Workshop 1 solutions/notes

Part A: Structures can be found under "Memorize" on blackboard.

Part B:
1) Radio-labelling: When the reaction has reached equilibrium, count radio-labelled molecules after filtering. Measured amounts are directly proportional to the amount of reactants/products in solution.
For PNP reaction, wait at least 48 hours for reaction to reach equilibrium, unless an enzyme is present, in which case wait appx. 30 seconds-1 minute.
2) If Keq>1, then there are more products than reactants (when reaction has reached equilibrium) -- Keq= [P]/[R] = [ADP][Pi]/[ATP]
Keq= 1.0x10^5 >> 1, so there are much more products than reactants, and delta G (dG) will be negative for this reacton.
Work (max) = delta G
dG=(-5.75 kJ/mol)log(Keq)

dG=(-5.75)log(1.0x10^5)
dG=-28.75 kJ/mol x 1 mol ATP= -28.75 kJ = W(max)

Part C:
1) Given W=10 J to lift weight once **Coupling: Rxn able to do work (release kJ/mol)
W(max) = dG = -50 kJ/mol --> -dG rxn does POSITIVE work
(5.0x10^4) J / 1 mol =10 J / x mol
x = 2.0x10^-4 mol ATP
2)(2.0x10^-4 mol ATP) x (6.02x10^23 molecules)/1 mol = 1.204x10^20 molecules ATP

Part D:
1 cal = 4.19 J, 10 J of work done to lift once
1 cal/4.19 J x 10 J/1 lift = 2.3 cal/lift
2000 Cal/1 day x 1 lift/2.3 cal x 1000 cal/1 Cal = 8.7 x 10^5 lifts/day!

Part E: Atoms in bonds in wood are holding the table together enough to exert Normal force and oppose gravity. Measure bond strength (in Newtons) by adding more mass until table breaks.