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?

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.

try using substitution:

x + y - 6 = x - y. (i'm not giving you the exact answer). since 0 = x - y, then sub 0 in the equation with x - y. combine your like terms, and then use inverse operations, and you'll have your answer.

: -)

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