Another consequence of tight binding (problem 9 ) is the free energy change for the binding process. Calculate \(\Delta G^{\circ}\) ' for an equilibrium with a \(K_{\mathrm{D}}\) of \(10^{-27} M .\) Compare this value to the free energies of the noncovalent and covalent bonds with which you are familiar. What are the implications of this number, in terms of the nature of the binding of a transition state to an enzyme active site?

Short Answer

Expert verified
1) The calculated \(\Delta G^{\circ}\) gives us an indication of the energy change during the binding process. 2) By comparison with known bond energies, we can infer whether the transition state binding to the enzyme active site resembles a covalent or a noncovalent bond or perhaps a bond strength in-between. 3) The implications of this analysis will help understand the nature, strength and potential longevity of the binding.

Step by step solution

01

Calculation of Free Energy Change (\(\Delta G^{\circ}\))

Use the formula \(\Delta G^{\circ} = -RT lnK_{\mathrm{D}}\), where \(R = 8.314 J/(mol·K)\) is the universal gas constant and \(T\) is the temperature in Kelvin. Assuming room temperature (298K), we substitute the values into the formula.
02

Comparison of \(\Delta G^{\circ}\) with known bond energies

The calculated \(\Delta G^{\circ}\) is compared with the free energies of known types of bonds. Normally, noncovalent bonds have \(\Delta G^{\circ}\) in the range of 20-40 kJ/mol while the energy for covalent bonds is usually higher than 150 kJ/mol.
03

Analysing the implications

Based on the calculated \(\Delta G^{\circ}\) and comparison, an inference is drawn about the nature of binding of a transition state to an enzyme active site.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Free Energy Change Calculation
Understanding the free energy change calculation is essential for various biochemical processes, including enzyme-substrate interactions. Free energy change, represented by \textbf{\(\Delta G^{\circ}\)}, is a quantitative measure of the spontaneity of a process or reaction at constant temperature and pressure.

Free energy change is pivotal in determining the equilibrium constant, \textbf{\(K_{\mathrm{D}}\)}, which is a measure of the strength of the interaction between an enzyme and its substrate. The formula \textbf{\(\Delta G^{\circ} = -RT lnK_{\mathrm{D}}\)} uses the universal gas constant \textbf{\(R\)} and the absolute temperature \textbf{\(T\)} to calculate this free energy. At room temperature (298 K), incorporating \textbf{\(K_{\mathrm{D}}\)} into this equation gives a concise description of how favorable the binding is.

For a highly favorable binding, such as a substrate with a low dissociation constant (\textbf{\(K_{\mathrm{D}}\)}) of \(10^{-27} M\), the resultant \textbf{\(\Delta G^{\circ}\)} would be significantly negative, indicating a strong and spontaneous binding process.
Noncovalent and Covalent Bonds
Bonds play a fundamental role in the molecular interactions that define biological processes. Noncovalent bonds, including hydrogen bonds, ionic bonds, van der Waals interactions, and hydrophobic interactions, are typically weaker than covalent bonds.

Free energies (\(\Delta G^{\circ}\)) for noncovalent bonds usually range between 20-40 kJ/mol, which allows them to be dynamic and reversible—qualities essential for biological regulation and signaling. These bonds are crucial for maintaining the three-dimensional structure of proteins and facilitating enzyme-substrate interactions.

In contrast, covalent bonds are much stronger, with free energies greater than 150 kJ/mol. These bonds form the backbone of organic molecules and create stable structures necessary for the integrity and function of biomolecules. While necessary for stability, their strength requires a significantly more energy to form or break, making them less versatile in transient biological interactions.
Transition State Theory
Transition state theory is central to understanding chemical reactions, including those catalyzed by enzymes. According to this theory, reactants must pass through a high-energy intermediate state—the transition state—before converting to products.

The enzyme stabilizes this transition state, reducing the activation energy required for the reaction to proceed. The tight binding of the transition state, as reflected in the problem's extremely low \textbf{\(K_{\mathrm{D}}\)} value, suggests that enzymes are incredibly efficient at recognizing and stabilizing the transition state compared to the normal substrate or product. This selective binding is vital for enzyme activity and catalysis.

The calculated large negative \textbf{\(\Delta G^{\circ}\)} value implies a very stable, enzyme-bound transition state, contributing to the enzyme's catalytic power and specificity. By embracing the transition state more tightly than the substrate, enzymes lower the activation energy, thereby accelerating the reaction rate and supporting life's biochemical processes.

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Most popular questions from this chapter

Tosyl-L-phenylalanine chloromethyl ketone (TPCK) specifically inhibits chymotrypsin by covalently labeling His \(^{57}\). Tosyl-L-phenylalanine chloromethyl ketone (TPCK) a. Propose a mechanism for the inactivation reaction, indicating the structure of the product(s). b. State why this inhibitor is specific for chymotrypsin. c. Propose a reagent based on the structure of TPCK that might be an effective inhibitor of trypsin.

An enzyme-substrate complex can form when the substrate \((\mathrm{s})\) bind (s) to the active site of the enzyme. Which environmental condition might alter the conformation of an enzyme to the extent that its substrate is unable to bind? a. Enzyme \(A\) at \(40^{\circ} \mathrm{C}\) b. Enzyme \(B\) at pH 2 c. Enzyme \(X\) at \(p H 4\) d. Enzyme \(Y\) at \(37^{\circ} \mathrm{C}\)

At \(35^{\circ} \mathrm{C},\) the rate of the reaction catalyzed by enzyme \(\mathrm{A}\) begins to level off. Which hypothesis best explains this observation? a. The temperature is too far below optimum. b. The enzyme has become saturated with substrate. c. Both \(A\) and \(B\). d. Neither A nor B.

As noted on page \(423,\) a true transition state can bind to an enzyme active site with a \(K_{\mathrm{T}}\) as low as \(7 \times 10^{-26} M .\) This is a remarkable number, with interesting consequences. Consider a hypothetical solution of an enzyme in equilibrium with a ligand that binds with a \(K_{\mathrm{D}}\) of \(10^{-27} M .\) If the concentration of free enzyme, \([\mathrm{E}],\) is equal to the concentration of the enzyme-ligand complex, [EL], what would \([\mathrm{L}],\) the concentration of free ligand, be? Calculate the volume of solution that would hold one molecule of free ligand at this concentration.

In which of the following environmental conditions would digestive enzyme Y be unable to bring its substrate(s) to the transition state? a. At any temperature below optimum b. At any pH where the rate of reaction is not maximum c. At any pH lower than 5.5 d. At any temperature higher than \(37^{\circ} \mathrm{C}\)

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