Rachel V. answered • 10/30/20

Biology, Math, and Music Tutor (B.S. in Biology and Music)

Hi Zainab! Not 100% sure this is what the question wants but I did get 1/6 ...

Let's say "A" is the dominant CF gene and "a" is the recessive gene.

We know her parents cannot be AA and aa because her brother would not have had CF and we know they cannot be aa and aa because then she would have CF. So her parents are either Aa and Aa OR Aa and aa (CF patients can live and reproduce and I guess since her parents are dead we cannot rule out the aa possibility, although I think that makes this a bit of a trick question). I think you were just missing the possibility that her parents could be Aa and aa and are probably good to go, but I'll work it out anyway.

Aa and Aa: I assume you got to 1/15 by deducting that both parents are carriers (Aa), did a Punnett square resulting in 1/3 chance the woman is AA and 2/3 chance that she is Aa (ignoring the chance of her being aa since we know she does not have CF). As you said, 2/3 times the 10% chance that it's one of the mutations they don't test for is a 1/15 chance of her being Aa

Aa and aa: If we do a Punnett square for one parent as Aa and one parent as aa, we get 50% Aa and 50% aa. Ruling out the aa option (because she does not have CF) gives us a 100% chance of Aa, or 1. 1 times 10% = 1/10

So there is a 1/15 chance if her parents are Aa and Aa and a 1/10 chance if Aa and aa. We add these chances together and 1/15 + 1/10 = 1/6 :) Hope this helps! Again, you knew what you were doing, the question was just a bit of a trick.