1 / 9

Mechanisms of enzyme inhibition

Mechanisms of enzyme inhibition. Michaelis-Menten mechanism E + S → ES k 1 ES → E + S k 2 ES → P + E k 3 Competitive inhibition: the inhibitor (I) binds only to the active site. EI ↔ E + I

astro
Download Presentation

Mechanisms of enzyme inhibition

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Mechanisms of enzyme inhibition • Michaelis-Menten mechanism E + S → ES k1 ES → E + S k2 ES → P + E k3 • Competitive inhibition: the inhibitor (I) binds only to the active site. EI ↔ E + I • Non-competitive inhibition: binds to a site away from the active site. It can take place on E and ES EI ↔ E + I ESI ↔ ES + I • Uncompetitive inhibition: binds to a site of the enzyme that is removed from the active site, but only if the substrates already present. ESI ↔ ES + I

  2. The efficiency of the inhibitor (as well as the type of inhibition) can be determined with controlled experiments

  3. Autocatalysis • Autocatalysis: the catalysis of a reaction by its products A + P → 2P The rate law is = k[A][P] To find the integrated solution for the above differential equation, it is convenient to use the following notations [A] = [A]0 - x; [P] = [P]0 + x One gets = k([A]0 - x)( [P]0 + x) integrating the above ODE by using the following relation gives or rearrange into with a=([A]0 + [P]0)k and b = [P]0/[A]0

  4. Exercise 23.12a The pKa of NH4+ is 9.25 at 25.0 oC for the reaction of NH4+ and OH- to form aqueous NH3 is 4.0 x 1010 dm3 mol-1 s-1. Calculate the rate constant for proton transfer to NH3. What relaxation time would be observed if a temperature jump was applied to a solution of 0.15 mol dm-3 NH3(aq) at 25.0 oC? • Solution: pKa corresponds: NH4+ + H2O(l) ↔ NH3(aq) + H3O+(aq) The rate constant to be calculated is from NH3 + H2O(l) ↔ NH4+(aq) + OH-(aq) Utilizing the relationship pKa + pKb = pKw • Time constant and relaxation time

  5. 22.14a The rate constant for the decomposition of a certain substance is 2.8 x 10-3 dm3 mol-1 s-1 at 30 oC and 1.38 x 10-2 dm3 mol-1 s-1 at 50 oC. Evaluate the Arrhenius parameters of the reaction. • Hint: Arrhenius parameters include A and Ea k = A e(-Ea/RT), where R is the universal constant

  6. Numerical Problem 25.1 (7th edition) The data below applies to the formation of urea from ammonium cyanate NH4CNO → NH2CONH2. Initially 22.9 g of ammonium cyanate was dissolved in enough water to prepare 1.00 L of solution. Determine the order of the reaction, the rate constant, and the mass of ammonium cyanate left after 300 min. t/min 0 20.0 50.0 65.0 150.0 M(urea)/g 0 7.0 12.1 13.8 17.7 • Solution

  7. Discussion problem: Bearing in mind distinctions between the mechanisms of stepwise and chain polymerization, describe ways in which it is possible to control the molar mass of a polymer by manipulating the kinetic parameters of poltmerization. • Answers: For stepwise polymerization the degree of polymerization is given by <n> = 1 + kt[A]0 For chain polymerization the kinetic chain length is calculated through <n> = kp(fki/kt)-1/2 [M] [I]-1/2

  8. Distinguish between competitive, non-competitive, and uncompetitive inhibition of enzymes. Discuss how these modes of inhibition may be detected experimentally. • Hint: Using Lineweaver-burk plot, look for differences in the intercept with y-axis and changes in the slope.

  9. Exercise 26.6b (7th edition) or 23.2b (8th edition) Consider the following mechanism for the thermal decomposition of R2 (1) R2→ R + R (k1) (2) R + R2 → PB + R’ (k2) (3) R’ → PA + R (k3) (4) R + R → PA + PB (k4) Where R2, PA and PB are stable hydrocarbons and R and R’ are radicals. Find the dependence of the rate of decomposition of R2 on the concentration of R2. • Solution

More Related