Lecture 7

Question 1

How do you simulate genetic drift?

Answer 1

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

Question 2

Wright-Fisher population:

Answer 2

• 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

Question 3

Genetic drift:

Answer 3

• 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

Question 4

Decay in heterozygosity by drift:

Answer 4

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

Question 5

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

Answer 5

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

Question 6

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

Answer 6

• 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.

Question 7

Mutation (u) and drift:

Answer 7

• 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

Question 8

A problem in Neutral Theory:

Answer 8

• 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

Question 9

Effective population size (Ne)

Answer 9

• 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

Question 10

Harmonic mean:

Answer 10

• Dominated by the smallest numbers in the sample

Question 11

Factors effecting Ne:

Answer 11

• Different number of males than females
• Y chromosome
• ## Mitochondria

Question 12

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

Answer 12

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

Question 13

Ne for Y: Ne=Nem/2

Answer 13

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

Question 14

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

Answer 14

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

Question 15

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

Answer 15

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

Question 16

Migration:

Answer 16

• Individuals that move from one area to a new area

Question 17

Gene flow:

Answer 17

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

Question 18

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

Answer 18

• 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.

Question 19

H=4Nu/(1+4Nu)

Answer 19

• 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)