Extending Mendel: Dominance, Multiple Alleles and Gene Interaction

Mate two yellow mice as often as you like, and you will never get a yellow mouse that breeds true.

Lucien Cuénot found this in his mouse crosses in the first years after Mendel’s rules were taken up. It looked like a failure of Mendel. It was not. In Mendel’s peas four assumptions sat quietly under the 3:1 and the 9:3:3:1: one allele fully hides the other, a gene has two alleles, every genotype survives to be counted, and one gene controls one character. Relax each in turn and the ratios change in ways you can predict. Segregation and independent assortment never stop working. What changes is the map from genotype to phenotype.

Dominance is a property of the heterozygote, not the allele

Cross red snapdragons (Antirrhinum majus) with white ones and the F1 flowers are pink. This is incomplete dominance: the heterozygote’s phenotype lies between the two homozygotes’. Self the pink plants and red and white come back, at 1 red : 2 pink : 1 white. Nothing blended. With incomplete dominance the phenotype ratio simply equals the genotype ratio, so the hidden 1:2:1 of Mendel’s peas is now in plain sight.

In codominance, the products of both alleles are detected in the heterozygote. People with the M and N alleles of the MN blood group carry both the M and the N antigens on their red cells, and an MN by MN cross gives 1 M : 2 MN : 1 N.

Which word applies depends on what you measure. A person heterozygous for the sickle-cell allele of β-globin makes both normal and sickle haemoglobin, and both show on a gel: codominant. The same person usually does not have sickle cell disease: for anaemia, the allele is close to recessive.

Why is one copy often enough? Think of a factory with two identical machines. Lose one and the other usually keeps up, because output rarely depends on running every machine at full speed. Most enzymes are like that (see how enzymes work), so an allele that makes no working enzyme is usually recessive. There are two exceptions.

• Haploinsufficiency: half the product is not enough. A fly with one working copy of Notch has nicked wings (Notch signaling).
• Dominant negative: the mutant product interferes with the normal one. If a protein works as a complex of four identical subunits and any bad subunit ruins the complex, a heterozygote that assembles subunits at random keeps only one complex in sixteen intact:

fraction of complexes with four normal subunits = (1/2) x (1/2) x (1/2) x (1/2) = 1/16, about 6%

A null allele in the same gene would leave about half. The dominant negative does worse than having no mutant protein at all.

ABO: three alleles, two sugars and codominance

A gene can have more than two alleles in a population, a set called multiple alleles. Each diploid person still carries only two. The ABO blood groups are the classic case. In the usual model there are three alleles at one locus:

• I^A encodes a glycosyltransferase that adds N-acetylgalactosamine to a sugar chain on the cell surface called the H antigen, making the A antigen.
• I^B encodes a nearly identical enzyme that adds galactose instead, making the B antigen. (The antigens are carbohydrates; see carbohydrates.)
• i makes no working enzyme, so H is left unchanged.

I^A and I^B are codominant: an I^A I^B person carries both antigens and is type AB. Each is dominant to i, because one working enzyme is enough to decorate the cells. So I^A I^A and I^A i are type A, I^B I^B and I^B i are type B, and ii is type O. Parents who are I^A i and I^B i can have children of all four types, each with probability 1/4.

The history came in pieces. Karl Landsteiner described groups A, B and a third he called C (now O) in 1901; AB was reported in 1902 by Decastello and Sturli; and Felix Bernstein’s analysis of population frequencies in the 1920s showed that three alleles at one locus fit. In 1990 Fumi-ichiro Yamamoto, Sen-itiroh Hakomori and colleagues reported that the A and B alleles they cloned differ at four amino acids, and that the O allele they studied lacks a single base, which shifts the reading frame.

Alleles that kill: the 2:1 ratio

Back to the yellow mouse. The yellow allele, A^y, is dominant for coat colour: A^y/a mice are yellow, a/a mice are non-agouti (black). It is recessive for survival: A^y/A^y embryos die before implantation. William Castle and Clarence Little proposed this in 1910, after finding that yellow by yellow matings gave about two yellow young for every non-yellow one.

Worked problem, part 1. Cross A^y/a x A^y/a. Assume equal transmission of the two alleles, random fertilization, and complete early death of A^y/A^y.

Step 1. Conceptions. 1/4 A^y/A^y, 1/2 A^y/a, 1/4 a/a.

Step 2. Remove the dead class. One quarter of conceptions die. The survivors are 1/2 and 1/4, in the ratio 2:1.

Step 3. Phenotypes among live pups. 2/3 yellow (A^y/a), 1/3 black (a/a). In 12 live pups, expect 8 yellow and 4 black.

Step 4. Litter size. A^y/a x a/a gives 1/2 A^y/a and 1/2 a/a, and none die. With the same number of conceptions, the yellow by yellow litters average one quarter smaller.

So every live yellow mouse is a heterozygote, which is why none breeds true. A recessive lethal allele is one that kills in two copies while one copy is survivable; the 2:1 is a ratio among survivors, and the 1:2:1 still holds at conception.

Two genes, one trait: epistasis and the 9:3:3:1 family

When two genes act on one trait, a genotype at one can hide the effect of the other. This is epistasis. The gametes and their sixteen combinations stay exactly as in a dihybrid cross; only the phenotype labels change.

Worked problem, part 2. In Labrador retrievers, the B allele (black pigment) is dominant to b (brown pigment), and the E allele (lets pigment be deposited in the hair) is dominant to e. An ee dog is yellow whatever its B genotype. Assume the two loci are unlinked, and that both parents carry alleles at the other coat-colour loci (K among them) that let B and E show these effects; the example holds only for these two loci. Cross BbEe x BbEe.

Step 1. Each locus alone. P(B_) = 3/4, P(bb) = 1/4, P(E_) = 3/4, P(ee) = 1/4.

Step 2. Multiply and label.

B_E_ = 3/4 x 3/4 = 9/16 black; bbE_ = 1/4 x 3/4 = 3/16 chocolate; B_ee + bbee = 3/16 + 1/16 = 4/16 yellow

The ratio is 9:3:4, a 9:3:3:1 with the last two classes merged. Because ee masks the B locus only when homozygous, this is recessive epistasis.

Step 3. Counts. In 64 pups, expect 64 x 9/16 = 36 black, 64 x 3/16 = 12 chocolate and 64 x 4/16 = 16 yellow, which sums to 64.

Step 4. A conditional question. Among yellow pups, what fraction is bbee? Of the 4 yellow sixteenths, 1 is bbee, so P(bbee given yellow) = 1/4.

Other interactions regroup the same sixteen classes:

Ratio	Model	Classic case
9:3:4	recessive epistasis	Labrador coat; mouse albino (cc)
9:7	both dominant alleles needed (complementary genes)	sweet-pea flower colour
12:3:1	a dominant allele at one gene masks the other	summer squash fruit colour
15:1	a dominant allele at either gene is enough (duplicate genes)	shepherd’s purse seed capsule shape

How the two genes act on the trait

Classes under the chosen rule: sixteenths, expected count and observed count in the sample

Class	Sixteenths	Expected of 64	Observed of 64

Resample 64 progeny

The gametes and their sixteen combinations never change; only the phenotype labels do. Expected counts are 64 x (sixteenths) / 16, and the observed counts scatter around them. This is a sketch of the logic, not data.

Humans have a recessive epistasis too. Rare hh people make no H antigen, so neither transferase has anything to modify, and their cells type as O whatever their ABO genotype.

Penetrance and expressivity: genotype does not fix phenotype

So far each genotype gave one phenotype. Often it does not. Penetrance is the fraction of carriers of a genotype who show the trait. If 70 of 100 carriers are affected, penetrance is 70%. It describes a group, never a person. Expressivity is how strongly the trait shows in those who have it; with variable expressivity, the same genotype gives mild disease in one carrier and severe disease in another.

Three things loosen the map. Other genes: in UK Biobank participants carrying variants for familial hypercholesterolemia, the modelled probability of coronary disease by age 75 ranged from 17.5% to 77.9% across polygenic-score percentiles. Chance in development. And environment, of which phenylketonuria (PKU) is the clearest case. People homozygous for loss-of-function alleles of the gene for phenylalanine hydroxylase cannot convert phenylalanine to tyrosine. On an ordinary diet phenylalanine accumulates and causes severe intellectual disability; in children kept on a low-phenylalanine diet from early infancy, that outcome is largely prevented. The genotype is the same in both cases. What differs is what the diet supplies.

Design problem

Two true-breeding white-flowered sweet-pea lines are crossed, and all the F1 plants are purple. Explain how two white parents give a purple child, predict the F2 from selfed F1 plants, and design crosses that show two genes are involved.

One way to do it

Suppose purple pigment is made in two enzymatic steps, each needing a dominant allele at a different gene: A for step 1, B for step 2. Each white line is homozygous recessive for a different step, aaBB and AAbb. The F1 is AaBb and has a working copy of both steps, so it is purple.

Selfing AaBb with independent assortment gives 9 A_B_ purple, and 3 A_bb + 3 aaB_ + 1 aabb = 7 white: a 9:7 ratio.

Tests. Cross the F1 back to each parent line. AaBb x aaBB gives all B_, half Aa, so 1 purple : 1 white; AaBb x AAbb gives 1 purple : 1 white as well. A one-gene model with a dominant purple allele cannot be made from two true-breeding white parents in the first place. Then intercross F2 whites: an aaB_ plant crossed with an A_bb plant can give purple offspring, while two whites blocked at the same step give only white. Whites therefore fall into classes that do or do not restore colour together. That is the logic of complementation and pathways, taken up later.