K1 K3
A + E ========= Complex -------> P + E
K2
The differential equations based on mass balances are:
dA/dt = - k1 [A] [E] + k2 [Complex]
dE/dt = - k1 [A] [E] + (k2+k3) [Complex]
d Complex/dt = k1 [A] [E] - (k2+k3) [Complex]
dP/dt = k3 [Complex]
where A is the concentration of some reactive biochemical, E is the concentration of an enzyme, [Complex] is the concentration of an intermediate formed by A and E, P is the concentration of product.
When the derivatives are set equal to zero and the reaction rate coefficients are consolidated, the Michaelis-Menten equation results (it was the inspiration for the Monod equation). This set of simultaneous differential equations was used with SIMBAS (Bungay, 1993) to generate Figure 1 that is shown in most textbooks that discuss enzyme kinetics. The rates of substrate decline and product formation are nearly linear throughout much of the simulation. This portion of the program alone has some teaching value because a student can verify that the textbook results make sense. The usual depiction is for rate constants that are equal. Additional insight comes from tinkering with the rate coefficients to see their effect.
It must be appreciated that one reaction with one enzyme in vitro is not typical of living biochemistry where there are complicated pathways and intermediates tend not to accumulate. Actual metabolic pathways have branches and many steps, and the over accumulation of intermediates is prevented by feedback inhibition where a concentration of a biochemical further along in the pathway inhibits or represses the activity of the enzyme for an early step. An option of the simulation treats a multi-enzyme path by allowing P in the above scheme to react further. Figure 2 shows that the former product is now an intermediate that peaks and declines as the substrate runs out. With no substrate, the complex cannot form, and the enzyme is released.
Another option considers feedback inhibition by letting the rate coefficient K1 depend on the final product concentration. In this simulation, the final product reacts further but at a very slow rate compared to the other reactions in the path. This is not the best way to handle feedback inhibition because it would be more realistic to control formation or activation of the enzyme. More differential equations would be needed, and deriving them and entering the computer code would not be overly difficult for a student. This simplified treatment gives Figure 3 that shows that the control is not very good. The later product overshoots, turns everything off, and a cycle starts again after its concentration drops enough to let the earlier reactions to operate. Note that the time scale is longer for Figure 3. The dependence of K1 on the concentration of final product makes the concentration of complex drop close to zero, and this could be quite different for a more realistic mechanism for feedback control.
The complete computer code that simulates all three cases is shown, but those familiar with SIMBAS will note that only about 20 lines must be changed from SIMBAS.BAS that is widely available. Learning from computer simulations is quite effective because the students are in control. They can experiment with the coefficients to understand their effects and to evaluate the importance of various terms.
Reference
Bungay, H.R., "Basic Biochemical Engineering", 2nd edition, BiLine Assoc., Troy, NY, 1993.
The following is new BASIC code to make better use of the colors and screen resolution of Qbasic. The file MICHAEL.BAS is in the same directory as this HTML page and can be downloaded. However, it is easy to use your browser to capture this text and to delete everything except the BASIC program.
1 REM SIMBAS, Michaelis-Menten Enzyme kinetics, Jan. 1994, Qbas 15-Apr-97 2 REM ORIG. BY D. STANZIONE, H. BUNGAY 3 REM MORE COMMENTS BY J. SAHLIN 5 DIM I(20), O(20), T0(20), T6(20) 6 SCREEN 9: xmax = 620: Ymax = 310 7 KEY OFF 8 CLS 9 NN = 2 11 REM THIS SECTION MAKES THE AXIS 13 LINE (20, 20)-(20, Ymax - 20) 14 LINE -(xmax - 10, Ymax - 20) 15 LOCATE 22, 4: PRINT "0 T I M E"; 17 NAM$ = "CONCENTRATION" 20 FOR I = 1 TO LEN(NAM$) 21 LOCATE 4 + I, 2: PRINT MID$(NAM$, I, 1) 22 NEXT I 27 REM PUT COEFF IN THIS REGION 30 k1 = .5: k2 = .5: k3 = .5: k4 = k1: k5 = k1: k6 = k1 39 LOCATE 1, 1 40 PRINT "Type 1 for one-enzyme, 2 for multi-enzyme " 41 INPUT "3 for feedback inhibition"; a 42 n = a + 3 43 COLOR 1: LOCATE 12, 50 44 IF a = 2 THEN PRINT "LATER PRODUCT" 55 REM T1 IS INTEGRATION STEP 56 T1 = .1 57 REM T2 IS TOTAL TIME 58 T2 = 30 59 LOCATE 22, 70: PRINT T2: REM T3 IS PRINT INTERVAL 60 T3 = 1 61 Tx = (xmax - 30) / T2: REM SCALE FACTOR FOR ABSCISSA 63 Ty1 = 16: REM SCALE FAC FOR O(1) 79 REM N IS THE NUMBER OF EQUATIONS. O(N) ARE INITIAL CONDITIONS 80 COLOR 4: LOCATE 8, 50: PRINT "ENZYME" 81 COLOR 2: LOCATE 9, 50: PRINT "COMPLEX" 82 COLOR 3: LOCATE 10, 50: PRINT "PRODUCT": COLOR 5: LOCATE 11, 50: PRINT "REACTANT" 83 O(5) = 0 84 O(4) = 0 86 O(1) = 15 87 IC = O(1) 88 O(2) = 2 96 REM DO NOT TOUCH LINES 98-620!!! THIS IS WHAT MAKES IT WORK!!! 98 T = INT(T2 / T1 + .5): T1 = T2 / T: T = INT(T3 / T1 + .5): T3 = T * T1: T = 0: T4 = 0 99 T8 = 1: GOTO 1000 100 IF (T - T4 + T1 / 10) < 0 THEN 125 105 REM PUT A ' BEFORE THE GOSUB IN LINE 110 TO SUPPRESS PRINTING NUMBERS." 110 T4 = T4 + T3: T5 = INT(T / T1 + .5): T = T5 * T1: 'GOSUB 2000 115 IF (T - T2 + T1 / 10) < 0 THEN 125 120 LINE (AX, Ymax - 20)-(AX, Ymax - 16) 121 END 125 ON T8 GOTO 300, 400, 500, 600 145 PRINT "** ERROR **": STOP 300 FOR T5 = 1 TO n 305 T0(T5) = T1 * I(T5): T6(T5) = O(T5): O(T5) = O(T5) + T0(T5) / 2 320 NEXT T5 330 T = T + T1 / 2: T8 = 2: GOTO 1000 400 FOR T5 = 1 TO n 410 T7 = T1 * I(T5): T0(T5) = T0(T5) + 2 * T7: O(T5) = T6(T5) + T7 / 2 420 NEXT T5 430 T8 = 3: GOTO 1000 500 FOR T5 = 1 TO n 510 T7 = T1 * I(T5): T0(T5) = T0(T5) + 2 * T7: O(T5) = T6(T5) + T7 520 NEXT T5 530 T8 = 4: T = T + T1 / 2: GOTO 1000 600 FOR T5 = 1 TO n 605 O(T5) = T6(T5) + (T0(T5) + T1 * I(T5)) / 6 610 NEXT T5 620 T8 = 1: GOTO 1000 990 REM DIFFERENTIAL EQUATIONS IN SIMBAS NOTATION 1000 I(1) = -k1 * O(1) * O(2) + k2 * O(3) 1010 I(2) = -k1 * O(1) * O(2) + (k2 + k3) * O(3) 1020 I(3) = k1 * O(1) * O(2) - (k2 + k3) * O(3) 1030 AX = 20 + T * Tx 1035 IF a > 1 THEN 1100 1040 I(4) = k3 * O(3) 1099 GOTO 1200 1100 I(4) = k3 * O(3) - k5 * O(4) 1140 I(5) = k5 * O(4) - k6 * O(5) 1145 Z = IC - O(1) - O(3) - O(5) 1150 IF a = 3 THEN I(6) = k6 * O(5) - .05 * O(6): Z = O(6) 1160 PL = Ymax - 20 - Z * Ty1 1170 CIRCLE (AX, PL), 3, 1 1200 PL = Ymax - 20 - Ty1 * O(1) 1215 CIRCLE (AX, PL), 3, 5 1220 PL = Ymax - 20 - Ty1 * O(2) 1225 CIRCLE (AX, PL), 3, 4 1230 PL = Ymax - 20 - Ty1 * O(3) 1235 CIRCLE (AX, PL), 3, 2 1240 PL = Ymax - 20 - Ty1 * O(4) 1245 CIRCLE (AX, PL), 3, 3 1250 IF a = 3 THEN k1 = .5 - Z: IF k1 < 0 THEN k1 = 0 1999 GOTO 100 2000 REM **DATA POINTS: SUPPRESSED AT LINE 110. SEND TO PRINTER WITH LPRINT 2005 REM **COMMAND ON 2020. 2010 LOCATE NN, 2 2020 PRINT T; O(1); O(2); O(3) 2030 NN = NN + 1 2040 IF NN > 8 THEN NN = 2: col = 15 2045 RETURN