题目描述:
某生物个体经减数分裂可产生的配子种类及比例为Yr:yR:YR:yr=3:3:2:2,若该生物自交,则其后代出现纯合体的概率是( )A. 6.25%
B. 13%
C. 25%
D. 26%
答案解析
【答案】由题意知,该生物个体产生的配子的类型及比例为Yr:yR:YR:yr=3:3:2:2,雌雄配子结合形成的子代中的纯合子是YYrr、yyRR、YYRR、yyrr,因此纯合子的比例是:YYrr+yyRR+YYRR+yyrr=310×310+310×310+210×210+210...
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