2. In rabbits, two genes control fur igmentation. At the color locus, the Grey a

Question

2. In rabbits, two genes control fur igmentation. At the color locus, the Grey allele is dominant to the tan allele and at the Albino locus, pigmented fur is dominant to pigment-less (albino) fur. Rabbits homozygous recessive at the albino locus have albino fur no matter what the genotype is at the color locus. The Grey and Albino loci are both on chromosome two-20 map units apart. A homozygous Tan rabbit was mated with a homozygous Albino rabbit. All the F1 rabbits had grey fur. What phenotypic ratio would you expect in the F2 generation? (Remember the genes are linked) Ocular albinism is a recessive X-linked trait that causes a lack of pigment in the iris of the eye. This results in very pale grey-blue eyes. The lack of pigment associated also causes partial blindness. Suppose a man with ocular albinism has children with a brown eyed woman whose father had ocular albinism. What is the likelihood their children will ack pigment in their retina? will the ratio be different between male and female children? 3. This pedigree shows the inheritances of an X linked dominant trait. Referring to the data in this pedigree. A. Describe is the clearest evidence that this trait is dominant? B. Describe is the clearest evidence that the trait is X linked? (Hint 4. the answer to this involves the trait being dominant)

Explanation / Answer

4)

a. This trait is clearly dominant as there is no skip in generation. In every generation, there is an individual who is affected by the disease which strongly suggests that the trait is dominant.

b. All fathers that are affected by an X-linked dominant trait have daughters who are affected by the same but not sons. When the mother is affected the sons will have a chance to be affected also. In this case, we can clearly see the same happening where affected fathers have affected daughters but not sons which implies that this is an X-linked dominant trait.

3) Ocular albinism is an X-linked recessive trait.

Hence-

Here, the man has ocular albinism, hence he is xY.

The woman is brown eyed but her father had ocular albinism, hence she must be a carrier of the disease, i.e. Xx

So, the cross becomes-
xY * Xx
The chance of x coming from the father is 1/2 and from the mother is also 1/2

so if it is a girl the chances are 1 * 1/2 = 1/2
for the boy, he will receive either x or X from his mother so the chances are 1/2

Affected Normal Male xY XY Female xx Xx / XX

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