Question

If you meet a man with (naturally) red hair, what is the probability that neither of his parents has red hair?

Hints: About 2% of the world population has red hair. You can assume that the alleles for red hair are purely recessive. Also, you can assume that the Red Hair Extinction theory is false, so you can apply the Hardy–Weinberg principle.

Answer 1

Remember the Hardy-Weinberg Equilibrium thing?

let the prevalence of the A allele be p and the prevalence of the a allele be q

p + q = 1
p**2 + 2*p*q + q**2 = 1

that means that P(AA) + P(Aa) + P(aa) = 1

Answer 2

that's better

Answer 3

We are given that P(aa) = 0.02

Now I will step back and let you geniuses solve it :-D

Answer 4

oh and P(Aa) = 2*p*q

Answer 5

Well, you know immediately what p is and then you just need to calculate, right?

Answer 6

right

Answer 7

so you already know the values of P(AA) and P(Aa)

Answer 8

is p**2=p^2 or 2p?

Answer 9

p**2 is p^2

Answer 10

so q=sqrt0.02
p^2+sqrt0.02p+(0.02-1)=0
quadratic equation and you're done

Answer 11

how do we use P(AA), P(Aa), and P(aa) to find the probability that neither parents have aa given that the child has aa?

Answer 12

I guess I get confused easily in probability sometimes, I'm not sure.

Answer 13

let R be the probability that the child has red hair, i.e. aa

according to mendelian genetics:

P(R | aa and aa) = 1
P(R | aa and Aa) = 0.5
P(R | Aa and Aa) = 0.25

Answer 14

If it helps the parents would have a 2/4 chance of carrying the recessive alleles each. I'd help more but I've not dealt with the Hardy-Weinberg principle before now.

Answer 15

yep. The biology part ends now. Now we must apply Bayes theorem.

I need help :(

Answer 16

oh and
P(R | AA and AA) = 0
P(R | AA and Aa) = 0 :-P

Answer 17

so the only way that a guy has red hair while his parents doesn't have red hair is in the Aa and Aa case.

Answer 18

I've never heard of this theorem before. I'm just looking up info on it.

Answer 19

therefore P(Aa and Aa | R) =

(P(Aa) * P(Aa)) * P(R | Aa and Aa)/((P(Aa) * P(Aa)) * P(R) + (not (P(Aa) * P(Aa))) * P(R | not Aa and Aa))

Answer 20

or just (P(R|Aa and Aa) * P(Aa) * P(Aa))/P(R)

Answer 21

the answer becomes equal to P(AA), why?

Answer 22

I'm dizzy.

Answer 23

the probability for both parents having Aa with the son having aa is about 14.286%. I get that through the following calculation:

- forget the 2%. fact is: the little bastard got the reds, that means his parents are both non-AA.
Hardy-Weinberg:
- Aa + Aa --0.25-> aa
- Aa + aa --0.50-> aa
- aa + aa --1.00-> aa

that being said, add the three probabilities (=1,75), and divide 0.25 (neither has red hair) by that value. --> 1/7 or 14.286% is the probability.

Answer 24

I'm afraid the answer is not 1/7 :-(

Answer 25

P(R) = 0.020000
P(AA) = 0.737157, P(Aa) = 0.242843
P(R | Aa and Aa) = 0.250000
P(R | aa and AA) = 0.500000
P(R | aa and aa) = 1.000000
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P(Aa and Aa | R) = (P(R | Aa and Aa) * P(Aa) * P(Aa))/P(R)

ans = 0.737157