Question

One model for the genetic determination of handedness in humans is the "random recessive" model proposed by Amar Klar. In this model, recessive homozygotes (rr) are equally likely to be right handed or left handed, whereas those with the dominant phenotype (RR or Rr) are always right handed. Suppose this model is true and that the allele frequency of r is 0.28 in a randomly-mating population.

What proportion of the population is right handed?  0.9608

Emily is right handed. What is the probability that she is a heterozygote?  0.41965

Emily's husband, Joel, is left handed and they have two children. If both children prove to be right handed, what would be your revised estimate of the probability that Emily is a heterozygote? (Hint: Remember that Emily could be any of the three genotypes. You want the probability that she is Rr given that she is right handed and that both children are also right handed) ?

Answer 1

Let r- frequency of left handed, R- frequency of right handed

In case of equilibrium R+r=1

R= 1-r= 1- 0.28= 0,72

Now r has an equal probability of being right or left handed, probability of right or left handed=0.5

probability of right handed =1 in a population that has Rr or RR

When RR X rr

F1 generation: Rr Rr

F2 generation: RR Rr Rr rr

When Rr X rr

F1: Rr rr

F2 Rr Rr rr rr

Probability of Emily being heterzygote is: (0.72*1*1 ) + 0.28* (1/4)= 0.79

1 is the probability of being heterozygote (RR, Rr) but if the genotype is rr probability of right or left handed is 1/2