MENDELIAN GENETICS in POPULATIONS I:
SELECTION and MUTATION as MECHANISMS of EVOLUTION

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Chapter 6 in the 4th edition, Chapter 5 in the 3rd.

review questions

INTRODUCTION
  1. An introduction to Hardy-Weinberg from Kimball's Biology Place
  2. Population genetics follows allele and genotypic frequencies across generations.
  3. population: a group of interbreeding organisms and their offspring [Fig. 6.1
  4. gene pool: the collection of alleles present within a population 
  5. allele frequency: the proportion of an individual allele within the gene pool 
  6. genotype frequency: the proportion of a genotype within the population 
  7. Hardy-Weinberg equilibrium: null model for gene frequencies in the absence of evolution.
    1. The null hypothesis is that the observed and expected values are not significantly different from one another for either allele frequencies or genotype frequencies
    2. Null hypothesis and HW.

6.1    THE HARDY-WEINBERG EQUILIBRIUM PRINCIPLE

A Numerical Example

  1. hypothetical population through one cycle from gametes in generation 1 to gametes in generation 2 (Fig. 6.1)
  2. Figure 6.2: A gene pool with allele frequencies of 0.6 for allele A and 0.4 for allele a.
  3. simulation showing allele frequencies change across generations as a result of chance for 100 sperm and 100 eggs (Fig. 6.3)
    1. For example, 34 AA, 57 Aa, and 9 aa adults
    2. The 34 AA adults produce 340 A gametes; the 57 Aa adults produce 285 A gametes and 285 a gametes; and the 9 aa adults produce 90 a gametes.
  4. Punnett square [Fig. 6.4]
  5. F1 gene pool with fr(A) = 0.6 and fr(a) = 0.4 (fig. 6.2) yields zygotes with AA = 0.36, Aa = 0.48, and aa = 0.16 [Fig. 6.5, Box 6.1]
  6. F2 gene pool has gametes with fr(A) = 0.6 and fr(a) = 0.4 (Figs. 6.6, 6.7)

The General Case

  1. fr(A1) = p
  2. fr(A2) = q
  3. p + q = 1
  4. the frequency of genotypes is [Figs. 6.6 - 6.9]:
    1. homozygote A1: fr(A1A1) = p2
    2. heterozygote A1A2: fr(A1A2) = pq + pq = 2pq
    3. homozygote A2: fr(A2A2) = q2
  5. p = p2 + (2pq) = p2 + pq
  6. q = (q2 + (2pq) = q2 + pq
  7. p + q = 1 = (p + q)2 = 1 = p2 + 2pq + q2 (the Hardy-Weinberg Principle)
  8. Same trick works with multiple loci: e.g., 3 alleles [box 6.2]
    1. p + q + r = 1 
    2. p2 + q2 + r2 + 2pq + 2pr + 2qr = 1

 Key points:

  1. Conclusion 1: At equilibrium, allele frequencies do not change from generation to generation
  2. Conclusion 2: allele frequencies predict genotype frequencies (p2 + 2pq + q2)
  3. if population displaced from equilibrium, return to equilibrium in one generation
  4. Assumptions for Hardy-Weinberg:
    1. No selection on any genotype
    2. No mutation
    3. No migration
    4. No random events affecting population
    5. Mating within population at random (panmixis)
    6. In other words, no evolution.(Fig. 6.10)

Changes in the Frequency of the CCR5-Δ32 Allele

6.2    SELECTION

Adding selection to the H-W analysis: Changes in allele frequencies

Empirical Research on Allele Frequency Change by Selection

Adding Selection to Hardy-Weinberg Analysis: Calculation of Genotype Frequencies

Changes in the Frequency of the CCR5-Δ32 allele revisited:

6.3    PATTERNS OF SELECTION

Selection on Dominant and Recessive Alleles

Selection on Heterozygotes and Homozygotes

  1. selection can promote increased heterozygosity [Fig.6.11]
  2. selection can promote reduced variability [Figs.6.11, 6.12, 6.13]
  3.   many other patterns such as

Frequency-Dependent Selection

  1. selection can maintain two different alleles in a population if each allele is advantageous when it is rare
    1. Third edition: frequency of left-handed scale-eating fish [Fig. 5.19] over time [Fig. 5.20].
      1. as left handed increase, prey become wary of being bit on the left
      2. left decrease and right increase leading to an increase of left.
      3. fluctuation around equilibrium value
    2. Elderflower orchids [Fig 6.21]

Compulsory Sterilization

6.4    MUTATION

Adding Mutation to the Hardy-Weinberg Analysis

Example:

  1. μ = 0.0001 per generation will change A=0.9, a=0.1 to 
  2. a =.10009 and 
  3. aa from .01 to .01002 in one generation [Fig. 6.23]
  4. over many generations there can be a small, but appreciable effect [Fig. 6.24, Box 6.9]
    1. pn =  poe-μn
    2. bacteria have mutation rates of 2 X 10-6
    3. after 5,000 generations: p5000 = (1)e-0.01 = .99 for A1 (a decrease of .01)

Mutation and Selection

Mutation-Selection Balance

  1. as selection removes deleterious alleles, mutation resupplies them.
  2. A balance between the two may explain the persistence of deleterious alleles in populations
  3. μ = rate at which A1   A2
  4. wxy = relative fitness of AxAy  = 1 - s
  5. s is the selection coefficient: 0<s<1; as s 1, relative fitness of wxy
  6. equilibrium frequency of q = (μ/s)1/2 [see Box 6.10 for derivation]
  7. if A2 is a lethal dominant, then = μ/s = μ [see Box 6.10 for derivation]
  8. spinal muscular atrophy
    1. a loss of function allele with q = .01 in Caucasian populations
    2. s is estimated to be 0.9
    3. μ = s* 2 = 0.9 x 10-4 per allele per generation
    4. this agrees with the observation that 7 of 340 affected individuals carried a new mutation not present in either parent.
    5. This is a mutation rate of 1.1 x 10-4 per allele per generation [Box 5.11]

Example: Cystic Fibrosis

  1. CF is caused by recessive loss-of-function mutations in the CFTR (cystic fibrosis transmembrane conductance regulator) gene
  2. gene plays a role in allowing cells of lung lining to ingest and destroy Pseudomonas aeruginosa bacteria
  3. inability to destroy bacteria means individuals homozygous for mutant CFTRs have chronic P. aeruginosa lung infections [Fig. 6.26]
  4. ultimately leads to severe lung damage and early death
  5. homozygous recessive (cc) lethal
  6. 1/2000 (0.0005) people in northern Europe have CF
    1.  if in H-W equilibrium, then q2 = 0.0005 and q = 0.0224 
    2. if c is maintained by mutation selection balance, and cc has 0 fitness, s = 1, then
    3. q = 0.0224 = (μ/1)
    4. μ = 0.0005 = 5 x 10-4, an unusually high mutation rate, suggesting incorrect assumptions
  7. actual mutation rate is 6.7 x 10-7
  8. perhaps c is maintained by heterozygote superiority
  9. cells heterozygous for CF have a substantial resistance to the bacteria that cause typhoid [Fig. 6.27]

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