Glycolysis and anaerobic respiration - AQA A-Level Biology (Paper 2) Exam Questions

Practise 17 exam-style questions on Glycolysis and anaerobic respiration for AQA A-Level Biology (Paper 2). Write your answer as you would in the exam and get it marked in seconds against the mark scheme, point by point, with the Study Helper on hand if you get stuck.

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1–8 marks each
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17 questions, marked as you go

Sample questions with model answers

Here are 4 of the questions in this topic, each with a model answer, an explanation of why it earns the marks and the mark scheme an examiner would use.

Question 1

2 marks

In glycolysis, each molecule of triose phosphate is converted to pyruvate.

Explain why this reaction is described as an oxidation.

Apart from pyruvate, name two products of this stage of glycolysis.

Model answer

Hydrogen is removed from triose phosphate, so it is oxidised. The other products are ATP and reduced NAD.

Explanation

Oxidation is the loss of hydrogen (or electrons). In glycolysis the hydrogen removed from triose phosphate is accepted by the coenzyme NAD, which becomes reduced NAD, and ATP is produced in the same stage, giving the net gain of ATP. Students often write NADP, which is the coenzyme used in photosynthesis, not respiration.

Mark scheme

1 mark
Hydrogen (atoms) / electrons are removed (from triose phosphate)
Accept 'triose phosphate loses hydrogen' or 'NAD accepts hydrogen from triose phosphate'. Reject 'oxygen is added'.
1 mark
ATP AND reduced NAD
Both needed. Accept NADH / NADH₂ / NADH + H⁺ for reduced NAD. List rule: if more than two products are given, a wrong extra product (e.g. reduced NADP, ADP, reduced FAD, carbon dioxide) cancels the mark. Ignore water.

Question 2

3 marks

Crucian carp are fish that live in small ponds. In winter the ponds can be covered by ice for several months, and the water then contains no oxygen. The carp survive by respiring anaerobically. Their muscle cells convert pyruvate to ethanol instead of lactate, and the ethanol diffuses out of the blood into the water across the gills. In autumn, the carp build up very large stores of glycogen in their liver.

Explain why the carp need very large stores of glycogen to survive the winter.

Suggest one advantage to the carp of producing ethanol instead of lactate.

Model answer

Anaerobic respiration produces very little ATP from each glucose molecule, because ATP is made only in glycolysis. To produce enough ATP to stay alive for several months, the carp must respire a very large amount of glucose, which comes from the hydrolysis of glycogen.

The ethanol diffuses out across the gills, so it does not build up in the body; lactate would accumulate and lower the pH of the tissues, because without oxygen it could not be removed.

Explanation

Without oxygen, pyruvate cannot be oxidised in the link reaction and Krebs cycle, and reduced NAD cannot be oxidised by the electron transfer chain, so only the small net gain of ATP from glycolysis is available. The fish therefore use glucose much faster than they would aerobically, and need large glycogen stores. Most vertebrates make lactate, which stays in the body until oxygen is available again; producing ethanol, which can diffuse away, avoids this.

Mark scheme

1 mark
Anaerobic respiration produces little ATP per glucose / only glycolysis produces ATP
Accept 'only 2 ATP per glucose' or 'much less ATP than aerobic respiration'. Reject the idea of energy being produced (energy is released).
1 mark
So a lot of glucose (from glycogen) is needed / glucose must be used at a faster rate to produce enough ATP (for several months)
Accept 'glycogen is hydrolysed to glucose for respiration' only when linked to the need for a large amount of glucose.
1 mark
Ethanol is removed from the body (across the gills) so it does not accumulate / lactate would accumulate
Accept 'lactate would lower the pH / cause fatigue / be toxic' or 'lactate cannot be removed without oxygen'. Ignore 'ethanol is less toxic' unless linked to its removal.

Question 3

4 marks

Yeast cells in bread dough quickly use up the oxygen in the dough and then respire anaerobically, producing ethanol and carbon dioxide.

The total concentration of NAD (oxidised NAD and reduced NAD together) in the yeast cells is 1.5 μmol per gram of yeast. During anaerobic respiration, the yeast cells produce ethanol at a rate of 1.8 μmol per gram of yeast per second. One molecule of reduced NAD is oxidised for each molecule of ethanol produced.

Calculate the mean number of times each NAD molecule is reduced and then oxidised again in one minute.

Use your answer to explain why the conversion of pyruvate to ethanol is essential for the yeast cells to continue producing ATP.

Model answer

Number of times per second = 1.8 ÷ 1.5 = 1.2
Number of times per minute = 1.2 × 60 = 72

Converting pyruvate to ethanol oxidises reduced NAD, regenerating NAD. The yeast cells contain only 1.5 μmol of NAD per gram, so each NAD molecule has to be reused about 72 times a minute; without regeneration, all of it would be reduced in less than a second. NAD is needed to accept hydrogen when triose phosphate is oxidised to pyruvate, so without it glycolysis would stop and the cells would produce no more ATP.

Explanation

Divide the rate at which reduced NAD is oxidised by the amount of NAD present to find how often each molecule is recycled each second, then multiply by 60. A cell holds very little NAD, so glycolysis can continue only if reduced NAD is oxidised again as fast as it is formed. In anaerobic respiration this happens when pyruvate is converted to ethanol (or lactate); this step makes no ATP itself but keeps glycolysis going.

Mark scheme

1 mark
Correct method, e.g. 1.8 ÷ 1.5 × 60; or a correct final answer of 72 with no working
Max 4. Points 1 and 2 are the calculation: a correct answer of 72 (accept 72.0) with no working scores 2 marks. Accept for 1 mark: 1.2 (number per second, not multiplied by 60) OR 50 (inverted: 1.5 ÷ 1.8 × 60) OR 0.83 (time in seconds for one cycle) OR 144 or 36 (multiplied or divided by 2 in error).
1 mark
72
1 mark
Converting pyruvate to ethanol oxidises reduced NAD / regenerates (oxidised) NAD
Points 3 to 5: max 2. Reject NADP and FAD.
1 mark
The cells contain only a small amount of NAD, so each molecule must be reused many times (72 times a minute) / without regeneration all the NAD would be reduced in less than a second
The use of the calculated value, or of the time (about 0.8 s), is needed for this mark. Allow ECF from an incorrect answer to the calculation.
1 mark
(Oxidised) NAD is needed to accept hydrogen when triose phosphate is oxidised (to pyruvate), so without it glycolysis / ATP production would stop
Accept 'NAD is needed for glycolysis to continue'. Reject the idea that converting pyruvate to ethanol itself produces ATP.

Question 4

6 marks

A student wrote this description of anaerobic respiration in a muscle cell.

'Glycolysis turns glucose into pyruvate. When there is no oxygen, pyruvate is turned into lactate, and this reaction makes the extra ATP that the muscle needs.'

Explain what is wrong with the student's description. Then describe how glycolysis and the conversion of pyruvate to lactate allow a muscle cell to continue producing ATP without oxygen.

Model answer

The student is wrong to say that converting pyruvate to lactate produces ATP: all the ATP is produced in glycolysis.

In glycolysis, glucose is phosphorylated using ATP to form glucose phosphate, which is split into two molecules of triose phosphate. Each triose phosphate is oxidised to pyruvate: hydrogen is removed and accepted by NAD, which is reduced, and ATP is produced. More ATP is produced than is used, so there is a net gain of ATP.

Without oxygen, the reduced NAD cannot be oxidised by the electron transfer chain. Instead, pyruvate is reduced to lactate by reduced NAD, which oxidises the reduced NAD and regenerates NAD. The NAD can then accept hydrogen from more triose phosphate, so glycolysis continues and the cell keeps producing a small amount of ATP.

Explanation

The key idea is that lactate production is not a source of ATP but a way of regenerating NAD, which glycolysis needs to keep oxidising triose phosphate. Describe glycolysis in AQA's terms: phosphorylation of glucose using ATP, triose phosphate, oxidation to pyruvate, net gain of ATP and reduced NAD. Saying that triose phosphate is 'converted' to pyruvate, rather than oxidised, does not gain the mark.

Mark scheme

1 mark
No ATP is produced when pyruvate is converted to lactate / all the ATP is produced in glycolysis
Max 5 if the answer does not state what is wrong with the student's description. Max 6. Accept 'the lactate step does not make ATP'.
1 mark
Glucose is phosphorylated (to glucose phosphate) using ATP
Reject 'substrate-level phosphorylation' or 'oxidative phosphorylation' for this step.
1 mark
(Glucose phosphate is split into) triose phosphate
Accept 'two three-carbon molecules / 2 TP'.
1 mark
Triose phosphate is oxidised to pyruvate / hydrogen is removed from triose phosphate
'Converted' or 'broken down' to pyruvate is insufficient.
1 mark
NAD is reduced (when triose phosphate is oxidised)
Accept 'reduced NAD is produced'. Reject NADP.
1 mark
Net gain of ATP / more ATP is produced than is used (in glycolysis)
Accept '4 ATP produced and 2 used' or 'net 2 ATP'.
1 mark
Pyruvate is reduced to lactate by reduced NAD / reduced NAD is oxidised (when lactate is made)
Accept lactic acid for lactate. Reject NADP.
1 mark
NAD is regenerated / reused, so triose phosphate can still be oxidised / glycolysis can continue (and keep producing ATP)
Reject FAD.

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