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Title: Hardy-Weinberg Principle help!? Post by: smile on Jan 23, 2012 I just need some help getting started.
Question: If 98 out of 200 individuals in a population express the recessive phenotype, what percent of the population would you predict would be heterozygotes? I know when the question is asked about individuals you use p^2 + 2pq + q^2 = 1 and that heterozygotes are the 2pq variable but am stuck with getting it set up. OK so I got that 98/200 = 0.49. So q^2 = 0.49 which means q = 0.7 and then p = 0.3 So 2pq = 2 * 0.7 * 0.3 which results in 0.42 0.42 * 200 = 82 heterozygotes. Sound right? Title: Hardy-Weinberg Principle help!? Post by: smilealways on Jan 23, 2012 well you got the recessive percentage, and q^2 is the recessive phenotype, so i think you should start with: q^2 = 98
and then square root 98, get the q value, and using that q value you plug that into the equation: p+q=1, and get p! FINALLY, by getting p, you plug the values of q and p into 2pq, because that's representative of heterozygous, and i think you'll get the percentage then :) i hope it helps :0 hmm i forgot that it was out of 200, so i guess it's right to divide that by 100 first -> 98/200 = .49 = q^2 okay i think that's right but it asked for percentage so isn't it just 42%? |