I. Michaelis-Menton Enzyme Kinetics

A. If [S] = KM then v = Vmax{s}/2[S] = Vmax/2 or KM is the substrate concentration at which the velocity of the reaction is half-maximal

note the units we are using here. KM = k-1 + k2/k1 where k-1 is first-order (s-1), k2 is first-order (s-1), and k1 is second-order thus KM has units of M.

B. If [S] <<< KM then [S] + KM approaches KM and v ~ Vmax[S]/KM which is the equation of a straight line which then explains the linearity of the graph at low[S]

C. Other Mechanisms

k1 k2 k3

Reaction E + S <---> ES <---> EP ---> E + P

k-1 k-2

The reaction still may follow Michaelis-Menton kinetics v = Vmax[S]/[S] + KM but the terms that define KM and Vmax may be more complex, and thus our assumptions may be incorrect when we derived the Michaelis-Menton equation.

KM = k-1k3 + k-1k-2 + k2k3/k1(k2 + k-2 + k3)

Vmax = k2k3[ET]/(k2 + k-2 + k3)

Similarly, just because we are able to measure a Vmax and KM does not mean that the mechanism we proposed is correct, it just means that it is possible.

Thus you can by careful application, use kinetic measurements to exclude a mechanism, but usually you can not prove a mechanism. Therefore you must always qualify a statement like: "Assuming the enzyme follows simple Michaelis-Menton kinetics, the values for KM and Vmax are X and Y."

II. Enzyme Effectiveness - lock & key vs. induced fit (figure 1)

1) KM refers to substrate binding but not enzyme efficiency, while Vmax refers to maximum velocity but not efficiency.

2) a term called kcat refers to the maximum # of moles of substrate converted to product per mole of enzyme per unit of time.

3) the maximum rate then is for the conversion of [ES] to [P] and thus it is the rate when all the E is filled or when [ES] = [ET] and thus kcat = Vmax/[ET] units are (mol-s-1-enz-1)

4) usually enzymes under physiological conditions don't have saturating [S] they work at low [S] thus v = Vmax[S]/KM because KM >> [S]

5) thus [S] converted to [P] per time per enzyme amount v = (Vmax[S]/KM)/[ET]

6) but kcat = Vmax/[ET] therefore v = (kcat/KM)[S]

7) (kcat/KM) is defined as the specificity constant and shows how well an enzyme can work at low [S]

8) the specificity constant is used to compare different substrates for the same enzyme, or rates of substrate conversion as compared to the diffusion constant which is the maximum rate of any reaction

III. Graphical Representation

A. Lineweaver-Burk (double-reciprocol)

A common way to write the Michaelis-Menton equation is:

1/v = 1/Vmax + KM/Vmax * 1/[S]

this is the equation for a straight line where 1/v = y, 1/[S] = x, the slope of the line is KM/Vmax, the yint = 1/Vmax, and the xint = -1/KM

this type of plot is referred to as the Linweaver-Burk plot or the double- recipricol plot (figure 2)

B. Eadie-Hofstee v = -KM(v/[S]) + Vmax

C. Why one is better than the other depending upon measurements. But the best is a direct fit to v vs. [S] if possible.

- at low substrate concentrations L-B can be off due to reciprocal values - at high substrate concentrations E-H can be off due to [S] denominator

IV. Enzyme Inhibition (figure 3)

A. Competitive (refer to handout)

1)

k1 k2

E + S <---> ES ---> E + P

k-1

ki

2) but E + I <---> EI

k-i

3) thus [E] = [ET] - [ES] - [EI]

4) and KI = [E][I]/[EI] = k-i/ki

5) therefore v = Vmax[S]/{[S] + (1 + KM([I]/KI))}

6) or 1/v = 1/Vmax + KM/Vmax{1 +([I]/KI)}1/[S]

7) thus in the Lineweaver-Burke the 1/Vmax is not changing but the slope is by the term 1 +([I]/KI). Thus it must be the KM portion of the slope that is changing.

8) with increasing [I] you get increasing slope which means the KM is being affected and getting larger because the -1/KM x-intercept is getting smaller (two graphs on board)

9) this makes sense because the inhibitor is taking up some of the binding sites and thus it takes a higher concentration of [S] to out-compete the [I]. A higher [S] is measured as a higher KM.

B) Noncompetitive (refer to handout)

kis

1) but ES + I <---> ESI

k-is

2) when we do the math

1/v = {(1/Vmax + (KM/Vmax)(1/[S])}{1 +([I]/KI)}

3) in this case KM is not affected, instead Vmax is decreased which means the slope again increases but the 1/Vmax term gets larger thus the Vmax value gets smaller. (two graphs on board)

4) this makes sense because the [I] is not competing with the [S] for binding site, it is only affecting the ability of the enzyme to convert [S] to [P] thus the velocity of the reaction. No increase in [S] can overcome a noncompetitve inhibitor unlike the case for a competitive inhibitor.

C) Uncompetitive (refer to handout)

1) in this case

1/v = (1/Vmax){1 +([I]/KI)} + (KM/Vmax)(1/[S])

2) in this case because the 1/Vmax term is changing, but the slope is not, then both Vmax and KM are changing (two graphs) V. Effect of pH and Temperature (two graphs on board)