Start with your lowest catechol concentration and record the colour change every 30 seconds after adding the enzyme. When plotted on a graph of Absorbance units against Time the linear region usually corresponding to the initial rate of the resulting graph will give you the velocity of the reaction in Absorbance units per second. Repeat this for each concentration of catechol but keeping the concentration of enzyme constant. For each substrate concentration, calculate the rate velocity of reaction Absorbance units produced per unit Time.
This will enable you to plot a graph of Velocity of reaction absorbance units per sec against Substrate concentration M.
From the graph find the maximum velocity and half it i. Draw a horizontal line from this point till you find the point on the graph that corresponds to it and read off the substrate concentration at that point.
This will give the value of Km. 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 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 - 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. 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:. 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:. This is the most widely used method of linearising the data, and generally gives the best precision for estimates of Km and Vmax.
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