つく 146 Bb 9146 Bb 10311 Class E Etsy Traps E Traps New Free Chat + ☆ 口出 Assignment/takeCovalentActivity.do?locator-assignment-take [References] You do an enzyme kinetic experiment and calculate a Vmax of 118 μmol per minute. If each assay used 0.10 mL of an enzyme solution that had a concentration of 0.20 mg/mL, what would be the turnover number if the enzyme had a molecular weight of 128,000 g/mol? (Enter your answer to two significant figures.) turnover number = sec1 D Submit Answer Try Another Version 2 item attempts remaining

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
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You do an enzyme kinetic experiment and calculate a Vmax of 118 μmol per minute. If each assay used 0.10 mL of an enzyme solution that had a concentration of 0.20 mg/mL, what would be the
turnover number if the enzyme had a molecular weight of 128,000 g/mol?
(Enter your answer to two significant figures.)
turnover number =
sec-1
D
1 pt
Submit Answer
Try Another Version
2 item attempts remaining
estion
stion 5
on 6
7
1pt
1 pt
1 pt
1pt
1pt
1pt
1 pt
1 pt
D
is the substrate concentration multiplied by the
catalytic constant. KM is equivalent to the
substrate concentration multiplied by the ratio
of rate constants for the formation and
dissociation of the enzyme-substrate complex. KM
is equivalent to the substrate concentration. KM
is equivalent to the substrate concentration
divided by 2
A: KM is equivalent to the substrate
concentration when the reaction rate is at half
of its maximum velocity (Vmax/2).
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Transcribed Image Text:1 pt pt 9146 Bb 9146 Bb 1031 Class Etsy E Traps E Traps New Free Chat + ☆ 出口 keAssignment/takeCovalentActivity.do?locator-assignment-take [References] You do an enzyme kinetic experiment and calculate a Vmax of 118 μmol per minute. If each assay used 0.10 mL of an enzyme solution that had a concentration of 0.20 mg/mL, what would be the turnover number if the enzyme had a molecular weight of 128,000 g/mol? (Enter your answer to two significant figures.) turnover number = sec-1 D 1 pt Submit Answer Try Another Version 2 item attempts remaining estion stion 5 on 6 7 1pt 1 pt 1 pt 1pt 1pt 1pt 1 pt 1 pt D is the substrate concentration multiplied by the catalytic constant. KM is equivalent to the substrate concentration multiplied by the ratio of rate constants for the formation and dissociation of the enzyme-substrate complex. KM is equivalent to the substrate concentration. KM is equivalent to the substrate concentration divided by 2 A: KM is equivalent to the substrate concentration when the reaction rate is at half of its maximum velocity (Vmax/2). Cengage Learning | Cengage Technical Support Previous New
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