Haplotypes and Linkage Disequilibrium in Pharmacogenes

Why haplotype structure matters

A pharmacogene may contain more than one functionally relevant variant. Knowing the frequency of each variant is not always enough to know how often they occur together on the same chromosome. Two variants can be common individually but rarely share a chromosome, or they can occur together more often than independence predicts. That chromosome-level arrangement can change the frequency of genotype combinations used as PBPK inputs.

A genotype records the alleles an individual has at a locus; it does not always reveal which alleles sit together on each of the two homologous chromosomes. A haplotype is the ordered set of alleles carried together on one chromosome across specified loci. The distinction between, for example, variants in cis (on the same chromosome) and in trans (on opposite chromosomes) is therefore central when combining variant effects.

Definition: For two biallelic loci with alleles $A/a$ and $B/b$, the possible haplotypes are $AB$, $Ab$, $aB$, and $ab$. If their population frequencies are $x_{AB}$, $x_{Ab}$, $x_{aB}$, and $x_{ab}$, respectively, then each frequency is nonnegative and $$ x_{AB}+x_{Ab}+x_{aB}+x_{ab}=1. $$ The haplotypes are said to be in linkage disequilibrium (LD) when the alleles at the two loci are associated on chromosomes rather than occurring in the proportions expected from their marginal frequencies alone.

This definition concerns statistical association, not proof that the loci are physically close or that one variant causes the other. Physical linkage can help preserve associations because recombination may separate nearby loci less often, but population history, selection, and sampling can also shape observed LD.

Estimating haplotype frequencies from phased data

Phased data identify the alleles on each chromosome. For $N$ diploid individuals with complete phase at both loci, there are $2N$ observed chromosome copies. The estimated frequency of a haplotype is its chromosome count divided by $2N$. The marginal frequency of allele $A$, denoted $p_A$, is the sum of frequencies of haplotypes carrying $A$; the corresponding frequency of $B$ is $p_B$.

Worked Example: Counting Phased Haplotypes

In a hypothetical sample of 40 individuals, both chromosome copies are phased at two loci. Across the 80 chromosome copies, the counts are $AB=32$, $Ab=8$, $aB=8$, and $ab=32$. Calculate the haplotype and marginal allele frequencies.

For each haplotype, divide its count by 80: $$ x_{AB}=\frac{32}{80}=0.400,\quad x_{Ab}=\frac{8}{80}=0.100,\quad x_{aB}=\frac{8}{80}=0.100,\quad x_{ab}=\frac{32}{80}=0.400. $$

The frequencies sum to $0.400+0.100+0.100+0.400=1.000$, as required. Allele $A$ occurs on $AB$ and $Ab$ chromosomes, so $p_A=0.400+0.100=0.500$. Allele $B$ occurs on $AB$ and $aB$, so $p_B=0.400+0.100=0.500$. The frequencies of $a$ and $b$ are each $1-0.500=0.500$.

With unphased genotypes, some individuals may have more than one compatible arrangement. For example, an individual heterozygous at both loci could have $AB/ab$ chromosomes or $Ab/aB$ chromosomes. Those alternatives have the same unphased genotype but different haplotypes. Directly observed phase or an explicit statistical phasing procedure is needed to resolve or represent that uncertainty. Treating an uncertain phase as known can produce misleading haplotype counts.

Quantifying linkage disequilibrium

The simplest signed measure of two-locus LD is $D$, the difference between the observed frequency of $AB$ and the frequency expected if the two alleles occurred independently. Under independence, the expected $AB$ frequency is $p_Ap_B$.

Definition: For two biallelic loci, $$ D=x_{AB}-p_Ap_B. $$ Equivalently, $D=x_{AB}x_{ab}-x_{Ab}x_{aB}$. Independence of alleles across loci corresponds to $D=0$, in which case each haplotype frequency equals the product of its two allele frequencies. Positive $D$ indicates excess $AB$ and $ab$ haplotypes relative to independence; negative $D$ indicates excess $Ab$ and $aB$ haplotypes.

The alternative expression follows by writing $p_A=x_{AB}+x_{Ab}$ and $p_B=x_{AB}+x_{aB}$. Then $x_{AB}-p_Ap_B=x_{AB}(1-x_{AB}-x_{Ab}-x_{aB})-x_{Ab}x_{aB}=x_{AB}x_{ab}-x_{Ab}x_{aB}$. This form makes clear that the association compares the two matching haplotype pairs with the two mismatching pairs.

The magnitude of $D$ depends on allele frequencies, so it is not directly comparable in all populations. Two commonly reported summaries address different questions. $r^2$ is a squared correlation measure: for polymorphic loci, $r^2=D^2/[p_A(1-p_A)p_B(1-p_B)]$. It ranges from 0 to 1 and describes how well an allele at one locus predicts the allele at the other. The magnitude of $D'$ scales $D$ by its largest possible magnitude given the allele frequencies; it also ranges from 0 to 1. High $|D'|$ indicates a strong association relative to the available frequency bounds, but does not necessarily mean one variant predicts the other well. If either locus is monomorphic, these standardized measures are not defined in their usual form.

Worked Example: Calculating D, r², and D'

A hypothetical set of 100 phased chromosome copies has frequencies $x_{AB}=0.080$, $x_{Ab}=0.120$, $x_{aB}=0.020$, and $x_{ab}=0.780$. Calculate the LD measures.

The marginal frequencies are $p_A=0.080+0.120=0.200$ and $p_B=0.080+0.020=0.100$. Independence predicts $p_Ap_B=0.200(0.100)=0.020$ for $AB$. Thus, $$ D=0.080-0.020=0.060. $$

For $D>0$, the maximum possible $D$ is the smaller of $p_A(1-p_B)$ and $(1-p_A)p_B$. Here those bounds are $0.200(0.900)=0.180$ and $0.800(0.100)=0.080$, so $D_{\max}=0.080$ and $|D'|=0.060/0.080=0.750$.

The denominator for $r^2$ is $0.200(0.800)0.100(0.900)=0.0144$, while $D^2=0.0036$. Therefore $r^2=0.0036/0.0144=0.250$. The loci show substantial scaled disequilibrium, but the allele at one locus does not perfectly predict the allele at the other: $r^2$ is 0.250, not 1.

From haplotypes to genotype frequencies

An important modeling step is to combine chromosome-level frequencies into individual genotypes. If individuals receive one chromosome from each of two randomly paired haplotype pools, the probability of a diplotype formed by distinct haplotypes $h_1$ and $h_2$ is $2x_{h_1}x_{h_2}$; the probability of two copies of the same haplotype $h$ is $x_h^2$. These are Hardy–Weinberg proportions for haplotypes. They require random union of haplotypes in the population being modeled; they should not be assumed automatically for every sample.

If linked loci are instead treated as independent, the model replaces observed haplotype frequencies with products of marginal allele frequencies. For example, the predicted $ab$ frequency becomes $(1-p_A)(1-p_B)$. This may preserve each locus’s allele frequency while changing how often variant alleles appear together. A functional genotype category defined by the combination or arrangement of variants can consequently be misestimated.

Worked Example: Independence Changes a Functional Genotype Frequency

In a hypothetical population, phased chromosome frequencies are $x_{AB}=0.350$, $x_{Ab}=0.150$, $x_{aB}=0.150$, and $x_{ab}=0.350$. The marginal frequencies are $p_A=p_B=0.500$. Suppose, only for this illustration, that the $ab$ haplotype is assigned a particular functional effect because it carries both specified variants. Compare the frequency of individuals with two copies of the $ab$ haplotype under observed linkage and under an independence assumption.

Assume random union of haplotypes. With the observed $x_{ab}=0.350$, the frequency of the $ab/ab$ diplotype is $$ x_{ab}^2=(0.350)^2=0.1225. $$ Thus, the modeled frequency is 12.25%.

Under independence, the predicted frequency of the $ab$ haplotype is $(1-p_A)(1-p_B)=(0.500)(0.500)=0.250$. The corresponding $ab/ab$ frequency is $(0.250)^2=0.0625$, or 6.25%. The independence model predicts a frequency lower by $0.1225-0.0625=0.0600$, which is 6.00 percentage points. This comparison concerns the specified genotype class, not a general claim about a clinical phenotype or a drug response.

The same issue appears for individuals carrying at least one $ab$ chromosome. Under random union, the observed-linkage frequency is $1-(1-0.350)^2=1-0.4225=0.5775$. Under independence it is $1-(1-0.250)^2=1-0.5625=0.4375$. The difference is 0.1400, or 14.00 percentage points. Both calculations depend on the assumed random pairing of haplotypes.

Assumptions, diagnostics, and limitations

A frequency calculation from phased chromosomes is descriptive of the sampled population. Applying it to a PBPK cohort requires the sample to represent the intended cohort, or a justified adjustment for differences in ancestry, recruitment, or other population structure. To convert haplotype frequencies into diplotype frequencies, additionally state the random-union or Hardy–Weinberg assumption. Deviations from that assumption, including inbreeding, assortative mating, or population mixture, can alter genotype frequencies even if the estimated haplotype frequencies are accurate.

Useful checks include verifying that haplotype counts sum to the number of observed chromosome copies, frequencies sum to one, marginal frequencies agree with allele counts, and computed $D$ agrees under both expressions. Investigate missing genotypes and phase uncertainty, especially when rare haplotypes are supported by very few observations. A large $|D'|$ alone is not evidence of strong predictive value: inspect $r^2$ and the allele frequencies as well. Apparent LD can also arise from combining subpopulations with different allele frequencies, so it need not reflect a within-group association.

Haplotype frequencies are useful when the functional interpretation depends on which variants occur together. They do not by themselves establish a variant’s functional effect, explain the mechanism of an association, or guarantee transferability to another cohort. Those conclusions require appropriate functional evidence and population-specific data. When the phase is unresolved, a single set of haplotype frequencies may also conceal uncertainty; results should reflect that uncertainty rather than silently treating estimated phase as certain.

The principal alternative is a model that assumes independence and uses marginal allele frequencies. It is simpler and can be reasonable when LD is negligible for the loci and population in question, but it can distort combined-variant frequencies when LD is material. A haplotype-based model preserves observed chromosome-level association but needs phased data and may be unstable for rare haplotypes. The appropriate choice depends on the data quality and the functional genotype classes that matter for the PBPK application.

Mistake: multiplying the two variant frequencies to estimate the frequency of their combined haplotype, even when phased data show an association.
Correction: use the observed phased haplotype frequency for the target population, quantify its departure from the product of marginal frequencies, and justify any independence assumption explicitly.
Mistake: interpreting high $|D'|$ as meaning that one variant reliably predicts the other.
Correction: examine $r^2$ as well; it reflects predictive correlation and can be low even when $|D'|$ is high.

Key takeaway: Marginal allele frequencies do not determine how variants are arranged across chromosomes. Phased haplotype frequencies quantify that arrangement, and accounting for it can materially change modeled functional genotype frequencies.

Check Your Understanding

Answer each question using the definitions and assumptions in this tutorial. Solutions follow the questions.

Phased chromosome counts are $AB=18$, $Ab=12$, $aB=12$, and $ab=18$. What is $p_A$, $p_B$, and $D$?
If $p_A=0.300$ and $p_B=0.200$, what $AB$ frequency is expected under independence?
Why can two individuals with the same unphased genotype at two heterozygous loci have different haplotypes?
Using the frequencies in the third worked example, what is the expected frequency of the $ab/ab$ diplotype under random union of haplotypes and independence?

Solutions. (1) There are 60 chromosome copies. $p_A=(18+12)/60=0.500$, and $p_B=(18+12)/60=0.500$. The $AB$ frequency is $18/60=0.300$, so $D=0.300-(0.500)(0.500)=0.0500$. (2) Independence predicts $p_Ap_B=(0.300)(0.200)=0.0600$. (3) The unphased double heterozygote can be $AB/ab$ or $Ab/aB$; these arrangements have the same genotype at each locus but different chromosome-level combinations. (4) Independence predicts $x_{ab}=(1-0.500)(1-0.500)=0.250$. Under random union, $P(ab/ab)=(0.250)^2=0.0625$, or 6.25%.