Question

Many biological reactions are very sensitive to \(\mathrm{pH}\). This can readily be incorporated into the rate laws because protolytic reactions can be assumed to be much faster than other rates in most cases. For example, in enzyme mechanisms the ionization states of a few key protein side chains are often critical. Suppose that two ionizable groups on the enzyme are critical for catalytic activity and that one of them needs to be protonated and the other deprotonated. The protolytic reactions can be written as $$ \mathrm{EH}_{2} \rightleftharpoons \mathrm{EH}+\mathrm{H}^{\prime} \rightleftharpoons \mathrm{E}+2 \mathrm{H}^{\prime} $$ If only the species \(\mathrm{EH}\) is catalytically active and the protolytic reactions are much more rapid than the other steps in the reaction, all of the rate constants that multiply the free enzyme concentration in the rate law have to be multiplied by the fraction of enzyme present as EH. A. Calculate the fraction of free enzyme present as EH at a given pH. Your answer should contain the concentration of \(\mathrm{H}^{+}\)and the ionization constants of the two side chains, \(\left.K_{\mathrm{E} 1}=[\mathrm{E}]\left[\mathrm{H}^{+}\right] / \mathrm{EH}\right]\) and \(\left.K_{\mathrm{E} 2}=[\mathrm{EH}]\left[\mathrm{H}^{+}\right] / \mathrm{EH}_{2}\right]\). B. Assume that ES in the Michaelis-Menten mechanism (Eq. 5-2) also exists in three protonation states, \(\mathrm{ESH}_{2}\), ESH, and ES, with only ESH being catalytically active. Calculate the fraction of the enzyme-substrate complex present as EHS. Designate the ionization constants as \(K_{\mathrm{ESI}}\) and \(K_{\mathrm{ES} 2}\). C. Use the results of parts \(\mathrm{A}\) and \(\mathrm{B}\) to derive equations for the \(\mathrm{pH}\) dependence of \(V_{\mathrm{m}^{\prime}} K_{\mathrm{M}}\), and \(V_{\mathrm{m} /} K_{\mathrm{M}}\). Measurement of the \(\mathrm{pH}\) dependence of the steady. state parameters permits determination of the ionization constants, and sometimes identification of the amino acid side chains.

Short answer

Fraction of EH: \( f_{\mathrm{EH}} = \frac{1}{1 + \frac{[\mathrm{H}^+]}{K_{\mathrm{E}1}} + \frac{K_{\mathrm{E}2}}{[\mathrm{H}^+]}} \). Fraction of ESH: \( f_{\mathrm{ESH}} = \frac{1}{1 + \frac{[\mathrm{H}^+]}{K_{\mathrm{ES1}}} + \frac{K_{\mathrm{ES2}}}{[\mathrm{H}^+]}} \).

Step-by-step solution

Understand the System of Equilibrium

We are given a sequence of equilibria involving the enzyme in different protonation states: \( \mathrm{EH}_2 \rightleftharpoons \mathrm{EH} + \mathrm{H}^+ \rightleftharpoons \mathrm{E} + 2 \mathrm{H}^+ \). \( \mathrm{EH} \) is catalytically active, meaning we need to find its fraction relative to other states.

Write Equilibrium Expressions for Protolytic Reactions

Using the ionization constants given, we have: - \( K_{\mathrm{E}1} = \frac{[\mathrm{E}][\mathrm{H}^+]}{[\mathrm{EH}]} \)- \( K_{\mathrm{E}2} = \frac{[\mathrm{EH}][\mathrm{H}^+]}{[\mathrm{EH}_2]} \).We need these to express concentrations in terms of \([\mathrm{EH}]\).

Convert to Fractional Concentration of EH

Given that all enzyme must be in one of the three forms, the total concentration of enzyme \( [E_T] = [\mathrm{E}] + [\mathrm{EH}] + [\mathrm{EH}_2] \). We express \([\mathrm{E}]\) and \([\mathrm{EH}_2]\) using the equilibrium constants and calculate the fraction of enzyme as EH: \[ f_{\mathrm{EH}} = \frac{[\mathrm{EH}]}{[E_T]} = \frac{1}{1 + \frac{[\mathrm{H}^+]}{K_{\mathrm{E}1}} + \frac{K_{\mathrm{E}2}}{[\mathrm{H}^+]}}. \]

Repeat Equilibrium Process for ES Complex

For ES, the sequence is \( \mathrm{ESH}_2 \rightleftharpoons \mathrm{ESH} + \mathrm{H}^+ \rightleftharpoons \mathrm{ES} + 2\mathrm{H}^+ \). We apply similar equilibrium expressions:- \( K_{\mathrm{ES1}} = \frac{[\mathrm{ES}][\mathrm{H}^+]}{[\mathrm{ESH}]} \)- \( K_{\mathrm{ES2}} = \frac{[\mathrm{ESH}][\mathrm{H}^+]}{[\mathrm{ESH}_2]} \).

Calculate Fraction of ESH

Using the total enzyme-substrate complex concentration \( [ES]_T = [\mathrm{ES}] + [\mathrm{ESH}] + [\mathrm{ESH}_2] \), the fraction of active ES is:\[ f_{\mathrm{ESH}} = \frac{[\mathrm{ESH}]}{[ES]_T} = \frac{1}{1 + \frac{[\mathrm{H}^+]}{K_{\mathrm{ES1}}} + \frac{K_{\mathrm{ES2}}}{[\mathrm{H}^+]}}. \]

Integrate Fractions into Kinetic Equations

The fractions affect the rate constants in the Michaelis-Menten equation. Thus, the \( V_{\mathrm{m}} \) and \( K_{\mathrm{M}} \) become functions of pH:\[ V_{\mathrm{m}} K_{\mathrm{M}} = (f_{\mathrm{EH}})(V_{\mathrm{m}^{\prime}}) (K_{\mathrm{M}}), \]\[ \frac{V_{\mathrm{m}}}{K_{\mathrm{M}}} = f_{\mathrm{ESH}} \cdot \frac{V_{\mathrm{m}}}{K_{\mathrm{M}}}. \]These allow determination of ionization constants and identification of specific amino acids.

Key concepts

Enzyme Catalysis

Enzyme catalysis is a fascinating process that enables the acceleration of biochemical reactions in the body. Enzymes are special proteins that act as biological catalysts, speeding up reactions by lowering the activation energy. This makes processes happen at a significantly faster rate than they would without the enzyme.
Enzymes work by binding to a substrate—the molecule upon which the enzyme acts—and converting it into a product. The specific region of the enzyme where this binding occurs is called the active site, and the shape and chemical environment here are uniquely suited to facilitate a particular reaction. The substrate fits into the active site much like a key fits into a lock, allowing the enzyme to precisely target what it is supposed to catalyze.

• Enzymes are highly specific, meaning they only catalyze particular reactions related to their unique active sites.
• They work under mild conditions of temperature and pH typical of biological systems.
• Catalytic efficiency is often affected by the concentration of enzymes and substrates, as well as by environmental conditions.

This process can be influenced by mechanisms such as temperature changes and pH levels, where the newer emphasis is on understanding how protolytic reactions can alter enzymatic activity by affecting protonation states.

Protolytic Reactions

Protolytic reactions involve the transfer of protons (H⁺ ions) between the reactant molecules, significantly influencing the reaction rates. They are essential in maintaining the delicate balance of biological systems since many enzymes rely on the protonation state for their function.
The ionization of molecules results in different forms according to the pH level. In biological systems, these states can significantly alter enzyme activity, as seen in the problem where EH is catalytically active only when correctly protonated. The equations for these reactions show that the presence of H⁺ ions, and thus the pH, is critical for determining the form that is present.
In the exercise, the following ionization equilibria are noted:

• \( \mathrm{EH}_2 \rightleftharpoons \mathrm{EH} + \mathrm{H}^+ \)
• \( \mathrm{EH} \rightleftharpoons \mathrm{E} + 2\mathrm{H}^+ \)

By examining the fractions of each state in the equilibrium expressions, it's clear that protolytic reactions are vital for enzyme catalysis as they control which forms of the molecular species are present and active under specific conditions.

Equilibrium Constant

In the context of biological thermodynamics, the equilibrium constant plays a vital role. It gives insight into the proportion of each reactant present in a reversible reaction when it has reached equilibrium. More specifically, it provides a measure of the favorability of specific states in protonation reactions, as seen in enzyme mechanisms.
For example, in the given exercise:

• The equilibrium constant \( K_{\mathrm{E}1} \) is expressed as:\[ K_{\mathrm{E}1} = \frac{[\mathrm{E}][\mathrm{H}^+]}{[\mathrm{EH}]}\]
• Another, \( K_{\mathrm{E}2} \), is expressed as:\[ K_{\mathrm{E}2} = \frac{[\mathrm{EH}][\mathrm{H}^+]}{[\mathrm{EH}_2]}\]

These expressions help predict the concentration of active enzyme forms like EH in different protonation states, showing how critical the equilibrium constant is in determining these distributions.
By understanding the equilibrium constant, scientists and students can predict how changes in conditions, such as pH, will shift the balance of these reactions. Essentially, the equilibrium constant is a bridge connecting the chemical conditions of the environment with the biochemical functionality of enzymes.

Enzyme Kinetics

Enzyme kinetics is all about understanding the rate at which enzymatic reactions proceed. The study of enzyme kinetics allows us to assess the efficiency and speed of biological reactions catalyzed by enzymes.
The Michaelis-Menten model, often illustrated in enzyme kinetics, describes how the reaction rate is affected by substrate concentration and the enzyme's affinity for the substrate.
The key parameters in this model are the maximum reaction rate ( \( V_{\mathrm{max}} \)) and the Michaelis constant ( \( K_M \)).

• \( V_{\mathrm{max}} \) represents the rate of reaction when the enzyme is saturated with substrate.
• \( K_M \) is the substrate concentration at which the reaction velocity is half of \( V_{\mathrm{max}} \).

The pH-dependence described in the exercise highlights how protonation states can alter \( V_{\mathrm{max}} \) and \( K_M \), thus affecting the overall reaction velocity within organisms.
By understanding these dynamics, it becomes easier to predict and manipulate the conditions under which an enzyme operates optimally. This ability is crucial for efforts ranging from drug design to the enhancement of biotechnological processes.