1. A has a lower K_M. The V_max for Enzyme X is over 40; probably it is 44 or so.

     A: K_M= 8.0 and V_max = 44.0

     B: K_M= 16.0 and V_max = 44.0

     Note: although you should get the same results for any two sets of data, I didn't find that to be the case, although my results were consistent if I ignore the tenths.

     C: The K_Mshould be either 8 or 16. Using 8 for the K_M gives V_max = 36, and a visual check
      of the chart looks consistent with that. That is, [S] = 8 does seem to give v₀ = 18.

           Using 16 for the K_Mgives V_max = 32.5 which looks less realistic. It seems a bit low for
      the V_max , and the [S] that gives v₀ = 16 for curve C is clearly less than 10.

2. The data you should have for a–c ([S] = 0 - 400) are on the left, and your chart should look much like the one on the right. I could include all of the data, but I don't expect you to check all of the numbers.

3. a. Not from the chart; b. Yes; c. Yes.

4. The additional data are on the left.

     a. Yes; b. Yes; c. No; d. K_Mbecause an enzyme in a cell is more likely to have [S] near the K_Mrather than a high [S].

5. Again, data are included for [S] = 0 to 600.

6. a. No–about all you can tell is that the y-intercept is between 0.001 and 0.002, which
     gives a very broad range for the reciprocal.
b. No. The chart above doesn't even show the x-intercept for the uninhibited plot.
c. This clearly shows that the three lines have the same V_max and different K_M's, which
    indicates competitive inhibition. That is possible but not clear from looking at the
    Michaelis-Menton chart.

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