Frequency Constraints and the Bounds on D

Why Frequency Constraints Bound $D$

At two biallelic loci, the marginal allele frequencies do not specify all four haplotype frequencies. They do, however, restrict them: every haplotype frequency must be at least zero, and the four frequencies must sum to one. Those basic requirements place exact lower and upper limits on linkage disequilibrium $D$.

The bounds depend on the allele frequencies at both loci. Consequently, the same numerical value of $D$ can be feasible for one pair of margins but impossible for another. The aim here is to find the full feasible range when the two marginal frequencies are fixed—not to estimate those frequencies from data or to interpret a standardized measure of LD.

Definition: Let the alleles be $A$ and $a$ at the first locus, and $B$ and $b$ at the second. Write $p_A$ and $p_B$ for the population frequencies of $A$ and $B$. Write $f_{AB}$, $f_{Ab}$, $f_{aB}$, and $f_{ab}$ for the frequencies of the four corresponding haplotypes. The linkage disequilibrium coefficient is $D=f_{AB}-p_Ap_B$. Each haplotype frequency is nonnegative, and their sum is one.

Because the loci are biallelic, the frequencies of $a$ and $b$ are $1-p_A$ and $1-p_B$. The haplotype frequencies must reproduce the margins:

\begin{aligned} f_{AB}+f_{Ab}&=p_A,\\ f_{AB}+f_{aB}&=p_B,\\ f_{AB}+f_{Ab}+f_{aB}+f_{ab}&=1. \end{aligned}

The definition of $D$ gives $f_{AB}=p_Ap_B+D$. Substituting this expression into the marginal and total-frequency equations determines each of the other three haplotype frequencies. This is the key step: all four frequencies can be checked using one quantity, $D$, once the margins are known.

\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}

Deriving the Feasible Range

A valid set of haplotype frequencies requires each expression above to be at least zero. The first and fourth expressions put lower limits on $D$; the second and third put upper limits on $D$. Both lower limits must hold, as must both upper limits. Taking the more restrictive limit on each side gives the complete bounds.

\begin{aligned} D&\geq -p_Ap_B,\\ D&\geq -(1-p_A)(1-p_B),\\ D&\leq p_A(1-p_B),\\ D&\leq (1-p_A)p_B. \end{aligned}

Theorem: For fixed marginal allele frequencies $p_A$ and $p_B$, the feasible values of $D$ are exactly $$ \max\{-p_Ap_B,\;-(1-p_A)(1-p_B)\} \leq D \leq \min\{p_A(1-p_B),\;(1-p_A)p_B\}. $$ Every value in this interval gives four nonnegative haplotype frequencies with the specified margins, and values outside it cannot do so.

The lower endpoint is the larger of the two lower limits, because $D$ must satisfy both. The upper endpoint is the smaller of the two upper limits for the same reason. These are not merely necessary inequalities: if $D$ lies between the endpoints, each of the four expressions for the haplotype frequencies is nonnegative. Their sum is one, since adding the four expressions gives

$$ p_Ap_B+p_A(1-p_B)+(1-p_A)p_B+(1-p_A)(1-p_B)=1, $$

with the $D$ terms cancelling. The first two frequencies sum to $p_A$, and the first and third sum to $p_B$. Thus the margins are preserved. This also shows why every value between the endpoints is feasible, rather than only the endpoints themselves.

Worked Examples

Worked Example: Bounds When One Allele Is Uncommon

Suppose $p_A=0.08$ and $p_B=0.70$. Find the minimum and maximum feasible values of $D$, and show haplotype frequencies at both endpoints.

Lower endpoint. The two lower limits are

-p_Ap_B=-(0.08)(0.70)=-0.056, \qquad -(1-p_A)(1-p_B)=-(0.92)(0.30)=-0.276.

The larger lower limit is $-0.056$. Upper endpoint. The two upper limits are

p_A(1-p_B)=(0.08)(0.30)=0.024, \qquad (1-p_A)p_B=(0.92)(0.70)=0.644.

The smaller upper limit is $0.024$, so the feasible range is $-0.056\leq D\leq0.024$.

At the lower endpoint, $f_{AB}=p_Ap_B+D=0.056-0.056=0$. The remaining frequencies are $f_{Ab}=0.08$, $f_{aB}=0.70$, and $f_{ab}=0.22$. They sum to $0+0.08+0.70+0.22=1$; their margins are $0+0.08=0.08$ for $A$ and $0+0.70=0.70$ for $B$.

At the upper endpoint, $f_{AB}=0.056+0.024=0.08$. The other frequencies are $f_{Ab}=0$, $f_{aB}=0.70-0.08=0.62$, and $f_{ab}=0.30$. They sum to $0.08+0+0.62+0.30=1$, and recover the margins $p_A=0.08$ and $p_B=0.08+0.62=0.70$. The marginal frequencies add to $0.78$; the haplotype frequencies, as required, add to $1$.

Worked Example: Comparing the Two Candidate Limits

Let $p_A=0.40$ and $p_B=0.30$. Calculate the bounds and verify the endpoint haplotype frequencies.

The lower candidates are $-(0.40)(0.30)=-0.12$ and $-(0.60)(0.70)=-0.42$. The larger is $-0.12$. The upper candidates are $(0.40)(0.70)=0.28$ and $(0.60)(0.30)=0.18$. The smaller is $0.18$. Therefore,

-0.12\leq D\leq0.18.

At $D=-0.12$, the four frequencies are $f_{AB}=0.12-0.12=0$, $f_{Ab}=0.40$, $f_{aB}=0.30$, and $f_{ab}=0.30$. They sum to $1$; the $A$ margin is $0+0.40=0.40$, and the $B$ margin is $0+0.30=0.30$.

At $D=0.18$, they are $f_{AB}=0.12+0.18=0.30$, $f_{Ab}=0.40-0.30=0.10$, $f_{aB}=0.30-0.30=0$, and $f_{ab}=0.60$. These sum to $1$, with margins $0.30+0.10=0.40$ and $0.30+0=0.30$. The endpoint tables confirm both feasibility and the original margins.

Worked Example: Checking a Proposed Value of $D$

Suppose $p_A=0.35$, $p_B=0.60$, and a proposed value is $D=0.10$. Determine whether it is feasible and calculate the implied haplotype frequencies.

First, $p_Ap_B=(0.35)(0.60)=0.21$, so $f_{AB}=0.21+0.10=0.31$. Using the remaining formulas gives

\begin{aligned} f_{Ab}&=0.35(0.40)-0.10=0.04,\\ f_{aB}&=0.65(0.60)-0.10=0.29,\\ f_{ab}&=0.65(0.40)+0.10=0.36. \end{aligned}

All four frequencies are nonnegative and sum to $0.31+0.04+0.29+0.36=1$. The margins are $0.31+0.04=0.35$ and $0.31+0.29=0.60$. Thus $D=0.10$ is feasible. In contrast, the upper bound here is $\min\{(0.35)(0.40),(0.65)(0.60)\}=\min\{0.14,0.39\}=0.14$; a proposed $D=0.15$ would be infeasible.

Interpreting the Endpoints

At a boundary, at least one haplotype frequency is zero. In the first worked example, the lower endpoint eliminates $AB$, while the upper endpoint eliminates $Ab$. Which haplotype reaches zero depends on the marginal allele frequencies: the most restrictive nonnegativity condition determines the endpoint.

The range is specific to the two margins. For example, the interval in the first example is asymmetric around zero: it extends to $-0.056$ below zero but only to $0.024$ above it. This does not indicate an arithmetic problem. It reflects the fact that the available haplotype-frequency combinations differ with the allele frequencies. A bound describes what is possible under the stated margins; it does not by itself establish which value occurs in a population or why.

These bounds concern population frequencies under the specified two-locus, biallelic setup. If frequencies are estimated from a sample, the same algebra can describe the constraints on a proposed set of sample haplotype frequencies, but it does not account for sampling uncertainty. Nor does a feasible value alone establish a biological mechanism or a causal relationship.

Common Mistake

A frequent error is to use only one lower or upper constraint—for example, to take $D\geq-p_Ap_B$ and $D\leq p_A(1-p_B)$ as the entire range. Those are only two of the four inequalities. The other haplotypes can impose stricter limits: in the second example, $(1-p_A)p_B=0.18$, not $p_A(1-p_B)=0.28$, sets the upper bound.

The correction is to write all four haplotype frequencies in terms of $D$, require each one to be nonnegative, and then take the larger lower limit and the smaller upper limit. Checking the resulting haplotype frequencies is a useful final safeguard: a negative frequency signals that a proposed value or a calculation is inconsistent with the margins.

Key takeaway: With fixed $p_A$ and $p_B$, the four haplotype frequencies are determined by $D$. Requiring each to be nonnegative yields a closed interval of feasible values, with lower endpoint $\max\{-p_Ap_B,-(1-p_A)(1-p_B)\}$ and upper endpoint $\min\{p_A(1-p_B),(1-p_A)p_B\}$.

Check Your Understanding

Use the four haplotype-frequency constraints to answer each question.

For fixed margins, why must both lower inequalities on $D$ hold, rather than just one?
Calculate the lower and upper bounds on $D$ when $p_A=0.20$ and $p_B=0.50$.
For $p_A=0.20$, $p_B=0.50$, and $D=0.08$, calculate all four haplotype frequencies and check their sum and margins.
What does it mean about the haplotype frequencies when $D$ is at a lower or upper endpoint?
A proposed value of $D$ makes $f_{aB}$ negative. What does that imply about the proposal, even if the other three calculated frequencies are nonnegative?

Frequency Constraints and the Bounds on D

What You'll Learn

Why Frequency Constraints Bound \(D\)

Deriving the Feasible Range

Worked Examples

Worked Example: Bounds When One Allele Is Uncommon

Worked Example: Comparing the Two Candidate Limits

Worked Example: Checking a Proposed Value of \(D\)

Interpreting the Endpoints

Common Mistake

Check Your Understanding

Frequency Constraints and the Bounds on D

What You'll Learn

Why Frequency Constraints Bound \(D\)

Deriving the Feasible Range

Worked Examples

Worked Example: Bounds When One Allele Is Uncommon

Worked Example: Comparing the Two Candidate Limits

Worked Example: Checking a Proposed Value of \(D\)

Interpreting the Endpoints

Common Mistake

Check Your Understanding

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