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Case Study Recessive Alleles

Decent Essays

1. a) Our expectation was that Allele A1 recurrence would increment after Some time since it is the most qualified. Yes, my expectation was right. b) The last recurrence expanded at a much slower rate than some time recently. 2. A1 is the prevailing allele. Since A1 is shown in A1 A2 and A1A1 genotype. These two genotype delineates one and only phenotype, which is constantly present in both of these genotypes. Additionally, this phenotype is preferential in natural selection. 3. The kind of natural selection that was happening in the cases above was directional in light of the fact that the diagram was leveling out asymptotically at 1, demonstrating that variety will be moving in one course. 4. Yes, in the long run after some time …show more content…

10. Heterozygote advantage of interest is said to keep up hereditary variety since it contains both dominant and recessive alleles. The allele recurrence for heterozygotes will increment and they will be defended from malaria and sickle cell sickness. The homozygotes will fall to either malaria or sickle cell illness. HWE questions a) • No Natural Selection • There is no migration • There is no genetic drift • There is no mutation b) 1. 4/10,000 = 0.0004, this is q^2 2. 0.0004 = 0.02. 3. p = 1 – 0.02 = 0.98. p^2 = 0.9604 2pq = 0.0392 q^2 = 0.0004 Accepting Hardy-Weinberg for this situation is not right on the grounds that there natural selection acting against the allele that causes the illness, given that kids with the ailment will die before they pass it on to their kids. c) • y frequency q = 0.8 • p = 1 – 0.8 = 0.2 • p^2 = 0.04, 2pq = 0.32, q^2 = 0.64 d) i. 336 + 64 = 400 animals in total, q^2 = 64/400 = .016, this is the homozygous recessive. ii. Since q^2 is 0.16, q = 0.4. p+q=1 so p is equal to 0.6 the dominant allele. iii. 2pq = (2)(0.6)(0.4) = 0.48 *(400) = 192, the heterozygous iv. p = 0.6, q =

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