Pooled sequencing and allele frequencies

Two rounds of sampling

Pool-seq mixes the DNA of n diploid individuals into one library and sequences it. The result is cheap allele frequencies, but no individual genotypes. The estimate of a population frequency p comes from two rounds of sampling.

In round one, the pool holds 2n chromosomes drawn from the population. The number of A copies among them, k, follows a Binomial(2n, p) distribution, so the pool frequency q = k/2n differs from p by chance. This is the same sampling error as drift. In round two, C reads are drawn from the pool, each showing A with probability q, plus a small misreading rate ε. The estimate is p̂ = a/C, where a counts reads showing A, corrected for ε.

At the defaults of the module (p = 0.3, n = 20, mean depth 30, ε = 0.005) a pool has 40 chromosomes. Reading them with 30 reads cannot reveal more than the pool holds.

Where the variance comes from

The variance of p̂ adds the two rounds: Var = pq/2n + [pq(1 − 1/2n) + error term] × E[1/C]. With p = 0.3, pq = 0.21. For a fixed depth C = 40 = 2n and no error, the pool term is 0.21/40 = 0.00525 and the read term is 0.21 × (1 − 1/40)/40 = 0.00512. Reads cost as much as the pool: 51% of the variance comes from the pool and 49% from the reads, and the SD is 0.102.

Deeper reads shrink only the second term. At C → ∞ the variance falls to pq/2n, so the SD reaches a floor of √(0.21/40) = 0.072 here. That floor is set by the 40 chromosomes in the pool, and no depth lowers it. The effective number of chromosomes, pq divided by the variance, never exceeds 2n. At the defaults it is 16.8 of 40, because with a Poisson depth of mean 30, reads and misreading still make 58% of the variance.

The practical rule: when C is below 2n, reads dominate and depth helps. When C is above 2n, the pool dominates and a bigger pool helps. At the defaults, doubling the depth to 60 lowers the SE from 0.112 to 0.094, and doubling the pool to 40 individuals lowers it to 0.100. Reads dominate here, so depth helps more.

Sampling noise looks like differentiation

Now take two populations with true differentiation F_ST = 0.05 and sequence one pool of each. F_ST measures how much of the allele-frequency variance lies between populations. A naive estimate takes the read frequencies of the two pools as if they were the population frequencies. But each pool frequency also carries pool-sampling and read-sampling noise, and noise in the difference between two pools adds to the between-population signal.

The expected naive value is F + (1 − F)[1/2n + (1 − 1/2n)E(1/C)] plus an error term. With F = 0.05, 2n = 40 and infinite depth it is 0.05 + 0.95/40 = 0.074. Depth removes the read part only. The pool part, (1 − F)/2n = 0.024, stays. A pool-corrected estimator subtracts the estimated sampling variance of each pool, and an analysis-of-variance estimator (Hivert et al. 2018) does the same from raw read counts.