Question

For four genes (all autosomal recessive), beginning with a AaBbCcdd x aabbCcDd cross, what is the probability of producing a child in the F1 with the A- bb C- dd phenotype?

Answer 1

Answer:

3/32

Explanation:

Given genotypes of parents: AaBbCcdd x aabbCcDd  

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

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

Cc x Cc = 3/4 C- : 1/4 cc

dd x Dd = 1/2 Dd : 1/2 dd

The total probability of a child with desired the genotype A-bbC-dd can be calculated by the rule of multiplication.

Therefore, the probability of the child with the genotype A- bb C- dd would be= 1/2 Aa x 1/2 bb x 3/4 C- x 1/2 dd = 3/32.