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.

Short Answer

Expert verified
The concentration of the free ligand (\[L\]) is \(10^{-27} M\). The volume of solution that would hold one molecule of the free ligand at this concentration is \(0.166 L\) or \(166 mL\).

Step by step solution

01

Understand the Concept of Dissociation Constant

The dissociation constant, \(KD\), is used to describe the affinity between the ligand and its receptor (in this case, the enzyme). It is the ratio of the rate constant for the dissociation of the complex to the rate constant for the formation of the complex. When \([E]\) equals \([EL]\), the concentration of free enzyme equals the concentration of the enzyme-ligand complex, the equation \([L]=[E]\) applies.
02

Calculate the Concentration of the Free Ligand

Since \([E]=[EL]\), we can use that to solve for \([L]\). That gives us \([L]= KD = 10^{-27}\ M\), because in equilibrium the product of the unbound components equals the product of the bound components times the \(KD\) value.
03

Calculate the Volume of Solution Holding One Molecule of Free Ligand

First, we need to convert the concentration to moles, as concentration is typically written in moles per litre. 1 molecule is \(1/Avogadro's\ number\) moles. Avogadro's number is \(6.022 x 10^{23} mol^{-1}\). So, the volume \(V\) that would hold one molecule of \[L] would be \(1/(L*NA)\) in litres. Substituting \(L=10^{-27} M\) and \( NA=6.022 x 10^{23} mol^{-1}\), we get \(V=1/(10^{-27}*6.022 x 10^{23})\). Therefore, \(V\) is around \(0.166 L\) or \(166 mL\) for one molecule.

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Key Concepts

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

Dissociation Constant
When studying enzyme-ligand interactions, the dissociation constant, denoted as \(K_D\), emerges as a pivotal metric. This value helps us understand just how tightly a ligand, which could be a small molecule or a substrate, binds to an enzyme. At its core, the dissociation constant is defined by the ratio of the disassociation rate of the enzyme-ligand complex to the rate at which they combine.

Put simply, a lower \(K_D\) signifies a stronger affinity—the ligand is not keen on leaving the cozy embrace of the enzyme. In stark contrast, a higher \(K_D\) value would suggest that the ligand frequently dissociates, indicating a weaker attraction. Here, the essential takeaway is that by knowing the \(K_D\) value, one can deduce the concentration of free ligand required to reach equilibrium with the enzyme, where half of the active sites are occupied.
Transition State
Moving on to the transition state, this is a temporary molecular structure representing a high-energy state that exists between the reactants (enzyme and ligand) and the products (enzyme-ligand complex). The transition state is akin to the peak of a mountain in a reaction landscape; it's the point where old bonds are on the brink of breaking and new ones are about to form.

Enzymes are exceptional in their ability to stabilize this fleeting state, thereby lowering the energy required to reach it—this is known as the transition state energy. In our context, a very low \(K_{\mathrm{T}}\) value reflects an enzyme's remarkable ability to stabilize the transition state. This consequently accelerates the reaction by lowering the energetic hurdle that must be overcome.
Equilibrium Calculation
Equilibrium calculation bears immense importance in biochemistry, providing insights into the concentrations of various species at a state where the rates of the forward and reverse reactions are equivalent. In our exercise, where the dissociation constant \(K_D\) and the enzyme's concentration are known, the equilibrium calculation allows us to pinpoint the exact levels of free ligand.

To perform this calculation, we draw on the fundamental principle that at equilibrium, the rate of formation of the enzyme-ligand complex and its dissociation are equal. This establishes a fixed ratio between the concentration of free enzyme \([E]\), free ligand \([L]\), and the enzyme-ligand complex \([EL]\), governed by the equation \([L]=[E]\) under the specific condition that \([E]\) equals \([EL]\). Ultimately, it is this balanced state of affairs that allows us to compute the desired ligand concentration.
Avogadro's Number
Entering the molecular realm, Avogadro's number is a constant that signifies the quantity of units—be they atoms, molecules, or ions—in one mole of substance. This staggering value is approximately \(6.022 \times 10^{23}\) per mole and serves as a bridge between the microscopic and macroscopic worlds.

When dealing with concentrations as infinitesimally tiny as those in our exercise, Avogadro's number becomes indispensable. It converts an almost incomprehensibly small concentration into an understandable, countable quantity of molecules—or in this case, ligand particles. In calculating the volume that contains a single ligand molecule, Avogadro's number enables the translation from an abstract concentration to a practical volume of solution, demystifying the scale of molecular interactions to something tangible.

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Most popular questions from this chapter

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.

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 \(37^{\circ} \mathrm{C}\)

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 \(37^{\circ} \mathrm{C}\)

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?

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.

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