Variation and Selection

Lessons

Senior 6 · Biology

Variation and Selection

5 lessons

Lesson 2 of 5

Continuous and discontinuous variation, and the t-test

35 min

Is the difference real?

A pupil measures ten leaves from a shrub growing in full sun and ten from the same kind of shrub in shade. The shade leaves are longer on average, by almost a centimetre. But leaves vary anyway: pick ten different leaves from the same bushes and the means would change. Is the difference between sun and shade real, or could it have arisen just by chance in which leaves were picked? Biologists answer such questions with statistical tests. This lesson explains why some characteristics vary continuously, how to describe that variation with a mean and a standard deviation, and how to use the t-test to decide whether two means really differ.

Definition

Discontinuous variation: variation in which individuals fall into a few distinct classes with no intermediates, such as ABO blood groups. It is usually controlled by one or two genes, with little effect of the environment.

Continuous variation: variation in which there is a smooth range of values between two extremes, such as height or leaf length. It is usually controlled by many genes, together with the environment.

Polygenes: several genes, often on different chromosomes, that each add a small effect to one characteristic.

Standard deviation (s): a measure of how spread out the values are around the mean.

Null hypothesis: the statement that there is no difference (for example, between two means) except that caused by chance.

t-test: a statistical test that compares the means of two samples of continuous data to decide whether the difference between them is significant.

Two kinds of variation

Feature	Discontinuous	Continuous
Classes	A few distinct categories	Any value within a range
Data and graph	Counts in categories; bar chart with gaps	Measurements; histogram with touching bars, often bell-shaped (a normal distribution)
Genetic control	One or a few genes with large effects (monogenic)	Many genes with small, additive effects (polygenic)
Effect of environment	Little or none	Usually large
Examples	ABO blood group, sickle-cell haemoglobin, shell banding in grove snails	Height, mass, leaf length, milk yield, number of seeds per pod

Why many genes give a smooth curve

Suppose plant height is controlled by three genes, A, B and C, on different chromosomes. Each has a contributing allele (A, B, C) that adds 5 cm to a base height of 20 cm, and a non-contributing allele (a, b, c) that adds nothing. A plant can have from 0 to 6 contributing alleles, so there are 7 height classes, from 20 cm to 50 cm.

Worked example 1: a polygenic cross

Two plants of genotype AaBbCc (35 cm tall) are crossed. Using the model above, find the proportion of offspring in each height class.

Each parent can make $2^{3} = 8$ kinds of gamete, so there are $8 \times 8 = 64$ equally likely combinations.
Each offspring receives 6 alleles, and each has a probability of $\frac{1}{2}$ of being contributing. The number of ways to get $k$ contributing alleles out of 6 follows the row of Pascal's triangle: 1, 6, 15, 20, 15, 6, 1.
So the offspring are 20, 25, 30, 35, 40, 45 and 50 cm tall in the ratio 1 : 6 : 15 : 20 : 15 : 6 : 1.
The commonest class is 35 cm: $\frac{20}{64} = 31\%$ (2 s.f.). Only $\frac{1}{64}$ , about 1.6%, are 50 cm tall.

Interpretation. Even three genes give seven classes with a bell-shaped spread. With more genes the classes become smaller and closer, and because the environment also adds or subtracts a little from each plant, the steps blur into a smooth, continuous distribution.

Describing continuous data

Two numbers summarise a sample: the mean, $\bar{x} = \frac{\sum x}{n}$ , and the standard deviation, which measures the spread:

s = \sqrt{\frac{\sum (x - \bar{x})^{2}}{n - 1}}

In a normal distribution about 68% of values lie within one standard deviation of the mean and about 95% within two. A small standard deviation means the values cluster tightly; a large one means they are widely spread.

The t-test

The t-test asks whether the means of two samples differ by more than chance would explain. It can be used when the data are continuous, roughly normally distributed, from two independent samples, with similar spreads.

State the null hypothesis: there is no significant difference between the two means.
Calculate the mean, $\bar{x}$ , and standard deviation, $s$ , of each sample.
Calculate $t$ :

t = \frac{\left| \bar{x}_{1} - \bar{x}_{2} \right|}{\sqrt{\frac{s_{1}^{2}}{n_{1}} + \frac{s_{2}^{2}}{n_{2}}}}

Find the degrees of freedom: $n_{1} + n_{2} - 2$ .
Compare $t$ with the critical value at the 0.05 (5%) probability level. If $t$ is greater than the critical value, the probability that the difference is due to chance is less than 5%: reject the null hypothesis; the difference is significant. If $t$ is smaller, accept the null hypothesis.

Degrees of freedom	10	12	14	16	18	20	22	24	26	28	30
Critical t at p = 0.05	2.23	2.18	2.14	2.12	2.10	2.09	2.07	2.06	2.06	2.05	2.04
Critical t at p = 0.01	3.17	3.05	2.98	2.92	2.88	2.85	2.82	2.80	2.78	2.76	2.75

A larger $t$ comes from a bigger difference between the means, smaller standard deviations, or bigger samples: each makes it less likely that the difference is due to chance.

Worked example 2: sun leaves and shade leaves

Illustrative data. The lengths (mm) of 10 leaves from shrubs in sun and 10 from the same kind of shrub in shade:

Sun	52	48	55	60	47	51	58	54	49	56
Shade	63	58	66	70	61	59	68	64	57	62

Is the difference in mean length significant at the 5% level?

Null hypothesis: there is no significant difference between the mean lengths of sun and shade leaves.
Means: sun $\bar{x}_{1} = \frac{530}{10} = 53.0$ mm; shade $\bar{x}_{2} = \frac{628}{10} = 62.8$ mm.
Sums of squared deviations: sun 170.0; shade 165.6. So $s_{1}^{2} = \frac{170.0}{9} = 18.9$ and $s_{2}^{2} = \frac{165.6}{9} = 18.4$ (so $s_{1} = 4.35$ mm, $s_{2} = 4.29$ mm).
$t = \frac{62.8 - 53.0}{\sqrt{\frac{18.9}{10} + \frac{18.4}{10}}} = \frac{9.8}{\sqrt{3.73}} = \frac{9.8}{1.93} = 5.1$ (2 s.f.).
Degrees of freedom $= 10 + 10 - 2 = 18$ . Critical value at p = 0.05 is 2.10.
$5.1 > 2.10$ , so reject the null hypothesis. It is even greater than 2.88, the value at p = 0.01, so the probability that a difference this large arose by chance is less than 1%.

Conclusion and limits. Shade leaves are significantly longer. The t-test says the difference is real, not why: it may be an environmental response to low light, but the shrubs could also differ genetically. Growing cuttings of the same plants in both conditions would separate the two causes.

Warning

Common mistakes

"The t-test proves the hypothesis." It gives the probability that the difference is due to chance; it never proves anything.
Wrong degrees of freedom. For two samples use $n_{1} + n_{2} - 2$ , not $n - 1$ .
Using the t-test on categories. The t-test is for continuous, roughly normal data. Counts in categories, such as blood groups, need a different test, such as chi-squared.
Dividing by n instead of n − 1 when calculating a sample's standard deviation.

Bozeman Science: "Student's t-test". Paul Andersen works through a t-test step by step; watch how the null hypothesis, degrees of freedom and critical value are used.

✦

Practice: variation and the t-test

Classify each as continuous or discontinuous: (a) mass of eggs, (b) ABO blood group, (c) number of leaves on a seedling, (d) sickle-cell haemoglobin genotype.
Explain why continuous variation usually shows a bell-shaped distribution.
In the three-gene model, what is the height of (a) aabbcc, (b) AABbCc? What fraction of the offspring of AaBbCc × AaBbCc are 20 cm tall?
Calculate the mean and standard deviation of: 4, 6, 7, 8, 10.
Two samples have $n_{1} = n_{2} = 8$ , $\bar{x}_{1} = 12.4$ , $s_{1} = 1.6$ , $\bar{x}_{2} = 11.2$ , $s_{2} = 1.8$ . Calculate t and decide whether the difference is significant at p = 0.05.
Explain what is meant by "the difference is significant at the 5% level".
Suggest why a t-test should not be used to compare the numbers of people with blood groups A and B in two villages.
A t-test shows that seedlings given a fertiliser are significantly taller than untreated seedlings. A pupil concludes that the fertiliser changes the seedlings' genes. Evaluate this conclusion.

Show answersHide answers

(a) Continuous. (b) Discontinuous. (c) Discontinuous in the strict sense, as whole numbers, though it is usually treated like continuous data because there are many values. (d) Discontinuous.
It is controlled by many genes with small additive effects, plus environmental effects; most individuals have intermediate combinations, and few have all contributing or all non-contributing alleles.
(a) 20 cm. (b) 4 contributing alleles: $20 + 4 \times 5 = 40$ cm. $\frac{1}{64}$ .
Mean $= \frac{35}{5} = 7$ . Squared deviations: 9, 1, 0, 1, 9; sum 20. $s = \sqrt{\frac{20}{4}} = \sqrt{5} = 2.24$ (3 s.f.).
$t = \frac{1.2}{\sqrt{\frac{1.6^{2}}{8} + \frac{1.8^{2}}{8}}} = \frac{1.2}{\sqrt{0.320 + 0.405}} = \frac{1.2}{0.851} = 1.41$ . df = 14; critical value 2.14. $1.41 < 2.14$ : accept the null hypothesis; the difference is not significant.
There is less than a 5% probability that a difference this large would arise by chance alone if the populations really had the same mean.
Blood groups are discontinuous categories, not continuous measurements with a mean and standard deviation; a chi-squared test is appropriate.
Wrong: the fertiliser changes the environment (nutrient supply), which changes the phenotype; it does not change the alleles. The significant difference shows an environmental effect on height, and these changes would not be inherited.

Key points

Discontinuous variation: distinct classes, few genes, little environmental effect, bar chart. Continuous: a range, polygenes plus environment, histogram, often normal.
Polygenes with additive effects give many classes (1 : 6 : 15 : 20 : 15 : 6 : 1 for three genes), blurred by the environment into a smooth curve.
Mean and standard deviation describe continuous data; $s$ uses $n - 1$ .
t-test: null hypothesis; $t = \frac{|\bar{x}_{1} - \bar{x}_{2}|}{\sqrt{\frac{s_{1}^{2}}{n_{1}} + \frac{s_{2}^{2}}{n_{2}}}}$ ; df $= n_{1} + n_{2} - 2$ ; if t exceeds the critical value at p = 0.05, the difference is significant.

Sources and credits

10.1 Two Population Means with Unknown Standard Deviations, from Introductory Statistics 2e by Barbara Illowsky and Susan Dean, OpenStax (Rice University), licensed under CC BY 4.0. Used for the two-sample t-test. Adapted by Ganzaa.
Video: "Student's t-test", Bozeman Science, embedded from YouTube.

Figures 1 and 2, the tables, the worked examples (with illustrative data) and the practice questions were written by Ganzaa. Critical values of t are standard values for a two-tailed test.

1Is the difference real?

2Two kinds of variation

3Why many genes give a smooth curve

4Describing continuous data

5The t-test

Is the difference real?

Two kinds of variation

Why many genes give a smooth curve

Describing continuous data

The t-test