As we discussed in “lecture”, Rh is the most complex of the blood group types, i

Question

As we discussed in “lecture”, Rh is the most complex of the blood group types, involving at least 45 different antigens. The most clinically important antigen, D or RhO, 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).

Side note: Here’s a current headline that addresses this condition. It’s well worth a read!

http://www.cnn.com/2015/06/09/health/james-harrison-golden-arm-blood-rhesus/index.html

            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?

Explanation / Answer

Given that the frequency of Rh- negative allele (d) = 0.3

Frequency of Rh-positive allele (D) = 1 - 0.3 = 0.7 (Since p+q = 1)

Frequency of heterozygous alleles (Dd) = (0.7) (0.3) = 0.21

Hardy and Weinberg also described all the possible genotypes for a gene with two alleles. The binomial expansion representing this is, p2 + 2pq + q2 = 1.0

Where,

p2 = proportion of homozygous dominant individuals

q2 = proportion of homozygous recessive individuals

2pq = proportion of heterozygotes.

Thus, (0.7)2 + 2 (0.21) + (0.3)2 = 0.49 + 0.42 + 0.09 = 1

Thus, the frequency of crosses that could potentially produce eryhtroblastosis fetalis is = Dd *dd = (0.21)(0.09) = 0.0189

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.