Assessing Genotype and Physiology Dependence

Why Assess Genotype and Physiology Dependence?

A virtual cohort can differ across genotype groups for more than one reason. A relationship may be deliberately built into the cohort—for example, to represent a specified association between genotype and age—or it may appear by chance because a finite set of subjects was sampled. The distinction matters: a simulated difference in physiology can alter predicted exposure even when genotype-specific pharmacokinetic parameters are held fixed.

The diagnostic question is not simply whether two groups look different. First specify the intended target population and sampling design. Then compare the cohort with that target: contingency tables are useful for categorical variables, while covariate plots reveal shifts, spread, skew, and unusual tails in continuous variables. A statistical test can quantify evidence against independence under its assumptions, but it cannot determine whether an association was intended.

Definition: Genotype and a categorical physiological covariate are independent when the probability of each covariate category is the same in every genotype group. They are associated when at least one genotype group has a different category distribution. For a continuous covariate, assess dependence by comparing its distribution across genotype groups; a difference in means alone does not describe the whole distribution.

Contingency Tables and the Chi-Square Diagnostic

Let \(G\) denote genotype group and \(C\) a categorical physiological variable. A contingency table records the number of virtual subjects in each combination. Under the null hypothesis of independence, the expected count in row \(i\), column \(j\) is the row total multiplied by the column total, divided by the grand total.

E_{ij}=\frac{(\text{row total}_i)(\text{column total}_j)}{n}

The Pearson chi-square statistic measures the overall discrepancy between observed counts \(O_{ij}\) and the counts expected under independence:

X^2=\sum_{i=1}^{r}\sum_{j=1}^{c}\frac{(O_{ij}-E_{ij})^2}{E_{ij}}

Here \(r\) and \(c\) are the numbers of genotype and covariate categories. With independent subjects and sufficiently large expected counts, \(X^2\) is approximately distributed as a chi-square random variable with \(df=(r-1)(c-1)\) under the null hypothesis. The p-value is the upper-tail probability \(P(\chi^2_{df}\geq X^2)\). It measures how unusual a statistic at least this large would be if independence held; it is not the probability that independence is true.

Why these degrees of freedom? An \(r\)-by-\(c\) table has \(rc\) cells, but its row and column totals constrain the possible cell counts. Once the margins are fixed, there are \(rc-r-c+1=(r-1)(c-1)\) free cell counts. This gives the reference degrees of freedom for the independence test.

The usual approximation assumes each subject contributes to exactly one cell, subjects are independent under the sampling design, and expected counts are not too small. A common practical check is that expected counts are at least 5; it is a guideline for the approximation, not a universal guarantee. Sparse tables may require an exact test or a carefully justified alternative. Do not combine categories solely to obtain a convenient p-value.

Worked Example: A Deliberate Age-Group Association

Suppose a hypothetical virtual cohort was deliberately constructed with 60 noncarriers and 40 variant carriers. Age was categorized as under 40 or 40 and older. The observed counts are 42 noncarriers and 18 carriers under 40, and 18 noncarriers and 22 carriers aged 40 or older. Test independence as a diagnostic, while remembering that the age association was specified by the cohort design.

The row totals are 60 and 40; the column totals are also 60 and 40, with \(n=100\). Under independence, the expected counts are \(60(60)/100=36\), \(60(40)/100=24\), \(40(60)/100=24\), and \(40(40)/100=16\). All expected counts exceed 5.

The test statistic is

X^2=\frac{(42-36)^2}{36}+\frac{(18-24)^2}{24} +\frac{(18-24)^2}{24}+\frac{(22-16)^2}{16} =1+1.5+1.5+2.25=6.25

There are \(df=(2-1)(2-1)=1\) degree of freedom. The upper-tail chi-square probability is approximately \(p=0.0124\). Under an independent-subject design and the null hypothesis of independence, this statistic would be relatively unusual. The observed proportions aged 40 or older are \(18/60=0.300\) among noncarriers and \(22/40=0.550\) among carriers. In this example, however, the difference is consistent with the deliberate design, not evidence of a construction error. The test does not tell us why the groups differ or whether the association is scientifically appropriate.

Read the Design Before Reading the P-Value

A cohort may be sampled with genotype-specific quotas or stratified by genotype. In that case, the genotype totals are fixed by design rather than arising from an unconstrained random sample. The purpose of the diagnostic is then to assess whether physiology within and across the strata matches the intended target, not to test whether the imposed genotype frequencies arose naturally. If the design specifies a conditional physiology distribution within each genotype, compare each stratum with that specification. Do not interpret a test against an irrelevant independence model as a design failure.

Conversely, an association can arise accidentally in a finite simulated cohort even when the target population assumes independence. A non-significant test does not prove that the groups are exactly balanced; it says the data do not provide strong evidence against independence at the chosen threshold. Inspect the observed proportions and their practical importance as well as the p-value.

Worked Example: A Small Categorical Difference

In a hypothetical cohort of 150 subjects, a categorical physiological covariate is sex. The table contains 32 males and 28 females among 60 subjects with genotype group AA; 29 males and 31 females among 60 with group AG; and 15 males and 15 females among 30 with group GG. Assess the table using the independence model.

The total counts are 76 males and 74 females. Expected counts are 30.4 and 29.6 for AA, 30.4 and 29.6 for AG, and 15.2 and 14.8 for GG. Each expected count is at least 5. The statistic is

\begin{aligned} X^2 &=\frac{(32-30.4)^2}{30.4}+\frac{(28-29.6)^2}{29.6} +\frac{(29-30.4)^2}{30.4}+\frac{(31-29.6)^2}{29.6}\\ &\quad+\frac{(15-15.2)^2}{15.2}+\frac{(15-14.8)^2}{14.8}\\ &=0.0842+0.0865+0.0645+0.0662+0.0026+0.0027\\ &\approx 0.307 \end{aligned}

There are \(df=(3-1)(2-1)=2\) degrees of freedom, giving an upper-tail p-value of approximately 0.858. This table does not provide strong evidence against independence at a 0.05 significance level. The observed male proportions are 0.533 for AA, 0.483 for AG, and 0.500 for GG, which are close in this sample. The result does not establish exact equality or rule out a practically meaningful difference in a larger cohort.

Continuous Covariates: Plot Distributions, Not Just Means

For a continuous physiological variable—such as age, body weight, or organ volume—make a plot with genotype groups displayed separately. Histograms or density plots show shape; box plots or violin plots can compare location and spread; empirical cumulative distribution plots make differences across the full range visible. Include units and use a common horizontal scale. Examine overlap, skew, tails, clusters, and potential outliers, not only the center.

A standardized mean difference can summarize a location difference in units of pooled standard deviation. For two groups with sample sizes \(n_1,n_2\), means \(\bar{x}_1,\bar{x}_2\), and standard deviations \(s_1,s_2\), the conventional pooled standard deviation and standardized mean difference are

s_{\text{pooled}}= \sqrt{\frac{(n_1-1)s_1^2+(n_2-1)s_2^2}{n_1+n_2-2}}, \qquad \text{SMD}=\frac{\bar{x}_1-\bar{x}_2}{s_{\text{pooled}}}

This is a descriptive scale comparison, not a hypothesis test. It does not capture differences in shape or tails, and a small SMD does not guarantee that the distributions are otherwise similar. The sign indicates which group has the larger mean; the magnitude describes the difference relative to the pooled spread.

Worked Example: Comparing Age Across Two Genotype Groups

Consider a hypothetical cohort with 40 subjects in genotype group A and 60 in group B. Their ages have means 45 and 47 years and sample standard deviations 10 and 11 years, respectively. Suppose plots show substantial overlap, with group B slightly shifted toward older ages and no obvious isolated extreme values.

The pooled standard deviation is

s_{\text{pooled}}= \sqrt{\frac{(40-1)(10^2)+(60-1)(11^2)}{40+60-2}} =\sqrt{\frac{3900+7139}{98}} =\sqrt{112.64} \approx 10.61\text{ years}

Thus

\text{SMD}=\frac{45-47}{10.61}\approx -0.1885

The mean age in group A is about 0.188 pooled standard deviations lower than in group B. The negative sign reflects the lower mean in A. The size alone does not show whether the difference was planned, nor does it describe distributional tails. The plots and the cohort specification are needed to interpret it.

A Practical Diagnostic Sequence

Specify the target.
Write down whether the covariate should be independent of genotype or should follow a planned genotype-specific distribution. State whether genotype totals or strata were fixed by design.

Choose a display that matches the variable.
Use a contingency table for categories and group-specific plots for continuous physiology. Keep category definitions, units, and axes consistent.

Quantify the observed discrepancy.
For categorical variables, report counts, proportions, expected counts, and—when appropriate—the chi-square statistic and upper-tail p-value. For continuous variables, describe location and spread and consider a standardized difference.

Compare with the design, not an automatic ideal of equality.
Classify a discrepancy as intended, compatible with sampling variation, or requiring investigation. A statistical test alone cannot make this classification.

Alternatives, Failure Modes, and Limits

For a sparse two-by-two table, Fisher’s exact test is a standard alternative to the large-sample chi-square approximation. For larger sparse tables, an exact or simulation-based procedure may be appropriate if its conditioning and sampling assumptions match the design. These alternatives address the reference distribution; they do not resolve whether the chosen null hypothesis reflects the cohort target. For continuous covariates, plots and descriptive summaries are often more informative for diagnosing distributional mismatch than a single test of means.

A further limitation is that marginal comparisons can conceal structure. For example, genotype groups may have different age distributions because ancestry composition differs, even if there is little association within ancestry strata. Assess the covariates and relationships relevant to the intended population model; a single marginal table does not establish the cause of an association. Likewise, a plot of one covariate at a time cannot reveal every multivariable dependency.

Calling a significant test a construction error. A low p-value indicates evidence against the stated independence model, not evidence that the model was the correct target. Check whether dependence was deliberately specified.
Calling a non-significant result proof of balance. Failure to reject independence is not proof of exact equality. Review observed differences and whether they matter for the intended simulation.
Ignoring sparse expected counts. A chi-square approximation can be unreliable when expected counts are small. Inspect them before interpreting its p-value.
Comparing means alone. Similar means can coexist with different spread, skew, or tails. Use plots and relevant distribution summaries.

Key takeaway: Diagnose genotype–physiology dependence by combining the cohort’s intended design with tables for categorical covariates and distribution plots for continuous covariates. Tests quantify discrepancies under stated assumptions; they do not decide whether those discrepancies are deliberate or biologically justified.

Check Your Understanding

Answer each question, then compare your reasoning with the solutions.

A chi-square test of independence gives \(X^2=5.0\) with 1 degree of freedom. Which tail determines its p-value, and what does that p-value describe?
A genotype-stratified cohort intentionally assigns different age distributions to genotype groups. A test of independence is significant. Does this alone show a sampling error?
Why can a non-significant test not establish that two genotype groups have identical physiological distributions?
A continuous covariate has the same mean in two genotype groups, but one group has a longer upper tail. What diagnostic could reveal the difference?
What should be checked before using the usual Pearson chi-square approximation for a genotype-by-category table?

Solutions. (1) Use the upper tail, \(P(\chi^2_1\geq5.0)\), because larger statistics indicate greater disagreement with independence; it describes how unusual that statistic or a larger one would be under the null and the test assumptions. (2) No. The significant result is consistent with the deliberately specified dependence; compare the observed distribution with the design target. (3) A non-significant result is not proof of equality, and limited sample size may fail to detect a real difference. Inspect proportions and practical differences as well. (4) Plot group-specific distributions, such as histograms or empirical cumulative distributions, and compare their spread and tails. (5) Check that subjects contribute to one cell, that the sampling design supports the independence assumption, and that expected counts are adequate for the approximation; use a suitable exact or simulation-based alternative when needed.