C H A P T E R

CHAPTER 8POPULATION GENETICS

8.1 IntroductionPopulation genetics is the study of the genetics of populations. We are less interested in the expression of genes, rather in the frequencies of the alleles in the population and how they change (or do not change) over time. For this discussion, we consider only codominant alleles. In practice, many population geneticists are interested in molecular markers—genes without function that will give us an understanding of how populations or species are related to each other, and are inherently codominant.

8.2 Hardy-Weinberg PopulationsConsider a population of some organism, shown in Figure 8-1. As we are interested in their genotypes, they are represented by their genotypes in the figure.Table 8-1 shows the summary numbers for the population. Often, we use an “equation,” the genotypic array, to help organize these numbers.In addition to the genotypic frequencies, we often are interested in the frequencies of the alleles. Recall the way we calculated the gametic frequencies in pedigree problems in Chapter 5 (page 5-61). The frequency of the A-allele is , and for the a-allele it is

. They can also be written as an arithmetic statement, the

gametic array, as is done in Table 8-1.Note: Population geneticists have long used the terms allele and gene interchangeably, and refer to allele frequencies as gene frequencies. While it is incorrect terminology, it is too late to change now. Unless I slip into old habits, we will use allele frequencies for this notebook.

P 12-Q+

R 12-Q+

103

8.2.1 Summary ArraysThe two summary arrays are the genotypic array (summarizing the genotypes and their frequencies) and the gametic array (summarizing the alleles and their frequencies).

Genotypic Array

The genotypic array (set of genotypic frequencies) is a summary statement showing the genotypic frequencies in the population:

Figure 8-1 Sample Population Showing Genotypes of Individuals

Table 8-1 Summary of Sample Population (Figure 8-1).

Genotypes

AA Aa aa TotalNumber of Individuals 18 24 8 50

Relative Frequency 0.36 0.48 0.16 1.00

Genotypic Array 0.36 AA + 0.48 Aa + 0.16 aaGametic Array 0.6 A + 0.4 a

AA

AA

AA

AA

AA

AA

AA

AA

AA

AA

AA

AAAA

AA

AAAA

AAAA

AaAa

Aa

Aa

Aa

Aa

AaAa

Aa

Aa

Aa

Aa Aa

AaAa

AaAa

Aa

Aa

Aa

Aa

Aa Aa

Aa

aa

aa

aa

aa aa

aa

aaaa

104

Gametic Array

Similarly, the gametic array (set of allelic frequencies) is the statement of the allele frequencies in the population:pA + qa,where

8.2.2 Hardy-Weinberg AssumptionsHardy and Weinberg independently came up with the Hardy-Weinberg Law. A Hardy-Weinberg population is a population that follows the Hardy-Weinberg Law. Such a population serves as a reference population for comparing natural populations.In order for a population to be in Hardy-Weinberg equilibrium, it must have several very restrictive assumptions:

• No mutation• No migration• No selection• Random mating (random pairing of gametes)• Population is infiniteThere are more assumptions, such as diploid organism or discrete generations, that we will not explore in this chapter.

Hardy-Weinberg LawThe Hardy-Weinberg Law states that after a single generation under the given assumptions, the genotypic array is

,

and these frequencies are in equilibrium (do not change over time).

Calculating Hardy-Weinberg ProportionsThe population has genotypic frequencies , so the population has allelic frequencies

.

and

PAA QAa Raa+ +

p P 12-Q+= q R 1

2-Q+=

p2AA 2pqAa q2aa+ +

PAA QAa Raa+ +

P 12-Q+⎝ ⎠

⎛ ⎞A R 12-Q+⎝ ⎠

⎛ ⎞ a+ pA qa+=

105

The population has Hardy-Weinberg proportions if

.

Because of the relationship between allele frequencies and genotypic frequencies, if any of these equations is not valid, none of them will be. You only have to check one of the equations.

8.2.3 Hardy-Weinberg Proportions

ExampleMN blood group in Eskimos.

Without doing a formal test (see Section 8.2.6 for the formal test), we can decide if this population seems to have Hardy-Weinberg proportions. To do this, we will look at what frequencies we would expect under Hardy-Weinberg proportions.First, compute the allele frequencies from the number of individuals with each genotype:

Next, we compute the expected frequencies if the population has Hardy-Weinberg frequencies:

Finally, we compare to the original population data as given in Table 8-3.

It seems that the MN blood group has Hardy-Weinberg proportions. In a later section, we will formally test this hypothesis (Section 8.2.6).

Table 8-2 Frequency of MN Blood Groups among 2500 Eskimos

MM MN NN Total

Number of Individuals 2,087 390 23 2,500Relative Frequency 0.8347 0.1560 0.0093 1.0000

M: (2087 + 390/2)/2500 = 0.9128 = pN: (23 + 390/2)/2500 = 0.0872 = q

MM: (0.9128)2 = 0.8332 = p2

MN: 2(0.9128)(0.0872) = 0.1592 = 2pqNN: (0.0872)2 = 0.0076 = q2

Table 8-3 Comparison of Genotypic Frequencies to Hardy-Weinberg

GenotypeObservedFrequency

ExpectedFrequency

MM 0.8347 0.8332MN 0.1560 0.1592NN 0.0093 0.0076

P p2= Q, 2pq and R, q2= =

106

Some Terms about Variability

Heterozygosity The heterozygosity is the fraction of heterozygotes in the population. If the study includes several loci, the heterozygosity is usually given as an average heterozygosity. It is related to homozygosity, the total fraction of homozygotes in the population.

Polymorphism When there is more than one allele with any significant frequency in the population for that locus. The cutoff frequency depends on the organism studied. The generally accepted frequency for humans is 99%, so a locus is said to be polymorphic if the frequency of the most frequent allele is less than 99%.

8.2.4 Hardy-Weinberg Proportions with More Than Two Alleles

Often in population genetics, we look at loci that have more than two alleles. In fact, most loci have many more alleles than that. The basis of Hardy-Weinberg proportions is that with random mating of parents from the same population, the probabilities of the genotypes can be derived by multiplying the gametic array from each parent as(p A + q a)2 = p2 AA + 2pq Aa + q2 aa.If we have, say, three alleles, the equation becomes(p1 A1 + p2 A2 + p3 A3)2 = p1

2 A1A1 +2p1p2 A1A2 + p22 A2A2 +

2p1p3 A1A3 + 2p2p3 A2A3 + p32 A3A3.

As with our original Hardy-Weinberg proportions, this occurs immediately (after a single generation of random mating). Table 8-4 presents an example of Hardy-Weinberg proportions for a situation with three alleles.

8.2.5 Hardy-Weinberg Proportions with a Sex Linked Trait

For a sex-linked trait, the Hardy-Weinberg proportions are not reached immediately. This is because of the unequal contribution of males and females. We can do a Punnett square with sex-linked traits that have the allele frequencies p and q.

Table 8-4 Hardy-Weinberg Proportions when p1 = 0.7, p2 = 0.2, p3 = 0.1

Genotype A1A1 A1A2 A2A2 A1A3 A2A3 A3A3

Frequency 0.49 0.28 0.04 0.14 0.04 0.01

pA qa Y

pA p2 AA pq Aa p AY

qa pq Aa q2 aa q aY

107

Or, you can simply multiply the gametic array from each of the parents as:(p A + q a) (p A + q a|| Y) = (p2 AA + 2pq Aa + q2 aa)||p AY + q aYFrom this, it is clear females have the normal Hardy-Weinberg proportions and males have the allele frequencies for their two possible genotypes.For our example, the first generations are given in Table 8-5. After a long time, the population will reach an equilibrium (the last row in Table 8-5), with frequencies that have the Hardy-Weinberg proportions.

8.2.6 Testing for Hardy-Weinberg RatiosChi Square Goodness of Fit tests are similar to what we found in Section 2.3. The basic formula for calculating the Chi Square is exactly the same,

.

Finding the expected numbers for the genotypes is somewhat more complicated, and the degrees of freedom follow a different rule.

HypothesisThe expected number follows directly from the hypothesis. The hypothesis for these problems are always of the type, “The trait under consideration for that population has Hardy-Weinberg proportions.” This means that the numbers of individuals with each of the genotypes follows the Hardy-Weinberg frequencies.

Expected NumbersThe expected proportions of the genotypes follows the Hardy-Weinberg frequencies, p2, 2pq, and q2. First, calculate the observed allele frequencies from the data. The expected numbers are then the Hardy-Weinberg proportions multiplied by the total.

Table 8-5 Random Mating with an X-Linked Trait

Generation Females Males

0 AA × aY

1 Aa × AY

2 1/2 AA + 1/2 Aa × 1/2 AY + 1/2 aY

3 3/8 AA + 1/2 Aa + 1/8 aa × 3/4 AY + 1/4 aY

Many GenerationsEquilibrium 4/9 AA + 4/9 Aa + 1/9 aa 2/3 AY + 1/3 aY

χ2 O E–( )2

E-----------∑=

108

Degrees of FreedomIn previous Chi Square tests, we knew the hypothesized frequencies before the experiment was run. For Hardy-Weinberg, the hypothesized frequencies are relative, not absolute. We need to use the data to compute the actual expected numbers. Because of this, the degrees of freedom are different:

df = Number of Genotypes - Number of Alleles

ExampleFor an electrophoretic survey, the following numbers of individuals were observed:

Is there evidence that this population does not have Hardy-Weinberg ratios?

The Hypothesis The hypothesis to be tested is, “This population has Hardy-Weinberg proportions for the A locus.”

The Test The Chi Square Goodness of Fit test is done to test this hypothesis.• First, compute the allele frequencies as

and

.

• Next, compute the expected Hardy-Weinberg proportions (Table 8-6).

• These genotypic frequencies are used to compute the expected values in the Chi Square table (Table 8-7).

Genotypes AA Aa aa TotalNumber Observed 40 25 50 115

Table 8-6 Hardy-Weinberg Frequencies

Genotype Hardy-Weinberg Computed Values Expected Frequency

AA p2 (0.45652)2 0.20841

Aa 2pq 2(0.45652)(0.54348) 0.49622

aa q2 (0.54348)2 0.29537

p 40 12- 25( )+

115-------------- 0.45652= =

q 50 12- 25( )+

115-------------- 0.54348= =

109

The value is 36.30, with 3-2=1 degrees of freedom (3 genotypes - 2 alleles).

The Statistical Conclusion The calculated Chi Square is greater than the tabled value, so we reject the hypothesis at the 5% significance level.

The Genetic Conclusion There is sufficient evidence at the 5% significance level to conclude that this locus do not have Hardy-Weinberg proportions for this population.

Problem: A population survey resulted in frequencies for different genotypes as given in Table 8-8.

This Chi Square Goodness of Fit test for this situation is very similar to the test we just did.1 Calculate the allelic frequencies for all the alleles.2 Calculate the expected genotypic frequencies for all the

genotypes as if the population had Hardy-Weinberg proportions, as given in the previous section.

3 Use the Chi Square test to see if the frequencies deviate significantly from Hardy-Weinberg proportions (df = 6 genotypes - 3 alleles = 3).

8.3 Some Population Genetics ProblemsEXAMPLE: Assume that the frequencies of A2 for an autosomal locus in males and females are 0.8 and 0.4, respectively. Calculate the genotypic frequencies for the next two generations.EXAMPLE: Assume that the frequencies of A1 for an X-linked trait are 0.0 in males and 0.3 in females. What are the allelic frequencies in the two sexes for the first four generations?

Table 8-7 Chi Square Table for Hardy-Weinberg Test

Genotypes Observed Expected O-E Chi-Square

AA 40 (0.20841)115 = 23.97 16.03 10.72

Aa 25 (0.49622)115 = 57.07 -32.07 18.02

aa 50 (0.29537)115 = 33.97 16.03 7.56

Total 115 115.01 -0.01 36.30

Table 8-8 Three Allele Goodness of FitGenotype AA AB AC BB BC CC

Number Observed 8 38 121 27 252 401Hardy-Weinberg

Proportions p2 2pq 2pr q2 2qr r2

110

EXAMPLE: In a survey of 100 individuals, the frequencies of five alleles at a single locus are estimated to be 0.2, 0.1, 0.6, 0.05, and 0.05. If the population is in Hardy-Weinberg proportions, what is the heterozygosity (sum of the frequency of all the heterozygotes)?

8.4 When Hardy-Weinberg Assumptions Are Not Valid

Hardy-Weinberg assumptions rarely hold true for any sort of interesting traits. In this section, we will look at the various assumptions and consider how the population changes when the assumptions are not valid. In real populations, many of the assumptions will be violated in unusual and interesting ways. Here, we consider each assumption separately of each other, so that only one assumption is violated at a time.

8.4.1 MutationThe key to understanding mutation is to consider where each A gamete originated. An A gamete could have been an A allele in the previous generation that did not mutate, or it could have been an a allele in the previous generation that did mutate. Mathematically, this is

,where the mutation rates are

With mutation in both directions, the population will eventually reach an equilibrium with frequency

EXAMPLE: Suppose the initial genotypic array is:0.3025 AA + 0.4950 Aa + 0.2025 aawith gametic array: 0.55 A + 0.45 a.

Mutation occurs at gamete formation, and has a rate of from A to a and a rate of from a to A. This modifies the gametic array, so the new frequency for A is

The genotypic array of the offspring is:(0.54999495 A + 0.45000505 a)2 = 0.30249445 AA + 0.49500101 Aa + 0.20250455 aa

p′ 1 μ–( )p ν 1 p–( )+=

μ Prob A a⇒[ ], and ν Prob a A⇒[ ].= =

p νμ ν+------- .=

μ 1 10 5–×=

ν 1 10 6–×=

1 1 10 5–×–( ) 0.55( )⋅ 1 10 6–×( ) 1 0.55–( )⋅+ 0.54999495=

111

Give the genotypic array after another generation of mutation.

Mutation in the Long RunMutation can change the allele frequencies in a population. The change is extremely small, but will accumulate over time. If there is only one-way mutation, then eventually one of the alleles will be lost from the population. If there is mutation going in both ways, the population will eventually reach an equilibrium, depending on the relative mutation rates.

8.4.2 Migration (Mainland-Island Model)The frequencies reflect frequencies in the contributing population. The parents come from two sources, the island and the mainland. The migration rate is given as m, which is the fraction of the mating pool who are from the mainland (each generation). The probability a parent will come from the mainland is m, so the probability a parent will come from the island is 1-m. The allele frequency of A in the next generation is

,where P = Prob(A) on Mainland and m = Fraction of mating pool who are immigrants.EXAMPLE: Assume an island population with genotype frequencies0.09 AA + 0.42 Aa + 0.49 aa,and a mainland population with allelic frequency for A of 0.8. The island has in influx from the mainland of 1% of its population per generation. The initial allele frequency of the A allele is p=0.3.What are the genotypic frequencies in the next generation?Solution:The initial genotypic array is0.09 AA + 0.42 Aa + 0.49 aa, with gametic array 0.3 A + 0.7 a.We can compute the gametic array from the equation:

The genotypic array is(0.3050 A + 0.6950 a)2 = 0.09303 AA + 0.42395 Aa + 0.48303 aa.Question: What is the genotypic array after another generation of migration?

p′ 1 m–( )p mP+=

p′ 1 m–( )p mP+0.99( ) 0.3( ) 0.01( ) 0.8( )⋅+⋅

0.3050

===

112

Migration in the Long Run

Regardless of the mutation rate (so long as it is positive), eventually the island population will look exactly like the mainland population (p = P). The rate at which this happens depends on the migration rate. This is generally a moderate amount, and certainly much faster than mutation.

8.4.3 Viability SelectionSelection can take several forms. It is basically that some individuals are better to survive and reproduce than others, based on their genotypes. Some general forms of selection you can find are:• Viability Selection When some individuals are more likely to

survive to reproduction than others.• Fertility Selection When some individuals have a higher

fertility than others.• Fecundity Selection When some individuals are more fecund

than others.There are more esoteric forms of selection as well, including kinship selection, density selection, cline selection, etc. We don’t have time to cover more than the basic viability selection in this Notebook.For viability selection, we consider the fitness of the genotypes. The fitness can be viewed as the relative chances for individuals with that genotype to survive and reproduce. Table 8-9 contains examples of common situations. By convention, AA is given a fitness of 1.0, the other fitnesses are relative to it. Usually, they are smaller than 1.0, but they are not constrained by that. In addition, dominance and recessiveness is determined by the fitness of the heterozygote.

Depending on the values of the fitnesses (w’s), there are three general classes of selection:• Directional Selection The favored allele eventually fixed in the

population (the other is lost). The speed that this happens depends on the fitnesses involved. The first three rows of Table 8-9 are examples of directional selection.

• Balancing Selection (Overdominance) All alleles remain in the population, the equilibrium frequencies depend on the

Table 8-9 Examples of Fitness Values

Genotypes AA AB BB

Relative Fitness wAA wAB wBB

Recessive Lethal 1.0 1.0 0Dominant Lethal 1.0 0 0Additive Effects 1.0 0.8 0.6Overdominance 1.0 1.3 0.2

Underdominance 1.0 0.3 0.7

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Tabl

G

S

US

Frafte

aa

selection coefficients. For this simple model, the only way this can occur is if the fitness of the heterozygote is greater than both homozygotes.

• Disruptive (Underdominance) Either allele could be fixed, depending on initial conditions and the selection coefficients. For this simple model, the only way this can occur is if the fitness of the heterozygote is smaller than either homozygotes.

ComputationsWe are interested in computing the allele frequencies in the next generation. To do this, we imagine the fitnesses are simply the probability of survival. We see who survives, then mate the survivors at random. These computations are given in Table 8-10.

The last row is the frequencies of the genotypes among the individuals who survive until reproduction. The individuals who survive selection will not have Hardy-Weinberg proportions. However, if we mate the survivors at random with each other, their offspring will have Hardy-Weinberg proportions (since there is random mating). The allele frequency among the offspring are the same as the allele frequency among the surviving parents, and can be calculated as 1/2 of the heterozygotes + all of the homozygotes.EXAMPLE: Assume we have a population with genotypic array0.16 AA + 0.38 Aa + 0.36 aaand with fitnesses of 1.0, 0.9 and 0.6 respectively. We are interested in the change in allele frequencies over one generation of natural selection., The calculations are given in Table 8-11.The original allele frequency for A is 0.4. The following table allows us to compute the allele frequency in the population after one generation of natural selection. Look at the surviving parents, and calculate the gametic frequencies as P[A] = 0.19802 + 1/2(0.53465) = 0.46535, and P[a] = 0.26733 + 1/2(0.53465) = 0.53465.

e 8-10 Computation for Viability Selection

enotype AA Aa aa

Before election p2 2pq q2

Fitness WAA = 1.0 WAa Waa

nadjusted urvivors

Mean Fitness =

equencies r Selection

p2WAA2pqWAa q2Waa W p2WAA 2pqWAa q2W++=

p2WAA

W----------

2pqWAa

W------------ q2Waa

W---------

114

Tab

G

S

US

Frafte

Question: How would this change if A were completely dominant? completely recessive?

Selection in the Long Run

What happens to the population in the long run depends on the type of selection.

Directional Selection With directional selection, eventually only one allele will remain. It will be the allele with the greatest fitness as a homozygote.

Balancing Selection With balancing selection, the heterozygote has a higher fitness than either of the two homozygotes. In this case, neither allele can be lost in the population and it will eventually reach an equilibrium, depending on the parameters.

Underdominance With underdominance, the heterozygote has a lower fitness than both of the two homozygotes. For this, the population will quickly lose one of the two alleles, depending on the initial conditions. It is possible to have the allele with the lower fitness fixed in the population.

8.4.4 Nonrandom MatingFor this class, we consider just two types of nonrandom mating, inbreeding and assortative mating. In inbreeding, mating is based on relatedness (mating among relatives). In assortative mating, mating is based on the similarity of phenotypes.

InbreedingInbreeding occurs when mating is based on relatedness, for example with first cousin matings. Allelic frequencies remain unchanged, but

le 8-11 Example of Viability Selection

enotype AA Aa aa

Before election 0.16 0.48 0.36

Allele Frequenciesp = 0.4q = 0.6

Fitness 1.00 0.90 0.60

nadjusted urvivors 0.16 0.432 0.216

Mean (Population) Fitness

equencies r Selection 0.19802 0.53465 0.26733

Allele Frequenciesp = 0.46535q = 0.53465

W 0.808=

115

genotypic frequencies are modified, with a decrease in heterozygosity. If a population is undergoing a systematic system of inbreeding, the genotypic array becomes

(p2+Fpq) AA + 2(1-F)pq Aa + (q2+Fpq) aa,where p is constant. The inbreeding coefficient, F, is a positive number. Notice that there will be an increase in frequency of both homozygous genotypes, with a corresponding decrease in frequency of the heterozygote. One result of this is an increase in the expression of recessive characteristics. Table 8-12 shows the increase in appearance of recessive characteristics due to inbreeding. Even with very small inbreeding, there is a significant increase in the appearance of autosomal recessive traits.

EXAMPLE: F increases each generation where consanguineous matings occur. If a population is undergoing a systematic inbreeding scheme, and has genotypic array:0.60 AA + 0.20 Aa + 0.20 aa,what is the inbreeding coefficient?SOLUTION:p = 0.60 + 0.20/2 = 0.70; q = 0.20 + 0.20/2 = 0.30.The frequency of the AA genotype is p2 + Fpq = 0.60. Solving for F, we get(0.70)2 + F(0.70)(0.30) = 0.60 ⇒ F = 0.52381.

Inbreeding in the Long Run

Inbreeding, by itself, does not change allele frequencies. However, in concert with selection, it intensifies directional selection. With inbreeding alone, the homozygosity of the population will rise, with a corresponding decrease in the heterozygosity. Eventually, the population will be entirely homozygous, with genotypic frequenciesp AA + q aa.

Table 8-12 Increase in Appearance of Recessive Traits Due to Inbreeding with q = 0.01

Inbreeding Coefficient (F)

Frequency of Appearance of aa genotype under

Inbreeding

Percentage Increase in Frequency of

Appearance over Hardy-Weinberg

0 (H-W) 0.0001 —.005 0.0001495 49.5%.01 0.000199 99%.02 0.000298 198%.05 0.000595 495%.1 0.00109 990%.4 0.00406 3960%1 0.01 9900%

116

This will be true for all loci.

Assortative MatingAssortative mating is when mates are chosen based on the similarity of their phenotypes. Under assortative mating, we still allow all individual in the population to mate, we just restrict their mates. As with inbreeding, allelic frequencies do not change with assortative mating (unless accompanied by selection), but there will be an increase in homozygosity. This increase will be for loci associate with the trait, and for those traits linked to them.

Assortative Mating in the Long Run

Assortative mating is similar to inbreeding in that homozygosity increases. The only difference is that inbreeding increases homozygosity for every locus, while assortative mating only increases homozygosity for those loci related to the mate similarity (or loci that are tightly linked to those loci).

8.4.5 Small Population SizeWhen the population size is small, the breeding pool is limited and the randomness of the gamete formation has several effects, including random drift and random loss and fixation of genes.Small population size leads to consanguinity, which causes inbreeding. The inbreeding effect is large if the population size is small. In addition, alleles are lost from the population due to random events, even if they are advantageous.

Effective Population Size Depending on the mating structure, the population may act as if it has a smaller number of individuals. Extreme examples are with species where the male has a harem. Many males do not mate, and half the genes in the next population come from a few individuals.

Founder PopulationA founder population is a small population colonizing a new area. This population shows several results of the manner in which it was selected.

Founder Populations Have Small Population Size

Because of the small size, it is likely to undergo dramatic genetic drift. This may help to further isolate the population, particularly if there are chromosomal mutations in the population.

117

Founder Populations Are Not Necessarily Representative of the Entire Population

The founder population is chosen from some population, but the allele frequencies may not match the allele frequencies in the original population. For example, in a human founder population, the founding parents may have had a higher frequency of red hair. This may not have been a selecting force in the founding population, another level of “randomness.”

Founder Populations May Undergo Strong Selection

Because the founding population is colonizing a new environment, it is unlikely the selective pressures are the same as in the parent population. Most likely, it is a harsher environment (which may be why it was not colonized earlier).All of these conditions lead to stronger, faster selection. Most founding populations eventually fail, but those that do survive are likely to be well on the way towards evolutionary differentiation.Small Population Size in the Long RunWith small population sizes, eventually all alleles but one will be lost in the population. However, which allele is fixed (the surviving one) is random.

8.5 Example of a Complex SystemAs an example of a more complex system, this section outlines research into the mechanisms of extinction. In this case, we consider mutation, selection, inbreeding, and a small population size.

Mutation and SelectionThere is a recurrent mutation to lethal alleles. Eventually, the loss of these alleles due to selection is equal to the gain due to mutation. They will reach an equilibrium. All populations have this, and we use the term genetic load to refer to the loss in population fitness due to lethal alleles.

Population Size and InbreedingA large population is able to deal with a large genetic load. With a small population, the inbreeding will increase the homozygosity. In this case, the homozygote mutant forms are lethal. This leads to a decrease in population size.

Bottleneck and ExtinctionConsider a population with a relatively large population. It will maintain a large genetic load. Assume that some natural catastrophe reduces the population size dramatically, as in Figure 8-2.

118

Table 8-13 Po

Situation ns

Hardy-Weinbe

Mutationro,

is ely

At this stage, the genetic load will have more of an effect. The increase in homozygosity will lead to more individuals with lethal genotypes, leading to an even smaller population. At the same time, the lethals will be purged. The question is whether the population becomes extinct before they are entirely purged. Figure 8-2 shows a population that has been through the bottleneck and has survived. At the end, they will have very little genetic load. They will also have very little genetic variability, which may lead to problems later.

8.6 Summary EquationsTable 8-13 gives the basic equations and some information about rate and longterm effects of each of the violations. The situation will be more complex if more than one violation is considered at the same time.

Figure 8-2 Population Size After Bottleneck

pulation Genetics Equations

Basic Equation Rate (Shortterm) Longterm Conditio

rg

Typically very slow

Equilibrium:

If one or other rate is zeeventually one allele

fixed, but rate is extremslow.

Bottleneck

Appearanceof Lethals

LethalsPurged

At RiskforExtinction

Time

Popu

latio

n Si

ze

p2AA 2pqAa q2aa+ +

1 μ–( )p ν 1 p–( )+p ν

μ ν+------- .=

119

Migration at

Selection

of

s . or t,

Nonrandom Mating

not

s 1.s 0.

l

cts rait

Small PopulatioSize

ne. to

Table 8-13 Po

Situation ns

Problems

8.1 Consider a population with the following genotypic frequencies:0.4225 DD + 0.4550 Dd + 0.1225 dda) Give the gene frequencies for alleles D and d.D________ d________b) We will take this population and apply a systematic system

of inbreeding until the inbreeding coefficient rises to 0.25. Give the genotypic frequencies when this experiment is completed.

DD ______ Dd ______ dd ______

Depends on m and difference between p and P

p = Pm only affects the rate

which this happens.

From Table Depends on WAa and Waa

Directional: Fixation most fit allele.

Balancing: Both alleleremain in population

Underdominance: Onethe other allele is losdepending on initial

conditions.

Inbreeding

(p2+Fpq) AA + 2(1-F)pq Aa + (q2+Fpq) aa

Depends on the inbreeding scheme

Allele frequency does change.

Homozygosity becomeHeterozygosity become

ALL ALLELES

Assortative MatingNone given

Depends on correlation between parents

Same as inbreeding.Inbreeding affects al

alleles.Assortative mating affeonly alleles related to t

& hitchhikers (linkedtraits).

n None given

Allele frequency change depends on effective

population size — larger effects for smaller

population size

Loss of all alleles but oIt is random, but related

the initial allele frequencies.

pulation Genetics Equations

Basic Equation Rate (Shortterm) Longterm Conditio

1 m–( )p mP+

120

ultsalleles the

c) Consider the original population in a). We will follow this population through a generation of natural selection. We know that the fitness of the dd individuals is only half of the DD individuals, and that the fitness of the heterozygote is three-fourths of the DD individuals. Give the gene frequencies after one generation.

D______ d______d) Consider the original population in a). Assume the mutation

rate is 1´10-5 and 3´10-7 for A to a and a to A, respectively. Give the gene frequencies after one generation.

D_______________ d_______________e) Consider the original population in a). Assume this is an

island population, migration from a large mainland population. The genotypic array on the mainland is 0.09DD+0.42Dd+0.49dd, and the migration rate is 1%. Give the gene frequencies after one generation.

D______ d______8.2 What is the long-term result of balancing selection in terms of

allele frequencies?8.3 We find increased frequencies of genetic diseases in isolated

populations (such as on Pitcairn Island). What is the cause of this increase?

8.4 What is assortative mating? What effect does it have on a population undergoing assortative mating?

8.5 List three effects of a founder population.8.6 Following are six of the violations of Hardy-Weinberg

populations. For each set of short term and long term results, indicate which mode of evolution best exemplifies them. If more than one seems to apply, choose the one that best fits the description.a) Assortative Matingb) Balanced Selectionc) Disruptive Selectiond) Inbreedinge) Migrationf) Mutationg) Small Population Size

Shortterm Results Longterm Res

____ Moderate change in allele frequencyLeads to both maintained in population

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ozy- loci

rsity

Resultsomozy-

ll lociations r

or evolu-

8.7 Consider a population with the following genotypic frequencies:0.64 BB + 0.32 Bb + 0.04 bba) Give the allele frequencies for alleles B and b.B____________b____________b) Assume that bb is fatal before reproduction, and Bb has a

fitness of W=0.5. Compute the allele frequency for B for the next generation. (Use at least 5 decimal places)

Freq(B) = __________c) Consider the original population given in part a). Ignore

selection, but assume there is migration into this population from a mainland with gametic array 0.3 B + 0.7 b with a rate of 7%. What is the allele frequency for our population after a single generation? (Use at least 5 decimal places)

Freq(B) = __________8.8 Following are six of the violations of Hardy-Weinberg

populations. For each set of short term and long term results, indicate which mode of evolution best exemplifies them. If more than one seems to apply, choose the one that best fits the description.a) Assortative Matingb) Directional Selectionc) Inbreedingd) Migratione) Mutationf) Small Population Size

____ No change in allele frequencies Increases homgosity at some

____ Very small change per generation Increases dive

Shortterm Results Longterm ____ No change in allele frequencies Increases h

gosity at a____ Can change allele frequencies, depending

on differences in populationsTwo popullook simila

____ Only very small changes Essential ftion

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a)b)c)d)e)

8.9 Consider a population with the following genotypic frequencies:0.29 BB + 0.50 Bb + 0.21 bba) Give the allele frequencies for alleles B and b.B____________b____________b) Assume that bb has a fitness of W=0.5, and Bb has a fitness

of W=0.8. Compute the allele frequency for B for the next generation. Use at least 5 decimal places.

Freq(B) = __________8.10 In the table are given the genotypic frequencies of several

populations. For each, determine whether they fit Hardy-Weinberg proportions. Do not do a formal Chi-Square test, look to see if it has exactly the right frequencies.

8.11 Following are some of the violations of Hardy-Weinberg populations we discussed in class. For each of the statements, choose the best answer. If more than one seems to apply, choose the one that best fits the description.a) Assortative Matingb) Inbreedingc) Migrationd) Mutatione) Selectionf) Small Population Size____It is anti-evolutionary.____Decreases heterozygosity for some traits.____The change in allele frequency is extremely slow.____An allele may increase in frequency in one generation,

but decrease in the next.8.12 Following are six of the violations of Hardy-Weinberg

populations. For each set of short term and long term results, indicate which mode of evolution best exemplifies them. If more than one seems to apply, choose the one that best fits the description.

AA Aa aaHardy-Weinberg

Proportions?0.49 0.42 0.09 Yes No0.36 0.39 0.25 Yes No

0.3025 0.4950 0.2025 Yes No0.4525 0.3555 0.1920 Yes No0.9025 0.0950 0.0025 Yes No

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Results heterozy-all loci.ly no allele All alleles e popula-

ly all but is lost from ation.

a) Assortative Matingb) Directional Selectionc) Inbreedingd) Migratione) Mutationf) Small Population Sizeg) Stabilizing Selection

Shortterm Results Longterm____ No change in allele frequencies Decrease

gosity at

____ Change in allele frequencies

Eventualchanges. stay in thtion.

____ Allele frequencies can sometimes increase and sometime decrease

Eventualone allelethe popul

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