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:
A. 2.82 eV
B. 439nm
C. 59.5 angstroms
Explanation:
A. To calculate the energy of the photon emitted you use the following formula:
(1)
n1: final state = 5
n2: initial state = 2
Where the energy is electron volts. You replace the values of n1 and n2 in the equation (1):
B. The energy of the emitted photon is given by the following formula:
(2)
h: Planck's constant = 6.62*10^{-34} kgm^2/s
c: speed of light = 3*10^8 m/s
λ: wavelength of the photon
You first convert the energy from eV to J:
Next, you use the equation (2) and solve for λ:
C. The radius of the orbit is given by:
(3)
where ao is the Bohr's radius = 2.380 Angstroms
You use the equation (3) with n=5:
hence, the radius of the atom in its 5-th state is 59.5 anstrongs