Except the allelic frequency affects the chances that the parents are heterozygous. The parents having a 1/2 chance of being heterozygous would only be true if the allelic frequency was fifty-fifty since:
2ab=1/2⇔ a=1/(4b) (Interaction term of Hardy Weinberg)
and
a+b=1⇔a=1-b (Sum of probabilities must equal 1)
only intersect at (1/2,1/2)
This is correct. That would be affirming the consequent, I believe.
You calculated the probability of two heterozygotes mating when nothing states that diseased individuals are unable to mate. You are neglecting Rrxrr and rrxrr. You are also assuming an allelic frequency of .5 which is not what the question states.
NOPE NOPE NOPE. I think you may be confusing the red letter "B" for an indication that the child is affected when it's merely a label. Additionally, as I stated above, you are assuming diseased individuals cannot mate and thus the
only way to get a homozygous diseased is to mate heterozygotes, which isn't stated in the question.
If the offspring were assumed to have the disease, the answer would be 1 and this would be a trick question. The question never states that the child
has the disease. The question asks what is the probability that they have it without giving us direct information about the zygosity of the parents. There are no labels saying that either the parents or the child are affected.
Let R = Normal allele
Let r = Diseased allele
Since we know the allelic frequency in the population (R=2/3, r=1/3) and we assume Hardy-Weinberg equilibrium, we know that each parent has a 4/9 chance to be RR homozygous normal, 1/9 chance to be rr homozygous recessive, and 4/9 chance to be Rr heterozygous.
PROBABILITY THAT DAD DONATES DISEASED ALLELE:
- 0% chance from RR
- (Chance of r from RR)*(Weight of RR branch)=0*4/9=0
- 50% chance from Rr
- (Chance of r from Rr)*(Weight of Rr branch)=1/2*4/9=2/9
- 100% chance from rr
- (Chance of r from rr)*(Weight of rr branch)=1*1/9=1/9
0+2/9+1/9=3/9=1/3 (We didn't
actually have to do this calculation since given a random allele in the population, it'd be r 1/3 of the time as stated in the question, but I wanted to be thorough)
PROBABILITY THAT MOM DONATES DISEASED ALLELE:
The problem states nothing about sex-linked so it is the same as Dad: 1/3
PROBABILITY THAT B HAS rr GENOTYPE:
1/3*1/3=1/9. The answer is incorrect.
---------------------
Another "well duh" way to do this problem is that given a HW equilibrium population,
any individual has a (1/3)^2 chance of having the disease, so without knowing the parents' genotypes
of course the chance of the kid having the disease would be 1/9.
---------------------
Another "well duh" way to see that the answer in the slide is wrong is that 3 and 16
ARE FREAKING RELATIVELY PRIME (or coprime if you prefer)
. The only way that you could get a 1/16 out of a problem starting with a 1/3 is to at one point multiply by 3, which in single-gene genetics, basically never happens.
EDIT:
I honestly think it was a typo and your prof meant to write "1 in 4 (.25)"