📖 ContentsChapter 12
Genetics · Chapter 12

Population and Evolutionary Genetics

Evolution, at its most basic, is a change in allele frequencies in a population over time. Population genetics measures those frequencies, predicts what happens to them when nothing disturbs them, and then calculates how mutation, migration, chance and natural selection push them around. The same few equations let you estimate how many people carry a recessive disease allele, why small populations lose diversity, how a harmful allele can persist for generations, and how new species arise.

By the end of this chapter you can

  • Calculate genotype frequencies and allele frequencies from counts for autosomal, multiple allele and X-linked loci.
  • State the Hardy Weinberg principle and its assumptions, and use p2 + 2pq + q2 to predict genotype frequencies and carrier frequencies.
  • Test a population for Hardy Weinberg equilibrium with a chi square test, using the correct degrees of freedom.
  • Calculate the inbreeding coefficient (F) from a pedigree and predict its effect on homozygosity and recessive disease.
  • Distinguish positive and negative assortative mating from inbreeding.
  • Calculate the effects of mutation, migration and natural selection on allele frequency, including heterozygote advantage and mutation selection balance.
  • Explain genetic drift, effective population size, the founder effect and bottlenecks.
  • Describe the neutral theory, the molecular clock and the use of dN/dS ratios.
  • Classify reproductive isolating mechanisms and modes of speciation.

Section 1Allele and genotype frequencies

Populations and gene pools

Transmission genetics follows alleles through individual families. Population genetics steps back and asks about all the alleles in a group of interbreeding individuals. A population in this sense is a group of individuals of the same species that live in the same area and can mate with one another, sometimes called a Mendelian population. The complete set of alleles carried by all its members at a given time is its gene pool. For a diploid population of N individuals, the gene pool at an autosomal locus contains 2N alleles.

Two kinds of frequencies describe a gene pool, and the distinction is fundamental:

Genotype frequencies are what we observe when we classify individuals; allele frequencies are what is transmitted from one generation to the next, because individuals pass on alleles, not genotypes. Evolution, in the population genetic definition, is a change in allele frequencies across generations.

Calculating allele frequencies from counts

There are two equivalent ways to calculate allele frequencies at an autosomal locus with codominant alleles, where every genotype can be recognized.

The counting method. Each homozygote carries two copies of its allele and each heterozygote carries one copy of each. Count all copies and divide by the total number of alleles, 2N:

p = (2 × number of AA + number of Aa) / 2N

The genotype frequency method. Half of a heterozygote's alleles are A, so

p = f(AA) + ½ f(Aa), and q = f(aa) + ½ f(Aa).

Both methods give the same answer, and you can use whichever is more convenient. Both require that you can identify heterozygotes. When one allele is completely dominant, AA and Aa look alike, and you cannot count alleles directly; you need the Hardy Weinberg relationships of Section 2 to estimate them.

500 people (genotypes) 1,000 alleles (gene pool) MM: 140 (0.28)MN: 260 (0.52)NN: 100 (0.20) M alleles: 540p = 540/1,000 = 0.54 N alleles: 460q = 460/1,000 = 0.46 2 × 140 = 2802602602 × 100 = 200 Each homozygote gives two copies of its allele; each heterozygote gives one of each
Figure 12.1 From genotypes to alleles. Counting the alleles carried by 500 people typed for the MN blood group gives p (M) = 0.54 and q (N) = 0.46. Notice that p + q = 1 and that the genotype frequencies also sum to 1.
Study tip. Worked problem: In a sample of 500 people typed for the codominant MN blood group, 140 are MM, 260 are MN and 100 are NN. Find the genotype and allele frequencies. Step 1: Genotype frequencies: MM = 140/500 = 0.28; MN = 260/500 = 0.52; NN = 100/500 = 0.20. Check: 0.28 + 0.52 + 0.20 = 1.00. Step 2: Counting method: M alleles = 2(140) + 260 = 540; N alleles = 2(100) + 260 = 460; total = 2 × 500 = 1,000. p = 540/1,000 = 0.54 and q = 460/1,000 = 0.46. Step 3: Genotype frequency method as a check: p = 0.28 + ½(0.52) = 0.28 + 0.26 = 0.54. Answer: f(MM) = 0.28, f(MN) = 0.52, f(NN) = 0.20; p = 0.54, q = 0.46. Always confirm p + q = 1.

Multiple alleles

The same logic extends to loci with more than two alleles. With three alleles A1, A2 and A3 there are six genotypes: three homozygotes and three heterozygotes. The frequency of A1 is the frequency of A1A1 plus half the frequency of each heterozygote containing A1:

p1 = f(A1A1) + ½f(A1A2) + ½f(A1A3)

In general, a locus with k alleles has k(k + 1)/2 genotypes, of which k are homozygous and k(k − 1)/2 are heterozygous. A locus with 10 alleles therefore has 55 genotypes. This is why highly polymorphic loci, such as forensic short tandem repeats and HLA genes, distinguish individuals so well.

X-linked loci

For genes on the X chromosome, females (XX) carry two copies and males (XY) carry one. Male genotype frequencies equal allele frequencies directly, because each male has exactly one allele. To find the population allele frequency, count every X chromosome: two per female and one per male.

Study tip. Worked problem: A population sample contains 100 females (81 XAXA, 18 XAXa, 1 XaXa) and 100 males (90 XAY, 10 XaY). Find q, the frequency of Xa. Step 1: Total X chromosomes = 2(100) + 1(100) = 300. Step 2: Copies of Xa: from heterozygous females 18, from homozygous females 2 × 1 = 2, from males 10. Total = 18 + 2 + 10 = 30. Step 3: q = 30/300 = 0.10, so p = 0.90. Answer: q = 0.10. As a check, the male frequency of XaY (10/100 = 0.10) equals q, and female frequencies match 0.902 = 0.81, 2(0.90)(0.10) = 0.18 and 0.102 = 0.01, so this population is in equilibrium.

Measuring genetic variation

Allele frequencies are the raw data for describing how much variation a population carries. Population geneticists commonly use three summary measures:

MeasureDefinitionExample of use
Proportion of polymorphic lociFraction of loci where the most common allele has frequency below 0.95 (or 0.99)Early protein electrophoresis surveys showed about a third of loci are polymorphic in many species
Observed heterozygosity (Ho)Proportion of individuals heterozygous at a locus, averaged across lociComparing diversity in endangered and healthy populations
Expected heterozygosity (He)1 − Σpi2, the heterozygosity predicted under random matingDetecting inbreeding when Ho < He
Allelic richnessNumber of distinct alleles at a locus, corrected for sample sizeDetecting bottlenecks, which remove rare alleles first
Nucleotide diversity (π)Average proportion of nucleotide sites that differ between two randomly chosen sequencesAbout 0.001 in humans: two copies of a genome differ at roughly 1 site per 1,000

Note one important point: allele frequencies alone do not determine genotype frequencies. Two populations can both have p = 0.5, one with 50 percent AA and 50 percent aa (no heterozygotes), and another with 25 percent AA, 50 percent Aa and 25 percent aa. Something must connect allele frequencies to genotype frequencies, and under random mating that connection is the Hardy Weinberg principle.

Key point

A gene pool is described by genotype frequencies (proportions of individuals) and allele frequencies (proportions of gene copies). For codominant autosomal alleles, p = f(AA) + ½f(Aa); for X-linked loci, count two X chromosomes per female and one per male. Multiple alleles produce k(k + 1)/2 genotypes, and heterozygosity summarizes a population's variation.

Section 2Hardy Weinberg equilibrium

The principle

The mathematician G. H. Hardy and the physician Wilhelm Weinberg independently showed that dominant alleles do not automatically spread. Their result, the Hardy Weinberg principle, states that in a large, randomly mating population with no mutation, migration or selection:

  1. Allele frequencies do not change from generation to generation.
  2. After one generation of random mating, genotype frequencies are given by p2 (AA), 2pq (Aa) and q2 (aa), and they stay at those values.

A population with these genotype frequencies is in Hardy Weinberg equilibrium. Dominance has nothing to do with it: a dominant allele at a frequency of 0.01 stays at 0.01. Dominance affects which phenotype a genotype shows, not how often alleles are transmitted.

Why it works

Random mating of individuals is equivalent to random union of gametes drawn from the gene pool. A fraction p of all gametes carries A and a fraction q carries a. The probability that an egg and a sperm both carry A is p × p = p2; both a, q2; and one of each, pq + qp = 2pq (an A egg with an a sperm or an a egg with an A sperm). These terms are the expansion of (p + q)2 = p2 + 2pq + q2 = 1.

Now check that allele frequencies are unchanged. In the offspring, the frequency of A is f(AA) + ½f(Aa) = p2 + ½(2pq) = p2 + pq = p(p + q) = p. Allele frequencies are preserved, so the same genotype frequencies are produced in every later generation. Meiosis and fertilization, by themselves, do not change the gene pool.

AAp² = 0.49 Aapq = 0.21 Aapq = 0.21 aaq² = 0.09 A sperm, p = 0.7a, q = 0.3 A eggsp = 0.7a eggsq = 0.3 Random union of gametes AA = p² = 0.49Aa = 2pq = 0.42aa = q² = 0.09Sum = 1.00
Figure 12.2 The Hardy Weinberg square. Each side is divided by the gamete frequencies p and q, so the area of each cell is the probability of that zygote. With p = 0.7 and q = 0.3, genotype frequencies are 0.49, 0.42 and 0.09.

The assumptions

The principle holds exactly only under idealized conditions. Each assumption corresponds to a force that can change the population, and much of the rest of this chapter is about those forces.

AssumptionIf violatedMain effect
Random matingInbreeding or assortative matingChanges genotype frequencies (often fewer heterozygotes) without changing allele frequencies
Very large populationGenetic driftRandom change in allele frequencies; loss of variation
No mutationNew alleles ariseVery slow change in allele frequencies; source of all new variation
No migrationGene flowAllele frequencies move toward those of the source population
No selectionGenotypes differ in survival or reproductionDirectional or balancing change in allele frequencies

The surprising lesson is how robust the principle is in practice. Many loci in large human populations fit Hardy Weinberg proportions closely, because mutation and migration are weak over a single generation, and selection on most variants is weak. Even when a population departs from equilibrium, one generation of random mating restores Hardy Weinberg proportions for an autosomal locus, provided allele frequencies are the same in both sexes. For X-linked loci the approach is gradual: if male and female allele frequencies differ, the difference is halved and changes sign each generation, so the population oscillates toward equilibrium.

Uses of the Hardy Weinberg equation

The most common practical use is estimating allele and carrier frequencies for a recessive trait, where heterozygotes cannot be recognized. If the population is assumed to be in equilibrium, the frequency of affected individuals equals q2, so q = √(frequency of affected). The carrier frequency is then 2pq. When q is small, p is close to 1 and 2pq ≈ 2q.

Study tip. Worked problem: Cystic fibrosis affects about 1 in 2,500 newborns of northern European ancestry. Estimate q, the carrier frequency, and the chance that two unrelated, unaffected people from this population have an affected child. Step 1: q2 = 1/2,500 = 0.0004, so q = √0.0004 = 0.02. Then p = 0.98. Step 2: Carrier frequency = 2pq = 2 × 0.98 × 0.02 = 0.0392, about 1 in 25. Step 3: Because each person is unaffected, use the conditional probability of being a carrier: 2pq/(p2 + 2pq) = 0.0392/0.9996 = 0.0392 (almost unchanged because q2 is tiny). Step 4: Probability both are carriers and the child is aa = 0.0392 × 0.0392 × ¼ = 0.00154 × 0.25 = 0.000384. Answer: q = 0.02, carriers about 1 in 25, and the couple's risk is about 1 in 2,600. Notice that carriers (1 in 25) outnumber affected people (1 in 2,500) about 100 to 1: rare recessive alleles hide mostly in heterozygotes.

That last observation follows from the ratio of heterozygotes to recessive homozygotes, 2pq/q2 = 2p/q. When q = 0.02, the ratio is 2(0.98)/0.02 = 98. This has an important consequence for selection: eliminating affected individuals barely changes the frequency of a rare recessive allele, because almost all copies are in carriers who are not selected against.

For X-linked recessive traits, the frequency of affected males equals q, and the frequency of affected females is q2. About 8 percent of males of northern European ancestry have red green color vision deficiency, so q ≈ 0.08, the expected frequency of affected females is 0.082 = 0.0064 (about 0.6 percent), and carrier females are 2(0.92)(0.08) = 0.147, about 15 percent. This is why X-linked recessive conditions are much more common in males: a male needs one copy, a female two.

00.250.500.751.00 00.250.500.751.0 Frequency of allele a (q); p = 1 − q Genotype frequency AA = p²Aa = 2pq (max 0.5)aa = q² When q is small,carriers (2pq)far outnumberaffected (q²)
Figure 12.3 Hardy Weinberg genotype frequencies as a function of allele frequency. Heterozygotes are most common at intermediate frequencies, never exceed 50 percent, and vastly outnumber recessive homozygotes when an allele is rare.

Multiple alleles: the ABO system

With three alleles at frequencies p, q and r (p + q + r = 1), Hardy Weinberg genotype frequencies come from (p + q + r)2: p2, q2, r2, 2pq, 2pr and 2qr. For ABO blood groups, with IA (p), IB (q) and i (r), the phenotypes are type A = p2 + 2pr, type B = q2 + 2qr, type AB = 2pq and type O = r2. Because type O is homozygous recessive, r = √f(O) is the natural starting point. For example, if 49 percent of a population is type O and 32 percent type A, then r = 0.7; solving p2 + 2(0.7)p = 0.32 gives p = 0.2; and q = 1 − 0.7 − 0.2 = 0.1. The model then predicts type B = 0.01 + 0.14 = 0.15 and type AB = 0.04, which sum with the others to 1.00.

Testing for equilibrium with chi square

To ask whether a population fits Hardy Weinberg proportions, calculate allele frequencies from the observed data, use them to calculate expected genotype numbers, and compare observed and expected with a chi square test: χ2 = Σ(O − E)2/E. The degrees of freedom are the number of genotype classes minus 1, minus the number of independent allele frequencies estimated from the data. For two alleles and three genotypes, df = 3 − 1 − 1 = 1, not 2. This is one of the most common errors on population genetics problems. With df = 1, the critical value at P = 0.05 is 3.84.

Study tip. Worked problem: Test the MN data from Section 1 (140 MM, 260 MN, 100 NN; p = 0.54, q = 0.46; N = 500) for Hardy Weinberg equilibrium. Step 1: Expected MM = p2N = 0.2916 × 500 = 145.8. Expected MN = 2pqN = 0.4968 × 500 = 248.4. Expected NN = q2N = 0.2116 × 500 = 105.8. Check: 145.8 + 248.4 + 105.8 = 500. Step 2: MM: (140 − 145.8)2/145.8 = 33.64/145.8 = 0.231. MN: (260 − 248.4)2/248.4 = 134.56/248.4 = 0.542. NN: (100 − 105.8)2/105.8 = 33.64/105.8 = 0.318. Step 3: χ2 = 0.231 + 0.542 + 0.318 = 1.09. Step 4: df = 3 − 1 − 1 = 1. Critical value = 3.84. Since 1.09 < 3.84 (0.2 < P < 0.3), we fail to reject the hypothesis. Answer: the population is consistent with Hardy Weinberg equilibrium. A slight excess of heterozygotes is within the range expected from sampling error.
Connection. Genetic counselors use Hardy Weinberg calculations every day. When a person with a family history of a recessive condition plans a pregnancy with an unrelated partner, the partner's risk of being a carrier is estimated from population carrier frequencies (about 1 in 25 for cystic fibrosis in people of northern European ancestry, lower in many other groups). The same equations explain why expanded carrier screening panels test both partners, and why population specific allele frequencies matter.
Key point

In a large, randomly mating population without mutation, migration or selection, allele frequencies stay constant and genotype frequencies settle at p2, 2pq and q2 after one generation. The equation estimates carrier frequencies (2pq ≈ 2q for rare alleles), shows that rare recessive alleles hide mostly in heterozygotes, extends to multiple and X-linked alleles, and is tested with chi square using 1 degree of freedom for two alleles.

Section 3Inbreeding and nonrandom mating

Kinds of nonrandom mating

Random mating means that the chance two individuals mate does not depend on their genotypes or their relationship. Real populations often break this rule in one of three ways:

The central point about all three is that nonrandom mating changes genotype frequencies but not allele frequencies. Inbreeding and positive assortative mating increase homozygosity and decrease heterozygosity; negative assortative mating does the reverse. On its own, nonrandom mating is therefore not an evolutionary force in the strict sense, but by exposing recessive alleles in homozygotes it changes how selection acts.

The inbreeding coefficient

Two alleles at a locus can be identical in two different senses. They can be identical by state, meaning they have the same DNA sequence, or identical by descent (IBD), meaning they are both copies of a single allele in a common ancestor. The inbreeding coefficient (F) is the probability that the two alleles at a locus in an individual are identical by descent. An individual whose two alleles are identical by descent is called autozygous. F ranges from 0 (no inbreeding beyond the reference population) to 1 (complete inbreeding, as in a fully inbred line).

With inbreeding coefficient F, genotype frequencies in a population become:

Heterozygotes are reduced by the fraction F, and the missing heterozygotes are added equally to the two homozygote classes. Allele frequencies stay the same: f(AA) + ½f(Aa) = p2 + Fpq + pq − Fpq = p. Rearranging the heterozygote equation gives a way to estimate F from data: F = 1 − Ho/He, where Ho is the observed heterozygosity and He = 2pq is the heterozygosity expected under random mating.

Study tip. Worked problem: A recessive disorder has q = 0.01. Compare the frequency of affected children from random mating with the frequency among children of first cousins (F = 1/16). Step 1: Random mating: q2 = 0.012 = 0.0001, or 1 in 10,000. Step 2: First cousin offspring: q2 + Fpq = 0.0001 + (1/16)(0.99)(0.01) = 0.0001 + 0.0625 × 0.0099 = 0.0001 + 0.000619 = 0.000719. Step 3: Ratio = 0.000719/0.0001 = 7.2. Answer: about 7 in 10,000 (1 in 1,390) versus 1 in 10,000, a roughly sevenfold increase. The rarer the allele, the larger the relative increase, because the Fpq term shrinks only in proportion to q while q2 shrinks in proportion to q squared.

Calculating F from a pedigree

F for an individual can be calculated by path counting. Find every common ancestor of the individual's two parents. For each common ancestor, trace the path that runs from one parent up to the common ancestor and back down to the other parent, and count n, the number of individuals in that path including both parents and the common ancestor (but not the inbred individual). Each step in a path is a 1/2 chance of passing on a particular allele, and the common ancestor's two alleles are themselves identical by descent with probability FA. Then

F = Σ (1/2)n (1 + FA)

summed over all paths. If the common ancestor is not itself inbred, FA = 0 and each path contributes (1/2)n.

ABDECFI common ancestors first cousins C and F(double line: related mates) inbred child I Paths: C D A E F and C D B E F, each n = 5 individuals F(I) = (1/2)⁵ + (1/2)⁵ = 1/32 + 1/32 = 1/16
Figure 12.4 Path counting for a first cousin mating. The parents of I (C and F) share two common ancestors, A and B. Each path through a common ancestor contains five individuals, so each contributes (1/2)5 = 1/32, and F = 1/16.
Study tip. Worked problem: Calculate F for the child of a half sibling mating, in which the parents share only their mother, who is herself the product of a first cousin mating (FA = 1/16). Step 1: There is one common ancestor (the shared mother). The path runs from one parent, to the mother, to the other parent: n = 3 individuals. Step 2: If the mother were not inbred, F = (1/2)3 = 1/8 = 0.125. Step 3: With FA = 1/16: F = (1/2)3(1 + 1/16) = (1/8)(17/16) = 17/128 = 0.133. Answer: F = 17/128, about 0.133. The ancestor's own inbreeding adds a small extra chance that the two alleles are identical by descent.
MatingCommon ancestors and path lengthF of offspring
Self fertilizationParent is the only ancestor (n = 1)1/2
Parent and offspringOne path, n = 21/4
Full siblingsTwo common ancestors, n = 3 each1/4
Half siblingsOne common ancestor, n = 31/8
Uncle and niece (or aunt and nephew)Two common ancestors, n = 4 each1/8
First cousinsTwo common ancestors, n = 5 each1/16
First cousins once removedTwo common ancestors, n = 6 each1/32
Second cousinsTwo common ancestors, n = 7 each1/64

Repeated inbreeding and inbred lines

Continued inbreeding pushes F toward 1. Self fertilization is the extreme case: each generation of selfing halves the remaining heterozygosity, so after t generations Ht = H0(1/2)t and F = 1 − (1/2)t. Starting from a fully heterozygous plant, five generations of selfing leave only (1/2)5 = 1/32, about 3 percent, of the original heterozygosity (F ≈ 0.97). Repeated full sibling mating approaches homozygosity more slowly, reaching F ≈ 0.99 after about 20 generations. This is how inbred strains of mice, rats and many crop lines are created, and it is why every member of such a strain is essentially genetically identical, a property exploited in Section 3 of Chapter 11 to estimate environmental variance.

Inbreeding depression and hybrid vigor

Inbreeding depression is the decline in fitness and in trait values such as growth, fertility and survival that follows inbreeding. It has two main genetic explanations. The more important is the dominance hypothesis: populations carry many rare, partially recessive deleterious alleles, mostly hidden in heterozygotes; inbreeding makes them homozygous and exposes them. The overdominance hypothesis proposes that heterozygotes at some loci are intrinsically superior, so losing heterozygosity lowers fitness. The reverse effect, heterosis or hybrid vigor, appears when two different inbred lines are crossed: the F1 is heterozygous at many loci, masking each line's recessive defects, and is often more vigorous than either parent. Hybrid corn, which dominates modern corn production, exploits heterosis, and farmers buy new hybrid seed every year because the F2 loses much of the advantage as segregation restores homozygosity.

Connection. In human populations with a tradition of consanguineous marriage, recessive disorders are noticeably more frequent, and many rare recessive conditions were first described in such communities or in isolated groups. In conservation biology, small zoo and wild populations suffer inbreeding depression; the Florida panther, reduced to a few dozen animals with heart defects, kinked tails and poor sperm quality, recovered after eight females from a related Texas population were introduced to restore heterozygosity, a strategy called genetic rescue.
Key point

Nonrandom mating changes genotype but not allele frequencies. The inbreeding coefficient F, the probability that two alleles are identical by descent, reduces heterozygotes to 2pq(1 − F) and adds Fpq to each homozygote, which greatly increases the frequency of rare recessive disorders. F is calculated by path counting, F = Σ(1/2)n(1 + FA); first cousins give 1/16. Inbreeding depression comes mainly from exposed recessive alleles, and crossing inbred lines produces heterosis.

Section 4Mutation, migration, drift and selection

Four processes change allele frequencies: mutation, migration, genetic drift and natural selection. They differ in speed, in direction and in whether they increase or decrease variation. The table below gives the overview; the rest of the section works through each force.

ForceChanges allele frequency?Variation within populationsVariation between populationsDirection
MutationYes, very slowlyIncreasesIncreasesToward mutation equilibrium
Migration (gene flow)YesUsually increasesDecreasesToward the source population
Genetic driftYes, randomDecreasesIncreasesRandom; fixation or loss
Natural selectionYesDirectional: decreases; balancing: maintainsIncreases or decreasesToward higher fitness
Inbreeding (for comparison)NoDecreases heterozygosityIncreases if subpopulations inbreedToward homozygosity

Mutation

Mutation is the ultimate source of all new alleles, but as a force changing allele frequencies it is extremely weak. Let A mutate to a at rate u per generation (forward mutation) and a mutate back to A at rate v. The change in the frequency of A in one generation is

Δp = vq − up

At equilibrium Δp = 0, so up = vq, which gives q̂ = u/(u + v) and p̂ = v/(u + v). Typical mutation rates are around 10−5 to 10−6 per gene per generation, so the approach is glacial. With forward mutation alone at u = 10−5, it takes about ln 2/u ≈ 0.693/10−5 ≈ 69,000 generations to halve p. Mutation supplies variation; other forces determine what happens to it.

Migration

Migration, or gene flow, is the movement of alleles between populations when individuals or gametes (such as pollen) move and reproduce. In the simplest model, a fraction m of a recipient population each generation consists of migrants from a source population with allele frequency qm. The new frequency is a weighted average:

q' = (1 − m)q + m qm, so Δq = m(qm − q)

Migration changes allele frequencies fastest when the populations differ most, and it makes populations more similar. Over t generations the difference shrinks geometrically: (qt − qm) = (q0 − qm)(1 − m)t. Even a modest amount of migration, on the order of one migrant per generation, is enough to prevent drift from making populations diverge completely at neutral loci.

Study tip. Worked problem: An island population has q = 0.20. Each generation, 10 percent of breeders are migrants from the mainland, where qm = 0.60. Find q after one generation, and after 5 generations. Step 1: q' = (1 − 0.1)(0.20) + 0.1(0.60) = 0.18 + 0.06 = 0.24. Equivalently Δq = 0.1(0.60 − 0.20) = 0.04. Step 2: After t generations, qt = qm + (q0 − qm)(1 − m)t = 0.60 + (−0.40)(0.9)5. Step 3: 0.95 = 0.59049, so q5 = 0.60 − 0.40 × 0.59049 = 0.60 − 0.236 = 0.364. Answer: q = 0.24 after one generation and about 0.36 after five. Check: the island frequency moves toward, but never overshoots, the mainland value.

Genetic drift

Genetic drift is random change in allele frequencies caused by sampling error. Each generation, the gametes that form the next generation are a finite sample of the parents' gene pool, and like any sample they rarely match it exactly. Flip a coin 10 times and you often get 7 heads; flip it 10,000 times and you get very close to half. In the same way, drift is strong in small populations and weak in large ones. The variance of the change in allele frequency in one generation is pq/2N.

Drift has four key consequences:

00.250.500.751.0 010203040 Generation Frequency of allele A fixedlost Eight simulated populations, N = 10 diploid individuals (2N = 20 alleles), starting at p = 0.5
Figure 12.5 Genetic drift in small populations. Each line is one population. Allele frequency wanders at random; in this run four populations fixed A, three lost it, and one was still polymorphic after 40 generations. In expectation, the probability of fixation equals the starting frequency, here 0.5.

Effective population size

The N in drift equations is not simply a census count. It is the effective population size (Ne), the size of an idealized population that would drift at the same rate as the real one. Ne is usually much smaller than the census size for three reasons. Unequal numbers of breeding males and females: Ne = 4NmNf/(Nm + Nf). Fluctuating population size: Ne over several generations is the harmonic mean of the sizes, which is dominated by the smallest generation. And variation in family size, since a few parents may leave most of the offspring.

Study tip. Worked problem: (a) A herd has 10 breeding males and 90 breeding females. What is Ne? (b) A population is 1,000, then 10, then 1,000 over three generations. What is Ne? (c) With Ne = 50, what fraction of heterozygosity remains after 10 generations? Step 1: (a) Ne = 4(10)(90)/(10 + 90) = 3,600/100 = 36, far below the census of 100. Step 2: (b) 1/Ne = (1/3)(1/1,000 + 1/10 + 1/1,000) = (1/3)(0.102) = 0.034, so Ne = 29.4. One bad generation drags Ne to about 29, although the arithmetic mean is 670. Step 3: (c) H10/H0 = (1 − 1/100)10 = 0.9910 = 0.904. Answer: Ne = 36; Ne ≈ 29; about 90 percent of heterozygosity remains (roughly 10 percent lost in 10 generations).

Founder effects and bottlenecks

Two special cases of drift have great practical importance. The founder effect occurs when a new population is started by a small number of individuals, whose alleles are a chance sample of the source population. A bottleneck occurs when an existing population is drastically reduced in size for one or more generations. In both cases, rare alleles are often lost, others rise by chance to high frequency, and the resulting population can differ markedly from its ancestors. Many human populations show founder effects: Ellis van Creveld syndrome (a form of dwarfism with extra fingers) is unusually common among the Old Order Amish of Pennsylvania, traced to a single founding couple, and several recessive disorders, such as Tay Sachs disease, are more frequent among Ashkenazi Jewish populations, partly because of historic bottlenecks. Northern elephant seals, hunted to perhaps 20 to 30 animals, have recovered to large numbers but retain extremely low genetic diversity.

Natural selection

Natural selection occurs when genotypes differ in their contribution to the next generation. That contribution, combining survival and reproduction, is fitness. In population genetics we use relative fitness (w), with the best genotype usually set to 1. The selection coefficient s = 1 − w measures how strongly a genotype is selected against. A lethal genotype that never reproduces has w = 0 and s = 1.

The general method for any selection problem is always the same, and it is worth memorizing as a procedure:

  1. Write the Hardy Weinberg genotype frequencies before selection: p2, 2pq, q2.
  2. Multiply each by its relative fitness.
  3. Add the products to get the mean fitness, w̄.
  4. Divide each product by w̄ to get genotype frequencies after selection.
  5. Calculate the new allele frequency: q' = f(aa) + ½f(Aa) after selection.
Study tip. Worked problem: Fitnesses are AA = 1, Aa = 1 and aa = 0.8 (s = 0.2 against a recessive phenotype). If q = 0.4, what is q in the next generation? Step 1: Before selection: AA = 0.62 = 0.36; Aa = 2(0.6)(0.4) = 0.48; aa = 0.42 = 0.16. Step 2: Multiply by fitness: 0.36 × 1 = 0.36; 0.48 × 1 = 0.48; 0.16 × 0.8 = 0.128. Step 3: w̄ = 0.36 + 0.48 + 0.128 = 0.968. Step 4: q' = (0.128 + ½ × 0.48)/0.968 = (0.128 + 0.24)/0.968 = 0.368/0.968 = 0.380. Answer: q falls from 0.400 to 0.380, a change of −0.020. Check with the formula for selection against a recessive, Δq = −spq2/(1 − sq2) = −(0.2)(0.6)(0.16)/(1 − 0.032) = −0.0192/0.968 = −0.0198.

The speed of selection depends on dominance and on allele frequency. Selection against a dominant allele is fast, because every copy is exposed. Selection against a recessive allele slows dramatically as the allele becomes rare, because nearly all copies hide in heterozygotes. For a recessive lethal (s = 1), qt = q0/(1 + tq0), so the number of generations to go from q0 to qt is t = 1/qt − 1/q0. Reducing q from 0.02 to 0.01 takes 100 − 50 = 50 generations, and from 0.01 to 0.005 another 100. This is why eugenic proposals to eliminate recessive disorders by preventing affected people from reproducing were not only unethical but biologically futile.

Modes of selection

Heterozygote advantage and the sickle cell allele

Let the fitnesses be AA = 1 − s, Aa = 1 and aa = 1 − t. Selection then removes both alleles from homozygotes and keeps them in heterozygotes, and the population reaches a stable equilibrium at

q̂ = s/(s + t), where q is the frequency of a.

The classic example is the sickle cell allele, HbS, in regions with endemic falciparum malaria. HbAHbA individuals are more susceptible to severe malaria, HbSHbS individuals have sickle cell disease, and HbAHbS carriers are partly protected from malaria while rarely having symptoms. With illustrative values s = 0.12 (malaria cost to HbAHbA) and t = 0.80 (cost of sickle cell disease), q̂ = 0.12/(0.12 + 0.80) = 0.12/0.92 = 0.13, similar to the HbS frequencies of 0.1 or more observed in parts of sub-Saharan Africa. Where malaria is absent, the advantage disappears and the allele slowly declines.

Mutation selection balance

Deleterious alleles persist because mutation keeps recreating them while selection removes them. At equilibrium the two rates match. For a recessive deleterious allele, q̂ ≈ √(u/s); for a dominant one, q̂ ≈ u/s. With u = 10−6 and s = 1, a recessive lethal settles at q ≈ √10−6 = 0.001, but a dominant lethal at only 10−6, essentially just the new mutations each generation. Recessive deleterious alleles can therefore be a thousand times more common than dominant ones with the same mutation rate.

Connection. Antibiotic resistance is natural selection in real time. In a patient taking an antibiotic, susceptible bacteria die and rare resistant mutants (s strongly favoring them) take over within days. Bacteria with large populations and short generation times, plus horizontal gene transfer that acts like migration of resistance genes between species, evolve resistance quickly. Completing prescribed courses, avoiding antibiotics for viral infections and limiting agricultural use all aim to reduce selection for resistance.
Key point

Mutation creates variation but changes frequencies very slowly (q̂ = u/[u + v]). Migration moves frequencies toward the source, Δq = m(qm − q). Drift is random, strongest in small populations, fixes a neutral allele with probability equal to its frequency, erodes heterozygosity by 1/2Ne per generation, and drives founder effects and bottlenecks. Selection is computed by weighting genotype frequencies by fitness and dividing by w̄; heterozygote advantage gives q̂ = s/(s + t), and mutation selection balance gives q̂ ≈ √(u/s) for recessives.

Section 5Molecular evolution and speciation

The neutral theory

Sequencing revealed far more molecular variation than selection alone seemed able to explain. Motoo Kimura's neutral theory of molecular evolution proposed that most of the differences we see at the DNA level are selectively neutral, or nearly so, and were fixed by genetic drift rather than by positive selection. The theory does not deny that selection shapes adaptations. It says that most mutations are either harmful, and quickly removed by purifying selection, or neutral, and left to drift; only a small fraction are advantageous.

The theory makes a striking prediction. In a diploid population of size N, about 2Nu new neutral mutations arise at a locus each generation, and each has a probability of 1/(2N) of eventually being fixed by drift. The rate of substitution is the product:

k = 2Nu × 1/(2N) = u

The rate at which neutral substitutions accumulate equals the neutral mutation rate, regardless of population size. Large populations produce more mutations, but each has a smaller chance of fixation, and the two effects cancel exactly.

The molecular clock

If neutral substitutions accumulate at a roughly constant rate, then the number of sequence differences between two species should be roughly proportional to the time since they shared a common ancestor. This is the molecular clock. Because both lineages accumulate changes independently after they split, the divergence between them is

K = 2μT

where K is the number of substitutions per site, μ the substitution rate per site per year, and T the time since divergence. A clock must be calibrated with fossils or dated geological events, and it ticks at different rates in different genes and lineages (for example, faster in rodents, with short generation times, than in primates), so modern methods use "relaxed" clocks that allow rates to vary.

Study tip. Worked problem: Two species differ at 2 percent of aligned neutral sites (K = 0.02 substitutions per site). The neutral substitution rate is 1 × 10−9 per site per year. When did they diverge? Then apply the Jukes Cantor correction to a more divergent pair with an observed difference of p = 0.20. Step 1: T = K/(2μ) = 0.02/(2 × 10−9) = 1 × 107 years. Step 2: For divergent sequences, some sites have changed more than once, so observed differences underestimate true substitutions. The Jukes Cantor correction is d = −(3/4) ln(1 − (4/3)p). Step 3: (4/3)(0.20) = 0.2667; 1 − 0.2667 = 0.7333; ln(0.7333) = −0.3102; d = −0.75 × −0.3102 = 0.233. Answer: about 10 million years for the first pair. For the second, the corrected distance is 0.233 substitutions per site, about 16 percent higher than the observed 0.20. The factor of 2 in K = 2μT is the most commonly forgotten step: both lineages have been changing.

Rates differ with functional constraint

The neutral theory explains why different parts of the genome evolve at very different rates. The more a change is likely to harm function, the more often it is removed by purifying selection, and the slower the observed rate. Changes in sequences with no function accumulate at the full mutation rate.

Sequence typeRelative rate of substitutionReason
PseudogenesFastestNo function, so no constraint; rate approaches mutation rate
Synonymous (silent) sites in codonsFastDo not change the amino acid
Introns and most intergenic DNAFastMostly weakly constrained, except regulatory elements
Nonsynonymous (replacement) sitesSlowChange the amino acid; often harmful
Highly constrained proteins (histones, ubiquitin)SlowestAlmost every residue is essential for function
FibrinopeptidesFast for a proteinCleaved off and discarded during clotting, little constraint

Comparing the two kinds of coding sites provides one of the most useful tests of selection. Let dN be the number of nonsynonymous substitutions per nonsynonymous site and dS the number of synonymous substitutions per synonymous site. Their ratio, ω = dN/dS, reveals the kind of selection:

Gene duplication and gene families

New genes most often arise by gene duplication. After a duplication, one copy can keep the original function while the other is free to accumulate changes. Most extra copies decay into pseudogenes, but some acquire new functions (neofunctionalization) or split the original functions between them (subfunctionalization). The globin family is the textbook case: duplications produced α and β globin clusters, myoglobin, and fetal and embryonic globins with different oxygen affinities. Genes related by duplication within a genome are paralogs; genes in different species descended from one gene in their common ancestor are orthologs.

What is a species?

Speciation connects population genetics to the diversity of life. Several species concepts are used. The biological species concept defines a species as a group of actually or potentially interbreeding natural populations reproductively isolated from other such groups. It captures why species stay distinct, but it cannot be applied to asexual organisms or fossils. The morphological species concept uses physical traits and is practical for fossils. The phylogenetic species concept defines a species as the smallest group of populations sharing a common ancestor, diagnosable by shared derived traits, and is widely used with DNA data.

Reproductive isolating mechanisms

Under the biological species concept, the key to speciation is the evolution of reproductive isolation. Barriers are classified by whether they act before or after the formation of a zygote.

Prezygotic barriers (before fertilization) Postzygotic barriers Habitatdifferent places Temporaldifferent breeding times Behavioralsongs, displays, scents Mechanicalparts do not fit Gameticsperm cannot fertilize egg Hybrid inviabilityhybrid dies in development Hybrid sterilityviable but sterile (mule) Hybrid breakdownF2 or backcross unfit fertilization (zygote forms) Barriers act in sequence; the earliest that works prevents wasted reproductive effort
Figure 12.6 Reproductive isolating mechanisms. Prezygotic barriers stop mating or fertilization; postzygotic barriers reduce the viability or fertility of hybrids. Selection against unfit hybrids can strengthen prezygotic barriers, a process called reinforcement.
BarrierTypeExample
Habitat isolationPrezygoticTwo garter snake species, one aquatic and one terrestrial, in the same area
Temporal isolationPrezygoticRelated frogs or plants that breed or flower in different seasons
Behavioral isolationPrezygoticSpecies specific bird songs and firefly flash patterns
Mechanical isolationPrezygoticIncompatible genitalia in insects; flower shapes that fit only certain pollinators
Gametic isolationPrezygoticSea urchin sperm bindin proteins recognize only their own species' eggs
Hybrid inviabilityPostzygoticMany frog hybrids fail during development
Hybrid sterilityPostzygoticMule (horse × donkey): 63 chromosomes that cannot pair properly in meiosis
Hybrid breakdownPostzygoticF1 rice hybrids are fertile but many F2 plants are weak or sterile

Modes of speciation

Speciation is classified by geography. In allopatric speciation, the most common mode, a population is divided by a geographic barrier, such as a rising mountain range, a new river course or the formation of an isthmus, and the separated populations diverge through mutation, drift and selection until they can no longer interbreed. Peripatric speciation is a special case in which a small peripheral population, such as a few colonists on an island, diverges rapidly under strong drift and new selective pressures; the founder effect plays a large role. In parapatric speciation, populations along a continuous range diverge across an environmental gradient despite some gene flow at the boundary. In sympatric speciation, a new species arises within the range of its parent species with no geographic separation. It is rare in animals, though host shifts in insects such as the apple maggot fly and diversification of cichlid fishes in African lakes are candidate examples. In plants, sympatric speciation by polyploidy is common: a tetraploid formed from a diploid ancestor produces sterile triploid offspring when it crosses back, so it is reproductively isolated in a single generation. Bread wheat is an allopolyploid (hexaploid) combining three ancestral genomes.

The genetics of speciation

How can two populations evolve hybrid incompatibility if every intermediate step must remain fit? The Dobzhansky Muller model answers this. Suppose the ancestral genotype is aabb. In one isolated population, allele A arises and becomes fixed, giving AAbb, which is perfectly fit. In the other, allele B arises and becomes fixed, giving aaBB, also fit. A and B have never been tested together. When the populations meet and hybridize, the hybrid carries A and B, and if those alleles interact badly, the hybrid is inviable or sterile. No population ever passed through an unfit stage; the incompatibility appears only in the hybrid. Several such interacting gene pairs have been identified in fruit flies, mice and plants.

A second regularity, Haldane's rule, states that when only one sex of a hybrid is inviable or sterile, it is usually the heterogametic sex: males in mammals and fruit flies (XY), females in birds and butterflies (ZW). A leading explanation is that recessive incompatibility alleles on the X (or Z) chromosome are exposed in the hemizygous sex, much as X-linked recessive diseases appear in males. When hybrids have reduced fitness and the incipient species come back into contact, selection can favor individuals that avoid mating with the other population, strengthening prezygotic isolation, a process called reinforcement.

Key point

The neutral theory holds that most molecular differences are neutral and fixed by drift, so neutral substitutions accumulate at the mutation rate and give a molecular clock (K = 2μT). Functionally constrained sites evolve slowly, and dN/dS distinguishes purifying (ω < 1), neutral (ω ≈ 1) and positive (ω > 1) selection. Species are maintained by prezygotic and postzygotic barriers; they usually arise allopatrically, sometimes sympatrically by polyploidy, and Dobzhansky Muller incompatibilities explain how hybrid problems evolve without unfit intermediates.

The chapter in one breath

  • Allele frequencies (p, q) describe the gene pool; for codominant alleles p = f(AA) + ½f(Aa), and X-linked loci count one X per male and two per female.
  • Hardy Weinberg: with random mating and no evolutionary forces, allele frequencies stay constant and genotypes settle at p2, 2pq, q2 in one generation.
  • For a rare recessive, q = √(affected) and carriers 2pq ≈ 2q vastly outnumber affected people.
  • A chi square test of Hardy Weinberg with two alleles has 1 degree of freedom (critical value 3.84).
  • Inbreeding raises homozygosity by Fpq per class without changing allele frequencies; first cousins give F = 1/16 by path counting.
  • Mutation is slow; migration makes populations alike (Δq = m[qm − q]).
  • Genetic drift is random, strongest when Ne is small, fixes neutral alleles with probability equal to their frequency, and causes founder effects and bottlenecks.
  • Selection problems: weight genotypes by fitness, divide by w̄; heterozygote advantage gives q̂ = s/(s + t), as for sickle cell.
  • Mutation selection balance: q̂ ≈ √(u/s) for recessive and u/s for dominant deleterious alleles.
  • The neutral theory predicts substitution rate = mutation rate and a molecular clock, K = 2μT; dN/dS reveals purifying or positive selection.
  • Species are kept apart by prezygotic and postzygotic barriers and arise mostly by allopatric speciation, with polyploidy common in plants.

Words worth knowing

TermWhat it means
Allele frequencyThe proportion of all copies of a gene in a population that are a given allele.
Genotype frequencyThe proportion of individuals in a population with a given genotype.
Hardy Weinberg equilibriumThe state in which genotype frequencies equal p2, 2pq and q2 and allele frequencies stay constant.
Carrier frequencyThe proportion of heterozygotes for a recessive allele, 2pq.
Identical by descentDescribes two alleles that are copies of the same ancestral allele.
Inbreeding coefficient (F)The probability that the two alleles at a locus in an individual are identical by descent.
Assortative matingNonrandom mating based on phenotype, positive for similar and negative for dissimilar partners.
Inbreeding depressionReduced fitness in inbred offspring, mainly from exposed recessive deleterious alleles.
HeterosisHybrid vigor; superior performance of offspring from crosses between inbred lines.
Gene flowMovement of alleles between populations through migration of individuals or gametes.
Genetic driftRandom change in allele frequencies caused by sampling in finite populations.
Effective population size (Ne)The size of an ideal population that would drift at the same rate as the real one.
Founder effectDrift that occurs when a few individuals start a new population.
BottleneckA drastic temporary reduction in population size that causes drift and loss of variation.
Relative fitness (w)The reproductive contribution of a genotype compared with the best genotype.
Selection coefficient (s)The proportional reduction in fitness of a genotype, s = 1 − w.
Heterozygote advantageHigher fitness of heterozygotes than of either homozygote, which maintains both alleles.
Mutation selection balanceThe equilibrium at which mutation creates deleterious alleles as fast as selection removes them.
Neutral theoryThe view that most molecular differences are selectively neutral and fixed by drift.
Molecular clockThe roughly constant accumulation of neutral substitutions, used to date divergences.
dN/dSThe ratio of nonsynonymous to synonymous substitution rates, used to detect selection.
Prezygotic barrierA reproductive barrier that prevents mating or fertilization.
Postzygotic barrierA reproductive barrier that reduces hybrid viability or fertility.
Allopatric speciationSpeciation after a population is divided by a geographic barrier.
Dobzhansky Muller incompatibilityA harmful interaction between alleles that evolved separately and meet only in hybrids.

Check yourself

In 200 people, 98 are AA, 84 are Aa and 18 are aa. Calculate p and q.
Answer. A alleles = 2(98) + 84 = 280 of 400, so p = 0.70 and q = 0.30. These counts match Hardy Weinberg exactly (0.49, 0.42, 0.09).
A recessive disease affects 1 in 10,000 people. Estimate the carrier frequency.
Answer. q = √0.0001 = 0.01, p = 0.99, so 2pq = 0.0198, about 1 in 50.
Why does a Hardy Weinberg chi square test with two alleles have 1 degree of freedom rather than 2?
Answer. Three genotype classes give 2 degrees of freedom, but one more is lost because the allele frequency used to compute expected values was estimated from the same data: 3 − 1 − 1 = 1.
If 1 in 20 males has an X-linked recessive trait, how many females are expected to be affected?
Answer. q = 0.05, so affected females = q2 = 0.0025, or 1 in 400.
Calculate F for the child of a mating between full siblings whose parents were unrelated.
Answer. Two common ancestors (the siblings' parents), each path with n = 3 individuals: F = 2 × (1/2)3 = 2/8 = 1/4.
Does inbreeding change allele frequencies? Explain.
Answer. No. It moves heterozygotes into the two homozygous classes equally (Fpq each), so f(AA) + ½f(Aa) still equals p. Only selection acting on the exposed homozygotes can then change allele frequencies.
A population of 20 breeding males and 20 breeding females drifts. What is the probability that a single new neutral mutation will eventually be fixed?
Answer. Ne = 4(20)(20)/40 = 40, so there are 80 gene copies and the fixation probability is 1/(2N) = 1/80 = 0.0125.
Fitnesses are AA 0.9, Aa 1.0, aa 0.6. What is the equilibrium frequency of a?
Answer. s = 0.1 and t = 0.4, so q̂ = s/(s + t) = 0.1/0.5 = 0.2.
A recessive lethal allele arises by mutation at u = 4 × 10−6. What is its equilibrium frequency?
Answer. q̂ ≈ √(u/s) = √(4 × 10−6/1) = 2 × 10−3 = 0.002.
A gene has dN/dS = 2.5 in a comparison of two primate species. What does this suggest, and what kind of gene might show it?
Answer. Amino acid changes have accumulated faster than silent ones, indicating positive selection. Immune genes such as MHC binding regions, and genes in arms races with pathogens, often show this pattern.