Question

Consider three independently assorting gene pairs, A/a, B/b, and C/c, where each demonstrates a typical dominance (A-, B-, C-) and recessiveness (aa, bb, cc) relationship. What is the probability of obtaining an offspring that is triply recessive from parents that are AaBbcc and AaBbCc? What is the probability that the offspring will be recessive for only two traits?

Answer 1

Answer:

AaBbcc x AaBbCc ------Parents

Aa x Aa = AA (1/4), Aa(1/2) & aa(1/4)

Bb x Bb = BB (1/4), Bb (1/2) & bb(1/4)

Cc x Cc = Cc (1/2) & cc(1/2)

a).

The probability of obtaining an offspring that is triply recessive from parents = aabbcc = ¼ * ¼ * ½ = 1/32

b).

aabbCc probability = ¼ * ¼ * ½ = 1/32

A_bbcc probability = ¾ * ¼ * ½ = 3/32

aaB_cc probability = ¼ * ¾ * ½ = 3/32

The probability that the offspring will be recessive for only two traits = 1/32 + 3/32 + 3/32= 7/32