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# How to balance molecular equations

### Derivations of Binding Equations and Graphs

These derivations can be made and interpreted using simple principles from General Chemistry, which you reviewed and strengthened in Analytical Chemistry. with some slight differences. Biochemists rarely talk about equilibrium or association constants ($$K_a$$), but rather their reciprocal: the dissociation constants, $$K_d$$. For the reverisble reaction $$M + L \rightleftharpoons ML$$, where $$M$$ is free macromolecule, $$L$$ is free ligand, and $$ML$$ is macromolecule-ligand complex (which is held together by intermolecular forces, not covalent forces):

$K_d= \dfrac <[M]_[L]_><[ML]_>$

Figure 1: $$M$$ is free macromolecule, $$L$$ is free ligand, and $$ML$$ is macromolecule-ligand complex

Notice the unit of $$K_d$$ is molarity, M. The lower the $$K_d$$ (i.e. the higher the [ML] at any given $$M$$ and $$L$$), the tighter the binding. The higher the $$K_d$$, the looser the binding. $$K_d$$ values for biological molecules are finely tuned to their environments and vary from about 1 mM (weak interactions) for some enzyme-substrate complexes, to pM to fM levels. Examples of very tight, non-covalent interactions include the avidin (an egg protein)-biotin (a vitamin) and thrombin (enzyme initiating clotting)-hirudin (a leech salivary protein) complexes.

For a simple macromolecule/ligand equilibrium:

$M + L \rightleftharpoons ML$

with

$$M$$ = free macromolecule, $$L$$ = free ligand, and $$ML$$ = bound complex of $$M$$ and $$L$$

Three equations can be then written:

Equation 1 - Dissociation constant (units of molarity):

Equation 2 - Mass Balance of $$M$$:

$M_o = M + ML$

Equation 3 - Mass Balance of $$L$$:

$L_o = L + ML$

We would like to derive equations which give $$ML$$ as a function of known or measurable values. The $$K_d$$ equations shows that $$ML$$ depends on both free $$M$$ and free $$L$$. From Equations 1-3, two different and equally valid equations can be derived for two different cases.

Case 1: used either when you can readily measure free $$L$$ or when experimental conditions are such the $$L_o \gg M_o$$, which is often encountered. Under these latter conditions, free $$L = L_o$$, which you know without measuring it, simply by knowing how much total ligand was added to the system. Case 2 (more general): used when you don't know free L or haven't measured it, and you just wish to calculate how much ML is present at equilibrium. These conditions imply that $$L_o$$ is not $$\gg M_o$$ (If $$L_o \gg M_o$$, we would know free $$L = L_o$$.)

### Experimental Case 1 (Approximate)

USE THIS FORM OF THE EQUATION WHEN L IS MEASURABLE OR WHEN $$L_o \gg M_o$$ (i.e. $$L= L_o$$)

Equation 4 - Substitute 2 into 1:

$[ML]K_d = M_o[L] - [ML][L]$

$[M_L]K_d + [ML][L] = M_o[L]$

$[M_L](K_d+[L]) = M_o[L]$

$ML = \dfrac \tag<5>$

This equation is ALWAYS TRUE for the chemical equation written above with $$L$$ representing the free ligand concentration at equilibrium. If $$L_o \gg M_o$$, then $$L \approx L_o$$ and equation 5 simplifies to:

$ML = \dfrac, \tag<6>$

whose graph is drawn below. Dividing Equation 5 by $$M_o$$ gives the fractional saturation of the macromolecule $$M$$, where

$Y = q = \dfrac<[ML]> = \dfrac \tag<7>)$

where $$Y$$ can vary from 0 (when $$L$$ = 0) to 1 (when $$L \gg K_d$$)

Graphs of $$ML$$ vs. $$L$$ (equation 5) and $$ML$$ vs. $$L_o$$ (equation 6), when $$L_o \gg M_o$$, and $$Y$$ vs. $$L$$ (equation 7) are all HYPERBOLIC functions.

Equations 5 and by analogy Equations 6 and 7 can be understood best by examining three limiting cases:

• Case 1. $$L = 0$$, $$ML = 0$$
• Case 2. $$L = K_d$$, $$ML = \frac<(L + L)>= \frac<2l> = \frac<2>$$, which indicates that $$M$$ is half saturated. In fact, the operational definition of $$K_d$$ is the ligand concentration at which $$M$$ population is half saturated.
• Case 3. $$L \gg K_d$$, $$ML$$ = $$M_o$$

### Experimental Case 2 (More General)

USE THIS FORM OF THE EQUATION WHEN FREE L IS UNKNOWN (such as when $$L_o$$ is not $$\gg M_o$$) OR YOU WISH TO CALCULATE $$M_L$$ FROM JUST $$L_o$$, $$M_o$$ AND $$K_d$$

Equation 8 - Substitute 2 AND 3 into equation 1:

$[ML]K_d = ([M_o] - [ML])([L_o] - [ML])$

$[M_L]K_d = [M_o][L_o] - [ML][L_o] - [ML][M_o] + [ML]^2$

or Equation 9:

$M_L]^2 - (L_o + M_o +K_d)[ML] + [M_o][L_o] = 0$

which is of the quadratic form $$ax^2 + bx + c = 0$$, where

a = 1 b = $$- ([L_o] + [M_o] +K_d)$$ c = $$[M_o][L_o]$$

which are all constants, and

A graph of ML calculated from this formula vs free L (or Lo if $$L_o \gg M_o$$) give a A HYPERBOLA

In the derivations, we came up with two equations for ML:

one (Equation 5) using mass conservation on $$M$$, which gave: $ML = \frac<[K_d +L]\) one (Equation 10) using mass conservation on $$M$$ and L, which gave ML = quadratic equation as function of $$M_o$$, $$L_o$$, and $$K_d$$: $$ML = <-(L_o+M_o+K_d) +/- ((L_o+M_o+K_d)^2 - 4M_oL_o)1/2>/2$$ Both equations are valid. In the first you must known free $$L$$, which is often $$L_o$$ if $$M_o \ll L_o$$. In the second, you do not need to know free M or L at all. At a given $$L_o$$, $$M_o$$, and $$K_d$$, you can calculate ML, which should be the same ML you get from the first equation if you know free L. Equations 5 and 10 are useful in several circumstances. They can be used to calculate the concentration of $$M_L$$ if $$K_d$$, $$M_o$$, and $$L$$ (for equation 5) or if $$K_d$$, $$M_o$$, and $$L_o$$ (for equation 10) are known. This is analogous to the use of the Henderson-Hasselbach approximation to calculate the protonation state (HA) and hence charge state of a weak acid at various pH values. In the former case we are measuring the concentration of bound ligand (ML) and in the later case, the concentration of bound protons (HA). calculate $$K_d$$ if $$ML$$, $$M_o$$, and $$L$$ (for equation 5) or if $$ML$$, $$M_o$$, and $$L_o$$ (for equation 10) are known. Techniques to extract the $$K_d$$ from binding data will be discussed in the next chapter section. ### Interpretation of Binding Equations and Graphs It is important to get a mathematical understanding of the binding equations and graphs. It is equally important to get an intuitive understanding of their properties. Just as we used the +/- 2 rule in determining at a glance the charge state of an acid, you need to be able to determine the extent of binding (how much of $$M$$ is bound with $$L$$) given their relative concentrations and the $$K_d$$. The usual situation is that $$[M_o] \ll [L_o]$$. What happens to the binding curves for $$M + L \rightleftharpoons ML$$ if the $$K_d$$ gets progressively lower? Intuitively, you should expect that binding will increase, especially as $$L$$ gets greater. The curves below should help you develop the intuition you need with respect to binding equilibria. Fig: ML vs. L at Even Lower $$K_d$$'s Fig: ML vs. L at a Very Low $$K_d$$! Note that in the last graph, given the same Mo and Lo concentrations, the "titration curves" for a binding equilibrium characterized by even tighter binding (for example, a $$K_d$$ = 0.5 pM or 0.05 pM) would be indistinguishable from the graph when $$K_d$$ = 5 pM. It should be apparent that for all of these $$K_d$$ values, all of the added ligand is bound until $$[L_o] > [M_o]$$. To differentiate these cases, much lower ligand concentrations would be required such that on addition of ligand, all is not bound. Also note that this curve is NOT hyperbolic, which makes sense since the graph is of $$Y$$ vs. $$L_o$$, not $$Y$$ vs. $$L$$, and since $$L_o$$ is not $$\gg M_o$$. It is quite interesting to compare graphs of Y (fractional saturation) vs. L (free) and $$Y$$ vs. $$L_o$$ (total L) in the special case when $$L_o$$ is not $$\gg M_o$$. Examples are shown below when Mo = 4 mM, $$K_d$$ = 0.19 mM. Under the ligand concentration used, it should be apparent the L can't be approximated by Lo. Two points should be evident from these graphs when L is not approximated by Lo: a graph of Y vs. Lo is not truly hyperbolic, but it does saturate a $$K_d$$ value (ligand concentration at half-maximal binding) can not be estimated by inspection from the Y vs. Lo, but it can be from the Y vs. L graph. ### A Special Case: Macromolecule Dimerization In the previous examples, we considered the case of a macromolecule M binding a ligand L at a single site, as described in the equation below: \[M + L \rightleftharpoons ML$

with $$K_d = \frac<[M][L]><[ML]>$$

We saw that the binding curves ($$ML$$ vs. $$L$$ or $$Y$$ vs. $$L$$) are hyperbolic, with a $$K_d = L$$ at half maximal binding.

A special, yet common example of this equilibrium occurs when a macromolecule binds itself to form a dimer, as shown below:

$M + M \rightleftharpoons M^2 = D$

where

$$D$$ is the dimer, and where

$K_d = \dfrac<[M][M]><[D]> = \dfrac<[M]^2><[D]> \;\;\;\; (11)$

At first glance you would expect a graph of $$[D]$$ vs. $$[M]$$ to be hyperbolic, with the $$K_d$$ again equaling the [M] at half-maximal dimer concentration. This turns out to be true, but a simple derivation is in order since in the previous derivation, it was assumed that $$M_o$$ was fixed and $$L_o$$ varied. In the case of dimer formation, $$M_o$$, which superficially represents both $$M$$ and $$L$$ in the earlier derived expression, are both changing.

One again a mass balance expression for $$M_o$$ can be written:

$[M_o] = [M] + 2[D] \;\;\;\; (12)$

where the coefficient 2 is necessary since their are 2 $$M$$ in each dimer.

More generally, for the case of formation of trimers (Tri), tetramers (Tetra), and higher oligomers,

$[M_o] = [M] + 2[D] + 3[Tri] + 4[Tetra] +. \;\;\;\; (13)$

Rearranging Equation 12 and solving for M gives

$[M] = [M_o] -2[D] \;\;\;\; (14)$

Substituting (14) into the $$K_d$$ expression (1) gives

where can be rearranged into quadratic form:

$4D2 - (4M_o+K_d)D + (M_o)^2 = 0 \;\;\;\; (15)$

which is of the form $$y = ax^2+bx+c$$. Solving the quadratic equation gives [D] at any given $$[M_o]$$. A value $$Y$$, similar to fractional saturation, can be calculated, where $$Y$$ is the fraction of total possible D, which can vary from 0 to 1.

$Y = \dfrac<2[D]><[M_o]> \;\;\;\; (16).$

A graph of $$Y$$ vs $$M_o$$ with a dimerization dissociation constant $$K_d$$ = 25 μM, is shown below.

Note that the curve appears hyperbolic with half-maximal dimer formation occurring at a total M concentration $$M_o = K_d$$. Also note, however, that even at $$M_o$$ = 1000 μM, which is 40-fold greater than $$K_d$$, only 90% of the total possible D is formed (Y = 0.90). For the simple $$M + L \rightleftharpoons ML$$ equilibrium, if $$L_o$$ = 40x the $$K_d$$ and $$M_o \ll L_o$$,

$Y = \dfrac = \dfrac<(L/40)+L> = 0.976$

The aggregation state of a protein monomer is closely linked with its biological activity. For proteins that can form dimers, some are active in the monomeric state, while others are active as a dimer. High concentrations, such as found under conditions when protein are crystallized for x-ray structure analysis, can drive proteins into the dimeric state, which may lead to the false conclusion that the active protein is a dimer. Determination of the actual physiological concentration of [Mo] and $$K_d$$ gives investigators knowledge of the $$Y$$ value which can be correlated with biological activity. For example, interleukin 8, a chemokine which binds certain immune cells, exists as a dimer in x-ray and NMR structural determinations, but as a monomer at physiological concentrations. Hence the monomer, not the dimer, binds its receptors on immune cells. Viral proteases (herpes viral protease, HIV protease) are active in dimeric form, in which the active site is formed at the dimer interface.

### How Strong Is Binding - The Continuum

Binding affinities give us a way to measure the relative strength of binding between two substances. But how "tight" is tight binding? Weak binding? Let us exam that issue by considering a binding continuum. Consider two substances, A and B that might interact. Over what range of strengths can they actually bind to each other? It would helpful to set up the extremes of the binding continuum. At one end is no binding at all. At the other end, consider two things that bind covalently. We have discussed how $$K_d$$ reflects binding strength. Remember, $$K_d = \frac<1>>$$. Also, we know that $$K_$$ is related to $$\Delta G_o$$, by the equations: $$\Delta G_o = - RT \ln K_ = RT \ln K_d$$. Given these simple equations, you should be able to interconvert between $$K_$$, $$K_d$$, and $$\Delta G_o$$. (Keep your units straight.).

NO INTERACTION: One end of the binding continuum represents no interaction. Let's assume that $$K_$$ is tiny (Kd large), for example $$K_$$

2.4 10 -72. Plugging this into the thermodynamic equation $$\Delta\,Go = -RT\, \ln K_$$, where R = 2.00 cal/mol.K, and T is about 300K, the $$\Delta\,G)_o$$

+100 kcal/mol. That is, if we add A + B, there is no drive to form AB. If AB did form, then it would immediately fall apart. COVALENT INTERACTION. At the other end of the continuum consider the interaction of 1H atom with another to form H2. From a general chemistry book we can get $$\Delta G_o$$ form. Using General Chem. thermodynamics, we can calculate $$\Delta G_o$$ for H-H formation. ( $$\delta G_o = \Sigma G_o$$ from prod. $$- \Sigma G_o$$ from react.) Doing this gives a value of -97 kcal/mol. SPECIFIC AND NONSPECIFIC BINDING. Consider the interaction of a protein, the lambda repressor (R), with a small oligonucleotide to which it binds tightly (called the operator DNA, O). This is an example of a biologically tight, but reversible interaction. R can bind to many short oligonucleotides due to electrostatic interactions and H bonds from the positively charged protein to the negatively charge nucleic acid backbone. The tight binding interaction, however, involves oligonucleotides of specific base sequence. Hence we can distinguish between tight binding, which usually involves specific DNA sequence and weak binding which involves nonspecific sequences. Likewise, we will speak of specific and nonspecific binding. R and O, which bind with a $$K_d$$ of 1 pM, is an example of specific binding, while R and nonspecific DNA (D), which bind mostly through electrostatic interactions with a $$K_d$$ of 1 mM, is an example of nonspecific binding. You might expect any positively charged protein, like mitochondrial cytochrome C, would bind negatively charged DNA. This nonspecific interaction would have no biological significance since the two are localized in different compartments of the cell. In contrast, the interaction between positively charged histone proteins, bound to DNA in the nucleus, would be specific.

#### Rate Constants for Association and Dissociation

When the reaction $$M + L \rightleftharpoons ML$$ is in a (dynamic) equilibrium, the rate of the forward reaction is equal to the rate of the reverse reaction. From General Chemistry, the forward reaction is biomolecular and second order. Hence the $$v_f$$, the rate in the forward direction is proportional to [M][L], or $$v_f = k_f [M][L]$$, where $$k_f$$ is the rate constant in the forward direction. The rate of the reverse reaction, $$v_r$$ is first order. proportional to [ML], and is given by $$v_r = k_r [ML]$$, where $$k_r$$ is the rate constant for the reverse reaction. Notice that the units of $$k_f$$ are M -1 s -1. while units of $$k_r$$ are s -1 .

At equilibrium, $v_f = v_r$ or $k_f [M][L] = k_r [ML].$

Rearranging the equation gives

$\dfrac<[ML]><[M][L]> = \dfrac = K_.$

Hence $$K_$$ is given by the ratio of rate constants. For tight binding interactions, $$K_ \gg 1$$, $$K_d \ll 1$$, and $$k_f$$ is very large (in the order of 10 8-9 ) and $$k_r$$ must be very small (10 -2 - 10 -4 s -1 ).

To get a more intuitive understanding of $$K_d$$'s, it is often easier to think about the rate constants which contribute to binding and dissociation. Let us assume that kr is the rate constant which describes the dissociation reaction. It is often times called $$k_$$. It can be shown mathematically that the rate at which two simple object associate depends on their radius and effective molecular weight. The maximal rate at which they will associate is the maximal rate at which diffusion will lead them together. Let us assume that the rate at which M and L associate is diffusion limited. The theoretical $$k_$$ is about 10 8 M -1 s -1. Knowing this, the $$K_d$$ and the fact that $\dfrac>> = K_ = \dfrac<1>$, we can calculate $$k_$$, which remember is a first order rate constant..

We can also determine $$k_$$ experimentally. Imagine the following example. Adjust the concentrations of M and L such that $$M_o \ll L_o$$ and $$L_o \gg K_d$$. Under these conditions of ligand excess, $$M$$ is entirely in the bound from, $$ML$$. Now at t = 0, dilute the solution so that $$L_o \ll K_d$$. The only process that will occur here is dissociation, since negligible association can occur given the new condition. If you can measure the biological activity of $$ML$$, then you could measure the rate of disappearance of ML with time, and get $$k_$$. Alternatively, if you could measure the biological activity of $$M$$, the rate at which activity returns will give you $$k_$$.

Now you will remember from General Chemistry that from a first order rate constant, the half-life of the reaction can be calculated by the expression: $k = \dfrac<0.693>>.$

Hence given $$k_$$, you can determine the $$t_<1>$$ for the associated species existence. That is, how long will a complex of $$ML$$ last before it dissociates. Given $$\Delta G_o$$ or $$K_d$$, and assuming a $$k_$$ (10 8 M -1 s -1 ), you should be able to calculate $$k_$$ and $$t_<1>$$. Or, you could be able to determine $$k_$$ experimentally, and then calculate $$t_<1>$$. Applying these principles, you can calculate the parameters below.

Calculated $$k_$$ and t1/2 for binary complexes assuming diffusion-controlled $$k_$$

http://biowiki.ucdavis.edu/Biochemistry/Binding/Reversible_Binding%3A_1._Equations_and_Curves how to balance molecular equations

Source: biowiki.ucdavis.edu
Category: Forex