Hardy-Weinberg Practice Problems Handout

  1. Practice questions
    1. Levy and Levin (1975) used electrophoresis to study the phosphoglucose isomerase-2 locus in the evening primrose Oenothera biennis. . . They observed two alleles affecting electrophoretic mobility of the enzyme, PGI-2a and PGI-2b. In 57 strains, they observed 35 PGI-2a/PGI-2a, 19 PGI-2a/PGI-2b, and 3 PGI-2b/PGI-2b. . . Calculate the expected numbers of the three genotypes . . . assuming that the genotypes occur in Hardy-Weinberg proportions.
    2. Kelus (cited in Mourant et al., 1976) reports a study of 3100 Poles, of which 1101 were MM, 1496 were MN and 503 were NN. Calculate the allele frequencies of M and N (and) the expected numbers of the three genotypic classes.
    3. Mourant et al. (1976) cite data on 400 Basques from Spain, of which 230 were Rh+ and 170 were Rh-. Calculate the allele frequencies of D (i.e., the allele which in homozyogous form results in the Rh+ phenotype) and d (the allele which in homozyogous form results in the Rh- phenotype; recall that the phenotype of Dd is Rh+). How many of the Rh+ individuals would be expected to be heterozygous?
    4. Phenylketonuria is a severe form of mental retardation due to a rare autosomal recessive. About one in 10,000 newborn Caucasians are affected with the disease. Calculate the frequency of the carriers (heterozygotes).
    5. The IA "allele" for the ABO blood groups actually consists of two subtypes, IA1 and IA2, either being considered "IA". In Caucasians, about 3/4 of the IA alelles are IA1 and 1/4 are IA2 (Cavalli-Sforza and Edwards, 1967). Among individuals of genotype IAIO, what fraction would be expected to be IA1IO? What fraction IA2IO? What would be the expected proportions of IA1IA1, IA1IA2, and IA2IA2 among IAIA individuals?
    6. If the frequency of the "green" form of red-green color blindness (due to an X-linked locus) is 5 percent among males, what fraction of females would be affected? What fraction of females would be heterozygous?
    7. Imagine an autosomal locus with four alleles, A1, A2, A3, and A4, at frequencies .1, .2, .3, and .4, respectively. Calculate the expected random mating frequencies of all possible genotypes.
    8. Consider a locus with two alleles (A1 and A2) and another locus with three alleles (B1, B2, B3). Let p1 = .3 be the allele frequency of A1, q1 = .2 be that of B1, and q2 = .3 be that of B2. Calculate the frequencies of all possible gametes, assuming that the loci are in linkage equilibrium.
    9. Suppose genotypes AA, Aa, and aa have frequencies in zygotes of 0.16, 0.48, and 0.36, respectively, and relative viabilities of w11 = 1.0, w12 = 0.8, and w22 = 0.6, respectively. Calculate the genotype frequencies in the zygotes in the next generation.
    10. The frequencies in per cent of the blood group alleles in a Scottish population were computed to be IA = 20.62, IB = 7.56, and i = 71.83 (i.e., these are the ABO blood groupings where IA is frequency of the A allele, IA is the frequency of the B allele, and i is the frequency of the O allele,). What are the expected phenotype frequencies in this population, on the assumption of random mating?
    11. (If mating is at random and red-green color blindness (which is sex/X linked) does not affect survival or fertility, what should be the proportion of color-blind women in a population at (Hardy-Weinberg) equilibrium in which 8 per cent of the men are color blind?
    12. Among a sample of 1000 Britishers, the number of individuals with each of the MN blood group phenotypes was as follows M: 298, MN: 489, N: 213. What is the M allele frequency? What is the N allele frequency? What number of each of the genotypes would be expected with random mating?
    13. A certain randomly mating population has a frequency of Rh- blood types of 16 percent. What is the frequency of the d allele (i.e., the Rh- allele)? What the frequency of the D allele (i.e., the Rh+ allele)? What are the expected genotype frequencies?
    14. The dry type of ear cerumen ("wax") is due to homozygosity for a simple Mendelian recessive. Among American Indians the frequency of dry-cerumen individual is 66 percent. What is the frequency of the recessive allele? What is the overall frequency of heterozygotes? Among individuals with the wet type of cerumen, what is the frequency of heterozygotes? (hint: to answer this question you must assume that the population is in Hardy-Weinberg equilibrium)
    15. In Zurich, Switzerland, the allele frequencies of IA, IB, and IO are 0.27, 0.06, and 0.67, respectively. What are the expected frequencies of blood types A, B, AB, and O?
  2. Practice question answers
    1. Make p the frequency of PGI-2a and q the frequency of PGI-2b. The frequency of PGI-2a is equal to 35/57 + 0.5*19/57 = 0.78. Thus, p = 0.78 and q = 1 - 0.78 = 0.22. Given Hardy-Weinberg proportions, the expected frequency of PGI-2a/PGI-2a = p2 = 0.61, of PGI-2a/PGI-2b = 2pq = 0.34, and of PGI-2b/PGI-2b = q2 = 0.05. As a check, 0.61 + 0.34 + 0.5 do indeed equal 1.0. 0.61 * 57 = 35, 0.34 * 57 is a little greater than 19, and 0.05 * 57 is a little less than 3. Though they didn't ask, you probably would, to a first approximation (i.e., it's always good form to do the statistics even on the obvious), assume that this population is in Hardy-Weinberg equilibrium.
    2. The frequency of allele M is equal to (1101 + 0.5 * 1496)/3100 = 0.60. Therefore the frequency of N is 0.40, of MM 0.36, of MN 0.48, and 0.16 for NN assuming Hardy-Weinberg equilibrium. Expected numbers are 0.36 * 3100 = 1116, 0.48 * 3100 = 1488, and 0.16 * 3100 = 496, respectively, which, of course, are pretty similar to the observed numbers.
    3. Since Rh- is the phenotype of the homozygous recessive, the d allele frequency is equal to the square root of 170/400 = 0.65. The expected frequency of heterozygotes is equal to 2 * 0.65 * (1 - 0.65) = 0.46 which is 182 individuals of 400.
    4. The frequency of the homozygous recessive is 1 in 10,000 = 0.0001 (= 10-4). Assuming simple genetics (e.g., all homozygotes are born and counted at the same rate as non-affected individuals), the frequency of the recessive allele is the square root of this, 0.01. From there the simple answer is about one in 50. Why, because when the recessive is sufficiently rare, the dominant allele is sufficiently close to 1 that the expression 2pq is essentially equal to 2q. Since q = 0.01, 2q = 0.02 = 1 / 50. Note that if you prefer to do things without taking short cuts, the answer would be 2 * 0.01 * (1 - 0.01) = 0.0198. Basically 0.02.
    5. This question is of a type that may be categorized as almost unfairly easy. That is because it is actually so simple that one wants to read far more into it than there actually is, and thus distracts oneself maximally, or at least reach unwanted levels of anxiety. The answers are: 3/4, 1/4, 9/16, 6/16, and 1/16. Why? First, the question was made rediculously easy simply by asking only for answers which are frequencies among individuals already carrying IA. Thus, among IAIO individuals, the fraction with the IA1 has to be the same (on average, of course) as its fraction in the entire population, which is 3/4. Similarly, the fraction of IA2 has to be equal to the frequency of this allele in the total population, which is 1/4. What is the fraction of the genotypes made up of only the IA1 and IA2 alleles, among IAIA individuals? Again, this is a far simpler question than what is the fraction of these genotypes among the total population? and is calculated simply using the familiar p2, 2pq, and q2 from the Hardy-Weinberg equation where p is the frequency of IA1 and q is the frequency of IA2, which are 3/4 and 1/4, respectively.
    6. The frequency of the phenotype among males is 0.05. Recall that males are haploid for the X chromosome. Therefore the rate at which they carry an X-linked allele is equal to the frequency of the allele in the population, which therefore is also 0.05. The probability that a female will carry one copy of this realtively rare allele is actually a little less than twice the allelic frequncy, or a little less than 0.10. Particularly, nearly twice the male probability because the female has two chances of carrying the allele, i.e., one chance for each X chromosome she carries, but, since the allele is relatively rare, a relatively low chance of picking up both alleles (the latter chance is part of the reason this value is a little less than twice the male rate, i.e., the odds of picking up one allele is some value less the odds of picking up two alleles). In fact, the more precise calculation of the frequency of the heterozygote in this case is simply 2pq or 2 * 0.05 * (1 - 0.05) which equals 0.095. The probability of female affliction is equal to the frequency of the homozygous recessive, q2 or 0.052 = 0.0025. Note, for the sake of checking these answers, that the rate at which females are neither afflicted nor carriers is p2 or (1 - 0.05)2 = 0.9025. These values should all add up to one and they do: 0.095 + 0.0025 + 0.9025 = 1.000.
    7. Answering this question is conceptionally easy, but a lot of work in practice. First figure out the possible genotypes, then multiply out the allele frequencies for each allele of a given genotype, then check yourself by making sure that frequences add up to 1.0 (did you remember to multiply the frequency of all of the heterozygotes by 2, i.e., as in 2pq?). Thus:
      1. A1A1, 0.1 * 0.1 = 0.01
      2. A1A2, 0.1 * 0.2 * 2 = 0.04
      3. A1A3, 0.1 * 0.3 * 2 = 0.06
      4. A1A4, 0.1 * 0.4 * 2= 0.08
      5. A2A2, 0.2 * 0.2 = 0.04
      6. A2A3, 0.2 * 0.3 * 2 = 0.12
      7. A2A4, 0.2 * 0.4 * 2 = 0.16
      8. A3A3, 0.3 * 0.3 = 0.09
      9. A3A4, 0.3 * 0.4 * 2 = 0.24
      10. A4A4, 0.4 * 0.4 = 0.16
    8. Stating that the loci are in linkage equilibrium simply means that we are assuming that there are no biases in allele combinations. It is very important in answering this question that you keep in mind that what we are looking for are gamete frequencies, not diploid frequencies! Here, then, the frequency of any given gamete is equal to the product of the frequencies of the constituting alleles (which also, by the way, need to be calculated to answer this question: p2 = 1 - p1 = 1 - 0.3 = 0.7 and q3 = 1 - q1 - q2 = 1 - 0.2 - 0.3 = 0.5). However, unlike above, we are not calculating heterozygote frequencies so avoid multiplying by 2! Finally, as usual, check yourself by making sure that all of the frequencies add up to 1.0. Thus:
      1. A1 B1, 0.3 * 0.2 = 0.06
      2. A1 B2, 0.3 * 0.3 = 0.09
      3. A1 B3, 0.3 * 0.5 = 0.15
      4. A2 B1, 0.7 * 0.2 = 0.14
      5. A2 B2, 0.7 * 0.3 = 0.21
      6. A2 B3, 0.7 * 0.5 = 0.35
    9. Don't let the notation throw you. w11 is just the relative fitness of AA, w12 the relative fitness of Aa, etc., though note that here relative fitness is being described solely in terms of survival. Regardless, multiply genotype frequencies by relative viabilities: (0.16)(1.0) = 0.16, (0.48)(0.8) = 0.384, and (0.36)(0.6) = 0.216. These sum to 0.76 and these numbers thus translate to frequencies of 0.211, 0.505, and 0.284 for AA, Aa, and aa, respectively. The question specifically asked for zygote frequencies. These are just allelic frequencies. So 0.211 + 0.505/2 = 0.464 is the frequency of the A gamete and 0.536 the frequency of the a gamete. Assuming Hardy-Weinberg-like reestablishment of the next generation's zygotes (i.e., random mating, large population, no additional evolutionary change in allele frequency) this implies frequencies of AA, Aa, and aa of (0.464)2, 2(0.464)(0.536), and (0.536)2 or 0.215, 0.497, and 0.287, respectively.
    10. The frequency of IAIA is 0.2062 * 0.2062 = 0.0425. The frequency of IBIB is 0.0756* 0.0756 = 0.0057. The frequency of II is 0.7183 * 0.7183 = 0.5160. The frequency of IAIB is 2 * 0.2062 * 0.0756 = 0.0312. The frequency of IAI is 2 * 0.2062 * 0.7183 = 0.2962. The frequency of IBI is 2 * 0.0756 * 0.7183 = 0.1086. 0.0425 + 0.0057 + 0.5160 + 0.0312 + 0.2962 + 0.1086 = 1.0002 which is close enough to one (assuming rounding error) to assume that I have not only listed all of the possible genotypes but have properly determined their frequencies assuming Hardy-Weinberg conditions. The frequency of the A phenotype is equal to the sum of the frequencies of the IAIA and IAI genotypes = 0.0425 + 0.2962 = 0.3387. The frequency of the B phenotype is equal to the sum of the frequencies of the IBIB and IBI genotypes = 0.0057 + 0.1086 = 0.1143. The frequency of the AB phenotype is equal to the frequency of the IAIB genotypes = 0.0312. The frequency of the O phenotype is equal to the frequency of the II genotypes = 0.5160. Once again, check to make sure that the frequencies add up to one. This will only happen with high likelihood if you have done all of the calculations correctly: 0.3387 + 0.1143 + 0.0312 + 0.5160 = 1.0002.
    11. Because male's are haploid for the X chromosome, q = 0.08. Since females are diploid for the X chromosome, their rate of color blindness in this population would be q2 = 0.0064 or 0.64%.
    12. Note that since the frequency of the three phenotypes equals 1.0 (i.e., 298 + 489 + 213 = 1000) that this implies that either this is a 1 locus, 2 allele system, or that this population is not in Hardy-Weinberg equilibrium (i.e., if there were a third, recessive allele, it apparently is not found in this population in the homozygous state). In addition, it would appear that the two alleles display codominance. Thus, the three genotypes associated with the three phenotypes, M, MN, and N, are MM, MN, and NN. The allelic frequency of M is (298 + 0.5 * 489) / 1000 = 0.5425. The allelic frequency of N is (213 + 0.5 * 489) / 1000 = 0.4575 = 1 - 0.5425. Given Hardy-Weinberg equilibrium, the expected genotype frequencies of MM, MN, and NN are 0.54252, 2 * 0.5425 * 0.4575, and 0.45752, respectively, which translates to 0.29, 0.50, and 0.21, respectively. These add up to 1.0. The associated numbers given a population size of 1000 are 290, 500, and 210, respectively, which his pretty close to the numbers observed.
    13. Given that this is a standard autosomal recessive allele, the frequency of d is the square root of 0.16 which is 0.4. Thus the frequency of D must be 0.6. The three possible genotypes are DD, Dd, dd and though the problem didn't state it, we will assume the null state and calculate the frequencies of these genotypes assuming Hardy-Weinberg equilibrium which are 0.36, 0.48, and, of course, 0.16 respectively. 0.36 + 0.48 + 0.16 = 1.0.
    14. The frequency of the recessive allele is simply the square root of 0.66 = 0.81. The frequency of heterozygotes must therefore be 2 * 0.81 * (1 - 0.81) = 0.30. 0.66 + 0.30 + (1 - 0.81)2 = 1.0. The last question is the tough one since it is not asking for the overall frequency but instead the frequency among a subgroup. Individuals with wet type cerumen include the homozygous dominant and the heterozygotes. Thus, the frequency of the heterozygotes among individual with wet type cerumen is equal to the frequency of heterozygotes divided by the sum of the frequency of heterozygotes and the frequency of homozygous dominants: 0.30 / (0.30 + 0.035) = 0.90, or 90 per cent.
    15. The frequency of type A blood = IA2 + 2 * IA * IO = 0.4347. The frequency of type B blood = IB2 + 2 * IB * IO = 0.084. The frequency of type AB blood = 2 * IA * IA = 0.0324. The frequency of type O blood = IO2 = 0.4489. 0.4347 + 0.084 + 0.0324 + 0.4489 = 1.000.
  3. References
    1. Campbell, N. A. (1996). Biology. Fourth Edition. Benjamin/Cummings Publishing, Menlo Park, California. pp. 434-435.
    2. Hartl, D. L. (1980). Principles of Population Genetics. Sinauer Ass., Inc., Sunderland, Massachusetts. pp. 78, 137-139.
    3. Hartl, D.L. (1983). Human Genetics. Harper & Row, Publishers, New York. pp. 484-485.
    4. Sinnott, E.W., Dunn, L.C., Dobzhansky, T. (1958). Principles of Genetics. Fifth Edition. McGraw-Hill Book Co., Inc. New York. p. 253.