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
- 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 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
- 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, A1, A2, A3,
and A4, at frequencies .1, .2, .3, and .4,
respectively. Calculate the expected random mating frequencies of all
- 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
- 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
- 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
- (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
- 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
- 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?
- 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
= 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.
- 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
- 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 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.
- 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
- 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.
- 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:
0.1 * 0.1 = 0.01
0.1 * 0.2 * 2 = 0.04
0.1 * 0.3 * 2 = 0.06
0.1 * 0.4 * 2= 0.08
0.2 * 0.2 = 0.04
0.2 * 0.3 * 2 = 0.12
0.2 * 0.4 * 2 = 0.16
0.3 * 0.3 = 0.09
0.3 * 0.4 * 2 = 0.24
0.4 * 0.4 = 0.16
- 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:
- A1 B1,
0.3 * 0.2 = 0.06
- A1 B2,
0.3 * 0.3 = 0.09
- A1 B3,
0.3 * 0.5 = 0.15
- A2 B1,
0.7 * 0.2 = 0.14
- A2 B2,
0.7 * 0.3 = 0.21
- A2 B3,
0.7 * 0.5 = 0.35
- 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,
- 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.
- 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%.
- 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.
- 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 = 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.
- 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.