Question

The Manx cat has abnormal spinal development and no tail. The mutation is a dominant allele (M). For notation purposes, m refers to the wild-type allele that results in a tail. If two Manx cats have a litter with 4 vigorous kittens, what is the probability of at least one of the four kittens having a normal tail?
a. 65/81 0
b. 16/81
c. 2/3
d. 1/3
e. 80/81

Answer 1

Answer:

d. 1/3

Explanation:

Given, M is a dominant allele which is responsible for tailless kittens and m is recessive wild type allele which leads to kittens with tail. M is a lethal allele which means that if it is present in homozygous condition, the kitten will not survive. So the Manx cats should be heterozygous i.e. Mm.

When they mate, Mm X Mm :

      M        m

M   MM    Mm

m   Mm    mm

1/4 kittens will have MM genotype so they will not survive. Out of the remaining kittens, 1/3 will be mm and hence they will have normal tail. Remaining 2/3 will have Mm genotype like their parents and will be born without tail. Hence, there is 1/3 probability that one of the four kittens will have a normal tail.