hardy
Population Genetics / Hardy-Weinberg Problems
Directions: Work out the following problems on a separate piece of paper. Show ALL work and circle your answers.
1) If the frequency of a recessive allele is 30% in a population of 100 people, how many would you predict would be carriers of this allele, but would not express the recessive phenotype? q= 0.30 p= 0.70
Carriers = 2pq = 2(.3)(.7) = .42
#= (.42)(100) = 42 individuals
2) From a sample of 278 American Indians, the following MN blood types were obtained: MM = 78, MN = 139, NN = 61. Calculate the allele frequency of M and N.
M= 0.53
N= 0.47
MM = 78/278 = 0.281 = p2 p = 0.530 = 53%
MN = 139/278 = 0.50 = 2pq q = 0.468 = 46.8%
NN = 61/139 = 0.219 = q2
3) Among 86 Indians from Central America, the frequencies of M and N were 0.78 and 0.22 respectively. Calculate the expected percentages of individuals with M, MN and N type blood. p = 0.78 q = 0.22
MM= p2 = 0.608 = 61%
MN= 2pq = 0.343 = 34%
NN= q2 = 0.0484 = 5%
4) In a population in Hardy-Weinberg equilibrium, if the genotype frequencies are 81% AA, 18% Aa, and 1% aa, what are the frequencies of the A and a alleles, respectively?
p2 = .81 A= 0.9
2pq = .18 a= 0.1 q2 = .01
5) Tay-Sachs disease is inherited as a Mendelian recessive and occurs in Ashkenazic Jews at a frequency of 1 per 3,600 births. Estimate the percentage of this population that are heterozygote carriers of the Tay-Sachs allele.
1/3600 = .0002778 = q2 q = .0167 p = 1-q = .9833
% of population with Aa (2pq)= 3.3%
6) In a population in Hardy-Weinberg equilibrium, there are two alleles, A and a, for a particular gene. If the frequency of the a allele is 0.4, what is the frequency of the genotype AA? q = .4 p = .6
AA = p2 = 0.36
7) In a population at Hardy-Weinberg equilibrium, what proportion of individuals are heterozygous for allele a if its frequency is 0.01?
q = 0.01 p = 0.99
Aa (2pq) = 0.02 or 2%
8) Phenylketonuria is a
Directions: Work out the following problems on a separate piece of paper. Show ALL work and circle your answers.
1) If the frequency of a recessive allele is 30% in a population of 100 people, how many would you predict would be carriers of this allele, but would not express the recessive phenotype? q= 0.30 p= 0.70
Carriers = 2pq = 2(.3)(.7) = .42
#= (.42)(100) = 42 individuals
2) From a sample of 278 American Indians, the following MN blood types were obtained: MM = 78, MN = 139, NN = 61. Calculate the allele frequency of M and N.
M= 0.53
N= 0.47
MM = 78/278 = 0.281 = p2 p = 0.530 = 53%
MN = 139/278 = 0.50 = 2pq q = 0.468 = 46.8%
NN = 61/139 = 0.219 = q2
3) Among 86 Indians from Central America, the frequencies of M and N were 0.78 and 0.22 respectively. Calculate the expected percentages of individuals with M, MN and N type blood. p = 0.78 q = 0.22
MM= p2 = 0.608 = 61%
MN= 2pq = 0.343 = 34%
NN= q2 = 0.0484 = 5%
4) In a population in Hardy-Weinberg equilibrium, if the genotype frequencies are 81% AA, 18% Aa, and 1% aa, what are the frequencies of the A and a alleles, respectively?
p2 = .81 A= 0.9
2pq = .18 a= 0.1 q2 = .01
5) Tay-Sachs disease is inherited as a Mendelian recessive and occurs in Ashkenazic Jews at a frequency of 1 per 3,600 births. Estimate the percentage of this population that are heterozygote carriers of the Tay-Sachs allele.
1/3600 = .0002778 = q2 q = .0167 p = 1-q = .9833
% of population with Aa (2pq)= 3.3%
6) In a population in Hardy-Weinberg equilibrium, there are two alleles, A and a, for a particular gene. If the frequency of the a allele is 0.4, what is the frequency of the genotype AA? q = .4 p = .6
AA = p2 = 0.36
7) In a population at Hardy-Weinberg equilibrium, what proportion of individuals are heterozygous for allele a if its frequency is 0.01?
q = 0.01 p = 0.99
Aa (2pq) = 0.02 or 2%
8) Phenylketonuria is a