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 30ºC for all of your experiments. It is usual to determine enzyme activity at 30ºC, although 37ºC is sometimes used for mammalian enzymes.
The time of incubation
For as long as the enzyme remains active, and there is substrate present to undergo reaction, the amount of product formed will increase with increasing time of reaction, levelling off as equilibrium is reached.
If the enzyme loses activity or is denatured, or if the supply of substrate is exhausted, then the formation of product will slow down and level off earlier during the incubation.
Obviously, in any studies in which you will be expressing the results /unit time (i.e. the rate of reaction) you have to know that the enzyme was acting at a constant rate over the time of incubation - i.e. that there was a (nearly) linear increase in the amount of product formed with increasing time.
The amount of enzyme present
As long as there is enough substrate present, you would expect the amount of product formed to increase in a linear fashion as the amount of enzyme increases. This is not always so.
Some enzymes form dimers or other aggregates at high concentration, and this can affect their catalytic activity, either increasing or decreasing it compared with the monomer at low concentration.
Other enzymes are unstable at low concentrations, and so are readily denatured. This means that there is less activity than you would expect at low concentrations of enzyme.
Sometimes there is also non-enzymic conversion of substrate to product, or there may be some product already present in the sample you are working with, so that there is some product present even when there is no enzyme present, or it has been incubated for zero time.
It is therefore useful to ensure, before you carry out too many experiments, that the enzyme you are working with behaves in a predictable manner as the amount of enzyme present in the incubation is varied.
The concentration of substrate
A simple chemical reaction generally shows a linear relationship between the rate of formation of product and the concentration of substrate available. This is not so for enzyme catalysed reaction.
At high concentrations of substrate (region [A] in the graph) there is enough substrate available to ensure that as soon as the catalytic site of the enzyme is empty (i.e. as soon as the product has left) more substrate is available to bind and undergo reaction. The rate of formation of product now depends on the innate activity of the enzyme itself. ...
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The concentration of substrate
A simple chemical reaction generally shows a linear relationship between the rate of formation of product and the concentration of substrate available. This is not so for enzyme catalysed reaction.
At high concentrations of substrate (region [A] in the graph) there is enough substrate available to ensure that as soon as the catalytic site of the enzyme is empty (i.e. as soon as the product has left) more substrate is available to bind and undergo reaction. The rate of formation of product now depends on the innate activity of the enzyme itself. Adding more substrate will not affect the rate of the reaction to any significant effect - the enzyme is saturated with its substrate.
At low concentrations of substrate (region [B] in the graph) the catalytic site of the enzyme is empty, waiting for substrate to bind, for much of the time. Therefore that rate at which product can be formed is limited by the amount of substrate which is available to bind to the enzyme. As the concentration of substrate increases from very low levels, so the rate of formation of product increases very steeply.
This means that a graph of the rate of reaction (i.e. amount of product formed /min) against the concentration of substrate is hyperbolic. At high concentrations of substrate, the curve tends towards the maximum rate of reaction, Vmax.
In order to characterise an enzyme, and understand how its activity will respond to changes in the concentration of substrate in tissues, and therefore understand how metabolic pathways are regulated, we need to know both the maximum rate at which it can act (Vmax) and also the concentration of substrate which is required to saturate the enzyme. An enzyme which is saturated with its substrate under normal conditions will act at a more or less constant rate, while the activity of an enzyme which is not saturated will vary as the concentration of substrate varies.
If we want to determine the amount of an enzyme present in a sample of tissue, obviously we have to determine its activity at a concentration of substrate which is high enough to ensure that it is saturated at therefore acting at Vmax.
Because the relationship between the rate of reaction and the concentration of substrate is hyperbolic, as shown in the diagram above, it is not possible to determine with any precision the concentration of substrate at which the enzyme becomes saturated. Although it is quite easy to extrapolate the curve above to see what is the Vmax that the enzyme will achieve, it is not possible to say at what concentration of substrate it achieves Vmax. What we do is to determine the concentration of substrate which gives a rate equal to half Vmax.
The original work which established the theoretical basis of our understanding of the relationship between enzyme activity and substrate concentration was performed by Michaelis and Menten, and the concentration of substrate to give half Vmax is called the Km (Michaelis constant) of the enzyme.
Km is an inverse measure of the affinity of the enzyme for its substrate, and the units of Km are units of concentration (normally mmol or ?mol /L).
The units of Vmax are amount of product formed /time - e.g. mmol /min or ?mol/ min).
The rate of reaction (v) is related to the concentration of substrate (S) by the following equation (the Michaelis equation): v = Vmax x S
(Km + S)
This means that if we know the values of Km and Vmax for an enzyme, we can predict its rate of reaction at any concentration of substrate.
It is difficult to fit a smooth curve to experimental points, and therefore it is difficult to determine Km and Vmax from the hyperbolic graph of v against S. It is usual to plot a derived graph which is linear, so that the best fitting line between the experimental points can be calculated or drawn with adequate precision.
The most commonly used such derived plot is the Lineweaver-Burk double reciprocal plot:
/v is plotted against 1/S.
The Lineweaver-Burk double reciprocal plot:
cuts the y axis (1/v) at a value of 1/Vmax and cuts the x axis (1/S) at a value of -1/Km.
It is thus a simple matter to determine the Km and Vmax of an enzyme from the results of a series of incubations at different concentrations of substrate, plotting 1/v against 1/S and extrapolating the graph back the x axis.
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.
A non-competitive inhibitor has no effect on the apparent Km, but decreases Vmax.
The inhibitor constant, Ki, is an indication of how potent an inhibitor is; it is the concentration to give half maximum inhibition.
Plotting 1/v against concentration of inhibitor at each concentration of substrate:
for a competitive inhibitor the lines converge above the x axis;
for a non-competitive inhibitor the lines converge on the x axis;
in both cases the value of [I] where they meet is - Ki.
The computer simulation exercise
Rather than performing a large number of incubations under different conditions in order to characterise an enzyme, you will use a computer simulation, based on the Michaelis equation, to perform a relatively large number of 'experiments' in a short time.
The program permits you to choose any one of five different enzymes, which have different characteristics. Your task is to characterise the enzyme you have chosen, and investigate its inhibition by two compounds.
In any one set of 'incubations' you may vary one of the incubation conditions:
* pH
* time of incubation
* incubation temperature
* amount of enzyme
* concentration of substrate added.
You will be asked for the minimum and maximum values you want to use (the permissible range is shown on the screen) and the 'incubations' will be performed using 11 different values over the range you have chosen.
If you are happy with the range, the click OK, otherwise you can change it.
You then have to set values for the other incubation conditions. Initially these are set to sensible 'guesses'. As you determine better values, so you can enter them. Future 'experiments' will use the new values that you have entered, unless you change them again.
Note that if you start a series of 'experiments' varying a new incubation condition, the box for the value of the condition that you were varying in the last set of 'experiments' will be red, and a value of '0' will appear. You will have to set the value for this incubation condition to what seems appropriate from your previous set of 'experiments'.
You have a choice of performing 'experiments' with or without inhibitors. Obviously, you should not waste time studying the effects of inhibitors until you have established the basic incubation conditions.
After you click the perform experiments button, your results appear , as a table and a graph. If the results are slow to appear, be patient - the program is a real-time simulation. The longer your incubation time, the longer it will take before you have any results.
You may save any of the results that you think will be useful. These results will be printed out when you finish running the program. Note that the printout only includes the tables of results, and not the graphs. You do not have to save all of your results, but you must click save these results to print if you want that set of data saved.
When you have set the incubation conditions, you will be asked whether you want to perform the 'incubations' with an inhibitor or not. You may perform each set of 'incubations' without added inhibitor or in the presence of one of two inhibitors ['A' and 'B']. You will be offered a choice of 4 ranges of concentration of inhibitor to add to your 'incubations'. If you select too low a range of concentrations of inhibitor, you will see little or no inhibition. If you select too high a range of concentrations of inhibitor, then there will be little or no enzyme activity to be measured.
You should carry out the following 'experiments' on your chosen enzyme:
) Determine the pH at which it has maximum activity
Initially, you have no idea of the range of pH over which the enzyme has significant activity, so you should perform the first experiment over a wide range (e.g. 1 - 14). You can refine this is subsequent experiments to get a more precise estimate of the pH optimum. Be realistic; do not try to determine the pH optimum to more than one decimal place.
Once you have determined the optimum pH, use this pH for all further experiments.
2) Determine the time over which the enzyme acts at a constant rate
You will need to express the results of later experiments in terms of the rate of enzyme activity - i.e. the amount of product formed /unit time. Therefore you have to be sure that the enzyme is acting at a constant rate - i.e. that there is a more or less linear increase in the amount of product formed with increasing time - over the incubation time that you will use in later experiments.
Using the minimum amount of enzyme and the maximum amount of substrate, and working at the optimum pH, vary the incubation time.
For future experiments, you should use an incubation time well within the range over which the enzyme shows a linear increase in formation of product with increasing time. You will need to be sure that when you divide product formed by time incubated you really have the rate of reaction, and that as near as possible the reaction has been continuing at a constant rate.
If the reaction continues steadily for (say) up to 40 min, use at most 10 or 20 min.
3) Determine the effect of varying the amount of enzyme
To make sure that the formation of product that you observe really is due to the enzyme, you should vary the amount of enzyme added - using the maximum amount of substrate, incubating at the optimum pH, and for a short time within the range over which you know the enzyme acts at a constant rate.
For future experiments there is no need to use more than the minimum amount of enzyme which gives you measurable formation of product. Remember that in real life the enzyme is likely to be the most valuable component of the incubation - you may only have a few hundred mg of tissue from a patient.
4) Determine the Km and Vmax of the enzyme
Once you have established the basic 'incubation' conditions, you can try the effect of varying the amount of substrate added. Initially you should try using a wide range (perhaps the whole of the permitted range). 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