下图为甲病（A或a）和乙病（B或b）两种遗传病的系谱图，已知Ⅰ2无乙病基因。请回答下列问题：
Ⅱ3是杂合子的几率是()，若与正常男性婚配，所生子女中患病的几率是()。
参考答案：5/6；17/24
解析：
Ⅱ3是甲病患者，AA可能性是1/3，Aa可能性是2/3，但是她不患乙病，是XBXB可能性是1/2(因为已经知道她是女性，所以只可能是或者XBXb)，所以纯合子概率就是1/3×1/2="1/6" ，杂合子的几率是5/6 。与正常男性婚配，所生子女患甲病的概率为2/3，患乙病的概率为1/8，所生子女中完全正常的概率是1/3×7/8＝7/24，患病的概率是17/24。
考点：本题考查人类遗传病及自由组合定律的知识。意在考查能理解所学知识的要点，把握知识间的内在联系，能用文字、图表以及数学方式等多种表达形式准确地描述生物学方面的内容。能运用所学知识与观点，通过比较、分析与综合等方法对某些生物学问题进行解释、推理，做出合理的判断或得出正确的结论的能力。
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