Why Construct a Pseudocontrol?
A family-based association analysis can compare an affected child’s genotype with a genotype assembled from parental alleles that were not transmitted. This constructed genotype is called a pseudocontrol. It is not an observed unaffected person. The comparison is useful because it focuses on which alleles were transmitted within a family, but its interpretation depends on how the pseudocontrol was constructed and on the disease-risk model.
When both parents are genotyped, the untransmitted alleles may be directly determined. With only one genotyped parent, the missing parent’s genotype and untransmitted allele are generally uncertain. A single completed pseudocontrol should not be invented in that setting: the compatible genotypes and transmission states must be handled explicitly, or a method that integrates over their uncertainty must be used.
Constructing a Pseudocontrol From a Complete Trio
Each parent transmits one allele to the child and retains one allele that was not transmitted. The two nontransmitted alleles together make the pseudocontrol genotype. At a biallelic locus, if the parents’ dosages are \(G_m\) and \(G_f\), and the child’s dosage is \(G_c\), then the pseudocontrol dosage is
This subtraction is valid because the child received exactly two alleles, one from each parent. The total parental dosage counts all four parental alleles; subtracting the two transmitted \(A\) alleles leaves the two nontransmitted \(A\) alleles. The result must be between 0 and 2. A result outside this range signals inconsistent genotype data or an incorrect transmission assignment.
Worked Example: A Complete Trio
Suppose the mother is \(A/a\), the father is \(a/a\), and their affected child is \(A/a\). The dosages are \(G_m=1\), \(G_f=0\), and \(G_c=1\).
The pseudocontrol dosage is
The child received \(A\) from the heterozygous mother and \(a\) from the father. The mother’s untransmitted allele is therefore \(a\), and the father’s untransmitted allele is also \(a\), giving pseudocontrol genotype \(a/a\). In this example the child and pseudocontrol are the two possible offspring genotypes obtained by changing which allele the heterozygous mother transmits.
That last feature is not guaranteed for every trio. For example, when both parents are heterozygous and the child is \(A/a\), the pseudocontrol is also \(A/a\). Such a pair provides no dosage contrast for an additive association model.
When One Parent Is Genotyped
With one parent’s genotype missing, the complete-trio subtraction cannot usually be applied. The observed child genotype does not always reveal which allele came from the genotyped parent, and the missing parent’s untransmitted allele is not observed. There may therefore be several compatible parental genotypes and transmission configurations, each implying a different pseudocontrol.
A principled analysis lists the compatible latent states and assigns them probabilities under stated assumptions. For example, those probabilities may use Mendelian transmission together with an allele-frequency model for the missing parent. The risk model then reweights the states when conditioning on an affected child. If parental genotype probabilities are used, their assumptions matter: population allele frequencies and mating assumptions can affect the analysis, so the resulting inference is not automatically protected from population structure merely because a family is involved.
Worked Example: An Uncertain Pseudocontrol With One Parent Missing
Suppose the genotyped parent is \(A/a\), the child is \(A/a\), and the other parent is ungenotyped. To illustrate the uncertainty, assume the missing parent’s genotype follows Hardy–Weinberg proportions with \(P(A)=0.30\) and \(P(a)=0.70\). Thus the probabilities of \(A/A\), \(A/a\), and \(a/a\) are \(0.09\), \(0.42\), and \(0.49\), respectively. Consider the Mendelian transmission distribution before any disease-risk weighting.
The genotyped parent could have transmitted \(A\) while the missing parent transmitted \(a\), or the genotyped parent could have transmitted \(a\) while the missing parent transmitted \(A\). The probability that the missing parent transmits \(A\), averaging over its genotype, is
Likewise, the probability it transmits \(a\) is \(0.49(1)+0.42(1/2)=0.70\). Since the genotyped parent transmits either allele with probability \(1/2\), the two compatible routes to the observed \(A/a\) child have unnormalized probabilities \(0.5(0.30)=0.15\) and \(0.5(0.70)=0.35\). Conditional on the observed child genotype, their probabilities are \(0.15/(0.15+0.35)=0.30\) and \(0.35/(0.15+0.35)=0.70\).
For the first route, the genotyped parent transmitted \(a\), leaving its untransmitted \(A\). Given that the missing parent transmitted \(A\), it was \(A/A\) with conditional probability \(0.09/0.30=0.30\), or \(A/a\) with probability \(0.21/0.30=0.70\). The pseudocontrol dosage is therefore 2 with probability 0.30 and 1 with probability 0.70 on this route.
For the second route, the genotyped parent transmitted \(A\), leaving its untransmitted \(a\). Given that the missing parent transmitted \(a\), it was \(a/a\) with conditional probability \(0.49/0.70=0.70\), or \(A/a\) with probability \(0.21/0.70=0.30\). The pseudocontrol dosage is 0 with probability 0.70 and 1 with probability 0.30 on this route. Combining the routes gives pseudocontrol dosage probabilities \(P(G_{\mathrm{pc}}=0)=0.70(0.70)=0.49\), \(P(G_{\mathrm{pc}}=1)=0.30(0.70)+0.70(0.30)=0.42\), and \(P(G_{\mathrm{pc}}=2)=0.30(0.30)=0.09\).
These are probabilities over possible pseudocontrols under the stated genotype model, not evidence that the actual pseudocontrol is known. An analysis must preserve this uncertainty rather than select one of the three dosages as though it had been observed.
Risk-Based Conditioning and Matched Conditional Logistic Models
Let \(f(g)=P(\text{affected}\mid G=g)\) denote disease risk for a genotype with dosage \(g\). If parental genotypes are fixed, let \(q_j\) be the Mendelian probability of offspring genotype \(g_j\). Conditioning on an offspring being affected gives the risk-weighted transmission probability
This expression is the key distinction between a pseudocontrol and a genuine observed unaffected control. The numerator weights each possible transmission by disease risk, not generally by disease odds. If candidate transmissions have unequal Mendelian probabilities, those probabilities also enter the likelihood.
A simple exact case illustrates when a two-genotype comparison has a log-linear risk interpretation. Suppose one parent is heterozygous, the other is homozygous, and the two equally likely transmissions from the heterozygote produce dosages 0 and 1. Under the log-linear risk model \(f(g)=C\exp(\beta g)\), where \(C\) is a constant over the genotypes being compared, the probability that the affected child has dosage 1 is
Here \(\exp(\beta)\) is the relative risk per additional \(A\) copy under the specified log-linear risk model. This calculation relies on the stated two alternatives, equal transmission probabilities, and log-linear risk; it is not a general derivation of a conditional-logistic likelihood for every pseudocontrol pair.
Conditional logistic regression is commonly used by treating each affected child and its pseudocontrol as a matched set. For a matched set with case genotype \(G_{i1}\) and pseudocontrol genotype \(G_{i0}\), the working conditional likelihood contribution is
This is the conditional case-assignment likelihood for a matched case-control model with a log-linear model for the odds of being a case. A pseudocontrol is constructed, not sampled as an unaffected person, so applying this likelihood to pseudocontrol pairs requires suitable assumptions about their sampling and the disease model. In particular, under a logistic penetrance model, interpreting this coefficient as a disease odds ratio generally relies on a rare-disease approximation for the pseudocontrol comparison. If disease is not rare, disease odds and disease risk need not be close.
For example, with logistic penetrance \(\operatorname{logit} f(g)=\alpha+\beta g\), the affected-transmission weights are \(f(g)\), not \(\exp(\alpha+\beta g)\) in general. When risks are small, \(f(g)\) is approximately \(\exp(\alpha+\beta g)\), making the risk ratio and odds ratio approximately equal. Without that approximation, an exact risk-based transmission likelihood should use \(q_j f(g_j)\); a standard conditional-logistic coefficient should not automatically be called a disease odds ratio or a relative risk.
Worked Example: A Conditional Logistic Matched-Pair Estimate
Suppose a hypothetical study has 10 informative matched sets at a biallelic locus. In 8 sets, the affected child has dosage 1 and the pseudocontrol dosage 0. In the other 2 sets, the affected child has dosage 0 and the pseudocontrol dosage 1. Assume the matched conditional-logistic model is appropriate for these constructed sets.
Each of the first 8 sets contributes \(\exp(\beta)/(1+\exp(\beta))\), and each of the last 2 contributes \(1/(1+\exp(\beta))\). The likelihood is
Writing \(\theta=\exp(\beta)\), the likelihood is proportional to \(\theta^8/(1+\theta)^{10}\). Its log-likelihood is \(8\log(\theta)-10\log(1+\theta)\); differentiating gives \(8/\theta-10/(1+\theta)=0\). Solving, \(8(1+\theta)=10\theta\), so \(\hat{\theta}=4\) and \(\hat{\beta}=\log(4)=1.386\).
The fitted conditional probability that the dosage-1 genotype is assigned to the case rather than the dosage-0 pseudocontrol is \(4/(1+4)=0.80\). Under a valid matched case-control interpretation, 4 is the estimated conditional odds ratio per additional \(A\) copy. If instead the data are interpreted through the exact log-linear disease-risk model described above, 4 is a relative risk per copy. Those are distinct interpretations; which one is justified depends on the model and sampling assumptions, not on the numerical estimate alone.
Handling Missing-Parent Uncertainty in an Analysis
When one parent is missing, a defensible analysis can enumerate compatible parental genotypes and transmissions, calculate their probabilities under an explicit model, and carry the uncertainty through the likelihood. In general, if \(O\) denotes observed genotypes and \(m\) indexes compatible missing-parent states, the probability of offspring genotype \(y\) among affected offspring has the form
Here \(P(m\mid O)\) is the assumed distribution of latent parental states given observed genotypes, and \(q(y\mid m)\) is the Mendelian probability of offspring genotype \(y\) for state \(m\). This expression makes clear where assumptions enter. If \(P(m\mid O)\) is estimated using population allele frequencies, misspecified frequencies can affect the result. If both parents are genotyped, this particular missing-parent integration is unnecessary, though risk and matched-model assumptions still require attention.
If neither parent is genotyped, the parental untransmitted alleles cannot be reconstructed from the trio’s parental data because there are no observed parental genotypes. Other family-based designs may use relatives’ genotypes, but that is a different construction. In all settings, genotyping consistency and the definition of the matched sets should be checked before interpreting an association estimate.
At the locus, identify the affected child’s genotype and which parental genotypes are available. Do not treat an unobserved parental allele as known.
With both parents observed, form the nontransmitted-allele genotype. With a parent missing, enumerate compatible latent states and retain their probabilities.
State whether the analysis uses a risk-based transmission likelihood or a matched conditional-logistic model, and give the assumptions needed for its interpretation.
Identify the effect measure, its units or genotype increment, and any rare-disease or allele-frequency assumptions. Do not describe a pseudocontrol as an observed unaffected individual.
Common Mistake and Key Takeaway
A common mistake is to say, “The child and pseudocontrol are matched, so the conditional logistic coefficient is automatically the disease odds ratio.” Matching alone does not establish that interpretation. The pseudocontrol is constructed from parental alleles, and the probability that a transmission appears among affected offspring is weighted by disease risk. A conditional-logistic odds-ratio interpretation needs an appropriate matched case-control sampling model; for pseudocontrols, a rare-disease approximation is often needed to connect disease odds to the risk-weighted transmission comparison. Under a specified log-linear risk model, the exponentiated coefficient instead has a relative-risk interpretation.
Check Your Understanding
Answer each question by stating the relevant assumption as well as the calculation or conclusion.
- Parents have dosages 2 and 1, and their affected child has dosage 2. What is the pseudocontrol dosage?
- Why can one observed heterozygous parent and a heterozygous child be insufficient to identify a unique pseudocontrol genotype?
- In the risk-based transmission formula, what quantities determine the probability of a particular genotype among affected offspring?
- Under what risk model does \(\exp(\beta)\) represent a relative risk per additional allele copy?
- Why is a conditional-logistic coefficient from a pseudocontrol analysis not automatically a disease odds ratio?