From Genotype Probabilities to an Expected Allele Count
An imputation result can represent uncertainty about a person’s genotype with probabilities for several possible genotype states. A single genotype dosage compresses those probabilities into an expected count of a specified allele. It is therefore a summary of the posterior distribution, not necessarily a genotype that was observed or even one of the most probable genotype states.
For a diploid individual at a biallelic variant, let \(G\) denote the number of copies of the allele being counted. If that allele is labeled A, the possible genotype states correspond to \(G=0\), \(G=1\), and \(G=2\) copies of A. Let \(D\) denote the imputation data and information used to form the genotype posterior probabilities. Write \(q_g=P(G=g\mid D)\) for the posterior probability of state \(g\).
The probabilities must describe mutually exclusive, exhaustive genotype states: each \(q_g\) is between 0 and 1, and \(q_0+q_1+q_2=1\). The dosage is calculated conditional on the imputation data and the model that produced those posterior probabilities. It does not require the genotype to be certain. Indeed, uncertainty is why a dosage need not be an integer.
Calculating and Interpreting Dosage
Worked Example: A Nearly Certain Heterozygous Genotype
Suppose a hypothetical diploid sample has posterior probabilities \(q_0=0.02\), \(q_1=0.96\), and \(q_2=0.02\) for zero, one, and two copies of allele A. The probabilities are valid because they are nonnegative and \(0.02+0.96+0.02=1.00\).
The dosage is the probability-weighted allele count:
The expected count is 1.00 copy of A. This does not assert that the genotype is known with certainty: the posterior assigns 4% probability in total to the two homozygous states. In this example, the one-copy state is also the most probable state, but that need not always be true.
Worked Example: A Fractional Dosage From Uncertain Probabilities
Consider a second hypothetical sample with \(q_0=0.20\), \(q_1=0.50\), and \(q_2=0.30\). These probabilities sum to \(1.00\), so they form a valid posterior distribution over the three diploid genotype states.
The dosage is
The dosage is 1.10 copies of A, an expected count under this posterior distribution. It is not a claim that the sample carries 1.10 physical copies: a diploid genotype has an integer number of copies. Here the most probable state has one copy, while the posterior expectation is 1.10 because the probability on the two-copy state contributes twice its probability to the expected count.
A dosage is tied to the allele being counted. If the counted allele is changed to the other allele at a diploid biallelic variant, the count becomes \(2-G\). Therefore the dosage for that other allele is \(E[2-G\mid D]=2-d\). Allele labels and orientation must be consistent when dosages are interpreted or compared.
Why the Posterior Mean Is the Dosage
The formula follows directly from the definition of expectation for a discrete random variable. Conditional on \(D\), the random variable \(G\) takes values 0, 1, and 2 with probabilities \(q_0,q_1,q_2\). Multiplying each possible count by its probability and summing gives its conditional mean:
The final equality follows because the zero-copy state contributes zero. This calculation uses the posterior probabilities as given; it does not require an additional Hardy–Weinberg assumption to turn them into a dosage. Any assumptions used to obtain the posterior probabilities belong to the imputation model that produced them.
The posterior mean also has a precise prediction property. Among all fixed numerical predictions \(a\) of the genotype count, it minimizes the posterior expected squared error. To see this, write \(d=E[G\mid D]\) and expand the squared difference:
The middle term is zero because \(E[G-d\mid D]=0\). Since \((d-a)^2\geq 0\), the expected squared error is smallest at \(a=d\). This explains a useful property of the posterior mean; it does not say that dosage preserves every feature of the posterior distribution.
Worked Example: The Same Dosage Can Hide Different Uncertainty
Compare two hypothetical posterior distributions for the number of copies of A. Distribution 1 has probabilities \((q_0,q_1,q_2)=(0.10,0.80,0.10)\). Distribution 2 has probabilities \((q_0,q_1,q_2)=(0.50,0,0.50)\). Each set sums to 1.
For Distribution 1, the dosage is \(0(0.10)+1(0.80)+2(0.10)=1.00\) copy. For Distribution 2, it is \(0(0.50)+1(0)+2(0.50)=1.00\) copy as well.
The posterior variances show that the distributions differ in uncertainty. Using \(\operatorname{Var}(G\mid D)=\sum_g (g-d)^2q_g\), Distribution 1 has \( (0-1)^2(0.10)+(1-1)^2(0.80)+(2-1)^2(0.10)=0.20\) copies squared. Distribution 2 has \( (0-1)^2(0.50)+(1-1)^2(0)+(2-1)^2(0.50)=1.00\) copy squared.
Both posteriors have the same expected count, but the first concentrates most of its probability on one copy, while the second places all its probability on the two homozygous states. Dosage alone does not reveal that difference.
Checks, Comparisons, and Limitations
Before using a dosage, check that the posterior probabilities refer to the intended genotype states, sum to 1 apart from possible rounding, and use a consistent allele orientation. If printed probabilities sum to a value slightly different from 1 because of rounding, the reported dosage may likewise be slightly affected. If they sum far from 1, verify the input or representation rather than treating the calculation as valid without question.
A common alternative summary is the genotype state with the largest posterior probability (the posterior mode). The mode identifies one state; dosage averages the allele count over all states using their posterior probabilities. They answer different questions and need not agree as summaries. For example, probabilities \(q_0=0.40\), \(q_1=0.35\), and \(q_2=0.25\) give dosage \(0(0.40)+1(0.35)+2(0.25)=0.85\) copies, even though zero copies is the most probable single state. A fractional dosage is not evidence of a fractional physical genotype.
The formula \(q_1+2q_2\) assumes a diploid individual and a biallelic locus with the three states defined above. It should not be transferred unchanged to a setting with a different ploidy or a different set of alleles. In such settings, define the possible allele counts for the relevant allele and take their posterior expectation by summing each count times its probability. Likewise, a dosage is only as well supported as the posterior probabilities supplied to it: an expectation does not repair errors in allele labels, model assumptions, or input data.
A practical pitfall is to read a dosage such as 1.10 as a literal genotype, or to assume that a dosage near 1 always means high certainty of one copy. The first interpretation confuses an expectation with a realized genotype count; the second ignores that different posterior distributions can have the same mean. Retaining or reviewing the genotype probabilities is necessary when the distribution’s uncertainty matters.
Identify which allele is being counted and verify that the three-state, diploid formula applies.
Match each probability to zero, one, or two copies of the counted allele and check that the probabilities are valid and sum to 1, allowing for rounding.
Calculate \(0q_0+1q_1+2q_2\), then report the result as an expected allele count rather than a literal genotype.
Check Your Understanding
Answer each question, then compare your reasoning with the solutions below.
- A hypothetical posterior is \((q_0,q_1,q_2)=(0.10,0.60,0.30)\). Calculate the dosage and state its interpretation.
- For \((0.45,0.45,0.10)\), calculate the dosage. Which genotype state has the largest posterior probability, and is that state’s allele count equal to the dosage?
- Explain why changing the counted allele to the other allele changes a diploid dosage from \(d\) to \(2-d\).
- Why do the posterior distributions \((0.10,0.80,0.10)\) and \((0.50,0,0.50)\) have the same dosage but represent different uncertainty?
- What should be checked before applying \(q_1+2q_2\) to a reported set of probabilities?
Solutions. (1) \(d=0(0.10)+1(0.60)+2(0.30)=1.20\) copies; this is the posterior expected number of copies, not a literal genotype. (2) \(d=0(0.45)+1(0.45)+2(0.10)=0.65\) copies. The zero-copy and one-copy states tie as the most probable, each at 0.45; neither state’s count equals 0.65. (3) If one allele has \(G\) copies, the other has \(2-G\) copies in a diploid genotype, so its expected count is \(E[2-G\mid D]=2-E[G\mid D]=2-d\). (4) Both expectations equal 1.00, but the first posterior places 0.80 probability on one copy, while the second places probability 0.50 on each of zero and two copies and none on one copy. (5) Check the counted allele, diploid ploidy and biallelic state definition, the mapping of probabilities to counts, and that the probabilities are valid and sum to 1 apart from rounding.