Question

AaBbCc*AaBbCc杂交后代中只含有3个显性基因的概率是多少?

Answer 1

5/16

首先A、B、C三个位点上每一个是显性纯合的几率都是1/4,是杂合的几率是1/2,隐性纯合的几率是1/4.

所以一个显性也没有的几率是(1/4)*(1/4)*(1/4)=1/64

有一个是(1/2)*(1/4)*(1/4)*3=3/32

有两个是(1/4)*(1/4)*(1/4)*3+(1/2)*(1/2)*(1/4)*3=15/64

有三个是(1/2)*(1/2)*(1/2)+(1/2)*(1/4)*(1/4)*6=5/16

剩下的以此类推.