Question 32

[admin]

mvenus929
Guest

Posted: 07 Apr 2007 00:21 Post subject: Questions on GS-4: 15, 17, 28, 32, 48, 50

[jvuofm1709]

I believe D is the best answer. It asks for what accounts for the shape of the graph...yes it is a enzyme concentration graph...that doesn't mean that enzyme saturation accounts for its shape. First, the passage and the answer D itself make it very clear that every enzyme has a distinct Km value. Second, the Km characterizes a specific point on the graph: the substrate concentration where Vmax/2. This means that not only does Km graphically account for the Vmax/2, it also defines the vmax:the y axis maximum where ENZYME SATURATION occurs ON THE GRAPH. Km is very important in determining the shape of a graph, this is exemplified by a competitive inhibitor's effect on the Km. It will shift the Km down the x axis, giving the graph a more gradual slope to the vmax. So the Km can define different graph shapes for varying reaction conditions. This is only the tip of the iceberg on the importance of km on the shape of enzyme activity graphs, a more complete description involves michaelis menten kinetics and double reciprocal plots which intertwine the variables of Km, vmax, and the slope more precisely. So, yes, enzyme saturation is depicted in the graph, but all that shows is a straight horizontal line...the Km actually defines where that line is and helps define the slope of the graph at low concentrations, defining the shape of the graph.

[admin]

The explanation you provide with respect to the Km value being the best answer is an excellent argument in particular I can understand your insight.

However, the only problem is the fact that how would the Km be calculated [?] from the curve itself as there are no numerical values etc. So the answer B would, I believe be best or correct as the answer D indicates the calculation of the Km value but the problem is that one may not really be able to calculate it directly from the curve and so, the answer would be simply: answer B as the curve shows clearly saturation.