An enzyme catalyzes a reaction in which substrateA is cleaved into two products, P and Q. In the catalytic mechanism, the enzyme converts A to a covalently-bound reaction intermediate X and product P, P then desorbs from the enzyme, and in a second chemical step, the enzyme converts the intermediate X in the EX complex to the final product Q (in EQ), which then desorbs from the enzyme E.You discover two inhibitors of this enzyme, I and J. I is a competitive inhibitor of the substrate A, and has nearly double the molecular weight of J. On the other hand, J is a mixed inhibitor of enzyme E, and its inhibitory effect on Km/ Vmax (the slope effect from the double reciprocal plot) is greater than that of 1 / Vmax (the intercept effect in a double reciprocal plot). That is Kis< Kii .At low pH, the conversion of EX to EQ is greatly slowed, kcat is decreased, and the intercept effect of inhibitor J is elevated, that is, the value of Kiiis diminished.When a high, fixed concentration of product P is added to the enzymatic reaction, the inhibition due to inhibitor I is unchanged, but inhibitor J becomes a competitive inhibitor vs. A.(1) To which enzyme forms do I and J bind? Explain how you got there.I cannot bind when A is at a concentration of infinity (no intercept effect). I is competitive vs. A and therefore binds to E. I may structurally resemble substrate A.J is a mixed inhibitor; it exerts both a slope effect and an intercept effect, but the slope effect is more significant than the intercept effect: Kis< Kii.J probably binds to E twice: to E to form EJ, which gives rise to the slope effect, and the intercept effects comes from binding to EA, or a downstream complex such as EXP, EX, or EQ. J has structural features similar to I, whichis why it can bind to E, but is smaller than I and can fit into another pocket during catalysis.

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Asked Nov 30, 2019
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An enzyme catalyzes a reaction in which substrateA is cleaved into two products, P and Q. In the catalytic mechanism, the enzyme converts A to a covalently-bound reaction intermediate X and product P, P then desorbs from the enzyme, and in a second chemical step, the enzyme converts the intermediate X in the EX complex to the final product Q (in EQ), which then desorbs from the enzyme E.You discover two inhibitors of this enzyme, I and J. I is a competitive inhibitor of the substrate A, and has nearly double the molecular weight of J. On the other hand, J is a mixed inhibitor of enzyme E, and its inhibitory effect on Km/ Vmax (the slope effect from the double reciprocal plot) is greater than that of 1 / Vmax (the intercept effect in a double reciprocal plot). That is Kis< Kii .

At low pH, the conversion of EX to EQ is greatly slowed, kcat is decreased, and the intercept effect of inhibitor J is elevated, that is, the value of Kiiis diminished.

When a high, fixed concentration of product P is added to the enzymatic reaction, the inhibition due to inhibitor I is unchanged, but inhibitor J becomes a competitive inhibitor vs. A.

(1) To which enzyme forms do I and J bind? Explain how you got there.

I cannot bind when A is at a concentration of infinity (no intercept effect). I is competitive vs. A and therefore binds to E. I may structurally resemble substrate A.

J is a mixed inhibitor; it exerts both a slope effect and an intercept effect, but the slope effect is more significant than the intercept effect: Kis< Kii.

J probably binds to E twice: to E to form EJ, which gives rise to the slope effect, and the intercept effects comes from binding to EA, or a downstream complex such as EXP, EX, or EQ. J has structural features similar to I, whichis why it can bind to E, but is smaller than I and can fit into another pocket during catalysis.

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Expert Answer

Step 1
  • A competitive inhibitor binds to active site of an enzyme and prevents the binding of actual substrate. It causes a reduction in the Km (Michaelis constant) value, however, the value of Vmax remains the same.
  • A mixed inhi...

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