13. An enzyme catalyzes the reaction A = B. The enzyme is present at a concentration...

Question

Question

science biology

Answer 1

Answer #1

Vo= Vmax [S]\Km+[S]

Vmax= 1.2microMS-1

Km= 10microM

a. When substrate con.= 2

Vo= 1.2×2/2+10= 2.4/12= . 2

b. When substrate con= 10

Vo= 1.2×10/10+10=12/20= .6

C when substrate con = 30

Vo= 1.2×30/10+30= 3.6/40= .09

> The answer for the substrate concentration of 30 is actually:
Vo= (1.2 x 30) / (10+30) = 36/40 = 0.9, not 0.09 as stated above.

Diana Libby-Billings Sun, Dec 5, 2021 1:51 PM

Answer 2

Similar Homework Help Questions

9. Applying the Michaelis-Menten Equation I An enzyme catalyzes the reaction A = B. The enzyme is present at a con- cen...

9. Applying the Michaelis-Menten Equation I An enzyme catalyzes the reaction A = B. The enzyme is present at a con- centration of 2 nm, and the Vmax is 1.2 ums". The Km for substrate A is 10 um. Calculate the initial velocity of the reaction, Vo, when the substrate concentration is (a) 2 um, (b) 10 um, (C) 30 um.
An enzyme catalyzes the reaction A ⇌ B. The enzyme is present at a concentration of...

An enzyme catalyzes the reaction A ⇌ B. The enzyme is present at a concentration of 2 nM, and the Vmax is 1.2 μM s−1. The Km for substrate A is 10 μM. Calculate the initial velocity of the reaction, V0, when the substrate concentration is (a) 2 μM, (b) 10 μM, (c) 30 μM.
An enzyme catalyzes the reaction M↽−−⇀N . An enzyme catalyzes the reaction M = N. The...

An enzyme catalyzes the reaction M↽−−⇀N . An enzyme catalyzes the reaction M = N. The enzyme is present at a concentration of 3.5 nM, and the Vmax is 1.9 uM -. The Km for substrate M is 2.9 uM. Calculate kcat kcat = 542.86 What values of Vmax and Km would be observed in the presence of sufficient amounts of an uncompetitive inhibitor to generate an a' of 1.5? apparent Vmax = 0.526 UM s-1 apparent Km = 1.2

An enzyme catalyzes the reaction M N. The enzyme is present at a concentration of 3.5...

An enzyme catalyzes the reaction M N. The enzyme is present at a concentration of 3.5 nM, and the Vmax is 2.9 PM s-. The Km for substrate M is 6.5 MM. Calculate kcat kcat = 1 What values of Vmax and Km would be observed in the presence of sufficient amounts of an uncompetitive inhibitor to generate an a' of 1.3? apparent Vmax = UM S-1 apparent Km = UM
10. An enzyme catalyzes the reaction of Y to Z. The enzyme is present at a...

10. An enzyme catalyzes the reaction of Y to Z. The enzyme is present at a concentration of 1 nM. and the Vmax is 2 M/s. The Km for substrate Y is 4 M. Calculate kot
An enzyme catalyzes a reaction with a K of 6.50 mM and a Vmax of 2.75...

An enzyme catalyzes a reaction with a K of 6.50 mM and a Vmax of 2.75 mM s velocity, vo. for the following substrate concentrations a) 2.50 mM Calculate the reaction Number mM. sl b) 6.50 mM Number c) 14.0 mM Number mM.sl

An enzyme catalyzes a reaction with a Km of 8.50 mM and a Vmax of 2.70...

An enzyme catalyzes a reaction with a Km of 8.50 mM and a Vmax of 2.70 mM 5-1. Calculate the reaction velocity, vo, for each substrate concentration. [S] = 2.25 mM Vo: 0.5 mM.s-1 [S] = 8.50 mM 1.70 mMs-1 [S] = 13.0 mM [S] = 13.0 mM Vo: 2.05 mm. s-1
Applying the Michaelis-Menten Equation II Another enzyme is found that catalyzes the reaction A <===> B...

Applying the Michaelis-Menten Equation II Another enzyme is found that catalyzes the reaction A <===> B Researchers find that the K for the substrate A is 4 uM, and the kcat is 20 min^-1 (a) In an experiment, [A] = 6 mM, and the initial velocity, Vo was 480 nM min^-1 What was the [Et] used in the experiment? (b) In another experiment, [Et] 0.5uM, and the measured Vo=5uM min^-1 What was the [A] used in the experiment? ( c)...
An enzyme catalyzes a reaction with a Km of 5.00 mM and a Vmax of 1.85...

An enzyme catalyzes a reaction with a Km of 5.00 mM and a Vmax of 1.85 mM s-1. Calculate the reaction velocity, vo. for the following substrate concentrations. a) 3.00 mM Number mM.s b) 5.00 mM Number mM.s c) 10.5 mM Number

An enzyme catalyzes a reaction with a Km of 8.00 mM and a Vmax of 4.45...

An enzyme catalyzes a reaction with a Km of 8.00 mM and a Vmax of 4.45 mM s-1. Calculate the reaction velocity, Vo, for the following substrate concentrations a) 1.00 mM Number mM.s-1 b) 8.00 mM Number mM s-1 c) 11.0 mM Number mM.s-1