Population Genetics Practice Problems and Answers

1. Consider the following population:   AA      Aa      aa

                                    Individuals         200    100     100

         - calculate the genotypic and gene frequencies.

                        Of the 400 individuals, 200 are AA = 200/400 = 0.5
                                                           100 are Aa =  100/400 = 0.25
                                                           100 are aa = 100/400 = 0.25

                        There are 400 "A" alleles in the 200 AA individuals and 100 "A" alleles in the 100 Aa individuals, for a total of 500/800 alleles = 0.625 there are 100 "a" alleles in the 100 aa individuals, and 100 "a" alleles in the 100 Aa individuals, for a total of 300/800 alleles = 0.375
 
 

2. What would the genotypic frequencies be in a population with p = 0.3, q = 0.7, if the population was in HWE?

f(AA) = p² = 0.09    f(Aa) = 2pq = 0.42        f(aa) = q² = 0.49
 

3. Consider this population:            AA  = 0.3        Aa = 0.2    aa = 0.5
    Is the population in HWE?

Well, the gene frequencies are f(A) = f(AA) + f(Aa)/2 = 0.3 + 0.2/2 = 0.4
                                              f(a)  = f(aa) + f(Aa)/2 = 0.5 + 0.2/2 = 0.6

IF the population were in HWE, THEN we should see genotypic frequencies of p², 2pq, and q², OR:  AA = 0.16, Aa = 0.48, aa = 0.36.  This isn't even close to what we have, so the population IS NOT in HWE.

4. If the assumptions of HWE do not actually apply to any real population, then how can it be useful?

               - We use it for COMPARISON.  This model describes what the genotypic frequencies should be IF the population was in equilibrium.  If the real genotypic frequencies are not close to these expectations, then:
                         - the population is not in HWE.... it is evolving.
                         - and if it IS NOT in HWE, then one of the assumptions must be violated.  Think about that.  The HWE is only 'true' if the assumptions are being met.  If your real population differs from the model, then one of the assumptions must not apply to your real population.  This narrows your focus on WHY the real populations isn't behaving randomly... and it might identify WHY the population is evolving.

                - In class, we used the coin analogy.  No REAL coin is probably exactly perfectly balanced. But, if I give you a coin and ask you how balanced it is, you flip it a few times and compare its behavior to WHAT YOU WOULD EXPECT FROM A PERFECT COIN (50:50 RATIO). Even though a perfectly balanced coin does not exist, we can use this theoretical model as a benchmark, to compare the behavior of real coins.  Many real coins act in a manner that is consistent enough with the expectations from a perfectly balanced coin that we are willing to use them AS IF they were perfectly balanced.

                - Hardy Weinberg Equilibrium Model is the same... it is a model of no change against which we can measure real populations.  Why would real populations differ from this model?  BECAUSE THEY VIOLATE ONE OR MORE OF THE ASSUMPTIONS!!  Know this logic.

5. What are the assumptions of HWE?

                      a. Infinitely large population (no random error)
                      b. Random mating (fertilization is probabilistic)
                      c. No selection, migration, or mutation
                                (gene frequencies do not change)

6.  Consider this population.  AA = 0.4, Aa = 0.2  aa = 0.4.  What are the gene frequencies?  What would the genotypic frequencies be after 1 generation of positive assortative mating?
 

                                                AA        Aa        aa

                                                0.4        0.2        0.4

                                   F1 = AA = 0.4 (from AA) + 0.05 (from Aa parents) = 0.45
                                            Aa = 0.1 (1/2 of the offspring of the 0.2 heterozygous parents)
                                            aa = 0.4 (from aa) +0.05 (from Aa parents) = 0.45

                                                0.45        0.1        0.45   = 1  (check!)

7.  How do the effects of inbreeding similar to the effects of Positive Assortative Mating?  How are the effects of  inbreeding different?  Why are the effects of inbreeding 'bad'?

    Both decrease heterozygosity and increase homozygosity.  However, this effect occurs across the whole genome in inbreeding, and only at the loci for which the population is selecting mates in P-A-M.  This increase in homozygosity across the whole genome will combine rare deleterious allels in the homozygous offspring - resulting in their death or disfunction.
 

8. Mutation: consider a population with p = .7 and q = 3. Suppose A mutates to a at a rate of 2 x 10^-6. What will the new gene frequencies be after one generation of mutation?

OK, so A is mutating to a... the rate of change will be "pm" in this case.... so (0.7) x (0.000002) = 0.0000014

Since A is mutating to a, this amount gets SUBTRACTED from A, and added to a:

p1 = 0.7 - 0.0000014 = 0.6999986

q1 = 0.3 + 0.0000014 = 0.3000014

9. Migration: suppose we have a population with gene frequencies equaling F(A) = 0.4 and f(a) = 0.6. Suppose migrants immigrate, with gene frequencies = f(A)= 0.8 and f(a) = 0.2, and the immigrants comprise 30% of the new population. What are the new gen frequencies?

p1 = (0.4)(0.7) + (0.8)(0.3) = 0.28 + 0.24 = 0.52

10. Consider the following population:

                                                     AA              Aa              aa

                Genotypic Frequency     0.3             0.3             0.4

                      Prob. of survival       0.4             0.2             0.1
 

        a. What are the initial gene frequencies?

                A = .45, a = .55

        b. Is the population in HWE?

No, because the actual genotypic frequencies do not match the predicted values of AA = (.45)², Aa = 2(.45)(.55) and aa = (.55)²)

        c. What are the relative fitness values?

                   0.4/0.4 = 1,   0.2/0.4 = .5,   0.1/0.4 = 0.25

        d. What are the genotypic frequencies in the breeding population?

                 - STEP 1: multiply initial frequencies by the relative fitness values:

                                        AA              Aa              aa

                                     (.3)(1) = .3     (.3)(.5) = .15    (.4)(.25) = .1

                        HEY! They don't sum to 1 any more.... No, they don't, because not all of the indivuduals survived to reproductive age... there was 'selection', and those who carried 1 'a' (Aa) or 2 (aa) were selected against.  Those with 2 A's (AA) were selected for.   So what do I do now?

                 - STEP 2: SUM these values = .55

                 - Divide the products by this sum:

                                        .3/.55 = .55   .15/.55 = .27    .1/.55 = .18

                 THESE ARE THE GENOTYPIC FREQUENCIES IN THE BREEDING POPULATION (AND THEY SHOULD SUM TO 1 NOW.)
 

        e. What are the gene frequencies in the breeding gene pool?

                        F(A) = f(AA) + f(Aa)/2 = .55 + .27/2 = .685
                        f(a) = f(aa) + f(Aa)/2 = .18 + .27/2 = .315

                        These sum to 1, so I've done the math correctly.

        f. If there is random mating, what will be the genotypic frequencies in the next generation?

                - So, the probability of AA = f(A)² = .47
                - the probability of Aa = 2[f(A)][f(a)] = .43
                 - the frequency of aa = f(a)² = .1

                These sum to 1, so my math is correct.

        g. What agent of evolutionary change is at work?.

                Selection.

        h. Why have the gene frequencies changed?

Because individuals with 'A' reproduce their 'A' genes at a higher frequency than individuals with 'a' genes.  So 'A' genes accumulate in the population relative to 'a' genes.  this differnential reproductive success occurs because of differential survival probabilities.