D-Prime, Allele-Frequency Bounds, and Its Limitations

Why Scale Linkage Disequilibrium?

The disequilibrium coefficient \(D\) measures the difference between an observed two-locus haplotype frequency and the frequency expected under independence. Its possible range depends on the allele frequencies at both loci. Consequently, the same numerical value of \(D\) can be close to the largest attainable disequilibrium in one population but far from the largest in another. D-prime addresses this by scaling \(D\) to the largest magnitude allowed by the allele frequencies.

This scaling is useful, but it answers a specific question: how close is the observed disequilibrium to its frequency-dependent bound? It does not, by itself, say that knowing one allele predicts the other accurately. In particular, \(|D'|\) can equal one even when the allelic correlation \(r^2\) is small.

Definition and Frequency-Dependent Bounds

Consider two biallelic loci with alleles \(A,a\) at the first locus and \(B,b\) at the second. Let \(f_{AB}, f_{Ab}, f_{aB}, f_{ab}\) denote the population frequencies of the four haplotypes. They are nonnegative and sum to one. Write \(p_A=f_{AB}+f_{Ab}\) and \(p_B=f_{AB}+f_{aB}\) for the frequencies of \(A\) and \(B\). The disequilibrium coefficient is

D=f_{AB}-p_Ap_B.

The frequency \(p_Ap_B\) is the expected frequency of \(AB\) under independence. To determine the possible values of \(D\) at fixed allele frequencies, express each haplotype frequency in terms of \(D\):

\begin{aligned} f_{AB}&=p_Ap_B+D,\\ f_{Ab}&=p_A(1-p_B)-D,\\ f_{aB}&=(1-p_A)p_B-D,\\ f_{ab}&=(1-p_A)(1-p_B)+D. \end{aligned}

Each expression must be at least zero. The two frequencies that decrease as \(D\) increases give upper bounds; the two that increase as \(D\) increases give lower bounds.

Definition: For \(D>0\), the largest feasible value is \(D_{\max}=\min\{p_A(1-p_B),(1-p_A)p_B\}\). For \(D<0\), the most negative feasible value is \(D_{\min}=-\min\{p_Ap_B,(1-p_A)(1-p_B)\}\). These bounds depend on the allele frequencies, not just on the fact that \(D\) is positive or negative.

For example, nonnegativity of \(f_{Ab}\) requires \(D\leq p_A(1-p_B)\), while nonnegativity of \(f_{aB}\) requires \(D\leq(1-p_A)p_B\). Together these give the stated upper bound. Similarly, \(f_{AB}\geq0\) requires \(D\geq-p_Ap_B\), and \(f_{ab}\geq0\) requires \(D\geq-(1-p_A)(1-p_B)\). The stronger lower bound is therefore \(D_{\min}\). These conditions are also sufficient: if \(D\) lies between the two bounds, all four displayed haplotype frequencies are nonnegative and sum to one.

The sign matters. Positive \(D\) means \(AB\) occurs more often than under independence, while negative \(D\) means it occurs less often. A signed version of D-prime retains that direction.

Definition: Signed D-prime is

D'= \begin{cases} D/D_{\max}, & D>0,\\[4pt] D/|D_{\min}|=-D/D_{\min}, & D<0,\\[4pt] 0, & D=0. \end{cases}

Thus \(-1\leq D'\leq1\). The denominator is a positive magnitude in either nonzero case, so \(D'\) has the same sign as \(D\). Some reports give only the magnitude \(|D'|\); when the sign is omitted, the direction of the disequilibrium is not conveyed.

If a locus is monomorphic, the allele frequencies force \(D=0\), and there is no two-allele variation with which to describe linkage disequilibrium. The convention \(D'=0\) at \(D=0\) is then a numerical convention rather than evidence of an informative absence of association.

Worked Calculations

Worked Example: Positive Disequilibrium

Suppose a hypothetical population has haplotype frequencies \(f_{AB}=0.18\), \(f_{Ab}=0.22\), \(f_{aB}=0.12\), and \(f_{ab}=0.48\). They are all nonnegative and sum to \(1.00\). Calculate signed \(D'\).

First find the allele frequencies:

p_A=0.18+0.22=0.40,\qquad p_B=0.18+0.12=0.30.

Then calculate \(D\):

D=f_{AB}-p_Ap_B =0.18-(0.40)(0.30) =0.18-0.12 =0.06.

Since \(D>0\), use the positive bound:

D_{\max}=\min\{(0.40)(0.70),(0.60)(0.30)\} =\min\{0.28,0.18\} =0.18.

Therefore \(D'=0.06/0.18=0.3333\). The positive sign indicates excess \(AB\) relative to independence; the magnitude says that \(D\) is one-third of its largest feasible positive value for these allele frequencies.

Worked Example: Negative Disequilibrium

For a different hypothetical population, let \(f_{AB}=0.02\), \(f_{Ab}=0.38\), \(f_{aB}=0.28\), and \(f_{ab}=0.32\). These frequencies sum to one. Find signed \(D'\), taking care to preserve the sign.

The allele frequencies are \(p_A=0.02+0.38=0.40\) and \(p_B=0.02+0.28=0.30\). Thus

$$ D=0.02-(0.40)(0.30)=0.02-0.12=-0.10. $$

Because \(D<0\), the lower bound is

D_{\min}=-\min\{(0.40)(0.30),(0.60)(0.70)\} =-\min\{0.12,0.42\} =-0.12.

The positive magnitude of this bound is \(|D_{\min}|=0.12\), so

D'=\frac{-0.10}{0.12}=-0.8333.

The negative sign is essential: the observed \(D\) is negative and the scaling denominator is positive. The value means the disequilibrium has about \(83.33\%\) of the maximum possible negative magnitude for these allele frequencies.

Why a Value Near One Does Not Ensure Strong Prediction

A value of \(|D'|\) near one means the observed disequilibrium is close to the extreme allowed by the allele-frequency margins. It does not mean that the two alleles have similar frequencies, nor that their correlation is close to one in magnitude. The allelic correlation measure \(r^2\) also depends on the allele-frequency variances:

r^2=\frac{D^2}{p_A(1-p_A)p_B(1-p_B)}.

The denominator can be large relative to \(D^2\), particularly when one allele is rare and the other is common. Then even a \(D\) at its feasible boundary can correspond to a modest \(r^2\). The following example makes that distinction explicit.

Worked Example: Maximum \(|D'|\) but Modest \(r^2\)

Consider hypothetical haplotype frequencies \(f_{AB}=0.05\), \(f_{Ab}=0\), \(f_{aB}=0.45\), and \(f_{ab}=0.50\). They sum to one, and all are nonnegative. Calculate \(D'\) and \(r^2\).

The allele frequencies are \(p_A=0.05\) and \(p_B=0.50\). The disequilibrium is

$$ D=0.05-(0.05)(0.50)=0.05-0.025=0.025. $$

Here \(D>0\), and

D_{\max}=\min\{(0.05)(0.50),(0.95)(0.50)\} =\min\{0.025,0.475\} =0.025.

Thus \(D'=0.025/0.025=1\). In this population, every chromosome carrying the rare allele \(A\) also carries \(B\); the \(AB\) haplotype is at its positive frequency bound. But \(B\) is common and occurs on many chromosomes without \(A\), so knowing \(B\) does not closely identify \(A\).

For the correlation measure,

r^2= \frac{(0.025)^2} {(0.05)(0.95)(0.50)(0.50)} = \frac{0.000625}{0.011875} \approx0.05263.

So \(|D'|=1\) coexists with \(r^2\approx0.05263\), a much smaller allelic correlation measure. The example is not contradictory: the measures normalize disequilibrium differently and answer different questions.

Interpretation and Practical Limits

D-prime is useful when the question concerns how close the observed disequilibrium is to what the allele-frequency constraints permit. It is dimensionless, bounded, and retains the direction of disequilibrium under the signed convention used here. It should not be substituted for \(r^2\) when the question concerns the strength of allelic correlation or how well one variant predicts another.

The definition is in terms of population frequencies. In an analysis, frequencies are estimated from data, so a calculated \(D'\) is an estimate and may vary across samples. This tutorial's calculations treat the supplied frequencies as known; they do not quantify sampling uncertainty. Interpretation also depends on the population being described: allele and haplotype frequencies, and therefore the relevant bounds, can differ between populations.

A boundary value deserves particular care. If \(|D'|=1\), one or more haplotype frequencies are at a feasible boundary, often zero. This says that a particular haplotype combination is absent at the stated frequencies. It does not establish a biological mechanism, prove that one variant causes another, or guarantee high predictive correlation.

Common Mistake: Losing the Sign for Negative \(D\)

A common error is to write \(D'=D/D_{\min}\) for \(D<0\) and then claim that the result is negative. Since both \(D\) and \(D_{\min}\) are negative, their ratio is positive. For instance, \((-0.10)/(-0.12)=+0.8333\), not \(-0.8333\). That ratio can represent the magnitude, but it is not signed D-prime under the convention here.

To retain the direction, divide by the positive magnitude of the lower bound: \(D'=D/|D_{\min}|\). Equivalently, use \(D'=-D/D_{\min}\) when \(D<0\). In the numerical case, \((-0.10)/0.12=-0.8333\). Always check that the sign of \(D'\) matches the sign of \(D\), and that its magnitude is no greater than one.

Find the allele frequencies.
Sum the haplotype frequencies carrying each allele to obtain \(p_A\) and \(p_B\).

Calculate disequilibrium.
Use \(D=f_{AB}-p_Ap_B\), then determine its sign.

Choose the matching bound.
For positive \(D\), use \(D_{\max}\). For negative \(D\), use the positive magnitude \(|D_{\min}|\).

Interpret both sign and magnitude.
The sign gives the direction of disequilibrium; the magnitude gives its fraction of the attainable bound, not necessarily its predictive correlation.

Key takeaway: Signed \(D'\) scales \(D\) by the appropriate frequency-dependent bound, using a positive denominator so its sign matches \(D\). A large \(|D'|\) indicates near-boundary disequilibrium, not automatically strong allelic correlation; examine \(r^2\) when correlation is the relevant quantity.

Check Your Understanding

Use the definitions and frequency constraints above to answer these questions.

If \(p_A=0.20\) and \(p_B=0.60\), calculate \(D_{\max}\) and \(D_{\min}\).
A hypothetical data set gives \(D=-0.04\) and \(D_{\min}=-0.05\). Calculate signed \(D'\) and explain why the denominator is positive in the signed calculation.
What does \(D'=1\) say about \(D\) relative to the feasible positive bound? What does it not guarantee about \(r^2\)?
Why must the haplotype frequencies used to calculate \(D\) be nonnegative for the frequency bounds to apply?
Explain in words why a rare allele can have \(|D'|=1\) with a common allele while their allelic correlation remains modest.