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10. Ahern's BB 350 at Oregon State University - Enzymes II


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مشاهده 246

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توسط cementbrain در 10 Jul 2018
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Email me at kgahern@davincipress.com
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Highlights Enzymes II

1. To account for the amount of enzyme in a reaction, Kcat (also called turnover number) is used. Kcat is equal to Vmax/[Enzyme]. Because the concentration of enzyme is taken into account in this equation, Kcat does NOT vary with the amount of enzyme used and is therefore a constant for an enzyme. Kcat is equal to the number of molecules of product made per enzyme per unit time. A Kcat of 5/second means that that enzyme makes five molecules of product per molecule of enzyme per second.

2. A very important number that does NOT vary according to the quantity of enzyme used (that is to say that it is a constant for a given enzyme) is the Km (the Michaelis constant). Km turns out to be the concentration of substrate required to get an enzymatic reaction to half maximum velocity. Km is a constant for any given enzyme and provides a measure of an enzyme's "affinity" for its substrate. An enzyme with a high Km has a low affinity for its substrate. An enzyme with a low Km has a high affinity for its substrate. Note that Km is NOT Vmax/2. Instead, it is the substrate concentration required to get a reaction to Vmax/2.

3. The active site of an enzyme is the place on the enzyme where the substrate binds and where the reaction is catalyzed. Temperature affects an enzymatic reaction in two ways. First, temperature is a factor in the Gibbs Free Energy, which determines the energy of a reaction. Second, temperature can denature an enzyme and thus negatively impact an enzyme's activity.

4. A very important number for an enzyme iis the Km (the Michaelis constant). Km turns out to be the concentration of substrate required to get an enzymatic reaction to half maximum velocity. Km provides a measure of an enzyme's "affinity" for its substrate. An enzyme with a high Km has a low affinity for its substrate. An enzyme with a low Km has a high affinity for its substrate. Note that Km is NOT Vmax/2. Instead, it is the substrate concentration required to get a reaction to Vmax/2.

5. A "perfect enzyme" is one that has a maximum value for Kcat/Km. These enzymes have maximaly velocities and minimal Km values (high affinity). Perfect enzymes are slowed only by the rate with which the substrate can diffuse to the active site of the enzyme.

6. Determining Vmax from a plot of V versus S is not easy. Consequently, an alteration of this plot is done to make the calculation simpler. The most common alteration is known as a Lineweaver-Burk (double-reciprocal) plot. In it, a double reciprocal plot is performed - 1/V versus 1/S. When this is plotted for an enzymatic reaction, a line is produced, with the x-intercept (place where the line intersects the x-axis) equaling -1/Km and the y-intercept (place where the line intersects the y-axis) equaling 1/Vmax.

7. Inhibition of enzymes occurs competitively when the inhibitor of the enzyme resembles the substrate and competes with the substrate for binding to the enzyme. Because they are competing with each other for the same site on the enzyme, the substrate can "overcome" the inhibitor at high concentrations (because reactons are set up with a fixed amount of inhibitor). Thus, although it will require more substrate in the presence of an inhibitor to get the same enzyme reaction velocity as when there is no inhibitor, the Vmax of a competitive inhibition is unchanged from the uninhibited reaction.

8. Note above that "it will require more substrate in the presence of an inhibitor to get the same enzyme reaction velocity as when there is no inhibitor" so the Km will change. It requires MORE substrate to get the reaction with the inhibitor to half maximum velocity (compared to uninhibited). Therefore, the Km for a reaction undergoing competitive inhibition INCREASES, meaning the affinity of the enzyme for substrate DECREASES.

9. In non-competitive inhibition, the inhibitor binds to a site on the enzyme that is NOT related to the substrate. Therefore, substrate and inhibitor are NOT competing for the enzyme. This means then that increasing the substrate concentration does NOT affect the inhibitor's ability to inhibit the enzyme. It also means that a fixed percentage of the enzyme is always inactivated by a non-competitive inhibitor. In non-competitive inhibition, the apparent Km is the same as the uninhibited reaction, whereas the Vmax decreases.

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Enzymes


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