More Hardy-Weinberg Problems

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Presentation transcript:

More Hardy-Weinberg Problems More Example Problems & Fun for all Calculators at the ready!!

Quick review of terms p + q = 1 p2 + 2pq + q2 = 1 p = % of dominant allele q = % of recessive allele p2 = ??? q2 = ??? 2pq = ???

Quick review of terms p + q = 1 p2 + 2pq + q2 = 1 p = % of dominant allele q = % of recessive allele p2 = % of individuals who are homozygous dominant q2 = % of individuals who are homozygous recessive 2pq = % of individuals who are heterozygous

Example #3 If 98 out of 200 individuals in a population express the recessive phenotype, what percent of the population would you predict would be heterozygotes? Work on this with a partner.

Example #3 If 98 out of 200 individuals in a population express the recessive phenotype, what percent of the population would you predict would be heterozygotes? This means that q2 = 98/200 or 0.49 and q=0.7. Then p=0.3 and 2pq (% heterozygotes) is 2(0.3)(0.7) = 0.42 or 42%

Example #4 Lets say that brown fur coloring is dominant to gray fur coloring in mice. Of 200 mice 168 brown mice. What is the predicted frequency of heterozygotes? What is the predicted frequency of homozygous dominant?

Example #4 Lets say that brown fur coloring is dominant to gray fur coloring in mice. Your mouse 200 mouse population has 168 brown mice. What is the predicted frequency of heterozygotes? q2 = 32/200 = 0.16; q = 0.4; p = 0.6; 2pq = 0.48 or 48% What is the predicted frequency of homozygous dominant? p = 0.6; p2 = 0.36 or 36%

Example #5 In a certain population of fish, black is the dominant color over white with striped black and white fish being the heterozygous form. 81% of a population of 5000 fish are the recessive coloration of white. How many striped fish are there?

Example #5 In a certain population of fish, black is the dominant color over white with striped black and white fish being the heterozygous form. 81% of a population of 5000 fish are the recessive coloration of white. How many striped fish are there? q2 = 0.81; q = 0.9 therefore p = 0.1 Striped fish are heterozygous 2pq=2(.9)(0.1) = 0.18 or 18% 18% of 5000 is 900 striped fish