Enzyme kinetics - Assignment Example

Enzyme Kinetics al Affiliation) Kinetic Properties of Enzymes a) i) The rate of the initial reaction The reaction rate of an enzyme-catalyzed reaction varies with substrate concentration. The common form of the equation is given by:Vo = Vmax (S) KM + (S)Where: Vo = initial reaction velocityVmax = maximal reaction velocity when all enzyme active sites are filled with substrate molecules(S) = Substrate concentrationKM = Michaels constant = K2+k3 K1Rate = change in product Change in timeInitial rate = 12-0 = 1.

33 µmol min-1 9-0ii) Effect of doubling the substrateThere is a hyperbolic relationship between the rate of reaction and the concentration of substrate.The doubling of the concentration of substrate leads to the saturation of the enzyme by the substrate. When the catalytic site becomes empty, more substrate is available to bore and undergo reaction. Vmax is the maximum rate of reaction and denotes the rate of reaction when the enzyme is saturated. The relationship between concentration of a substrate and rate of reaction depends on the affinity of the enzyme for its substrate.

This is expressed as the michaelis constant (Km) of the enzyme, which is an inverse measure of affinity.iii) Time in minutes before 12 µmol of product forms in second reactionInitially 1 = 1.33 2 = 3.99Recall at the rate of 1.33, t=9Therefore, t = (3.99/1.33) * 9mins = 27minsPart C Enzyme InhibitorsEnzyme inhibitors are molecules that interact with the enzyme preventing it from working in a typical way. A competitive inhibitor is a compound that resembles the molecular geometry and the chemical structure of the substance.

Such inhibitors compete with the substrate molecule for the same site. We can measure the enzyme inhibitor ki through isothermal titration calorimeter. This involves titrating the inhibitor into a solution of enzyme and then measuring the heat released. Another way of measuring inhibitors is through Michaels Menten equation. Here, we need to observe an enzyme activity under a variety of substrates and concentrations. We can then plot the fitting data on graphs using Line weaver-Burk (Leskovac, 2003).

Part D (Short answer question)100 mm = 1 L (def of molarity)100mm in 1000 ml of the solutionY = 10 mlY = (10*100)/1000 =1mol75 g of glycerin =1 mol glycerinTherefore, the solution contains 75 mg of glycerin.ReferencesLeskovac, V. (2003). Comprehensive enzyme kinetics. New York: Kluwer Academic/Plenum Pub.