Survival and response - AQA A-Level Biology (Paper 2) Exam Questions

Practise 18 exam-style questions on Survival and response for AQA A-Level Biology (Paper 2). Write your answer as you would in the exam and get it marked in seconds against the mark scheme, point by point, with the Study Helper on hand if you get stuck.

18 questions
1–8 marks each
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18 questions, marked as you go

Sample questions with model answers

Here are 4 of the questions in this topic, each with a model answer, an explanation of why it earns the marks and the mark scheme an examiner would use.

Question 1

2 marks

Flatworms live on the undersides of stones in streams. A student observed flatworms in two dishes.

Dish 1: a lamp shone from one end of the dish. Most flatworms moved in fairly straight paths to the far end of the dish.

Dish 2: the dish was lit evenly from above. When the light intensity was increased, the flatworms changed direction more often, but they did not move towards any particular part of the dish.

For each dish, name the type of response shown and give a reason for your answer.

Model answer

Dish 1: a taxis (negative phototaxis), because the flatworms moved in a particular direction - directly away from the light source.

Dish 2: a kinesis, because the light came evenly from above, so the response was not directional; only the rate of turning changed with the intensity of the light.

Explanation

The key difference is direction: in a taxis the organism moves towards or away from a stimulus that comes from one direction, whereas in a kinesis it changes its speed or rate of turning in response to the intensity of a stimulus. Both responses can keep flatworms in dark places such as under stones, where they are hidden from predators.

Mark scheme

1 mark
Dish 1: (negative photo)taxis AND the movement is directional / away from the direction of the light (stimulus)
Both the type of response and the reason are needed for each mark. Accept 'negative phototaxis'.
1 mark
Dish 2: kinesis AND the movement is not directional / the rate of turning changes with the intensity of the stimulus
Accept klinokinesis. Reject taxis for dish 2.

Question 2

3 marks

Some people whose spinal cord has been damaged in the neck still show a withdrawal reflex when the sole of the foot is pricked, but they do not feel the prick.

Use your knowledge of a three-neurone reflex to explain these observations.

Model answer

The three neurones of the withdrawal reflex (sensory, relay and motor) connect in the spinal cord below the damage, so the reflex arc is intact. The reflex does not need impulses to reach the brain, so the leg is still withdrawn. However, impulses from the foot cannot pass up the damaged spinal cord to the brain, and the prick is only felt when impulses reach the brain, so the person does not feel it.

Explanation

A simple reflex is coordinated by a relay neurone in the spinal cord, so it happens without the brain. Feeling a stimulus (perception) depends on impulses reaching the brain, which is a separate pathway. Use 'impulses' or 'action potentials', not 'signals' or 'messages'.

Mark scheme

1 mark
The reflex arc (sensory, relay and motor neurones) is in the spinal cord below the damage / is not damaged
Accept 'the reflex arc passes through the lower part of the spinal cord'.
1 mark
The reflex does not need the brain / impulses do not have to reach the brain for the reflex (so it still happens)
Accept 'it is involuntary / automatic'.
1 mark
Impulses cannot pass up (the spinal cord) to the brain, so the prick is not felt / sensation (pain) is felt only when impulses reach the brain
Accept action potentials for impulses. Reject 'signals' or 'messages' for impulses.

Question 3

4 marks

Scientists measured how much segments cut from the roots and shoots of a species of plant elongated in solutions of IAA of different concentrations. They compared the elongation of each segment with that of segments kept in water.

The graph shows their results. A negative value means that the segments elongated less than those in water.

The concentration of IAA that gives the greatest elongation of shoot segments is greater than the concentration that gives the greatest elongation of root segments. Calculate how many times greater.

A student concluded that 'IAA stimulates the elongation of shoot cells and inhibits the elongation of root cells'. Use the graph to evaluate this conclusion.

A line graph. The horizontal axis is the concentration of IAA in mol dm⁻³ on a scale in which each mark is ten times the one before, from 10⁻¹² to 10⁻². The vertical axis is the change in elongation compared with water, in %, from −100 to +160, with a line at 0. A dashed line for roots rises to a peak at low concentrations and then falls below zero towards −100. A solid line for shoots stays near zero at low concentrations, rises to a peak at higher concentrations and then falls below zero at the highest concentration.

Model answer

Greatest elongation: shoots at 10⁻⁵ mol dm⁻³, roots at 10⁻¹⁰ mol dm⁻³.
10⁻⁵ ÷ 10⁻¹⁰ = 10⁵ = 100 000 times greater.

The conclusion is only partly supported. Between 10⁻⁸ and 10⁻³ mol dm⁻³, IAA increases the elongation of shoots (for example +100% at 10⁻⁶) and decreases the elongation of roots (−65% at 10⁻⁶). However, below 10⁻⁸ mol dm⁻³ IAA stimulates root elongation (up to +35% at 10⁻¹⁰), and above 10⁻³ mol dm⁻³ it inhibits shoot elongation (−62% at 10⁻²). So the effect depends on the concentration, and roots are far more sensitive to IAA than shoots.

Explanation

On a log scale each division is ten times the one before, so the number of divisions between the two peaks (5) gives 10⁵, not 5. Whether IAA stimulates or inhibits elongation depends on its concentration and on the organ: root cells are stimulated by very low concentrations and inhibited by the concentrations that stimulate shoot cells. This is why the same accumulation of IAA on the lower side makes a horizontal shoot bend up and a horizontal root bend down.

Mark scheme

1 mark
100 000 / 10⁵ / 1 × 10⁵ (times); a correct answer with no working scores this mark
Max 4. Point 1 is the calculation: a correct answer of 100 000 (accept 10⁵ or 1 × 10⁵) with no working scores 1 mark (10⁻⁵ ÷ 10⁻¹⁰). Reject 5 (the difference between the powers) and 10⁻⁵. Points 2 to 6 (the evaluation): max 3.
1 mark
Supports: between 10⁻⁸ and 10⁻³ mol dm⁻³ IAA increases elongation of shoots AND decreases elongation of roots
Accept any concentration in this range used correctly, e.g. at 10⁻⁶ mol dm⁻³ shoots +100% and roots −65%. Accept 'at intermediate concentrations' only with a value from this range.
1 mark
Does not support: at low concentrations / below 10⁻⁸ mol dm⁻³ IAA increases / stimulates elongation of roots
Accept the maximum of about +35% at 10⁻¹⁰ mol dm⁻³. Accept 'very low concentrations' without a value.
1 mark
Does not support: at high concentrations / above 10⁻³ mol dm⁻³ IAA decreases / inhibits elongation of shoots
Accept about −60% at 10⁻² mol dm⁻³. Accept 'very high concentrations' without a value.
1 mark
Does not support: at the lowest concentrations / below about 10⁻⁹ mol dm⁻³ IAA has (almost) no effect on shoots
Accept 'shoots do not respond to very low concentrations'.
1 mark
The effect depends on the concentration / roots are (much) more sensitive to IAA than shoots
Accept 'roots respond to much lower concentrations than shoots'.

Question 4

6 marks

A student wanted to find out whether maggots (blowfly larvae) respond to humidity. She had a choice chamber, silica gel (which absorbs water vapour), cotton wool, water, cobalt chloride paper (blue when dry, pink when damp) and 20 maggots.

Describe how she could carry out this investigation. Include how she would make sure that her results are valid and how she would treat the maggots ethically.

Model answer

Put silica gel in two sections (one half) of the base of the choice chamber and damp cotton wool in the other half. Place the gauze over the base so the maggots cannot touch the silica gel or the water, and put on the lid. Leave it for about five minutes, then use forceps to place cobalt chloride paper on each side to check that one side is dry (blue) and the other humid (pink).

Keep the light and temperature the same on both sides, for example by covering the whole chamber. Use a soft paintbrush to put the 20 maggots in through the central hole, so they start at the same distance from both sides. After 10 minutes, count the maggots on each side. Repeat with the humid and dry sides swapped, and set up a control chamber with nothing in the base to check that the maggots spread out evenly when there is no humidity difference.

Handle the maggots gently, keep them in the chamber for as short a time as possible and return them to their container afterwards.

Explanation

A valid choice chamber investigation changes only the independent variable (humidity), checks that the difference really exists (cobalt chloride paper), keeps other factors such as light and temperature the same, and includes a control and repeats. Swapping the humid and dry sides rules out any effect of the chamber's position. Required practical 10 also requires animals to be used safely and ethically: minimal handling, short exposure and returning them afterwards.

Mark scheme

1 mark
Silica gel in one half (of the base) AND damp cotton wool / water in the other half
Max 5 if the answer does not give any way of treating the maggots ethically. Otherwise max 6. Accept 'one side dry and one side humid' only if how this is done is given.
1 mark
Gauze / mesh over the base so the maggots cannot touch the silica gel or water / gauze kept dry
Accept 'maggots on the gauze above the substances'.
1 mark
Check with cobalt chloride paper that one side is dry (blue) and the other humid (pink) before adding the maggots
Accept 'use cobalt chloride paper to show there is a humidity difference'. Ignore 'handle the paper with forceps' on its own.
1 mark
Keep other conditions the same on both sides, e.g. light (cover the chamber / even light) or temperature
A named condition is needed. Ignore 'keep everything the same'.
1 mark
Put the maggots in at the centre / through the central hole
Accept 'equal distance from each side'.
1 mark
Leave for a set time, e.g. 10 minutes, then count the number of maggots on each side
Accept any stated time from 2 to 30 minutes.
1 mark
Repeat with the humid and dry sides swapped OR repeat several times (and calculate a mean)
Accept 'to remove any effect of the position of the chamber'.
1 mark
Control with both halves the same / no silica gel or water, to check that maggots spread out evenly
Accept 'a chamber with dry conditions on both sides'.
1 mark
Ethics: handle maggots gently (e.g. with a soft paintbrush or spoon) OR keep them in the chamber for a short time only OR return them to their container / habitat afterwards OR avoid extreme conditions
Any one ethical point.

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