μmol\mu\text{mol}, that could be synthesised. Show all working. [2 marks]
(d)
Explain why the actual ATP yield in a living cell is lower than the value calculated in (c). [2 marks]
diagram

Solutions

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