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# The virtual laboratory: Enzyme assay

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### copyright © 1982 - 2006 David A Bender

### Virtual Laboratory main menu

### Click here to run the program

### Click here to see theory pages

The program will then run in a separate window, and at any time you can minimise the program window to check the theory from this page.

As you run the program, you are asked after each simulated experiment whether you wish to save the results to print out. If you save results, they are saved in a file called enzassout.csv in your temporary file area. The program automatically locates your temporary file area, and displays the path on the opening screen. When you close the program you are given the option of printing out the results you have chosen to save for printing. You cannot print out results until you end the program.

You can open enzassout.csv in spreadsheet software to perform further calculations on the data. This will be especially valuable when you come to determine Km and Vmax of your chosen enzyme, and investigate the effects of inhibitors.

If you use Microsoft Excel to plot the graphs, remember that you have to plot the points as xy scatter graphs, not line graphs. Once you have plotted the graphs, right click on the line to extrapolate – use the command “add trendline”, and use the option to display the equation on the chart. This will display the equation in the form y = mx +c. Calculation of Km and Vmax from this linear equation depends on which method you have used to linearise the data (see the notes on substrate dependence):

In this simulation you can investigate the effects of pH, time, amount of enzyme, incubation temperature and substrate concentration on the activity of five different enzymes. You can then investigate the effects of adding two different inhibitors.

The five enzymes differ in:

> pH optimum
> maximum rate of activity (hence Vmax)
> affinity for substrate (hence Km)
> temperature dependence and thermal denaturation
> inhibitor sensitivity

In each set of experiments with your chosen enzyme, you may vary one of the following:

> the pH of the incubation buffer (between 1 - 14)
> the time of incubation (between 0 - 60 min)
> the volume of enzyme added (between 0 - 250 µL)
> the incubation temperature (between 0 - 100 °C)
> the concentration of substrate (between 0 - 250 mmol /L)

In each case, a series of 11 incubations will be set up over your chosen range. You then have to set the other incubation conditions.

When you first run the program, you are presented with a set of more or less standard conditions; you may change these or not, as you wish, and in future experiments your changed values will be shown on the screen. You may change these or not as you wish.

You may perform all the experiments with or without an added inhibitor. There are two inhibitors available, which differ in their affinity and type of inhibition. You are offered a series of different ranges of inhibitor concentration; whichever each range you select you will see results for four concentrations of inhibitor, as well as a control with no added inhibitor.

If your chosen range of inhibitor seems to make little difference to the results, you have used too little, and should choose a higher range.
If your chosen range of inhibitor results in more or less complete inhibition of the enzyme, then you have used too much, and should choose a higher range.

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Click the links below to jump to different pages of theory, or just browse down

> Effect of pH
> Effect of incubation time
> Effect of the amount of enzyme added
> Effect of the incubation temperature
> Substrate dependence of enzyme activity - the importance of Km and Vmax
> How to determine Km and Vmax
> Inhibition of enzyme activity

# pH of the incubation

Any enzyme has an optimum pH, at which it shows maximum activity.

Binding of the substrate to the active site of an enzyme involves interaction with reactive groups provided by the side-chains of amino acids at the binding site. The pH of the incubation medium may affect the ionisation of both the substrate and the amino acid side-chains and will therefore this will affect binding. It may also affect the ionisation of reactive groups that catalyse the reaction, although in the micro-environment of the catalytic site, when it is occupied by the substrate, this is less likely. Extreme values of pH may also disrupt the tertiary structure of the enzyme, and so distort the active site, or even denature the enzyme protein. Different proteins will have different sensitivity to extreme values of pH.

When it is mainly uncharged reactive groups that are involved in the interaction between an enzyme and its substrate, there will be a relatively broad range of pH over which the enzyme has activity. This graph shows the pH dependence of an enzyme that has maximum activity at pH 7.1 (i.e. an optimum pH of 7.1), and a relatively broad range of pH around that optimum at which it has activity. There would still be measurable activity at pH 6 or pH 8.

By contrast, when it is mainly charged groups that are involved in the interaction between an enzyme and its substrate, there will be a relatively narrow range of pH over which the enzyme has activity. This graph shows the pH dependence of an enzyme that also has a pH optimum of 7.1, but a narrower range of pH at which it is active. There is negligible activity at pH 6 or pH 8.

The pH optimum of an enzyme may be very different from the normal plasma or intracellular average pH of 7.35 - 7.45; some enzymes have pH optima as low as 2 - 3, or as high as 9 - 10, and therefore little or no activity around pH 7. In vivo, subcellular compartmentation means that enzymes that have pH optima very different from the intracellular average still act at or near their optimum pH.

This graph shows the pH dependence of two different enzymes, both found in blood plasma, which catalyse the same reaction: hydrolysis of phosphate esters. The enzyme shown in red has a pH optimum of about 3.8, and is known as acid phosphatase, while that shown in blue has a pH optimum around 9.5, and is known as alkaline phosphatase. Neither has any significant activity at the pH of plasma (7.35 - 7.45), and indeed neither has any physiological function in plasma. However, measurement of alkaline phosphatase in plasma can give valuable information about liver function and metabolic bone disease, while measurement of acid phosphatase in plasma can be useful in diagnosis of prostate disease.

If your results are to be meaningful, the activity of an enzyme must be determined at or near its optimum pH. Obviously, in the example of these phosphatases, you would work at around pH 3.8 to measure acid phosphatase, and around pH 9.5 to measure alkaline phosphatase. If you worked at pH 7 then neither enzyme would show any activity.

When you come to vary the incubation pH in this simulation you will be offered, initially, a broad range of pH, from 1 - 14. Using this range will give you an approximate idea of the optimum for your chosen enzyme, then you should repeat the experiments with a narrower range of pH to give a more precise estimate of the pH optimum. However, do not try to be too precise. With a computer simulation, it is tempting to attempt to determine an extremely precise value for the pH optimum, but in the laboratory you would be unlikely to achieve precision of better than 0.1 pH unit when preparing the incubation buffer, and an error of ± 0.1 pH unit will not have any significant effect on the activity of the enzyme.

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# Incubation time

The longer an enzyme is incubated with its substrate, the greater the amount of product that will be formed. However, the rate of formation of product is not a simple linear function of the time of incubation.

All proteins suffer denaturation, and hence loss of catalytic activity, with time. Some enzymes, especially in partially purified preparations, may be noticeably unstable, losing a significant amount of activity over the period of incubation.

If the activity of the enzyme is such that much of the substrate is used up during the incubation, then, even if the concentration of substrate added was great enough to ensure saturation of the enzyme at the beginning of the experiment, it will become inadequate as the incubation proceeds, and the formation of product will decrease.

Enzyme catalysed reactions are reversible. Initially, there is little or no product present, and therefore the reaction proceeds only in the forward direction. However, as the reaction continues, so there is a significant accumulation of product, and there is a significant rate of back reaction. As a result, the rate of formation of product slows down as the incubation proceeds, and if the incubation time is too long, then the measured activity of the enzyme is falsely low.

In some studies, especially when investigating the substrate dependence of the rate of reaction, it is usual to make measurements of the formation of product at relatively short time intervals (say every 10 seconds) for the first minute or so, then plot a graph of the amount of product formed against time, and determine the initial rate of reaction by drawing the tangent to the steepest part of the rate curve. However, in short incubations there can be a considerable error of timing; 1 second is a significant error in a short incubation, but negligible when the incubation time is several minutes. Similarly, in short incubations only a small amount of product has been formed, and analytical errors are magnified when the amount of product is extremely small.

Selecting an appropriate incubation time depends on a compromise between these various factors. As a general rule, the incubation should be long enough to permit a moderate amount of product to be formed, and long enough that the error in timing is insignificant, but not so long that there is detectable levelling off of the curve.

You need to be sure that when you determine the rate of reaction (in mol of product formed / minute) the enzyme has been active at a more or less constant rate throughout your incubation.

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# Effect of the amount of enzyme present

As long as there is substrate present, the amount of product formed might be expected to increase in a linear fashion as the amount of enzyme increases, as shown in the graph on the left.

This is not always so.

Some enzymes form dimers or other aggregates at high concentration. This may affect the catalytic activity, either increasing or decreasing it compared with the monomer in a dilute solution. Similarly, dimers, or other aggregates that are the normal form of the enzyme may disaggregate when the preparation is diluted, again affecting the catalytic activity of the sample.

Some enzymes are unstable in dilute solution, and so are readily denatured, resulting in lower activity in dilute solutions than would otherwise be expected. This is especially a problem with purified enzyme preparations, where there is very little protein present in the sample. One common way of overcoming the instability of purified preparations is to add a relatively large amount of an inert protein, such as albumin, to act as a chaperone to the purified enzyme.

This graph shows the effect of increasing the concentration of an enzyme that has low activity as the monomer and higher activity when it has formed a dimer at a higher concentration of enzyme.

The graph below shows the opposite - an enzyme that has higher activity as the monomer, and decreased activity as it associates to form a dimer, at higher concentrations.

Sometimes there may be some non-enzymic conversion of substrate to product, or there may be some product already present in the tissue sample being used, so that there is some product present even when there is no enzyme present, or it has been incubated for zero time. Investigation of. The graph below shows the results you would observer if there were formation of the product other than as a result of enzyme activity.

In general, the enzyme preparation is the most valuable constituent of the incubation; it may have taken several days to produce a purified preparation, or you may be working with a very small sample of tissue from a patient. Therefore, it is usual to set up incubations using the lowest amount of enzyme that can be pipetted with acceptable precision, and which leads to the formation of a readily detectable amount of product.

If you are working with a relatively unpurified preparation, or do not know the molecular mass of your enzyme, then you can express the activity of the enzyme as:

> mol of product formed / unit time / volume of preparation
> mol of product formed / unit time / gram of original tissue
> mol of product formed / unit time as the total amount in the original tissue

If you are working with a partially purified enzyme preparation you should express its activity as the specific activity:

> specific activity = mol of product formed / unit time / mg protein

If you are working with a purified enzyme preparation, and know its molecular mass, or if there is some other way of determining the molar concentration of the enzyme in the sample, e.g. by measurement of a prosthetic group bound to the enzyme protein, then it is possible to express the activity of the enzyme as the catalytic rate constant:

> kcat = mol of product formed / sec / mol of enzyme.

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# The incubation temperature

The effect of temperature on the rate of an enzyme-catalysed reaction is the result of two opposing factors:

As with any chemical reaction, the rate increases as the temperature increases, since the activation energy of the reaction can more readily be provided at a higher temperature. This means, as shown in the graph below, that there is a sharp increase in the formation of product between about 5 - 50°C.

Because enzymes are proteins, they are denatured by heat. Therefore, at higher temperatures (over about 55°C in the graph below) there is a rapid loss of activity as the protein suffers irreversible denaturation.

In this graph , the enzyme was incubated at various temperatures for 10 minutes, and the amount of product formed was plotted against temperature. The enzyme showed maximum activity at about 55 °C.

In this graph, the same enzyme was incubated at various temperatures for just 1 minute and the amount of product formed was again plotted against temperature. Now the increased activity with increasing temperature is more important than the loss of activity due to denaturation and the enzyme shows maximum activity at 80 °C.

This graph shows the results of incubating the same enzyme at various temperatures for different times ranging from 1 minute to 10 minutes - the longer the incubation time the lower the temperature at which there is maximum formation of product, because of the greater effect of denaturation of the enzyme.

This means that it is not useful to attempt to determine an 'optimum' temperature for an enzyme-catalysed reaction.

By convention, enzyme activity is determined at 30°C; this is a compromise between mammalian and clinical biochemists, who would expect to work at 37°C, and microbial biochemists, most of whom would expect to work at 20°C.

Investigation of the temperature dependence of an enzyme can be useful, for example in biotechnology and biochemical engineering, where there may be operational reasons for working at a relatively high temperature, so that enzymes with a higher thermal stability are advantageous.

## Determination of the activation energy of the reaction

The effect of increasing temperature on the rate of formation of product up to the point at which denaturation begins to reduce the activity of the enzyme can be investigated to determine the activation energy of the reaction. The equation relating the rate of a chemical reaction to temperature was derived empirically by Arrhenius in 1889:

> where:
> k = the rate constant for the reaction
> A = a constant for the reaction
> E = the activation energy
> R = gas constant = 1.987 cal / deg / mol = 8.31434 joules / deg / mol
> T = temperature (°K, = °C + 273)

T aking logs gives:

log = log A - E / 2.303 * R * T

For temperatures below that at which there is significant denaturation of the enzyme, if the enzyme is saturated with substrate, the rate constant is = Vmax, and a graph of log Vmax against 1 / T will be a straight line with gradient = - E / 2.303 * R

Activation energy
E = -4.576 * gradient (cal / deg / mol)
E = -19.148 * gradient (joules / deg / mol)

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# The effect of substrate concentration on enzyme activity

Skip the theory and go straight to: How to determine Km and Vmax

A simple chemical reaction with a single substrate shows a linear relationship between the rate of formation of product and the concentration of substrate.

For an enzyme-catalysed reaction, there is usually a hyperbolic relationship between the rate of reaction and the concentration of substrate.

(A) At low concentration of substrate, there is a steep increase in the rate of reaction with increasing substrate concentration. The catalytic site of the enzyme is empty, waiting for substrate to bind, for much of the time, and the rate at which product can be formed is limited by the concentration of substrate which is available.

(B) As the concentration of substrate increases, the enzyme becomes saturated with substrate. As soon as the catalytic site is empty, more substrate is available to bind and undergo reaction. The rate of formation of product now depends on the activity of the enzyme itself, and adding more substrate will not affect the rate of the reaction to any significant extent.

The rate of reaction when the enzyme is saturated with substrate is the maximum rate of reaction, Vmax.
The relationship between rate of reaction and concentration of substrate depends on the affinity of the enzyme for its substrate. This is usually expressed as the Km (Michaelis constant) of the enzyme, an inverse measure of affinity.

For practical purposes, Km is the concentration of substrate which permits the enzyme to achieve half Vmax. An enzyme with a high Km has a low affinity for its substrate, and requires a greater concentration of substrate to achieve Vmax."

## The importance of determining Km and Vmax

The Km of an enzyme, relative to the concentration of its substrate under normal conditions permits prediction of whether or not the rate of formation of product will be affected by the availability of substrate.

An enzyme with a low Km relative to the physiological concentration of substrate, as shown above, is normally saturated with substrate, and will act at a more or less constant rate, regardless of variations in the concentration of substrate within the physiological range.

An enzyme with a high Km relative to the physiological concentration of substrate, as shown above, is not normally saturated with substrate, and its activity will vary as the concentration of substrate varies, so that the rate of formation of product will depend on the availability of substrate.

If two enzymes, in different pathways, compete for the same substrate, then knowing the values of Km and Vmax for both enzymes permits prediction of the metabolic fate of the substrate and the relative amount that will flow through each pathway under various conditions.

In order to determine the amount of an enzyme present in a sample of tissue, it is obviously essential to ensure that the limiting factor is the activity of the enzyme itself, and not the amount of substrate available. This means that the concentration of substrate must be high enough to ensure that the enzyme is acting at Vmax. In practice, it is usual to use a concentration of substrate about 10 - 20-fold higher than the Km in order to determine the activity of an enzyme in a sample.

If an enzyme is to be used to determine the concentration of substrate in a sample (e.g. glucose oxidase is used to measure plasma glucose), then the substrate must be the limiting factor, and the concentration of substrate must be below Km, so that the rate of formation of product increases steeply with increasing concentration of substrate, so providing a sensitive assay for the substrate."

## How to determine Km and Vmax

Km and Vmax are determined by incubating the enzyme with varying concentrations of substrate; the results can be plotted as a graph of rate of reaction (v) against concentration of substrate ([S], and will normally yield a hyperbolic curve, as shown in the graphs above.
The relationship is defined by the Michaelis-Menten equation:

v = Vmax / (1 + (Km/[S]))

I t is difficult to fit the best hyperbola through the experimental points, and difficult to determine Vmax with any precision by estimating the limit of the hyperbola at infinite substrate concentration. A number of ways of re-arranging the Michaelis-Menten equation have been devised to obtain linear relationships which permit more precise fitting to the experimental points, and estimation of the values of Km and Vmax. There are advantages and disadvantages associated with all three main methods of linearising the data.

The Lineweaver-Burk double reciprocal plot rearranges the Michaelis-Menten equation as:

1 / v = 1 / Vmax + Km / Vmax x 1 / [S]

plotting 1/v against 1/[S] give a straight line:

> y intercept = 1 / Vmax
> gradient = Km / Vmax
> x intercept = -1/ Km

> This is the most widely used method of linearising the data, and generally gives the best precision for estimates of Km and Vmax. However, it has the disadvantage of placing undue weight on the points obtained at low concentrations of substrate (the highest values of 1/[S] and 1/v). These are the points at which the precision of determining the rate of reaction is lowest, because the smallest amount of product has been formed.

The Eadie-Hofstee plot rearranges the Michaelis-Menten equation as:

v = Vmax - Km x v / [S]

plotting v against v / [S] gives a straight line:

> y intercept = Vmax
> gradient = -Km
> x intercept = Vmax / Km

This plot overcomes the problem of uneven spacing of points, and undue weight given to points at low concentrations of substrate. However, it has the disadvantage that v, which is a dependent variable, is used on both axes, and hence errors in measuring the rate of reaction are multiplied, resulting in lower precision of the estimates of Km and Vmax

The Hanes plot rearranges the Michaelis-Menten equation as:

[S] / v = Km / Vmax + [S] / Vmax

plotting [S] / v against [S] gives a straight line:

> y intercept = Km / Vmax
> gradient = 1 / Vmax
> x intercept = -Km

This plot overcomes the problem of uneven spacing of points, and undue weight given to points at low concentrations of substrate. However, it has the disadvantage that [S] is used on both axes, and hence pipetting errors, which lead to errors in the true concentration of substrate available, are multiplied, resulting in lower precision of the estimates of Km and Vmax.

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# Enzyme inhibitors

Various compounds can reduce the activity of enzymes. They may act in a variety of different ways, and indeed may be reversible or irreversible inhibitors of the enzyme.

On this page there are notes about:

> > Competitive inhibition
> > Non-competitive inhibition
> > Uncompetitive inhibition
> > The choice of a competitive or non-competitive inhibitor as a drug
> > Ki, the inhibitor constant

An irreversible inhibitor causes covalent modification of the enzyme, so that its activity is permanently reduced. Compounds that act as irreversible inhibitors are often useful as drugs that need be taken only every few days, although adjusting the dose to suit the patient’s response is a lengthy process with such compounds. By contrast, the effect of a reversible inhibitor can be reversed by removing the inhibitor, e.g. by dialysis or gel filtration.

The normal sequence of an enzyme reaction can be represented as:

> where:
> E = enzyme
> S = substrate
> E-S = enzyme-substrate complex
> E-P = enzyme-product complex
> P = product

There are three main types of reversible inhibitor:

> competitive inhibitor
> non-competitive inhibitor
> uncompetitive inhibitor

They interact with the enzyme or enzyme-substrate complex at different stages in the sequence

## Competitive inhibition

A competitive inhibitor competes with the substrate for the active site of the enzyme:

This means that increasing the concentration of substrate will decrease the chance of inhibitor binding to the enzyme. Hence, if the substrate concentration is high enough the enzyme will reach the same Vmax as without the inhibitor. However, it will require a higher concentration of substrate to achieve this and so the Km of the enzyme will also be higher. Reacting the enzyme with a range of concentrations of substrate at different concentrations of a competitive inhibitor will give a family of curves as shown below:

The Lineweaver-Burk double reciprocal plot for this set of data shows a series of lines crossing the y (1/v) axis at the same point - i.e. Vmax is unchanged, but with a decreasing value of 1/Km (and hence a higher Km) in the presence of the inhibitor:

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## Non-competitive inhibition

A non-competitive inhibitor reacts with the enzyme-substrate complex, and slows the rate of reaction to form the enzyme-product complex.

This means that increasing the concentration of substrate will not relieve the inhibition, since the inhibitor reacts with the enzyme-substrate complex. Reacting the enzyme with a range of concentrations of substrate at different concentrations of a non-competitive inhibitor will give a family of curves as shown below:

The Lineweaver-Burk double reciprocal plot for this set of data shows a series of lines converging on the same point on the X (1/S) axis - i,.e. Km is unchanged, but Vmax is reduced:

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## Uncompetitive inhibition

This is a very rare class of inhibition. An uncompetitive inhibitor binds to the enzyme and enhances the binding of substrate (so reducing Km), but the resultant enzyme-inhibitor-substrate complex only undergoes reaction to form the product slowly, so that Vmax is also reduced:

Reacting the enzyme with a range of concentrations of substrate at different concentrations of an uncompetitive inhibitor will give a family of curves as shown below:

The Lineweaver-Burk double reciprocal plot for this set of data shows a series of parallel lines - both Km and Vmax are reduced:

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## The choice of a competitive or non-competitive inhibitor as a drug

If the requirement is to increase the intracellular concentration of the substrate, then either a competitive or non-competitive inhibitor will serve, since both will inhibit the utilisation of substrate, so that it accumulates.

However, if the requirement is to decrease the intracellular concentration of the product, then the inhibitor must be non-competitive. As unused substrate accumulates, so it will compete with a competitive inhibitor, and the final result will be a more or less normal rate of formation of product, but with a larger pool of substrate. Increasing the concentration of substrate does not affect a non-competitive inhibitor.

## Ki, the inhibitor constant

The inhibitor constant, Ki, is an indication of how potent an inhibitor is; it is the concentration required to produce half maximum inhibition.

Plotting 1/v against concentration of inhibitor at each concentration of substrate (the Dixon plot) gives a family of intersecting lines.

For a competitive inhibitor, the lines converge above the x axis, and the value of [I] where they intersect is -Ki

For a non-competitive inhibitor, the lines converge on x axis, and the value of [I] where they intersect is -Ki

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