Inheritance: Question 5
Syllabus 16.2
A gardener is investigating two genes in a species of ornamental plant called the moonflower. Petal colour is controlled by gene P, with allele P (purple, dominant) and allele p (white, recessive). Leaf shape is controlled by a separate gene, gene L, with allele L (broad leaves, dominant) and allele l (narrow leaves, recessive).
The gardener crosses two plants that are both heterozygous at both loci (genotype PpLl). If the two genes assort completely independently, a dihybrid cross of this kind is expected to produce offspring in the ratio "9 purple, broad-leaved : 3 purple, narrow-leaved : 3 white, broad-leaved : 1 white, narrow-leaved".
The gardener grows 160 offspring from this cross and records the following numbers in each phenotype category:
- purple, broad leaves: 100
- purple, narrow leaves: 15
- white, broad leaves: 15
- white, narrow leaves: 30
The gardener carries out a chi-squared test to compare these observed numbers with the numbers expected if the two genes assort independently.
(a) State a suitable null hypothesis for this chi-squared test. [1] (b) (i) Calculate the expected number of offspring in each of the four phenotype categories, assuming a "9:3:3:1" ratio for a total of 160 offspring. [1] (b) (ii) Using the formula "chi-squared = the sum of (observed minus expected) squared, divided by expected", calculate the value of chi-squared for the gardener's data. Show your working. [3] (c) The critical value of chi-squared at the p = 0.05 significance level, for the appropriate number of degrees of freedom, is 7.815. Compare this critical value with your calculated value, and state what you conclude about the two genes. [2] (d) In some pairs of genes, the alleles of one gene mask or alter the phenotypic effect of alleles at a completely different gene locus; this is called epistasis. State one way in which epistasis differs from the type of gene interaction identified in part (c). [2]
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Worked solution
Part (a): Null hypothesis
A chi-squared test compares observed data with the numbers expected under a stated hypothesis, to see whether any difference between them is likely to be due to chance alone. Here, the expected numbers are based on the assumption that the two genes (for petal colour and leaf shape) assort completely independently, giving a 9:3:3:1 ratio.
A suitable null hypothesis is: “There is no significant difference between the observed and expected numbers of offspring in each phenotype category”. In other words, any difference between the gardener’s observed numbers and the expected 9:3:3:1 numbers is due to chance, and the two genes assort independently.
Part (b)(i): Expected numbers
If the two genes assort independently, the dihybrid cross PpLl x PpLl is expected to give offspring in the ratio 9 purple broad : 3 purple narrow : 3 white broad : 1 white narrow, out of 16 equal parts. For a total of 160 offspring, each part of the ratio represents 160 ÷ 16 = 10 offspring.
| Phenotype | Ratio | Expected number |
|---|---|---|
| Purple, broad leaves | 9 | 9 × 10 = 90 |
| Purple, narrow leaves | 3 | 3 × 10 = 30 |
| White, broad leaves | 3 | 3 × 10 = 30 |
| White, narrow leaves | 1 | 1 × 10 = 10 |
Check: 90 + 30 + 30 + 10 = 160, which matches the total number of offspring grown.
Part (b)(ii): Calculating chi-squared
Using chi-squared = the sum of (observed minus expected) squared, divided by expected, for each category:
| Phenotype | Observed (O) | Expected (E) | O - E | (O - E)^2 | (O - E)^2 / E |
|---|---|---|---|---|---|
| Purple, broad | 100 | 90 | 10 | 100 | 100/90 = 1.11 |
| Purple, narrow | 15 | 30 | -15 | 225 | 225/30 = 7.50 |
| White, broad | 15 | 30 | -15 | 225 | 225/30 = 7.50 |
| White, narrow | 30 | 10 | 20 | 400 | 400/10 = 40.00 |
Summing the final column:
chi-squared = 1.11 + 7.50 + 7.50 + 40.00 = 56.11 (to 2 decimal places)
Part (c): Comparing with the critical value and drawing a conclusion
There are 4 phenotype categories, so the degrees of freedom = number of categories - 1 = 4 - 1 = 3.
The calculated chi-squared value, 56.11, is much greater than the given critical value of 7.815 (at p = 0.05, 3 degrees of freedom).
Because the calculated value exceeds the critical value, the null hypothesis is rejected: the difference between the observed and expected numbers of offspring is too large to be explained by chance alone, so it is statistically significant. This means the two genes are not assorting independently as a simple 9:3:3:1 dihybrid cross would predict. Instead, the pattern, more offspring than expected in the two “parental-looking” categories (purple broad and white narrow) and fewer than expected in the other two categories (purple narrow and white broad), supports the conclusion that the genes for petal colour and leaf shape are linked, most likely located close together on the same autosome, so their alleles tend to be inherited together rather than assorting independently.
Part (d): Epistasis compared with linkage
Both linkage and epistasis can cause offspring ratios to depart from the standard 9:3:3:1 dihybrid ratio, but the underlying cause is different in each case.
Linkage (as identified in part (c)) happens because two genes are physically close together on the same chromosome. This is a matter of chromosomal position: the alleles at the two loci tend to travel together into the same gamete because they are rarely separated by crossing over, so the observed ratio departs from 9:3:3:1 even though the two genes have no direct biochemical effect on one another.
Epistasis, in contrast, happens when the protein product of one gene interacts with, masks, or otherwise alters the phenotypic expression of a different gene, for example, by acting earlier in the same biochemical pathway. This is a functional interaction between the two genes’ products, and it can occur whether the two genes are on the same chromosome or on completely different chromosomes; it does not depend on the genes being physically close together.
So, one key difference is: linkage is caused by the chromosomal position of the two genes, whereas epistasis is caused by a biochemical interaction between the products of the two genes, regardless of where those genes are located.
Final answers
- (a) Null hypothesis: there is no significant difference between the observed and expected numbers of offspring; the two genes assort independently.
- (b)(i) Expected numbers: 90 : 30 : 30 : 10 (purple broad : purple narrow : white broad : white narrow).
- (b)(ii) Chi-squared = 56.11 (to 2 decimal places).
- (c) With 3 degrees of freedom, the calculated value (56.11) is far greater than the critical value (7.815), so the null hypothesis is rejected. The genes are linked, not assorting independently.
- (d) Linkage is due to the chromosomal position of two genes; epistasis is due to a biochemical interaction between gene products, independent of their chromosomal location.