Enzymes lower the activation energy of a chemical reaction (without changing the

Question

Enzymes lower the activation energy of a chemical reaction (without changing the overall free energy change). In other words, they promote or catalyze reactions. To describe how effective an enzyme is, we consider the speed of the reaction, v for velocity. As a substrate concentration increases, the reaction rate also increases in a hyperbolic way, up to a maximum speed , Vmax. The amount of substrate needed to reach 1/2 Vmax is called the Michaelis-Menton constant, Km. A simplified model for understanding this relationship uses the reciprocals to create a linear equation: 1/v = (Km/Vmax)(1/[S]) + I/Vmax and this can be plotted in an x-y coordinate plane such that y = (Km/Vmax) x + 1/Vmax 4. What is the slope of this line? This kind of graph is called a Lineweaver-Burk plot: 5. As the concentration of substrate [S] gets smaller, which direction do you ''move'' on the graph above?

Explanation / Answer

line weaver burk plot is a double reciprocal plot ; the above graph is an example of uncompetitive inhibhition , in the absence of inhibhitor slope is increased . the intercept is unchanged and it doent alter vmax unchanged . but there will be increase in km

slope = km /v max

5. as concentration of the substrate gets smaller , the graph move down wards towards the left side down in the above graph

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