Chapter 3: Enzymes
results to plot the two types of graph. The table in Question 3.8 gives you some results and it is worth spending some time answering the question before proceeding. Figure 3.14a shows a double-reciprocal plot. Note that it is a straight line. Using this graph, we can find Vmax in the following way. First, we find 1/Vmax . This is the point where the line crosses (intersects) the y-axis because this is where 1/[S] is zero (and therefore [S] is infinite). Once we know 1/Vmax we can calculate Vmax. Another useful value can be obtained from the double-reciprocal plot, namely the Michaelis–Menten constant, Km. The Michaelis–Menten constant is the substrate concentration at which an enzyme works at half its maximum rate (½Vmax). At this point half the active sites of the enzyme are occupied by substrate. The higher the affinity of the enzyme for the substrate, the lower the substrate concentration needed for this to happen. Thus the Michaelis–Menten constant is a measure of the affinity of the enzyme for its substrate. The higher the affinity, the lower the Michaelis–Menten constant and the quicker the reaction will proceed to its maximum rate, although the maximum rate itself is not affected by the Michaelis– Menten constant. Vmax and Km therefore provide two different ways of comparing the efficiency of different enzymes. Vmax gives information about the maximum rate of reaction that is possible (though not necessarily the rate
under cell conditions) while Km measures the affinity of the enzyme for the substrate. The higher the affinity, the more likely the product will be formed when a substrate molecule enters the active site, rather than the substrate simply leaving the active site again before a reaction takes place. These two aspects of efficiency are rather like using the maximum speed and acceleration to measure the efficiency of a car. How can we find Km from a double-reciprocal plot? The answer is that the point where the line of the graph intersects the x-axis is −1/Km (note that it is in the negative region of the x-axis). Figure 3.14a shows this point. From the value for −1/Km, we can calculate Km. Figure 3.14b shows the normal plot (as in Figure 3.7 and your first graph in Question 3.8). The relationship between ½Vmax and Km is shown in Figure 3.14b. The value of Km for a particular enzyme can vary, depending on a number of factors. These include the identity of the substrate, temperature, pH, presence of particular ions, overall ion concentration, and the presence of poisons, pollutants or inhibitors. Turnover numbers, which are related to Vmax and Km values, for four enzymes are shown in Table 3.2. This shows the great variation in efficiency that is possible between enzymes, and the fact that Vmax and Km are independent of each other.
Vmax 1 Vmax
= intercept on y-axis 1 V 2 max
1 v
Initial rate
–1 = intercept on x-axis Km
1 [S]
Km Substrate concentration
Figure 3.14 a A double-reciprocal plot of substrate concentration against initial rate: b A graph showing the effect of substrate concentration on initial rate, with Vmax, ½ Vmax and Km values shown.
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