Assignment - Population Genetics

Selection

Question 1

If the genotype fitnesses of $A A$ , $A a$ , $a a$ are $1.0, 0.9, 0.6$ and $p_{0} = 0.7$ , calculate $p_{1}, p_{2}$ and $p_{3}$ , the allele frequencies of allele $A$ after 1, 2, and 3 generations of selection.
$w_{A A} = 1.0$
$w_{A a} = 0.9$
$w_{a a} = 0.6$
$p_{0} = 0.7$
$q_{0} = 1 - p_{0} = 0.3$

First, I will calculate the mean fitness at generation 0 $\bar{w_{0}}$ :

\bar{w_{0}} = p_{0}^{2} w_{A A} + 2 p_{0} q_{0} w_{A a} + q_{0}^{2} w_{a a}

\bar{w_{0}} = (0.7)^{2} (1.0) + 2 (0.7) (0.3) (0.9) + (0.3)^{2} (0.6)

\bar{w_{0}} = 0.49 + 0.378 + 0.054 = 0.922

Now, we can find the change in allele frequency after selection:

p_{1} = \frac{p_{0} (p_{0} w_{A A} + q_{0} w_{A a})}{\bar{w_{0}}}

p_{1} = \frac{(0.7) ((0.7) (1.0) + (0.3) (0.9))}{0.922} = 0.736

Now we repeat for the following two generations:

\bar{w_{1}} = p_{1}^{2} w_{A A} + 2 p_{1} q_{1} w_{A a} + q_{1}^{2} w_{a a} = 0.933396

p_{2} = \frac{(0.736) ((0.736) (1.0) + (1 - 0.736) (0.9))}{0.933396} = 0.7681981

\bar{w_{2}} = p_{2}^{2} w_{A A} + 2 p_{2} q_{2} w_{A a} + q_{2}^{2} w_{a a} = 0.943

p_{3} = \frac{(0.777) ((0.777) (1.0) + (1 - 0.777) (0.9))}{0.943} = 0.796

So, the values are: $p_{1} = 0.736, p_{2} = 0.768, p_{3} = 0.796$

The probability increases since this is a positive mutation with relative fitness ( $w_{A A} / \bar{w} = 1.0 / 0.922$ ) > 1.

Question 2

Genotype	Survival Rate $s$	Fecundity (eggs per female) $f$	Absolute fitness $W$
$A_{1} A_{1}$	0.90	50	45
$A_{1} A_{2}$	0.85	55	46.75
$A_{2} A_{2}$	0.75	70	52.5
Since the absolute fitness is the number of surviving offspring per female, and we know the number of eggs and egg-to-adult survival rate, we simply multiply the two to get the absolute fitness $W$ . In this case, the most fit genotype is $A_{2} A_{2}$ . Despite having the lowest survival rate, it produces enough young to compensate for this and has highest absolute fitness.

Question 3

What are the fitnesses of the two other genotypes relative to $w_{11} = 1$ ?
Since we set $w_{11}$ to 1, we will get the other relative fitnesses from $W_{12} / W_{11}$ and $W_{22} / W_{11}$ . See the table below

Genotype	Absolute fitness $W$	Relative Fitness to $W_{11}$
$A_{1} A_{1}$	$W_{11} = 45$	$w_{11} = 1$
$A_{1} A_{2}$	$W_{12} = 46.75$	$w_{12} = 1.03889$
$A_{2} A_{2}$	$W_{22} = 52.5$	$w_{22} = 1.166667$

If the frequency of the A2 allele is p = 0.5, what will be its frequency after one generation of selection? (use the relative fitness values $w_{i j}$ in your calculations)
First, we need to calculate the mean fitness for the population.

Note that since $p + q = 1$ and $p = 0.5$ , $p = q$ . So I will only use $p$ in the equation:

\bar{w} = p^{2} w_{11} + 2 p^{2} w_{12} + p^{2} w_{22}

\bar{w} = (0.5)^{2} * (1) + 2 (0.5)^{2} * (1.03889) + (0.5)^{2} * (1.16667) = 1.0611

Now, we use this mean fitness to find the change in $p$ over one generation:

p^{'} = \frac{p^{2} (w_{22} + w_{12})}{\bar{w}}

p^{'} = \frac{(0.5)^{2} * ((1.16667) + (1.038889))}{(1.0611)} = 0.5196

The allele frequency of $A_{2}$ increases slightly from 0.5 to 0.5196.

What will be the allele frequency when it reaches equilibrium?
Since the only two equilibrium states are $p = 0$ and $p = 1$ , and this is a beneficial mutation that increases in frequency each generation, the frequency at equilibrium will be $p = 1$ (fixed mutation).

Hardy-Weinberg Equilibrium

Question 1

The table shows the number of individuals affected for several recessive traits (diseases). Assuming random mating proportions, what is expected frequency of heterozygotes for each recessive allele?

If these are affected by this recessive gene, they must be heterozygotes for the recessive trait. So the allele frequency $q$ will be given by the square root of the homozygote frequency $R$ . Then, we can calculate $p$ from $1 - q$ , then get the heterozygote frequency from $2 p q$ .

Trait	Recessive homozygotes $R$ (per million)	$q = \sqrt{R}$	$p = 1 - q$	Heterozygote frequency $H = 2 p q$
a	2786	0.052783	0.947217	0.099993
b	658	0.025652	0.974348	0.049987
c	287	0.016941	0.983059	0.033308
d	160	0.012649	0.987351	0.024978
e	102	0.010100	0.989900	0.019995

Question 2

Given the values of the heterozygotes, we can calculate the allele frequencies $p$ and $q$ , based on $p = D + H / 2$ where $D$ is dominant heterozygote genotype frequency and $H$ is homozygote frequency. The reverse case is $q = R + H / 2$ .

	A allele frequency $p$	a allele frequency $q$
a	0.345	0.655
b	0.395	0.605
c	0.42	0.58
d	0.355	0.645
Now, we can calculate the expected genotype frequencies based on Hardy-Weinberg formulas:

	Expected AA frequency $p^{2}$	Expected Aa frequency $2 p q$	Expected aa frequency $q^{2}$
a	0.119025	0.45195	0.429025
b	0.156025	0.47795	0.366025
c	0.1764	0.4872	0.3364
d	0.126025	0.45795	0.416025
Now, we can calculate the difference between the observed and expected frequencies. This tells us how close to H-W a population is.

	AA	Aa	aa
a	-0.039025	0.07805	-0.039025
b	-0.066025	0.13205	-0.066025
c	-0.0464	0.0928	-0.0464
d	0.053975	-0.10795	0.053975

This shows that population b has the most deviation from H-W (they measured 13 more heterozygotes than expected) and the one most like H-W is population a (only 7.8% more heterozygotes than expected). All the populations except d have more heterozygotes than expected, while d has fewer than expected.

Genetic Drift

Question 1

Suppose that a diploid population of size $N = 50$ undergoes a change in heterozygosity from 0.50 to 0.42 in a single generation. Is it plausible to attribute this magnitude of change to random genetic drift?

We expect heterozygosity to decrease by $1 / 2 N$ each generation. In our population of N=50, we expect it to decrease by $0.01$ each generation (0.50 to 0.49). So this change is not likely to be caused by genetic drift alone. Perhaps this is being affected by selection or gene flow (migration).

Question 2

A colony of N=28 mice gets a new neutral allele. Probability of fixation is 1/2N = 0.01786 and expected time to fixation for a new mutation is 4N = 112 generations.

Question 3

What is the effective population size of a herd of 10 dairy cows and 1 bull? What is it for 40 cows and 1 bull? For 10 cows and 2 bulls?

We use the formula for calculating effective population size with a variation in number of reproducing adults:

N_{e} = \frac{4 N_{m} N_{f}}{N_{m} + N_{f}}

Number of cows $N_{f}$	Number of bulls $N_{m}$	Effective population size $N_{e}$
10	1	3.636
40	1	3.902
10	2	6.667

Question 4

What is the effective population size in a population of African lions, in which each breeding male controls a harem of five females and the total population consists of 200 males and 200 females?

If the reproducing males control the entire population of females and the remaining males do not mate at all, that means that only 40 (200 / 5) males contribute to the effective population size. Therefore, using the same formula from question 3:

N_{e} = \frac{4 N_{m} N_{f}}{N_{m} + N_{f}}

N_{e} = \frac{4 * (40) * (200)}{40 + 200} = 133.33

So, the effective population size is 133.33.

Migration and population structure

Question 1

$A_{1}$ Allele frequency in generation 0 of population $i$ : $p_{i 0} = 0.4$
In population $j$ , the $A_{1}$ allele is constant at $p_{j} = 0.6$
Migration rate between $i$ and $j$ : $m = 0.10$

Calculate the frequency of $A_{1}$ in population $i$ in the next two generations
$Δ p_{i} = m_{i j} (p_{j} - p_{i 0})$
$p_{i 1} = m_{i j} (p_{j} - p_{i 0}) + p_{i 0}$
$p_{i 1} = (0.10) (0.6 - 0.4) + 0.4 = 0.42$
Generation 1: $p_{i 1} = 0.42$

$p_{i 2} = m_{i j} (p_{j} - p_{i 1}) + p_{i 1}$
$p_{i 1} = (0.10) (0.6 - 0.42) + 0.42 = 0.438$
Generation 2: $p_{i 2} = 0.438$

Is the change in allele frequency in generation 2 less than the change in generation 1? Why?
The change between generations 2 and 1 is less than the change between 1 and 0. Allele frequency change is faster when there are large differences in allele frequencies between the two populations, and by generation 1 the populations have become more similar (there is less genetic difference between them).
What will the allele frequency become in the population after many generations?
Since the $A_{1}$ allele is constant in population $j$ , population $i$ will also eventually reach this level ( $p_{i} = 0.6$ ) due to gene flow.

In a more realistic scenario, we would expect allele frequency to fluctuate due to selective pressure, genetic drift, and the effect of individuals from $i$ migrating to $j$ .

Question 2

We have two genotyped populations of snails. We want to estimate the number of migrants per generation between the populations.

Calculate $F_{S T}$
The index of differentiation $F_{S T}$ is calculated with the following formula:

F_{S T} = \frac{H_{T} - {\bar{H}}_{S}}{H_{T}}

This measures the difference between expected heterozygosity in the Total population ( $H_{T}$ ) and the average of expected heterozygosity between the subpopulations. In order to calculate these values, we first need to find the allele frequencies $p_{1}$ , $p_{2}$ (for allele $A$ ) and $q_{1}$ , $q_{2}$ (for allele $a$ ), calculated from our genotype frequencies in the two populations:

$p_{1} = A A_{1} + A a_{1} / 2$
$= 0.056 + 0.288 / 2 = 0.200$

$q_{1} = a a_{1} + A a_{1} / 2$
$= 0.656 + 0.288 / 2 = 0.800$

$p_{2} = A A_{2} + A a_{2} / 2$
$= 0.672 + 0.256 / 2 = 0.800$

$q_{2} = a a_{2} + A a_{2} / 2$
$= 0.072 + 0.256 / 2 = 0.200$

Now, we can use the given formulas to find $H_{T}$ (expected total population heterozygosity):
$H_{T} = 2 \cdot \frac{1}{n} \sum_{i}^{n} p_{i} \cdot \frac{1}{n} \sum_{i}^{n} q_{i}$
$H_{T} = 2 \cdot \frac{1}{2} (p_{1} + p_{2}) \cdot \frac{1}{2} (q_{1} + q_{2})$
$H_{T} = \frac{1}{2} (0.200 + 0.800) \cdot (0.800 + 0.200) = 0.500$

And $\bar{H_{S}}$ (average subpopulation heterozygosity):
$\bar{H_{S}} = \frac{1}{n} \sum_{i}^{n} 2 p_{i} q_{i}$
$\bar{H_{S}} = \frac{1}{2} (2 p_{1} q_{1} + 2 p_{2} q_{2})$
$\bar{H_{S}} = (0.200 * 0.800) + (0.800 * 0.200) = 0.320$

So, $F_{S T} = (0.500 - 0.320) / 0.500 = 0.360$

Using the following relationship valid for two populations and your $F_{S T}$ value, provide an estimate of effective number of migrants per generation $N_{e} m$ between the two populations.

F_{S T} = \frac{1}{1 + 16 N_{e} m}

16 N_{e} m = \frac{1}{F_{S T}} - 1

N_{e} m = \frac{1 - F_{S T}}{16 F_{S T}} = \frac{1 - 0.360}{16 (0.360)} = 0.111

There is effectively 0.111 migrants per generation - meaning in 10 generations there will be around one migrant that affects the other's gene pool.