Question

17. (8 points) Below is a pedigree from a family that lives in a population at equilibrium where the frequency of an autosomal recessive genetic allele is 1%. II-1 is affected by the disease. What is the probability that the child (III-1) will have this disease? Show your work and circle your answer. =p(dad is Aa) x p(mom is Aa) x 4 (2 points) = 2/3 x (2pq/p^2 + 2pq) x 74 (2 points) = 2/3 x (2*0.01) x 14 = 0.00333=0.333% (4 points)

Hi can someone explain this answer to me? What is the p and q values and what is canceling out to just give us that 2*0.01 part?

Answer 1

III I-2 TOI-3 -2 for I - I to be an UT-I to be an affected recessive honozygote be she must in bedt an recessive allele ca from father as well as mother, Given that I-I must be aa b Parents (I-1 & I-2) must be Aa 1 - 2 shows dominant phenotype therefore atleast bets he has one A (dominant) allele. I 3 is from outside of affected pedigree and can be assumed as AA The probability that I-2 is heterozygous : - Pedigree and 4 jous 5 Aa x Aa T-2 is ag cannol Fas we know L not affected be considered Ala VAATA a taa faa that III - 1 is heteroz y qous ; A so probability o a * probability that father is heterozygouy 1 a 0.666 XO:S 2 3333 what he will inhesit defective allele form Father's side) 0.3333 % . Answer is