Why Multicategory Phenotypes Need a Probability Model
A phenotype is not always recorded as simply present or absent. A clinical outcome might be classified as mild, moderate, or severe; a laboratory result might fall into one of several ordered or unordered groups. For each genotype, the relevant description is then a set of probabilities, one for each possible category. Those probabilities must collectively describe all possible outcomes.
Let \(Y\) denote a phenotype with categories \(1,\ldots,K\), where \(K\) is greater than two, and let \(G\) denote genotype. For a biallelic locus, we can write \(g=0,1,2\) for the number of copies of a specified allele. The quantity of interest for category \(k\) is \(P(Y=k\mid G=g)\): the probability of that phenotype category among individuals with genotype value \(g\). These are conditional probabilities, not proportions of the population with each genotype.
The conditions “mutually exclusive” and “exhaustive” matter. Each individual must belong to one and only one category, and the listed categories must cover every possible outcome being modeled. If categories overlap or an outcome can fall outside them, the probability model needs to account for that rather than silently treating the listed probabilities as a complete distribution.
Worked Example: Observed Category Probabilities by Genotype
Suppose a hypothetical data set classifies a phenotype as A, B, or C. The table gives category counts among individuals at three genotype values. Estimate the genotype-specific probabilities by dividing each category count by the total for that genotype.
| Genotype value \(g\) | A count | B count | C count | Total |
|---|---|---|---|---|
| 0 | 8 | 12 | 20 | 40 |
| 1 | 14 | 10 | 16 | 40 |
| 2 | 18 | 8 | 14 | 40 |
For \(g=0\), the estimated probabilities are \(8/40=0.20\) for A, \(12/40=0.30\) for B, and \(20/40=0.50\) for C. They sum to \(0.20+0.30+0.50=1.00\).
For \(g=1\), they are \(14/40=0.35\), \(10/40=0.25\), and \(16/40=0.40\), which sum to \(1.00\). For \(g=2\), they are \(18/40=0.45\), \(8/40=0.20\), and \(14/40=0.35\), which also sum to \(1.00\).
These are observed proportions for this hypothetical sample. They describe how the sample counts are distributed across categories within each genotype group; by themselves, they do not establish that genotype causes differences in phenotype.
Reference Categories and Multinomial Logits
A convenient way to model several category probabilities is to select one phenotype category as the reference category. Denote it by \(r\). For every other category \(k\), the model describes the log of the probability for category \(k\) relative to the probability for category \(r\). This is a log ratio of probabilities, often called a reference-category logit.
For example, if C is the reference category, the model needs one equation for A versus C and another for B versus C. A simple genotype-dosage specification might use \(\eta_A(g)=\alpha_A+\beta_Ag\) and \(\eta_B(g)=\alpha_B+\beta_Bg\). This is a model for the log probability ratios; it does not assume that the probabilities themselves change linearly with \(g\). Other choices of genotype coding can be used by defining different linear predictors.
The reference category is a comparison baseline, not a category with probability fixed at zero or one. Its probability is determined along with the other category probabilities. The choice of reference changes the form and interpretation of the model coefficients, but it does not change the underlying phenotype categories.
Converting Logits to Probabilities
Exponentiating a reference-category logit gives a probability ratio: \(\exp\{\eta_k(g)\}=\pi_k(g)/\pi_r(g)\). To turn these ratios into probabilities, set the reference category’s relative weight to 1. The other categories have weights \(\exp\{\eta_k(g)\}\). Dividing each weight by the sum of all weights gives a valid probability for each category.
The denominator includes the reference category’s weight, which is 1, as well as the exponentiated predictors for all other categories. Omitting any category from this denominator would generally make the resulting probabilities fail to sum to one.
Worked Example: Computing Probabilities From Genotype-Specific Logits
Consider a hypothetical model with phenotype categories A, B, and C, using C as the reference. Its genotype-dosage predictors are \(\eta_A(g)=-1+0.5g\) and \(\eta_B(g)=-0.4-0.2g\). Calculate the probabilities at \(g=1\).
Substituting \(g=1\) gives \(\eta_A(1)=-1+0.5=-0.5\) and \(\eta_B(1)=-0.4-0.2=-0.6\). Relative to C, the weights are \(\exp(-0.5)\approx0.6065\) for A, \(\exp(-0.6)\approx0.5488\) for B, and 1 for C.
The common denominator is \(1+0.6065+0.5488=2.1553\), using unrounded exponential values for the final calculations. Therefore, \(\pi_A(1)=\exp(-0.5)/[1+\exp(-0.5)+\exp(-0.6)]\approx0.2814\); \(\pi_B(1)=\exp(-0.6)/[1+\exp(-0.5)+\exp(-0.6)]\approx0.2546\); and \(\pi_C(1)=1/[1+\exp(-0.5)+\exp(-0.6)]\approx0.4640\).
The rounded probabilities sum to \(0.2814+0.2546+0.4640=1.0000\). In this model, at genotype value 1, the predicted probability of category A is about 0.2814, of B about 0.2546, and of C about 0.4640. These are model predictions conditional on genotype, not the observed proportions in a particular sample.
Interpreting Genotype Effects
In the dosage model, increasing \(g\) by one unit changes the log probability ratio for category \(k\) versus the reference by \(\beta_k\). Equivalently, the ratio \(\pi_k(g)/\pi_r(g)\) is multiplied by \(\exp(\beta_k)\) for each one-unit increase in genotype dosage. This interpretation is specific to the category-versus-reference comparison.
A positive \(\beta_k\) means that the category \(k\)-to-reference probability ratio increases with dosage under this model; it does not necessarily mean that \(\pi_k(g)\) itself increases at every genotype value. Other category probabilities are changing together, and all probabilities must share the same normalizing denominator. Likewise, a coefficient for A versus C does not directly describe A versus B.
Worked Example: Interpreting the Reference and Re-expressing Probabilities
Use the logits from the previous example at \(g=1\), where the probabilities for A, B, and C are approximately \(0.2814\), \(0.2546\), and \(0.4640\). First interpret the A-versus-C logit, then express the same probabilities using A as the reference.
At \(g=1\), the logit for A versus C is \(-0.5\), so the probability ratio is \(\pi_A(1)/\pi_C(1)=\exp(-0.5)\approx0.6065\). Thus, the predicted probability of A is about 0.6065 times the predicted probability of C at this genotype. This is a probability ratio between phenotype categories within genotype \(g=1\), not an odds ratio comparing different genotypes.
If A becomes the reference, the new logit for B versus A is the difference between the old B-versus-C and A-versus-C logits: \(-0.6-(-0.5)=-0.1\). The new logit for C versus A is \(0-(-0.5)=0.5\), since the reference category C had log weight zero in the original representation. The new relative weights for A, B, and C are therefore \(1\), \(\exp(-0.1)\approx0.9048\), and \(\exp(0.5)\approx1.6487\).
Their sum is \(1+0.9048+1.6487=3.5535\), approximately. Dividing each weight by the common sum gives probabilities approximately \(0.2814\) for A, \(0.2546\) for B, and \(0.4640\) for C. The probabilities are unchanged: choosing a different reference changes the logit representation, not the probability distribution.
Checks and Common Pitfalls
For any genotype value, verify that all category probabilities are nonnegative and that their total is one. In computations, retain sufficient precision in exponentials and the denominator, then round the final probabilities. Small discrepancies in a sum caused solely by rounding are expected; a substantial discrepancy suggests an omitted category or a calculation error.
- Confusing conditional and population probabilities. \(P(Y=k\mid G=g)\) describes phenotype category probabilities among individuals with genotype \(g\). It is not the probability of genotype \(g\), nor the proportion of all individuals with phenotype \(k\).
- Leaving out the reference weight. The denominator must include 1 for the reference category. Otherwise the computed values generally will not sum to one.
- Reading a coefficient as a probability change. A coefficient is a change in a log probability ratio per genotype unit. Its exponential is a multiplicative change in the category-versus-reference probability ratio, not a fixed percentage-point change in category probability.
- Assuming a reference category is intrinsically special. It is a chosen comparison baseline. Changing it alters which ratios are directly represented but should leave the predicted category probabilities unchanged.
- Claiming causation from conditional probabilities. A genotype-specific association described by these probabilities is not, by itself, proof that genotype causes the phenotype pattern.
A useful final check is to state the categories and reference explicitly, write one predictor for every nonreference category, and calculate all probabilities with a shared denominator. This makes both the comparisons and the normalization visible.
Check Your Understanding
Answer each question using the definitions and probability calculations above.
- A phenotype has four mutually exclusive, exhaustive categories. How many reference-category logits are needed when one category is selected as the reference?
- At a genotype value, the logits versus category D are \(\eta_A=-0.2\), \(\eta_B=0.4\), and \(\eta_C=-0.7\). Write the denominator used to calculate the four category probabilities.
- Why does a positive dosage coefficient for category A versus the reference not guarantee that the probability of A rises at every genotype value?
- If the reference category changes from C to A, should the predicted probabilities for A, B, and C change? Explain.
- In a sample of 50 individuals with genotype \(g=2\), 15 have category A, 10 have category B, and 25 have category C. Calculate the three observed within-genotype proportions and check their sum.