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Score: 10/10 Points 100 %

Hardy-Weinberg Equilibrium Hardy-Weinberg equilibrium can describe the gene frequencies of a population of organisms that meet five criteria for stability. Explore this interactive graph to discover how the Hardy-Weinberg equation can be used to predict genotypes and phenotypes of a population when allele frequencies are known, or vice versa. Enter frequency values or use the drag buttons to visualize the data three different ways: a frequency graph, a Punnett square, and a population simulation. All three model a population that is in Hardy- Weinberg equilibrium. Click the Info button (i) to review the Hardy-Weinberg equilibrium criteria and equation. Hardy-Weinberg Equilibrium GRAPH Frequency Punnett Population Frequency (units) p q 1 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 0 0.2 0.4 0.6 0.8 1 AA aa Aa Reset SETTINGS p q AA aa Aa 0.50 0 1 0.50 0 1 0.25 0 1 0.25 0 1 0.50 References

1. Award: 2 out of 2.00 points Hardy-Weinberg: Fungal population A scientist is studying a population of fungi. She finds that the recessive allele for small spores has a frequency of 0.75. What frequency should she expect for the dominant allele? 0.25 Assuming the population is in Hardy-Weinberg equilibrium, what percentage should she expect to see for the following genotypes? aa 0.56 Aa 0.38 AA 0.06 References Worksheet Hardy-Weinberg: Fungal population Hardy-Weinberg: Fungal population A scientist is studying a population of fungi. She finds that the recessive allele for small spores has a frequency of 0.75. What frequency should she expect for the dominant allele? 0.25 Assuming the population is in Hardy-Weinberg equilibrium, what percentage should she expect to see for the following genotypes? aa 0.56 Aa 0.38 AA 0.06 Explanation: The interactive performs the Hardy-Weinberg calculations, and also shows three visual representations of genotype frequencies. If you set the value of q to 0.75, you should discover that p , the frequency of the dominant allele, is 0.25 (1 – q = p ). Under these conditions the frequency of the homozygous

recessive genotype aa is 0.56, the frequency of the heterozygous genotype Aa is 0.38, and the frequency of the homozygous dominant genotype AA is 0.06. These are derived from the Hardy- Weinberg Equation, p 2 +2 pq + q 2 = 1.