Question

You are following segregation of five unlinked genes in a cross – A, B, C, D and E. Dominant alleles are capital letters and recessive are lower case. Being homozygous recessive for all five genes produces a multigenetic trait such as bipolar disorder. Two unaffected parents with the genotypes – aa bb Cc Dd Ee x aa Bb Cc dd Ee mate. What is the frequency expected for one child to have the affected genotype of being all homozygous recessive genotype – aa bb cc dd ee?

Answer 1

Answer:

1/64 = 0.015625

Explanation:

Genotype of parents:  aa bb Cc Dd Ee x aa Bb Cc dd Ee

To calculate the frequency of the child with desired genotype, individual frequency of desired genotype can be calculated followed by application of product rule.

aa x aa = All aa

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

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

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

Ee x Ee= 1/4 EE: 1/2 Ee: 1/4 ee

Expected frequency of the child with desired genotype= 1 aa x 1/2 bb x 1/4 cc x 1/2 dd x 1/4 ee = 1/2 x 1/4 x 1/2 x 1/4 = 1/64 = 0.015625