The following table indicates the rates at which a substrate...

The following table indicates the rates at which a substrate reacts as catalyzed by an enzyme that follows the Michaelis-Menten mechanism: (1) in the absence of inhibitor; (2) and (3) in the presence of $10 \mathrm{m} M$ concentration, respectively, of either of two inhibitors. Assume $[\mathrm{E}]_{\mathrm{T}}$ is the same for all reactions. $$\begin{array}{rccc} \begin{array}{c} {[\mathrm{S}]} \\ (\mathrm{m} M) \end{array} & \begin{array}{c} (1) v_{\mathrm{o}} \\ \left(\mu M \cdot \mathrm{s}^{-1}\right) \end{array} & \begin{array}{c} (2) v_{\mathrm{o}} \\ \left(\mu M \cdot \mathrm{s}^{-1}\right) \end{array} & \begin{array}{c} (3) v_{\mathrm{o}} \\ \left(\mu M \cdot \mathrm{s}^{-1}\right) \end{array} \\ \hline 1 & 2.5 & 1.17 & 0.77 \\ 2 & 4.0 & 2.10 & 1.25 \\ 5 & 6.3 & 4.00 & 2.00 \\ 10 & 7.6 & 5.7 & 2.50 \\ 20 & 9.0 & 7.2 & 2.86 \end{array}$$ (a) Determine $K_{M}$ and $V_{\max }$ for the enzyme. For each inhibitor determine the type of inhibition and $K_{\mathrm{I}}$ and/or $K_{\mathrm{I}}^{\prime} .$ What additional information would be required to calculate the turnover number of the enzyme? (b) For $[\mathrm{S}]=5 \mathrm{m} M,$ what fraction of the enzyme molecules have a bound substrate in the absence of inhibitor, in the presence of $10 \mathrm{m} M$ inhibitor of type $(2),$ and in the presence of $10 \mathrm{m} M$ inhibitor of type (3)$?$

The following table indicates the rates at which a substrate reacts as catalyzed by an enzyme that follows the Michaelis-Menten mechanism: (1) in the absence of inhibitor; (2) and
(3) in the presence of $10 \mathrm{m} M$ concentration, respectively, of either of two inhibitors. Assume $[\mathrm{E}]_{\mathrm{T}}$ is the same for all reactions.
$$\begin{array}{rccc}
\begin{array}{c}
{[\mathrm{S}]} \\
(\mathrm{m} M)
\end{array} & \begin{array}{c}
(1) v_{\mathrm{o}} \\
\left(\mu M \cdot \mathrm{s}^{-1}\right)
\end{array} & \begin{array}{c}
(2) v_{\mathrm{o}} \\
\left(\mu M \cdot \mathrm{s}^{-1}\right)
\end{array} & \begin{array}{c}
(3) v_{\mathrm{o}} \\
\left(\mu M \cdot \mathrm{s}^{-1}\right)
\end{array} \\
\hline 1 & 2.5 & 1.17 & 0.77 \\
2 & 4.0 & 2.10 & 1.25 \\
5 & 6.3 & 4.00 & 2.00 \\
10 & 7.6 & 5.7 & 2.50 \\
20 & 9.0 & 7.2 & 2.86
\end{array}$$
(a) Determine $K_{M}$ and $V_{\max }$ for the enzyme. For each inhibitor determine the type of inhibition and $K_{\mathrm{I}}$ and/or $K_{\mathrm{I}}^{\prime} .$ What additional information would be required to calculate the turnover number
of the enzyme? (b) For $[\mathrm{S}]=5 \mathrm{m} M,$ what fraction of the enzyme molecules have a bound substrate in the absence of inhibitor, in the presence of $10 \mathrm{m} M$ inhibitor of type $(2),$ and in the presence of $10 \mathrm{m} M$ inhibitor of type (3)$?$

Key Concepts

Turnover Number (kcat)

Turnover number represents the number of substrate molecules converted to product per enzyme molecule per unit time when the enzyme is fully saturated with substrate. kcat is calculated using Vmax and the total enzyme concentration, and it is a key measure of catalytic efficiency and enzyme performance under optimal conditions.

Michaelis-Menten Kinetics

This concept describes the relationship between the substrate concentration and the rate of an enzyme?catalyzed reaction. Key parameters include Km, the substrate concentration at which the reaction rate is half of Vmax, and Vmax, the maximum reaction velocity. Understanding these parameters allows one to characterize enzyme efficiency and affinity for the substrate under steady?state conditions.

Enzyme Inhibition Mechanisms

This concept covers the ways in which different inhibitors affect enzyme activity. Inhibition can be competitive, noncompetitive, uncompetitive, or mixed, each altering Km and/or Vmax differently. In enzyme kinetics, the inhibition constant (Ki or Ki') quantifies the inhibitor’s potency, and differentiating between types of inhibition helps in understanding how the inhibitor interacts with the enzyme and/or the enzyme–substrate complex.

Fractional Enzyme Occupancy

This concept refers to the fraction of the enzyme molecules that have a bound substrate at any given moment, often represented by the ratio [ES]/[E]?. Under Michaelis-Menten kinetics, this fraction can be determined by the relationship between substrate concentration and the Km value, providing insights into enzyme saturation and the dynamic equilibrium of enzyme-substrate binding.

Transcript

00:01 Okay, this is a long one to read.

00:03 And if you're looking at this problem, you can read it yourself.

00:06 In order to solve the problem, we're going to use the line weaver berk equation and its variance.

00:12 In first order to calculate kinetic parameters and afterwards to calculate the data required in the problem.

00:20 So the equation is 1 over vo equals plus 1 over vmax.

00:44 And this is the initial velocity.

00:54 M -k -m is a constant.

01:14 V -max is the maximum velocity, velocity, and s is a substrate concentration.

01:40 And then we're going to plot 1 over v -o versus 1 over s.

01:50 K -m will be the y intercept.

02:00 And then i'm going to get a table here.

02:03 So first of all, reference this table, and then i'm going to add several rows.

02:11 So i'm going to add for my s, this one i'm going to keep here.

02:19 We had 1, 2, 5, 10, 20.

02:27 And now i'm going to do 1 over s.

02:38 We'll be 1 over millimolar.

02:42 And this next one will be, we'll have 1, 2, and 3, 1 over v0, over v0.

02:58 And one over v o and this would be micromolar s to the minus one micromolar s to the minus one micromolar oopsie s to the minus one okay so i'm sort of running out a room there but that's okay so let me get started on writing these down and and that was one let's get over to two and this one is 1 .30 and 0 .35 okay, so i'm just going to put these numbers, putting calculator.

04:53 So one, we will get a line of y equals 0 .3017 x plus 0 .087.

05:07 For 2, we get y equals 0 .7541 x plus 0 .987.

05:19 And for 3, we get y equals 0 .9989x plus 0 .301.

05:36 Okay, so then we can do the switch colors to blue.

05:43 So let's do line 1 .0 .0907 equals 1 over v max.

05:55 So vmax will equal 10 .13 micromole, centred meters to the minus one, seconds to the minus one.

06:17 There we go.

06:20 And 0 .302 equals km over vmax.

06:34 So km will equal 0 .032 times.

06:49 10 .13.

06:57 I don't know why this isn't working over here.

06:59 Micromolar over s to the minus one there.

07:05 And this will equal my km will equal 3 .06.

07:17 And then let's do number two.

07:28 And this will be a competitive inhibitor since its plot intersects the one plot.

08:03 And our equation will be where alpha is equal to 1 plus i over k i, where this is the inhibitor concentration, and k1 is the, or k i is an inhibitor constant.

08:55 Okay, so let's do the calculation here.

09:06 So we're going to have 0 .754.

09:11 I'm going to write something else down here.

09:24 So 0 .754 equals kf over vmax, and that'll equal 10 .13 micromolar s to the minus 1 times 0 .754.

09:47 And this my kf will equal 7 .64 millimolar...

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