Metabolism III (thermodynamics and Gibbs energies

Question 1

Biochemical thermodynamics

Answer 1

• we can predict what respiratory metabolisms are actually possible.
• we have a given electron donor (which can be the C source if a heterotroph) and a given electron acceptor and we know what the product of the latter would be e.g. O2
  is reduced to 2H2O.
• we write out a balanced equation for the oxidation of the donor (in any assessment I will provide this for you!).
• we can then determine the change in Gibbs energy under standard conditions (ΔG°) at 25 °C OR the change in Gibbs energy at a given temperature (ΔG).
• two methods for determination.
• if the ΔG for the reaction is positive, it cannot provide energy for ATP biosynthesis and cannot be a valid metabolism.
• if the ΔG for the reaction is negative, it can in theory provide energy for ATP biosynthesis and can be a valid metabolism BUT the value given is a theoretical maximum – we’ll never see that much energy normally!

negative means its giving energy out to the universe
positive means in needs energy from the universe and therefore would not be a good respiratory metabolism.

Question 2

Describe method 1 for finding gibbs energy

Answer 2

Method I
* determine ΔG based on:
ΔG = [ΣGf]products – [ΣGf]reactants
* easiest option – works as long as ΔGf values are known for each chemical involved AND that you’re working at 25 °C.
EXAMPLE:
CH4 + 2O2 → CO2 + 2H2O
if: ΔGf (gibbs energy of formation)
* methane = -50.79 kJ/mol
* oxygen = 0 kJ/mol
* carbon dioxide = -394.38 kJ/mol
* water = -237.18 kJ/mol
* so [-394.38 + 2(-237.18)] – [-50.79 + 2(0)] = -817.95 kJ/mol CH4 oxidised
(could also write it as -408.98 kJ/mol O2 reduced this is because in the equation there are 2 oxygen molecules and they are being reduced

Question 3

Describe method 2 for finding gibbs energy

Answer 3

Method II
* determine ΔG based on:
ΔG = ΔH-TΔS
where:
ΔH = [ΣHf]products – [ΣHf]reactants = change in standard enthalpy of formation
ΔS = [ΣS]products – [ΣS]reactants = change in standard entropy
* longwinded option – but the only option if you don’t have ΔGf values for ALL reactants and products OR you need to change temperature.
COMMON ERRORS!
1) T is in K not °C.
2) S values are always in J/mol/K not kJ/mol so you need to convert the units from J to kJ.
EXAMPLE:
CH4 + 2O2 → CO2 + 2H2O
if: ΔH °f
* methane = -74.85 kJ/mol
* oxygen = -22.70 kJ/mol
* carbon dioxide = -393.51 kJ/mol
* water = -285.83 kJ/mol
if: S°f
* methane = +186.2 J/mol/K
* oxygen = 0 J/mol/K
* carbon dioxide = +213.64 J/mol/K
* water = +69.91 J/mol/K
* [-393.51 + 2(-285.83)] – [-74.85 + 2(-22.70)] = ΔH = -844.92 kJ/mol CH4 oxidised
* [+0.21364 + 2(+0.06991)] – [+0.1862 + 2(0)] = ΔS = +0.16726 kJ/mol/K CH4 oxidised
* 25 °C = 298.15 K
* SO: ΔG = ΔH-TΔS = -844.92 – (298.15 * +0.16726) = -894.79 kJ/mol CH4 oxidised
(not the same as with Method I – the data are all determined in different ways)

Question 4

what can we do with it?

Answer 4

1) comment on if a reaction can occur at all (-ve value) or not (+ve value).
2) work out how much ATP could be made in theory.
3) work out how much NAD(P)H could be made in theory.

ATP formation ΔG° = +29.4 kJ/mol ATP formed (sometimes higher values given – relates to speciation of ATP and phosphate at pH 7) – divide ΔG° of reaction by this to approximate how much ATP can be made in theory. Not
more can be made than this, not ever.
NAD(P)H formation ΔG° are both around +55 kJ/mol NAD(P)H formed –divide ΔG° of reaction by this to approximate how much NAD(P)H can be made in theory. Not more can be made than this, not ever.
“ATP” in vivo is a mixture of MgATP2- and ATP4-
.

Question 5

How can i work out the moles of ATP from a

Answer 5

you will get ATP value e.g 29.4
you ignore signs as long it was negative to begin with.
ΔG° of reaction/ cost of making ATP this is moles of ATP from oxidation

be careful if you need to find how much oxygen was produced or other stuff look at equation. Round downwards NOT upwards