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 **if any one of this conditions is not met, the population will evolve (allele frequencies will change from generation to generation)

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

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

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.

Genotype frequency % 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 Genotype frequency determines the % of Homo dominant (MM), heterozygous (MN) and homo recessive (NN) MM= p2 MN = 2pq NN = q2

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)

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

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

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?

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

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

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.

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)

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?