4) Duchenne muscular dystrophy is due to a sex-linked recessive gene. A normal m

Question

4) Duchenne muscular dystrophy is due to a sex-linked recessive gene. A normal male and a normal temale have children. Their first child (a boy) has Duchenne muscular dystrophy a) What are the possible genotypes of a second son (if they had one)? b) What are the chances their third son, if they had one, would have the disorder? c) What are the possible genotypes for the mother and their daughter who does not have the disorder? 5) In some cats the gene for tail length shows incomplete dominance. Cats with long tails and cats with no tails are homozygous for their respective alleles. Cats with one long tail allele and one no tail allele have short tails. For each of the following construct a punnett square and give phenotypic and genotypes and the ratios of the offspring. a) a long tail cat and a cat with no tail b) a long tail cat and a short tail cat c) a short tail cat and a cat with no tail d) two short tail cats 6) Clouded leopards are a medium sized, endangened srecies of cat, living in the very wet cloud forests of Central America. Assume that the normal spot Xare a dominant, sex-linked trait and that darik spots are the recessive counterpart. Suppose as s Couservation Biologist, you are involved in a clouded leopard breeding program. One year you cross a male with dark spots and a female with normal spots. She has four cubs and, conveniently, two are male and two female. One each of the male and female cubs has normal spots and one each has dark spots. a) What is the genotype of the mother? Suppose a few years later, you cross the female cub that has normal spots with a male that also has normal spots. What are the possible genotypes for their female and male offspring? b)

Explanation / Answer

1)The sex linked recessive gene is DMD situated in X chromosome

Females are carrier , being a recessive disorder it is only possible to transmit when the mother is affected as the son will get his X chromosome from his mother only

so the father was XY and the mother was XdX

The mother was a carrier the daughter can be a carrier XdX or normal XX

The first son had XdY, the second and third son have 25% chance of being affected.

2) Cat with long tail have genotype LL , no tail ll

Short tail Ll

A long tail cat and no tail cat

LL x ll= Ll ( short tail) (all)

A long tail cat and ssort tail cat

LL x Ll = LL and Ll ( one long tail and one short tail) (1:1)

A short tail and no tail cat

Llx ll= Ll and ll ( short tail and no tail) (1:1)

Two short tail cats

LlxLl= LL ,LI, ll ( long tail. short tail, no tail) (2:1:1)

3)As they had four cubs with one cub of each gender carrying the dominant trait

So the female cubs have genotype XNX and XX

Male cuba have genotype XNY and XY

So the mother had genotype XNX

If we cross XY with XNX we get the above results

The normal cubs will have normal children because this trait was a dominant one so the affected will show the results.

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