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Which mathematical skills are tested across the Eduqas Biology papers, and how do I apply them?

Mathematical skills: magnification and scale; surface-area-to-volume ratio; percentages and percentage change; standard form and units; the index of diversity; the Hardy-Weinberg equation; and rates from graphs.

A focused answer to the mathematical-skills requirements of Eduqas A-Level Biology. Covers magnification and scale, surface-area-to-volume ratio, percentages and percentage change, standard form and units, the index of diversity, the Hardy-Weinberg equation, and finding rates from graphs.

Generated by Claude Opus 4.813 min answer

Reviewed by: AI editorial process; not yet individually human-reviewed

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  1. What this dot point is asking
  2. Magnification, scale and units
  3. Surface-area-to-volume ratio
  4. Percentages and percentage change
  5. The index of diversity and Hardy-Weinberg
  6. Rates from graphs
  7. Examples in context
  8. Try this

What this dot point is asking

At least 10 percent of the Eduqas Biology marks assess mathematical skills at Level 2 (GCSE higher and above). You must handle magnification, ratios, percentages, standard form, the index of diversity, the Hardy-Weinberg equation and rates from graphs. This page drills the methods that recur across all three papers.

Magnification, scale and units

The magnification triangle links three quantities:

magnification=image sizeactual size\text{magnification} = \dfrac{\text{image size}}{\text{actual size}}

so actual size=image sizemagnification\text{actual size} = \dfrac{\text{image size}}{\text{magnification}}. Always convert to the same units first. Conversions go in steps of 1000: 1 mm=10001 \text{ mm} = 1000 micrometres and 11 micrometre =1000= 1000 nanometres. Standard form is essential for these small sizes, for example 55 micrometres =5×10−6= 5 \times 10^{-6} m.

Surface-area-to-volume ratio

As an object gets bigger, its volume grows faster than its surface area, so the SA:V ratio falls. For a sphere of radius rr:

SA=4πr2V=43πr3SAV=3r\text{SA} = 4\pi r^2 \qquad V = \tfrac{4}{3}\pi r^3 \qquad \dfrac{\text{SA}}{V} = \dfrac{3}{r}

This is why small organisms can rely on diffusion but large ones need exchange surfaces.

Percentages and percentage change

A percentage is a fraction out of 100. Percentage change is:

percentage change=change in valueoriginal value×100\text{percentage change} = \dfrac{\text{change in value}}{\text{original value}} \times 100

Always divide by the original value, not the final one. A positive answer is an increase, a negative answer a decrease.

The index of diversity and Hardy-Weinberg

Rates from graphs

A rate is how fast something changes, found as the gradient of a graph:

rate=change in ychange in x\text{rate} = \dfrac{\text{change in } y}{\text{change in } x}

For a curve, draw a tangent at the point of interest and find the gradient of the tangent. Watch the units (for example cm³ of oxygen per minute).

Examples in context

Example 1. Why a graph tangent is needed for an enzyme curve. Because the rate of an enzyme reaction changes over time (it slows as substrate runs out), a single gradient does not describe it; a tangent gives the rate at one instant, such as the initial rate.

Example 2. Hardy-Weinberg detecting evolution. If repeated sampling shows the genotype frequencies drifting away from the Hardy-Weinberg prediction, it is evidence that one of the assumptions has broken (for example selection), so the population is evolving.

Try this

Q1. Write the equation for percentage change. [1 mark]

  • Cue. change in valueoriginal value×100\dfrac{\text{change in value}}{\text{original value}} \times 100.

Q2. A cell image is 30 mm long at a magnification of times 1500. Calculate the actual length in micrometres. [2 marks]

  • Cue. 30 mm=30 00030 \text{ mm} = 30\,000 micrometres; 30 0001500=20\dfrac{30\,000}{1500} = 20 micrometres.

Q3. In a Hardy-Weinberg population, q2=0.04q^2 = 0.04. Calculate the frequency of the dominant allele pp. [2 marks]

  • Cue. q=0.04=0.2q = \sqrt{0.04} = 0.2; p=1−0.2=0.8p = 1 - 0.2 = 0.8.

Exam-style practice questions

Practice questions written in the style of WJEC Eduqas exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

Eduqas 20204 marksA spherical cell has a radius of 10 micrometres. Calculate its surface-area-to-volume ratio. (Surface area equals 4 pi r squared; volume equals four-thirds pi r cubed.)
Show worked answer →

Surface area equals 4 pi r squared equals 4 times pi times 10 squared equals 400 pi equals about 1257 square micrometres.

Volume equals four-thirds pi r cubed equals four-thirds times pi times 10 cubed equals about 4189 cubic micrometres.

Surface-area-to-volume ratio equals 1257 divided by 4189 equals about 0.3 (or 3 to 10).

A neat way is to note that for a sphere the ratio simplifies to 3 over r equals 3 over 10 equals 0.3.

Markers reward the correct surface area, the correct volume, and the ratio of about 0.3 (3 over r).

Eduqas 20213 marksThe mass of a seedling increased from 0.8 g to 1.4 g over one week. Calculate the percentage increase in mass.
Show worked answer →

The increase in mass is 1.4 minus 0.8 equals 0.6 g.

Percentage increase equals (increase divided by original) times 100 equals (0.6 divided by 0.8) times 100.

Equals 0.75 times 100 equals 75 percent.

Markers reward the change of 0.6 g, dividing by the original value (0.8), and the answer of 75 percent.

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