Energy & Control: A Level Biology questions and answers
10 free A Level Biology Energy & Control questions written in the style of Cambridge Biology papers, each with its answer and mark scheme. 4 of them are A* challenge questions, built to catch the mistakes strong students still make.
This topic covers Energy & respiration, Photosynthesis, Homeostasis, Control & coordination. Work through the questions below, then generate a full Energy & Control paper with a fresh set every time.
Germinating seeds in a respirometer take in 30 cm3 of oxygen and give out 25 cm3 of carbon dioxide in one hour.
(a)
Calculate the respiratory quotient and suggest the substrate being respired.
[2]
(b)
Describe how ATP is produced by oxidative phosphorylation.
[5]
(c)
Explain why the ATP yield from anaerobic respiration is much lower than from aerobic respiration.
[3]
Show answers and mark scheme
Part
Answer and marking points
Marks
(a)
0.83; protein
CO2 ÷ O2M1 0.83 with correct substrate A1
2
(b)
NADH/FADH2 deliver electrons to electron transport chain on inner membrane; energy used to pump H+ into intermembrane space; proton gradient; H+ diffuse back through ATP synthase (chemiosmosis); oxygen final electron acceptor forming water
reduced NAD/FAD donate electrons B1 electron carriers in inner membrane B1 protons pumped into intermembrane space B1 chemiosmosis through ATP synthase B1 oxygen as final acceptor B1
5
(c)
only glycolysis produces ATP (2 per glucose); no Krebs cycle or oxidative phosphorylation; NAD regenerated by producing lactate/ethanol
only glycolysis occurs B1 no electron transport chain/oxidative phosphorylation B1 NAD regenerated by lactate/ethanol pathway B1
3
Question 2Structured · 4 marksA* challenge
The Calvin cycle uses products of the light-dependent stage.
(a)
Describe the role of rubisco.
[2]
(b)
Explain why the concentration of RuBP increases when CO₂ concentration falls.
[2]
Show answers and mark scheme
Part
Answer and marking points
Marks
(a)
fixes CO₂ with RuBP to form GP
rubisco catalyses the combination of CO₂ with RuBP B1 forming two molecules of GP B1
2
(b)
RuBP not used but still regenerated
less RuBP combines with CO₂ B1 but RuBP is still regenerated from TP B1
2
Question 3Structured · 4 marksA* challenge
Anaerobic respiration in yeast and mammals produces different products.
(a)
Explain why anaerobic respiration produces much less ATP than aerobic respiration.
[2]
(b)
Explain why reduced NAD must be reoxidised in anaerobic conditions.
[2]
Show answers and mark scheme
Part
Answer and marking points
Marks
(a)
only glycolysis; no ETC
only glycolysis occurs (2 ATP) B1 Krebs cycle and oxidative phosphorylation cannot occur without oxygen B1
2
(b)
NAD needed for glycolysis to continue
NAD is needed to accept hydrogen in glycolysis B1 so glycolysis can continue B1
2
Question 4Structured · 3 marksA* challenge
Plants respond to their environment using hormones.
Explain the role of abscisic acid in the closing of stomata.
[3]
Show answers and mark scheme
Part
Answer and marking points
Marks
ABA binds guard cells; K⁺ leave; water leaves; flaccid
ABA binds receptors on guard cells B1 K⁺ ions leave, raising water potential B1 water leaves by osmosis and guard cells become flaccid, closing the stoma B1
3
Question 5Structured · 11 marks
This question is about glycolysis and the Krebs cycle.
(a)
State where glycolysis occurs and name the three-carbon compound produced at the end of it.
[2]
(b)
Outline the main stages of glycolysis.
[4]
(c)
Describe what happens to the acetyl group in the Krebs cycle.
[3]
(d)
The theoretical maximum yield is about 38 ATP per glucose, but the actual yield is about 30–32. Explain why it is less.
[2]
Show answers and mark scheme
Part
Answer and marking points
Marks
(a)
cytoplasm; pyruvate
cytoplasm B1 pyruvate B1
2
(b)
glucose phosphorylated using 2 ATP; splits into 2 triose phosphate; TP oxidised, NAD reduced; 4 ATP made, net 2
glucose is phosphorylated using two ATP (hexose bisphosphate) B1 it splits into two triose phosphates B1 each triose phosphate is oxidised (dehydrogenated), reducing NAD B1 four ATP are made by substrate-linked phosphorylation, net gain of two B1
4
(c)
combines with oxaloacetate (4C) to form citrate (6C); decarboxylation and dehydrogenation; NAD, FAD reduced, ATP made
acetyl (2C) combines with oxaloacetate (4C) to form citrate (6C) B1 citrate is decarboxylated and dehydrogenated back to oxaloacetate B1 producing reduced NAD and FAD, CO₂ and one ATP per turn B1
3
(d)
some H⁺ leak; energy lost as heat; ATP used to transport pyruvate/ADP
some protons leak across the inner membrane B1 energy is used to transport pyruvate, ADP and phosphate into the mitochondrion / lost as heat B1
2
Question 6Structured · 13 marks
The glomerular filtration rate of a healthy adult is 120 cm³ per minute. The person produces 1.2 dm³ of urine per day.
(a)
Calculate the volume of filtrate produced per day, in dm³.
[2]
(b)
Use your answer to (a) to calculate the percentage of the filtrate that is reabsorbed.
[2]
(c)
Explain how the structure of the glomerulus and Bowman’s capsule produces ultrafiltration.
[3]
(d)
Describe how glucose is reabsorbed in the proximal convoluted tubule.
[3]
(e)
Explain how ADH increases the reabsorption of water in the collecting duct.
[3]
Show answers and mark scheme
Part
Answer and marking points
Marks
(a)
173 dm³
120 × 60 × 24 = 172800 cm³ C1 173 dm³ A1
2
(b)
99.3 %
(173 − 1.2) ÷ 173 × 100 C1 99.3 % A1
2
(c)
afferent wider than efferent → high pressure; fenestrated endothelium; basement membrane filter; podocytes
Any 3 of: afferent arteriole wider than efferent, giving high hydrostatic pressure; endothelium has pores (fenestrations); basement membrane acts as the filter, stopping large proteins; podocytes have gaps between their processes B3
3
(d)
Na⁺ pumped out of cells into blood; Na⁺ and glucose co-transported in; glucose diffuses into blood
Any 3 of: Na⁺ actively pumped out of the PCT cells into the blood; lowers Na⁺ concentration in the cells; Na⁺ and glucose enter together by co-transport (facilitated); glucose then diffuses into the blood B3
3
(e)
ADH binds receptors; aquaporins inserted; more water by osmosis
ADH binds to receptors on collecting duct cells B1 vesicles containing aquaporins fuse with the membrane B1 more water moves out by osmosis into the medulla / blood B1
3
Question 7Structured · 8 marks
This question is about the control of stomatal aperture and water balance in plants.
(a)
Describe the structure of guard cells and how they open the stoma.
[3]
(b)
Explain why stomata usually open in light and close in darkness.
[2]
(c)
Explain how abscisic acid causes stomatal closure during drought.
[3]
Show answers and mark scheme
Part
Answer and marking points
Marks
(a)
thick inner walls, cellulose microfibrils; K⁺ pumped in, water potential falls; water enters, turgid, bend apart
guard cells have thicker inner walls and radially arranged cellulose microfibrils B1 H⁺ pumped out, K⁺ enter, lowering water potential B1 water enters by osmosis and the cells become turgid and bow apart B1
3
(b)
CO₂ for photosynthesis in light; reduce water loss at night
open in light to allow CO₂ in for photosynthesis B1 close at night when photosynthesis cannot occur, reducing water loss B1
2
(c)
ABA receptors; Ca²⁺ second messenger; K⁺ out, water out, flaccid
ABA binds receptors on guard cell membranes B1 Ca²⁺ enters, acting as a second messenger B1 K⁺ leaves, water potential rises, water leaves and guard cells become flaccid, closing the stoma B1
3
Question 8Structured · 5 marks
The light-dependent reactions occur in the thylakoids.
(a)
Describe the role of photolysis of water.
[2]
(b)
Explain how ATP is produced by photophosphorylation.
[3]
Show answers and mark scheme
Part
Answer and marking points
Marks
(a)
provides electrons, H⁺; O₂ by-product
Any 2 of: replaces electrons lost from photosystem II; provides H⁺ (for reduced NADP); oxygen is released B2
2
(b)
electron transport; proton gradient; ATP synthase
electrons pass along the electron transport chain B1 energy pumps H⁺ into the thylakoid space B1 H⁺ diffuse through ATP synthase producing ATP B1
3
Question 9Structured · 8 marks
This question is about cell signalling.
(a)
Describe the stages of cell signalling by a hormone such as glucagon.
[4]
(b)
Explain why only certain cells respond to a particular hormone.
[2]
(c)
Explain the advantage of an enzyme cascade.
[2]
Show answers and mark scheme
Part
Answer and marking points
Marks
(a)
hormone binds specific receptor; G protein activated; adenylyl cyclase makes cAMP; kinase cascade amplifies; response
Any 4 of: glucagon binds to a specific receptor on the target cell membrane; this activates a G protein; adenylyl cyclase converts ATP to cyclic AMP (second messenger); cAMP activates protein kinases in a cascade that amplifies the signal; leading to the response (glycogen breakdown) B4
4
(b)
only target cells have complementary receptors
only target cells have receptors with a shape complementary to the hormone B2
2
(c)
amplification; one hormone molecule activates many enzymes
amplification: each activated enzyme activates many others, so one hormone molecule produces a large response quickly B2
2
Question 10Structured · 8 marks
A student uses a simple respirometer containing germinating peas and soda lime. The movement of coloured liquid in the capillary tube is recorded.
time / min
0
5
10
15
20
distance moved / cm
0.00
0.20
0.42
0.60
0.81
Table 10.1
(a)
Explain why the liquid moves towards the peas.
[2]
(b)
The capillary has a radius of 0.50 mm. Calculate the rate of oxygen uptake in mm³ min−1.
[3]
(c)
Describe a suitable control and explain its purpose.
[2]
(d)
Explain why the apparatus is kept in a water bath.
[1]
Show answers and mark scheme
Part
Answer and marking points
Marks
(a)
O₂ used; CO₂ absorbed by soda lime; volume/pressure decreases
the peas take in oxygen for aerobic respiration B1 the carbon dioxide released is absorbed by soda lime, so the gas volume (pressure) inside decreases B1
2
(b)
0.318 mm³ min⁻¹
distance per minute = 0.405 mm M1 volume = πr2l = π × 0.502 × 0.405 M1 0.318 mm3 min−1A1
3
(c)
same apparatus with boiled peas / glass beads of same volume
identical tube with an equal volume of glass beads (or boiled peas) B1 shows changes in temperature or pressure not due to respiration B1
2
(d)
temperature affects gas volume and enzyme activity
temperature changes alter gas volume and the rate of respiration B1
In the 9700 syllabus, Energy & Control covers Energy & respiration, Photosynthesis, Homeostasis, Control & coordination.
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Original Cambridge-style practice questions written for mentorriyaz.in. These are not Cambridge past paper questions, and this site is not affiliated with or endorsed by Cambridge University Press & Assessment. Updated 2026-11-04. Paper generator · mentorriyaz.in