OxfordAQA International AS and A-level Biology: Reflexes, taxes and kineses, and sensory receptors (9610) – 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 sections 3.4.1, 3.4.2.1 and 3.4.2.2 of the OxfordAQA International AS and A-level Biology (9610) specification, Version 5.1. Unit 4 is International A-level only. Read the study guide first, or the revision notes. The course hub and checklist list the other units, and the diagnostics are a quick check.

Questions

1. A person treads on a sharp stone and the foot lifts before they feel pain. Name, in order, the three neurones in this reflex arc, and explain why the response protects the body. [4]

2. In a reflex the impulse pathway is 1.40 m long and impulses travel at 56 m s⁻¹. There are two synapses, and each adds a delay of 0.9 ms.

(a) Calculate the time from receptor to effector in milliseconds. [3] (b) Suggest why a response needing a decision by the brain would take longer. [1]

3. Brine shrimp larvae were placed in a horizontal tube with a lamp at one end. The tube was divided into five equal sections, A (nearest the lamp) to E. After 10 minutes the counts were: A 25, B 15, C 10, D 6, E 4.

(a) Calculate the percentage of larvae in sections A and B together. [2] (b) Name the type of response shown, and say whether it is positive or negative. [1] (c) Explain how this type of response increases the chance of survival. [1]

4. A small invertebrate was timed for 60 s in an arena of humid air and then in dry air. Data supplied: humid air, 138 mm travelled and 9 turns; dry air, 420 mm travelled and 34 turns.

(a) Calculate the mean speed in each condition in mm s⁻¹. [2] (b) Explain why these data show a kinesis, not a taxis. [2] (c) Explain how this behaviour keeps the animal in humid air. [2]

5. Describe how pressure applied to a Pacinian corpuscle leads to an action potential in the sensory neurone. [5]

6. The resting potential of a Pacinian corpuscle is −70 mV and the threshold is −55 mV. The table gives data supplied.

Pressure (kPa)	Membrane potential (mV)	Impulses per second
1.0	−66	0
2.0	−58	0
3.0	−52	12
4.0	−48	26

(a) Calculate the depolarisation at 3.0 kPa. [1] (b) Between which two pressures is the threshold reached? [1] (c) Explain why there are no impulses at 2.0 kPa. [2] (d) Calculate the percentage increase in impulse frequency from 3.0 to 4.0 kPa. [2]

7. Explain why a Pacinian corpuscle responds to pressure but not to light. [2]

8. In dim light each rod produces a generator potential of 0.5 mV. The threshold of the bipolar neurone it connects to is 2.5 mV.

(a) Calculate the minimum number of rods needed to reach the threshold. [1] (b) Explain why a single cone connected to its own bipolar neurone would not trigger an impulse in dim light. [3]

9. One region of a retina has 210 rods connected to 7 bipolar neurones. In another region each cone connects to one bipolar neurone.

(a) Calculate the mean number of rods per bipolar neurone. [1] (b) Explain how the difference in connections leads to a difference in visual acuity. [3]

10. The table gives relative absorbance of the three types of cone at two wavelengths (data supplied).

Wavelength (nm)	Short-wavelength cone	Medium-wavelength cone	Long-wavelength cone
540	0.04	0.95	0.62
600	0.00	0.20	0.80

(a) State which cone type is stimulated most at 540 nm. [1] (b) Calculate the ratio of long-wavelength to medium-wavelength absorbance at each wavelength, to two significant figures. [2] (c) Explain how the brain uses these data to tell the two lights apart, and why rods cannot do this. [2]

11. The centre of a human retina has 150 000 cones per mm² and few rods. A region at the edge has 6 000 cones per mm² and many rods.

(a) Calculate how many times greater the cone density is at the centre. [1] (b) A faint star is easier to see when you look slightly to one side of it. Explain why. [4] (c) Explain why you look directly at a word to read it. [3] (d) Explain why the star appears colourless. [2]

Answers

1. Sensory neurone [1], relay neurone [1], motor neurone [1]. The pathway is short with few synapses, so the response is fast and automatic, and the foot is lifted before the damage increases [1]. [4] Examiner insight: “In order” matters, so a list of three neurone names in the wrong order is unlikely to gain full credit. A reason for protection needs to say why the reflex is fast, not just that it is.

2. (a) Conduction time = 1.40 ÷ 56 [1] = 0.025 s = 25.0 ms. Synaptic delay = 2 × 0.9 = 1.8 ms [1]. Total = 25.0 + 1.8 = 26.8 ms [1]. [3] (b) The impulse must pass through the brain, which adds more neurones and synapses (each with a delay) [1]. [1] Examiner insight: Show each stage and the unit. A final answer in seconds, or with the synaptic delay left out, is the usual slip.

3. (a) A + B = 25 + 15 = 40 [1]; 40 ÷ 60 × 100 = 66.7% [1]. [2] (b) Positive phototaxis [1]. [1] (c) The larvae move to the light, so they are more likely to reach the lit surface water, where their algal food is found [1]. [1] Examiner insight: Use the word “phototaxis” with the sign (positive or negative). “Moves to light” alone describes the behaviour without naming it.

4. (a) Humid: 138 ÷ 60 = 2.3 mm s⁻¹ [1]. Dry: 420 ÷ 60 = 7.0 mm s⁻¹ [1]. [2] (b) The speed and the number of turns both change with the humidity [1], and there is no movement in a particular direction relative to the stimulus [1]. [2] (c) In dry air the animal moves fast and turns often, so it is likely to leave by chance [1]. In humid air it moves slowly and turns less, so it tends to stay [1]. [2] Examiner insight: Answers about a kinesis gain credit by tying speed and turning to stimulus intensity. Writing “moves towards humidity” describes a taxis.

5. Pressure deforms the lamellae around the sensory neurone ending [1]. The membrane of the ending is stretched [1], so stretch-mediated sodium ion channels widen [1]. Sodium ions diffuse in, depolarising the membrane to give a generator potential [1]. If the generator potential reaches the threshold, an action potential is set up [1]. [5] Examiner insight: A clear sequence from deformation to channel to ion movement reads well. Mention the threshold, because it is the step that separates a generator potential from an action potential.

6. (a) −52 − (−70) = 18 mV [1]. [1] (b) Between 2.0 and 3.0 kPa [1]. [1] (c) At 2.0 kPa the potential is −58 mV, a depolarisation of 12 mV [1], which is less than the 15 mV needed to reach the threshold at −55 mV, so no action potential is set up [1]. [2] (d) (26 − 12) ÷ 12 × 100 [1] = 117% [1]. [2] Examiner insight: Quote the numbers when you compare a value with the threshold. Percentage change is worked out from the starting value (12), not the final value.

7. The corpuscle’s membrane contains stretch-mediated sodium ion channels that respond to deformation [1]. It contains nothing that is changed by light, so light produces no generator potential [1]. [2] Examiner insight: Both halves are needed: what the receptor does respond to and why light has no effect.

8. (a) 2.5 ÷ 0.5 = 5 rods [1]. [1] (b) Cones need a higher light intensity to change their pigment, so they give a small or no generator potential in dim light [1]. A single cone connects to its own bipolar neurone, so there is no summation with other cells [1]. One cone alone cannot reach the threshold [1]. [3] Examiner insight: Rod sensitivity answers earn credit when they mention both the pigment and summation. Pigment alone gives half the explanation.

9. (a) 210 ÷ 7 = 30 rods [1]. [1] (b) Many rods share a bipolar neurone, so two points of light on different rods in the same group send a single set of impulses [1], and the brain sees one point [1]. Each cone has its own neurone, so two stimulated cones give two sets of impulses and the points are seen apart [1]. [3] Examiner insight: Acuity answers should compare both photoreceptors. Describing cones only leaves the contrast incomplete.

10. (a) The medium-wavelength cone [1]. [1] (b) 540 nm: 0.62 ÷ 0.95 = 0.65 [1]. 600 nm: 0.80 ÷ 0.20 = 4.0 [1]. [2] (c) The brain compares the relative stimulation of the three cone types, which differs between the two wavelengths [1]. Rods contain only one pigment, so there is no comparison [1]. [2] Examiner insight: “The cones detect different colours” is too vague. The brain compares the relative response of the types.

11. (a) 150 000 ÷ 6 000 = 25 times [1]. [1] (b) A faint star gives dim light [1]. Looking to one side puts its image on the edge, where rods are dense [1]. Many rods share a neurone, so their generator potentials add up by summation [1] and reach the threshold, which the few cones at the centre could not do [1]. [4] (c) The centre of the retina has a high density of cones [1], each connected to its own bipolar neurone [1], so there is high acuity and the letters are seen separately [1]. [3] (d) Only rods respond in dim light [1], and they have one pigment, so the brain cannot compare different types [1]. [2] Examiner insight: In long explanations, link each feature to its result. The strongest answers use the words summation, separate connections and the density of the photoreceptor.

Where marks are usually lost

• Naming neurones in the wrong order, or leaving out the relay neurone.
• Saying a kinesis is steered towards or away from the stimulus.
• Giving a reflex-time answer in the wrong unit, or forgetting the synaptic delay.
• Writing that sodium ions are pumped in, rather than diffusing through opened channels.
• Treating a generator potential as an action potential.
• Calculating percentage increase using the final value as the base.
• Explaining sensitivity without summation, or acuity without separate connections.
• Stating that cones “see colour” without the idea that the brain compares the three types.

Next steps

• Re-read the study guide or the revision notes.
• Open the OxfordAQA A-level Biology hub and the checklist.
• Take the free 10-minute diagnostics: all free 10-minute diagnostics.
• Book a free trial class.

Official syllabus

OxfordAQA International AS and A-level Biology (9610), Version 5.1, for International AS exams May/June 2017 onwards and International A-level exams May/June 2018 onwards, from OxfordAQA, covering specification content 3.4.1, 3.4.2.1 and 3.4.2.2.