Derive the kinetic equations for Vmax and KM using the transit time and net rate constant method as discussed in class and follo
wing the attached handout. (Hint: solve for ET/v by summing the inverse net rate constants, then invert to obtain v/ET and put it in the form of kcatS/[KM S]). Remember that if a step is irreversible, the net rate constant for that step is just the rate constant for that step.
1 answer:
Answer:
Rate = vmax k3/k2+k3
Explanation:
The rate of reaction when the enzyme is saturated with substrate is the maximum rate of reaction, is referred to as Vmax.
This is usually expressed as the Km ie. Michaelis constant of the enzyme, an inverse measure of affinity. For practical purposes, Km is the concentration of substrate which permits the enzyme to achieve half Vmax.
Please kindly check attachment for the step by step solution of the given problem.
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Answer:
When the ball goes to first base it will be 4.23 m high.
Explanation:
Horizontal velocity = 30 cos17.3 = 28.64 m/s
Horizontal displacement = 40.5 m
Time
Time to reach the goal posts 40.5 m away = 1.41 seconds
Vertical velocity = 30 sin17.3 = 8.92 m/s
Time to reach the goal posts 40.5 m away = 1.41 seconds
Acceleration = -9.81m/s²
Substituting in s = ut + 0.5at²
s = 8.92 x 1.41 - 0.5 x 9.81 x 1.41²= 2.83 m
Height of throw = 1.4 m
Height traveled by ball = 2.83 m
Total height = 2.83 + 1.4 = 4.23 m
When the ball goes to first base it will be 4.23 m high.
Answer:
Detailed step wise solution is attached below
Explanation:
(a) wavelength of the initial note 2.34 meters
(b) wavelength of the final note 0.389 meters
(d) pressure amplitude of the final note 0.09 Pa
(e) displacement amplitude of the initial note 4.78*10^(-7) meters
(f) displacement amplitude of the final note 3.95*10^(-8) meters
