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Chapter 12. Gene frequency

Useful Calculations

Useful Calculations

The purpose of this work is to promote fluency in the language of mathematics among students who have had little experience (or perhaps an unpleasant experience) with standard math courses. Every educated person needs to become "numerate", not only to solve practical problems, but also to develop paths of communication with scientists, physicians, and engineers, whose work requires that they think in a quantitative manner.

Gene Frequencies

The best-studied human genes are those that produce medical problems, such as hemolytic anemia in newborn. The Rhesus system is a set of antigens on red blood cells, of which the D agglutinogen is the strongest. It causes trouble when the mother is Rh-negative, and the fetus has inherited the D factor from the father. The immune system of the mother identifies the fetus as foreign, and produces antibodies that cause destruction of the fetal red cells. People may marry without inquiring into each other's blood types.

The symbol DD represents the genotype of someone who has received the D gene from both parents; the symbol dd indicates the genotype of a person lacking the antigen, known as Rh-negative (Rh-); the symbol Dd stands for the heterozygous or hybrid individual. In the blood donation center, DD and Dd people are lumped together as positives (Rh+). In the US, Rh+ people make up 85 percent of the population, and Rh- the other 15 percent.

The frequency of the genes D and d in a population can be calculated from the phenotypic numbers by means of the Hardy-Weinberg rule. This principle was discovered independently by G. H. Hardy, a mathematician, and S. Weinberg, a physician. Call the frequency of the dominant gene p, and the frequency of the recessive allele q. Then, by random mating, the frequencies of the genotypes are:

(p + q)2 = p2 + 2pq + q2

This is called the expansion of a binomial. You learned this in Algebra 1. In the case of the Rh factor, q2 is the same as the fraction of Rh- people, stated above as 0.15. The gene frequencies are calculated as follows:

q2 = 0.15       q = √0.15  = 0.387298334

p = 1 - q = 0.612701665

This result allows calculation of the fractions of the two Rh+ genotypes.

DD = p2 = ( 0.6127)2 = 0.3754

Dd = 2 pq = 2 X 0.6127 X 0.3873 = 0.4746

Adding the dd fraction 0.15 makes the total 1.0, or 100 percent, as expected.

Exercise 12-1.

In some Asian populations, the fraction of dd individuals has been reported as only 1 percent. Assuming exactly 0.010, calculate the gene frequencies p and q.

Fig. 12-1

image credit: Wikipedia

Gene frequencies may change as a result of mutation or selection or migration. As W. F. Ganong wrote, “In general, harmful mutations tend to die out, but mutant genes that confer traits with survival value persist and spread in the population.”  In effect, he stated the essence of human evolution.

The gene for sickle-cell anemia produces a modified hemoglobin molecule, which causes red blood cells to assume an abnormal sickle or crescent shape at low oxygen concentrations (Fig. 12-1). People with genotype ss are severely affected, those with Ss have few symptoms, and those with the dominant genotype SS are normal.  In some populations of equatorial Africa, sickle-cell anemia patients (ss) amount to 1 in 64 or more (0.015625), and people with the trait (Ss) up to 40%. Call the frequency of S, the normal allele, a, and call the frequency of s, the harmful mutant gene, b.  Applying the Hardy-Weinberg rule:

b2 = 0.01562,   b = √0.01562 = 0.12498,

a = 1 - b = 0.8750

These frequencies can be used to calculate the fractions of each class:

Phenotype Genotype Frequency Observed Expected
normal SS a2 0.58 0.765625
carrier Ss 2ab 0.40 0.218715
afflicted ss b2 0.01562

This result suggests that the observed fraction of carriers (Ss) is too high, and normal (SS) too low. Something is disturbing the genetic equilibrium.

Exercise 12-2.

Calculate the expected gene frequencies in the African-American population, with the disease 1 in 500,  and the trait 1 in 12, using the reasoning and steps above.

In this case, the observed and expected are not far apart. What is different in Africa and America? Many things, but the explanation seems to lie in the prevalence of malaria.  The abnormal hemoglobin hampers the rapid growth of the malarial parasite in the bloodstream, conferring a selective advantage on those with the sickle-cell trait, when they live in a region where malaria is prevalent. Such regions are common in central Africa, uncommon in USA.

Blood Type Frequencies

A successful blood transfusion requires matching ABO blood types of donor and recipient. Everyone has one of the four types, A, B, O, or AB, genetically determined, and unaltered through life. In an incompatible transfusion, the donor is not harmed, but his red cells encounter antibody in the recipient's bloodstream, which causes them to agglutinate and block capillary circulation. Inheritance gives each person two out of the set of three genes, two of which, A and B, are co-dominant. The genes are designated IA, IB, and i, for their role in producing isoagglutinogens.

Phenotypes Genotypes Fraction in USA
A IAIA, IAi  0.41
B IBIB, IBi 0.10
O ii 0.45
AB IAIB 0.04

If the gene frequencies are symbolized by p, q, and r, and the Hardy-Weinberg rule applies, then the fractions of the genotypes in a population should correspond to the expansion of a trinomial:

(p + q + r)2 = p2 + 2 pq + q2 + 2 pr + 2 qr + r2

r2 = (frequency of ii) = fraction of type O in population = 45% in USA

r2 = 0.45    r = √0.45 = 0.670820393

Type A corresponds to the terms of the expansion p2 + 2 pr, and occurs in about 41% of people in the US. But we already have r, and we recognize, because of our familiarity with basic algebra, that p can be found, since:

(p + r)2 = p2 + 2pr + r2 = 0.41 +  0.45 = 0.86 = total of type A and type O.

p + r = √0.86 = 0.927361849        p = 0.92736  - 0.67082 = 0.25654

Since the sum must equal 1:

q = 1 - (p + r) = 1 - 0.92736 = 0.07263815

With the frequencies p, q, and r known, the expected fractions of types B and AB can be calculated, and compared with the reported fractions.

Type B: q2 + 2 qr = (0.07263815)2 + 2 X 0.072638 X 0.67082 = 0.10273

Type AB: 2pq = 2 X 0.25654 X 0.072638 = 0.03727

These results agree with the 10% and 4% listed above, and indicate a genetic equilibrium, sometimes called a balanced polymorphism. The fraction of people who are both AB and Rh-negative is:

0.03727 X 0.15 = 0.0055905, or about 56 in 10 000.

Exercise 12-3.

The percentages of blood types in Japan are: O = 30, A = 38, B = 22, and AB = 10. Find the gene frequencies by the procedure shown here, and determine whether the fractions B and AB are consistent with the data.

Young couples in Japan have been reported to compare blood types when they first meet. Do not laugh; some of our own young people consider signs of the zodiac when choosing a potential mate.