下图为某家族甲、乙两种遗传病的系谱图。甲遗传病由一对等位基因(A，a)控制，乙遗传病由另一对等位基因(B，b)控制，这两对等位基因独立遗传。已知Ⅲ-4携带甲遗传病的致病基因，但不携带乙遗传病的致病基因。
(1)甲遗传病的致病基因位于_____(X／Y／常)染色体上。乙遗传病的致病基因位于_____染色体上。
(2)Ⅱ-2的基因型是_____。Ⅲ-3的基因型是_____。
(3)若Ⅲ-3和Ⅲ-4两人生一个孩子，则此孩是同时患甲乙两种遗传病男孩的概率是_____。
(4)若Ⅳ-1与一个正常男性结婚，则他们生一个患乙遗传病男孩的概率为_____。
正确答案及解析
正确答案
解析
(1)常：X
(3)1/24
(4)1/8
(1)根据“无中生有”“有中生无”判断甲、乙两种病都是隐性遗传，Ⅰ-1为患甲病女性，而其儿子正常，所以甲病为常染色体隐性遗传病；Ⅲ-3和Ⅲ-4均不患乙病，并且Ⅲ-4不携带乙遗传病的致病基因，因此所生儿子的致病基因只能来自Ⅲ-3，由此判断乙病为X染色体隐性遗传病。
(2)由遗传系谱图可知，Ⅱ-2不患甲病，其母亲患甲病，对甲病来说基因型为Aa，Ⅱ-2不患乙病，其外孙患乙病，因此Ⅱ-2对乙病来说基因型为XBXb，故Ⅱ一2的基因型为AaXBXb；Ⅲ-3及双亲不患甲病，但有一患甲病的兄弟，因此对甲病来说基因型为Aa或AA，Ⅲ-3是不患乙病，但有患病的儿子，对乙病来说基因型是XBXb，因此Ⅲ-3的基因型为AAXBXb或AaXBXb。
(3)分析遗传系谱图可知，Ⅲ-3基因型为2/3AaXBXb，由题意和遗传系谱图可知Ⅲ-4的基因型是AaXBY，他们的后代同时患两病的概率是2/3×1/4×1/4=1/24
(4)对乙病来说，Ⅲ-3基因型XBXb，Ⅲ-4的基因型是XBY，Ⅳ-l的基因型为XBXb的概率是1/2，正常男性XBY婚配后，他们生一个患乙遗传病男孩的概率是1/2×1/4=1/8。
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