PROBLEM SET
Chapter 19 -- Population Genetics: Hardy-Weinberg Equilibrium...

P. 548, #1.  One hundred persons from a small town in Pennsylvania
were tested for their MN blood types.  Is the population they
represent in Hardy-Weinberg proportions?  The genotypic data are:
MM, 41; MN, 38; and NN, 21.
The first step is to calculate the allele frequencies for M and N.
            2*(# of MM) + (# of MN)         2*(41) + (38)
freq(M) = ---------------------------- =  -----------------
           2*(total # of individuals)     2*(41 + 38 + 21)
freq(M) = 120/200 = 0.6
freq(N) = 1 - freq(M) = 1 - 0.6 = 0.4

The next step is to calculate the expected frequencies of each of
the genotypes given that the population is in Hardy-Weinberg
proportions.
freq(MM) = 0.6^2 = 0.36
freq(MN) = 2*0.6*0.4 = 0.48
freq(NN) = 0.4^2 = 0.16
The expected numbers of each genotype are determined by
multiplying the expected frequency of each genotype by the number
of individuals in the sample.
expected number of MM = 0.36 * 100 = 36
expected number of MN = 0.48 * 100 = 48
expected number of NN = 0.16 * 100 = 16

The final step is to compare the expected and observed
distributions of numbers with chi-square analysis.  This will test
the null hypothesis that the population is in Hardy-Weinberg
proportions.
                       MM            MN          NN             
observed numbers       41            38          21
expected numbers       36            48          16
((obs-exp)^2)/exp     0.69          2.08        1.56

chi-square value = 0.69 + 2.08 + 1.56 = 4.33
The degrees of freedom for the critical value is 1
(number of phenotypes - number of alleles) = (3-2) = 1
Thus the critical value to reject is 3.814.
As the calculated value is greater than the critical value, we
reject the null hypothesis that our population is in Hardy-Weinberg
proportions.  
 
P. 548, #2.  In the following two sets of data, calculate allelic
and genotypic frequencies and determine whether the populations
are in Hardy-Weinberg proportions.  Do a statistical test if one
is appropriate.
  a. Allele A is dominant to a:  A-, 91; aa, 9.
In this problem we have to assume that the population is in
Hardy-Weinberg to be able to estimate the frequencies of the alleles,
thus we will not be able to test if the population is in Hardy-Weinberg.
First, freq(aa) = freq(a) x freq(a) = q^2 = 9/100 = 0.09.
As q^2 = 0.09, taking the square root of both sides yields
 q = 0.30.  As q = 0.30, then p = 1 - q = 1 - 0.3 = 0.7.
Thus freq(A) = p = 0.7.
Second, given that p = 0.7 and q = 0.3, then
freq(AA) = p^2 = 0.7^2 = 0.49
freq(Aa) = 2pq = 2 x 0.7 x 0.3 = 0.42
freq(aa) = 9/100 = q^2 = 0.09
As the number of phenotypes is 2 and the number of alleles is 2,
the degrees of freedom is (2-2) 0.  Thus, we can not do a test for
Hardy-Weinberg proportions because we need at least 1 degree of
freedom to do a chi-square test.
  b. Electrophoretic alleles F and S are codominant at the alcohol
the dehydrogenase locus in Drosophila: FF, 137; FS, 196; SS, 87.
As we have three phenotypes and two alleles, we can do a chi-square test
for Hardy-Weinberg proportions, thus this problem is
just like problem #1.

The first step is to calculate the allele frequencies for F and S.
            2*(# of FF) + (# of FS)        2*(137) + (196)
freq(F) = ---------------------------- =  -----------------
           2*(total # of individuals)     2*(137 + 196+ 87)
freq(F) = 470/840 = 0.56
freq(S) = 1 - freq(S) = 1 - 0.56 = 0.44

The next step is to calculate the expected frequencies of each of
the genotypes given that the populations is in Hardy-Weinberg
proportions.
freq(MM) = 0.56^2 = 0.3136
freq(MN) = 2*0.56*0.44 = 0.4928
freq(NN) = 0.44^2 = 0.1936
The expected numbers of each genotype is determined by multiplying
the expected frequency of each genotype by the number of
individuals in the sample.
expected number of FF = 0.3136 * 420 = 131.7
expected number of FS = 0.4928 * 420 = 207.0
expected number of NN = 0.1936 * 420 = 81.3

The final step is to compare the expected and observed
distributions of numbers with chi-square analysis.  This will test
the null hypothesis that the population is in Hardy-Weinberg
proportions.
                       MM            MN          NN             
observed numbers      137           196          87
expected numbers      131.7         207.0        81.3
((obs-exp)^2)/exp     0.213         0.585        0.400

chi-square value = 0.213 + 0.585 + 0.400 = 1.198
The degrees of freedom for the critical value is 1
(number of phenotypes - number of alleles) = (3-2) = 1
Thus the critical value to reject is 3.814.
As the calculated value is less than the critical value, we fail
to reject the null hypothesis that our population is in Hardy-Weinberg
proportions.  

P. 548, #3.  The dominant ability to taste PTC is due to the
allele T.  Among a sample of 215 individuals from a population,
150 could detect the taste of PTC and 65 could not.  Calculate the
frequencies of T and t.  Is the population in Hardy-Weinberg proportions?
This problem is very similar to Problem 2a because we have only
two phenotypes and two alleles and thus we have to assume that the
population is in Hardy-Weinberg to be able to estimate the
frequencies of the alleles, thus we will not be able to test if
the population is in Hardy-Weinberg.
First, freq(tt) = freq(t) x freq(t) = q^2 = 65/215 = 0.302.
As q^2 = 0.302, taking the square root of both sides yields
 q = 0.55.  As q = 0.55, then p = 1 - q = 1 - 0.55 = 0.45.
Thus freq(T) = p = 0.45.
Second, given that p = 0.45 and q = 0.55, then
freq(TT) = p^2 = 0.45^2 = 0.2025
freq(Tt) = 2pq = 2 x 0.45 x 0.55 = 0.495
freq(tt) = 65/215 = q^2 = 0.302
As the number of phenotypes is 2 and the number of alleles is 2,
the degrees of freedom is (2-2) 0.  Thus, we can not do a test for
Hardy-Weinberg proportions because we need at least 1 degree of
freedom to do a chi-square test.

P. 548, #4.  The frequency of children homozygous for the
recessive allele for cystic fibrosis is about one in 2,500.  What
is the percentage of the heterozygotes in the population?
This problem is very similar to Problem 2a because we have only
two phenotypes (normal and with cystic-fibrosis) and two alleles
(the cystic fibrosis allele (c) and the normal allele (C)) and
thus we have to assume that the population is in Hardy-Weinberg to
be able to estimate the frequencies of the alleles.  Thus we will
not be able to test if the population is in Hardy-Weinberg.
First, freq(cc) = freq(c) x freq(c) = q^2 = 1/2500 = 0.0004.
As q^2 = 0.0004, taking the square root of both sides yields
 q = 0.02.  As q = 0.02, then p = 1 - q = 1 - 0.02 = 0.98.
Thus freq(C) = p = 0.98.
Second, given that p = 0.98 and q = 0.02, then
freq(CC) = p^2 = 0.98^2 = 0.9604
freq(Cc) = 2pq = 2 x 0.98 x 0.02 = 0.0392
freq(cc) = 1/2500 = q^2 = 0.0004

P. 548, #8.  PTC tasting is dominant in human beings.
a. Should most human populations be heading toward a 3:1 ratio of
tasters to nontasters? Explain.
You can assume that most human populations are in Hardy-Weinberg
equilibrium at this locus and thus that allele or genotype
frequencies are not changing and thus not heading towards any
given ratio.  In addition, we would expect a 3:1 ratio only if the
alleles were in equal frequency (p = q = 0.5).  There is no reason
to suspect that this might be true.
b. Confronted with a population sample of human beings of unknown
origin, would you expect more or less than half the sample would
be tasters.
Given that we do not know what the frequency of the T allele is
and that T is dominant, we would expect more individuals to be
tasters than nontasters because the frequency of tasters would be
p^2 + 2pq and this is less than 0.5 only when p is less than 0.30. 
If we assume a uniform distribution of allele frequencies across
all human populations, then we could assume that populations with
more tasters would be more likely to occur than populations with
more nontasters.


P. 549, #16. What allelic frequency will generate twice as many
recessive homozygotes as heterozygotes?
Algebraically, we need to solve the equation
 freq(aa) = 2 * freq(Aa)
Given that the population is in Hardy-Weinberg proportions
then,  q^2 = 2*(2*p*q)
which we then solve for q
q^2 = 4*p*q, substitute 1-q for p
q^2 = 4*(1-q)*q
q^2 = 4*q - 4*q^2, dividing through by q
q = 4 - 4*q, subtracting 4 from both sides and q from both
             sides gives
-4 = -5*q, dividing through by -5
-4/-5 = q
 0.8 = q.

P. 549, #20.  On a small island, 235 mating individuals are all
true-breeding for brown eyes.  An epidemic eliminates all the
population except 10 young women, two young men, and four older
(postmenopausal)women.  A boatload of foreigners arrives: the foreign
population consists of six heterozygous brown-eyed females, four 
homozygous brown-eyed males, and 10 blue-eyed males.  Assuming that eye 
color is controlled by one locus, that mating is random with respect to 
eye color, and that each male and female capable of breeding does so, 
calculate the allelic frequencies of their offspring.
After the epidemic (the elderly survivors of the epidemic are not part of the breeding populations) and with the arrival of the foreigners, we have in the resulting breeding population,
     survivors                foreigners
    10 BB women                 10 bb men
     2 BB men                    4 BB men
                                 6 Bb women
We need to determine the allele frequencies in the breeding males and females. Thus
                         2(10 for BB) + (6 for Bb)
freq(B) in the women is --------------------------- = 0.8125
                            2(total women)

freq(b) in the women is 1 - 0.8125 = 0.1875
and
                         2(6 for BB) + (0 for Bb)
freq(B) in the men is --------------------------- = 0.3750
                            2(total men)

freq(b) in the men is 1 - 0.3750 = 0.6250
Now to calculate the expected frequency of each genotype in the offspring of these individuals, we need to assume that Hardy-Weinberg will hold. Thus the frequency of each genotype is given by the frequency of each gamete from the male gamete pool and from the female gamete pool. Thus
  freq(BB) = freq(B) for males x freq(B) for females
           = 0.3750 * 0.8125 = 0.3047

  freq(Bb) = freq(B) for males x freq(b) for females plus
             freq(b) for males x freq(B) for females
           = 0.3750 * 0.1875 + 0.6250 * 0.8125 = 0.0703 + 0.5078
           = 0.5781

  freq(bb) = freq(b) for males x freq(b) for females
           = 0.6250 * 0.1875 = 0.1171
Having now determined the frequency of each genotype, we can use that information to calculate the allele frequency in the populations.
  freq(B) = freq(BB) + 0.5 * freq(Bb)
          = 0.3047 + 0.5 * 0.5781
          = 0.5938

  freq(b) = freq(bb) + 0.5 * freq(Bb)
          = 0.1171 + 0.5 * 0.5781
          = 0.4062
P. 549, #22. In a given population, only the A and B alleles are present in the ABO system; there are no individuals with type O blood or with O alleles. If two hundred people have type A blood, seventy-five have type AB blood, and twenty-five have type B blood, what are the alleleic frequencies of this population?
To calculate the allele frequencies for A and B, we need to remember that the individuals with type A blood are homozygous AA, individuals with type AB blood are heterozygous AB, and individuals with type B blood are homozygous BB.
            2*(# of AA) + (# of AB)         2*(200) + (75)
freq(A) = ---------------------------- =  -----------------
           2*(total # of individuals)     2*(200 + 75 + 25)

freq(A) = 475/600 =  0.792

freq(B) = 1 - freq(B) = 1 - 0.792 = 0.208
P. 550, #24. In a sample of 100 people, there are 14 MM, 32 MN, and 54 NN individuals. Calculate the inbreeding coefficient.
We need to use the formula for the inbreeding coefficient, which is
         expected # of heterozygotes - observed # of heterozygotes
    F =  --------------------------------------------------------
                    expected # of heterozygotes
where the expected # of heterozygotes is given by Hardy-Weinberg proportions. So the first thing that we need to do is to calculate the frequency of each allele and then the expected number of heterozygotes. To calculate the allele frequencies for M and N,
            2*(# of MM) + (# of MN)         2*(14) + (32)
freq(M) = ---------------------------- =  -----------------
           2*(total # of individuals)     2*(14 + 32 + 54)

freq(M) = 60/200 = 0.3

freq(N) = 1 - freq(M) = 1 - 0.4 = 0.7
The next step is to calculate the expected frequencies of each of the genotypes given that the population is in Hardy-Weinberg proportions.
freq(MM) = 0.3^2 = 0.09
freq(MN) = 2*0.7*0.3 = 0.42
freq(NN) = 0.7^2 = 0.49
The expected numbers of each genotype is determined by multiplying the expected frequency of each genotype by the number of individuals in the sample.
expected number of MM = 0.09 * 100 = 9
expected number of MN = 0.42 * 100 = 42
expected number of NN = 0.49 * 100 = 49
Now we use the formula for the inbreeding coefficient that we introduced above.
      expected # of heterozygotes - observed # of heterozygotes
 F =  --------------------------------------------------------
                   expected # of heterozygotes

 F = (42-32)/42 = 0.238
P. 550, #25. If, in a population with two alleles at an autosomal locus, p = 0.8, q = 0.2, and the frequency of heterozygotes is 0.20, what is the inbreeding coefficient?
We need to use the formula for the inbreeding coefficient, which is
       exp. freq of heterozygotes - obs. freq of heterozygotes
F = --------------------------------------------------------------
               expected freq of heterozygotes
where the expected freq of heterozygotes is given by Hardy-Weinberg
proportions.  Thus,

F = (2pq - 0.20)/2pq = (0.32 - 0.20)/0.32
F = 0.375

Last updated on 18 November 1996.
Provide comments to Dwight Moore at mooredwi@esumail.emporia.edu.
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