An investigator plots 1/v as a function of 1/(concentration of substrate) for da

Question

An investigator plots 1/v as a function of 1/(concentration of substrate) for data collected with and without an inhibitor. The best-fit line of the data collected in the absence of the inhibitor had a slope of 0.29 (arbitrary units of time) and a y-intercept of 0.02 (arbitrary units of time per concentration). The best-fit line of the data collected in the presence of the inhibitor had a slope that was 3.1-times as steep (arbitrary units of time) and a y-intercept of 0.02 (arbitrary units of time per concentration). If the concentration of inhibitor was 53 nM, what is the value of the inhibition constant Ki in terms of nM to the nearest tenth?

Explanation / Answer

Lineweaver-Burk equation for enzyme-substrate reaction in the absence of inhibitor:

1/V=Km/Vmax(1/[S]+1/Vmax

where V=reaction velocity

Km=michaelis-menten constant

[S]=substrate concentration

Vmax=maximum reaction velocity

So ,for plot of 1/V vs 1/[S], a straight line plot of the form y=mx+c,m=slope=Km/Vmx and c=y-intercept=1/Vmax

Given :slope=0.29=km/Vmax

y-intercept=1/Vmax=0.02

Vmax=1/0.02=50

km/50=0.29

km=50*0.29=14.5

Lineweaver-Burk equation for enzyme-substrate reaction in the presence of inhibitor:

As y-intercept=0.02, so Vmax is same in the presence and absence of indicator. So it is competitive inhibition.

Equation for competitive inhibition:

1/V=Km'(app)/Vmax(1/[S]+1/Vmax

with km(apparant)=Km(1+[I]/Ki)

Ki=inhibitor dissociation constant

slope=3.1*0.29=0.889=km(app)/Vmax

km(app)=0.889*Vmax=0.889*0.02=0.018

km(apparant)=Km(1+[I]/Ki)

0.018=14.8(1+53nM/Ki)

Ki=53.064 nM

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