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Enzyme Assay.

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Introduction

Biochemical Simulations. Copyright (c) David A Bender, 1982 - 1997 Enzyme Assay Units of enzyme activity Enzymes are quantified in terms of their catalytic activity - the amount of substrate that is converted to product /unit time, under specified conditions. 1 unit of enzyme activity = 1 ?mol of product formed / minute A number of factors affect the activity of an enzyme, and you will investigate most of these during this exercise: The pH of the incubation The pH of the incubation medium affects the ionisation of both the substrate and the amino acid side chains in the enzyme. Obviously, this will affect both the binding of the substrate to the enzyme and also the catalytic activity of the enzyme. Any enzyme has an optimum pH at which it shows maximum activity. For some enzymes there is a broad range of pH around this optimum where activity is nearly maximum; for others there is only a narrow range of pH over which the enzyme has significant activity. The pH optimum of an enzyme may be very different from the normal plasma pH of 7.4; some enzymes have pH optima as low as 2 - 3, or as high as 9 - 10. If your results are to be meaningful, you have to determine the activity of an enzyme at or near its pH optimum. The temperature of incubation Like all chemical reactions, the rate of enzyme-catalysed reactions increases as the temperature increases. However, because enzymes are proteins, they are also denatured by heat. This means that above a critical temperature (which differs for different enzymes) there is a rapid loss of activity as the enzyme is denatured. Although the program permits you to experiment with incubation temperature, this is only really useful if you want to determine the activation energy of the reaction. You should simply work at 30�C for all of your experiments. ...read more.

Middle

The presence of inhibitors Various compounds reduce the activity of enzymes. They may act in a variety of different ways, and indeed may be reversible or irreversible (i.e. permanent) inhibitors of the enzyme. For the present exercise, we are concerned with only two mechanisms of reversible inhibition of enzymes: competitive and non-competitive inhibition. The stages in an enzyme catalysed reaction can be represented as follows: E + S 3/4 E-S 3/4 E-P 3/4 E + P Where E is the enzyme, S the substrate and P the product. The first stage is the formation of the enzyme-substrate complex, E-S, which then undergoes chemical reaction to the enzyme-product complex, E-P. This then dissociates to release the product and leave the enzyme ready to undergo a further cycle of reaction. Competitive inhibition A competitive inhibitor (I) forms an enzyme-inhibitor complex E-I. This means that it competes with the substrate for the formation of the enzyme-substrate complex: E-I 3/4 E + S + I 3/4 E-S 3/4 E-P 3/4 E + P The relative amounts of E-I and E-S which are formed will depend on the relative concentrations of the inhibitor and the substrate, as well as the affinity of the enzyme for each of them. As the concentration of substrate is increased, so more E-S will be formed (and hence more product will be formed). At a high enough concentration of substrate, the enzyme will achieve the same Vmax as in the absence of inhibitor. A competitive inhibitor has no effect on Vmax, but increases the apparent Km. Non-competitive inhibition A non-competitive inhibitor forms an enzyme-substrate-inhibitor complex (E-S-I) which only proceeds to formation of product slowly: E + S + I 3/4 E-S + I 3/4 E-S-I 3/4 E-P-I 3/4 E + P + I This means that a non-competitive inhibitor does not compete with the substrate for the enzyme, but reduces the rate of reaction of the enzyme at all concentrations of substrate. ...read more.

Conclusion

This will give you some idea of the range of concentrations of substrate over which you can see a hyperbolic relationship between the rate of product formation and the concentration of substrate, as shown on page 3. Then refine the range of concentrations used, so as to get most of your points on the steep part of the curve - don't worry about trying to saturate the enzyme to achieve Vmax. Although you want most of your points on the steep part of the curve, you must be sure that you don't have a straight line relationship between v and S because you have exhausted the substrate! Once you have a set of results showing a hyperbolic relationship between v and S, you can determine Km and Vmax by plotting a graph of 1/v against 1/S (see page 4). When you are varying the concentration of substrate added, the printed table of results includes a table of values of 1/s and 1/v, although these are not shown on the screen. 5) Investigate inhibition of the enzyme There are two inhibitors available. You may use either of them. You should carry out the inhibitor studies using the same range of concentrations of substrate, and the same incubation conditions, as you used to determine the values of Km and Vmax. You should plot a series of graphs of 1/v against 1/S at each concentration of the inhibitor, to decide whether it is competitive or non-competitive (see pages 5 - 6). You should then plot a series of graphs of 1/v against concentration of inhibitor at each concentration of substrate, and use these to determine the value of Ki (see page 6). Plot such graphs as you think are necessary on the following pages and fill in your results in this table: enzyme number pH optimum incubating mL of enzyme with mmol of substrate the formation of product was linear for minutes incubating at pH with mL of enzyme for minutes the Km was mmol/L the Vmax was ?mol/min Inhibitor _____ was ( competitive ( non-competitive Ki of inhibitor ____ mmol/L 1 ...read more.

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