# TopScore #34 Test 3

Discussion in 'DAT Discussions' started by drtoothy, May 31, 2008.

1. ### drtoothy 2+ Year Member

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A woman caries a sex-linked lethal gene that causes spontaneous abortions. She has six children. How many of her children would you expect to be boys?

A. None
B. One
C. Two
D. Three
E. Four

Can someone explain please? I don't really understand their explanation. The answer is C, by the way.

2. ### daviddav 5+ Year Member

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What is the chance of a child being a boy? 50 %

What is the chance that the boy will live? 50 %

Total chance of having a boy for this woman = 25%

Basically this means that half of the boys will die.

All girls will live. In a punnett square the ratio of surviving girls to boys would be 2:1.

SO 1+2=3 and to solve for this problem RATIO OF BOYS TO TOTAL CHILDREN is 1:3. Now just solve.

1/3 = X/6 X=6/3=2 X= Unknown number of boys.

Hope I didn't confuse you!

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3. ### luaxel

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The fact that the woman is a carrier means that the gene is recessive, since expression of the gene would mean she was homozygous and that would not be possible because then she would have been aborted spontaneously at birth. Now the father of the children has to be normal because since a male only has 1 X chromosome if the allele was present it would be here and expression would occur and he also would have been spontaneously aborted at birth. Now if you do a punnet square crossing Xs X (x) X Y you see that of the four possible out comes 2 are boys and 1 of those 2 would be spontaneously aborted. So of 4 pregnancies 3 children would be alive(1 normal female, 1 female carrier and 1 normal male) Multiply by 2 to get 6 total children and you end up with 4 fems and 2 males. I think the tricky part is when they ask you what we should expect. I hope this helped.

4. OP

### drtoothy 2+ Year Member

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Yeah, that definitely cleared it up! Thanks a lot!

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