Question

In an enzyme catalyzed reaction, Km -
4x10 umol/l, and the rate of reaction (V) at
substrate concentration [S] - 1.2 x 10 Mis
80 pmol/l-min Assume no inhibitor is
present. Vmax is practically equal to
(1) 40 umoll-min​

Answer 1

Explanation:

Vmax is the inverse of the (1/V0) -intercept on the Lineweaver-Burke plot:

Vmax = 1/(4.6155 [L-min/mmol S]) = 0.217 mmol S/L-min

The slope of the Lineweaver-Burke plot is equal to Km/Vmax, so Km = slope x Vmax:

Km = (0.378 [L-min/mmol S]/[L/mmol S]) x (0.217 [mmol S/L-min]) = 0.0820 mmol S/L

Vmax = kcat x [E]total so kcat = Vmax/[E]total

kcat = (0.217 [mmol S/L-min]/(1.2 x 10-4 [mmol E/L] = 1.81 x 103 [(mmol S/mmol E)/min] x (1 min/60 sec)

kcat = 30.1 sec-1

Experiment 2:

Using the same methods as in experiment 1,

Vmax(apparent) = 0.183 mmol S/L.min

Km(apparent) = 0.152 mmol S/L

Inhibition is competitive, but note that with realistic data like this, there is some uncertainty in this conclusion. I conclude that inhibition is competitive because the inhibitor raises Km by almost 85% and lowers Vmax by only 16%. Given the poor data (look at the R2 values), the change is Vmax is probably not significant. You may judge otherwise.KI = Km[I]/(Kmapp - Km)

KI = (0.0820 [mmol S/L] x 0.033 [mmol I/L])/(0.152 [mmol S/L] - 0.0820 [mmol S/L])

KI = 0.039 mM

Compare KI with Km (not Kmapp). The inhibitor binds about twice as tightly as the substrate.