【高中生物】计算两种以上遗传病概率问题搞不懂？看完这个图你可以给别人讲题了
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(3)针对丙病来说,Ⅲ-15基因型为Dd的概率为1/100,Ⅲ-16基因型为1/3DD、2/3Dd,所以其子女患丙病的概率为1/100×2/3×1/4=1/600,针对乙病来说,Ⅲ-15的基因型为XBXb,Ⅲ-16的基因型为XBY,所以他们的后代患乙病的概率为1/4,只患一种病的概率是1/600×(1-1/4)+(1-1/600)×1/4=602/2 400=301/1200,患丙病女孩的概率为1/100×2/3×1/4×1/2=1/1 200。
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