Question

reting Data: Hardy-Weinberg Equation 2 of 10 you use the Hardy Weinberg equation to answer questions about a hypotheticalcat population Part A A hypothetical population of 500 cats has two wees, Tandt for a gene that codes for tail length (Tis completely dominantot) The table below presents the phenotype of cats with each possible genotype, as well as the number of individuals in the population with each genotype. Assume that this population is in Hardy-Weinberg equilibrium Recall that the Hardy Weinberg equation is p2+2q+92-1. Phenotype Genotype Number of individuals long tail long tail short til Drag the numbers on the left to the appropriate blanks on the right to answer these questions about a vypothetical cat population. Answers can be used once more than once, or not at all. Reset Help 1. What is the frequency of cats with long tails in the population? 2. What is the frequency of cats with short tails in the population? 3. What is the frequency of cats that are homozygous dominant in the population? 4. What is the frequency of the Talele in the gene pool of this population? 5. What is the frequency of the tallele in the gene pool of this population? 6. Use the Hardy-Weinberg equation to predict the frequency of heterozygous cats in the next generation 7. Use the Hardy-Weinberg equation to predict the frequency of homorygous recessive cats in the next generation 8. Use the Hardy-Weinberg equation and your answer to question 7 to estimate the frequency of the recessive alleet in the next generation Rest

Answer 1

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180+240-420 mm Frequency of cat with Long tail(+1)= 420 = 0-84 > l l ll ! li 500 ll short tet - 80 -0.16 500 " Homozygous dominant 180–0.36 500 i l 54% Frequency of Tallele = TT +1x Tit = 0.36+ 2x 300 +0.36+0.24 = 0.60 35 Frequency oft allele - I-T= 1-0.60=0:40 56) Frequency of heterozygote = 2 *pxq=20.480-6 = 0.48 Loro no 240 oC 57) Frequency of homozygous recessively)2=(0.42 -0.16 8) Frequency of recessive Allele = 10.16= 0.40

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