sex-linked gene question

[USCdh]

I can't quite figure out how to get the answer 2 boys from this question.

A woman carries a sex-linked lethal gene that causes spontaneous abortions. She has 6 children. How many of her children would you expect to be boys?

a. none
b. one
c. two
d. three
e. four

Answer is 2. I realize that when you cross a heterozygous female with a normal male you get 50% carries, 50% are aborted. However I cant figure out what to put into a punett square after to determine that 8 total offspring, 4 females and 4 males are found, however 2 of the males will be aborted leaving 2 normal.


thanks!

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[DDS4tehwin]

I can't quite figure out how to get the answer 2 boys from this question.

A woman carries a sex-linked lethal gene that causes spontaneous abortions. She has 6 children. How many of her children would you expect to be boys?

a. none
b. one
c. two
d. three
e. four

Answer is 2. I realize that when you cross a heterozygous female with a normal male you get 50% carries, 50% are aborted. However I cant figure out what to put into a punett square after to determine that 8 total offspring, 4 females and 4 males are found, however 2 of the males will be aborted leaving 2 normal.


thanks!

i could have sword the answer would be 0

 

[G-Money13]

Based on the punet square only 25% of the possible genotypes would be spontaneously aborted. Based on the punet square the following results would be possible:
25%-Carrier Female
25%-Normal Female
25%-Aborted Males
25%-Normal Males

The problem I'm having with the question is how you calculate the number of boys out of 6 children since each fertilization could result in any of the 4 genotypes. There is only a 33% chance of having a boy for each birth. I guess if you used the 33% value and multiplied it into the total number of children you would get two boys; but that doesn't seem very scientific

 

[Streetwolf]

One must assume the woman is impregnated 8 times and loses 2 males, leaving her with 6 children, 2 male.

 

[DDS4tehwin]

Search is your friend.

Negative, no typo.

Woman is XX(c) where X(c) denotes mutation and X is wild-type.
Man is XY (normal).

Possible children are:

XX
X(c)X
XY
X(c)Y

Of which 3 live and 1 dies (X(c)Y). Multiply by 2 and you get 6 live and 2 die. Of the 6 children, 2 are boys.

So you expect 2 boys, 4 girls, and 2 aborted boys = 8 total. But the question asks for living boys so the answer is 2.

 

[tRNA]

Search is your friend.

yea that's what i first thought about when i saw this problem, but, here you're assuming only 2 fertilization events and there might be 6 fertilization events, so how do u know, for example, that for the first 3 fertilization events she's going to have 1normal female, 1carrier female, and one normal male, she could have 2 normal females and 1carrier females, or 2 normal males and 1carrier female,...etc. ? or is this the only way this question could be answered? hopefully u understand my point

 

[Streetwolf]

Key word is 'expect'. Take another example:

A person tosses a standard die (6 'fair' sides) 6 times in a row. How many tosses do you expect to be the number 3? How about an odd number? A number divisible by 3?

Number 3: 1 time
Odd number: 3 times
Divisible by 3: 2 times (3 and 6)

It doesn't necessarily mean it will happen but it's the expected value.

Now take your problem. You have 6 kids. How many do you expect to be boys? The point is that you look at what probability dictates and go by that. You aren't supposed to think 'oh maybe she'll have 6 normal girls this time'. You have to think 'If she has 4 children then they will all be of a different genotype, so out of every 4 children, 3 will live. Thus, out of 8 children 6 will live and that's how many she had... so since she lost 2 boys, there must be 2 boys left'.

 

[tRNA]

thanks alot streetwolf, now it makes sense