Question

1. If AaBbCcDd is mated to AaBbCcDd (no gene interactions among all genes), what is the probability that the offspring will have either the phenotype aaB_CcDD or the genotype AAbbCcD_? (assuming complete dominance for genes) (Show procedure, NO score for direct result)

Answer 1

For such questions we calculate probability of each single genotype and multiply the probabilities.

When AaBbCcDd is mated with AaBbCcDd,

Then probablity of aa will be 1/4

Probability of B_ ( BB and Bb ) will be 3/4

Probability of Cc will be 2/4

And probability of DD will be 1/4

So, probability of aaB_CcDD = 1/4 X 3/4 X 2/4 X 1/4 = 3/128

Similarly,

Probability of AA will be 1/4

Probability of bb will be 1/4

Probability of Cc will be 2/4

Probability of D_ ( DD and Dd ) will be 3/4

So, probability od AAbbCcD_ = 1/4 X 1/4 X 2/4 X 3/4 = 3/128

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