Question

Rh is the most complex of the blood group types, involving at least 45 different antigens. The most clinically important antigen, D or Rh_O, is encoded by the gene RhD which is found on chromosome 1. Individuals that are Rh-positive have either one or two RhD genes, whereas the Rh-negative phenotype is caused by the absence of the RhD gene. (The antithetical allele d does not exist, however the letter "d" is used to indicate the D-negative phenotype). For the purpose of this homework, we will simplify things. Assume that the Rh blood group has only two alleles: the Rh-positive allele (D) and the Rh-negative allele (d).

            Erythroblastosis fetalis (EF) is a condition that causes the mother's red blood cells to attack those of the baby as if they were any foreign invaders. It is referred to as hemolytic anemia of the newborn. It is caused by anti-Rh antibodies from the mother which pass through the placenta and attack fetal blood cells that happen to be Rh-positive. Babies that are at risk for this condition are those with Rh-positive blood, whose mothers are Rh-negative (dd).

            Consider a population under Hardy-Weinberg equilibrium, where the frequency of the Rh-negative allele, d, is 0.3. What is the frequency of crosses that could potentially produce children with erythroblastosis fetalis?

Answer 1

Answer:

The frequency of Rh-negative allele (d) = q= 0.3

p+q=1

p = 1-q = 1-0.3 = 0.7

Hardy=Weinberg law = p^2 + q^2 + 2pq = 1

The frequency of Rh-positive allele (D) = p= 0.7

Therefore Rh-positive blood group children will be died with erythroblastosis fetalis. And theri genotype is DD or Dd

The frequency of DD = p^2 = 0.7 * 0.7 = 0.49

The frequency of Dd = 2pq = 2 * 0.7 * 0.3 = 0.42

The frequency of crosses that could potentially produce children with erythroblastosis fetalis =.0.49 + 0.42 = 0.91

Normal baby frequency= dd = q^2 = 0.3 * 0.3 = 0.09