Supplemental Lecture (97/05/13 update) by Stephen T. Abedon (abedon.1@osu.edu)

  1. Chapter title: Population Genetics
    1. A list of vocabulary words is found toward the end of this document
    2. Population genetics is the following of allele frequencies in populations, real or theoretical, through time and space. Since a frequency is a number, population genetics inevitably, and very early on delves deeply into algabreic manipulation. This number jockying is relatively straightforward, though ultimately can reach impressive levels of difficulty. However, it all begins with a simple statement which basically says that evolution does not occur in the absence of evolution. In this lecture we will begin with this statement, also know as Hardy-Weinberg equilibrium, before delving into the question of how gene frequencies might change in the face of simple evolutionary processes.
  2. Hardy-Weinberg equilibrium
    1. No evolution:
      1. Given the existence of genetic polymorphism, evolution can occur. However, in a given situation, is it?
      2. A population that shows an absence of evolution (as well as random mating) is said to be in Hardy-Weinberg equilibrium.
      3. That is, while there may be an input and removal of alleles over time, there is similarly no occurrence of change in allele frequency (equilibrium).
    2. Criteria for establishment:
      1. A touch more formally stated, Hardy-Weinberg equilibrium can occur only given a number of properties of a population:
        1. the population must be infinite
        2. there must be completely random mating
        3. there must be a lack of differential survival and reproductive success
        4. there must be no mutation (or any other genetic change)
        5. there must be no movement of individuals into or out of the population
      2. These properties translate to (respectively):
        1. no genetic drift
        2. no mating biases
        3. no selection
        4. no mutation
        5. no migration
    3. Deviations from equilibrium:
      1. Note that none of these properties likely occurs in real populations.
      2. However, over short periods a large, isolated population in a benign, stable, well mixed habitat to which members are well adapted can at least approximate such an equilibrium.
      3. In all other populations Hardy-Weinberg equilibrium is, by default, expected not to hold due to violation of one or many of the properties listed above.
      4. The study of evolution, therefore, at least in part is a study of conditions which lead to deviations from Hardy-Weinberg equilibrium.
  3. Hardy-Weinberg equation
    1. History:
      1. G.H. Hardy and G. Weinberg, both in 1908, made their mark on evolutionary biology by pointing out that genotype frequency could vary in the absence of any change in allelic frequency. Up until that point in time scholars were under the mistaken impression that mating within a population will lead to allelic change as a consequence of the algebra of genetics. Such as is not the case (i.e., instead Hardy-Weinberg equilibrium is expected) as illustrated upon deriving and employing the Hardy-Weinberg equation (below).
      2. If you have ever wondered why evolutionary theory is so controversial among the lay public, consider that the ideas that make it up are sufficiently complicated even to working scientists that it took nearly a decade, from the rediscovery of Mendelian principles, for the occurrence of one absurdly simple concept: That evolution cannot occur within a simple one loci, two allele system in an infinite, randomly mating population in the absence of natural selection, migration, and mutation. That is, in the absence of evolution, algebraically speaking, there is no evolution.
    2. Definitions:
      1. Assume a two allele, one loci system, e.g., A and a. Call the frequency of A, p, and the frequency of a, q. Recall by the definition of frequency (and of one loci, two allele systems) that p + q = 1. In addition:
        1. p = frequency of A
        2. q = frequency of a
        3. frequency of A plus frequency of a = frequence of all alleles at that loci in the population: p + q = 1
      2. A diploid population consisting of A and a alleles consists of the genotypes AA, Aa, and aa. The frequencies of each of these genotypes are constrained only by:
        1. that the sum of all of the frequencies must be equal to 1
        2. the frequency of all of the A alleles must be equal to p
        3. the frequency of the a alleles must be equal to q
    3. Allele frequency does not necessarily define genotype frequency:
      1. Note that allele frequency constrains but otherwise does not define genotype frequency. That is, any number of genotype frequencies is possible for a given set of allele frequencies.
      2. Prove to yourself that this is the case, that it is possible to vary genotype frequency without varying allele frequency.
      3. For example: Genotype frequencies of 0.2:0.5:0.3 (for AA:Aa:aa) has the same allelic frequencies as 0.3, 0.3, 0.4 and 0.4:0.1:0.5 (prove this to yourself, recalling how to calculate allelic frequency data from genotype frequency data).
    4. Allele frequency can sometimes define genotype frequency:
      1. Given:
        1. random mating
        2. no evolution (including infinite population)
        3. no differences in allelic frequencies between males and females
      2. then the frequency of genotypes in the next generation, regardless of the frequency of genotypes in the parental generation, will be:
        1. pp (p2) for AA
        2. 2pq for Aa
        3. qq (q2) for aa
      3. Huh?
        1. Any occurrence of an AA genotype implies the fertilization of a egg (or equivalent) having an A genotype by a sperm (or equivalent) also having an A genotype. Among eggs the frequency of A genotypes is p and among sperms the frequency of an A genotype is also p. Thus, the occurrence of an AA individuals is equal to the frequency p multiplied by the frequency p: p x p = p2.
          Restated: the frequency of AA individuals (p2) is equal to the produt of the frequency of A female gametes (p) and A male gametes (p).
        2. The calculation of the frequency of aa is exactly the same except that the frequency of the a allele is equal to q rather than p, thus resulting in the frequency of aa being equal to q x q or q2.
          Restated: the frequency of aa individuals (q2) is equal to the produt of the frequency of a female gametes (q) and a male gametes (q).
        3. The formation of an Aa individual, on the other hand, results from the fertilization of an A egg by an a sperm. Since the respective frequencies are p and q, the likelihood of such a fertilization is p x q or pq. However, note that an aA individual, resulting from the fertilization of an a egg by an A sperm, is genotypically equivalent to an Aa individual. However, the frequency of the occurrence of such an aA individual is q x p which equals qp which equals pq. Thus, there are two processes which can lead to an Aa individual, both occurring with a frequency of pq and thus the overall frequency of the occurrence of Aa individuals is 2 x pq or 2pq.
    5. Invariant genotype:
      1. Note, then, that no matter what the frequencies of original genotypes, for any given allelic frequency, one will end up with only a single, typical genotype frequency.
      2. Given this occurrence, it should not be surprising that subsequent generations will continue with this genotype frequency so long as the same conditions are maintained, i.e., random mating and no evolution.
      3. Thus, Hardy-Weinberg equilibrium is the indefinite maintenance of a specific genotype frequency, and its occurrence implies a lack of evolution.
    6. The Hardy-Weinberg equation, by the way, looks like this: (p + q)(p + q) = (p + q)2 = p2 + 2pq + q2 Can you derive this equation using a Punnett square?
  4. Example, change in allele frequency via selection
    1. Assumptions:
      1. Assume a one locus, two allele system where one allele is completely dominant over the other.
      2. The three genotypes are AA, Aa, and aa.
      3. Assume a relative fitness for the dominant homozygote of 1.0, the heterozygote also of 1.0 (this, actually, is equal to that seen with the dominant homozygote by definition since we're assuming complete dominance), and the recessive homozygote of 0.5.
      4. Assume that the population displays random mating, is infinitely large, and that the frequency of each allele is 0.5.
      5. Assume A is dominant to a.
    2. What is the frequency of the dominant allele after three generations (i.e., three rounds of replication)?
    3. Calculating first generation allele frequency:
      1. Since we assumed random mating with equiprevalent alleles, the population starts off with a frequency of the AA individual of 0.25, the frequency of the Aa individual 0.5, and the frequency of the aa individual 0.25.
      2. The relative contribution by each genotype of each allele to the gene pool is equal to the product of the genotype frequency, the allele frequency in a given genotype (i.e., either 1.0, 0.5, or 0.0), and the relative fitness of the genotype. Thus:
        1. f(A from AA) = 0.25 * 1.0 * 1.0 = 0.25
        2. f(A from Aa) = 0.50 * 0.5 * 1.0 = 0.25
        3. f(A from aa) = 0.25 * 0.0 * 0.5 = 0.0
        4. f(A from all) = 0.25 + 0.25 = 0.50 / (f(A) + f(a))
        5. f(a from AA) = 0.25 * 0.0 * 1.0 = 0.00
        6. f(a from Aa) = 0.50 * 0.5 * 1.0 = 0.25
        7. f(a from aa) = 0.25 * 1.0 * 0.5 = 0.125
        8. f(a from all) = 0.25 + 0.125 = 0.375 / (f(A) + f(a))
        9. f(A) = 0.50 / (0.50 + 0.375) = 0.57
        10. f(a) = 0.375 / (0.50 + 0.375) = 0.43
    4. Calculating first generation genotype frequency:
      1. Starting with allelic frequencies of 0.57 (i.e., p) and 0.43 (i.e., q) for allele A and allele a, respectively, we can calculate what the expected genotypic frequency would be:
      2. f(AA) = p2 = 0.57 * 0.57 = 0.325
      3. f(Aa) = 2pq = 2 * 0.57 * 0.43 = 0.49
      4. f(aa) = q2 = 0.43 * 0.43 = 0.185
    5. Calculating second generation allele frequency:
      1. The next round of selection is calculated as follows:
      2. f(A from AA) = 0.325 * 1.0 * 1.0 = 0.325
      3. f(A from Aa) = 0.49 * 0.5 * 1.0 = 0.245
      4. f(A from aa) = 0.185 * 0.0 * 0.5 = 0.0
      5. f(A from all) = 0.325 + 0.245 = 0.570 / (f(A) + f(a))
      6. f(a from AA) = 0.325 * 0.0 * 1.0 = 0.00
      7. f(a from Aa) = 0.49 * 0.5 * 1.0 = 0.245
      8. f(a from aa) = 0.185 * 1.0 * 0.5 = 0.093
      9. f(a from all) = 0.245 + 0.093 = 0.338 / (f(A) + f(a))
      10. f(A) = 0.570 / (0.570 + 0.338) = 0.63
      11. f(a) = 0.338 / (0.570 + 0.338) = 0.37
    6. Calculating second generation genotype frequency:
      1. Starting with allelic frequencies of 0.63 (i.e., p) and 0.37 (i.e., q) for allele A and allele a, respectively, we can calculate what the expected genotypic frequency:
        1. f(AA) = p2 = 0.63 * 0.63 = 0.40
        2. f(Aa) = 2pq = 2 * 0.63 * 0.37 = 0.47
        3. f(aa) = q2 = 0.37 * 0.37 = 0.13
    7. Calculating third generation allele frequency:
      1. The final round of selection we are considering is calculated as follows:
        1. f(A from AA) = 0.40 * 1.0 * 1.0 = 0.40
        2. f(A from Aa) = 0.47 * 0.5 * 1.0 = 0.24
        3. f(A from aa) = 0.13 * 0.0 * 0.5 = 0.0
        4. f(A from all) = 0.40 + 0.24 = 0.64 / (f(A) + f(a))
        5. f(a from AA) = 0.40 * 0.0 * 1.0 = 0.00
        6. f(a from Aa) = 0.47 * 0.5 * 1.0 = 0.24
        7. f(a from aa) = 0.13 * 1.0 * 0.5 = 0.065
        8. f(a from all) = 0.24 + 0.065 = 0.31 / (f(A) + f(a))
        9. f(A) = 0.64 / (0.64 + 0.31) = 0.67
        10. f(a) = 0.31 / (0.64 + 0.31) = 0.33
    8. Conclusion:
      1. Thus, after three rounds of selection allele A has increased its representation in the population from 50% to 67%.
      2. Although not terribly obvious from this example, selection against the recessive allele a will decline in intensity over time. This decline is due to less and less of the recessive present in the population being found in homozygotes (where it is amenable to selection) and more and more found in heterozygotes (where it is invisible to selection.
  5. Vocabulary
    1. Hardy-Weinberg equation
    2. Hardy-Weinberg equilibrium
  6. Practice questions
    1. Give five conditions that must be met in order for Hardy-Weinberg equilibrium to hold? Bonus points for a sixth condition which makes such equilibria meaningful: [PEEK]
    2. Given (i) two allele, one loci, diploid genetics (A and a), (ii) conditions appropriate for the establishment of Hardy-Weinberg equilibrium, and (iii) a frequency of A of 0.15, what are the genotype frequencies? [PEEK]
    3. Given a frequency of Aa of 0.70 and of AA of 0.30. If the selection coefficients associated with AA, Aa, and aa are 0.50, 0.60, and 1.0, respectively, what (i) are allelic frequencies in the current generation and (ii) the expected allelic frequencies in the next generation? [PEEK]
    4. Given frequencies of Aa of 0.30, of AA of 0.60, and aa of 0.10. If the selection coefficients associated with AA, Aa, and aa are 1.00, 1.00, and 0.50, respectively, what (i) are allelic frequencies in the current generation and (ii) the expected allelic frequencies in the next generation? [PEEK]
    5. How, if at all, does the mathematical treatment of the survival component of fitness differ from the mathematical treatment of the replication component of fitness? [PEEK]
    6. A large, isolated, randomly mating population of brown and blue eyed individuals is 20% blue eyed. Keeping in mind that blue eyes is the homozygous recessive condition, in this population what is the frequency of the brown eyed allele (i.e., brown yes results from both the homozygous dominant or heterozygous condition)? (assume that the effects of selection and mutation are negligible) (choose best answer) [PEEK]
      1. 0.45
      2. 0.55
      3. 0.65
      4. 0.75
      5. 0.85
      6. 0.95
    7. A recessive allele is present at a rate of 1 in 5000. What is the frequency of the heterozygote? (choose best answer) [PEEK]
      1. one in 5000
      2. one in 10,000
      3. one in 25,000
      4. 0.0002
      5. 0.0001
      6. 0.0004
    8. Given a one locus, three allele system (with allele frequencies of 0.1, 0.3, and 0.6), what is the frequency of the most prevalent heterozygote? (choose best answer) [PEEK]
      1. 0.12.
      2. 0.24.
      3. 0.36.
      4. 0.48.
      5. 0.60.
      6. 0.72.
    9. Given a one locus, three allele system (with allele frequencies of 0.1, 0.3, and 0.6), what is the frequency of the most prevalent genotype? (choose best answer) [PEEK]
      1. 0.12.
      2. 0.24.
      3. 0.36.
      4. 0.48.
      5. 0.60.
      6. 0.72.
    10. Given a one locus, two allele system displaying codominance where selection acts on diploids, the initial allelic frequencies are equal, is in Hardy-Weinberg equilibrium other than its being under selection, and diplays selection coefficients of 0.5, 1.0, and 0.1 where the selection coefficient of 1 is associated with the heterozygote. What is the gamete frequency following one round of selection? [PEEK]
  7. Practice question answers
    1. (i) infinite population, (ii) random mating, (iii) no selection, (iv) no migration, (v) no mutation, (vi-bonus) genetic polymorphism. Note that no genetic drift and infinite population mean the same thing.
    2. f(A) = 0.15, f(a) = 0.85, f(AA) = f(A) * f(A) = 0.023, f(Aa) = 2 * f(A) * f(a) = 0.255, f(aa) = f(a) * f(a) = 0.723.
    3. Current generation: f(A) = (2 * f(AA) + f(Aa)) / 2 = 0.65; f(a) = 1 - f(A) = 0.35; Next generation: f'(A) = ((f(Aa) * 0.60) + (2 * f(AA) * 0.5)) / 2 = .36; f'(a) = ((f(Aa) * 0.60) + (2 * f(aa) * 1.0)) / 2 = .21; f(A) = f'(A) / (f'(A) + f'(a)) = 0.36 / (0.36 + 0.21) = .63; f(a) = 0.37
    4. Current generation: f(A) = (2 * f(AA) + f(Aa)) / 2 = 0.75; f(a) = 1 - f(A) = 0.25; Next generation: f'(A) = ((f(Aa) * 1.00) + (2 * f(AA) * 1.00)) / 2 = 0.75; f'(a) = ((f(Aa) * 1.00) + (2 * f(aa) * 0.50)) / 2 = 0.20; f(A) = f'(A) / (f'(A) + f'(a)) = 0.75 / (0.75 + 0.20) = .79; f(a) = 0.21.
    5. there is no difference.
    6. ii, 0.55. That is, the frequency of the brown eyed allele is equal to p which is equal to 1 - q. We are given q2 as 0.2 (i.e., 20% of the population has blue eyes). q thus is equal to the square root of 0.2, which is approximately 0.45 (which is the frequency of the blue eyed allele). 1 - 0.45 = 0.55, which, therefore, is the frequency of the brown eyed allele.
    7. vi, 0.0004.
    8. iii, 0.36.
    9. iii, 0.36.
    10. Note first that the allele frequencies are 0.5 for both alleles (since they are equally frequent). Name the alleles A and a (or whatever turns you on). This means the genotypes are (with selection coefficients denoted parenthetically): AA (0.5), Aa (1.0), and aa (0.1). Note that whether AA or aa received the 0.5 or 0.1 selection coefficient was decided arbitrarily. Since this population can be described in terms of Hardy-Weinberg equilibrium the frequencies of the genotypes are, respectively, (0.5)2, 2(0.5)(0.5), and (0.5)2, or 0.25, 0.5, and 0.25. To find the frequency of genotypes following selection, multiply the present genotype frequency by the associated selection coefficients: (0.25)(0.5) = 0.125, (0.5)(1.0) = 0.5, (0.25)(0.1) = 0.025. Note that 0.125 + 0.5 + 0.025 equals 0.65, not 1.0. This is because the net effect of selection is to decrease the size of the population. To determine the post-selection genotypic frequencies you have to multiply by a value which results in their sum being equal to 1.0. This value is 1/(0.65). Multiplying by this value results in genotype frequencies for AA, Aa, and aa of 0.19, 0.77, and 0.04, respectively (note that these sum to 1.0). This, however, is not the answer. Instead, you are asked to find the gamete frequency. Note that gamete frequency is equal to allele frequency since gametes are haploid. You find allele frequency in the standard manner. That is, the frequency of A = 0.19 + (0.77)/2 = (2(0.19) + 0.77)/2 = 0.58 (possibly with some rounding error). This makes the frequency of a gamete equal to 1 - 0.58 = 0.42 and thus you have solved this problem.
  8. References
    1. Hartl, D.L. (1980). Principles of Population Genetics. Sinauer Associates, Inc., Sudnerland, Mass. p. 256.
    2. Raven, P.H., Johnson, G.B. (1995). Biology (updated version). Third Edition. Wm. C. Brown publishers, Dubuque, Iowa. pp. 7-14, 370-384, 419.