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?
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.
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.
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?
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.
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?
7) In a population at Hardy-Weinberg equilibrium, what proportion of individuals are heterozygous for allele a if its frequency is 0.01?
8) Phenylketonuria is a disease in which the affected individuals lack an enzyme required for the metabolism of the amino acid phenylalanine. It is inherited as a Mendelian recessive and appears at the rate of one in every 15,000 births in the U.S. What percentage of this population is homozygous NORMAL at the locus where the PKU allele occurs?
9) A recent study has shown that 9.0% of the natives on the south pacific island of Pago Fuago possess an allergic reaction to coconuts. This reaction, which appears to be an immune response similar to hives, is thought to be due to the action of a recessive gene (h). The population of this tropical paradise is 5000. complete the following chart based on this information.
The frequency of the (H) allele = / The total number of homozygous nonallergic Pago Fuagens =The frequency of the (h) allele = / The total number of Pago Fuagans that are carriers of the (g) allele and phenotypically normal =
The total number of (h) alleles on Pago Fuago = / The total number of Pago Fuagens that suffer from this allergic condition =
10) In Drosophila, the allele for normal length wings is dominant over the allele for vestigial wings (vestigial wings are stubby little curls that cannot be used for flight). In a population of 1,000 individuals, 360 show the recessive phenotype. How many individuals would you expect to be homozygous dominant and heterozygous for this trait?
11) The allele for the ability to roll one’s tongue is dominant over the allele for the lack of this ability. In a population of 500 individuals, 25% show the recessive phenotype. How many individuals would you expect to be homozygous dominant and heterozygous for this trait?
12) The allele for the hair pattern called “widow’s peak” is dominant over the allele for no “widow’s peak.” In a population of 1,000 individuals, 510 show the dominant phenotype. How many individuals would you expect of each of the possible three genotypes for this trait?
13) In the United States, about 16% of the population is Rh negative. The allele for Rh negative is recessive to the allele for Rh positive. If the student population of a high school in the U.S. is 2,000, how many students would you expect for each of the three possible genotypes?
14) In certain African countries, 4% of the newborn babies have sickle-cell anemia, which is a recessive trait. Out of a random population of 1,000 newborn babies, how many would you expect for each of the three possible genotypes?
15) In a certain population, the dominant phenotype of a certain trait occurs 91% of the time. What is the frequency of the dominant allele?
16) Which of the following populations are in Hardy-Weinberg equilibrium?
GenotypesPopulation # / AA / Aa / aa
I- 1980 (rodent gene) / 0.430 / 0.451 / 0.119
I- 1983 / 0.521 / .279 / 0.20
II Insect gene (July) / 0.64 / 0.32 / 0.04
II Insect (September) / 0.66 / 0.31 / 0.03
III- generation “1” (gene “A”) / 0.4225 / 0.4550 / 0.1225
III- generation “6” / 0.1225 / 0.4550 / 0.4225
IV- generation “1” (gene “B”) / 0.0025 / 0.095 / 0.9025
IV- generation “100” / 0.0025 / 0.095 / 0.9025