Genotype is the genetic constitution of an organism. Phenotype is the expression of this genetic constitution and its interaction with the environment. Most phenotypes are affected by more than one gene. The genotype, phenotype and ratio of offspring can be predicted for monohybrid and dihybrid crosses involving dominant, recessive, codominant and multiple alleles, sex-linkage, autosomal linkage and epistasis. The chi-squared (X2) test can be used to test the significance of the difference between observed and expected results.

What this dot point is asking

AQA wants you to distinguish genotype from phenotype, predict offspring ratios for every cross type named in the specification (monohybrid, dihybrid, codominance, multiple alleles, sex-linkage, autosomal linkage and epistasis), and use the chi-squared test to decide whether observed data fit an expected genetic ratio.

Monohybrid crosses

A monohybrid cross follows one gene with two alleles. Use a capital letter for the dominant allele and the same letter in lower case for the recessive allele.

A cross of two heterozygotes $Aa \times Aa$ gives a genotypic ratio of $1\ AA : 2\ Aa : 1\ aa$ and a phenotypic ratio of $3$ dominant $: 1$ recessive. A test cross ($Aa \times aa$ or $AA \times aa$) reveals an unknown genotype: a $1:1$ phenotypic ratio shows the unknown parent was heterozygous.

Dihybrid crosses

A dihybrid cross follows two genes on different chromosomes. The classic Mendelian result from $AaBb \times AaBb$ is a $9:3:3:1$ phenotypic ratio (9 both dominant : 3 dominant A recessive b : 3 recessive a dominant B : 1 both recessive). This ratio assumes the two genes assort independently.

Codominance and multiple alleles

Codominance: both alleles are expressed fully in the heterozygote. Example: cattle coat colour, where $C^R C^W$ produces roan (red and white hairs together), not pink.

Multiple alleles: more than two alleles exist for a gene in the population, though any diploid individual carries only two. Example: human ABO blood groups, with alleles $I^A$, $I^B$ (codominant to each other) and $I^O$ (recessive). Genotype $I^A I^B$ gives blood group AB; $I^A I^O$ gives group A.

Sex-linkage

Genes on the X chromosome are sex-linked. Males (XY) are hemizygous: a single recessive allele on their one X chromosome is expressed because the Y carries no corresponding allele. Females (XX) need two copies to show a recessive trait, so X-linked recessive conditions (haemophilia, red-green colour blindness) are more common in males.

A carrier mother $X^H X^h$ crossed with an unaffected father $X^H Y$ gives sons that are 50 percent affected ($X^h Y$) and daughters that are all unaffected but 50 percent carriers.

Autosomal linkage

Two genes on the same autosome are linked: they tend to be inherited together because they do not assort independently. Linkage reduces the number of recombinant offspring, so a dihybrid cross of linked genes does not give the expected $9:3:3:1$ ratio. Parental combinations dominate; recombinants arise only when crossing over separates the linked alleles, and the closer the loci, the rarer the recombinants.

Epistasis

Epistasis is the interaction of genes at different loci where one gene masks or modifies the expression of another.

• Recessive epistasis modifies the dihybrid ratio to $9:3:4$ (the recessive genotype at the epistatic locus masks the second gene). Example: coat colour in mice where a homozygous recessive at one locus gives albino regardless of the colour gene.
• Dominant epistasis can give a $12:3:1$ ratio (a dominant allele at the epistatic locus masks the second gene).

Recognising a modified ratio (9:3:4, 12:3:1, 9:7) is the exam signal that epistasis is operating rather than simple independent assortment.

The chi-squared test

The chi-squared ($X^2$) test asks whether the difference between observed and expected results is significant or just chance. Use it on categorical (counted) data with a clear expected ratio.

Steps:

1. State a null hypothesis: there is no significant difference between observed and expected results.
2. Calculate the expected counts from the predicted ratio and the total sample size.
3. Compute $X^2$ using the formula.
4. Find degrees of freedom = (number of categories) - 1.
5. Compare $X^2$ to the critical value at $p = 0.05$.

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

Try this

Q1. Distinguish between genotype and phenotype, and explain why most human phenotypes cannot be predicted from a single gene. [3 marks]

• Cue. Genotype = alleles carried; phenotype = observable trait from genotype plus environment; most traits are polygenic and environmentally modified.

Q2. A plant heterozygous for two unlinked genes ($AaBb$) is self-pollinated. State the expected phenotypic ratio and explain what a deviation to a 9:3:4 ratio would indicate. [3 marks]

• Cue. Expected 9:3:3:1; a 9:3:4 ratio indicates recessive epistasis, where one gene masks the expression of the other.

Q3. In a cross, the expected ratio was 1:1 but the observed counts were 58 and 42 (total 100). Carry out a chi-squared test and state your conclusion. Critical value at p = 0.05, 1 degree of freedom = 3.841. [4 marks]

• Cue. Expected 50:50. $X^2 = (58-50)^2/50 + (42-50)^2/50 = 1.28 + 1.28 = 2.56$. 2.56 < 3.841, so accept the null hypothesis; the difference is not significant.