Calculations in biology - AQA A-Level Biology (Paper 3) Exam Questions

Practise 18 exam-style questions on Calculations in biology for AQA A-Level Biology (Paper 3). 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
1–7 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

A student photographed a tardigrade (water bear), a tiny invertebrate, through an optical microscope. In the photograph, the tardigrade is 51 mm long. The magnification of the photograph is × 150.

Calculate the actual length of the tardigrade. Give your answer in μm.

The teacher then made an enlarged copy of the photograph on a photocopier. Another student measured the length of the tardigrade on the enlarged copy and divided it by 150. Explain how this would affect the actual length that the second student calculated.

Model answer

Image length = 51 mm = 51 × 1000 = 51 000 μm
Actual length = image size ÷ magnification = 51 000 ÷ 150 = 340 μm

The second student's length would be too large. Enlarging the photograph makes the image of the tardigrade longer, so the magnification of the copy is greater than × 150; dividing the longer image length by 150 gives an actual length larger than the true length.

Explanation

Magnification = image size ÷ actual size, so actual size = image size ÷ magnification, with both lengths in the same unit: convert the image length to μm first (1 mm = 1000 μm), then divide. A stated magnification applies only to the image it was worked out for: enlarging or reducing a photograph changes its magnification, so dividing by the original value gives a wrong actual size (a scale bar avoids this, because it is enlarged with the image). Multiplying by the magnification instead of dividing is a common error.

Mark scheme

1 mark
Correct method: 51 000 ÷ 150 (image length in μm divided by magnification); or a correct final answer of 340 μm with no working
A correct answer of 340 μm with no working scores 2 marks (points 1 and 2). Accept for 1 mark: 0.34 (correct length in mm, not converted to μm) OR evidence of 51 000 μm. Do not award for 7650 (51 × 150) or 2.9 (150 ÷ 51).
1 mark
340 (μm)
1 mark
The calculated length would be too large / an overestimate, because the enlarged copy has a greater magnification than × 150 (the image is longer but the same magnification was used)
Both the effect (too large) and the reason are needed. Accept 'the magnification of the copy is × 150 multiplied by the enlargement, so dividing by 150 gives too large a length'. Reject 'too small'. Ignore 'it would be inaccurate' or 'it would be wrong' with no direction.

Question 2

4 marks

A sample of cooked rice contained 3.0 × 10³ bacteria per gram. After 4.0 hours at 30 °C, it contained 1.5 × 10⁶ bacteria per gram. Assume that all the bacteria divided at the same rate.

N = N₀ × 2ⁿ, where N₀ is the starting number of bacteria, N is the final number and n is the number of generations (divisions).

Calculate the number of generations in the 4.0 hours.

Calculate the mean generation time in minutes.

Suggest one reason why the real generation time of the dividing bacteria is likely to be shorter than your calculated value.

Model answer

N ÷ N₀ = 1.5 × 10⁶ ÷ 3.0 × 10³ = 500, so 2ⁿ = 500
n = log 500 ÷ log 2 = 2.699 ÷ 0.301 = 8.97 generations

Mean generation time = (4.0 × 60) ÷ 8.97 = 240 ÷ 8.97 = 26.8 minutes

There was probably a lag phase at the start, when the bacteria were not yet dividing, so during the rest of the time they divided more often than every 26.8 minutes.

Explanation

If N = N₀ × 2ⁿ, then 2ⁿ = N ÷ N₀ and n = log(N ÷ N₀) ÷ log 2 (logarithms to any base work, as long as the same base is used). The mean generation time is the total time divided by the number of generations. The calculation assumes every cell divided throughout and none died; a lag phase or cell death means that the cells that did divide must have done so faster.

Mark scheme

1 mark
Correct method: n = log(1.5 × 10⁶ ÷ 3.0 × 10³) ÷ log 2 = log 500 ÷ log 2; or a correct final answer of 8.97 (generations) with no working
A correct answer of 8.97 (accept 8.9 to 9.0) with no working scores 2 marks (points 1 and 2). Accept for 1 mark: evidence of 500 (N ÷ N₀) OR 2.7 (log 500 not divided by log 2). Accept a trial-and-improvement method showing 2⁹ = 512 is close to 500.
1 mark
8.97 (generations)
Accept 8.9 to 9.0, including 9.
1 mark
Generation time = 26.8 (minutes)
Accept 26.6 to 27. ECF: 240 ÷ their number of generations. A correct answer with no working scores this point.
1 mark
Some bacteria died / not all the bacteria were dividing / there was a lag phase at the start when the bacteria did not divide, so the bacteria that were dividing must have divided more often
Accept 'the bacteria took time to adjust to the rice before dividing'. Ignore 'the temperature changed' unqualified.

Question 3

4 marks

An ecologist used a map with a scale of 1 : 2500 to find the area of a wood. On the map, the wood is a rectangle 8.4 cm long and 5.2 cm wide.

1 hectare (ha) = 10 000 m²

Calculate the area of the wood in hectares.

The ecologist found 41 nests of one species of bird in the wood. Calculate the number of nests per hectare.

Suggest one reason why your answer may not be the true density of nests in the wood.

Model answer

Actual length = 8.4 cm × 2500 = 21 000 cm = 210 m
Actual width = 5.2 cm × 2500 = 13 000 cm = 130 m
Area = 210 × 130 = 27 300 m² = 27 300 ÷ 10 000 = 2.73 ha

Density = 41 ÷ 2.73 = 15 nests per hectare

The wood is unlikely to be an exact rectangle, and some nests high in the trees may have been missed, so the true density could be different.

Explanation

A scale of 1 : 2500 means that every length on the map is 2500 times larger in reality, so convert each length before multiplying them. Multiplying the map area by 2500 is a common error: areas scale by 2500², not 2500. Then convert m² to hectares by dividing by 10 000.

Mark scheme

1 mark
Correct method: (8.4 × 2500 = 21 000 cm = 210 m) × (5.2 × 2500 = 13 000 cm = 130 m) = 27 300 m² ÷ 10 000; or a correct final answer of 2.73 ha with no working
Max 4. A correct answer of 2.73 ha (accept 2.7) with no working scores 2 marks (points 1 and 2). Accept for 1 mark: evidence of 210 m and 130 m OR 27 300 m². Do not award for answers in which only the area on the map is multiplied by 2500 (e.g. 0.0011 ha).
1 mark
2.73 (ha)
Accept 2.7.
1 mark
15 (nests per hectare)
Accept 15.0 or 15.2 (using 2.7 ha). ECF: 41 ÷ their area in ha. A correct answer with no working scores this point.
1 mark
The wood is not exactly a rectangle (so the area is not exact)
Points 4 to 6: max 1.
1 mark
Some nests may have been missed (e.g. hidden high in trees) / old or unused nests may have been counted
1 mark
The lengths measured on the map are uncertain (a small error is multiplied by 2500) / the map may be out of date
Accept 'uncertainty of measuring with a ruler on the map'.

Question 4

5 marks

At rest, a greyhound has a heart rate of 90 beats per minute and a stroke volume of 32 cm³. Its total volume of blood is 2.4 dm³.

Calculate its cardiac output at rest, in cm³ per minute.

Calculate the time, in seconds, for its heart to pump a volume of blood equal to its total blood volume out of the left ventricle.

During a race, the greyhound's cardiac output increases greatly. Give two changes in the heart that cause this increase.

Model answer

Cardiac output = stroke volume × heart rate = 32 × 90 = 2880 cm³ per minute

Total blood volume = 2.4 dm³ = 2400 cm³
Time = 2400 ÷ 2880 = 0.833 minutes = 0.833 × 60 = 50 s

During the race, the heart rate increases and the stroke volume increases (the ventricles contract more strongly).

Explanation

Cardiac output = stroke volume × heart rate; students must recall this. Convert 2.4 dm³ to 2400 cm³ so that the units match, then divide the blood volume by the volume pumped per minute and convert minutes to seconds. Cardiac output can increase only by increasing heart rate, stroke volume or both.

Mark scheme

1 mark
2880 (cm³ per minute); a correct answer with no working scores this mark
Points 1 to 3 are calculations: correct answers of 2880 cm³ per minute and 50 s with no working score 3 marks. Accept 2.88 dm³ per minute if the unit is given. Do not accept 0.36 (stroke volume ÷ heart rate) for this point.
1 mark
Correct method: 2400 ÷ 2880 (= 0.833 minutes) × 60; or a correct final answer of 50 (s) with no working
ECF from their cardiac output: 2400 ÷ their value × 60. Accept for this point 0.83 minutes (time correct but not converted to seconds).
1 mark
50 (s)
1 mark
Heart rate increases
Max 2 for points 4 and 5. List rule: if more than two changes are given, each wrong or contradictory extra answer cancels one correct answer. Accept 'the SAN sends impulses more often'.
1 mark
Stroke volume increases / the ventricles contract more strongly
Accept 'more blood is pumped out with each beat'. Ignore 'the heart beats harder' unless linked to more blood per beat.

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