CHM ENG 170A Study Guide - Midterm Guide: Phosphocreatine, Hyperthermia, Flux Balance Analysis
8 Jan 2019
School
Department
Course
Professor
Chemical Engineering 170B
Spring 2009
Midterm #2
Metabolism. The free energies of hydrolysis (in kcal/mol) for several phosphorylated
compounds are given: (12 points. 1 point each answer plus 2 points bonus for
actually putting it in a table.)
Compound ΔG°’
Phosphoenolpyruvate -14.8
Carbamoyl phosphate -12.3
Acetyl phosphate -10.3
Creatine phosphate -10.3
Pyrophosphate -8.0
ATP -7.3
Glucose 1-phosphate -5.0
Glucose 6-phosphate -3.3
Glycerol 3-phosphate -2.2
Assume the cellular ATP/ADP ratio is 20. (a) Calculate the equilibrium constant K’eq,
(b) predict the direction, and (c) determine the equilibrium ratios of reactants to products
of the following reactions. Express your answer as a small table.
ATP + carbamate ! carbamoyl phosphate + ADP
ATP + acetate ! acetylphosphate + ADP
ATP + inorganic phosphate ! pyrophosphate + ADP
ATP + glucose ! glucose 1-phosphate + ADP
We work this problem out like the homework.
ATP + carbmate ! carbamoyl phosphate + ADP
ATP ! ADP + Pi -7.3
carbamate + Pi ! carbamoyl phosphate 12.3
Sum: dG = 5
dG = -RT ln Keq
Keq = exp(-dG/RT) = exp (-5 kcal mol-1/(0.001985 kcal mol-1 K-1*310 K)
= predict LEFT, Reactants
Equilibrium ratio of 5.92e-3 (20*Keq)
Table:
Reaction i ii iii iv
a 2.96e-4 7.63e-3 3.21e-1 4.20e+1
b Reactants Reactants Reactants* Products
c 5.92e-3 1.53e-1 6.41 8.40e+2
iii: The direction is actually changed by the number of ATP/ADP!
Partial credit was assigned if your answer was self-consistent. Here’s my
calculations:
Flux balance analysis. Below is a simple metabolic network in E. coli. (18 pts)
Write the stoichiometric balances for each of the numbered reactions. (5 points)
For instance,
1 – ½ glucose + PEP + NADH = 0
2 – PEP – CO2 + oxaloacetate = 0
3 – PEP + pyruvate + ATP = 0
4 – pyruvate + lactate = 0
5 – pyruvate + acetyl–CoA + formate =0
6 – formate + CO2 + H2 = 0
7 – acetyl–CoA + acetate + ATP = 0
8 – acetyl–CoA – NADH + ethanol = 0
9 – oxaloacetate – NADH – NH3 + aspartate = 0
10 – aspartate – NH4 + asparagine = 0
11 – aspartate – NADH + NH3 + succinate = 0
