1Genotype, phenotype and dominance

An organism's genotype is the set of alleles it has for one or more genes: for one gene, AA, Aa or aa. If both alleles are the same it is homozygous for that gene; if they differ, heterozygous. Its phenotype is the observable trait: purple flowers, a wrinkled seed, a disease.

When a heterozygote shows only one allele's phenotype, that allele is dominant (written with a capital, A) and the hidden one recessive (a). So AA and Aa have the same phenotype, and only aa shows the recessive trait. Dominant does not mean common, and it does not mean better: it describes only what a heterozygote looks like.

2The two laws of probability

The formula sheet gives both.

• Product rule. If two events are independent, the probability that both happen is $P(A \text{ and } B) = P(A) \times P(B)$.
• Sum rule. If two events are mutually exclusive (they cannot both happen), the probability that one or the other happens is $P(A \text{ or } B) = P(A) + P(B)$.

Worked example 1, one gene. Two heterozygotes, Aa × Aa. Each parent's gamete carries A or a with probability $\frac{1}{2}$. For an aa child both gametes must carry a: $\frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$. For a heterozygote, the A can come from the mother or from the father, two mutually exclusive routes: $\frac{1}{4} + \frac{1}{4} = \frac{1}{2}$. The full result is a genotypic ratio of 1 AA : 2 Aa : 1 aa, and a phenotypic ratio of 3 dominant : 1 recessive. Keep the two ratios apart: the unit is examined on exactly this.

Worked example 2, three genes. AaBbCc × AaBbCc, all three genes on different chromosomes. What is the probability of an offspring that is aabbcc? Treat each gene as its own monohybrid cross and multiply: $\frac{1}{4} \times \frac{1}{4} \times \frac{1}{4} = \frac{1}{64}$. No 64-box Punnett square is needed.

Worked example 3, two genes with different parents. AaBb × Aabb. Probability of a child showing both dominant traits? Gene A: Aa × Aa gives a dominant phenotype with probability $\frac{3}{4}$. Gene B: Bb × bb gives Bb (dominant) with probability $\frac{1}{2}$. Both: $\frac{3}{4} \times \frac{1}{2} = \frac{3}{8}$.

3Monohybrid, dihybrid and test crosses

A monohybrid cross follows one gene. Cross two true-breeding (homozygous) parents with different traits: the F1 all show one trait, which tells you it is dominant. Self or intercross the F1 (Aa × Aa) and the recessive trait reappears in about a quarter of the F2, a 3:1 phenotypic ratio.

A dihybrid cross follows two genes. Figure 1 uses two of Mendel's own pea traits: yellow seeds (Y) are dominant to green (y), and round (R) to wrinkled (r). A plant that is YyRr makes four kinds of gamete, YR, Yr, yR and yr, in equal numbers, because the two genes are on different chromosomes and assort independently. Crossing two such plants gives sixteen equally likely boxes and a phenotypic ratio of 9 : 3 : 3 : 1: 9 showing both dominant traits, 3 and 3 showing one each, 1 showing both recessive. That ratio is the product rule at work: $\frac{3}{4} \times \frac{3}{4} = \frac{9}{16}$ for both dominant, and $\frac{1}{4} \times \frac{1}{4} = \frac{1}{16}$ for both recessive.

A test cross finds the genotype of an individual showing the dominant phenotype. Cross it with a homozygous recessive (aa), whose gametes all carry a, so every offspring's phenotype shows which allele came from the unknown parent. All offspring dominant means the parent is AA; about half recessive means it is Aa. For two genes, a test cross of AaBb × aabb gives a 1 : 1 : 1 : 1 ratio if the genes assort independently. In 5.4 that ratio is how linked genes give themselves away.

4Reading a pedigree

A pedigree is a family tree for one trait: squares are males, circles females, shaded symbols affected individuals. Work through it with two questions.

Dominant or recessive? If two unaffected parents have an affected child, the trait must be recessive: both parents carried a hidden allele. If an affected child always has at least one affected parent, and two affected parents can have an unaffected child, it is dominant.

Autosomal or sex-linked? For a recessive trait, an affected daughter with an unaffected father rules out an X-linked recessive trait, because a father passes his only X to every daughter (5.4). If affected individuals are mostly males, think X-linked.

Figure 2 is an invented pedigree. II-3 and II-4 are both unaffected but have an affected daughter, III-2, so the trait is recessive. III-2's father, II-3, is unaffected, so it cannot be X-linked recessive: it is autosomal recessive. Every unaffected child of the affected I-1 must be Dd, because I-1 could only give them d.

A harder reading: what is the probability that III-1 is a carrier? III-1 is unaffected, so dd is ruled out. Of the three remaining equally likely outcomes of Dd × Dd (DD, Dd, dD), two are carriers: the probability is $\frac{2}{3}$, not $\frac{1}{2}$.

5The chi-square test

A cross predicts a ratio; real offspring never match it exactly, because fertilisation is random. The chi-square ($\chi^2$) test asks whether the difference between observed and expected counts is small enough to be chance.

1. State the null hypothesis. It always says there is no real difference: "the offspring fit the expected 9:3:3:1 ratio; any difference is due to chance." An alternative hypothesis ("the genes do not assort independently") is not the null. Write the null; it is where students most often drop a mark.
2. Work out the expected numbers (e) from the total and the predicted ratio. Use counts, never percentages.
3. Calculate $\chi^2 = \sum \frac{(o - e)^2}{e}$, the formula on the sheet, one term per category.
4. Degrees of freedom = number of categories − 1.
5. Compare with the critical value at p = 0.05 in the sheet's table. If $\chi^2$ is greater than the critical value, reject the null hypothesis: the difference is unlikely to be chance alone. If it is smaller, fail to reject it.

Worked example. An invented dihybrid cross in beetles: shiny wing cases (G) dominant to matt (g), red bodies (R) dominant to black (r). Two GgRr beetles produce 320 offspring. Null hypothesis: the genes assort independently and the offspring fit a 9 : 3 : 3 : 1 ratio.

There are 4 categories, so 3 degrees of freedom; the critical value at p = 0.05 is 7.81. Since 0.89 < 7.81, fail to reject the null hypothesis. The data are consistent with independent assortment. Note what this does not say: it does not prove the null hypothesis true. It says the data give no reason to doubt it.

A chi-square larger than the critical value means reject the null hypothesis; smaller means fail to reject it. Never "accept" it, never "prove" it.

Chi-square is not only for genetics. It works on any counts in categories: how many pill bugs chose a damp chamber over a dry one, how many seeds germinated in each treatment. It is the wrong test for measurements such as heights or masses; for those, compare means using the standard error.