b. How many cats have the heterozy gous (LI) genotype? 4. In humans, the gene fo
ID: 143834 • Letter: B
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b. How many cats have the heterozy gous (LI) genotype? 4. In humans, the gene for sickle-cell trait has two alleles, A and a. People with the aa genotype have the blood disease sickle cell anemia. People with the genotype Aa are healthy, and additionally are resistant to the disease malaria. People with the genotype AA are healthy but are not resistant to malaria. In a human population, 14 people are found to have sickle cell anemia, while 6,024 people are healthy and not malaria resistant, and 3,046 people are healthy and resistant to malana. Is this population in Hardy-Weinberg Equilibrium for this gene? Show how you know. What does your answer say about evolution with respect to this gene? In humans, a gene for hair texture has two alleles, T and T. Individuals with the genotype TT have curly hair, individuals with the genoype TT' have straight hair, and heterozygotes with the genotype TT" have wavy hair. A human population has 961 straight-haired people, 361 curly- haired people, and 1178 wavy-haired people. Is this population in Hardy-Weinberg Equilibrium for this gene? Show how you know. What does your answer say about evolution with respect to this gene? 5.Explanation / Answer
4) The total population is,
= 14 + 6024 + 3046
= 9084
Number of individuals with aa genotype = 14
Number of individuals with AA genotype = 6024
Number of people with Aa genotype = 3046
The proportion of A allele (p) = (6024 + (3046/2))/9084 = 0.831
The proportion of a allele (q) = (14 + (3046/2))/9084 = 0.169
p + q = 0.831 + 0.169 = 1
p^2 + 2pq + q^2
= (0.831)(0.831) + 2(0.831)(0.169) + (0.169)(0.169)
= 0.691 + 0.281 + 0.029
The number of AA individuals should be = (0.691)(9084) = 6277
The number of Aa individuals should be = (0.281)(9084) = 2553
The number of aa individuals should be = (0.029)(9084) = 264
So, the population is not in Hardy Weinberg equilibrium.
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