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Mendelian populations consist of a group of interbreeding individuals

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Mendelian populations consist of a group of interbreeding individuals that form a gene pool. To describe the population, genotypic and allele (gene) frequencies are often computed under the assumptions of the Hardy-Weinberg Law. Utilizing this assumption, let A and a represent dominant and recessive alleles whose respective frequencies are p and q in a given interbreeding population at equilibrium. a. If 16% of the individual in the population have recessive phenotypes, what percentage of the total number of recessive genes exist in the heterozygous condition? b. If 1% of the individuals were homozygous recessive what percentage of the recessive genes would occur in heterozygotes? c. What is the allele frequency of p and q in b?

Mendelian populations consist of a group of interbreeding individuals that form a gene pool. To describe the population, genotypic and allele (gene) frequencies are often computed under the assumptions of the Hardy-Weinberg Law. Utilizing this assumption, let A and a represent dominant and recessive alleles whose respective frequencies are p and q in a given interbreeding population at equilibrium. a. If 16% of the individual in the population have recessive phenotypes, what percentage of the total number of recessive genes exist in the heterozygous condition? p2 + 2pq + q2 = 1 and p + q = 1 p = frequency of the dominant allele in the population q = frequency of the recessive allele in the population p2 = percentage of homozygous dominant individuals q2 = percentage of homozygous recessive individuals 2pq = percentage of heterozygous individuals 1. You have sampled a population in which you know that the percentage of the homozygous recessive genotype (aa) is 16%. That means that q2 = 16% or 0.16 Therefore q = the square root of 0.16 = 0. 4 (40%) = the frequency of the recessive allele Meaning that if p + q = 1, p=0.6 and q=0.4 Meaning that q2 = 0.16 and 2pq = 2 (0.6*0.4) = 0.48 Total number of recessive genes = 2pq +q2 = 0.48 + 0.16 = 0.64 Percent of total number of recessive genes in heterozygous condition = 0.48/0.64 = 75% 2. b. If 1% of the individuals were homozygous recessive what percentage of the recessive genes would occur in heterozygotes?