Title: Population Genetics

Q: A site with different nucleotides in independently sampled alleles is called [...].
A: a segregating (less often polymorphic) site

Q: [...] is called [...].
A: A nucleotide polymorphism that causes an amino acid polymorphism; a replacement polymorphism

Front: What percentage of random changes in a typical DNA sequence will cause an amino acid change?
Back: ~75%.

Front: What does the McDonald-Kreitman test do?
Back: It tests whether both silent and replacement variations are neutral (i.e., they have no discernible effect on the organism).
It does so by comparing variation between and among different species.

Front: What is a locus?
Back: It is the place on a chromosome where an allele resides.

Front: What is an allele?
Back: An allele is a bit of DNA at a locus, an instantiation of a locus.

Q: The three ways in which alleles at the same locus may differ are
• [...]
• [...]
• [...]
A: By origin: if they come from the same locus on different choromosomes.; By state: if they have different DNA sequences (in the context of the DNA sequence), or if they result in different amino acid sequences (in the context of protein evolution).; By descent: if they don't share a common ancestor allele (in the recent past, e.g. 10 generations).

Diploid individuals are said to be heterozygous at a locus iff the two alleles at that locus are different by state, and homozygous iff the two alleles are identical by state.

Front: With the frequency of tha \(A_i A_j\) genotype being \(x_{ij}\), how does one calculate the frequency of the \(i\)th allele?
Back: The frequency of having \(i\) on both alleles, plus the frequency where \(i\) occurs on any of the either upper or lower locus (e.g. chromosome).
\(p_i=x_{ii}+\frac{1}{2} \sum_{j=1}^{i-1} x_{ji} + \frac{1}{2} \sum_{j=i+1}^{n} x_{ij}\).

Front: What does the Hardy-Weinberg law describe (not yet how it describes it)?
Back: The equilibrium state of a single locus in a randomly mating infinitely big diploid population that is free of other evolutionary forces such as mutation, migration, and genetic drift.

Front: In a hermaphroditic species with an allele \(A_1\) (occuring with probability \(p\)) and an allele \(A_2\) (occuring with probability \(q\)), what are the Hardy-Weinberg genotype frequencies of different genotypes after the first round of random mating?
Back: \(A_1 A_1: p^2, A_1 A_2: 2pq, A_2 A_2: q^2\).

Front: How many generations of mating in species with separate sexes does it take to achieve Hardy-Weinberg equilibrium?
Back: 2.

Given the frequencies \((p_1, \dots, p_k)\) of \(k\) alleles \((A_1, \dots, A_k)\), the probability of the \(A_i A_i\) homozygote is \(p_i^2\) and the probability of the \(A_i A_j\) heterozygote is \(2 p_i p_j\).

Given the frequencies \((p_1, \dots, p_k)\) of \(k\) alleles \((A_1, \dots, A_k)\), the total frequency of homozygotes (the homozygosity) is given by \(G=\sum_{i=1}^{k} p_i^2\), the heterozygosity of the locus is given by \(H=1-G\).

Front: Given an initial probability \(\mathcal{H}_0\) of being a heretozygote, and a population size \(N\), what is the probability that an individual is going to be a heterozygote after \(t\) generations?
Back: \(\mathcal{H}_t=\mathcal{H}_0(1-\frac{1}{2N})^t\)

Front: When are neutral alleles?
Back: Alleles that are different by origin and identical by state (and therefore equally fit in the eyes of natural selection).

The variable \(\mathcal{G}\) is a measure of the genetic variation in the population, very similar to homozygosity. When there is no genetic variation, \(\mathcal{G}\)\(=1\), when every allele is different by state from every other allele, \(\mathcal{G}\)\(=0\).

Q: The value of \(\mathcal{G}\) after one round of random mating, \(\mathcal{G}'\), in a population with \(N\) individuals, is [...].
A: \(\mathcal{G}'=\frac{1}{2N}+(1-\frac{1}{2N})\mathcal{G}\)

Front: What is the formula for the change \(Δ_N\) of \(\mathcal{H}\) in a single generation, given \(N\) individuals?
Back: \(Δ_N \mathcal{H}=-\frac{1}{2N}\mathcal{H}\).
Here, the \(N\) subscript only indicates that this is in a finite population with \(N\) individuals.

Q: The probability that two alleles are different by state in different individuals decreases at a rate of [...] [...].
A: \(\frac{1}{2N}\); per generation

Q: The probability of ultimate fixation of a neutral allele with frequency \(p\) is [...].
A: \(π(p)=p\), in other words, the probability of fixation is the initial frequency

Q: The neutral theory of population genetics claims that [...].
A: most differences in alleles are due to neutral mutations that are neither beneficial nor detrimental, and evolution is due to drift and mutation alone

Q: Given a population with \(N\) individuals and a mutation rate to neutral alleles \(u\), mutation introduces variation into the population at the rate [...].
A: \(2Nu\)

Front: What is the formula for obtaining the equilibrium value of \(\mathcal{G}\) (\(\hat{\mathcal{G}}\)) in a population with \(N\) individuals and a mutation rate \(u\)?
Back: \(\hat{\mathcal{G}}=\frac{1}{1+4Nu}\).

Front: What does \(Δ\mathcal{H}\) stand for?
Back: The change in \(\mathcal{H}\) in a single generation.

\(Δ\mathcal{H}\) can be decomposed into two parts: \(Δ\mathcal{H} \approx Δ_N \mathcal{H}+Δ_u \mathcal{H}\).
The left component \(Δ_N \mathcal{H}\)\(=-\frac{1}{2N}\mathcal{H}\) describes the change in \(\mathcal{H}\) due to genetic drift.
The right component \(Δ_u \mathcal{H}\)\(=2u(1-\mathcal{H})\) describes the change in \(\mathcal{H}\) due to mutation with a rate \(u\).

Q: The change to \(\mathcal{H}\) due to drift is [...], while the change due to mutation is [...].
Equilibrium is reached when [...].
A: always negative (drift decreases variation); always positive (mutations introduce new variation); mutation and drift are in balance: \(-Δ_N \mathcal{H}=Δ_u \mathcal{H}\)

Q: The time it takes genetic drift to [...] is [...].
A: halve heterozygosity; proportional to the population size \(N\)

Front: What is the process of substitution?
Back: In substitution, a previously non-existent allele arises by mutation and undergoes fixation by spreading through a population.

Q: We use [...] to denote [...], in the unit of [...].
A: \(ρ\); the rate of substitution; substitutions per year (often, though, also measured in substitutions per generation)

Front: What is the relation between the substitution rate \(ρ\) and the mutation rate \(u\)?
Back: \(ρ=u\).

Front: Why is it the fact that \(ρ=u\), that is, the rate of substitution is the mutation rate?
Back: In a population with \(N\) individual, in each generation there are \(2Nu\) mutations (\(2N\) because that's the number of gametes produced).
On average, \(\frac{1}{2N}\) of those reach fixation, because in neutral evolution, the probability of an allele reaching fixation is equal to its frequency.
Then, \(ρ=2Nu \cdot \frac{1}{2N}=u\).

Front: What does the parameter \(θ\) stand for?
Back: The amount of variation under the neutral theory.

Front: How is \(θ\) calculated?
Back: \(θ=4Nu\)

Q: A [...] is [...].
A: coalescent; the lineage of alleles in a sample traced backward in time to their common ancestor

Q: [...] stands for [...].
A: \(S_n\); the number of segregating sites in a sample of \(n\) alleles

Q: [...] stands for [...].
A: \(T_c\); the total time of the coalescent

Q: For a neutral mutation rate \(u\), the [...] is [...].
A: expected number of mutations in the coalescent; \(T_cu\)

Front: What is the probability that the first coalescence occurs in any particular generation, with \(n\) alleles in a population with \(N\) individuals?
Back: \(\frac{n(n-1)}{4N}\)

Front: What is the intuition by which one can derive the probability that the last generation was a coalescence, in a population with \(N\) individuals?
Back: For a generation not to be a coalescence, all \(n\) alleles in the current generation need to have different ancestors.
The first allele has an ancestor allele. The probability that the second allele has a different ancestor from the first one is \(1-\frac{1}{2N}\), for the third one the probability is \(1-\frac{2}{2N}\).
Generalizing, the probability that all alleles in the previous generation were different is
\(P(\lnot C)=\prod_{i=1}^{n-1} 1-\frac{i}{2N}\)
where \(P(C)\) is the probability that the last generation was a coalescence.

Q: The distribution of coalescences is [...], with the mean [...] (\(n\) standing for the number of alleles, \(N\) for the size of the population, and \(T_n\) for the time until the first coalescence).
A: geometric; \(\mathbb{E}[T_n]=\frac{4N}{n(n-1)}\)

Front: How can one derive \(4Nu\) from \(S_n\)?
Back: \(4Nu=θ=\frac{S_n}{\mathcal{H}_{n-1}}\)
where \(\mathcal{H}_{n-1}=\sum_{i=1}^{n-1} \frac{1}{i}\) is the harmonic number.

Q: [...] are [...].
A: \(θ\) and \(π\); estimators of \(4Nu\)

Q: The formula for the estimator \(\hat{π}\) of \(π\) is [...], where [...].
A: \(\hat{π}=\frac{n}{n-1} \sum_{i=1}^{S_n} 2 \hat{p_i} (1-\hat{p_i})\); \(S_n\) is the number of segregating sites in the allele, \(n\) is the number of alleles, and \(\hat{p_i}\) is the frequency of one of the two alleles at the \(i\)th segregating site in the sample

Q: The relation between \(\hat{π}\) and \(θ\) is [...].
A: \(\mathbb{E}[\hat{π}]=θ\)

Q: Given [...], [...] is calculated by [...].
A: estimators \(\hat{π}, \hat{θ}\) for \(π, θ\), and a normalizing constant \(C\); Tajima's \(D\); \(D_T=\frac{\hat{π}-\hat{θ} }{C}\)

Q: The variable [...] stands for [...].
A: \(π\); the mean number of nucleotide differences between a pair of alleles drawn at random from the population

Q: [...] stands for [...].
A: \(\bar{t}\); the mean time back to the common ancestor of a random pair of alleles

Q: For [...], \(\bar{t}=\)[...] for \(N\) individuals in the population.
A: neutral loci; \(2N\)

Front: In a population with fluctuating size, \(N_1\) with probability \(r\) and \(N_2\) with probability \(1-r\), how does one calculate the effective population size \(N_e\)?
Back: The harmonic mean of the population size:
\(N_e=\frac{1}{\frac{r}{N_1}+\frac{1-r}{N_2}}\)

Q: In a randomly mating dioecious species with \(N_m\) males and \(N_f\) females, the effective population size \(N_e\) is [...].
A: \(N_e=\frac{4N_m N_f}{N_f+N_m}\)

Front: In the context of selection, what are viabilities?
Back: The probabilities of survival of individuals with certain genomes, given that they have been born.

Front: What is the relation of the viability and the frequency after selection?
Back: frequency after selection ∝ newborn-frequency × viability
For example, with the A₁A₁ allele, the frequency after selection is proportional to \(p^2w_{11}\).

Front: In population genetics, what does \(\bar{w}\) stand for?
Back: The mean fitness of the population.

Front: What is the Wright-Fisher model about?
Back: The Wright-Fisher model specifies how allele frequencies change during generations.

The Wright-Fisher model gives a probability distribution for the number of copies of a given allele \(A_1\) in the next generation:
If the current frequency of \(A_1\) is \(p\), and the current number of individuals is \(N\), then the number \(X=i\) of copies of \(A_1\) in the next generation is distributed binomially, given by the formula
\(P(X=i)={2N \choose i}p^i(1-p)^{2N-i}\)

Front: In mutation with selection, what does \(w_{11}\) stand for?
Back: The viability of the \(A_1 A_1\) genotype.

Front: In a population with two different alleles \(A_1, A_2\), how can one calculate the mean fitness of the population \(\bar{w}\)?
Back: \(\bar{w}=p^2w_{11}+2pqw_{12}+q^2w_{22}\)

In evolution with selection, \(Δ_s p\)\(=\)\(p'-p\) stands for the change of the frequency of an allele in a single generation, where \(p'\) is the frequency after selection, and \(p\) is the frequency before selection.

The change in frequency of an allele with the frequency \(p\) is calculated by the formula
\(Δ_s p=\)\(\frac{pq(p(w_{11}-w_{12})+q(w_{12}-w_{22}))}{p^2w_{11}+2pqw_{12}+q^2w_{22} }\),
where \(w_{ij}\) stands for the viability of the \(A_i A_j\) genotype, and \(p\) and \(q\) stand for the pre-selection frequencies of \(A_i\) and \(A_j\).

Q: In evolution with selection, the post-selection frequency [...] is calculated from [...] with the formula
[...].
A: \(p'\); the viabilities \(w_{11}\) and \(w_{12}\), as well as the pre-selection frequencies \(p\) and \(q\) and the mean fitness of the population \(\bar{w}\); \(p'=\frac{p^2w_{11}+pqw_{12} }{\bar{w} }\)

Q: In a lineage with \(t\) generations and a mutation rate \(u\), the probability distribution over the number of mutations can be described by [...] with mean [...], where the full distribution for the random variable \(M\) of mutations is
[...]
A: a Poisson distribution; \(tu\); \(P(M=i)=\frac{e^{-tu}(tu)^i}{i!}\)