What is genetic variation and why is it important in the context of evolution?

Genetic variation refers to the differences in DNA sequences among individuals within a population. It is crucial for evolution as it provides the raw material for natural selection to act upon. In the context of Cambridge International A Levels Biology, understanding genetic variation helps explain how populations adapt to changing environments, leading to evolutionary changes over time.

How does mutation contribute to variation in a population?

Mutations are changes in the DNA sequence that can introduce new genetic variations within a population. These changes can be beneficial, neutral, or harmful, and they occur randomly. In the context of Selection and Evolution, mutations are a primary source of genetic diversity, which is essential for natural selection to drive evolutionary processes.

What is the difference between continuous and discontinuous variation?

Continuous variation refers to traits that show a range of phenotypes, such as height or skin color, and are usually influenced by multiple genes and environmental factors. Discontinuous variation, on the other hand, involves distinct categories, such as blood type, and is typically controlled by a single gene. Understanding these concepts is important in Cambridge International A Levels Biology to grasp how different traits are inherited and expressed.

How does sexual reproduction contribute to genetic variation?

Sexual reproduction contributes to genetic variation through the processes of meiosis and fertilization. During meiosis, homologous chromosomes undergo recombination and independent assortment, creating new combinations of alleles. Fertilization further increases variation by combining genetic material from two parents. This genetic diversity is vital for the adaptability and evolution of species, a key concept in Cambridge International A Levels Biology.

What role does genetic drift play in variation and evolution?

Genetic drift is a mechanism of evolution that involves random changes in allele frequencies within a population. Unlike natural selection, which is driven by environmental pressures, genetic drift occurs by chance and can lead to significant changes in small populations. It can reduce genetic variation and potentially lead to the fixation or loss of alleles, influencing the evolutionary trajectory of a population, an important aspect of the Cambridge International A Levels curriculum.

Why is "Giving 'eye colour' as an example of discontinuous variation" a common mistake in Variation?

Why it happens: Eye colour seems to come in discrete categories (blue, green, brown). How to avoid it: Eye colour is actually **continuous** — it is polygenic (multiple genes including OCA2/HERC2) and there is a continuum of shades. Use **ABO blood group, tongue rolling, or pea seed shape** as safe examples of discontinuous variation. Source: 9700 Examiner Reports 2023-2024

Why is "Dividing by n (rather than n − 1) when calculating standard deviation" a common mistake in Variation?

Why it happens: Confusion between population SD (divides by n) and sample SD (divides by n − 1). How to avoid it: Cambridge uses the **sample standard deviation** formula: s = √((x - x))^2n - 1. The denominator n - 1 corrects for the fact that the sample mean itself is an estimate. Source: 9700 Examiner Reports 2024

Why is "Confusing crossing over with independent assortment as sources of variation" a common mistake in Variation?

Why it happens: Both occur during meiosis I and both shuffle alleles. How to avoid it: **Crossing over** = exchange of segments **between non-sister chromatids** (within a bivalent), in prophase I. **Independent assortment** = random orientation of **whole bivalents** (or chromatid pairs) at the equator in metaphase I (and II). Both contribute to variation but by different mechanisms. Source: 9700 Examiner Reports 2023

Why is "Forgetting to state the null hypothesis when describing a t-test" a common mistake in Variation?

Why it happens: Candidates jump straight to calculations. How to avoid it: Always begin with: 'H₀: there is no significant difference between the two means.' This is a discrete mark in Paper 4 and Paper 5 mark schemes. Source: 9700 Examiner Reports 2024

Why is "Claiming that environmentally acquired traits contribute to evolution" a common mistake in Variation?

Why it happens: Confusion with Lamarck's discredited 'inheritance of acquired characteristics'. How to avoid it: Only **genetic** variation is heritable, so only genetic variation can drive evolution. A trait gained from the environment (e.g. tanned skin, learnt behaviour, muscle from exercise) **does not change the alleles** and is not passed to offspring. Source: 9700 Examiner Reports 2023-2024

Why is "Drawing a bell-curve histogram for discontinuous data, or a bar chart for continuous data" a common mistake in Variation?

Why it happens: Confusion about how to display each type of variation. How to avoid it: **Continuous variation → histogram with bars touching → bell-shaped normal distribution**. **Discontinuous variation → bar chart with separate bars** (because the categories are discrete, not a continuum). Source: 9700 Examiner Reports 2024

Selection and EvolutionCambridge International A Levels Biology (9700)

Variation

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Variation — Cambridge International A Level Biology 9700 Study Notes (2025-2027 syllabus)

Continuous vs discontinuous variation, their genetic (polygenic vs monogenic) and environmental causes, the sources of genetic variation (mutation, meiosis, random fertilisation), and the statistical tools (mean, SD, SE, t-test) used to analyse variation — foundation for selection and evolution in spec 17.1.

At a glance

Continuous variation = quantitative range, polygenic, environment-influenced, normal distribution (e.g. human height).
Discontinuous variation = discrete categories, monogenic, little environmental effect, bar chart (e.g. ABO blood group).
Phenotype = genotype + environment ( $P = G + E$ ). Only the genetic component is heritable.
Sources of genetic variation: mutation, crossing over, independent assortment (meiosis I and II), random fertilisation.
Standard deviation $s = \sqrt{\sum(x - \bar{x})^2 / (n-1)}$ measures spread around the mean.
Standard error $SE = s/\sqrt{n}$ — used to compare means.
t-test compares two means of normally distributed samples; $df = n_A + n_B - 2$ .
Mark scheme cue: 'Polygenic / normal distribution' for continuous; 'monogenic / discrete categories' for discontinuous.

What you’ll learn

Mapped to the Cambridge IGCSE 9700 syllabus (2025-2027).

17.1.1 — Explain that phenotypic variation is the result of genotype, environment, or a combination of the two ( $P = G + E$ ).
17.1.2 — Explain the differences between continuous and discontinuous variation in terms of their genetic basis and the type of distribution they produce.
17.1.3 — Describe the genetic basis of continuous variation (polygenic — many genes, each with small additive effect) and discontinuous variation (monogenic).
17.1.4 — Explain that genetic variation in a population arises from mutation, meiosis (crossing over and independent assortment) and random fertilisation in sexual reproduction.
17.1.5 — Use Student's t-test to compare the means of two samples drawn from normally distributed populations.
17.1.6 — Calculate the standard deviation $s$ and the standard error of the mean $SE$ for a sample of continuous data.

Continuous and discontinuous variation

Continuous = quantitative range (height); discontinuous = discrete categories (blood group).

Variation in a phenotype falls into two broad types.

Continuous variation. The trait varies along a quantitative range with all values possible between two extremes. When the frequency of individuals is plotted against the trait value, the result is a normal (bell-shaped) distribution — most individuals cluster near the mean, with progressively fewer at the extremes.

Examples of continuous variation: human height, body mass in mice, leaf length in hedge plants, milk yield in dairy cows, skin colour, fish body length, blood pressure.

Discontinuous variation. The trait falls into discrete, non-overlapping categories with no intermediate values. When frequency is plotted, the result is a bar chart with separate categories.

Examples of discontinuous variation: human ABO blood group (A / B / AB / O), tongue rolling (yes / no), pea seed shape (round / wrinkled), pea flower colour (purple / white), the rhesus factor (Rh⁺ / Rh⁻).

Feature	Continuous	Discontinuous
Distribution	Normal (bell curve)	Bar chart (discrete bars)
Number of categories	Infinite (range)	Few (2–5 typically)
Genetic basis	Polygenic (many genes)	Monogenic (one gene)
Environmental influence	Strong	Weak or none
Example	Human height	ABO blood group

Why the difference? Continuous traits are controlled by many genes, each contributing a small additive effect. The total phenotype is the sum of these small contributions, plus environmental modulation. Because there are many combinations and continuous environmental variation, the result is a continuum. Discontinuous traits are controlled by one or a few genes with strongly distinct alleles — there are only a few possible genotypes, so only a few phenotypes. The environment cannot fine-tune them.

Continuous: range, polygenic, normal distribution.
Discontinuous: categories, monogenic, bar chart.
Continuous strongly affected by environment; discontinuous mostly not.
Eye colour is continuous (polygenic) — common exam trap.

See the full worked example for variation →

Causes of variation: genetic vs environmental

Phenotype = genotype + environment. Only the genetic component is heritable.

The phenotype of an individual is the product of its genotype (alleles inherited) and its environment (conditions experienced during life):

$P = G + E$

Genetic variation arises from differences in the alleles each individual inherits. It is heritable — passed from parent to offspring via the gametes. Three main sources contribute (see next section):

Mutation — creates new alleles (the ultimate source of all variation).
Meiosis — shuffles existing alleles into new combinations via crossing over and independent assortment.
Random fertilisation — multiplies variation by random fusion of genetically different gametes.

Environmental variation arises from differences in growth conditions — nutrition, temperature, light, disease, exercise, maternal effects during pregnancy. The environment changes the phenotype of the individual but does not alter the genotype, so it is not inherited.

Most traits are both. Almost every continuous trait reflects both contributions. Human height is determined by many genes (so genotype sets the potential) and modulated by childhood nutrition and health (so environment determines whether that potential is reached). Identical twins (same genotype) raised together usually have very similar heights; raised in very different conditions they can differ by several centimetres.

Why this matters for evolution. Natural selection acts on phenotype, but only the genetic component of phenotypic variation can be passed on. Environmentally acquired traits — a sun tan, a developed muscle, learned behaviour — die with the individual. This was Darwin's key insight against Lamarck's earlier 'inheritance of acquired characteristics': only heritable variation can fuel evolution.

Plant case study — hydrangea flower colour. Hydrangea petal colour can be blue, purple or pink depending on soil pH (acidic → blue, neutral → purple, alkaline → pink). This is purely environmental variation: cuttings from the same plant grown in different soils produce different colours. Such environmental colour change cannot drive evolution.

$P = G + E$ : phenotype = genotype + environment.
Genetic variation is heritable; environmental variation is not.
Mutation creates new alleles; meiosis + fertilisation shuffle them.
Only the genetic component drives evolution.
Hydrangea petal colour — classic example of environmental variation.

See the full worked example for variation →

Sources of genetic variation

Mutation creates new alleles; meiosis and random fertilisation shuffle them into new combinations.

Genetic variation has four main sources.

1. Mutation. Random changes in DNA base sequence (point mutations, deletions, insertions, frameshifts) or in chromosome structure / number. Mutation is the ultimate source of all genetic variation — it is the only mechanism that creates new alleles. Most mutations are neutral or harmful, but rare beneficial mutations provide the raw material for adaptation.

2. Crossing over (meiosis I). In prophase I of meiosis, homologous chromosomes pair up to form bivalents. Non-sister chromatids of the two homologues exchange equivalent segments at points called chiasmata. The result is recombinant chromatids carrying new combinations of alleles — alleles previously on different chromosomes are now linked together on one chromatid, and vice versa.

3. Independent assortment (meiosis I and II).

In metaphase I, bivalents line up at the equator with the orientation of each bivalent independent of every other. With $n$ pairs of chromosomes, this gives $2^n$ possible combinations of maternal and paternal chromosomes in the daughter cells. In humans ( $n = 23$ ), this is $2^{23} \approx 8.4 \times 10^6$ combinations.
In metaphase II, sister chromatids — now different because of crossing over — separate, and they too assort independently. This adds further variation.

4. Random fertilisation. Although not strictly part of meiosis, the random fusion of any one of many genetically different sperm with any one of many genetically different eggs multiplies the variation enormously. The total combination of variation in human gametes plus random fertilisation makes every zygote (except identical twins) genetically unique.

Outcome. Every gamete and every zygote is genetically different. The population's gene pool therefore harbours enormous variation, providing the raw material on which natural selection — and other evolutionary forces such as drift — can act.

Mutation = ultimate source of new alleles.
Crossing over = exchange between non-sister chromatids in prophase I → new allele combinations.
Independent assortment = random orientation of bivalents (metaphase I) and chromatids (metaphase II) → $2^n$ combinations.
Random fertilisation = any sperm × any egg multiplies variation.
Every zygote is genetically unique.

See the full worked example for variation →

Statistics: standard deviation and standard error

SD = spread; SE = precision of the mean. SE = SD / √n.

Continuous variation requires statistics to summarise and compare populations. Cambridge 9700 expects familiarity with the mean, standard deviation, standard error of the mean and Student's t-test.

Mean. The arithmetic average: $\bar{x} = \frac{\sum x}{n}$

Standard deviation ( $s$ ). A measure of the spread of data around the mean: $s = \sqrt{\frac{\sum (x - \bar{x})^2}{n - 1}}$ The denominator $n - 1$ (not $n$ ) gives the sample SD — Cambridge always uses this version. In a normal distribution, about 68% of values lie within $\pm 1\,s$ of the mean, and about 95% within $\pm 2\,s$ .

Standard error of the mean ( $SE$ ). An estimate of how precisely the sample mean estimates the population mean: $SE = \frac{s}{\sqrt{n}}$ SE falls as $n$ increases — larger samples give more precise estimates. Error bars on graphs are usually $\pm 1\,SE$ or $\pm 2\,SE$ (the latter approximates the 95% confidence interval). If 95% CIs of two means do not overlap, the means are usually significantly different.

Worked example (5 leaf lengths: 5, 7, 8, 6, 9 cm).

Step	Calculation	Value
Mean	$(5+7+8+6+9)/5$	7.0 cm
Deviations $(x - \bar{x})$	$-2, 0, +1, -1, +2$	—
Squared deviations	$4, 0, 1, 1, 4$	sum = 10
Variance	$10 / (n - 1) = 10 / 4$	2.5
Standard deviation	$\sqrt{2.5}$	≈ 1.58 cm
Standard error	$1.58 / \sqrt{5}$	≈ 0.71 cm

The mean leaf length is therefore $7.0 \pm 0.71$ cm (mean ± SE), and about two-thirds of leaves lie in $7 \pm 1.58$ cm (mean ± SD).

Mean = sum / n.
SD = $\sqrt{\sum(x - \bar{x})^2 / (n - 1)}$ . Sample SD uses $n - 1$ .
SE = SD / $\sqrt{n}$ — precision of the mean.
Error bars ±2 SE ≈ 95% confidence interval.
Larger n → smaller SE → more precise mean.

See the full worked example for variation →

Student's t-test

Compares two means; reject H₀ if calculated t exceeds critical t at p = 0.05.

When to use the t-test. Use when comparing the means of two samples of continuous data, where each sample is drawn from a normally distributed population (e.g. leaf lengths from two trees, wing-spans from two field sites).

Step 1 — State the null hypothesis (H₀). 'There is no significant difference between the mean of sample A and the mean of sample B; any observed difference is due to chance.' Cambridge mark schemes always credit this explicitly.

Step 2 — Calculate t. $t = \frac{|\bar{x}_A - \bar{x}_B|}{\sqrt{\dfrac{s_A^2}{n_A} + \dfrac{s_B^2}{n_B}}}$ The numerator is the difference between the means; the denominator is the combined standard error. Large t means the means differ much more than the natural variation in the data.

Step 3 — Calculate degrees of freedom (df). $df = n_A + n_B - 2$ .

Step 4 — Look up the critical value of t in a statistical table at the chosen significance level (usually $p = 0.05$ , two-tailed). For example, at $df = 38$ , critical $t \approx 2.02$ .

Step 5 — Compare and conclude.

If calculated $t$ > critical $t$ : reject H₀. The means differ significantly at the 5% level (i.e. there is less than a 5% chance that the observed difference is due to chance alone).
If calculated $t$ < critical $t$ : fail to reject H₀. There is no significant difference at the 5% level.

Worked example. Site A: $\bar{x} = 5.2$ , $s = 0.4$ , $n = 20$ . Site B: $\bar{x} = 4.8$ , $s = 0.5$ , $n = 20$ .

$t = \frac{|5.2 - 4.8|}{\sqrt{\dfrac{0.16}{20} + \dfrac{0.25}{20}}} = \frac{0.4}{\sqrt{0.0205}} = \frac{0.4}{0.143} \approx 2.80$

$df = 38$ ; critical $t$ at $p = 0.05$ ≈ 2.02. Since 2.80 > 2.02, reject H₀: the means differ significantly at the 5% level.

Use for two means of normally distributed continuous data.
Always state H₀ first ('no significant difference').
$t$ = (difference of means) / (combined SE).
$df = n_A + n_B - 2$ .
Reject H₀ if calculated $t$ > critical $t$ at $p = 0.05$ .

See the full worked example for variation →

How it’s examined

Cambridge 9700 examines this on Paper 4 (A Level structured questions, 4–6 marks per part) and on Paper 5 (Practical PAE). Frequent questions: (a) distinguish continuous and discontinuous variation with named examples (5–6 marks); (b) explain why a named trait shows continuous variation (3–4 marks); (c) describe how meiosis generates variation (4–5 marks); (d) calculate SD or SE from raw data (3–4 marks); (e) describe how to use a t-test to compare two sample means (4–5 marks). Paper 5 frequently asks candidates to plan an investigation with appropriate controls and statistical analysis.

Sources: Cambridge International AS & A Level Biology 9700 syllabus (2025-2027); 9700 Examiner Reports 2022-2024; 9700/42 May/Jun 2024 question paper and mark scheme; 9700/42 Oct/Nov 2024 question paper and mark scheme; 9700/52 Practical Paper 5 (PAE) — statistical analysis tasks. Last reviewed 2026-05-12.

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Step-by-step worked examples — Variation

Step-by-step solutions to past-paper-style questions on variation, written exactly the way a tutor would explain them at the board.

1Distinguishing continuous and discontinuous variation (5 marks)
Extended• Adapted from 9700/42 May/Jun 2024• variation, continuous, discontinuous, Paper 4
Question
Distinguish between continuous and discontinuous variation, giving a named example of each. (5 marks)
Step-by-step solution
1. Step 1
  Continuous variation describes a phenotype that varies along a quantitative range between two extremes, with all values possible. Examples: human height, body mass in mice, leaf length in a hedge plant, milk yield in dairy cows.
2. Step 2
  Discontinuous variation describes a phenotype that falls into discrete, non-overlapping categories, with no intermediates. Examples: human ABO blood group (A / B / AB / O), tongue rolling (yes/no), pea seed shape (round/wrinkled).
3. Step 3
  Genetic basis. Continuous variation is polygenic — controlled by many genes, each adding a small effect. Discontinuous variation is usually monogenic (one gene) or controlled by very few genes.
4. Step 4
  Environmental influence. Continuous traits are strongly influenced by environment (e.g. nutrition affects height). Discontinuous traits show little or no environmental effect (e.g. blood group is fixed by genotype).
5. Step 5
  Distribution. Continuous variation gives a normal (bell-curve) distribution when frequency is plotted against the trait. Discontinuous variation gives a bar chart with distinct categories.
Answer
Continuous: quantitative range, polygenic, environment-influenced, normal distribution (e.g. human height). Discontinuous: discrete categories, monogenic, little/no environmental effect, bar-chart distribution (e.g. ABO blood group).
Examiner tip
Mark scheme: (1) continuous = range / quantitative + example; (2) discontinuous = discrete categories + example; (3) polygenic vs monogenic; (4) environment affects continuous more; (5) normal distribution vs bar chart. Examiner Reports flag candidates who give 'eye colour' as discontinuous — eye colour is actually continuous (polygenic).
2How meiosis generates genetic variation (5 marks)
Extended• Adapted from 9700/42 Oct/Nov 2024• meiosis, crossing over, independent assortment, Paper 4
Question
Describe how meiosis is a source of genetic variation in sexually reproducing organisms. (5 marks)
Step-by-step solution
1. Step 1
  Crossing over (prophase I). Homologous chromosomes pair up to form bivalents; non-sister chromatids exchange segments at chiasmata. This produces new combinations of alleles on each chromatid — different from either parental chromosome.
2. Step 2
  Independent assortment (metaphase I). Bivalents line up on the equator; each pair orientates independently of the others. With 23 pairs in humans, there are $2^{23} \approx 8.4 \times 10^6$ possible combinations of maternal and paternal chromosomes in each gamete.
3. Step 3
  Independent assortment of chromatids (metaphase II). After the first division, sister chromatids (now different because of crossing over) also assort independently when they separate in meiosis II.
4. Step 4
  Random fertilisation. Although fertilisation is not part of meiosis itself, the random fusion of any one of many genetically different sperm with any one of many genetically different eggs multiplies the variation enormously.
5. Step 5
  Outcome. Each gamete (and therefore each zygote) is genetically unique — providing the raw material on which natural selection can act.
Answer
Crossing over in prophase I → new allele combinations on chromatids. Independent assortment in metaphase I → 2^23 ≈ 8.4 × 10⁶ chromosome combinations. Independent assortment of chromatids in metaphase II adds more variation. Random fertilisation further multiplies variation. Each gamete is genetically unique.
Examiner tip
Mark scheme: (1) crossing over / chiasmata; (2) new allele combinations / recombinants; (3) independent assortment in metaphase I; (4) 2ⁿ combinations stated; (5) random fertilisation. Examiner Reports note candidates often confuse 'crossing over' with 'independent assortment'.
3Standard deviation calculation (4 marks)
Extended• standard deviation, statistics, Paper 4, Paper 5
Question
A student measures the lengths of five leaves from one plant: 5, 7, 8, 6, 9 cm. Calculate the mean and standard deviation. (4 marks)
Step-by-step solution
1. Step 1
  Mean. $\bar{x} = \dfrac{5 + 7 + 8 + 6 + 9}{5} = \dfrac{35}{5} = 7.0$ cm.
2. Step 2
  Deviations from the mean. $(x - \bar{x})$ : $5 - 7 = -2$ ; $7 - 7 = 0$ ; $8 - 7 = +1$ ; $6 - 7 = -1$ ; $9 - 7 = +2$ .
3. Step 3
  Squared deviations. $(x - \bar{x})^2$ : $4 + 0 + 1 + 1 + 4 = 10$ . Divide by $n - 1 = 4$ to give the sample variance: $10 / 4 = 2.5$ .
4. Step 4
  Standard deviation. $s = \sqrt{2.5} \approx 1.58$ cm. So the leaf-length data are $7.0 \pm 1.58$ cm — about two-thirds of the sample lies within $\pm 1$ SD of the mean.
Answer
Mean = 7.0 cm. Deviations squared sum to 10; sample variance = 10/(n−1) = 10/4 = 2.5; SD = √2.5 ≈ 1.58 cm.
Examiner tip
Mark scheme: (1) mean = 7.0; (2) sum of squared deviations = 10; (3) divide by (n − 1) = 4; (4) SD ≈ 1.58 cm. Examiner Reports flag candidates who divide by n instead of (n − 1) — that gives the population SD; Cambridge wants the sample SD.
4Applying the t-test (5 marks)
Extended• Adapted from 9700/52 May/Jun 2024• t-test, statistics, Paper 5
Question
Two student groups measured the wing-spans of a fly population at two field sites. Site A: mean = 5.2 mm, s = 0.4, n = 20. Site B: mean = 4.8 mm, s = 0.5, n = 20. Describe how a t-test would be used to determine whether the means differ significantly. (5 marks)
Step-by-step solution
1. Step 1
  State the null hypothesis (H₀). 'There is no significant difference between the mean wing-span at Site A and at Site B.' Any difference observed is due to chance.
2. Step 2
  Check assumptions. Both samples must be normally distributed (continuous data — wing-span is) and independent. With $n \geq 20$ , the t-test is appropriate.
3. Step 3
  Calculate t. $t = \dfrac{|\bar{x}_A - \bar{x}_B|}{\sqrt{\dfrac{s_A^2}{n_A} + \dfrac{s_B^2}{n_B}}} = \dfrac{|5.2 - 4.8|}{\sqrt{\dfrac{0.16}{20} + \dfrac{0.25}{20}}} = \dfrac{0.4}{\sqrt{0.0205}} = \dfrac{0.4}{0.143} \approx 2.80$ .
4. Step 4
  Find degrees of freedom and critical t. $df = n_A + n_B - 2 = 38$ . At $p = 0.05$ (two-tailed), the critical value of t is ≈ 2.02.
5. Step 5
  Conclusion. Calculated $t$ (2.80) exceeds the critical t (2.02), so we reject H₀. There is a significant difference (p < 0.05) between the mean wing-spans — the probability that the observed difference is due to chance alone is less than 5%.
Answer
Null hypothesis: no significant difference between means. Calculate t = |x̄_A − x̄_B| / √(s_A²/n_A + s_B²/n_B) ≈ 2.80. df = 38; critical t at p = 0.05 ≈ 2.02. 2.80 > 2.02 so reject H₀ — significant difference at the 5% level.
Examiner tip
Mark scheme: (1) null hypothesis; (2) formula stated / values plugged in; (3) df = n_A + n_B − 2; (4) critical t looked up at p = 0.05; (5) compare and conclude (reject/accept H₀). Examiner Reports note candidates often forget to state the null hypothesis or quote p = 0.05.
5Genetic vs environmental variation (4 marks)
Extended• phenotype, genotype, environment, Paper 4
Question
Explain the difference between genetic variation and environmental variation, using human height to illustrate your answer. (4 marks)
Step-by-step solution
1. Step 1
  Phenotype equation. Phenotype = genotype + environment ( $P = G + E$ ). Variation in a phenotype can therefore be split into a genetic component (variation in the alleles inherited) and an environmental component (variation in growth conditions).
2. Step 2
  Genetic variation in height. Human height is polygenic — many genes (each with multiple alleles) each contribute a small additive effect to height. Identical twins (same genotype) show that genetic factors set the height potential.
3. Step 3
  Environmental variation in height. Nutrition (especially protein and calcium intake in childhood), illness, and even maternal health during pregnancy affect realised adult height. Two genetically identical twins raised in different conditions can finish at different heights.
4. Step 4
  Selection. Only the genetic component can be passed to offspring — and only this component can drive evolutionary change. The environmental component changes phenotype within a generation but is not heritable.
Answer
Phenotype = genotype + environment. Genetic variation = differences in inherited alleles (heritable); polygenic for height. Environmental variation = differences in growth conditions (e.g. nutrition); not heritable. Only the genetic component is selected upon in evolution.
Examiner tip
Mark scheme: (1) P = G + E or equivalent; (2) genetic = polygenic / heritable; (3) environment = e.g. nutrition / non-heritable; (4) only genetic component contributes to evolution. Examiner Reports note that asking 'why are siblings different heights?' usually wants BOTH causes — different gene combinations AND different environments.

Model Answers — Variation

High-scoring sample answers for variation on the Cambridge International A Level 9700 paper, with examiner-style notes mapping each response to the mark scheme and assessment objectives.

Question
9700/42 May/Jun 2024 Q3 (adapted)5 marks
Q (5 marks). Distinguish between continuous and discontinuous variation, using named examples to illustrate your answer.
Model answer
Continuous variation describes a phenotype that varies along a quantitative range between two extremes, with intermediate values possible at any point. Examples include human height, leaf length in a hedge plant, body mass in mice and milk yield in cattle. When the frequency of individuals is plotted against the trait, a normal (bell-shaped) distribution is obtained.

Discontinuous variation describes a phenotype that falls into discrete, non-overlapping categories, with no intermediates. Examples include human ABO blood group (only A, B, AB or O — no intermediates), tongue rolling (yes or no) and pea seed shape (round or wrinkled). When the frequency of individuals is plotted, a bar chart with separate categories is obtained.

Genetic basis. Continuous variation is normally polygenic — controlled by many genes (often 10 or more), each contributing a small additive effect. Discontinuous variation is usually monogenic or controlled by very few genes; one gene with two or three alleles is enough to produce the small number of categories.

Environmental influence. Continuous traits are strongly influenced by the environment (e.g. nutrition affects height; light affects leaf length). Discontinuous traits are usually unaffected by environment — ABO blood group, for instance, is fixed by genotype alone.
Why this scores
Why this scores 5/5. Mark scheme: (1) continuous = range / quantitative; (2) discontinuous = discrete categories; (3) named example of each; (4) polygenic vs monogenic; (5) environment affects continuous more. Examiner Reports stress the importance of naming specific examples — vague terms like "size" or "colour" are not credited.
Question
9700/42 Oct/Nov 2024 Q3(b) (adapted)4 marks
Q (4 marks). Explain why human height shows continuous variation.
Model answer
Human height is determined by many different genes (a polygenic trait), each of which has two or more alleles. The total height is the sum of the small effects of each contributing allele — so the more 'tall' alleles a person inherits, the taller they tend to be. With many independently inherited genes, the possible combinations are very numerous, producing a near-continuous range of heights.

In addition, height is strongly affected by the environment during development, particularly nutrition (protein and calcium availability), childhood illness and even maternal health during pregnancy. Two people with identical genotypes (e.g. monozygotic twins) raised in different conditions may finish at slightly different heights.

Because both the genetic component (many gene combinations) and the environmental component vary continuously, the phenotype itself varies continuously. When the heights of a population are plotted, a normal (bell-curve) distribution is observed, with most people clustered near the mean and progressively fewer at the extremes.
Why this scores
Why this scores 4/4. Mark scheme: (1) polygenic / many genes contribute; (2) each gene has small additive effect; (3) environmental influence (nutrition); (4) normal distribution / bell curve. Examiner Reports flag candidates who just write 'genes and environment' without explaining the polygenic mechanism.
Question
9700/42 May/Jun 2024 Q4 (adapted)5 marks
Q (5 marks). Describe how meiosis is a source of genetic variation between gametes.
Model answer
Meiosis generates genetic variation between gametes by two main mechanisms: crossing over and independent assortment. A third mechanism — random fertilisation — is not part of meiosis itself but multiplies the variation when gametes fuse.

1. Crossing over (prophase I). Homologous chromosomes pair up to form bivalents. Non-sister chromatids of the two homologues come into contact at points called chiasmata and exchange equivalent segments. This produces recombinant chromatids carrying new combinations of alleles — alleles that were previously on different chromosomes are now linked together on one chromatid, and vice versa.

2. Independent assortment of homologous chromosomes (metaphase I). Bivalents line up at the equator with the orientation of each bivalent independent of all the others. In humans (23 pairs), this gives $2^{23} \approx 8.4 \times 10^6$ different combinations of maternal and paternal chromosomes in each daughter cell.

3. Independent assortment of chromatids (metaphase II). After meiosis I, sister chromatids — now different from each other because of crossing over — also separate independently. This adds a further multiplicative factor to the variation.

4. Random fertilisation. Any one of many genetically different sperm may fuse with any one of many genetically different eggs. Combining this with the variation already generated by meiosis means each zygote is essentially unique.

Together these mechanisms ensure that, although the chromosome number is conserved, the allele combinations in offspring are different from the parents and from siblings — providing the raw material on which natural selection can act.
Why this scores
Why this scores 5/5. Mark scheme: (1) crossing over / chiasmata in prophase I; (2) recombinant chromatids / new allele combinations; (3) independent assortment in metaphase I; (4) 2ⁿ combinations (with figure or formula); (5) reference to random fertilisation OR independent assortment in meiosis II. Examiner Reports note candidates often confuse crossing over (within bivalents) with independent assortment (between bivalents).
Question
9700/42 Oct/Nov 2024 Q4(c) (adapted)5 marks
Q (5 marks). Describe how a t-test could be used to determine whether the mean leaf length of plants growing in two different soils is significantly different.
Model answer
The t-test is used to compare the means of two samples drawn from normally distributed populations of continuous data — leaf length fits these requirements.

Step 1 — State the null hypothesis (H₀). There is no significant difference between the mean leaf length of plants growing in Soil A and Soil B; any observed difference is due to chance alone.

Step 2 — Collect random samples. Measure the lengths of a sufficient number of leaves (typically $n \geq 20$ ) from each group. Calculate the mean $\bar{x}$ and sample standard deviation $s$ for each group.

Step 3 — Calculate the test statistic. $t = \dfrac{|\bar{x}_A - \bar{x}_B|}{\sqrt{\dfrac{s_A^2}{n_A} + \dfrac{s_B^2}{n_B}}}$ . The numerator is the difference between means; the denominator is the combined standard error.

Step 4 — Determine degrees of freedom. $df = n_A + n_B - 2$ . Look up the critical value of t in a statistical table at the chosen significance level (conventionally $p = 0.05$ , i.e. 5%).

Step 5 — Compare and conclude. If the calculated $t$ is greater than the critical value, the probability of obtaining this difference by chance alone is less than 5%, so we reject H₀ and accept that the means differ significantly. If $t$ is less than the critical value, we fail to reject H₀ and conclude there is no significant difference at this level.
Why this scores
Why this scores 5/5. Mark scheme: (1) null hypothesis stated; (2) sample large enough / take random samples; (3) formula for t (or 'difference / SE'); (4) df = n_A + n_B − 2 / refer to table at p = 0.05; (5) compare and conclude (reject / accept H₀). Examiner Reports note candidates often write 't = (x̄ − x̄) / s' — missing the standard error denominator entirely.
Question
Practice question — recurrent Paper 4 style4 marks
Q (4 marks). Explain the difference between genetic variation and environmental variation, and explain why only genetic variation contributes to evolution.
Model answer
Genetic variation arises from differences in the alleles inherited from parents. It is generated by mutation, meiosis (crossing over + independent assortment) and random fertilisation, and is passed from parent to offspring in the gametes.

Environmental variation arises from differences in the conditions an organism experiences during its life — nutrition, temperature, light, disease, exercise and so on. It changes the phenotype of the individual but does not alter the genotype, so it is not inherited.

The overall phenotype is the sum of both: $P = G + E$ . Two individuals with the same genotype can have different phenotypes if raised in different environments (e.g. monozygotic twins of different heights due to different diets), and two individuals with different genotypes can have the same phenotype if environmental effects mask the genetic difference.

Only genetic variation contributes to evolution because natural selection requires heritability. If an advantageous trait is purely environmental, it dies with the individual and cannot be passed on; if it is genetic, the allele can be passed to offspring and its frequency can increase in the population over generations. Environmentally acquired traits (Lamarck's discredited inheritance of acquired characteristics) are not heritable.
Why this scores
Why this scores 4/4. Mark scheme: (1) genetic = differences in alleles / heritable; (2) sources of genetic variation (mutation / meiosis); (3) environmental = non-heritable / e.g. nutrition; (4) only genetic component selected upon / heritability needed for evolution. Examiner Reports flag candidates who claim environmental adaptations (e.g. tanned skin from sun exposure) drive evolution — they don't.

Key Definitions and Keywords — Variation

Definitions to memorise and the exact keywords mark schemes credit for variation answers — sharpened from recent examiner reports for the 2026 Cambridge International A Level 9700 sitting.

Variation
Examiner keyword
Differences in the characteristics (phenotypes) of individuals within or between populations of the same species.
Continuous variation
Examiner keyword
Variation in a phenotype along a quantitative range between two extremes, with intermediate values possible. Typically polygenic and influenced by the environment. Distribution is normal (bell-shaped). Example: human height.
Discontinuous variation
Examiner keyword
Variation in a phenotype that falls into discrete, non-overlapping categories with no intermediates. Usually monogenic and little affected by environment. Distribution is a bar chart. Example: human ABO blood group.
Polygenic inheritance
Examiner keyword
Inheritance in which a phenotype is controlled by many genes, each contributing a small additive effect. Underlies continuous variation. Example: human skin colour, height, milk yield.
Monogenic inheritance
Inheritance in which a phenotype is controlled by a single gene (with two or more alleles). Underlies most discontinuous variation. Example: ABO blood group, pea seed shape.
Genetic variation
Examiner keyword
Variation in phenotype that is caused by differences in inherited alleles. Heritable. Sources: mutation, meiosis (crossing over, independent assortment), random fertilisation.
Environmental variation
Examiner keyword
Variation in phenotype that is caused by differences in growth conditions (e.g. nutrition, temperature, light) rather than differences in alleles. Not heritable.
Mutation
Examiner keyword
A random change in the base sequence of DNA (point mutation, deletion, insertion) or in chromosome structure / number. The ultimate source of new alleles and therefore of all genetic variation.
Crossing over
Examiner keyword
The exchange of segments between non-sister chromatids of homologous chromosomes at chiasmata during prophase I of meiosis. Produces new allele combinations on recombinant chromatids.
Independent assortment
Examiner keyword
The orientation of each bivalent at metaphase I (and each pair of chromatids at metaphase II) is independent of every other. Generates $2^n$ chromosome combinations in gametes ( $n$ = number of pairs).
Standard deviation (s)
Examiner keyword
A measure of the spread of data around the mean: $s = \sqrt{\dfrac{\sum(x - \bar{x})^2}{n - 1}}$ . About 68% of normally distributed data lies within $\pm 1\,s$ of the mean.
Standard error of the mean (SE)
Examiner keyword
An estimate of how precisely the sample mean estimates the population mean: $SE = \dfrac{s}{\sqrt{n}}$ . Used to compare means and to plot error bars (mean ± 2 SE ≈ 95% confidence interval).
t-test
Examiner keyword
A statistical test that compares the means of two normally distributed samples to decide whether they are significantly different. The calculated t is compared with a critical value at $p = 0.05$ with $df = n_A + n_B - 2$ .
Null hypothesis (H₀)
Examiner keyword
A statement that there is no significant difference (or no association) between the variables being tested. The statistical test seeks to reject (or fail to reject) H₀.
Normal distribution
A symmetrical, bell-shaped probability distribution centred on the mean. Most natural continuous variables (e.g. human height) approximate a normal distribution.

Common Mistakes and Misconceptions — Variation

The traps other students keep falling into on variation questions — taken from recent Cambridge International A Level 9700 examiner reports and mark schemes — and how to avoid them.

✕Giving 'eye colour' as an example of discontinuous variation
9700 Examiner Reports 2023-2024
Why it happens
Eye colour seems to come in discrete categories (blue, green, brown).
How to avoid it
Eye colour is actually continuous — it is polygenic (multiple genes including OCA2/HERC2) and there is a continuum of shades. Use ABO blood group, tongue rolling, or pea seed shape as safe examples of discontinuous variation.
✕Dividing by n (rather than n − 1) when calculating standard deviation
9700 Examiner Reports 2024
Why it happens
Confusion between population SD (divides by n) and sample SD (divides by n − 1).
How to avoid it
Cambridge uses the sample standard deviation formula: $s = \sqrt{\dfrac{\sum(x - \bar{x})^2}{n - 1}}$ . The denominator $n - 1$ corrects for the fact that the sample mean itself is an estimate.
✕Confusing crossing over with independent assortment as sources of variation
9700 Examiner Reports 2023
Why it happens
Both occur during meiosis I and both shuffle alleles.
How to avoid it
Crossing over = exchange of segments between non-sister chromatids (within a bivalent), in prophase I. Independent assortment = random orientation of whole bivalents (or chromatid pairs) at the equator in metaphase I (and II). Both contribute to variation but by different mechanisms.
✕Forgetting to state the null hypothesis when describing a t-test
9700 Examiner Reports 2024
Why it happens
Candidates jump straight to calculations.
How to avoid it
Always begin with: 'H₀: there is no significant difference between the two means.' This is a discrete mark in Paper 4 and Paper 5 mark schemes.
✕Claiming that environmentally acquired traits contribute to evolution
9700 Examiner Reports 2023-2024
Why it happens
Confusion with Lamarck's discredited 'inheritance of acquired characteristics'.
How to avoid it
Only genetic variation is heritable, so only genetic variation can drive evolution. A trait gained from the environment (e.g. tanned skin, learnt behaviour, muscle from exercise) does not change the alleles and is not passed to offspring.
✕Drawing a bell-curve histogram for discontinuous data, or a bar chart for continuous data
9700 Examiner Reports 2024
Why it happens
Confusion about how to display each type of variation.
How to avoid it
Continuous variation → histogram with bars touching → bell-shaped normal distribution. Discontinuous variation → bar chart with separate bars (because the categories are discrete, not a continuum).

Practice questions

Exam-style questions with step-by-step worked solutions. Try one before checking the method.

Past paper style quiz

Get a report showing which sub-topics you've nailed and which ones still need work.

Variation — frequently asked questions

The things students keep getting wrong in this sub-topic, answered.

Variation

Short Study Notes in the form of Slides

Detailed Study Notes

What you’ll learn

How it’s examined

1Distinguishing continuous and discontinuous variation (5 marks)

2How meiosis generates genetic variation (5 marks)

3Standard deviation calculation (4 marks)

4Applying the t-test (5 marks)

5Genetic vs environmental variation (4 marks)

Variation

Continuous variation

Discontinuous variation

Polygenic inheritance

Monogenic inheritance

Genetic variation

Environmental variation

Mutation

Crossing over

Independent assortment

Standard deviation (s)

Standard error of the mean (SE)

t-test

Null hypothesis (H₀)

Normal distribution

✕Giving 'eye colour' as an example of discontinuous variation

✕Dividing by n (rather than n − 1) when calculating standard deviation

✕Confusing crossing over with independent assortment as sources of variation

✕Forgetting to state the null hypothesis when describing a t-test

✕Claiming that environmentally acquired traits contribute to evolution

✕Drawing a bell-curve histogram for discontinuous data, or a bar chart for continuous data

Practice questions

Past paper style quiz

Assess your understanding

Variation — frequently asked questions