Inheritance: monohybrid and dihybrid crosses; codominance and multiple alleles; sex linkage; epistasis; the use of genetic diagrams; and the chi-squared test.

What this dot point is asking

Eduqas wants you to carry out monohybrid and dihybrid crosses, explain codominance, multiple alleles, sex linkage and epistasis, use genetic diagrams to predict ratios, and use the chi-squared test to compare observed and expected results. This is the most quantitative topic in Component 2.

Key terms and the basic crosses

A gene codes for a characteristic; an allele is a version of it. Genotype is the alleles present; phenotype is the observable feature. Homozygous is two identical alleles; heterozygous two different. A dominant allele shows in the heterozygote; a recessive allele only in the homozygote.

• A monohybrid cross follows one gene. $Tt \times Tt$ gives a 3:1 ratio of dominant to recessive.
• A dihybrid cross follows two genes on different chromosomes. $TtRr \times TtRr$ gives 9:3:3:1, because the genes assort independently.

Always define your allele symbols, give the parental genotypes and gametes, draw a Punnett square, and read off the phenotype ratio.

Codominance, multiple alleles, sex linkage and epistasis

• Codominance: both alleles are fully expressed in the heterozygote (for example the AB blood group, where both A and B antigens appear).
• Multiple alleles: a gene with more than two alleles in the population (the ABO blood group has $I^A$, $I^B$ and $I^O$).
• Sex linkage: genes on the X chromosome. Males (XY) express a single recessive X-linked allele, so haemophilia and red-green colour blindness are commoner in males; females can be unaffected carriers.
• Epistasis: one gene affects the expression of another (for example one gene must make a pigment precursor before a second can act), distorting the expected ratio.

The chi-squared test

The chi-squared test tests whether observed results differ significantly from those expected by a genetic ratio, or whether any difference is just chance:

State a null hypothesis (no significant difference from the ratio), find the degrees of freedom (number of categories minus 1), and compare $\chi^2$ with the critical value at $p = 0.05$:

• if $\chi^2$ is less than the critical value, the difference is not significant (accept the null hypothesis; the data fit the ratio);
• if $\chi^2$ is greater than or equal to the critical value, the difference is significant (reject the null hypothesis).

Examples in context

Example 1. ABO blood groups. The ABO system shows both multiple alleles ($I^A$, $I^B$, $I^O$) and codominance ($I^A$ and $I^B$ are codominant, giving group AB), one locus illustrating two patterns at once.

Example 2. Haemophilia in royal families. Haemophilia, an X-linked recessive condition, passed through European royal families via unaffected female carriers to affected sons, a classic demonstration of sex-linked inheritance.

Try this

Q1. State the expected phenotypic ratio from crossing two organisms heterozygous for two independently assorting genes. [1 mark]

• Cue. 9:3:3:1.

Q2. Explain why X-linked recessive conditions are more common in males than females. [2 marks]

• Cue. Males have only one X chromosome, so a single recessive allele is expressed; females have two X chromosomes and would need two recessive alleles to be affected (otherwise they are carriers).

Q3. Write the formula for the chi-squared statistic. [1 mark]

• Cue. $\chi^2 = \sum \dfrac{(O - E)^2}{E}$.