Unformatted text preview: R and C W in this population? C R = 200+740 / 2000= 0.47 C W = 1 0.47 = 0.53 c. what are the expected frequencies of the genotypes if the population is at Hardy Weinberg equilibrium? P2 = (0.47) 2 = 0.22 C R C R Q2= (0.53) 2 = 0.28 C W C W 2PQ= 2 (0.47)(0.53) = 0.50 C R C W Problem 8. Assuming that the population is at HardyWeinberg equilibrium, what proportion of the males in this population will also have the condition? X s = dominant allele for condition. X= recessive, doesn’t have the condition. Females with X s _ have the condition and 36% are X s X s and X s X genotypes. Thus, 64% is XX. Since 64% is XX (p 2 ), the X (p) allele frequency is 0.8. Using p + q =1, X s (q) allele frequency is 0.2 For male population: Since male has only one X, the genotype frequencies are equal to the allele frequencies. . Thus males with X s Y have the condition and there are 20% since the X s allele frequency is 0.2....
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 Spring '09
 Freeling
 Genetics, Evolution, Population Genetics, Xs allele frequency, Allele Frequency M=0.6225

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