Why Genotype Representation Matters
A genotype can be written as a pair of alleles, as two binary indicators, or as one number counting copies of an allele. These representations describe the same biological state when the genotype is known, but they are not interchangeable in every data set. In particular, a genotype recorded as missing is not a genotype with dosage zero, and a probability-based dosage is not necessarily an observed genotype.
The focus here is one biallelic locus in a diploid sample: each individual has two chromosome copies at the locus, and there are two possible alleles. This setting makes the coding transparent. The representations introduced here are tied to the allele labels established for a variant; changing which allele is counted changes the dosage coding.
Allele Pairs and Binary Allele Indicators
The three possible unordered allele pairs are \(R/R\), \(R/A\), and \(A/A\). Because the two copies are unordered here, \(R/A\) and \(A/R\) denote the same genotype. At a single biallelic locus, there are no other allele-pair categories under the diploid assumption.
One way to encode the pair is to give each allele an indicator: code \(R\) as 0 and \(A\) as 1. The two alleles then have a pair of binary codes. For example, \(R/A\) becomes \((0,1)\). Adding those two indicators gives the alternate-allele dosage. Thus a pair of allele codes and a single dosage are related, but they contain different forms: the pair records an indicator for each chromosome copy, while the sum records only the number of alternate copies.
| Allele pair | Binary allele indicators | Alternate-allele dosage |
|---|---|---|
| \(R/R\) | \((0,0)\) | 0 |
| \(R/A\) | \((0,1)\) or \((1,0)\) | 1 |
| \(A/A\) | \((1,1)\) | 2 |
In this table the binary indicators are assigned to the two copies, whereas dosage is their sum. The heterozygote has two possible orders if the chromosome copies are listed, but both orders give dosage 1. The dosage therefore does not retain an ordering of the alleles, nor does it identify which chromosome copy carries the alternate allele.
Here \(G\) is the alternate-allele dosage and \(I_1,I_2\) are the binary indicators for the two copies. A common shorthand writes the genotype categories as 0, 1, and 2. These are dosage codes, not the two allele indicators: the code 1 represents the heterozygote, not a single allele written by itself.
Worked Example: Converting Allele Pairs to Dosage
Suppose the reference allele at a hypothetical diploid biallelic locus is C and the alternate allele is T. Three hypothetical individuals have called genotypes C/C, C/T, and T/T. Convert each pair to binary allele indicators and alternate-allele dosage.
For C/C, both alleles are reference. The indicators are \((0,0)\), and the dosage is \(0+0=0\) copies of T.
For C/T, one allele is reference and one is alternate. The indicators are \((0,1)\) or \((1,0)\), and the dosage is \(0+1=1\) copy of T.
For T/T, both alleles are alternate. The indicators are \((1,1)\), and the dosage is \(1+1=2\) copies of T.
These calculations assume the locus is diploid, has exactly the two stated alleles, and has a genotype call for each individual. They do not infer a genotype from other data.
Hard Calls and Missing Calls
A hard call assigns an individual to exactly one genotype category. At this locus, a hard call might be the allele pair \(R/A\) or its dosage code 1. The call is categorical at the genotype level, even though its dosage representation is a number. If a data set reports a hard-called dosage, 1 means the selected genotype is the heterozygote under this coding; it is not a statement that the individual has a fractional allele.
A missing call is different from all three genotype categories. It means that the genotype is unavailable or not assigned, not that the individual carries zero alternate alleles. In a data table, missingness should therefore be represented with an explicit missing value or missing-data marker, rather than with the dosage 0. Coding missing as 0 would incorrectly assert the genotype \(R/R\).
Worked Example: Reading a Small Hard-Call Data Set
For a hypothetical locus with \(R\) as reference and \(A\) as alternate, a small data set contains hard-call dosages \(0, 1, 2,\) and missing for four hypothetical individuals. Translate each entry into its meaning.
The first individual has dosage 0, so the called pair is \(R/R\): \(0+0=0\) alternate copies.
The second has dosage 1, so the called pair is \(R/A\): \(0+1=1\) alternate copy.
The third has dosage 2, so the called pair is \(A/A\): \(1+1=2\) alternate copies.
The fourth has a missing call. No allele pair or dosage can be concluded from that entry. The data do not establish whether the genotype is \(R/R\), \(R/A\), or \(A/A\). In particular, replacing missing with 0 would add a false \(R/R\) call.
Probability-Based Dosage
Sometimes the genotype is not represented by one certain category. Instead, probabilities are assigned to the three possible genotypes. Let \(P(R/R)\), \(P(R/A)\), and \(P(A/A)\) denote the probabilities for the individual at the locus. If these are a complete probability distribution over the three categories, each is nonnegative and their sum is 1. The probability-based alternate-allele dosage is the expected number of alternate copies under this distribution.
The formula weights each possible dosage by its probability. A genotype assigned probability 1 contributes its ordinary hard-call dosage. When probability is spread across categories, the weighted average can lie between 0 and 2 without corresponding to any possible diploid genotype category. For example, 0.7 is a possible expected dosage, but no individual can carry seven-tenths of an alternate allele at this diploid locus.
Worked Example: Calculating a Probability-Based Dosage
For a hypothetical individual, suppose the probabilities are \(P(R/R)=0.20\), \(P(R/A)=0.50\), and \(P(A/A)=0.30\). Verify that they form a probability distribution and calculate the expected alternate-allele dosage.
The probabilities sum to \(0.20+0.50+0.30=1.00\), and each is between 0 and 1, so they form a valid distribution over the three genotype categories.
Using the dosage values 0, 1, and 2, $$ D = 0(0.20) + 1(0.50) + 2(0.30) = 0 + 0.50 + 0.60 = 1.10. $$
The probability-based dosage is 1.10 alternate copies. This is the expected dosage under the stated genotype probabilities. It does not mean the genotype is known to be \(R/A\), and it does not identify a single allele pair. A hard-call rule, if one is used, would be a separate step requiring a stated decision rule.
Why the Expected Dosage Formula Works
The definition follows directly from the expectation of a discrete random variable. Let \(G\) be the alternate-copy count for the individual, taking values 0 for \(R/R\), 1 for \(R/A\), and 2 for \(A/A\). The expectation of a discrete random variable is the sum of each possible value multiplied by its probability. Applying that definition gives the dosage formula.
Under diploidy and biallelic coding, \(G\) can be 0, 1, or 2 alternate copies.
The probabilities for \(G=0,1,2\) are \(P(R/R), P(R/A), P(A/A)\), respectively.
By the definition of expectation, \(E(G)=0P(R/R)+1P(R/A)+2P(A/A)\), which is the probability-based dosage \(D\).
Since the probabilities are nonnegative and sum to 1, this weighted average cannot be less than 0 or greater than 2. Specifically, each possible value of \(G\) lies in the interval from 0 to 2, so their probability-weighted average also lies in that interval. If one genotype probability is 1, the dosage equals that genotype's count; if more than one category has positive probability, a fractional dosage is possible.
This result explains the interpretation, not the source or calibration of genotype probabilities. A probability-based dosage is only as meaningful as the probabilities supplied for the categories. The probabilities may encode uncertainty about the genotype, but the dosage alone does not preserve that uncertainty: different probability distributions can produce the same expectation.
Worked Example: Same Dosage, Different Uncertainty
Compare two hypothetical probability distributions at the same diploid biallelic locus. Distribution 1 has probabilities 0.50 for \(R/R\), 0 for \(R/A\), and 0.50 for \(A/A\). Distribution 2 has probabilities 0 for \(R/R\), 1 for \(R/A\), and 0 for \(A/A\). Calculate their dosages and interpret the difference.
For Distribution 1, $$ D_1 = 0(0.50)+1(0)+2(0.50)=1.00. $$
For Distribution 2, $$ D_2 = 0(0)+1(1)+2(0)=1.00. $$
Both expected dosages equal 1.00 alternate copies. But Distribution 1 places probability on the two homozygous categories, whereas Distribution 2 assigns certainty to the heterozygote. The single dosage value does not show this distinction. Therefore, when uncertainty itself matters, the three genotype probabilities contain information that the expected dosage alone does not.
Allele Coding Determines the Meaning of Dosage
Dosage is always defined relative to a counted allele. With \(A\) as the alternate allele, the dosage is the number of \(A\) copies. If the allele being counted is switched to \(R\), the count becomes \(2-G\), because every diploid individual has exactly two copies in total. Thus \(R/R\), \(R/A\), and \(A/A\) have counts 2, 1, and 0 for \(R\), respectively.
This matters when comparing data sets or interpreting a coefficient downstream: a dosage value of 2 is meaningful only when the counted allele is known. The reference and alternate labels depend on the variant representation. The earlier material on variant records and allele normalization establishes why allele labels must be read consistently; the coding conversion itself is simply a count of copies of the designated allele.
Common Mistakes
- Confusing allele indicators with dosage. The pair \((0,1)\) contains two allele indicators; its sum, 1, is the genotype dosage. The code 1 is not one allele on its own.
- Treating missing as reference homozygous. Dosage 0 means a known \(R/R\) hard call. A missing call means no genotype category is available.
- Calling fractional dosage a genotype. A value such as 1.10 is an expectation over possible genotype categories, not a possible allele-pair category for an individual.
- Ignoring which allele is counted. Dosage 2 means two copies of the designated allele. If the counted allele changes, the dosage changes to \(2-G\) at a diploid locus.
- Assuming one dosage fully describes probability uncertainty. Different genotype-probability distributions can have the same expected dosage, so dosage does not preserve the full distribution.
Check Your Understanding
Answer each question using the diploid, biallelic setting and the allele coding stated.
- If \(R\) is reference and \(A\) is alternate, what are the binary allele indicators and dosage for genotype \(R/A\)?
- Why is a missing genotype call not equivalent to dosage 0?
- For probabilities 0.10, 0.60, and 0.30 for \(R/R\), \(R/A\), and \(A/A\), respectively, calculate the probability-based alternate-allele dosage.
- Can an individual have a hard-called dosage of 1.4 at this locus? Explain the difference between a hard call and probability-based dosage.
- If \(G\) counts alternate alleles, express the reference-allele count in terms of \(G\) and explain why.