1.3.7: Calorimeter- Titration Microcalorimetry- Enzyme-Substrate Interaction
The technique of titration microcalorimetry was developed with the aim of probing enzyme-substrate interactions.[1,2] At the start of the experiment, the sample cell contains an aqueous solution containing a known amount of a macromolecular enzyme \(\mathrm{M}(\mathrm{aq})\). The injected aliquots contain a known amount of substrate \(\mathrm{X}\) such that during the experiment the composition of the sample cell is described in terms of the following chemical equilibrium.
\[\mathrm{M}(\mathrm{aq})+\mathrm{X}(\mathrm{aq}) \Leftrightarrow \mathrm{MX}(\mathrm{aq}) \nonumber \]
[It is important to note that the term substrate refers to the chemical substance which is bound (adsorbed?) by the macromolecular enzyme. In treatments of adsorption, the macromolecular substrate adsorbs small molecules.] In the limit of strong binding (see below), most of the injected substrate at the start of the experiment is bound by the enzyme. But gradually as more substrate is added, the number of free binding sites decreases until eventually all sites are occupied and hence no heat \(q\) is recorded. The plot of \(\left[\mathrm{q} / \mathrm{dn}_{\mathrm{X}}^{0}\right]\) against injection number in the textbook case is sigmoidal.
The equilibrium established in the sample cell is described as follows.
\[\begin{array}{llccc}
& \mathrm{M}(\mathrm{aq}) & \mathrm{X}(\mathrm{aq}) \quad<==> & \mathrm{MX}(\mathrm{aq}) & \\
\text { At } \mathrm{t}=0 & \mathrm{n}^{0}(\mathrm{M}) & \mathrm{n}^{0}(\mathrm{X}) & 0 & \mathrm{~mol} \\
\text { At } \mathrm{t}=\infty & \mathrm{n}^{0}(\mathrm{M})-\xi & \mathrm{n}^{0}(\mathrm{X})-\xi & \xi & \mathrm{mol} \\
& {\left[\mathrm{n}^{0}(\mathrm{M})-\xi\right] / \mathrm{V}} & {\left[\mathrm{n}^{0}(\mathrm{X})-\xi\right] / \mathrm{V}} & \xi / \mathrm{V} & \mathrm{mol} \mathrm{dm}
\end{array} \nonumber \]
\(\mathrm{V}\) is the volume of the sample cell. The analysis uses equilibrium constants defined in terms of the concentrations of chemical substances in the system. There are advantages in using equilibrium constants having the following form where \(\mathrm{c}_{\mathrm{r} = 1 \mathrm{~mol dm}^{-3}\).
\[\mathrm{K}=(\xi / \mathrm{V}) \, \mathrm{c}_{\mathrm{r}} /\left\{\left[\mathrm{n}^{0}(\mathrm{M})-\xi\right] / \mathrm{V}\right\} \,\left\{\left[\mathrm{n}^{0}(\mathrm{X})-\xi\right] / \mathrm{V}\right\} \nonumber \]
The latter is a quadratic in the extent of reaction, \(\xi\). [3]
\[\xi^{2}+b \, \xi+c=0 \nonumber \]
where
\[ b=-n^{0}(M)-n^{0}(X)-V \, c_{r} \, K^{-1} \nonumber \]
and
\[\mathrm{c}=\mathrm{n}^{0}(\mathrm{M}) \, \mathrm{n}^{0}(\mathrm{X}) \nonumber \]
Therefore
\[\xi=-(b / 2) \pm(1 / 2) \,\left(b^{2}-4 \, c\right)^{1 / 2} \nonumber \]
The negative root of the quadratic yields the required solution on the grounds that, with increase in \(n^{0}(\mathrm{X}\)\) in the sample cell, more substrate is bound by the enzyme. The required quantity is \(\left(\mathrm{d} \xi / \operatorname{dn}_{\mathrm{X}}^{0}\right)\). We note that from equations (g) and (h), \[\mathrm{db} / \mathrm{dn}_{\mathrm{X}}^{0}=-1 ; \quad \mathrm{dc} / \mathrm{dn}_{\mathrm{X}}^{0}=\mathrm{n}_{\mathrm{M}}^{0} \nonumber \]
In the experiment we control the ratio of total amounts of substrate to enzyme, \(\mathrm{n}^{0}(\mathrm{X}) / \mathrm{n}^{0}(\mathrm{M})\), which increases as more substrate is added to the sample cell. A measure of the ‘tightness of binding’ is the fraction of substrate bound when this ratio is unity. \[\text { By definition, } \quad X_{r}=n^{0}(X) / n^{0}(M)=\left[X_{\text {total }}\right] /\left[M_{\text {total }}\right] \nonumber \]
We define two variables \(\mathrm{r}\) and \(\mathrm{C}\); (note uppercase). \[\mathrm{V} \, \mathrm{c}_{\mathrm{r}} / \mathrm{K} \, \mathrm{n}^{0}(\mathrm{M})=\mathrm{c}_{\mathrm{r}} / \mathrm{K} \,\left[\mathrm{M}_{\text {total }}\right]=\mathrm{r}=1 / \mathrm{C} \nonumber \]
\[\text { From equation (e), }[\mathrm{X}]^{\mathrm{eq}} /[\mathrm{MX}]^{\mathrm{eq}}=\mathrm{c}_{\mathrm{r}} / \mathrm{K} \,[\mathrm{M}]^{\mathrm{eq}} \nonumber \]
If \(\mathrm{K}\) is small, then \(\mathrm{r}\) is large and only a small amount of the injected substrate is bound to the enzyme. \([\mathrm{X}]^{\mathrm{eq}}\) is, in relative terms, large and \([\mathrm{MX}]^{\mathrm{eq}\) is small. If \(\mathrm{r}\) is large, \(\mathrm{C}\) is small.
If on the other hand \(\mathrm{K}\) is large, \(\mathrm{r}\) is small, \(\mathrm{C}\) is large and in the limit all substance \(\mathrm{X}\) is bound to the enzyme \(\mathrm{M}\).
We return to equation (i) because in order to calculate \(\xi\) we require \(b\). [4] \[\mathrm{b}^{2}-4 \, \mathrm{c}=\left[\mathrm{n}^{0}(\mathrm{M})\right]^{2} \,\left[\mathrm{X}_{\mathrm{r}}^{2}-2 \, \mathrm{X}_{\mathrm{r}} \,(1-\mathrm{r})+(1+\mathrm{r})^{2}\right] \nonumber \]
We also require \(\left[\mathrm{d} \xi / \operatorname{dn}^{0}(\mathrm{X})\right]\) which describes the dependence of extent of substrate binding on total amount of \(\mathrm{X}\) in the sample cell, noting that we can control the latter through the concentration of the injected aliquots. We return to equation (i) making use of equation (n). [5] \[\frac{\mathrm{d} \xi}{\mathrm{dn}_{\mathrm{X}}^{0}}=\frac{1}{2}+\frac{\left[1-(1 / 2) \,(1+\mathrm{r})-\mathrm{X}_{\mathrm{r}} / 2\right]}{\left[\mathrm{X}_{\mathrm{r}}^{2}-2 \, \mathrm{X}_{\mathrm{r}} \,(1-\mathrm{r})+(1+\mathrm{r})^{2}\right]^{1 / 2}} \nonumber \]
The enthalpy of the solution in the sample cell (assuming the thermodynamic properties of the solution are ideal) is given by equation (p). \[\begin{aligned}
\mathrm{H}(\mathrm{aq} ; \mathrm{id})=\mathrm{n}_{1}(\lambda) \, \mathrm{H}_{1}^{*}(\lambda)+\left[\mathrm{n}^{0}(\mathrm{M})-\xi\right] \, \mathrm{H}^{\infty}(\mathrm{M} ; \mathrm{aq}) \\
&+\left[\mathrm{n}_{\mathrm{X}}^{0}-\xi\right] \, \mathrm{H}^{\infty}(\mathrm{X} ; \mathrm{aq})+\xi \, \mathrm{H}^{\infty}(\mathrm{MX} ; \mathrm{aq})
\end{aligned} \nonumber \]
\[[\partial \mathrm{H}(\mathrm{aq} ; \mathrm{id}) / \partial \xi \xi]=-\mathrm{H}^{\infty}(\mathrm{M} ; \mathrm{aq})-\mathrm{H}^{\infty}(\mathrm{X} ; \mathrm{aq})+\mathrm{H}^{\infty}(\mathrm{MX} ; \mathrm{aq})=\Delta_{\mathrm{B}} \mathrm{H}^{\infty} \nonumber \]
Hence the dependence of the ratio \(\left[\mathrm{q} / \mathrm{dn}_{\mathrm{X}}^{0}\right]\) on composition of the sample cell is given by equation (r). \[\frac{\mathrm{q}}{\mathrm{dn}_{\mathrm{X}}^{0}}=\Delta_{\mathrm{B}} \mathrm{H}^{\infty} \,\left[\frac{1}{2}+\frac{\left[1-(1 / 2) \,(1+\mathrm{r})-\mathrm{X}_{\mathrm{r}} / 2\right]}{\left[\mathrm{X}_{\mathrm{r}}^{2}-2 \, \mathrm{X}_{\mathrm{r}} \,(1-\mathrm{r})+(1+\mathrm{r})^{2}\right]^{1 / 2}}\right] \nonumber \]
The latter key equation describes the recorded enthalpogram. The latter falls into one of three general classes determined by the quantity \(\mathrm{C}\) in equation (\(\lambda\)).
- \(\mathrm{C} = \infty\). All injected substrate is bound to the enzyme over the first few injections such that the ratio \(\left[\mathrm{q} / \mathrm{dn}_{\mathrm{X}}^{0}\right]\) equals \(\Delta_{\mathrm{B}} \mathrm{H}^{\infty}\). When all sites are occupied the ratio \(\left[\mathrm{q} / \mathrm{dn}_{\mathrm{X}}^{0}\right]\) drops to zero and remains at zero for all further injections.
- \(40 < \mathrm{C} < 500\). This is the textbook case where the recorded enthalpogram has a sigmoidal shape. [6] The ratio \(\left[\mathrm{q} / \mathrm{dn}_{\mathrm{X}}^{0}\right]\) at low injection numbers is close to \(\Delta_{\mathrm{B}} \mathrm{H}^{\infty}\). The recorded enthalpogram is fitted using a non-linear least squares technique to yield estimates of \(\mathrm{K}\) and \(\Delta_{\mathrm{B}} \mathrm{H}^{\infty}\).
- \(0.1 < \mathrm{C} < 20\). Only a small fraction of the injected substrate is bound to the enzyme such that only poor estimates of \(\mathrm{K}\) and \(\Delta_{\mathrm{B}} \mathrm{H}^{\infty}\) are obtained.
Footnotes
[1] T. S. Wiseman, S. Williston, J.F.Brandts and Z.-N. Lim, Anal. Biochem., 1979, 179 ,131.
[2] M. J. Blandamer, in Biocalorimetry, ed. J. E. Ladbury and B. Z. Chowdhry, Wiley Chichester, 1998, p5.
[3] \(\mathrm{n}^{0}(\mathrm{M}) \, \mathrm{n}^{0}(\mathrm{X})-\xi \,\left[\mathrm{n}^{0}(\mathrm{M})+\mathrm{n}^{0}(\mathrm{X})\right]+\xi^{2}=\mathrm{V} \, \mathrm{c}_{\mathrm{r}} \, \mathrm{K}^{-1} \, \xi\)
[4] \[b^{2}=\left[-n^{0}(M)-n^{0}(X)-V \, c_{r} \, K^{-1}\right]^{2} \nonumber \]
\[\text { or, } b^{2}=\left[n^{0}(M)\right]^{2} \,\left[1+n^{0}(X) / n^{0}(M)+V \, c_{r} / K \, n^{0}(M)\right]^{2} \nonumber \]
\[\begin{aligned}
\mathrm{b}^{2}-4 \, \mathrm{c}=\left[\mathrm{n}^{0}(\mathrm{M})\right]^{2} \, & {\left[\mathrm{n}^{0}(\mathrm{X}) / \mathrm{n}^{0}(\mathrm{M})+1\right.} \\
&\left.+\mathrm{V} \, \mathrm{c}_{\mathrm{r}} / \mathrm{K} \, \mathrm{n}^{0}(\mathrm{M})\right]^{2}-4 \, \mathrm{n}^{0}(\mathrm{M}) \, \mathrm{n}^{0}(\mathrm{X})
\end{aligned} \nonumber \]
\[\text { Or, } b^{2}-4 \, c=\left[n^{0}(M)\right]^{2} \,\left[X_{r}+1+r\right]^{2}-4 \, n^{0}(M) \, n^{0}(X) \nonumber \]
\[\mathrm{b}^{2}-4 \, \mathrm{c}=\left[\mathrm{n}^{0}(\mathrm{M})\right]^{2} \,\left[\left(\mathrm{X}_{\mathrm{r}}+1\right)^{2}+\mathrm{r}^{2}+2 \, \mathrm{r} \,\left(\mathrm{X}_{\mathrm{r}}+1\right)\right]-4 \, \mathrm{n}^{0}(\mathrm{M}) \, \mathrm{n}^{0}(\mathrm{X}) \nonumber \]
\[\mathrm{b}^{2}-4 \, \mathrm{c}=\left[\mathrm{n}^{0}(\mathrm{M})\right]^{2} \,\left[\mathrm{X}_{\mathrm{r}}^{2}+2 \, \mathrm{X}_{\mathrm{r}}+1+\mathrm{r}^{2}+2 \, \mathrm{r} \, \mathrm{X}_{\mathrm{r}}+2 \, \mathrm{r}-4 \, \mathrm{X}_{\mathrm{r}}\right] \nonumber \]
[5] From equation (i) \[\frac{\mathrm{d} \xi}{\mathrm{dn}_{\mathrm{x}}^{0}}=-\frac{1}{2} \frac{\mathrm{db}}{\mathrm{dn}_{\mathrm{x}}^{0}}-\frac{1}{2} \, \frac{1}{2} \, \frac{1}{\left(\mathrm{~b}^{2}-4 \, \mathrm{c}\right)^{1 / 2}} \,\left[2 \, \mathrm{b} \, \frac{\mathrm{db}}{\mathrm{dn}_{\mathrm{x}}^{0}}-4 \, \frac{\mathrm{dc}}{\mathrm{dn}_{\mathrm{x}}^{0}}\right] \nonumber \]
\[\text { Or, } \quad \frac{\mathrm{d} \xi}{\mathrm{dn}_{\mathrm{X}}^{0}}=-\frac{1}{2} \frac{\mathrm{db}}{\mathrm{dn}_{\mathrm{X}}^{0}}-\frac{1}{2} \, \frac{1}{\left(\mathrm{~b}^{2}-4 \, \mathrm{c}\right)^{1 / 2}} \,\left[\mathrm{b} \, \frac{\mathrm{db}}{\mathrm{dn}_{\mathrm{X}}^{0}}-2 \, \frac{\mathrm{dc}}{\mathrm{dn}_{\mathrm{X}}^{0}}\right] \nonumber \]
We ignore for the moment the term \(\left(b^{2}-4 \, c\right)^{1 / 2}\) and concentrate attention on the two derivatives \(\frac{\mathrm{db}}{\mathrm{dn}_{\mathrm{X}}^{0}}\) and \(\frac{\mathrm{dc}}{\mathrm{dn}_{\mathrm{X}}^{0}}\); equation (j). \[\frac{\mathrm{d} \xi}{\operatorname{dn}_{\mathrm{X}}^{0}}=\frac{1}{2}-\frac{1}{2} \, \frac{1}{\left(\mathrm{~b}^{2}-4 \, \mathrm{c}\right)^{1 / 2}} \,\left[-1 \,\left\{-\mathrm{n}^{0}(\mathrm{M})-\mathrm{n}^{0}(\mathrm{X})-\mathrm{V} \, \mathrm{c}_{\mathrm{r}} \, \mathrm{K}^{-1}\right\}-2 \, \mathrm{n}^{0}(\mathrm{M})\right] \nonumber \]
\[\text { Or, } \frac{\mathrm{d} \xi}{\mathrm{dn}_{\mathrm{X}}^{0}}=\frac{1}{2}+\frac{1}{2} \, \frac{1}{\left(\mathrm{~b}^{2}-4 \, \mathrm{c}\right)^{1 / 2}} \,\left[\mathrm{n}^{0}(\mathrm{M})-\mathrm{n}^{0}(\mathrm{X})-\mathrm{V} \, \mathrm{c}_{\mathrm{r}} \, \mathrm{K}^{-1}\right] \nonumber \]
\[\begin{aligned}
&\text { Or, } \\
&\frac{\mathrm{d} \xi}{\operatorname{dn}_{\mathrm{X}}^{0}}=\frac{1}{2}+\frac{1}{2} \, \frac{1}{\left(\mathrm{~b}^{2}-4 \, \mathrm{c}\right)^{1 / 2}} \, \mathrm{n}^{0}(\mathrm{M}) \,\left[1-\mathrm{n}^{0}(\mathrm{X}) / \mathrm{n}^{0}(\mathrm{M})-\mathrm{V} \, \mathrm{c}_{\mathrm{r}} / \mathrm{K} \, \mathrm{n}^{0}(\mathrm{M})\right]
\end{aligned} \nonumber \]
\[\begin{aligned}
&\text { Or, } \\
&\frac{\mathrm{d} \xi}{\mathrm{dn}_{\mathrm{x}}^{0}}=\frac{1}{2}+\frac{1}{2} \, \frac{1}{\left(\mathrm{~b}^{2}-4 \, \mathrm{c}\right)^{1 / 2}} \, \mathrm{n}^{0}(\mathrm{M}) \,\left[2-\left\{1+\mathrm{V} \, \mathrm{c}_{\mathrm{r}} / \mathrm{K} \, \mathrm{n}^{0}(\mathrm{M})\right\}-\mathrm{n}^{0}(\mathrm{X}) / \mathrm{n}^{0}(\mathrm{M})\right]
\end{aligned} \nonumber \]
\[\begin{aligned}
&\text { Or } \\
&\frac{d \xi}{\operatorname{dn}_{\mathrm{X}}^{0}}=\frac{1}{2} \\
&+\frac{1}{\left(\mathrm{~b}^{2}-4 \, \mathrm{c}\right)^{1 / 2}} \, \mathrm{n}^{0}(\mathrm{M}) \,\left[1-(1 / 2) \,\left\{1+\mathrm{V} \, \mathrm{c}_{\mathrm{r}} / \mathrm{K} \, \mathrm{n}^{0}(\mathrm{M})\right\}-\left\{\mathrm{n}^{0}(\mathrm{X}) / 2 \, \mathrm{n}^{0}(\mathrm{M})\right\}\right]
\end{aligned} \nonumber \]
Now consider the term \(\left(b^{2}-4 \, c\right)^{1 / 2}\). From equation (n) \[b^{2}-4 \, c=\left[n^{0}(M)\right]^{2} \,\left[X_{r}^{2}-2 \, X_{r} \,(1-r)+(1+r)^{2}\right] \nonumber \]
\[\text { But } \frac{\mathrm{d} \xi}{\mathrm{dn}_{\mathrm{x}}^{0}}=\frac{1}{2}+\frac{1}{\left(\mathrm{~b}^{2}-4 \, \mathrm{c}\right)^{1 / 2}} \, \mathrm{n}^{0}(\mathrm{M}) \,\left[1-(1 / 2) \,(1+\mathrm{r})-\mathrm{X}_{\mathrm{r}} / 2\right] \nonumber \]
[6] J. E. Ladbury and B. Z.Chowdhry, Chemistry and Biology, 1996, 3 , 791.