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HARDY WEINBERG

HARDY WEINBERG. What is Hardy- weinberg Population?. a population that is in genetic equilibrium (not evolving) due to (five conditions) … no mutations large population size no migration random mating no natural selection

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HARDY WEINBERG

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  1. HARDY WEINBERG

  2. What is Hardy-weinberg Population? • a population that is in genetic equilibrium (not evolving) due to (five conditions) … • no mutations • large population size • no migration • random mating • no natural selection **if any one of this conditions is not met, the population will evolve (allele frequencies will change from generation to generation)

  3. What is Hardy-weinberg Population? • is represented by two mathematical equations… p2 + 2pq + q2 =1 and p + q = 1 p= frequency of dominant allele q= frequency of recessive allele p2= frequency of homozygous dominant 2pq= frequency of heterozygous q2= frequency of homozygous recessive

  4. Problem one • A study on blood types (M and N are codominant) in a population found the following genotypic distribution among the people sampled: 1101 were MM, 1496 were MN and 503 were NN. Calculate the allele frequencies of M and N and the three genotypic frequencies. • Start with … • How many people total are in the population? 1101 + 1496 + 503 = 3100

  5. Allele frequency • Shows the percentage of alleles in the group • p = % of M alleles • q = % of N alleles the total alleles in the population = number of individuals in the population X 2 • p = 1101 X 2 + 1469/ 3100 X 2 = .59 • q = 503 X 2 + 1469 / 3100 X 2 = .40 Now… Calculate the genotypic frequency for MM, MN and NN.

  6. Genotype frequency • Genotype frequency determines the % of Homo dominant (MM), heterozygous (MN) and homo recessive (NN) • MM= p2 • MN = 2pq • NN = q2 • % of MM= 1101/3100= .36 or (.59)2 =.35 • % of MN= 1496/3100= .48 or 2(.59)(.40) = .47 • % of NN= 503/3100 = .162 or (.40)2 = .16

  7. Problem 2 • The compound phenylthiocarbamide (PTC) tastes very bitter to most persons. The inability to taste PTC is controlled by a single recessive gene. In America, about 70% can taste PTC while 30% cannot (are non-tasters). Estimate the frequencies of the Taster (T) and nontaster (t) alleles in this population as well as the frequencies of the genotypes. • Hint: start by calculating q2 (think about why)

  8. Allele frequency • 30% cannot taste PTC, so 30% of population is aa • aa= q2 = .30 • so q = .5477 • p + q = 1 • p = 1 - .5477 • p = .4523

  9. Genotype frequency • p = .4523 • So, AA = p2 = .2045 • We know that aa is .30 • So, Aa = 2pq = .4956 • 20.5% of population is AA • 30% of population is aa • 49.6% of population is Aa

  10. Problem 3 • In another study of human blood groups, it was found that among a population of 400 individuals, 230 were Rh+ and 170 were Rh-. Assuming that this trait (i.e., being Rh+) is controlled by a dominant allele (D), calculate the allele frequencies of D and d. How many of the Rh+ individuals would be expected to be heterozygous?

  11. Allele Frequency • p + q = 1 • We know that Rh- is recessive • There are 170 people who are Rh- out of 400 • So, 170/400 = 42.5% of population are aa • So q2 = .425 • So q = .652 • So p = 1-.652 = .348 • So 65.2% of the alleles are recessive while 34.8% of the alleles are dominant

  12. How many of the Rh+ individuals would be expected to be heterozygous • 2pq= Aa • 2 (.348) (.652)= Aa • So 2pq = .453 • So 45.3% of population is Aa • .453 (400 people) = 181 people in the population are Aa

  13. Problem one • A study on blood types (M and N are codominant) in a population found the following genotypic distribution among the people sampled: 1101 were MM, 1496 were MN and 503 were NN. Calculate the allele frequencies of M and N and the three genotypic frequencies.

  14. Problem 2 • The compound phenylthiocarbamide (PTC) tastes very bitter to most persons. The inability to taste PTC is controlled by a single recessive gene. In America, about 70% can taste PTC while 30% cannot (are non-tasters). Estimate the frequencies of the Taster (T) and nontaster (t) alleles in this population as well as the frequencies of the genotypes. • Hint: start by calculating q2 (think about why)

  15. Problem 3 • In another study of human blood groups, it was found that among a population of 400 individuals, 230 were Rh+ and 170 were Rh-. Assuming that this trait (i.e., being Rh+) is controlled by a dominant allele (D), calculate the allele frequencies of D and d. How many of the Rh+ individuals would be expected to be heterozygous?

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