Answered: How will a graph of reaction rate (V)…

How will a graph of reaction rate (V) vs. [substrate] for an allosteric enzyme differ from the hyperbolic plot characteristic of enzymes having Michaelis-Menten kinetics?

Biochemistry

9th Edition

ISBN:9781319114671

Author:Lubert Stryer, Jeremy M. Berg, John L. Tymoczko, Gregory J. Gatto Jr.

Publisher:Lubert Stryer, Jeremy M. Berg, John L. Tymoczko, Gregory J. Gatto Jr.

Chapter1: Biochemistry: An Evolving Science

Section: Chapter Questions

Problem 1P

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Question

How will a graph of reaction rate (V) vs. [substrate] for an
allosteric enzyme differ from the hyperbolic plot characteristic
of enzymes having Michaelis-Menten kinetics?

Expert Solution

Step 1

The enzymes are biochemical catalysts that enhances the rate of biological reactions.

Most enzymes are protein in nature although some NA molecules also sow enzymatic activity.

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82.16
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a particular enzyme catalyzes a single reactant S to a single product P, following michaelis-menten kinetics rp=(VmaxCs) / (Km + Cs) 1. A reaction with this enzyme is carried out at very low substrate concentrations. Draw and label a curve on the plot that describes the reaction kinetics under those conditions.
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The Michaelis-Menten equation is often used to describe the kinetic characteristics of an enzyme-catalyzed reaction. Vmax [S] Km + [S] where v is the velocity, or rate, Vmax is the maximum velocity. K is the Michaelis-Menten constant, and [S] is the substrate concentration. A graph of the Michaelis-Menten equation is a plot of a reaction's initial velocity (ro) at different substrate concentrations ([S]). First, move the line labeled Vmax to a position that represents the maximum velocity of the enzyme. Next, move the line labeled 1/2 Vmax to its correct position. Then, move the line labeled Km to its correct position. Estimate the values for Vmax and Km- Vmax= µM/min v (µM/min) 300 275 250 225 200 175 150 125 100 75 50 Km = 25 K 0 10 20 30 V max 40 50 [S] (M) 1/2 V max Michaelis-Menten curve 60 70 80 90 100 HM
The total concentration of enzyme in a reaction, [E], is made up of the concentration of enzyme bound to substrate, [ES], and the concentration of enzyme still free in solution, [Ef]. Similarly, the total amount of substrate is made up of [Sf] and [ES]. We can assume that the concentration of enzyme is much less than that of the substrate, [E] << [S]. Assuming the steady state condition and the relationships between [E], [Ef] and [ES], and similar ones for S, given in lectures, derive an expression for the saturation factor, , in terms of [S] and . (Note that [E] and [S] denote the total amounts of enzyme and substrate added to the reaction, respectively. You may assume that [S]>>[E].)
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