BCH380 Homework 3 Name: 12. Hexokinase is an enzyme that obeys Michaelis-Menten

Question

BCH380 Homework 3 Name: 12. Hexokinase is an enzyme that obeys Michaelis-Menten kinetics. [glucose] (mM) 0.20 0.40 0.80 (A) Draw a double-reciprocal plot for hexokinase using the data below (label the axes!) (B) What is the Vmax for the uninhibited enzyme (remember the correct units)? (C) What is the KM for the uninhibited enzyme (remember the correct units)? (D) If the amount of enzyme added was 0.10 nM, what is the kecat for the uninhibited enzyme? (E) What type of inhibitor is fructose (CsH120s)? (F) If you assayed the same enzyme with fructose alone (which has a lower affinity than glucose), would you expect KM to be greater or smaller? Vo, no inhibitor (mM/min) Vo, with 5 mM fructose 1.67 2.50 3.33 1.00 1.67 2.50

Explanation / Answer

Michaelis- Menten kinetics is given by

1/V =KM/Vmax)*1/S+ 1/Vmax (1), so a plot of 1/ V vs 1/S gives straight line whose slope is KM/Vmax and intercept is 1/Vmax.

for the case of inhibition, KM becomes KMapp and Vmax becomes Vmaxapp.

from the plot of equation of best fit. for no inhibitor

1/Vmax= intercept = 0.2, Vmax= 1/0.2= 5 mM/min =5*10-3 M/minand slope is KM/Vmax= 0.079 and KM= 0.079*5= 0.395 mM=0.395*10-3M

given Et= Enzyme concentration = 0.1nM= 0.1*10-9 M and Kcat= Vmax/Et= 5*10-3/ (0.1*10-9) /min=5*107/min

for the presence of inhibitor, from the equation of best fit, 1/Vmaxapp= 0.2, Vmaxapp=1/0.2= 5mM/min and Kmapp/Vmaxapp =0.159, Kmapp=0.159*5mM=0.795 mM=0.795*10-3M

since Vmax remain the same in the presence and absence of an enzyme, the type of inhibition is competitive inhibition.
inhibitior is a molecule similar to the substrate competes with the substrate for the active site. For competitive inhibition, KM increases.

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