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Notes on Population Genetics: Nonrandom Mating, Genetic Drift, & Mutation, Exams of Genetics

University of New Mexico (UNM) - GallupGenetics

An overview of nonrandom mating, genetic drift, and mutation in population genetics. It covers the concepts of assortative mating and inbreeding, and discusses the Wahlund Effect, which can lead to non-conformity to Hardy-Weinberg expectations. The document also introduces the concept of inbreeding coefficients and their relationship to kinship coefficients.

Typology: Exams

2021/2022

Uploaded on 09/27/2022

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“Coarse” Notes Population Genetics
II-1
NONRANDOM MATING, GENETIC DRIFT, & MUTATION
NONRANDOM MATING/INBREEDING
READING: Nielsen & Slatkin, pp. 13–16, 59-63, 198-205
•Will distinguish two types of nonrandom mating:
(1) Assortative mating: mating between individuals with similar phenotypes or among
individuals that occur in a particular location.
(2) Inbreeding: mating between related individuals.
– Both types of nonrandom mating may have similar consequences since individuals with
similar phenotypes often have similar genotypes.
– It is often difficult to separate cause from effect.
• E.g., individuals with similar phenotypes may mate because
a) phenotypic assortative mating occurs;
b) mating with relatives is preferred;
c) matings are primarily based on proximity.
• Population subdivision: The Wahlund Effect
– It turns out that population subdivision per se can effect the distribution of genotypes in the
entire population.
– Consider a locus with 2 alleles (A and a) and a collection of isolated subpopulations
numbered 1, 2, 3, ...
– Let the frequencies of A and a in subpopulation i be and .
– Assuming random mating within each (isolated) subpopulation:
• Freq(AA) in subpopulation i =
• Freq(Aa) in subpopulation i =
• Freq(aa) in subpopulation i =
– Let = average freq. of A across all subpopulations.
• Likewise, let = 1 -
– What is the average frequency of each genotype over all the subpopulations?
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NONRANDOM MATING, GENETIC DRIFT, & MUTATION

NONRANDOM MATING/INBREEDING

READING: Nielsen & Slatkin, pp. 13 – 16, 59 - 63 , 198- 205

  • Will distinguish two types of nonrandom mating: (1) Assortative mating: mating between individuals with similar phenotypes or among individuals that occur in a particular location. (2) Inbreeding: mating between related individuals.
    • Both types of nonrandom mating may have similar consequences since individuals with similar phenotypes often have similar genotypes.
    • It is often difficult to separate cause from effect.
      • E.g., individuals with similar phenotypes may mate because a) phenotypic assortative mating occurs; b) mating with relatives is preferred; c) matings are primarily based on proximity. - Population subdivision: The Wahlund Effect
    • It turns out that population subdivision per se can effect the distribution of genotypes in the entire population.
    • Consider a locus with 2 alleles ( A and a ) and a collection of isolated subpopulations numbered 1, 2, 3, ...
    • Let the frequencies of A and a in subpopulation i be and.
    • Assuming random mating within each (isolated) subpopulation:
      • Freq( AA ) in subpopulation i =
      • Freq( Aa ) in subpopulation i =
      • Freq( aa ) in subpopulation i =
    • Let = average freq. of A across all subpopulations.
      • Likewise, let = 1 -
    • What is the average frequency of each genotype over all the subpopulations?
  • Consider AA homozygotes first:
    • From Fun Facts: so
    • So,
  • Similarly, for aa homozygotes:
    • since Var( q ) = Var(1 – p ) = Var( p ).
  • Finally, for Aa heterozygotes:
    • =.
  • A Thought Experiment:
  • Suppose genotypes were randomly sampled from a population whose substructure was unknown.
  • The frequencies of A and a in sample would be and.
  • With random mating, would then expect to find genotypes in proportions =.
  • But, the genotype frequencies observed would be = = .
  • I.e., would find an excess of homozygotes and a deficit of heterozygotes, compared to expectations.
  • Why? Simply because of population subdivision and, in particular, variance in allele frequencies across subpopulations.
  • Given across-subpopulations differences in allele frequencies, the apparent excess in homozygotes and deficit of heterozygotes from what is expected were the entire population to mate at random defines what is called the Wahlund Effect.
  • The Wahlund effect is a common “cause” of non-conformity to Hardy-Weinberg expectations in population samples.
  • Assuming the copies are made independently, then with probability , the copied alleles are both A (genotype AA ), etc.
  • Putting this all together, have:
  • Great! So now only need to determine. Thanks to Wright, this is very easy to do.
  • E.g., Let's find in the pedigree above.
  • Need only concentrate on the central part of the pedigree:
  • So the expected genotype frequencies for this pedigree are:
  • ASIDE: What is the average frequency of A among individuals with inbreeding coefficient ?
  • Inbreeding does not affect allele frequencies on average, but does affect the probabilities that 2 A 's or 2 a 's co-occur in an individual.

E G

C

J

c c' e g

  • Computing Inbreeding Coefficients (in general)
    • Say we want to find in the pedigree at right:
    • Rules : (1) Enumerate each loop (2) Each loop must a) go through each individual no more than once b) only change from up to down once (3) Multiply by for each passage through an individual - If the passage through an individual involves a change of direction (up/down) multiply by instead of 1/2, where is the inbreeding coefficient for that individual. (4) Add the probabilities of each loop.
    • For the above example: =

=

- Some forbidden loops: IGECBDEGI (goes through G twice) IJDBCEGI (already counted) IGECBDEJI (loop ECBDE already accounted for in )
  • Evolutionary Application: Kin Selection
    • Probability that two individuals share an allele descended from a common ancestor is called the kinship coefficient or coefficient of consanguinity.
    • Kinship coefficient between individuals A and B is denoted.
    • What is in the last example?

B

E

C D

G

J

I