AQA A-level Biology: Homeostasis and Control of Blood Glucose (7402) – Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – examination boards hold copyright in their own papers. Use these alongside the official past papers from your board or school.

These questions follow AQA A-level Biology (7402), sections 3.6.4.1 and 3.6.4.2 of the AQA AS and A-level Biology specification for first teaching 2015, including Required practical 11. Both sections are A-level only. Explanations are in the study guide and the revision notes. Course pages: the AQA A-level Biology hub and the printable checklist. Where data are given, use only those data. The marks shown are for practice and are not an official tariff.

Questions

1. State what is meant by homeostasis and explain why separate mechanisms for raising and lowering a variable give greater control than a single mechanism. [4]

2. Explain why a sustained change in core temperature, and a fall in blood pH, each reduce the rate of enzyme-controlled reactions in the body. [4]

3. Explain why it is important that blood glucose concentration is neither too high nor too low. [3]

4. A person has not eaten for 14 hours. Name the process by which the liver releases glucose from its store, the process by which it makes glucose from non-carbohydrate sources, and the hormone that stimulates both. [3]

5. Explain how insulin lowers blood glucose concentration. [4]

6. Describe how the binding of adrenaline to a liver cell leads to an increase in the release of glucose, including the second messenger model. [5]

7. Healthy volunteers drank a glucose solution. Mean values were recorded.

Time / min	0	30	60	90	120
Blood glucose / mmol dm⁻³	4.6	8.2	6.4	5.1	4.7
Insulin / arbitrary units	8	62	45	20	10

(a) Calculate the percentage increase in blood glucose in the first 30 minutes. [2] (b) Calculate the mean rate of decrease in blood glucose between 30 and 90 minutes, in mmol dm⁻³ min⁻¹, to 2 significant figures. [2] (c) Explain the changes in blood glucose and insulin between 30 and 120 minutes. [2]

8. Two patients were tested after fasting. Patient X had a blood glucose of 11.5 mmol dm⁻³ and an insulin level of 2 arbitrary units. Patient Y had a blood glucose of 9.8 mmol dm⁻³ and an insulin level of 38 arbitrary units. A healthy fasting person has about 8 arbitrary units of insulin and 4.6 mmol dm⁻³ of glucose.

(a) Suggest which type of diabetes each patient most probably has, and give the reasoning. [4] (b) State how patient Y could reduce the symptoms without taking insulin. [1]

9. A student makes a dilution series from a glucose stock solution of 125 mmol dm⁻³ to calibrate a colorimeter, using Benedict’s reagent.

(a) Calculate the volume of stock and of distilled water needed to make 20 cm³ of the 75 mmol dm⁻³ solution, and the volume of stock needed for the 25 mmol dm⁻³ solution. [2] (b) The absorbance readings, using a red filter, were:

Glucose / mmol dm⁻³	0	25	50	75	100	125
Absorbance	0.80	0.69	0.57	0.47	0.35	0.24

The line of best fit is absorbance = 0.80 − 0.00448 × concentration. A urine sample gave an absorbance of 0.53. Calculate its glucose concentration, to the nearest whole number. [2] (c) A second sample was diluted 1 in 2 with distilled water before testing and gave an absorbance of 0.41. Calculate the glucose concentration in the original sample, to 3 significant figures. [2] (d) Explain why absorbance falls as glucose concentration rises. [1] (e) Suggest one variable the student should control. [1]

10. Evaluate the positions of health advisers and the food industry on the increased incidence of type II diabetes. [6]

11. Oxytocin and uterine contractions were monitored during labour.

Time / h	0	1	2	3	4
Oxytocin / arbitrary units	2	3	6	13	30
Contractions per 10 min	1	2	3	5	7

(a) Calculate how many times greater the oxytocin level was at 4 h than at 0 h. [1] (b) Explain why these data show positive feedback. [2] (c) Explain why this feedback is not an example of homeostasis. [1]

12. Isolated liver cells were incubated for 10 minutes. Glucose released was measured in micromoles per million cells.

Treatment	Glucose released
None (control)	0.4
Adrenaline	5.2
Adrenaline and an inhibitor of adenylate cyclase	0.5
A cell-permeable form of cAMP	5.0
cAMP form and the inhibitor	4.9

(a) Calculate how many times greater glucose release was with adrenaline than in the control. [1] (b) Explain what the results with adrenaline and the inhibitor show. [2] (c) Explain why the inhibitor had almost no effect when the cAMP form was used. [2] (d) Explain how cAMP leads to glucose release. [3]

Answers

1. Homeostasis is the maintenance of a stable internal environment within restricted limits [1] by physiological control systems [1]. With two mechanisms, a departure in either direction can be corrected [1], so an overshoot can itself be reversed and the level is held more closely [1]. [4] Examiner insight: A definition earns credit when it includes “restricted limits” or an equivalent, and the second half needs a reason that mentions both directions.

2. Temperature: above the optimum, hydrogen bonds break and the tertiary structure and active site change, so substrates no longer fit and the enzyme denatures [2]. Below it, there is less kinetic energy, so fewer collisions and fewer enzyme-substrate complexes [1]. A pH change alters the charges on R groups, disrupting ionic and hydrogen bonds and the shape of the active site [1]. [4] Examiner insight: Naming the type of bond and the active site makes an answer specific; “the enzyme stops working” alone gains little.

3. High glucose makes the blood water potential more negative [1], so cells lose water by osmosis [1]. Low glucose means too little respiratory substrate for ATP production [1]. [3] Examiner insight: Both sides of the question are needed; many answers describe only high glucose.

4. Glycogenolysis [1]; gluconeogenesis [1]; glucagon [1]. [3] Examiner insight: Spell the process names accurately, because similar words (glycogenesis, glycogenolysis) mean opposite things.

5. Insulin binds to receptors on the surface of target cells [1]. This leads to more glucose channel proteins in the surface membrane [1], so more glucose enters by facilitated diffusion [1]. Insulin also activates enzymes that convert glucose to glycogen [1]. [4] Examiner insight: State that insulin binds at the cell surface; an answer that says insulin enters the cell contradicts the specification.

6. Adrenaline (first messenger) binds to a receptor on the cell surface membrane [1]. This activates adenylate cyclase [1], which converts ATP to cyclic AMP (cAMP) [1]. cAMP (second messenger) activates protein kinase [1], which activates enzymes that convert glycogen to glucose, so glucose is released [1]. [5] Examiner insight: Give the steps in sequence and name each enzyme or molecule; answers gain credit by naming each link in the chain.

7. (a) Increase = 8.2 − 4.6 = 3.6 [1]. 3.6 ÷ 4.6 × 100 = 78% [1]. (b) (8.2 − 5.1) ÷ 60 [1] = 0.052 mmol dm⁻³ min⁻¹ [1]. (c) Insulin is high at 30 and 60 minutes, so cells take up glucose and the liver converts it to glycogen, lowering blood glucose [1]. As glucose approaches the starting value, insulin secretion falls (negative feedback) [1]. [6] Examiner insight: Show the subtraction and division, give the unit, and in (c) refer to the figures in the table rather than only restating the trend.

8. (a) Patient X: low insulin (below the healthy 8) with high glucose [1], so most probably type I, in which the immune system has destroyed beta cells [1]. Patient Y: high insulin but high glucose [1], so most probably type II, where target cells have lost responsiveness to insulin [1]. (b) Manipulate the diet (control carbohydrate and energy intake), with exercise [1]. [5] Examiner insight: Use both the insulin and the glucose values; reasoning from glucose alone cannot separate the types.

9. (a) Stock = 75 × 20 ÷ 125 = 12.0 cm³, with 8.0 cm³ water [1]. For 25 mmol dm⁻³, stock = 25 × 20 ÷ 125 = 4.0 cm³ [1]. (b) Concentration = (0.53 − 0.80) ÷ (−0.00448) [1] = 60.27, so 60 mmol dm⁻³ to the nearest whole number [1]. (c) (0.41 − 0.80) ÷ (−0.00448) = 87.05 [1]. Multiply by 2 to give 174 mmol dm⁻³ [1]. (d) Glucose reduces the blue copper(II) ions in the Benedict’s reagent, so less blue colour remains to absorb red light [1]. (e) Any one: the volume of Benedict’s reagent, the heating time, the temperature of the water bath, or the same filter and cuvette [1]. [8] Examiner insight: Rounding the intermediate value early can shift the final answer, so carry the unrounded value and round once at the end; remember the dilution factor.

10. Health advisers link diets high in sugar, fat and total energy, and low activity, to obesity and type II diabetes [1]. They advise reduced sugar intake, clear food labelling and more exercise [1]. Their evidence is largely from population studies, which show correlation and not necessarily cause [1]. The food industry stresses personal choice and a balanced diet [1] and the industry has a commercial interest in sales, so its evidence may be biased [1]. A balanced judgement weighs the size and independence of the studies and concludes that evidence supports reducing sugar and energy intake, while accepting that other factors such as inactivity and genetics also contribute [1]. [6] Examiner insight: To score on “evaluate”, reach a supported judgement; two lists of opinions with no weighing of the evidence usually gain little.

11. (a) 30 ÷ 2 = 15 times [1]. (b) The oxytocin level rises faster and faster [1], and the contractions increase with it, so the change causes more of the same change [1]. (c) Positive feedback moves the system away from its starting value; it does not keep the internal environment stable [1]. [4] Examiner insight: Quote values from the table in (b); an answer that only says “it gets bigger” does not show how the data were used.

12. (a) 5.2 ÷ 0.4 = 13 times [1]. (b) Glucose release was almost the same as the control [1], so adrenaline needs adenylate cyclase to produce its effect [1]. (c) The cAMP form enters the cell and acts after the step blocked by the inhibitor [1], so adenylate cyclase is not needed [1]. (d) cAMP activates protein kinase [1], which activates enzymes [1] that convert glycogen to glucose [1]. [8] Examiner insight: In (b) and (c), link the data to the position of the block in the pathway and compare with the control, not only the highest value.

Where marks are usually lost

• Mixing up glycogenesis, glycogenolysis and gluconeogenesis.
• Writing that insulin enters cells, or that the pancreas stores glucose.
• Describing the second messenger steps without adenylate cyclase or without ATP to cAMP.
• Saying glucagon and adrenaline “break down” glucose.
• Using “sugar” instead of glucose, or “blood sugar levels” without a reference to concentration.
• Forgetting the dilution factor when reading a calibration curve.
• Rounding too early in multi-step calculations.
• Giving an evaluation that lists views and does not reach a judgement.

Next steps

• Read the revision notes and the study guide.
• Return to the AQA A-level Biology hub and the printable checklist.
• Try all the free 10-minute diagnostics.
• Book a free trial class.

Official syllabus

AQA AS and A-level Biology specification for first teaching 2015, published by AQA (sections 3.6.4.1 and 3.6.4.2, with Required practical 11).