Variation

1. Overview

Members of a species are not identical: they show phenotypic variation in their observable characteristics. This variation comes from genetic factors, from environmental factors, or from a combination of the two. Variation is classified into two patterns: discontinuous variation, where individuals fall into a few distinct categories, and continuous variation, where individuals show a complete range of values. These patterns reflect their genetic basis — a small number of genes for discontinuous variation, and many genes (polygenic inheritance) plus the environment for continuous variation. Because continuous data vary, biologists use the t-test to decide whether the difference between the means of two samples is statistically significant or just due to chance.

Key Definitions

• Phenotypic variation: the differences in observable characteristics (the phenotype) between individuals of the same species.
• Genetic variation: differences in phenotype caused by differences in genotype, arising from mutation, meiosis and random fertilisation.
• Environmental variation: differences in phenotype caused by the conditions in which an organism develops and lives, with no change to the genotype.
• Discontinuous variation: variation in which individuals fall into a small number of distinct, non-overlapping categories with no intermediates.
• Continuous variation: variation in which individuals show a complete range of values between two extremes, with no distinct categories.
• Polygenic inheritance: the control of a single characteristic by many genes at different loci, each making a small additive contribution.
• t-test: a statistical test used to decide whether the difference between the means of two samples of normally distributed continuous data is significant.
• Null hypothesis: the statement that there is no significant difference between the two sample means and that any observed difference is due to chance.

Content

Causes of phenotypic variation

The phenotype of an organism is determined by its genotype, by the environment, and by the interaction between them. This is often summarised as a memory aid: $$\text{phenotype} = \text{genotype} + \text{environment}$$

This is not a literal sum — it simply reminds you that the genetic and environmental contributions combine to produce the final phenotype.

Genetic factors produce inherited differences. The main sources are:

• Mutation — random changes to genes (new alleles) and chromosomes; this is the only source of completely new alleles.
• Meiosis — reshuffles existing alleles into new combinations (see the two distinct mechanisms below).
• Random fertilisation — any of many genetically different gametes can fuse, so offspring receive a novel mixture of alleles from two parents.

Meiosis itself produces variation in two separate ways that are easy to mix up, so keep them clearly apart:

Mechanism	What happens	When
Independent assortment	Homologous pairs line up randomly at the equator, so maternal and paternal chromosomes are distributed into gametes in random combinations	Metaphase I
Crossing over	Non-sister chromatids of a homologous pair exchange sections, swapping alleles between them	Prophase I

Together, meiosis and random fertilisation generate variety by recombining existing alleles, while mutation supplies the new alleles in the first place. Variation from genetic factors is inherited and passed to offspring.

Environmental factors affect the phenotype without changing the genotype, so these effects are not inherited. Examples include a plant grown in low light producing less chlorophyll and appearing pale, or a person's body mass changing with diet. Environmental variation cannot be passed on through gametes.

Most variation results from a combination of both. For example, human height has a strong genetic component but is also influenced by childhood nutrition; the colour of a hydrangea flower depends on its alleles but also on soil pH. A useful way to remember the distinction: genetic variation sets the potential range, and the environment determines where within that range the phenotype falls.

Discontinuous variation

In discontinuous variation, individuals fall into a few distinct, non-overlapping categories with no intermediates. Examples include human ABO blood groups (A, B, AB or O) and the ability or inability to roll the tongue. When plotted, the data form a bar chart of separate categories rather than a smooth curve.

Discontinuous variation is controlled by one or a small number of genes. Each genotype produces a clearly different phenotype, and the environment has little or no effect. Because only genetic factors are involved, the categories stay sharp and distinct.

Continuous variation

In continuous variation, individuals show a complete range of values between two extremes, with no distinct categories. Examples include height, body mass and leaf length. When the data are grouped, they typically form a normal distribution (a symmetrical bell-shaped curve), as the graph below shows, usually drawn as a histogram.

Continuous variation is controlled by many genes at different loci — this is polygenic inheritance. Each gene makes a small additive contribution to the characteristic, so the more "value-increasing" alleles an individual inherits, the larger the value of the feature. Because so many genotype combinations are possible, the phenotypes blend into a continuous range rather than separating into categories. In addition, continuous characteristics are usually influenced by the environment, which further smooths the variation into a continuous spread.

Comparing the two types

Feature	Discontinuous variation	Continuous variation
Categories	a few distinct, non-overlapping types	a complete range of values
Genetic control	one or few genes	many genes (polygenic)
Environmental effect	little or none	usually significant
Typical graph	bar chart of separate categories	histogram / normal distribution curve
Example	ABO blood group	height

Using the t-test to compare two means

When you have two samples of continuous data and want to know whether their means are genuinely different, you use the t-test. The test applies when the data are continuous, approximately normally distributed, and the samples are drawn at random.

Begin by stating a null hypothesis: that there is no significant difference between the two means and any difference is due to chance. Calculate the value of $t$, then compare it with the critical value at the 0.05 (5%) probability level for the correct degrees of freedom.

• If the calculated $t$ is greater than or equal to the critical value, the probability that the difference is due to chance is less than 5%, so you reject the null hypothesis — the difference is significant.
• If the calculated $t$ is less than the critical value, you accept (do not reject) the null hypothesis — the difference is not significant.

The degrees of freedom are calculated from the two sample sizes ($n_1$ and $n_2$) as $df = n_1 + n_2 - 2$.

Worked example

Exam-style question: A student measured the wing length (in mm) of mayflies from two separate ponds. Pond 1 gave a mean of 13.0 mm and Pond 2 gave a mean of 17.0 mm; each variance ($s^2$) was 2.50 mm² and each sample contained 5 insects. Using the formula provided, calculate $t$, state the degrees of freedom, and use the critical value of 2.31 (at $p = 0.05$) to decide whether the difference in means is significant. [4]

Model answer:

• State the null hypothesis: there is no significant difference between the mean wing lengths of the two populations.
• Substitute into the formula: $$t = \frac{|\,\bar{x}_1 - \bar{x}_2\,|}{\sqrt{\dfrac{s_1^{\,2}}{n_1} + \dfrac{s_2^{\,2}}{n_2}}} = \frac{|13.0 - 17.0|}{\sqrt{\dfrac{2.50}{5} + \dfrac{2.50}{5}}} = \frac{4.0}{\sqrt{0.50 + 0.50}} = \frac{4.0}{1.0} = 4.00$$
• Degrees of freedom: $df = n_1 + n_2 - 2 = 5 + 5 - 2 = 8$.
• Conclusion: the calculated $t$ (4.00) is greater than the critical value (2.31), so the probability that the difference is due to chance is less than 5%. We reject the null hypothesis: the difference between the two mean wing lengths is significant.

Worked example

Exam-style question: A student measured the leaf length (in cm) of a plant species growing on two hillsides. The raw readings were:

	Leaf lengths (cm)
Site A	6, 7, 8, 9
Site B	7, 8, 9, 10

Using the formula provided, calculate the mean and variance for each site, work out $t$, state the degrees of freedom, and use the critical value of 2.45 (at $p = 0.05$) to decide whether the difference in means is significant. [5]

Model answer:

• State the null hypothesis: there is no significant difference between the mean leaf lengths at the two sites.
• Calculate the means: $\bar{x}_A = \frac{6+7+8+9}{4} = 7.5$ cm and $\bar{x}_B = \frac{7+8+9+10}{4} = 8.5$ cm.
• Calculate each variance using $s^2 = \dfrac{\sum (x - \bar{x})^2}{n - 1}$. For both sites the deviations from the mean are $\pm 1.5$ and $\pm 0.5$, so $\sum (x - \bar{x})^2 = 1.5^2 + 0.5^2 + 0.5^2 + 1.5^2 = 5.0$, giving $s^2 = \dfrac{5.0}{4-1} = 1.67$ cm² for each site.
• Substitute into the formula: $$t = \frac{|7.5 - 8.5|}{\sqrt{\dfrac{1.67}{4} + \dfrac{1.67}{4}}} = \frac{1.0}{\sqrt{0.417 + 0.417}} = \frac{1.0}{0.913} = 1.10$$
• Degrees of freedom: $df = n_1 + n_2 - 2 = 4 + 4 - 2 = 6$.
• Conclusion: the calculated $t$ (1.10) is less than the critical value at 6 degrees of freedom (2.45), so the probability that the difference is due to chance is greater than 5%. We accept (do not reject) the null hypothesis: the difference between the two mean leaf lengths is not significant.

Key Equations

The t-test compares two sample means $\bar{x}_1$ and $\bar{x}_2$ with sample variances $s_1^{\,2}$ and $s_2^{\,2}$ and sample sizes $n_1$ and $n_2$ (formula provided in the exam): $$t = \frac{|\,\bar{x}_1 - \bar{x}_2\,|}{\sqrt{\dfrac{s_1^{\,2}}{n_1} + \dfrac{s_2^{\,2}}{n_2}}}$$

Sample variance (needed when you are given raw data rather than $s^2$): $$s^2 = \frac{\sum (x - \bar{x})^2}{n - 1}$$

Degrees of freedom: $$df = n_1 + n_2 - 2$$

Common Mistakes to Avoid

• Mixing up independent assortment and crossing over. Use independent assortment for the random way homologous pairs line up at the equator in metaphase I; use crossing over for the exchange of alleles between non-sister chromatids. They are two separate sources of genetic variation.
• Saying the environment or selection "causes" mutations. Mutations are random events; the environment does not direct them. Selection pressures only decide which existing alleles give an advantage, raising or lowering their frequency.
• Describing low genetic variation as "alleles with low frequency". Low genetic variation means a small number of different alleles present at a gene locus in the population, regardless of how common each one is.
• Treating continuous variation as purely genetic. Continuous variation is polygenic and strongly influenced by the environment; mention both when explaining its genetic basis.
• Confusing the graph types. Discontinuous data give a bar chart of separate categories; continuous data give a histogram / normal distribution.
• Forgetting the null hypothesis or a clear conclusion in the t-test. Always state the null hypothesis first, then compare $t$ with the critical value at $p = 0.05$, and end by clearly accepting or rejecting it.
• Comparing $t$ at the wrong degrees of freedom. Calculate $df = n_1 + n_2 - 2$ and read the critical value at that line of the table, not at a fixed value.

Exam Tips

• When asked to explain variation, name the specific source (mutation, meiosis with independent assortment / crossing over, or random fertilisation) rather than writing a vague statement about "differences in genes".
• Distinguish clearly between inherited (genetic) variation and non-inherited (environmental) variation — environmental effects are not passed to offspring because the genotype is unchanged.
• In t-test calculations, show every step: null hypothesis, substitution, the value of $t$, the degrees of freedom, and the comparison with the critical value. Marks are awarded for the method even if an arithmetic slip occurs.
• State conclusions in the language of probability — say the difference is "significant at $p = 0.05$", not just "the populations are different".
• For an "explain the genetic basis" question on continuous variation, the key marking points are many genes (polygenic), additive small effects, and environmental influence.
• Quote means and measurements with their units and to a sensible number of decimal places when answering data-handling questions.