BioWeb
Glossary of biochemical terms
Frequently Asked Questions (FAQs)
Enzyme Problems Set
1. What is the direction of the following reactions when the reactants are present at equimolar amounts?
a.     ATP  +  Creatine ---------------->   Creatine-P + ADP
b.     ATP   + Glycerol ---------------->   Glycerol-3-P +  ADP
c.     ATP  + Pyruvate ----------------->  Phosphoenolpyruvate  +  ADP
The standard free energies of hydrolysis of ATP, creatine-P, glycerol-3-P and phosphoenolpyruvate are -7.3, -10.3, -2.2 and -14.8 Kcal/mole respectively.
To solve such problems, remember that the direction of reaction can only be predicted by the sign of free energy change i.e deltaG. However, since the concentrations of reactants and products are equal, deltaG is equal to deltaG0' applying the equation:
deltaG = delta G0'  + RT log {P1}{P2}/{S1}{S2}
The ratio {P1}{P2}/{S1}{S2} is equal to 1 and log1=0. Therefore, if deltaG0' comes out to be negative, the reaction will be exergonic and proceed to the right. If deltaG0' is positive then the reaction as written is endergonic and proceeds in the opposite direction.

2. Consider the reaction: ATP + Pyruvate  ------------->  ADP  + Phosphoenolpyruvate
    This can be written as two half-reactions:
    ATP -------------->  ADP  +  P                        deltaG0'  = -7.3 Kcal/mole
    Pyruvate + P  ------>  Phosphoenolpyruvate    deltaG0' = + 14.8 Kcal/mole
These two reactions can be added, because of a common intermediate, P. So final reaction has deltaG0' of 
7.5 Kcal/mole. Therefore,
deltaG0' = -RTlog Keq
7.5x1000 cal/mole = -2cal/mole/degK x 298deg K log Keq
Calculate Keq as 3.16x10-6 
Keq = {PEP}{ADP}/{ATP}{Pyr}
3.16x10-6 = 1/10 x {PEP}/{Pyr}
If you do the math now, the ratio
{pyr}/{PEP} will be 3.16x104.

3. The Michaelis-Menten equation can be arranged in many ways. One way to do it is:
    v = VmaxS/Km + S
    v(Km + S ) = Vmax S
vKm  + vS      = Vmax S
vS  = Vmax S - vKm........divide this equation by S
v   =  Vmax  - vKm/S
A plot of v vs v/S should be linear with slope equal to -Km and intercept on Y-axis equal to Vmax.
 
Glucose Homeostsis Problem
You can do this problem two ways:
1. Use equation for two substrates: v= Vmax/(1+ Km1/S1 + Km2/S2) 
For brain metabolism KmATP/ATP = .1/1.0 and KmGlu/Glucose = .01/5
v = Vmax/ (1+.01 + .002) or v ~Vmax; you can do the same for liver
2. You could also use simpler MM equation. In this case [ATP]/Km ATP= 1/.1 = 10 so  the enzyme is saturated with ATP. Therefore, the reaction rate will only depend on glucose concentration. Use equation: v = Vmax S/ Km + S. Substitute values for Km for enzymes in the liver and brain and determine the relationship between v and Vmax.