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

Question

Biology

Alaina Bradley

14 July, 13:13

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

+4

Answers (1)

Know the Answer?

Zaire Curtis · Answer 1

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.