Hardy-Weinberg Practice Problems Handout

**Practice questions**- 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. - 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. - 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? - 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).
- The
*I*"allele" for the ABO blood groups actually consists of two subtypes,^{A}*I*and^{A1}*I*, either being considered "^{A2}*I*". In Caucasians, about 3/4 of the^{A}*I*alelles are^{A}*I*and 1/4 are^{A1}*I*(Cavalli-Sforza and Edwards, 1967). Among individuals of genotype^{A2}*I*, what fraction would be expected to be^{A}I^{O}*I*? What fraction^{A1}I^{O}*I*? What would be the expected proportions of^{A2}I^{O}*I*,^{A1}I^{A1}*I*, and^{A1}I^{A2}*I*among^{A2}I^{A2}*I*individuals?^{A}I^{A} - 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?
- Imagine an autosomal
locus with four alleles,
*A*,_{1}*A*,_{2}*A*, and_{3}*A*, at frequencies .1, .2, .3, and .4, respectively. Calculate the expected random mating frequencies of all possible genotypes._{4} - Consider a locus with
two alleles (
*A*and_{1}*A*) and another locus with three alleles (_{2}*B*,_{1}*B*,_{2}*B*). Let_{3}*p*= .3 be the allele frequency of_{1}*A*,_{1}*q*= .2 be that of_{1}*B*, and_{1}*q*= .3 be that of_{2}*B*. Calculate the frequencies of all possible_{2}*gametes*, assuming that the loci are in linkage equilibrium. - Suppose genotypes
*AA*,*Aa*, and*aa*have frequencies in zygotes of 0.16, 0.48, and 0.36, respectively, and relative viabilities of*w*_{11}= 1.0,*w*_{12}= 0.8, and*w*_{22}= 0.6, respectively. Calculate the genotype frequencies in the zygotes in the next generation. - 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? - (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?
- 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? - 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? - 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)
- In Zurich,
Switzerland, the allele frequencies of
*I*,^{A}*I*, and^{B}*I*are 0.27, 0.06, and 0.67, respectively. What are the expected frequencies of blood types A, B, AB, and O?^{O} **Practice question answers**- 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*=*p*^{2}= 0.61, of*PGI-2a/PGI-2b*= 2*pq*= 0.34, and of*PGI-2b/PGI-2b*=*q*^{2}= 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. - 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. - 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. - 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 2*pq*is essentially equal to 2*q*. Since*q*= 0.01, 2*q*= 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. - 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
*I*. Thus, among^{A}*I*individuals, the fraction with the^{A}I^{O}*I*has to be the same (on average, of course) as its fraction in the entire population, which is 3/4. Similarly, the fraction of^{A1}*I*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^{A2}*I*and^{A1}*I*alleles,^{A2}*among**I*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^{A}I^{A}*p*^{2}, 2*pq*, and*q*^{2}from the Hardy-Weinberg equation where*p*is the frequency of*I*and^{A1}*q*is the frequency of*I*, which are 3/4 and 1/4, respectively.^{A2} - 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 2*pq*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,*q*^{2}or 0.05^{2}= 0.0025. Note, for the sake of checking these answers, that the rate at which females are neither afflicted nor carriers is*p*^{2}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. - 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 2
*pq*?). Thus: *A*, 0.1 * 0.1 = 0.01_{1}A_{1}*A*, 0.1 * 0.2 * 2 = 0.04_{1}A_{2}*A*, 0.1 * 0.3 * 2 = 0.06_{1}A_{3}*A*, 0.1 * 0.4 * 2= 0.08_{1}A_{4}*A*, 0.2 * 0.2 = 0.04_{2}A_{2}*A*, 0.2 * 0.3 * 2 = 0.12_{2}A_{3}*A*, 0.2 * 0.4 * 2 = 0.16_{2}A_{4}*A*, 0.3 * 0.3 = 0.09_{3}A_{3}*A*, 0.3 * 0.4 * 2 = 0.24_{3}A_{4}*A*, 0.4 * 0.4 = 0.16_{4}A_{4}- 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:*p*= 1 -_{2}*p*= 1 - 0.3 = 0.7 and_{1}*q*= 1 -_{3}*q*-_{1}*q*= 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:_{2} *A*_{1}*B*, 0.3 * 0.2 = 0.06_{1}*A*_{1}*B*, 0.3 * 0.3 = 0.09_{2}*A*_{1}*B*, 0.3 * 0.5 = 0.15_{3}*A*_{2}*B*, 0.7 * 0.2 = 0.14_{1}*A*_{2}*B*, 0.7 * 0.3 = 0.21_{2}*A*_{2}*B*, 0.7 * 0.5 = 0.35_{3}- Don't let the notation
throw you.
*w*_{11}is just the relative fitness of*AA*,*w*_{12}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. - The frequency of
*I*is 0.2062 * 0.2062 = 0.0425. The frequency of^{A}I^{A}*I*is 0.0756* 0.0756 = 0.0057. The frequency of^{B}I^{B}*II*is 0.7183 * 0.7183 = 0.5160. The frequency of*I*is 2 * 0.2062 * 0.0756 = 0.0312. The frequency of^{A}I^{B}*I*is 2 * 0.2062 * 0.7183 = 0.2962. The frequency of^{A}I*I*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^{B}I*A*phenotype is equal to the sum of the frequencies of the*I*and^{A}I^{A}*I*genotypes = 0.0425 + 0.2962 = 0.3387. The frequency of the^{A}I*B*phenotype is equal to the sum of the frequencies of the*I*and^{B}I^{B}*I*genotypes = 0.0057 + 0.1086 = 0.1143. The frequency of the^{B}I*AB*phenotype is equal to the frequency of the*I*genotypes = 0.0312. The frequency of the^{A}I^{B}*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. - 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*q*^{2}= 0.0064 or 0.64%. - 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 0.5425^{2}, 2 * 0.5425 * 0.4575, and 0.4575^{2}, 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. - 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. - 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. - The frequency of type
A blood =
*I*^{A}^{2}+ 2 **I**^{A}*I*= 0.4347. The frequency of type B blood =^{O}*I*^{B}^{2}+ 2 **I**^{B}*I*= 0.084. The frequency of type AB blood = 2 *^{O}*I**^{A}*I*= 0.0324. The frequency of type O blood =^{A}*I*^{O}^{2}= 0.4489. 0.4347 + 0.084 + 0.0324 + 0.4489 = 1.000. **References**- Campbell, N. A.
(1996).
*Biology***.**Fourth Edition. Benjamin/Cummings Publishing, Menlo Park, California. pp. 434-435. - Hartl, D. L. (1980).
*Principles of Population Genetics.*Sinauer Ass., Inc., Sunderland, Massachusetts. pp. 78, 137-139. - Hartl, D.L. (1983).
*Human Genetics.*Harper & Row, Publishers, New York. pp. 484-485. - Sinnott, E.W., Dunn,
L.C., Dobzhansky, T. (1958).
*Principles of Genetics.*Fifth Edition. McGraw-Hill Book Co., Inc. New York. p. 253.