Question

An enzyme is discovered that catalyzes the chemical reaction

SAD --> <-- HAPPY

A team of motivated researchers set out to study the enzyme, which they call happyase. Under a constant enzyme concentration, [E] of 2.0 uM, they studied the relationship between substrate concentration [SAD] and reaction velocity V₀. Their data is shown below:

[SAD] (uM)                V₀ (uMs^-1)      

20                    3.4 x 10³

40                   5.4 x 10³

60                   6.6 x 10³

80                    7.4 x 10³

Can you determine K_cat and K_m from the above data?

2. While [E] remains 2.0 um and [SAD] = 12 uM, what would the reaction velocity V₀ be?

3. In a separate happyase experiment using [E] = 1.0 uM, the reaction velocity, V_0, is measured as 3.6 x 10³ uMs^-1. What is the [SAD] used in this experiment?

4.The researchers working on happyase discovered that the compound STRESS is a potent inhibitor of happyase. A kinetics study of happyase in the presence of         1 nM STRESS, gives the following data:

[SAD] (uM)                V₀ (uMs^-1)      

20                    1.7 x 10³

40                   2.7 x 10³

60                    3.3 x 10³               

80                    3.7 x 10³

Which class of inhibitors does STRESS belong to?

Answer 1

(1) The values are calculated below

[SAD] (uM)	V0 (uM s-1)	1/[SAD](uM-1)	1/V0(s/uM)
20	3.40E+03	0.05	0.000294118
40	5.40E+03	0.025	0.000185185
60	6.60E+03	0.016666667	0.000151515
80	7.40E+03	0.0125	0.000135135

The plot is shown below

From lineweaver burk plot,

1/V0=(Km/Vmax) (1/[S]) + 1/Vmax

y=mx+c

The plot 1/V0 vs 1/[S] gives a slope=Km/Vmax and y-intercept=1/Vmax

From the above plot, y-intercept=8.06188x10^-5=1/Vmax

Therefore Vmax=1/8.06188x10^-5=12404.05 uM/s.

Slope=0.00426=Km/Vmax

Therefor Km=0.00426 x Vmax=0.00426 x 12404.05

Km=52.84 uM.

K_cat=Vmax/[E] where [E]=enzyme concentration

Given [E]=2 uM

K_cat=(12404.05 uM/s)/2 uM

K_cat=6202.03 s^-1.

(2) The Michaelis–Menten equation is

V₀=(K_cat x [E]_tot x [S])/(Km+[S])

Given [E]tot=2 uM, [SAD]=[S]=12 uM, Kcat=6202.03 s^-1, Km=52.84 uM

V₀=(6202.03 s^-1 x 2 uM x 12 uM)/(52.84 uM+12 uM)

V₀=2295.63 uM/s.

(30) Given [E]=1 uM and V0=3.6x10³ uM/s, the [SAD]=?

The Michaelis–Menten equation is

V₀=(K_cat x [E]_tot x [S])/(Km+[S])

3.6x10³ uM/s=(6202.03 s^-1 x 1 uM x [SAD])/(52.84 uM+[SAD])

3.6x10³ uM/s x (52.84 uM+[SAD])=(6202.03 s^-1 x 1 uM x [SAD])

190224 uM/s + 3.6x10³ uM/s [SAD]=6202.03 uM/s [SAD]

190224=2602.03 uM/s [SAD]

[SAD]=190224/2602.03

[SAD]=73.1 uM.

As per our guidelines I have to answer first question but I answered three for you. Thanks.