Lecture 7 Flashcards Preview

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Flashcards in Lecture 7 Deck (19):
1

How do you simulate genetic drift?

1. Starting frequency is 0.2
2. Choose an allele at random, copy it and put it into the population of the next generation
3. Repeat until population size is reached
4. Repeat sampling through generations

2

Wright-Fisher population:

- Each allele is equally likely to be sampled in the next generation
- Typically assumes a diploid and that self fertilisation is possible
- Binomial sampling
- Variance greatest at p=0..5
- Variance is greater in smaller pppulations

3

Genetic drift:

- In each population heterozygosity decays with time
- The starting frequency gives you the probability of fixation
- Very weak/slow in large populations
- Gene flow counteracts fixation
- Eliminates variation

4

Decay in heterozygosity by drift:

H=1-G, where
H = frequency of the heterozygotes in the population
G = frequency of the homozygotes in a population
Ht = Ho(1-1/2N)to the power of t

5

Ht = Ho(1-1/2N)to the power of t

- Ht is expected heterozygosity at time t
- Ho: initial heterozygosity
- N = population size
- The decay in any single population will be much more stochastic

6

If N is really high (eg, 1,000, 000)

- t0.5=1.38x10to the power of 6 generations (the time it takes to lose half the heterozygosity via drift)..
- This means that it will take around 28 million years to loose half the heterozygosity in humans (with a generation time of 20 years.

7

Mutation (u) and drift:

- H=4Ni/(1+4Nu)
- Heterozygosity in the population is determined by the mutation rate and the population size
- This is the equilibrium value of heterozygosity

8

A problem in Neutral Theory:

- The generation time effect
- Expected and Observed are so different
- Expect: many organisms have extreme levels of heterozygosity
- Observe: correlation between population size and heterozygosity is weak

9

Effective population size (Ne)

- The size of the idealised (wright-Fisher) population whose decay of heterozygosity equals that of the real populations
- May be smaller than the breeding population
- Allows us to accommodate the dramatic fluctuations of population sizes over time
- The harmonic mean of the population sizes

10

Harmonic mean:

- Dominated by the smallest numbers in the sample

11

Factors effecting Ne:

- Different number of males than females
- Y chromosome
- Mitochondria
-

12

Different number of males than females: Ne = 4NmNf/Nm+Nf

eg) Elephant seals have harems, 5 males and 20 females (ie. Nc=25, Ne = 16)

13

Ne for Y: Ne=Nem/2

- Y is only in males, and only one copy per male

14

Ne for mitochondria: Ne(y)=Ne(mitochondria)

- Mitochondria is in al individuals but is almost always passed from mothers to children, and mothers almost always have a single version.

15

The X chromosome or haplodiploidy: Ne=9NmNf/4Nm+2nf

eg) a bee hive may have 20,000 bess, a singel queen and a dozen drones, Ne=9x12/4x12+2 = 2.16

16

Migration:

- Individuals that move from one area to a new area

17

Gene flow:

- To move to a new area and interbreed with resident populations

18

Continent-Island model is one way flow from a very large stable population:

- Pc -m-> Pi
- Pc is the frequency on the continent, last generation migrants
- Pi is the island, last generation residents
- m is the proportion of alleles that has come from another population in that generation
- Eventually, because the population is finite, they will all fix for one allele.

19

H=4Nu/(1+4Nu)

- Calculates the equilibrium heterozygosity, in a finite population.
- M is the number of migrants instead of the mutation rate
- You do not need much gene flow to remove population structure (sometimes only 1 or 2 individuals are needed to rescue a population)