...a)和乙病(显性基因B,隐性基因b)两种遗传病的系谱.请据图回答:(1...
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发布时间：2024-10-23 21:26
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（1）由以上分析可知，甲病为常染色体显性遗传病．
（2）①由以上分析可知乙病是隐性遗传病，若Ⅱ3和Ⅱ8两者的家庭均无乙病史，说明乙病为伴X染色体隐性遗传病．Ⅱ5的基因型为aaXBXb；Ⅲ12的基因型为AAXBY或AaXBY．
②Ⅲ9的基因型及概率为aaXBXB（12）、aaXBXb（12），Ⅲ12的基因型及概率为13AAXBY、23AaXBY，他们的子女不患甲病的概率为23×12＝13，不患乙病的概率为1-12×14＝78，因此后代正常的概率为13×78＝724，患病的概率为1-724＝1724．
（3）①若乙病为白化病，则Ⅱ3、Ⅱ4、Ⅲ9和Ⅲ10的基因型分别为Bb、Bb、B_、bb，其中只有Ⅲ10号没有B基因，再根据电泳结果图可知，7.0和2.7两个电泳带来自白化病基因．
②个体d即为Ⅲ10，其基因型为bb；根据题意，人群中每10000个人当中有1个白化病患者，即bb的基因型频率为110000，则b基因频率为1100，B的基因频率为99100，根据遗传平衡定律，BB的基因型频率为99100×99100，Bb的基因型频率为2×1100×99100，因此正常人中基因型为Bb的概率为BbBB+Bb=2101．因此，d与一个表现型正常的人结婚，生育一个患白化病的概率为2101×12＝1101．
③白化基因上有酶E的切割位点，而正常基因上不存在，因此切割结果不同．
故答案为：
（1）常染色体显性
（2）①aaXBXb AAXBY或AaXBY ②1724
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