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The constant K has the same units as s. Evaluate R(K), the rate of production when s = K, and use the result to explain why K is the initial enzyme concentration that gives a production rate of V/2 (half of the maximum rate). %3D

The constant K has the same units as s. Evaluate R(K), the rate of production when s = K, and use the result to explain why K is the initial enzyme concentration that gives a production rate of V/2 (half of the maximum rate). %3D

Biochemistry
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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V = 10 μm/s

K = 5 μm

Enzymes are catalysts that facilitate the biochemical reactions that occur within all living organisms. One of the fundamental laws of enzyme kinetics was proposed by Leonor Michaelis and Maud Menten in 1913. The law has been supported by laboratory experiments and explained through mathematical modeling. Today, Michaelis-Menten kinetics are used in many biological models. An enzyme molecule is designed to "fit" another molecule called a substrate. The substrate (S) and enzyme (E) form an intermediate complex (ES), which then dissociates to form the final end-product of the reaction (P) and the original enzyme (which can be re-used; see figure). An important question concerns the rate at which product molecules are formed. Under certain assumptions, Michaelis-Menten kinetics relates the rate of production of P to the amount of S present. **Diagram Explanation:** The diagram illustrates the sequence of enzyme reactions: 1. **E + S**: Represents the enzyme (E) and substrate (S) combining. 2. **ES**: Denotes the enzyme-substrate complex. 3. **E + P**: Shows the dissociation into enzyme (E) and product (P). **Equation and Explanation:** 1. Let \( R \) be the rate of production of the final product \( P \) and let \( s \) be the concentration of the substrate \( S \) initially present. Concentrations are measured in units such as micro-moles (µM), while \( R \) is measured in µM/s. (Time is usually measured in seconds.) The Michaelis-Menten law states that: \[ R(s) = \frac{V s}{K + s} \] Where: - \( V \): Maximum reaction rate. - \( K \): Michaelis constant. - \( s \): Substrate concentration.
Transcribed Image Text:Enzymes are catalysts that facilitate the biochemical reactions that occur within all living organisms. One of the fundamental laws of enzyme kinetics was proposed by Leonor Michaelis and Maud Menten in 1913. The law has been supported by laboratory experiments and explained through mathematical modeling. Today, Michaelis-Menten kinetics are used in many biological models. An enzyme molecule is designed to "fit" another molecule called a substrate. The substrate (S) and enzyme (E) form an intermediate complex (ES), which then dissociates to form the final end-product of the reaction (P) and the original enzyme (which can be re-used; see figure). An important question concerns the rate at which product molecules are formed. Under certain assumptions, Michaelis-Menten kinetics relates the rate of production of P to the amount of S present. **Diagram Explanation:** The diagram illustrates the sequence of enzyme reactions: 1. **E + S**: Represents the enzyme (E) and substrate (S) combining. 2. **ES**: Denotes the enzyme-substrate complex. 3. **E + P**: Shows the dissociation into enzyme (E) and product (P). **Equation and Explanation:** 1. Let \( R \) be the rate of production of the final product \( P \) and let \( s \) be the concentration of the substrate \( S \) initially present. Concentrations are measured in units such as micro-moles (µM), while \( R \) is measured in µM/s. (Time is usually measured in seconds.) The Michaelis-Menten law states that: \[ R(s) = \frac{V s}{K + s} \] Where: - \( V \): Maximum reaction rate. - \( K \): Michaelis constant. - \( s \): Substrate concentration.
**Question 7** The constant \( K \) has the same units as \( s \). Evaluate \( R(K) \), the rate of production when \( s = K \), and use the result to explain why \( K \) is the initial enzyme concentration that gives a production rate of \( V/2 \) (half of the maximum rate). *Explanation for Educational Context*: The problem statement deals with enzyme kinetics, specifically examining a scenario where the substrate concentration \( s \) equals the constant \( K \). When \( s = K \), it represents a point in enzyme reactions where the rate of production \( R(K) \) is at half the maximum rate, \( V/2 \). Understanding this relationship is crucial in biochemical studies to determine how enzymes function under varying substrate concentrations.
Transcribed Image Text:**Question 7** The constant \( K \) has the same units as \( s \). Evaluate \( R(K) \), the rate of production when \( s = K \), and use the result to explain why \( K \) is the initial enzyme concentration that gives a production rate of \( V/2 \) (half of the maximum rate). *Explanation for Educational Context*: The problem statement deals with enzyme kinetics, specifically examining a scenario where the substrate concentration \( s \) equals the constant \( K \). When \( s = K \), it represents a point in enzyme reactions where the rate of production \( R(K) \) is at half the maximum rate, \( V/2 \). Understanding this relationship is crucial in biochemical studies to determine how enzymes function under varying substrate concentrations.
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