Question 1: You are working on understanding the mechanism of a newly discovered

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Question 1: You are working on understanding the mechanism of a newly discovered aminopeptidase. You model it using simple Michaelis-Menton kinetics.  

You obtain the following microscopic rate constants:

k1

1.8 x 107

M-1s-1

k-1

2.1 x 105

s-1

k2

8.7 x 105

s-1

k-2

4.4 x 102

M-1s-1

a.Write out the general form of the M-M equation showing E, S, ES, P and placing the 4 microscopic rate constants in their appropriate positions.

For [enzyme] = 3.0 nM and [substrate] = 1.2 uM, calculate the macroscopic kinetic parameters:

b.Km
c.kcat (turnover number)
d.Vmax
e.Specificity constant (kcat/Km)
f.Initial reaction rate
g.What assumptions do you make when you use simple Michaelis-Menton kinetics?
h.Describe (in words, not equations) what the turnover number and the specificity constant mean, and the difference between them.
i.Use the specificity constant to compare this enzyme’s efficiency to those of the enzymes described in table 12-1 of the text.
j.If this enzyme fails to match up to the efficiency of the best enzymes, in which kinetic parameter(s) does it fail (kcat, Km, or both)?
k.What would an “aminopeptidase” do?

Explanation / Answer

a). Reaction Rate v = d[P]/dt = k2{[E] [S]} / {km + [S]}

Where Km = (k2 + k-1) / k1

   Therefore, v = {k1 k2 ([E] [S])} / {k2 + k-1 + k1 [S]}

b). km = (k2 + k-1)/ k1

km = {(8.7* 105) + (2.1 * 105)} / (1.8 * 107)

km = (10.8 * 105) / (1.8 * 107)

km = 6 * 102 M

c). turnover number = kcat = k-1 = 2.1 * 105 s-1

d). vmax = k-1 [E]

= (2.1 * 105)s-1 * (3 * 10-9) M

= 6.3 * 10-4 M s-1

e). Specificity Constant = kcat / km
= (k-1 k1)/ ( k2 + k-1)

= (2.1 * 105 ) * (1.8 * 107) /{(8.7 * 105) + (2.1 * 105)}

= (3.78 * 1012 )/ (10.8 * 105)

= 0.35 * 107 or 3.5 * 10

g). Assumptions to make while using simple Michaelis-Menton kinetics:

1). The concentration of the substrate molecules is greater than the concentration of the product molecules.

[E] >> [P]

2). The reactants Enzyme [E] and the Substrate [S] are in rapid equillibrium with the Enzyme substrate complex [ES]

3). Steady State Approximation : The intermediate in the reaction is very short lived species. i.e Enzyme-Substrate complex is consumed as soon as it is generated.

h). Turnover number ( kcat) is the rate of conversion of Substrate molecules per second for a given enzyme concentration.

Specificity Constant is actually the kinetic efficiency of the enzyme, which means how efficiently an enzyme is converting substrates into products

The difference between the two is as follows :

Specificity Constant = m mol of substrate / (minutes * mg of Enzyme )

Turnover Number = m mol of substrate / (minutes * m mol of Enzyme )

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