Ternary Complex Formation With Competing Targets

Why Competition Changes Ternary-Complex Abundance

A bispecific antibody can form a ternary complex when one arm binds target A and the other binds target B. In a biological system, however, a molecule of the antibody may encounter soluble molecules that bind one of those arms. Such a competitor occupies the same arm and therefore prevents that antibody molecule from binding the corresponding target at the same time. Competition on either arm can reduce the number of antibodies available to form the A–antibody–B complex.

A useful first model treats the two arms as independent binding sites at equilibrium. It makes the effect of each competitor explicit and gives a direct calculation of ternary-complex concentration. Its assumptions matter: the calculation below uses specified free concentrations of targets and competitors, not their total concentrations in a closed system.

Definition: Let D be a bispecific antibody with one arm that binds target A and one arm that binds target B. Let S_A and S_B be soluble competitors that bind the A- and B-directed arms, respectively. Define the dissociation constants $K_A$, $K_B$, $K_{S_A}$, and $K_{S_B}$ for D–A, D–B, D–S_A, and D–S_B binding. The ternary complex is DAB: one antibody bound to one A and one B.

For a reversible 1:1 interaction between D and a ligand L, the dissociation constant is $K_D=[D][L]/[DL]$, where brackets denote concentrations at equilibrium. Smaller $K_D$ means stronger binding, all else equal. In the model here, each arm can be unoccupied or occupied by its target or its soluble competitor, but cannot bind both at once. Define the dimensionless free-ligand ratios $a=[A]/K_A$, $b=[B]/K_B$, $c_A=[S_A]/K_{S_A}$, and $c_B=[S_B]/K_{S_B}$.

Binding States and the Equilibrium Model

The A-directed arm has three possible states: free, bound to A, or bound to S_A. Their relative statistical weights are $1$, $a$, and $c_A$. The corresponding weights for the B-directed arm are $1$, $b$, and $c_B$. If the arms bind independently, the weight of a joint state is the product of the weights for its two arms. Thus the full binding polynomial, the sum of the weights of all nine joint states, is

$$ Z=(1+a+c_A)(1+b+c_B) $$

The all-unbound state has weight 1. If $[D]_{\mathrm{tot}}$ is the total concentration of antibody across all states, the free antibody concentration is $[D]_{\mathrm{free}}=[D]_{\mathrm{tot}}/Z$. The DAB state has weight $ab$, so its concentration is $[D]_{\mathrm{free}}ab$. These relationships give the central result.

Result: When the free concentrations of A, B, S_A, and S_B are specified and the arms bind independently at equilibrium, $$ [DAB]=[D]_{\mathrm{tot}} \frac{a}{1+a+c_A} \frac{b}{1+b+c_B} $$ The two fractions are the probabilities that the respective arms are occupied by their targets under this equilibrium model. Their product is the fraction of antibody molecules in the ternary state.

The expression also shows how to compare a condition with competitors to one without them. If the free target concentrations are unchanged between conditions, the ratio of ternary abundance with competitors to abundance without competitors is

\frac{[DAB]_{\mathrm{with}}}{[DAB]_{\mathrm{without}}} = \frac{1+a}{1+a+c_A} \frac{1+b}{1+b+c_B}

Each factor is at most 1. A competitor has a larger effect when its concentration relative to its dissociation constant is large compared with the target’s corresponding ratio. Competition on both arms compounds: the reductions multiply, rather than add. This comparison assumes the target free concentrations and total antibody concentration are the same in both conditions.

Worked Calculations

Worked Example: Ternary Complex Without Soluble Competitors

Suppose a hypothetical equilibrium system contains 30 nM total bispecific antibody. The specified free target concentrations are $[A]=4$ nM and $[B]=3$ nM; $K_A=2$ nM and $K_B=1.5$ nM. There are no soluble competitors. Calculate $[DAB]$.

The free-ligand ratios are $a=4/2=2$ and $b=3/1.5=2$, while $c_A=c_B=0$. The independent-arm equilibrium assumption means that each arm’s target occupancy is computed from its own target ratio:

[DAB]=30\left(\frac{2}{1+2}\right) \left(\frac{2}{1+2}\right) =30\left(\frac{4}{9}\right) =13.33\ \mathrm{nM}

Thus 4/9 of the antibody is in the ternary state, giving 13.33 nM DAB. The specified target concentrations are treated as maintained free concentrations, for example in an idealized reservoir. They are not interpreted as total target concentrations in a closed system; that distinction is essential because bound target can be depleted in a closed system.

Worked Example: Competitors Occupy Both Arms

Consider a separate hypothetical system with $[D]_{\mathrm{tot}}=60$ nM, $[A]=6$ nM, $K_A=2$ nM, $[B]=4$ nM, and $K_B=2$ nM. The free competitor concentrations and dissociation constants are $[S_A]=6$ nM, $K_{S_A}=3$ nM, $[S_B]=3$ nM, and $K_{S_B}=1.5$ nM. Calculate the ternary abundance with and without competitors and the relative change.

First, $a=6/2=3$, $b=4/2=2$, $c_A=6/3=2$, and $c_B=3/1.5=2$. Without competitors, the ternary fraction is

\frac{3}{1+3}\frac{2}{1+2} =\frac{3}{4}\frac{2}{3} =\frac{1}{2}

The no-competitor abundance is $60(1/2)=30$ nM. With competitors, the target-occupancy fractions are $3/(1+3+2)=1/2$ and $2/(1+2+2)=2/5$. Therefore,

[DAB]_{\mathrm{with}} =60\left(\frac{1}{2}\right)\left(\frac{2}{5}\right) =12\ \mathrm{nM}

The ratio is $12/30=0.4000$. Under these specified free concentrations, the competitors leave 40.00% of the no-competitor ternary abundance, a 60.00% reduction. The reduction can also be checked arm by arm: $(1+3)/(1+3+2)=4/6$ on the A arm and $(1+2)/(1+2+2)=3/5$ on the B arm; their product is $(4/6)(3/5)=0.4000$. As in the model definition, these calculations presume that the stated free concentrations are maintained and that binding on one arm does not alter binding on the other.

Worked Example: A Competitor on One Arm Only

In another hypothetical system, $[D]_{\mathrm{tot}}=40$ nM, $a=0.5$, and $b=2$. A soluble competitor on the A arm has $c_A=1$, and no competitor binds the B arm, so $c_B=0$. Calculate ternary abundance before and after adding the A-arm competitor.

Without the competitor, the ternary fraction is

\left(\frac{0.5}{1+0.5}\right) \left(\frac{2}{1+2}\right) =\frac{1}{3}\frac{2}{3} =\frac{2}{9}

Using that unrounded fraction, the no-competitor concentration is $[DAB]=40(2/9)=80/9=8.888\ldots$ nM, or approximately 8.889 nM. With the competitor, the A-arm fraction is $0.5/(1+0.5+1)=0.5/2.5=1/5$; the B-arm fraction remains $2/3$. Hence,

[DAB]_{\mathrm{with}} =40\left(\frac{1}{5}\right)\left(\frac{2}{3}\right) =40\left(\frac{2}{15}\right) =\frac{16}{3} =5.333\ldots\ \mathrm{nM}

The relative abundance is $(2/15)/(2/9)=3/5=0.6000$, so this competitor leaves 60.00% of the original ternary abundance. The concentration calculation uses the exact fraction $2/15$, rather than multiplying a rounded decimal occupancy fraction.

Why the Formula Works and When to Use Mass Balance

The binding-polynomial derivation follows directly from equilibrium ratios. For example, for the A arm, the ratio of antibody bound to A to antibody with that arm free is $[A]/K_A=a$. The corresponding ratio for binding S_A is $c_A$. Taking the free-arm state to have weight 1, the three weights sum to $1+a+c_A$; therefore the fraction of A arms occupied by A is $a/(1+a+c_A)$. The same reasoning gives $b/(1+b+c_B)$ for the B arm. Under independence, the joint probability of both target-bound states is the product of these marginal fractions. Multiplying by total antibody gives the stated complex concentration.

This derivation depends on specific assumptions: binding is at equilibrium; each arm binds its target and its corresponding competitor reversibly and mutually exclusively; there is one binding site of each type per antibody; binding at one arm does not change the affinity or accessibility of the other; and the free concentrations used in the ratios are known or maintained. A separate assumption is that DAB is the state of interest: other states, including D bound to a target on one arm and a competitor on the other, contribute to total antibody but are not counted as ternary complex.

If the system is closed and ligand depletion is appreciable, the free concentrations cannot simply be set equal to the supplied total concentrations. With one binding site per ligand and the same independent-arm model, the total A balance, for example, is

[A]_{\mathrm{tot}}=[A]_{\mathrm{free}} +[D]_{\mathrm{free}}a(1+b+c_B)

The second term counts all antibody states in which the A arm binds A, whatever occupies or does not occupy the other arm. Analogous balances apply to B and both competitors; together with $[D]_{\mathrm{tot}}=[D]_{\mathrm{free}}Z$, these equations determine the free concentrations and complex abundance. In general, the coupled equations must be solved together, often numerically. Substituting total ligand concentrations into the fixed-free-concentration formula can then give an incorrect result.

Specify the binding network.
Identify which ligand binds each arm, whether each pair is mutually exclusive, and whether there is one site of each type per antibody.

Choose the concentration formulation.
Use specified free concentrations only when they are known or depletion is negligible. If totals are supplied for a closed system, write the corresponding mass balances.

Calculate and check the state fractions.
Evaluate each target fraction, multiply the fractions for DAB, and verify that the resulting concentration is between zero and $[D]_{\mathrm{tot}}$.

Diagnostics, Alternatives, and Limitations

A few checks can expose errors before interpreting the result. Every ratio $a$, $b$, $c_A$, and $c_B$ must be dimensionless: a concentration divided by a dissociation constant expressed in the same units. Each target-occupancy fraction must lie between zero and one, and $[DAB]$ cannot exceed $[D]_{\mathrm{tot}}$. Setting both competitor ratios to zero must recover the no-competitor result. Increasing a competitor concentration while holding everything else fixed must not increase ternary abundance in this model. If a calculation violates any of these checks, revisit the denominators, units, binding-site assignments, and use of free versus total concentrations.

The factorized formula is a compact alternative to listing and solving for every antibody state individually. It is appropriate when free ligand concentrations are available and arm independence is defensible. For a closed system with substantial ligand depletion, solving the mass-balance equations is the appropriate alternative. If the arms interact cooperatively, a state model with state-specific equilibrium constants or interaction factors is needed; the simple product of two occupancy fractions no longer follows. If measurements are made before equilibrium, a kinetic model with association and dissociation rates is required instead.

The calculation predicts the abundance of a molecular binding state, not a downstream biological response. It does not by itself establish whether DAB is functional, where it forms, or how long it persists. In addition, effective affinities can depend on geometry and rebinding, as considered in the earlier material on effective avidity. Treating such an effective affinity as a fixed $K_D$ is an approximation that should be checked against the binding context and data.

Common Pitfall

A common error is to calculate the no-competitor ternary fraction and then subtract the competitor occupancy from it. That reasoning treats arm occupancy changes as additive. The correct calculation uses the target fraction on each arm with its competitor in that arm’s denominator, then multiplies the two fractions. In particular, a competitor on the A arm changes the A-arm occupancy fraction; it does not directly enter the B-arm denominator under the independent-arm model. Another frequent error is to use total target concentration in place of free concentration without checking for depletion. In a closed system, this breaks the equilibrium ratios and requires the mass-balance formulation.

Key takeaway: With independent arms and specified free ligand concentrations, ternary abundance is total antibody multiplied by the A-arm target fraction and the B-arm target fraction. Soluble competitors reduce those fractions through their arm-specific terms in the denominators; depletion, cooperativity, or non-equilibrium binding requires a more detailed model.

Check Your Understanding

Answer each question using the independent-arm equilibrium model. The solutions are included below so that the calculations can be checked.

If $a=1$, $b=3$, and there are no competitors, what fraction of total antibody is DAB?
For $a=2$, $b=1$, $c_A=2$, and $c_B=0$, what is the ratio of ternary abundance with the competitor to abundance without it, assuming free targets are unchanged?
With $[D]_{\mathrm{tot}}=25$ nM, $a=1$, $b=1$, $c_A=1$, and $c_B=0$, calculate $[DAB]$.
Why is it generally incorrect to substitute total target concentration for free target concentration in a closed system?

Solutions. (1) The fraction is $(1/2)(3/4)=3/8=0.3750$. (2) The ratio is $(1+2)/(1+2+2)\times(1+1)/(1+1)=3/5=0.6000$. (3) The fraction is $(1/3)(1/2)=1/6$, so $[DAB]=25/6=4.167$ nM. (4) Binding removes free target from solution; total concentration includes both free and antibody-bound target. The equilibrium ratios require free concentration, so a closed system with depletion requires the target mass-balance equations to be solved.