Question

Suppose two individuals with the genotype AaBbCc are mated. Assuming that the genes are not linked, what fraction of the offsprings is expected to be homozygous recessive for the three traits?

A) 1/4
B) 1/8
C) 1/16
D) 1/64

Answer 1

D. 1/64

Step-by-step explanation:

The desired genotype for the progeny = aabbcc (homozygous recessive for all the three loci). The probability of getting a progeny with a homozygous recessive genotype can be calculated during the product rule.

The parent genotypes= AaBbCc x AaBbCc

Aa x Aa = 1/4 AA: 1/2 Aa: 1/4 aa

Bb x Bb = 1/4 BB: 1/2 Bb: 1/4 bb

Cc x Cc = 1/4 CC: 1/2 Cc: 1/4 cc

Therefore, the probability of getting a progeny which is homozygous recessive for all the three traits (aabbcc)= 1/4 aa x 1/4 bb x 1/4 cc = 1/64