Measurement of the rate constants for a simple enzymatic reaction obeying Michaelis-Menten kinetics gave the following results: \(k_{1}=2 \times 10^{8} M^{-1} \sec ^{-1}\) \(k_{-1}=1 \times 10^{3} \sec ^{-1}\) \(k_{2}=5 \times 10^{3} \mathrm{sec}^{-1}\) a. What is \(K_{\mathrm{S}},\) the dissociation constant for the enzyme- substrate complex? b. What is \(K_{m},\) the Michaelis constant for this enzyme? c. What is \(k_{\text {cat }}\) (the turnover number) for this enzyme? d. What is the catalytic efficiency \(\left(k_{\mathrm{cat}} / K_{m}\right)\) for this enzyme? e. Does this enzyme approach "kinetic perfection"? (That is, does \(k_{\mathrm{cat}} / K_{m}\) approach the diffusion-controlled rate of enzyme association with substrate? f. If a kinetic measurement was made using 2 nanomoles of enzyme per \(\mathrm{mL}\) and saturating amounts of substrate, what would \(V_{\max }\) equal? g. Again, using 2 nanomoles of enzyme per mL of reaction mixture, what concentration of substrate would give \(v=0.75 V_{\max } ?\) h. If a kinetic measurement was made using 4 nanomoles of enzyme per \(\mathrm{mL}\) and saturating amounts of substrate, what would \(V_{\max }\) equal? What would \(K_{m}\) equal under these conditions?

Step by step solution

01

Calculate the dissociation constant for the enzyme-substrate complex

To determine \(Ks\), use the formula \(Ks = k_{-1}/k_1\), from the given values: \(Ks = (1 \times 10^{3}\ sec^{-1})/(2 \times 10^{8} M^{-1} sec^{-1}) = 5 \times 10^{-6} M\)

02

Calculate the Michaelis constant for this enzyme

\(Km\) is calculated using the formula \(Km = (k_{-1} + k_2)/k_1\). Substituting in the given values, we have: \(Km = (1 \times 10^{3}\ sec^{-1} + 5 \times 10^{3} sec^{-1}) / (2 \times 10^{8} M^{-1} sec^{-1}) = 3 \times 10^{-6} M\).

03

Determine the turnover number for this enzyme

The turnover number, often denoted as \(k_{cat}\) or \(k_2\), is the number of substrate molecules an enzyme can convert into product per unit time when the enzyme is fully saturated with substrate. From the exercise's data it is clear that \(k_{cat} = k_2 = 5 \times 10^{3}\ sec^{-1}\).

04

Calculate the catalytic efficiency

Catalytic efficiency is computed as the ratio of \(k_{cat} / Km\). Substituting the values we calculated before: \(k_{cat}/ Km = (5 \times 10^{3}\ sec^{-1})/(3 \times 10^{-6} M) = 1.67 \times 10^{9} M^{-1} sec^{-1}\).

05

Discuss the enzyme's kinetic perfection

Enzymes with higher \(k_{cat} / Km\) ratios are considered more 'perfect' because they can convert substrate to product more rapidly. Moreover, the maximum known rate is \(~10^{10} M^{-1} sec^{-1}\). Since this enzyme's ratio is \(1.67 \times 10^{9} M^{-1} sec^{-1}\), it is in fact approaching kinetic perfection.

06

Calculate Vmax for 2 nanomoles of enzyme per mL

Vmax is calculated using the formula \(Vmax = k_{cat} \times [E]\). Given \(E = 2 \times 10^{-9} M\) (since 1 mole = 10^9 nanomoles) and \(k_{cat} = 5 \times 10^{3} sec^{-1}\), we have: \(Vmax = 5 \times 10^{3} sec^{-1} \times 2 \times 10^{-9} M = 1 \times 10^{-5} M sec^{-1}\).

07

Find substrate concentration at 75% Vmax

The Michaelis-Menten equation defines \(v=0.75 V_{max}\), or the substrate concentration necessary to reach three-quarters of the maximal rate of a reaction. It states: \(v = V_{max}[S]/(K_m + [S])\). Setting \(v = 0.75 V_{max}\) and isolating \([S]\), you get \([S] = Km(0.75 V_{max}/(V_{max}-0.75 V_{max})) = 3 \times 10^{-6} M\).

08

Calculate Vmax and Km for 4 nanomoles of enzyme per mL

If doubling the amount of enzyme to \(4 \times 10^{-9} M\), then \(V_{max}\) also doubles to \(2 \times 10^{-5} M sec^{-1}\) because \(V_{max}\) is proportional to \([E]\). However, \(K_m\) remains unchanged at \(3 \times 10^{-6} M\) as it is inherent to the enzyme and is not directly affected by the enzyme concentration.

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Key Concepts

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

Enzymatic Reaction Rate Constants

Enzymatic reaction rate constants, represented by symbols like k1, k-1, and k2, are fundamental parameters that describe the rates at which an enzyme binds to a substrate (k1), the complex dissociates (k-1), and the enzyme catalyzes the reaction to form a product (k2 or kcat). Understanding these constants is crucial for characterizing enzyme kinetics because they relate directly to how efficiently an enzyme can convert a substrate into a product.

These constants can be determined experimentally by measuring reaction rates under various substrate concentrations. Additionally, the balance between the constants k1 and k-1 provides insight into the stability of the enzyme-substrate complex—a critical factor for the efficiency of the enzymatic reaction.

Dissociation Constant (Ks)

The dissociation constant, denoted as Ks, is an expression of the affinity between an enzyme and its substrate: a low Ks value means high affinity because the enzyme holds on to the substrate tightly, and vice versa. It is calculated as the ratio of the rate constants for dissociation (k-1) and association (k1). In the given exercise, the calculated Ks indicates how readily the enzyme-substrate complex falls apart, providing an insight into the first step of the enzymatic reaction—a crucial aspect since the formation and breakdown of this complex dictate the overall reaction rate.

Michaelis Constant (Km)

The Michaelis constant, Km, is a key parameter in enzyme kinetics. It is defined as the substrate concentration at which the reaction rate is half of Vmax (the maximum rate achievable by the system). It gives a sense of how much substrate is needed to significantly engage the enzyme in catalysis. A small Km implies high enzyme affinity for the substrate, indicating that the enzyme is effective even at low substrate concentration. In the provided problem, Km was calculated using the rate constants to help gauge the enzyme's performance under physiological conditions.

Turnover Number (kcat)

The turnover number, referred to as kcat, indicates the number of substrate molecules converted to product by an enzyme molecule in a given time when the enzyme is saturated with substrate. It serves as a measure of the enzyme's catalytic activity, or how 'fast' an enzyme can operate. The calculation of kcat in our exercise reflects the enzyme's capacity to facilitate a reaction, which is important for understanding how effective an enzyme could be within a biological system.

Catalytic Efficiency

Catalytic efficiency combines the concepts of kcat and Km to evaluate an enzyme's proficiency. It's expressed as the ratio of these two values (kcat/Km), providing a single number that encapsulates both the speed and affinity of an enzyme for its substrate. The higher the catalytic efficiency, the better the enzyme performs at lower substrate concentrations. This efficiency can identify potential rate-limiting steps in metabolic pathways and help in designing drugs that target specific enzymes.

Vmax Calculation

The calculation of Vmax involves determining the maximum rate of the enzymatic reaction when all enzyme active sites are occupied by substrate, referred to as saturation. Vmax is directly proportional to the enzyme concentration. It represents the theoretical maximum rate of the reaction; however, it is seldom reached in live systems. The computed Vmax in the exercise helps predict how changes in enzyme concentration could affect the reaction rate, an important consideration in biochemical and pharmacological studies.

Substrate Concentration for Enzyme Kinetics

Substrate concentration plays a significant role in enzyme kinetics, impacting the reaction rate until all the enzyme's active sites are saturated. By analyzing various substrate concentrations, we can determine the Km and Vmax values, which then allow us to use the Michaelis-Menten equation to calculate the rate of an enzymatic reaction at any given substrate concentration. In the exercise, this principle is utilised to ascertain at what substrate level 75% of Vmax is achieved, revealing practical implications for enzyme functionality in different cellular environments.