BMEN E3500, fall 1999

Lecture Notes

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Lecture Number 4a

Date: 9/27/99

The general rule for the "automatic" solution of enzyme kinetics problems is:

1. Write an expression for the velocity of the reaction.  In the first example given below (no reversion of product to complex) the rate, v, is formulated as v - k2*es = 0 where k2 is the rate of decomposition of complex es to product.
2. Relate the total amount of enzyme present, e0, to its two forms, free enzyme, e, and complexed enzyme, es.  Thus e + es - e0 = 0.
3. Formulate a "balance" on es (and in more complicated problems other 'intermediates').  Its rate of formation must equal its rate of decomposition.  Thus e*k1ps-(k1m+k2)*es = 0 where k1ps is k₁₊*s, k1m is k­_1- , and both are multiplied by es.
4. Solve.  If s is treated as a parameter, the system of equations 1 -- 3, above, is linear in v, e, and the intermediates, and can be solved for each of these, as shown.
5. Rearrange the expression to a "standard" form.  Here one can obtain  v=(k₂e₀s/((k₂+k_-1)/k₊₁ + s). This is the Michaelis-Menten form if one recognizes (k₂+k_-1)/k₊₁ as K_m.

The second solution given below is for the reversible case, where product p (treated as a parameter, like s, see 4. above) can also combine with enzyme to form complex.  It is left as an exercise for the student to put this expression in "standard" form.  (Computer algebra does not always arrange results in the clearest way.)

linsolve([v - k2*es, e + es - e0, e*k1ps-(k1m+k2)*es],[v,e,es])

[v = ((e0 * k1ps * k2)/(k2 + k1ps + k1m)),

e = ((e0 * k2 + e0 * k1m)/(k2 + k1ps + k1m)),

es = ((e0 * k1ps)/(k2 + k1ps + k1m))]

linsolve([v-k2*es+km2p*e,e+es-e0, (k1ps+k2mp)*e-(k1m+k2)*es],[v,e,es])

[v =  - ((e0 * k2 + e0 * k1m) * km2p

 + e0 * k2 * ( - k2mp - k1ps)/(k2mp + k2 + k1ps + k1m)),

e = ((e0 * k2 + e0 * k1m)/(k2mp + k2 + k1ps + k1m)),

es = ((e0 * (k2mp + k1ps))/(k2mp + k2 + k1ps + k1m))]

Very complicated problems with multiple substrates can be solved in this way.  The key is that the variables to be solved for always appear as linear factors, allowing the equations to be treated as a linear set.

Inhibition problems are often solved assuming the complex always to be in equilibrium with substrate and inhibitor and free enzyme.  The substrate can bind inhibitor or substrate, but not both at the same time.  Thus for totally competitive inhibition we may write:

linsolve([es-kss*e, ei-kii*e,es+ei+e-e0,v-k*es],[es,ei,e,v])

where ks denotes the equilibrium (here, not the rate) constant for the ratio (es)/(e*s) and ki  denotes the equilibrium constant for the ratio (ei)/(e*i).  (Thus kss is ks*s, and kii is ki*i.) The solution is:

[es = ((e0 * kss)/(kss + kii + 1)),

ei = ((e0 * kii)/(kss + kii + 1)), e = (e0/(kss + kii + 1)),

v = ((e0 * k * kss)/(kss + kii + 1))]

The last expression is easier to interpret if it is rewritten as:

v= k * e₀ * s/((k_i*i + 1)/k_s + s)

The inhibition occurs because less enzyme exists in the ES form, es is lower, and the rate of conversion to p is correspondingly less.  If one plots 1/v vs. 1/s  competitive inhibition will increase the slope of the curve by the factor  k_i*i + 1 but the intercept will be the same as in the uninhibited case:  (k * e₀₎^-1.  (The confusing situation discussed in class arose because the k_i defined there was the dissociation constant of the complex, the reciprocal of the constant considered here.)  

Noncompetitive inhibition assumes that the enzyme can form three complexes, ES, EI, and EIS.   Inhibition occurs when it is assumed that EIS cannot decompose into P.   This situation gives rise either to the equation set:

linsolve([es-kss*e, ei-kii*e,eis+es+ei+e-e0,v-k*es,eis-kss*ei],[eis,es,ei,e,v])

giving the solution:

[eis = ((e0 * kii * kss)/((kii + 1) * kss + kii + 1)),

es = ((e0 * kss)/((kii + 1) * kss + kii + 1)),

ei = ((e0 * kii)/((kii + 1) * kss + kii + 1)),

e = (e0/((kii + 1) * kss + kii + 1)),

v = ((e0 * k * kss)/((kii + 1) * kss + kii + 1))]

The rate can be rewritten as:

v = (e0 * k * kss)/((kii + 1) * (1+kss))

In a plot of 1/v vs. 1/s, both the slope and intercept of the straight line will be increased by the factor k_ii + 1.  Unlike a competitive inhibitor, a noncompetitive inhibitor decreases the rate even at infinite substrate concentration.

[See Stryer, L., Biochemistry, pp. 187-195 in 3^rd  Ed.]

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