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Nt1310 Unit 6 Lab Report

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For the first round of the stimulation was random mating. After the 20 students drew out two new cards from starting genotype, the new genetic pool had five pairs of dominant homozygous, ten pairs of heterozygous, and five pairs of recessive homozygous. The new generation had the same number of the starting gene pool of the known allele frequency. Random selection frequency of alleles does not change over time, due to which all individuals have an equal chance of being selected. The Hardy-Weinberg equilibrium equation, p2 + 2pq + q2 = 1, is used to define a population in which to understand both allele and genotype frequencies. The equation only occurs when mating is at random on a large population, and the relative genotype and allele frequencies persist in being constant. In all possible combination defined by the Punnett square, 60% or 0.6 of the gene pool is the A allele, and 40% or 0.4 of the a allele, totaling in for 100%. When adding the probability, the results are 0.36 for AA, 0.24 for aA, 0.24 for Aa, and 0.16 aa. In total, will equal 1. The AA homozygotes will have the frequency of p2, and similarly, aa homozygotes will have the frequency of q2 when the population is in equilibrium. …show more content…
One half representing the population remaining, and the other half as the separated population. This phenomenon is called gene flow when there is a transfer of alleles into a new population between two populations. For the population remaining, there were two AA, ten Aa, and one aa, in total of 13. The ratio of genotype has not changed from the parental generation, because there are still more heterozygotes than both dominant and recessive homozygotes. Though there was one more dominant than recessive. As for the separated population, there were zero AA, six Aa, and one aa, in total of seven. The ratio, like population remaining, is the same, but with one more recessive than

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