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v (micromol/sec) 120 100 80 60 40 20 0 <--X ...... Y 20 40 [A] (micromolar) 60 Enzyme X must have 2 active sites, whereas enzyme Y has only 1. The Kd for enzyme X is 25 micromolar. 80 At low substrate concentrations, enzyme Y is more effective than enzyme X at converting substrate A into product. 100 Enzyme Y reaches Vmax faster than enzyme X. At saturating substrate concentrations, enzyme X produces more product per unit time than does enzyme Y.

v (micromol/sec) 120 100 80 60 40 20 0 <--X ...... Y 20 40 [A] (micromolar) 60 Enzyme X must have 2 active sites, whereas enzyme Y has only 1. The Kd for enzyme X is 25 micromolar. 80 At low substrate concentrations, enzyme Y is more effective than enzyme X at converting substrate A into product. 100 Enzyme Y reaches Vmax faster than enzyme X. At saturating substrate concentrations, enzyme X produces more product per unit time than does enzyme Y.

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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The 2 eynzymes are affected by a substrate. Choose all correct answers.

### Graph Explanation The graph displays the reaction velocity (\(v\) in micromoles per second) as a function of substrate concentration \([A]\) in micromolar for two enzymes, X and Y. The lines represent the relationship between substrate concentration and reaction rate for each enzyme. - The solid line (\(\boldsymbol{-}\)) represents enzyme X. - The dotted line (\(\cdots\)) represents enzyme Y. #### Key Observations: - Enzyme X shows a steep increase in velocity with increasing substrate concentration and eventually reaches a plateau, indicating a maximum velocity (Vmax). - Enzyme Y initially increases more rapidly at low substrate concentrations but reaches a plateau at a lower velocity compared to enzyme X. ### Statements for Evaluation 1. **Enzyme X must have 2 active sites, whereas enzyme Y has only 1.** 2. **At low substrate concentrations, enzyme Y is more effective than enzyme X at converting substrate A into product.** 3. **The \(K_d\) for enzyme X is 25 micromolar.** 4. **At saturating substrate concentrations, enzyme X produces more product per unit time than does enzyme Y.** 5. **Enzyme Y reaches Vmax faster than enzyme X.** These statements are based on the graphical data provided and aim to analyze the behavior and characteristics of the two enzymes in different substrate conditions.
Transcribed Image Text:### Graph Explanation The graph displays the reaction velocity (\(v\) in micromoles per second) as a function of substrate concentration \([A]\) in micromolar for two enzymes, X and Y. The lines represent the relationship between substrate concentration and reaction rate for each enzyme. - The solid line (\(\boldsymbol{-}\)) represents enzyme X. - The dotted line (\(\cdots\)) represents enzyme Y. #### Key Observations: - Enzyme X shows a steep increase in velocity with increasing substrate concentration and eventually reaches a plateau, indicating a maximum velocity (Vmax). - Enzyme Y initially increases more rapidly at low substrate concentrations but reaches a plateau at a lower velocity compared to enzyme X. ### Statements for Evaluation 1. **Enzyme X must have 2 active sites, whereas enzyme Y has only 1.** 2. **At low substrate concentrations, enzyme Y is more effective than enzyme X at converting substrate A into product.** 3. **The \(K_d\) for enzyme X is 25 micromolar.** 4. **At saturating substrate concentrations, enzyme X produces more product per unit time than does enzyme Y.** 5. **Enzyme Y reaches Vmax faster than enzyme X.** These statements are based on the graphical data provided and aim to analyze the behavior and characteristics of the two enzymes in different substrate conditions.
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