Question

The gene for manx tail in cats is an embryonic lethal in homozygous state, such that: MM = lethal (progeny not observed at birth)

Mm = manx tail

mm = normal tail

Another trait in cats, the short hair (S) is dominant over long hair (s) so that:

S_ = short hair

ss = long hair

The two traits are controlled by genes that assort independently.

A) What is the genotypic and phenotypic distribution from an inter se mating?

B) What is the probability of a kitten with long hair given it has manx tail?

C) What is the probability that kittens in a litter are born in the following order: manx tail long hair, normal tail long hair, manx tail short hair?

D) What is the probability that in a litter of 7, 3 kittens will have manx tail.

Answer 1

A) What is the genotypic and phenotypic distribution from an inter se mating?
MmSs x MmSs (Manx tail, short hair)
2/3 Mm + 1/3 mm
3/4 S_ + 1/4 ss

(1/2)    6/12    MmS_ manx, short
(1/6)    2/12    Mmss manx, long
(1/4)    3/12    mmS_ normal, short
(1/12)    1/12    mmss normal, long

B). What is the probability of kitten with long hair given it has manx tail?
P(ss / Mm) = P(ss and Mm) / P(Mm) = (1/6) / (2/3) = 1/4

C) What is the probability that kittens in a litter are born in the following order: manx tail
long hair, normal tail long hair, manx tail short hair?
Independent events: = 1/6 x 1/12 x 1/2 = 1/144 = 0.0069

D) What is the probability that in a litter of 7, 3 kittens will have manx tail.
N = total, M = manx, NT = normal tail
N = 7, M = 3, NT = 4

$\frac{N!}{M!T!}[P(M)^3 \times P(T)^4]=\frac{7!}{3!4!} = (8/12)^3 \times (4/12)^4 = 0.12$