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If AaBbCcDd is mated to AaBbCcDd (no gene interactions among all genes), what is the probability...

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_?

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Answer #1

AaBbCcDd × AaBbCcDd

This is a tetrahybrid cross. Convert it to 4 monohybrid crosses.

Aa × Aa = 3/4 A_, 1/4 aa = 1/4 AA, 2/4 Aa, 1/4 aa

Bb × Bb = 3/4 B_, 1/4 bb = 1/4 BB, 2/4 Bb, 1/4 bb

Cc × Cc = 3/4 C_, 1/4 cc = 1/4 CC, 2/4 Cc, 1/4 cc

Dd × Dd = 3/4 D_, 1/4 dd = 1/4 DD, 2/4 Dd, 1/4 dd

Probability of phenotype aaB_CcDd = 1/4 × 3/4 × 2/4 × 1/4 = 6/256

Probability of genotype AAbbCcD_ = 1/4 × 1/4 × 2/4 × 3/4 = 6/256

Final answer = 6/256 + 6/256 = 12/256 = 6/128 = 3/64

Answer is 3/64

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