What is the genotype of the woman?
Posted by adminFeb 28
“A woman whose brother had cystic fibrosis marries a man who had a child with cystic fibrosis from a previous marriage.”
That is the information I got.
Cystic Fibrosis is autosomal recessive. I figured out that the wife’s mom and dad genotypes were Aa. And the brother’s genotype is aa, because he had cystic fibrosis. (Check me on this)
But isnt there not enough information? Couldnt the wife have an: AA or Aa genotype?
3 comments
Comment by kiddietyte on 02/28/2010 at 7:03 pm
Yes. it is not enough information.
Assuming the woman does not have cystic fibrosis (if she did, she would probably be dead), and using the fact that her parents were both heterozygous for the trait, the woman would have AA or Aa.
Cross the parents:
……A…..a
A…AA..Aa
a….Aa…aa
25% AA, 50% Aa, 25% aa.
If you need a BEST answer, the best answer would be that the woman is Aa genotype. This is because there is a greater possibility of her being Aa than AA. Refer to the punnett square for proof. 25% chance of AA while 50% chance of Aa.
Comment by xochi_13 on 02/28/2010 at 7:57 pm
You are correct. The wife’s parents are both Aa because they didn’t have CF but they had a kid who did. The wife has a 2/3 probability of being Aa and a 1/3 probability of being AA.
The woman’s husband must be an Aa heterozygote. You can’t get any further than that unless you are told something about their kids.
The probability of them having a kid with CF would be
probability of her being heterozygote * probability of two heterozygotes having a kid with CF, or 2/3 * 1/4 = 1/6.
Comment by ecolink on 02/28/2010 at 8:36 pm
You’re right about the wife, her parents, and her brother.
Wife is either AA or Aa.
Husband is Aa because he has a child with CF from previous marriage.
If the wife is AA, then none of their children will have CF, but they might be heterozygous and be carriers.
If the wife is Aa, then we would expect 3 normal: 1 CF among their offspring. Half of their children (2/3 of the normal ones) would be expected to be carriers.