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The Krebs cycle and oxidative phosphorylation questions

One turn of the Krebs cycle and what happens to every carbon; the products per turn and per glucose; the electron transport chain on the inner membrane; chemiosmosis, with protons pumped into the intermembrane space and returning through ATP synthase; oxygen as the final electron acceptor forming water; and why the textbook total of 38 ATP is a ceiling rather than a measurement.

6 original questions · 23 marks · the krebs cycle and oxidative phosphorylation notes · Respiration and cellular energy

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• 5 marks · Explain

  Explain how the reduced NAD produced by the earlier stages of respiration leads to the synthesis of ATP on the inner mitochondrial membrane.

  Mark scheme

  1. B1 reduced NAD is oxidised at the first carrier of the chain, handing over its electrons and releasing its protons into solution
  2. B1 the electrons pass from carrier to carrier along the inner membrane, releasing energy at each transfer because each carrier holds them a little less tightly than the one before
  3. B1 that energy is used to pump protons out of the matrix into the intermembrane space, and since the membrane is impermeable to protons an electrochemical gradient builds across it
  4. B1 protons return to the matrix down that gradient through the channel in ATP synthase, and the flow drives the enzyme to join ADP and inorganic phosphate into ATP, which is chemiosmosis
  5. B1 oxygen acts as the final electron acceptor at the end of the chain, taking up electrons and combining with protons to form water, so electrons do not pile up and the chain keeps running

• 4 marks · Describe

  Describe one turn of the Krebs cycle, and account for every carbon atom that enters it.

  Mark scheme

  1. B1 the two-carbon acetyl group is transferred from coenzyme A onto the four-carbon acceptor oxaloacetate, giving the six-carbon compound citrate, and coenzyme A leaves at once to collect another acetyl group
  2. B1 citrate is decarboxylated and dehydrogenated to a five-carbon compound, and again to a four-carbon compound, so two molecules of carbon dioxide are released per turn
  3. B1 a series of further changes regenerates oxaloacetate, so the acceptor ends the turn exactly as it began it and the cycle can turn again
  4. B1 six carbons are present at the start of the turn, two leave as carbon dioxide, and the four handed back are the four the cycle borrowed, so the two carbons released are the two that arrived on the acetyl group

• 4 marks · Calculate

  One molecule of glucose respired aerobically yields 10 reduced NAD, 2 reduced FAD and 4 ATP made directly. Calculate the theoretical ATP yield per glucose using the textbook ratios of 3 ATP per reduced NAD and 2 ATP per reduced FAD, and calculate it again using the better ratios of 2.5 and 1.5.

  Mark scheme

  1. M1 multiply each coenzyme by its ratio and add the ATP made directly by substrate-level phosphorylation
  2. M1 10 × 3 = 30 from reduced NAD and 2 × 2 = 4 from reduced FAD, plus the 4 made directly
  3. A1 a theoretical total of 30 + 4 + 4 = 38 ATP per molecule of glucose
  4. A1 with the better ratios, 10 × 2.5 = 25 and 2 × 1.5 = 3, giving 25 + 3 + 4 = 32 ATP per molecule of glucose

• 4 marks · Explain

  A working cell is measured to yield around 30 molecules of ATP per molecule of glucose, rather than the 38 quoted in textbooks. Explain why the measured yield falls short of the theoretical one.

• 4 marks · Suggest

  A drug blocks the proton channel through ATP synthase in the inner mitochondrial membrane but does not affect the electron carriers themselves. Suggest what happens to the proton gradient, to ATP production, to the rate of electron transport and to the rate at which the tissue consumes oxygen.

• 2 marks · State

  State the products of one complete turn of the Krebs cycle, and state how many turns take place per molecule of glucose.

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