$4-CO-DOMINANT ALLELES - HbA, HbS, & PROTECTION FROM MALARIA: During class, I ou
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$4-CO-DOMINANT ALLELES - HbA, HbS, & PROTECTION FROM MALARIA: During class, I outlined the changes in the frequency of hemoglobin A (HbA) and Hemoglobin S (HbS) alleles in a West African village with a breeding population of 50 (25 couples) over three generations. For this exercise, I will use the symbol A for HbA alleles and & S for HbS alleles. During the period, malaria became prevalent and killed a large proportion of the children born who had the genotype AA and were thus unprotected from malaria. During the same period, homozygous individuals (genotype AS) became much more common in the population. In another, similar village (breeding population of 50) where malaria was also killing large numbers of children, a slightly different pattern emerged over 4 generations. Calculate the gene frequencies of each of the two alleles, A and S, for the following four generations (G1 - G4). During that period, no SS individuals lived long enough to reproduce. After UNESCO eradicated the malaria mosquitoes from the village area during generation Gen 4, no villagers succumbed to malaria. AA individuals also had a slight advantage over heterozygous individuals because their blood worked better during heavy exertion. The following pattern occurred during the next five generations. Using the numbers provided, the frequencies of A and S alleles in village population for generations G5-G9. Comment on the evolutionary changes (changes in gene frequency, population size, etc.) that occurred in this village over the 9 generations represented on the chart.Explanation / Answer
Answer:
4. a. Gen 3:
p2 (AA)= 33/50 = 0.66. So, p = 0.812 (Frequency of allele A)
q = 1 - 0.812 = 0.188 (Frequency of allele S)
Gen 4:
p2 (AA)= 26/50 = 0.52. So, p = 0.72 (Frequency of allele A)
q = 1 - 0.72 = 0.28 (Frequency of allele S)
b. Gen 5:
p2 (AA)= 53/75 = 0.70.
Frequency of allele A = p2 + 1/2 (2pq) = 0.7 + 1/2 (22/75) = 0.7 + 0.1466 = 0.8466
q = 1 - 0.8466 = 0.1534 (Frequency of allele S)
Gen 6:
p2 (AA)= 81/100 = 0.81.
Frequency of allele A = p2 + 1/2 (2pq) = 0.81 + 1/2 (19/100) = 0.81 + 0.095 = 0.905
q = 1 - 0.905 = 0.095 (Frequency of allele S)
Gen 8:
p2 (AA)= 137/150 = 0.91.
Frequency of allele A = p2 + 1/2 (2pq) = 0.91 + 1/2 (13/150) = 0.91 + 0.0433 = 0.95
q = 1 - 0.95 = 0.05 (Frequency of allele S)
Gen 9:
p2 (AA)= 191/200 = 0.955.
Frequency of allele A = p2 + 1/2 (2pq) = 0.955 + 1/2 (9/200) = 0.955 + 0.0225 = 0.9775
q = 1 - 0.9775 = 0.0225 (Frequency of allele S)
Comment on the evolutionary changes:
When the mosquitoes were present (Gen 1-4), we can see an increase in the frequency of allele S and a decrease in allele A. The allele S was getting selected on the course of evolution due to its survival advantage.
When no mosquitoes were present (Gen 5-9), we can see an increase in the frequency of allele A and a decrease in allele S. The allele S was no more required due to the absence of mosquitoes (malaria) in the population.
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