AQA A-level Biology: Survival, Response and Receptors (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 3.6.1.1 and 3.6.1.2, in the AQA AS and A-level Biology specification for first teaching 2015. Both sections are A-level only. Explanations are in the study guide and the revision notes. Course pages: AQA A-level Biology hub and the printable checklist. Where a question gives data, use only that data. The marks shown are for practice and are not an official tariff.

Questions

1. Distinguish between a taxis and a kinesis, and give one example of either. [3]

2. In 30 seconds a woodlouse travelled 66 mm in damp air and 150 mm in dry air. (a) Calculate its speed in each condition, in mm s⁻¹. [2] (b) Calculate the percentage increase in speed in dry air compared with damp air, to the nearest whole number. [1] (c) Explain how this response helps woodlice to stay in damp areas. [2]

3. Forty animals were placed in the centre of a choice chamber with a dry half and a damp half. After 10 minutes, 29 were in the damp half and 11 in the dry half. (a) State a null hypothesis for the investigation. [1] (b) Calculate χ². The expected number in each half is 20. [2] (c) There is one degree of freedom and the critical value at p = 0.05 is 3.84. State and explain your conclusion. [2]

4. A student investigates the effect of light intensity on the movement of fly larvae using a choice chamber. (a) Describe how the student should carry out the investigation so that the results are reliable. [4] (b) Give one ethical precaution. [1]

5. A cyclist’s eye is struck by a flying insect and the eyelid closes before the cyclist has registered what happened. (a) Name the three neurones in this reflex arc, in order, and state the role of each. [3] (b) Explain why this type of reflex has a protective effect. [2]

6. A reflex arc is 1.2 m long and impulses travel along it at 40 m s⁻¹. There are two synapses, each adding a delay of 0.8 ms. Calculate the total response time in milliseconds. [3]

7. Sections of shoot and root were kept for 24 hours in water or in a high concentration of IAA. Shoot sections lengthened by 3.0 mm in water and 7.5 mm in IAA. Root sections lengthened by 8.0 mm in water and 2.0 mm in IAA. (a) Calculate the percentage change in lengthening caused by IAA for the shoot and for the root. [2] (b) Explain what the results show about the effect of IAA on cell elongation. [2] (c) Suggest one way to improve the investigation. [1]

8. A seedling is laid on its side. Explain how its root bends downward. [4]

9. Describe how pressure on a Pacinian corpuscle can lead to an action potential in a sensory neurone. [5]

10. A Pacinian corpuscle has a resting potential of −70 mV and a threshold of −55 mV. Pressure of 1.0 kPa gives a membrane potential of −64 mV, 1.5 kPa gives −58 mV and 2.0 kPa gives −51 mV. (a) Calculate the depolarisation at 1.0 kPa. [1] (b) State whether 1.5 kPa triggers an action potential and explain your answer. [2] (c) Explain why shining a bright light on the corpuscle gives no response. [1]

11. (a) Explain why you see fine detail in daylight. [2] (b) Explain why you cannot see colour in very dim light. [2] (c) Explain why you can see dim objects at night. [2]

12. The table gives data supplied for two regions of a human retina.

Region	Cones per mm²	Rods per mm²
A	160 000	0
B	8 000	90 000

(a) Calculate the ratio of rods to cones in region B, to the nearest whole number. [1] (b) A person has damage to region A. Predict and explain the effect on reading small print in bright light. [3] (c) Astronomers sometimes see a faint star better by looking slightly to one side of it. Explain this. [3]

Answers

1. A taxis is a directional movement in response to the direction of a stimulus [1]; a kinesis is a change in speed of movement or rate of turning in response to the intensity of a stimulus, not its direction [1]; example, such as woodlice moving faster in dry air (kinesis) or Euglena swimming towards light (taxis) [1]. [3] Examiner insight: Define the pair by the word “direction” versus “intensity”; answers that only say “movement in response to a stimulus” for both do not separate them.

2. (a) Damp: 66 / 30 = 2.2 mm s⁻¹ [1]. Dry: 150 / 30 = 5.0 mm s⁻¹ [1]. (b) (5.0 − 2.2) / 2.2 × 100 = 127.3, so 127% [1]. (c) In dry air it moves faster, so it is more likely to leave the dry area [1]; in damp air it moves slowly, so it tends to remain there [1]. [5] Examiner insight: Show the division before the answer and include units; a percentage change should be divided by the starting (damp) value.

3. (a) The animals are distributed equally between the damp and dry halves, so any difference is due to chance [1]. (b) (29 − 20)² / 20 = 4.05 and (11 − 20)² / 20 = 4.05 [1]; χ² = 8.1 [1]. (c) 8.1 is greater than 3.84, so the null hypothesis is rejected [1]; the difference is significant at p = 0.05, so the animals prefer the damp half [1]. [5] Examiner insight: Conclusions that compare the calculated value with the critical value and then refer back to the biology tend to gain credit; stating only “significant” leaves the link to the animals unsaid.

4. (a) Cover one half with black card or foil and leave the other half lit [1]. Put the same number of larvae in the centre [1]. Count the number in each half at fixed times and keep other variables such as temperature the same [1]. Repeat with new larvae and calculate a mean [1]. (b) Handle larvae gently, limit exposure time, and return them to where they were collected [1]. [5] Examiner insight: Naming the control variables that matter and the repeat with new animals usually separates a strong method from a list of apparatus.

5. (a) Sensory neurone carries impulses from the receptor to the central nervous system [1]; relay neurone connects the sensory and motor neurones [1]; motor neurone carries impulses to the effector (muscle) [1]. (b) The pathway is short with few synapses, so the response is very rapid [1]; it is automatic, so the eyelid closes before any decision, reducing damage to the eye [1]. [5] Examiner insight: Pair each neurone with its role in order; “fast” alone is weaker than “fast because few synapses”.

6. Conduction time = 1.2 / 40 = 0.030 s = 30 ms [1]. Synaptic delay = 2 × 0.8 = 1.6 ms [1]. Total = 30 + 1.6 = 31.6 ms [1]. [3] Examiner insight: Convert seconds to milliseconds before adding the synaptic delays; mixing units is a common loss.

7. (a) Shoot: (7.5 − 3.0) / 3.0 × 100 = +150% [1]. Root: (2.0 − 8.0) / 8.0 × 100 = −75% [1]. (b) IAA stimulates cell elongation in the shoot [1]; it inhibits cell elongation in the root [1]. (c) Test a range of IAA concentrations, or repeat and use a mean [1]. [5] Examiner insight: State the sign or direction of each percentage change and interpret each organ separately; one comment covering both organs hides the difference.

8. IAA moves to the lower side of the root [1]. A high concentration of IAA inhibits cell elongation in roots [1]. The upper side elongates faster than the lower side [1]. The root therefore curves downward, which is positive gravitropism [1]. [4] Examiner insight: The same sequence applied to a shoot gives the opposite bend; name the organ in every sentence.

9. Pressure deforms the lamellae and stretches the neurone membrane [1]. Stretch-mediated sodium ion channels widen [1]. Sodium ions diffuse into the neurone [1]. The membrane depolarises, producing a generator potential [1]. If the generator potential reaches threshold, an action potential is set up [1]. [5] Examiner insight: Keep the events in order and name the channels as stretch-mediated; “channels open because of the voltage” describes a different mechanism.

10. (a) −64 − (−70) = 6 mV [1]. (b) No [1]. The potential reaches −58 mV, which is a depolarisation of 12 mV, and a threshold of −55 mV needs 15 mV [1]. (c) A Pacinian corpuscle responds only to its specific stimulus, which is pressure [1]. [4] Examiner insight: Use the numbers from the question in the explanation; a statement that only repeats “below threshold” is thinner than one that shows the gap.

11. (a) Cones each connect to their own bipolar neurone [1], so two nearby points of light send separate impulses to the brain [1]. (b) Cones need higher light intensity to be stimulated [1]; rods contain one pigment, so rods give no colour discrimination [1]. (c) Rhodopsin in rods is broken down by low-intensity light [1]; many rods connect to one bipolar neurone, so summation of generator potentials reaches threshold [1]. [6] Examiner insight: Link each feature (connection, pigment) to its effect on the impulse sent to the brain, not just to the name of the cell.

12. (a) 90 000 / 8 000 = 11.25, so about 11 : 1 [1]. (b) Reading would be impaired [1]. Region A has the highest density of cones, each with its own bipolar neurone [1], so loss of region A reduces the acuity needed to see fine detail [1]. (c) Region B has many rods and few cones [1]. Many rods connect to one bipolar neurone, so summation of generator potentials reaches threshold in faint light [1]. Region A has no rods, so looking directly at the star does not allow this [1]. [7] Examiner insight: Use the data (cone density, rod presence) in your explanation and name the property being explained, sensitivity or acuity.

Where marks are usually lost

• Describing a kinesis as directional movement.
• Working out percentage change relative to the final figure rather than the starting one.
• Omitting units or the conversion from seconds to milliseconds.
• Stating a chi-squared result without the null hypothesis, degrees of freedom and comparison with the critical value.
• Writing a vague method for a choice chamber with no controls and no repeats.
• Saying IAA has the same effect in roots and shoots.
• Writing that a stimulus “makes an action potential” without naming the generator potential and threshold.
• Explaining sensitivity or acuity without linking it to how rods and cones connect to bipolar neurones.

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 3.6.1.1 and 3.6.1.2).