Question

Sickle-cell anemia is an interesting genetic disease. Normal homozygous individuals (SS) have normal blood cells that are easily infected with the malarial parasite. Thus, many of these individuals become very ill from the parasite and many die. Individuals homozygous for the sickle-cell trait (ss) have red blood cells that readily collapse when deoxygenated. Although malaria cannot grow in these red blood cells, individuals often die because of the genetic defect. However, individuals with the heterozygous condition (Ss) have some sickling of red blood cells, but generally not enough to cause mortality. In addition, malaria cannot survive well within these "partially defective" red blood cells. Fitness of different genotypes is as follows: 50% of the SS individuals survive and reproduce; 100% of the Ss individuals survive and reproduce while only 5% of the ss individuals survive and reproduce. Thus, heterozygotes tend to survive better than either of the homozygous conditions. If 9% of an African population is born with a severe form of sickle-cell anemia (ss), what percentage of the population will be more resistant to malaria in the second generation because they are heterozygous (Ss) for the sickle-cell gene?

Answer 1

If 9% of an African population is born with a severe form of sickle-cell anaemia (ss), what percentage of the population will be more resistant to malaria in the second generation because they are heterozygous (Ss) for the sickle-cell gene?

genotype frequency of ss is = 9% = 0.09

that means q² = 0.09

so q= recessive allele(s) frequency of = 0.3

so dominant allele (S) frequency p= 1-q = 1-0.3 = 0.7

so frequency of heterozygous genotype = 2pq = 2 x 0.7 x 0.3 = 0.42

That means 42% of people will become more resistant to malaria due to heterozygous condition (Ss).