When n = 1, the Hill equation reduces to the MichaelisMenten equation,
Recall that we can make a double reciprocal plot of the MichaelisMenten equation using the LineweaverBurk equation,
Notic that 1/V is a linear function of 1/[S] in the LineweaverBurk equation. We can also take the reciprocal of both sides of the Hill equation as,
Notice that 1/V is now a power function of 1/[S]. When there is no cooperativity, n = 1, the graph of 1/V versus 1/[S] will be a straight line. On the other hand, when there is positive cooperativity, i.e. n > 1, then the graph of 1/V versus 1/[S] will be concave up; and when there is positive cooperativity, i.e. n < 1 then the graph of 1/V versus 1/[S] will be concave down.
Using the above information, does the enzyme pictured in the cartoon plot below exhibit positive, negative, or no cooperativity?
