Hardy-Weinberg practice and solutions

• Here are the answers to #3 on page 650 from your text:
• a) i) SA 0 0.6 and SB 0 0.4 (because if mating is random, the frequencies shouldn´t change)
• ii) SASA =0.36 x 175 = 63, SASB = 0.48 x 175 = 84 individuals, SBSB = 0.16 x 175 = 28 individuals
• b) i) they have misshapen red blood cells which cannot properly bind oxygen, therefore they have very little energy
• ii) The allele remains in the population because of the heterozygous advantage, which means that heterozygotes for sickle cell anemia are protected against malaria.
• c) i) Sickle cell anemia is caused by a point mutation, affecting only one base! This mutation is located in the gene that codes for hemoglobin.
• ii) Due to independent segregation, there is 50 percent chance that a heterozygote will pass on the sickle cell allele to their offspring.
• d) don´t worry about part d yet..we will cover the topic of directional selection in a future class