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?

Question

Abraham Potter

Biology

16 August, 12:34

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?

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Answers (1)

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Rachael Mayer · Answer 1

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.