OCR A Level Biology A: Cell communication, homeostasis and the liver (H420) – 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 cover learning outcomes 5.1.1(a) to 5.1.1(d) and 5.1.2(a) to 5.1.2(b) of the OCR A Level Biology A (H420) specification, Version 4.1 (April 2026), for first assessment in 2025. Learn the content in the study guide and revision notes first. The course hub, the checklist and the free diagnostics show the wider course. Values in the data questions are illustrative.

Questions

1. Explain why a large multicellular animal needs communication systems. [3]

2. Describe how cells communicate by signalling. (a) Between adjacent cells, using a named example. [2] (b) Between distant cells, using a named example. [2]

3. State the difference between a receptor and an effector, and name one receptor and one effector involved in controlling temperature in a human. [3]

4. Distinguish between negative feedback and positive feedback, giving one example of each. [4]

5. The table shows body temperature in a lizard and a mouse as air temperature was raised from 15 °C to 35 °C over one hour.

Time (min)	0	30	60
Lizard (°C)	15.5	26.0	34.0
Mouse (°C)	37.1	37.0	37.3

(a) Calculate the mean rate of change of the lizard’s body temperature over the hour, to 2 significant figures. [2] (b) State the range of the mouse’s body temperature. [1] (c) Explain the difference between the two animals. [3]

6. In a warm room the diameter of a skin arteriole in a volunteer is 0.30 mm. After moving to a cold room it is 0.18 mm. (a) Find the percentage change in the diameter, to 2 significant figures. [2] (b) Explain how the change helps to maintain core temperature. [3]

7. A runner loses 0.45 kg of sweat in one hour and all of it evaporates. Assume that evaporation of 1.0 g of sweat removes 2.4 kJ of heat. (a) Calculate the heat removed in the hour, in kJ. [2] (b) Calculate the mean rate of heat removal in watts (1 W = 1 J s⁻¹). [1]

8. A person runs in hot weather. Describe and explain the physiological and behavioural responses that help to keep the core temperature near its set point. [6]

9. A drawing of a liver lobule is 54 mm wide. The lobule is 1.2 mm wide in the tissue. (a) Work out how many times larger the drawing is than the real lobule. [2] (b) Describe the structure of a lobule as seen in a stained section. [4]

10. Explain the roles of the liver in: (a) storing glycogen [2] (b) detoxification, using ethanol as an example. [3]

11. Concentrations in blood entering and leaving the liver of a healthy person (mmol dm⁻³):

Substance	Hepatic portal vein	Hepatic vein
Ammonia	0.24	0.03
Urea	3.1	3.9

(a) Calculate the percentage decrease in ammonia concentration. [2] (b) Calculate the percentage increase in urea concentration, to 3 significant figures. [2] (c) Explain these changes. [3]

12. Excretion is important for metabolism and homeostasis. (a) Define excretion. [1] (b) Explain why carbon dioxide and nitrogenous waste must be removed from the body. [4] (c) Describe how the liver forms urea. [3]

Answers

1. Cells are specialised and many are far from the outside, so they cannot detect changes themselves [1]. The animal must respond to changes in the external environment, such as temperature or predators, and the internal environment, such as blood glucose [1]. Organs must be coordinated, for example heart rate and breathing rate rising together during exercise [1]. [3] Examiner insight: “Explain” needs a reason, not just a list, so link each point to what would go wrong without communication.

2. (a) A neurone releases neurotransmitter into the synapse [1], which binds to receptors on the next cell [1]. (Gap junctions or T helper cell contact are alternatives.) (b) An endocrine gland secretes a hormone into the blood [1], which reaches all cells but only target cells with a complementary receptor respond [1]. [4] Examiner insight: Name the signal molecule and the receptor. Saying only “a message is sent” does not describe signalling.

3. A receptor detects a stimulus, whereas an effector carries out a response [1]. Receptor: a temperature receptor in the skin (or the hypothalamus) [1]. Effector: a sweat gland, skin arteriole muscle or skeletal muscle [1]. [3] Examiner insight: Pick a specific structure. “The brain” is too vague, and the hypothalamus is a coordination centre.

4. Negative feedback: the response reverses the change and returns the factor towards its set point [1]. Example: sweating when core temperature is too high [1]. Positive feedback: the response increases the change [1]. Example: platelets attracting more platelets at a wound (or oxytocin and uterine contractions in birth) [1]. [4] Examiner insight: Define the term in terms of the direction of the change, and then give the example, since a correct example alone does not show you know the difference.

5. (a) Change = 34.0 − 15.5 = 18.5 °C [1]. Rate = 18.5 ÷ 60 = 0.31 °C per minute [1]. (b) 37.3 − 37.0 = 0.3 °C [1]. (c) The lizard is an ectotherm and gains heat from its surroundings, so its temperature follows the air temperature [1]. The mouse is an endotherm [1] and uses negative feedback, such as changing blood flow through skin arterioles and shivering or behaviour, to hold its core temperature near the set point [1]. [6] Examiner insight: Write out the subtraction and the division, then give the unit and the requested significant figures.

6. (a) Change = 0.18 − 0.30 = −0.12 mm [1]. Percentage change = −0.12 ÷ 0.30 × 100 = −40% [1]. (b) The narrower arteriole is vasoconstriction, caused by contraction of smooth muscle in its wall [1]. Less blood flows into skin capillaries [1]. Less heat is lost from the skin to the surroundings, so core temperature is maintained [1]. [5] Examiner insight: Say that arterioles change diameter. Writing that capillaries move, or that blood vessels “close”, is imprecise.

7. (a) Mass = 450 g [1]. Heat = 450 × 2.4 = 1080 kJ [1]. (b) 1 080 000 J ÷ 3600 s = 300 W [1]. [3] Examiner insight: Convert units before dividing: kJ to J and hours to seconds. Sweat that runs off the skin without evaporating removes less heat than the question assumes.

8. Muscle respiration releases heat and core temperature rises [1]. Receptors in the skin and the hypothalamus detect the rise [1]. The hypothalamus sends impulses to effectors [1]. Skin arterioles dilate, so more heat is lost by radiation [1]. Sweat glands secrete sweat, and evaporation removes heat from the skin [1]. Behavioural: slow down, take shade, remove clothing or drink water [1]. This is negative feedback, as the temperature returns towards the set point. Metabolic rate reduction or erector pili hairs lying flat are also acceptable. [6] Examiner insight: A good answer follows the sequence stimulus, receptor, coordinator, effector, response. Marks are lost when responses are listed with no explanation of why each lowers temperature.

9. (a) Magnification = image size ÷ actual size [1]. = 54 ÷ 1.2 = ×45 [1]. (b) A roughly hexagonal unit with a central vein (a branch of the hepatic vein) [1]. At the edges, branches of the hepatic artery, hepatic portal vein and bile duct [1]. Sinusoids run between rows of hepatocytes towards the central vein [1]. Kupffer cells lie in the sinusoid walls (or hepatocytes have a large, round nucleus) [1]. [6] Examiner insight: Use matching units in magnification calculations. In a description of tissue, state where each structure is, since a list of names alone gives little.

10. (a) When blood glucose is high, hepatocytes convert glucose to glycogen [1]. Glycogen is broken down to glucose when blood glucose falls [1]. (b) Hepatocytes contain enzymes that break down ethanol [1] to ethanal and then ethanoate [1]. This turns a toxic substance into less harmful ones, which can be used or removed [1]. [5] Examiner insight: Name the product of each step. “Detoxifies it” without the chemical change gains little.

11. (a) Decrease = 0.24 − 0.03 = 0.21 [1]. 0.21 ÷ 0.24 × 100 = 87.5% [1]. (b) Increase = 3.9 − 3.1 = 0.8 [1]. 0.8 ÷ 3.1 × 100 = 25.8% [1]. (c) Hepatocytes take up ammonia from the blood [1]. They combine it with carbon dioxide in the ornithine cycle to form urea [1]. The urea enters the hepatic vein, so its concentration there is higher [1]. [7] Examiner insight: Divide by the original value. For “explain”, link each change to a process in the hepatocyte.

12. (a) Getting rid of metabolic waste products from the body [1]. (b) Carbon dioxide dissolves to form an acid and lowers blood pH [1]. This changes the shape and activity of enzymes [1]. Nitrogenous waste starts as ammonia, which is toxic [1]. It must be converted to urea and removed to keep conditions suitable for metabolism [1]. (c) Excess amino acids are deaminated in the liver, releasing ammonia [1]. Ammonia reacts with carbon dioxide in the ornithine cycle to form urea [1]. Urea is carried in the blood to the kidneys [1]. [8] Examiner insight: Say “metabolic” waste in the definition, since “waste” alone could include undigested food.

Where marks are usually lost

• Describing a feedback loop without naming its receptor and effector.
• Writing that the hypothalamus itself sweats or shivers.
• Writing “vasoconstriction of capillaries” instead of arterioles.
• Forgetting that sweat cools only when it evaporates.
• Using the final value, not the original, as the divisor in a percentage change.
• Not converting units in magnification and power calculations.
• Saying positive feedback keeps conditions stable.
• Omitting carbon dioxide when describing how urea forms.
• Mixing up the vessels supplying the liver.

Next steps

• Revise with the revision notes and re-read the study guide.
• Check the wider course on the OCR A Level Biology hub and the checklist.
• Try all free 10-minute diagnostics at /diagnostics/.
• Book a free trial class.

Official syllabus

OCR, A Level in Biology A (H420), Version 4.1 (April 2026), first assessment 2025, Cambridge OCR. Sections 5.1.1 and 5.1.2.