Imagine that you are a genetic counselor,and a couple planning to start a family comes to you for information.Charles was married once before,and he and his first wife had a child with cystic fibrosis.The brother of his current wife,Elaine,died of cy

Imagine that you are a genetic counselor,and a couple planning to start a family comes to you for information.Charles was married once before,and he and his first wife had a child with cystic fibrosis.The brother of his current wife,Elaine,died of cy
Imagine that you are a genetic counselor,and a couple planning to start a family comes to you for information.Charles was married once before,and he and his first wife had a child with cystic fibrosis.The brother of his current wife,Elaine,died of cystic fibrosis.What is the probability that Charles and Elaine will have a baby with cystic fibrosis?(Neither Charles nor Elaine has cystic fibrosis.)

Imagine that you are a genetic counselor,and a couple planning to start a family comes to you for information.Charles was married once before,and he and his first wife had a child with cystic fibrosis.The brother of his current wife,Elaine,died of cy
题目大致意思：假若你是个遗传顾问,那么一对准夫妻向你咨询一些问题.男方Charles结果一次婚,和前期有过一个患病（cystic fibrosis）的孩子.女方elaine的哥哥死于该病.问他们如果生一个孩子患病的可能有多大?（前提是男女方都没患该病）.
1、首先男方必然带有该基因,孩子才可能患病.那么可以先确定的是男方的基因型为Cc.
2、女方的哥哥死于该病,说明父母都有该基因.但是有一个问题,该病会导致不孕.如果考虑这个因素的话,说明女方父母都不会是纯合,应该都是Cc.那么女方的基因型有可能是CC或者Cc.不考虑的话你就按一楼的要考虑女方父母的基因型的问题,但是一楼有一个明显的失误是女方不换病,所以不会有可能是dd.
3、也就是说男方贡献c的几率是1/2,女方贡献c的几率是1/4.所以应该是1/8.或者你分情况讨论,女方是显性纯合的话患病几率为0,是杂合的话,几率为1/4,所以患病的几率为(0+1/4)*1/2=1/8,这个1/2就是说女方是Cc的几率.

cystic fibrosis 是常染色体上的隐形疾病，也就是说DD是正常人，Dd是携带者，dd是患者。
本题中因为charles无疾病，但他的第一个孩子是患者，所以他的基因型为Dd
charles现任妻子的哥哥是患者，所以哥哥的基因型为dd。因为是dd，所以charles现任妻子的父母不可能是DD，也不可能同时是dd。由于题目没说妻子父母的患病情况，所以妻子父母为Dd,Dd的可能...

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cystic fibrosis 是常染色体上的隐形疾病，也就是说DD是正常人，Dd是携带者，dd是患者。
本题中因为charles无疾病，但他的第一个孩子是患者，所以他的基因型为Dd
charles现任妻子的哥哥是患者，所以哥哥的基因型为dd。因为是dd，所以charles现任妻子的父母不可能是DD，也不可能同时是dd。由于题目没说妻子父母的患病情况，所以妻子父母为Dd,Dd的可能性是50%，Dd,dd的可能性是50%。
按照妻子父母基因推断，妻子的基因型为DD的可能性是50%*1/4=1/8，为Dd的可能性是50%*50%+50%*50%=1/2，为dd的可能性是50%*1/4+50%*50%=3/8。
若妻子为DD，孩子不会得病。若为Dd，得病率为50%*1/4=1/8，若为dd，概率是3/8*1/2=3/16。
所以孩子得病总概率是1/8+3/16=5/16。
打字好痛苦。。。

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The answer is 1/6.
We get some basic knowledge or, say, common sense that people with cystic fibrosis are sterile, that means they could not have children. Thus, we decude Elaine's parents are bot...

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The answer is 1/6.
We get some basic knowledge or, say, common sense that people with cystic fibrosis are sterile, that means they could not have children. Thus, we decude Elaine's parents are both Dd genotype, and Elaine herself is 1/3 DD, 2/3Dd. Charles is Dd for sure. So Charles provide 1/2 "d" containing gamete, and Elaine produce (1/2)*(2/3)=1/3 "d" containing gamete. So their presumed baby would be 1/6 cystic fibrosis patient.
Thanks.

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