Genetics Problem Set

1.

a) The mutation is dominant and the mice are heterozygous so the ratio for the offspring should be ¾ short ear and ¼ normal ear.

b) It is possible that the homozygous mutant genotype is lethal in which case only the heterozygous mutants would have short ears. Our new ratio will be 2:1 for short ear to normal.

c)

For chi-squared we have a sum from 1 to N of ((observed-expected)^2)/(expected). For our equation we want to have observed = 0.66n and expected is 0.75n we also need to sum this with a second equation where observed is 0.33n and expected is 0.25n. Since we are working with two alleles we have one degree of freedom so this must add up to 3.814 for 0.05 confidence. In the end:

[(0.33n-0.25n)^2]/0.25n + [(0.66n-0.75n)^2]/0.75n = 3.841

n = 106

d) If the inheritance pattern is the same it is a dominant mutation that causes the short-eared phenotype. If we assume that the two mutations are on the same gene we can cross a het from each and now the phenotypic ratio will be 1/3 normal and 2/3 short ear since the homozygous mutant is lethal. If there are two genes we can again cross the two hets and will see ¾ short ear and ¼ long ear because it is not possible to have a homozygous mutant in this cross since they are on different genes and we only need one copy of the mutation to have the short-eared phenotype.

2.

a) The F2 generation is in the ratio of 9:6:1 which is similar to a standard dihybrid cross which gives 9:3:3:1. This ratio implies the coat pigment is controlled by 2 genes with alleles which we can refer to as R and Y. So the initial population 1 is RRyy and population 2 is rrYY since they are true breeding but give a different phenotype when crossed with each other. With this cross all of the F1 are dark brown with genotype RrYy. For the F2 we have RRYY, RrYY, RrYy which give the dark brown phenotypes; rrYy, rrYY, RRyy, and Rryy which give the light coat phenotype, and rryy which gives the albino phenotype. The punnet square gives all of these ratios as 9:6:1. This ratio is a form of epistasis known as duplicate interaction where the presence of a dominant allele on one gene but not the other gives the same phenotype as the presence of the opposite case.

b) Continuing the logic of two genes controlling coat color from the previous section we can say the standard genotype would be RRyy for the light colored population 1. The question now is whether the revertant is RRYy or RRYY to give it the dark brown color. Since subsequent generations still have the light brown phenotype the recessive allele still exists in the population and thus the revertant must be RRYy to pass on the recessive allele. The observed ratio is a form of epistasis with ratio 13:3 known as dominant and recessive epistasis.

c) In order for a phenotypic ratio to give 15:1 as seen there must be a double homozygous genotype that gives the 1 phenotype while the presence of any dominant allele present gives the more common phenotype. This is a form of epistasis known as duplicate dominant epistasis. Essentially the presence of one dominant allele from either gene expresses the same phenotype.

d) In this example we have another dihybrid cross with two genes with alleles we’ll call R and G and we have a phenotypic ratio of 9:4:3. If both dominant alleles are present we have a dark brown phenotype, if only G is present it will give a gray phenotype, if only R is present it will give light brown, and if neither dominant allele is present it gives gray.

Po rrgg (gray) X RRgg (light brown)

F1 RrGg x RrGg

F2 R_G_ (dark brown) rrG_ (gray) R_gg (light brown) rrgg (gray)

RG Rg rG rg

RG RRGG RRGg RrGG RrGg

Rg RRGg RRgg RrGg Rrgg

rG RrGG RrGg rrGG rrGg

rg RrGg Rrgg rrGg rrgg

We can see this cross gives the 9:4:3 ratio observed which is a form of epistasis known as recessive epistasis. From a biochemical perspective it appears that these genes represent pigments, which result in specific colors. R is a light brown pigment which combined with G gives a dark brown phenotype because of the additive effect of the pigments. r is an absence of pigment causing rrG_ to have a gray phenotype, but g is a gray pigment rather than an absence of pigment as well, but it is not visible in the presence of the R allele either due to the epistasis effects. Finally, R_gg gives light brown because the brown phenotype overrules the gray from the g allele.