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Hardy – Weinberg

Hardy – Weinberg. AP Bio Chapter 23. The Equation. p 2 + 2 pq + q 2 =1 Where p 2 =AA p=A q 2 =aa q=a 2pq= Aa And p + q = 1. 1 in 1,700 US Caucasian newborns have cystic fibrosis (CF) which is a recessive trait.

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Hardy – Weinberg

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  1. Hardy – Weinberg AP Bio Chapter 23

  2. The Equation • p2 + 2 pq + q2=1 • Where • p2=AA • p=A • q2=aa • q=a • 2pq= Aa • And p + q = 1

  3. 1 in 1,700 US Caucasian newborns have cystic fibrosis (CF) which is a recessive trait. • What is the frequency of heterozygotes in our population (heterozygotes are carriers, but ARE NOT sick with CF)? • What is the frequency of homozygotes for CF?

  4. Examples: Calculating allele frequencies • 1 in 1,700 US Caucasian newborns have cystic fibrosis (CF) which means that the frequency of homozygotes for this trait is • q2= 1/1,700 = 0.00059 • q = the square root of 0.00059 = 0.024

  5. Examples continued • The frequency of the normal allele (p) is equal to 1 – the frequency of the CF allele • p = 1 – 0.024 = 0.976 • The frequency of carriers (heterozygotes) for the CF allele is • 2pq = 2(0.976)(0.024) = 0.047

  6. Another Example • After graduation, you and 19 friends build a raft, sail to a deserted island, and start a new population, totally isolated from the world. Two of your friends carry the allele (that is, are heterozygous for the recessive PKU allele, which in homozygotes causes PKU). • Assuming that the frequency of this allele does not change as the population grows, what will be the instance of PKU on your island?

  7. Solution • In the population you have 40 alleles – of which 2 alleles are PKU causing • 2/40 = 0.05 = frequency of the PKU allele (q) • q2= (0.05)2 = 0.0025 or .25% of the pop’n

  8. Another Example • This is a classic data set on wing coloration in the scarlet tiger moth (Panaxia dominula). Coloration in this species had been previously shown to behave as a single-locus, two-allele system with incomplete dominance. Data for 1612 individuals are given below: • White-spotted (AA) = 1469 • Intermediate (Aa) = 138 • Little spotting (aa) = 5

  9. White-spotted (AA) = 1469 • Intermediate (Aa) = 138 • Little spotting (aa) = 5 • Calculate the frequency of: • A • a • AA • Aa • aa

  10. Solution • Calculate the frequency of: • A = (2 x (1469) + (138))/(2x(1469 +138 + 5)) = 0.954 or 95% • a = 1 - .954 = 0.046 or 4.6% • AA = (0.954)2=0.910 or 91% • Aa = 2(0.954)(0.046) = 0.087 or 8.7% • aa = (0.046)2 = 0.002 or 0.2%

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