1Spotting a deviation: let the statistics decide

A single cross never matches a ratio exactly, so "it looks a bit off" is not evidence. Run the chi-square test from 5.3 against the Mendelian prediction. If $\chi^2$ is larger than the critical value, reject the null hypothesis: something other than chance, and other than simple Mendelian inheritance, is at work. Then look at how the data differ, which points to the explanation.

2Linked genes and gene mapping

Genes on the same chromosome are genetically linked. Look at Figure 1. A heterozygote has A and B on one homologue and a and b on the other. Without a crossover between the two loci, its gametes carry the parental combinations AB or ab. Only when a crossover happens between the two genes (5.2) does a gamete get a recombinant combination, Ab or aB.

So a test cross of AaBb × aabb with linked genes does not give 1 : 1 : 1 : 1. It gives two large parental classes and two small recombinant classes.

The further apart two genes are, the more likely a crossover falls between them, so the proportion of recombinant offspring measures the distance. This is gene mapping:

$$\text{recombination frequency} = \frac{\text{number of recombinant offspring}}{\text{total offspring}} \times 100\%$$

and 1% recombination = 1 map unit.

Worked example. An invented test cross: a fly heterozygous for two genes, with AB on one homologue and ab on the other, is crossed with an aabb fly. Of 1,000 offspring, 412 show AB, 398 ab, 94 Ab and 96 aB.

• Expected for independent assortment: 250 in each class. $\chi^2 = 384.8$ with 3 degrees of freedom, far above 7.81: reject independent assortment.
• The pattern points to linkage: the two parental classes (412 and 398) are large, the two recombinant classes (94 and 96) small.
• Recombination frequency: $\frac{94 + 96}{1000} \times 100\% = 19\%$, so the genes are 19 map units apart.

Map distances roughly add up, which lets you order genes. If A–B is 19 units, A–C 12 units and C–B 7 units, then C lies between A and B, because 12 + 7 = 19.

Two limits to remember. Recombination frequency can never usefully exceed 50%, because genes so far apart that a crossover almost always falls between them give 50% recombinants, the same as genes on different chromosomes. And a recombination frequency tells you which genes are close; it does not tell you the distance in base pairs.

3Codominance and incomplete dominance

Both describe a heterozygote whose phenotype differs from both homozygotes. They differ in what the heterozygote looks like.

• Codominance: both alleles are fully expressed, side by side. Human ABO blood groups: the allele $I^A$ makes the A antigen and $I^B$ the B antigen on red blood cells. A person who is $I^A I^B$ has both antigens, blood group AB: not a mixture of the two, but both at once.
• Incomplete dominance: neither allele masks the other, and the heterozygote is a blend, between the two. In snapdragons, red-flowered × white-flowered gives pink F1 plants.

In both, crossing two heterozygotes gives a 1 : 2 : 1 phenotypic ratio, because each genotype has its own phenotype. Pink × pink snapdragons give 1 red : 2 pink : 1 white. That is the giveaway in data: a 1 : 2 : 1 where 3 : 1 was expected.

4Sex-linked traits

In humans and many other animals, females are XX and males XY. Genes on the X chromosome are X-linked; the Y carries few genes, and those on it are Y-linked.

A male has only one X, so whatever allele it carries is expressed: he cannot be a carrier of an X-linked recessive allele. A female needs two copies of the recessive allele to show the trait. So X-linked recessive traits are much more common in males (XY) than females (XX). Haemophilia A and red–green colour blindness in humans are examples.

Write X-linked genotypes with the allele as a superscript: $X^H X^h$ is a carrier female, $X^h Y$ an affected male. Figure 2 shows a carrier mother and an unaffected father. Each daughter gets her father's $X^H$, so no daughter is affected, though half are carriers. Each son gets his Y from his father and one of his mother's X chromosomes, so each son has a probability of $\frac{1}{2}$ of having haemophilia.

Signs of X-linked recessive inheritance in a pedigree: mostly males affected; affected sons of unaffected (carrier) mothers; no father-to-son passing, because a father gives his son a Y, not an X. Y-linked traits pass from father to every son and never to daughters.

Not every species uses X and Y. In birds, females are ZW and males ZZ, so a Z-linked recessive trait is more common in females. In bees, sex is set by haplodiploidy: females develop from fertilised eggs and are diploid, males from unfertilised eggs and are haploid.

5Pleiotropy

Pleiotropy is when one gene affects several traits. Because all the effects come from the same gene, they are inherited together: they cannot segregate independently, so they appear as one package in the offspring. The allele for sickle cell haemoglobin is an example: one change in one gene produces misshapen red blood cells, anaemia, pain from blocked vessels and damage to several organs, and in heterozygotes, some protection against malaria.

In data, pleiotropy looks like two "traits" that always appear together, in every individual, which could be mistaken for tight linkage. With linkage, a few recombinants eventually appear; with pleiotropy, never.

6Non-nuclear inheritance

Mitochondria and chloroplasts have their own DNA. These organelles are not shared out by the spindle; they are randomly assorted into gametes and daughter cells, so their traits do not follow Mendel's rules.

• In animals, the egg passes on its mitochondria and the sperm, in general, does not. Traits coded by mitochondrial DNA are therefore maternally inherited: an affected mother passes the trait to all or most of her children; an affected father passes it to none.
• In plants, mitochondria and chloroplasts are passed on in the ovule, not the pollen, so traits coded by their DNA are also typically maternal. A plant with variegated leaves caused by faulty chloroplasts can produce green, white or variegated seedlings when it is the seed parent, depending on which chloroplasts each embryo receives; when it is the pollen parent, the seedlings simply match the seed parent.