Eukaryotes and prokaryotes - AQA GCSE Biology Exam Questions

Practise 18 exam-style questions on Eukaryotes and prokaryotes for AQA GCSE Biology. Write your answer as you would in the exam and get it marked in seconds against the mark scheme, point by point, with an AI tutor on hand if you get stuck.

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1–10 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

3 marks

Table shows how a blockage (%) in a coronary artery affects blood flow (cm3/min):

0% ? 100
10% ? 64
20% ? 42
50% ? 8
80% ? 2

Estimate the blood flow (in cm3/min) when the coronary artery is 35% blocked. Show clearly how you work out your answer.

Model answer

Approximately 25 cm3/min. Linear interpolation between 20% (42) and 50% (8): change = 8 - 42 = -34 over 30% ? slope ˜ -34/30 ˜ -1.13 cm3·min-1 per 1%. At 35% (15% above 20%): 42 + 15×(-1.13) ˜ 42 - 17 ˜ 25 cm3/min.

Explanation

Use linear interpolation between the two nearest table points. Between 20% and 50% blockage blood flow falls from 42 to 8 (a drop of 34) over 30 percentage points, giving a fall of about 1.13 cm3/min per 1% blockage. Thirty-five percent is 15% above 20%, so the flow is about 42 - 15×1.13 ˜ 25 cm3/min.

Mark scheme

1 mark
Identify the correct interval to interpolate (between 20% and 50%)
Award if student states they use the 20%–50% pair or uses values 42 and 8.
1 mark
Calculate gradient / rate of change: (8 - 42) / (50 - 20) ˜ -34/30 ˜ -1.13 cm3·min-1 per 1%
Allow equivalent correct fractions or decimals.
1 mark
Compute flow at 35% ˜ 42 + 15×(-1.13) ˜ 25 cm3/min (allow ˜ 24–26)
Accept answers ˜24–26 cm3/min with supporting working.

Question 2

4 marks

In a leaf-disc photosynthesis experiment the times for 10 leaf discs to rise were: Blue light 115 s, Green light 831 s, Red light 397 s. Explain why the discs rose much faster under blue light than under green light. In your answer refer to chlorophyll absorption.

Model answer

Chlorophyll absorbs blue (and red) wavelengths strongly but absorbs very little green (green is mostly reflected). Blue light is therefore absorbed efficiently, providing more energy for the light-dependent reactions so photosynthesis (oxygen production) is faster; more O2 collects in the discs and they float sooner. Green light gives little absorbed energy so much slower O2 production and much longer times.

Explanation

Chlorophyll a and b strongly absorb in the blue (and red) parts of the spectrum but absorb little green light. When the leaf discs are illuminated with blue light, chlorophyll absorbs more photons, driving the light reactions faster and generating oxygen more quickly. The accumulated oxygen trapped in the leaf discs makes them buoyant, so they rise rapidly under blue light; with green light (poorly absorbed) oxygen production is slow and the discs take much longer to float.

Mark scheme

1 mark
State that chlorophyll absorbs blue (and red) light strongly but not green light
Reference to absorption spectra or 'green is reflected' acceptable.
1 mark
More light energy absorbed (under blue) ? faster photosynthesis / faster oxygen production
Must link absorption to rate of photosynthesis / O2 production.
1 mark
More O2 produced fills discs ? discs become buoyant and rise sooner
Award for linking O2 production to discs floating.
1 mark
Green light is poorly absorbed ? little energy for photosynthesis ? slow O2 production ? much longer time for discs to rise
Reverse argument to explain long time under green light.

Question 3

6 marks

Polydactyly (extra fingers or toes) is caused by a dominant allele D. A woman with polydactyly is heterozygous (D d) and the man is homozygous recessive (d d).

Draw a Punnett square to show the possible genotypes of their children and state the probability that a child will have polydactyly

Model answer

Punnett square: parents Dd (mother) × dd (father). Gametes: mother D and d; father d and d. Offspring genotypes: Dd and dd in a 1 : 1 ratio (two boxes: Dd, Dd, dd, dd if using two d gametes for father). Therefore probability of a child having polydactyly (phenotype) = 1/2 = 50%.

(If drawn as a 2×2 with columns d and d and rows D and d ? boxes: Dd, Dd, dd, dd.)

Explanation

The heterozygous parent (D d) produces gametes D and d; the homozygous recessive parent (d d) produces only d gametes. Combining these gives genotypes Dd (polydactyly) or dd (normal) in equal numbers. So each child has a 50% chance of having polydactyly.

Mark scheme

2 marks
Correct gametes shown: mother D and d; father d and d
Award 2 marks for both parents' gametes correctly identified.
2 marks
Correct Punnett square with offspring genotypes Dd, Dd, dd, dd (or equivalent 1:1 Dd : dd)
Award 2 marks for correctly filling the 2×2 square or stating the genotypic ratio.
2 marks
Probability that a child has polydactyly = 1/2 or 50%
Award 2 marks for the correct probability and equivalent percentage or fraction.

Question 4

9 marks

A student investigated the effect of concentration of sucrose solution on the mean percentage change in mass of potato cylinders after 30 minutes.

Plan a method to obtain reliable data for this investigation. Your plan should include: what you would measure and how, one control variable and one step to improve reliability (6 marks).

Explain the result you would expect for a potato cylinder placed in 0.6 mol/dm³ sucrose solution compared with in water (0 mol/dm³)

Model answer

Plan (6 marks): Use potato cylinders of equal diameter and length (cut with cork borer and ruler). Blot dry and measure initial mass with a balance to 0.01 g. Place equal numbers (e.g. 3–5) of cylinders into different sucrose concentrations (e.g. 0, 0.2, 0.4, 0.6, 1.0 mol/dm³) for 30 minutes. Remove, blot dry, measure final mass and calculate percentage change = (final - initial)/initial × 100. Control variable: same temperature (use water bath) or same cylinder size or same time. To improve reliability: repeat and calculate mean; use several replicates and discard anomalous results. Expected result (3 marks): In 0.6 mol/dm³ solution potato will lose water by osmosis to the hypertonic solution (water potential lower outside), so mass decreases; in water (hypotonic) potato gains water so mass increases; at 0.6 mol/dm³ the percentage change will be negative compared with positive change in water.

Explanation

Method: use identical potato cylinders, measure initial mass, immerse in known sucrose concentrations for fixed time and temperature, blot dry and measure final mass, calculate percentage change, repeat for replicates and take mean. Explanation: Osmosis moves water from higher water potential to lower water potential across a partially permeable membrane. Water leaves the potato in the 0.6 mol/dm³ (hypertonic) solution so mass falls; in pure water (hypotonic) water enters the potato so mass rises.

Mark scheme

1 mark
Measure initial and final mass with balance and calculate percentage change
Award for stating mass measurement and formula for % change or idea of comparing initial and final mass.
1 mark
Use potato cylinders of equal size (cork borer and ruler)
Size standardisation is a valid control step.
1 mark
Use series of sucrose concentrations including 0 mol/dm³ and 0.6 mol/dm³
Accept specific concentrations and number of samples.
1 mark
Control variable: same temperature OR same time in solution OR same volume of solution
Any one valid control variable.
1 mark
Reliability improvement: use replicates and calculate mean (repeat)
Or discard anomalies / more repeats.
1 mark
Blot dry before measuring to remove surface liquid
Accept use of calibrated balance to 0.01 g.
1 mark
0.6 mol/dm³: mass decreases (water leaves by osmosis) because solution outside is hypertonic / lower water potential
Award for stating water leaves and mass decreases; must mention osmosis or water potential.
1 mark
0 mol/dm³ (water): mass increases (water enters by osmosis) because solution outside is hypotonic / higher water potential
Dependent on understanding osmosis direction; accept mass increases in water.
1 mark
Compare signs: negative percentage change at 0.6 mol/dm³ vs positive percentage change in water
Award for clear comparison of expected outcomes.

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