Questions on the Krebs cycle and oxidative phosphorylation

Question 1

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

Accepted Answer

B1 reduced NAD is oxidised at the first carrier of the chain, handing over its electrons and releasing its protons into solution B1 the electrons pass from carrier to carrier along the inner membrane, releasing energy at each transfer because each carrier holds them a little more tightly than the one before 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 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 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

Question 2

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

Accepted Answer

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 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 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 B1 six carbons are present at the start of the turn, two leave as carbon dioxide, and four are handed back, as many as the cycle borrowed, so the carbon count balances: two in on the acetyl group, two out as carbon dioxide

Question 3

One molecule of glucose respired aerobically yields 10 reduced NAD, 2 reduced FAD and a net 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.

Accepted Answer

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

Question 4

Taking 2.5 ATP for each reduced NAD and 1.5 for each reduced FAD, the yield from one glucose molecule is 32 ATP. This counts the 2 reduced NAD made in glycolysis at 2.5 ATP each. They are made in the cytoplasm and cannot cross the inner mitochondrial membrane, and in some cells a shuttle system passes their electrons into the mitochondrion by way of FAD instead, so each then yields 1.5 ATP. Calculate the yield per glucose in such a cell, and suggest two reasons why the yield in a working cell may be lower still.

Accepted Answer

M1 each of the two reduced NAD from glycolysis yields 1.5 instead of 2.5 ATP, so the total falls by 2 × 1 = 2 A1 32 − 2 = 30 ATP per glucose B1 the inner membrane leaks protons, so some return to the matrix without passing through ATP synthase and their energy is released as heat B1 the proton gradient also drives the transport of other substances, such as pyruvate, into the matrix, so not all of it is used to make ATP

Question 5

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.

Accepted Answer

B1 protons can no longer return to the matrix through ATP synthase, so they accumulate in the intermembrane space and the gradient across the membrane becomes steeper B1 ATP production by oxidative phosphorylation stops, because there is no flow of protons through the enzyme to drive it B1 pumping further protons against an ever steeper gradient needs more energy than the transfers can supply, so electron transport slows and then stops B1 with few electrons reaching the end of the chain, little oxygen is needed as the final electron acceptor, so oxygen consumption falls even though the carriers themselves are untouched

Question 6

One molecule of glucose is respired aerobically. For glycolysis, the link reaction and the Krebs cycle combined, before oxidative phosphorylation, calculate the net gain in ATP made by substrate-level phosphorylation and the total numbers of molecules of reduced NAD and of reduced FAD produced.

Accepted Answer

M1 glycolysis gives a net 2 ATP and 2 reduced NAD; the link reaction gives 0 ATP and 2 reduced NAD, since it runs twice per glucose; the Krebs cycle gives 2 ATP, 6 reduced NAD and 2 reduced FAD, since one turn gives 1 ATP, 3 reduced NAD and 1 reduced FAD and it turns twice A1 net gain in ATP made directly = 2 + 0 + 2 = 4 M1 sum the reduced NAD from all three stages: 2 + 2 + 6 A1 total reduced NAD = 10, and total reduced FAD = 0 + 0 + 2 = 2

Question 7

In brown adipose tissue, an uncoupling protein in the inner mitochondrial membrane allows protons to leak back into the matrix without passing through ATP synthase. Suggest the effect this has on the rate of electron transport, on the amount of ATP made, and on the temperature of the tissue.

Accepted Answer

B1 protons leaking back through the uncoupling protein continually reduce the proton gradient, so the rate of electron transport speeds up rather than slowing, because pumping into a gradient that keeps leaking away meets less resistance B1 ATP production falls, because much of the proton gradient is dissipated through the uncoupling protein rather than driving ATP synthase B1 the energy that would otherwise have been captured as ATP is instead released as heat, since the proton gradient represents stored energy that has to go somewhere when it is not used to make ATP B1 the tissue's temperature rises as a result, which is the physiological function of brown adipose tissue: generating heat rather than ATP, useful in newborn mammals and in hibernators

Question 8

A cell has a supply of acetyl coenzyme A that has not come from glucose. Calculate the number of molecules of carbon dioxide and of reduced NAD produced when 5 molecules of acetyl coenzyme A pass through the Krebs cycle, one turn per molecule.

Accepted Answer

M1 one turn of the cycle, driven by one acetyl group, gives 2 CO₂ and 3 reduced NAD (as well as 1 reduced FAD and 1 ATP) M1 multiply each figure by 5, since each of the 5 acetyl CoA molecules drives one turn of the cycle A1 CO₂ = 5 × 2 = 10 molecules A1 reduced NAD = 5 × 3 = 15 molecules

Question 9

A mutation in mitochondrial DNA produces a faulty version of one of the electron carrier proteins partway along the chain, so it can still accept electrons from the carrier before it but cannot pass them on to the next carrier. Suggest the effect this would have on ATP production in cells carrying this mutation, and suggest why muscle and nerve cells are often especially badly affected by mitochondrial DNA mutations of this kind.

Accepted Answer

B1 electron transport would stop at the faulty carrier: it stays reduced, so the carriers before it cannot pass their electrons on either and the carriers after it receive none, so proton pumping along the chain stops and the proton gradient falls B1 with a smaller proton gradient, less ATP could be made by ATP synthase, so oxidative phosphorylation in affected mitochondria would be greatly reduced B1 the cell would be limited mostly to the small ATP yield of glycolysis, far short of normal aerobic respiration's output B1 muscle and nerve cells have very high, continuous ATP demands, for contraction and for maintaining electrical excitability respectively, and contain large numbers of mitochondria to meet that demand, so they are especially reliant on oxidative phosphorylation working normally

Question 10

A revision guide states: 'the Krebs cycle is the stage that makes most of a cell's ATP.' Evaluate this statement.

Accepted Answer

B1 for the statement: the Krebs cycle is essential to the great majority of ATP eventually made during aerobic respiration, because it produces most of the reduced NAD and reduced FAD, 6 reduced NAD and 2 reduced FAD per glucose, that oxidative phosphorylation later uses to make most of the cell's ATP B1 against the statement: the Krebs cycle itself makes only a small amount of ATP directly, 1 per turn by substrate-level phosphorylation, 2 per glucose, similar to the net 2 ATP made directly by glycolysis and far less than the roughly 30 ATP eventually made at the inner membrane B1 the stage that actually makes most of the ATP directly is oxidative phosphorylation at the inner mitochondrial membrane, not the Krebs cycle itself B1 a judgement: the statement is misleading as worded, since it credits the Krebs cycle with ATP that is actually made at a later, separate stage; a more accurate statement is that the Krebs cycle is essential for the ATP later made by oxidative phosphorylation, rather than that it makes that ATP itself

Question 11

Explain why a cell deprived of oxygen cannot make ATP by oxidative phosphorylation within a very short time, even though the mitochondria and all the electron carriers are still present and undamaged.

Accepted Answer

B1 without oxygen there is no final electron acceptor at the end of the chain, so the last carrier cannot pass on its electrons and stays reduced B1 with the last carrier unable to accept further electrons, the carrier before it cannot pass its own electrons on either, and this blockage extends back along the whole chain within seconds, so no carrier can pass on electrons and no protons are pumped B1 with pumping stopped, protons continue to leak back and the existing gradient collapses, so ATP synthase has no gradient left to drive it and oxidative phosphorylation stops

Question 12

Explain why it is essential that oxaloacetate is regenerated at the end of each turn of the Krebs cycle.

Accepted Answer

B1 oxaloacetate is the four-carbon molecule that accepts each new acetyl group arriving from the link reaction, so without regenerating it there would be nothing for the next acetyl group to combine with B1 the cell's total supply of Krebs cycle intermediates is small, so oxaloacetate has to be reused rather than a fresh molecule being available for every turn B1 if oxaloacetate were not regenerated, the cycle would stop after a single turn once its starting supply ran out, and most acetyl groups arriving from the link reaction, and therefore most of the useful energy in the original glucose, would not be able to enter the cycle for further oxidation at all

Question 13

Compare substrate-level phosphorylation with oxidative phosphorylation as ways of making ATP during aerobic respiration.

Accepted Answer

B1 substrate-level phosphorylation is the direct transfer of a phosphate group from a substrate molecule onto ADP, and happens in glycolysis and in the Krebs cycle, whereas oxidative phosphorylation makes ATP using energy from a proton gradient built by the electron transport chain, at the inner mitochondrial membrane B1 substrate-level phosphorylation does not require oxygen and makes only a small amount of ATP directly, 4 from glycolysis and 2 from the Krebs cycle per glucose, whereas oxidative phosphorylation requires oxygen and makes the great majority of the ATP produced during aerobic respiration B1 substrate-level phosphorylation happens in a single enzyme-catalysed step each time, whereas oxidative phosphorylation depends on a whole series of electron carriers building a gradient before ATP synthase can use it, a much more indirect route to the same product

Question 14

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

Accepted Answer

B1 one turn gives two molecules of carbon dioxide and one molecule of ATP B1 one turn also gives three reduced NAD and one reduced FAD, and the cycle turns twice per molecule of glucose

Question 15

State one similarity and one difference between FAD and NAD as coenzymes in respiration.

Accepted Answer

B1 similarity: both are coenzymes that accept hydrogen, and therefore electrons, removed from a substrate during respiration, becoming reduced B1 difference: reduced FAD hands its electrons to the transport chain at a later point than reduced NAD does, so it yields less ATP per molecule