Question

The results for enzyme activity analysis are presented in Table 5 (below). In this case, do not use the graph to determine: (1.5 pts each) (Hint- use M-M equation) a) Vmax; b) Km; c) initial velocity at [S] 1 x 10-1 M; d) the amount of product formed during the first 5 minutes at [S] 2x10 3M; at a [S of 2x10 6M? e) what is Km and Vmax if the free [E is increased 3-fold? 3. v (umol/min 0.25 0.25 0.25 0.20 0.07 0.01 x 10-2 5x 10-3 5x 10-4 x 10-5 5 x 10-6 5 x 10 Table 5. Enzyme Activity Data 4. Consider the subtilisin' substrates A and B (pay attention to binding interaction) Phe-Ala-GIn-Phe-X Phe-Ala-His-Phe-X These substrates are cleaved (between Phe and X) by native subtilisin at essentially the same rate However, the His64-to-Ala subtilisin mutant activity data indicates that the substrate B is cleaved more than 1000-fold as rapidly as substrate A is cleaved. Assuming the data is correct, please provide an explanation. (3 pts)

Answer 1

We are asked to use Michelis-Menten equation to find out the parameters. The equation is:

V= V_max X [S] / Km + [S], Where V is initial velocity or rate of the enzyme catalyzed reaction, V_max is the maximum velocity or maximum rate of the reaction attained under saturating substrate concentration, Km is the substrate concentration at half the maximum rate, [S] is concentration of substrate

With this information we can now, find out

a. Vmax- It is very clear from the table that the rate does not increase more than 0.25 Micromole/ min even when the substrate concentration is increased. Therefore, the Vmax is 0.25 micromole/min.

b. We can use the M-M equation and the data from any of the point in the table to find out Km. Let us take the point when the [S] is 5 x 10^-5. At this concentration, the initial rate (v) is  0.20. Now, putting the values for v, [S] and Vmax, we can calculate the km. It comes to 1.25 X 10^-5 moles/ l. The Km value may be checked at other points too like when [S] is 5 X 10 ^-6 ,and V is 0.07. The value comes to approx the same that is 1.2 X 10^-5. Therefore it is rechecked.

c.  The initial velocity at [S] = 1 X 10^-1 will be same as Vmax because it is a very high substrate concentration and at this concentration the initial velocity will not become more than the Vmax. Also, if the M-M equation is used to calculate the initial velocity at such a high substrate concentration, the km is very very less to [S] so that denominator term becomes [S] and therefore v is equal to Vmax.

d. To find out the product formed in the first 5 min with the given substrate concentration, we first need to find out the initial rate of reaction. When the [S] is 2 X 10^-3, which is again a high substrate concentration so that the initial velocity is Vmax i.e.0.25 micromoles/min. Therefore, amount of product formed in 5 min will be 0.25 X 5 = 1.25 micromoles

For [s] = 2 X 10^-6 , we can use the MM equation to find out the initial rate of reaction. By putting the values it comes to around 0.017 micromoles/min. Therefore in 5 min 0.085 micromoles of product will be formed.

e. If the enzyme concentration is increased 3-fold the Vmax will also increase three fold as more enzyme will be available to bind to substrate as it increases. Therefore the Vmax will become 0.75 micromoles/min. Km is the substrate concentration to achieve half this rate.