The turnover number is defined as the maximum number of substrate molecules that can be converted into product molecules per unit time by an enzyme molecule. The concentration of enzyme active sites is not necessarily equal to the concentration of enzyme molecules, because some enzyme molecules have more than one active site. If the enzyme molecule has one active site, the turnover number is given by turnover number = Rmax [E]t = k₂ (Rmax is often written as Vmax) If the enzyme molecule has more than one active site, then [E], is multiplied by the number of active sites to determine its effective concentration. Determine the value of the turnover number of the enzyme carbonic anhydrase, given that Rmax for carbonic anhydrase equals 249 μmol Ls¹ and [E], = 2.62 nmol · L-¹. Carbonic anhydrase has a single active site.
The turnover number is defined as the maximum number of substrate molecules that can be converted into product molecules per unit time by an enzyme molecule. The concentration of enzyme active sites is not necessarily equal to the concentration of enzyme molecules, because some enzyme molecules have more than one active site. If the enzyme molecule has one active site, the turnover number is given by turnover number = Rmax [E]t = k₂ (Rmax is often written as Vmax) If the enzyme molecule has more than one active site, then [E], is multiplied by the number of active sites to determine its effective concentration. Determine the value of the turnover number of the enzyme carbonic anhydrase, given that Rmax for carbonic anhydrase equals 249 μmol Ls¹ and [E], = 2.62 nmol · L-¹. Carbonic anhydrase has a single active site.
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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Q14:
![The turnover number is defined as the maximum number of substrate molecules that can be converted into product molecules per unit time by an enzyme molecule. The concentration of enzyme active sites is not necessarily equal to the concentration of enzyme molecules, because some enzyme molecules have more than one active site. If the enzyme molecule has one active site, the turnover number is given by:
\[
\text{turnover number} = \frac{R_{\text{max}}}{[E]_t} = k_2 \quad (R_{\text{max}} \text{ is often written as } V_{\text{max}})
\]
If the enzyme molecule has more than one active site, then \([E]_t\) is multiplied by the number of active sites to determine its effective concentration.
Determine the value of the turnover number of the enzyme carbonic anhydrase, given that \( R_{\text{max}} \) for carbonic anhydrase equals 249 µmol · L\(^{-1}\) · s\(^{-1}\) and \([E]_t = 2.62 \, \text{nmol} \cdot \text{L}^{-1}\). Carbonic anhydrase has a single active site.
The problem offers a calculated turnover number as:
\[
\text{turnover number} = 652.38 \times 10^{-3} \;\text{s}^{-1}
\]
This result is marked as incorrect.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fc94cb5d1-997a-49d0-b0ef-9ec02bf0a187%2Ffd0e165a-33a8-4152-afee-8c5f6d113bed%2Fg7yfr9_processed.png&w=3840&q=75)
Transcribed Image Text:The turnover number is defined as the maximum number of substrate molecules that can be converted into product molecules per unit time by an enzyme molecule. The concentration of enzyme active sites is not necessarily equal to the concentration of enzyme molecules, because some enzyme molecules have more than one active site. If the enzyme molecule has one active site, the turnover number is given by:
\[
\text{turnover number} = \frac{R_{\text{max}}}{[E]_t} = k_2 \quad (R_{\text{max}} \text{ is often written as } V_{\text{max}})
\]
If the enzyme molecule has more than one active site, then \([E]_t\) is multiplied by the number of active sites to determine its effective concentration.
Determine the value of the turnover number of the enzyme carbonic anhydrase, given that \( R_{\text{max}} \) for carbonic anhydrase equals 249 µmol · L\(^{-1}\) · s\(^{-1}\) and \([E]_t = 2.62 \, \text{nmol} \cdot \text{L}^{-1}\). Carbonic anhydrase has a single active site.
The problem offers a calculated turnover number as:
\[
\text{turnover number} = 652.38 \times 10^{-3} \;\text{s}^{-1}
\]
This result is marked as incorrect.
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