What is the physical significance of ? We have said that the is inversely proportional to the affinity of E for S only when << . What about a more general case? If we set the rate at half of the maximal rate (v = /2) then it can be shown that = [S], i.e. the Michaelis constant is the substrate concentration when the rate is half of the . Furthermore, note that has the same units as [S]. The also tells something about the efficiency of catalysis and control. When [S]>> 10 the reaction is efficient but control (by altering the [S]) is poor. When [S] << 0.1 the efficiency of the reaction is poor but there is good control of the rate. When [S] is within one order of magnitude of the both control and efficiency are reasonable. Above or below this range they are not. Note that the rate changes as the relationship between [S] and changes.
|at [S] =10||v/ = 0.91|
|at [S] =||v/ = 0.50|
|at [S] = 0.1||v/ = 0.09|
Try calculating other values of [S].
At the high end nearly all of the enzyme is saturated an therefore efficiently utilized. Below this range the enzyme is mostly free, with no bound substrate and is therefore not being used efficiently. So,physiologically, if an enzyme is to work efficiently it should be operating in a substrate concentration of or better. But if control by substrate concentration is important, the [S] should ideally be below about 5 .
If you are setting up an assay, either in the clinical lab or the research lab, you must have the enzyme saturated with substrate. Knowing the lets you estimate the substrate concentration necessary to ensure saturation. You need at least twice the .
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