Question

For her undergraduate project, Jessica studied an enzyme that catalyzes the reaction A↽−−⇀B. For substrate A, she determined that ?m=2.5 μM and ?cat=35 min−1. Jessica graduated and her project has been passed on to you. Unfortunately, Jessica was so busy that she sometimes forgot to record all of the details of an assay in her lab notebook. Your mentor suggests that you try to back calculate some of the missing concentration values. Assume that the enzyme follows Michaelis–Menten kinetics.

1) In experiment A, where [A]=5 mM, she found that the initial velocity, ?0, was 144 nM min−1. What was the [Et] in experiment A?

[Et]=????nM

2) In experiment B, where [Et]=0.6 μM, she found that ?0=6 μM min−1. What was the [A] in experiment B?

[A]=????uM

3)The compound Z was found to be a very strong competitive inhibitor of the enzyme, with an α of 10. In experiment C, in which the same [Et] was used as in experiment A but a different [A] was used, an amount of Z was added that reduced the rate ?0 to 72.0 nM min−1. What was the [A] in experiment C?

[A]=????uM

4)A measure of the catalytic efficiency of an enzyme is the initial slope of the Michaelis–Menten curve at low [S], the second‑order rate constant ?cat/?m. A perfect enzyme, limited only by diffusion, would have a value of 10^8 or 10^9 M−1s−1.Calculate ?cat/?m for this enzyme.

Kcat/Km=?????M^-1s^-1

Answer 1

We know that

Turnover number, Kcat = Vmax/Et

Michaelis-Menten equation,   V_o = (Vmax[S]) / (Km+[S])

For our problem,

Kcat = Vmax/Et ------ Eq (1)

Et = Vmax/Kcat ------- Eq (2)

V_o =(Vmax[A]) / (Km+[A]) ------- Eq(3)

Vmax = Vo (Km+[A]) / [A] --------- Eq (4)

1) Given that  ?m = 2.5 μM = 2.5 x 1000 nM = 2500 nM [ 1 μM = 1000 nM ]

?cat =35 min−1

[A]= 5 mM = 5000000 nM [ 1 mM = 1000000 nM]

Vo = 144 nM min⁻¹

  Et = ?

Then, From Eq(4)

  Vmax = Vo (Km+[A]) / [A]

= 144 nM min⁻¹ [ 2500 nM + 5000000 nM] / [5000000 nM]

= 144.072 nM min⁻¹

Vmax = 144.072 nM min⁻¹

Then, From Eq(2)

Et = Vmax/Kcat

= 144.072 nM min⁻¹/ 35 min⁻¹

= 4.116 nM

Therefore,

Et = 4.116 nM

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2)

Given that  ?m = 2.5 μM

?cat =35 min−1

[Et]= 0.6 μM

Vo = 6 μM min−1

[A] = ?

Kcat = Vmax/Et ------ Eq (1)

35 min−1 = Vmax/0.6 μM

Vmax = 21 μM min-1

Then,  Vmax = Vo (Km+[A]) / [A] --------- Eq (4)

21 μM min-1 = 6 μM min−1 { 2.5 μM + [A] }/[A]

3.5 = { 2.5 μM + [A] }/[A]

3.5 [A] = 2.5 μM + [A]

2.5 [A] = 2.5 μM

[A] = 1 μM

Therefore,  [A] = 1 μM (or) 1 uM

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3) Given that same [Et] was used as in experiment A.

Hence, Et = 4.116 nM

Given data is

?o =   72.0 nM min−1

?m = 2.5 μM = 2.5 x 1000 nM = 2500 nM [ 1 μM = 1000 nM ]

?cat =35 min−1

[A] = ?

Kcat = Vmax/Et ------ Eq (1)

35 min−1 = Vmax/4.116 nM

Vmax = 144 nM min-1

Then,  Vmax = Vo (Km+[A]) / [A] --------- Eq (4)

144.06 nM min-1 =  72.0 nM min−1 { 2500 nM + [A] }/[A]

2 = { 2500 nM + [A] }/[A]

2 [A] = 2500 nM + [A]

[A] = 2500 nM

[A] = 2500 x 0.001  μM [ 1 nm = 0.001 μM]

= 2.5  μM

Therefore,  

[A] = 2.5  μM (or) 2.5 uM

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4) For perfect enzymes, the upper limit of Kcat/Km is between 10^8 and 10^9 M−1s−1.

Therefore.

Kcat/Km = 10^8 to 10^9 M−1s−1.