Question

For problem 1, to compute probabilities, the probability of event 1 AND event 2 is obtained...

For problem 1, to compute probabilities, the probability of event 1 AND event 2 is obtained by multiplying the two probabilities together, whereas the probability of event 1 OR event 2 is obtained by adding the two probabilities.

1.Duchenne muscular dystrophy (DMD) is a condition that causes muscle degeneration and then death. Most people with the condition are wheelchair-bound by age 12 and dead before age 30.

The condition is due to a mutation in the gene for a protein, dystrophin, that helps to hold in place the thin filaments of a muscle cell. The dystrophin gene is located on the X chromosome, so it is sex-linked; because the gene is found on the X chromosome, and because the mutant (= DMD) allele is recessive, when DMD occurs it usually occurs in males.

a.If your mother does not have DMD but her brother did – he’s dead now – then what is the probability that you have received the DMD allele?  (You can assume that your mom’s parents – your grandparents – are normal, and that your father is normal as well.)

b.If your father does not have DMD but his late brother did, then what is the probability that you have received the DMD allele?  For this problem, assume that your mother is homozygous for the normal allele.

2.Male house cats are either black or orange; females are black, orange, or tortoiseshell (mixed black and orange).

These colors result because coat color is governed by a sex-linked gene. If we use B for black and O for orange, then males can be XBY (black) or XOY (orange), whereas females can be XBXB (black), XOXO(orange), or XBXO (tortoiseshell, a mosaic with black and orange patches).

a.Determine the phenotypes expected in the offspring of a cross between an orange female and a black male.

b.Determine the phenotypes expected in the offspring of the reciprocal cross – between a black female and an orange male.

c.Half of the females produced by a certain kind of mating are tortoiseshell and half are black; half of the males are orange and half are black. For this mating, what colors are the male and female parents?

d.Another kind of mating produces offspring in the following proportions: 1/4 orange males, 1/4 orange females, 1/4 black males, and 1/4 tortoiseshell females. For this mating, what colors are the male and female parents?

For problem 3, the genes are unlinked and autosomal, and dominance is complete.

3.In corn with purple kernels, the production of purple pigment requires two genes, the A gene and the B gene. If an individual corn plant is homozygous dominant or heterozygous for the A geneAND homozygous dominant or heterozygousfor the B gene, then purple pigment is produced. Conversely, if one or both of the genes is homozygous recessive, then no pigment is made and the kernels are white.

a.In the P generation, there are two varieties of corn – one with genotype AAbb and the other with genotype aaBB. What are the phenotypes of these two strains?

(1)phenotype of AAbb

(2)phenotype of aaBB

b.If these two strains are crossed (AAbb x aaBB) then the offspring – the F1 generation – are expected to have what genotype(s) and phenotype(s)?

c.If F1 individuals are allowed to self-pollinate, then the offspring – the F2 generation – are expected to have what genotype(s) and phenotype(s)? Be sure to use a Punnett square to solve this part of the problem.

d.This is an example of epistasis. If there had NOT been epistasis – if this had, instead, been a typical dihybrid cross with complete dominance – then what ratio of phenotypes would you have expected to see in the F2generation?


Homework Answers

Answer #1

father is normal. Since Duchenne muscular dystrophy (DMD) is an X linked disorder, we can write down the genotypes of the parents as,

XdX X XY

F1 generation: XdX ; XX ; XdY ; XY

Thus the probability of having an affected kid is 1/4 or 25% and the probability of having an affected son is 1/2 or 50%.

b) For this scenario, the father's genotype would be XY since he is not affected. The mother's genotype would be XX. Since there is no carrier and none affected, the probability of the F1 generation having affected children is 0%.

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