Enzyme E catalyzes the conversion of substrate A to product B. The Lineweaver-Bu
ID: 98898 • Letter: E
Question
Enzyme E catalyzes the conversion of substrate A to product B. The Lineweaver-Burk plot depicting this conversion is shown as a solid line. The kinetics after addition of compound X to the reaction mixture are also plotted as a dashed line. The kinetics of a mutant version of E called E* are also depicted on the plot. You measure the kinetics for the reaction with E* in the presence of X and find that they are identical to the kinetics of E* without X (data not shown). Based on all these data, answer the following questions: a) What is the MOST LIKELY conclusion about the effect of X on E? Explain. b) What conclusion can you make about the effect of the mutation on E? Explain. c) What is the BEST explanation for the effect of X on E*? Explain.Explanation / Answer
The enzyme , also called a biocatalyst, accelerates the rate of a reaction by enhancing the rate of conversion of substrates into products without altering the equilibrium constant of the reaction.
The effect of enzyme on velocity of reaction with respect to varying levels of substrate concentration was first devised by Michaelis-Menten and was plotted in a graph with substrate concentration as X-axis and velocity V as Y-axis. It gave a hyperbolic curve.
Later for more precise calculations Line-Weaver Burk double reciprocal plot was devised where 1/V0 is the Y-axis and 1/[S] is the X-axis. The slope is Km/Vmax which is extrapolated to give - 1/Km.
1/V0 = Km/Vmax ×1/[S] + 1/Vmax
(a) A competitive inhibitor bears close structural similarity with the substrate and competes with the substrate for binding to the enzyme. It binds to the substrate binding site of the enzyme and tends to decrease the substrate affinity of the enzyme i. e. the Kmincreases and Vmax remains same as it can be achieved by increasing the substrate concentration. As is seen in the graph E+X has decreased 1/Km and thereby increased Km value. 1/Vmax is same. So, X is a competitive inhibitor.
(b) The product of mutation of E created an enzyme which has slightly lowered 1/Km and thereby slight increased Km or, substrate affinity and increased Vmax . So, mutated E does not bind the substrate so readily as E and has less potential to accelerate the reaction.
(c) As mutated E has developed different structural conformation due to mutation, so the competitive inhibitor X does not hold structural analogy any longer with mutated E.
Therefore, it cannot affect the action of mutated E. So, the graph is same for action of mutated E on substrate whether in presence or absence of X.
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