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Title: In a hypothetical population in Hardy-Weinberg equilibrium? Post by: RJW on Sep 10, 2012 In a hypothetical population in Hardy-Weinberg equilibrium, 65 percent of alleles are of the dominant type while the remaining 35 percent are of the recessive type. Calculate the proportion of heterozygous individuals you would expect to see in the F1 generation.
Title: In a hypothetical population in Hardy-Weinberg equilibrium? Post by: buhlig on Sep 10, 2012 p^2+2pq+q^2=0
p^2=0.65 q^2=0.35 p=sqrt(0.65) q=sqrt(0.35) heterozygous = 2pq Title: In a hypothetical population in Hardy-Weinberg equilibrium? Post by: lex6909 on Sep 10, 2012 let p represent the dominant allele and q the recessive allele frequencies. (pq is heterozygous)
the hardy weinberg equations are as follows: p+q=1 and p^2 + 2pq +q^2 =1 (NOT 0!!!) given: p= 0.65 and q=0.35, you would plug those values into the 2pq of the second equation to find the number of heterozygotes. you would get 0.455 *the values given are values of p and q, not p squared or q squared because they are the actual allele frequencies of the population. Title: In a hypothetical population in Hardy-Weinberg equilibrium? Post by: rizkallah15 on Sep 10, 2012 p=.65 q=.35
2pq which is the heterozygotes in the population would equal .455. |