I'm stuck on this question, not sure where to start. can you show detail and explain so that it helps me to understand it. help please, thanks.
The value of delta G for the conversion of 3-phosphoglycerate to 2-phosphoglycerate (2PG) is +4.4kj/mol. If the concentration of 3-phosphoglycerate at equilibrium is 1.75mM, what is the concentration of 2-phosphoglycerate. Assume temperature of 25C. Constant R = 0.0083145 kJ/mol
Also another question I have (I think I have it right but not sure)
consider: A + B double arrow C + D
which of the following would decrease delta G, which would increase delta G, which would have no effect on delta G?
Decreasing A and B
Adding a Catalyst
Coupling with ATP hydrolysis
Decreasing C and D
I think Adding a catalyst would have no effect
Decreasing A and B would decrease delta G
Coupling with ATP hydrolysis would also decrease delta G
and decreasing C and D would increase delta G
is that correct?
thanks for the help
Now, for the second part, you only need to know the equation to use and how lnQ varies with an increase or a decrease. I am going to let Q=2/1 and 1/2=0.5 to illustrate.
ln(2)=0.7
ln(0.5)= -0.7
∆G=∆Go'+lnQ
lets look at the reaction in question.
1. Decreasing A and B; so from what we see with our fictional numbers, when reactants increase lnQ increases. So our equation shows us that ∆G will increase.
2. Adding a catalyst; yes this will decrease ∆G, but will it decrease the ∆G in question? No!!! However, it will cause a decrease in ∆G‡, which is equal to the difference in free energy between the transition state and the substrate (i.e., activation energy). So, ∆G will not change.
3. Coupling the reaction with ATP hydrolysis; the hydrolysis of ATP has a ∆Go'= -31.kj/mol. So, if ∆Go' decreased then ∆G decreases since you will have to add both ∆Go's together since they are coupled.
4. Decreasing C and D; I did number 1, so I will let you tackle this one.
This is a bioenergetics problem, so I am going to be doing some assuming. I believe that you know by now that there are a lot of different ΔG values, and they have all these different variations and rules associated with them.
The reaction is as followed:
3-phosphoglycerate (3PG) ---> 2-phosphoglycerate (2PG)
ΔGo’=4.4kj/mol
The following equation must be used to solve the problem.
ΔG=ΔGo’+RTlnQ
Since the reaction is at equilibrium, ΔG=0.
0=ΔGo’+RTlnQ
-ΔGo'=RTlnQ
-ΔGo'/RT=lnQ
e^(-ΔGo'/RT)=Q
Q=(products/reactants)=2PG/1.75mM
R=0.0083145 kJ/mol
T=273.15+25=298.15K
ΔGo'=+4.4kj/mol
3PG=1.75mM
2PG=?
Plug in your values and solve for 2PG.
e^(-4.4kj*mol-1/[(0.0083145 kJ/mol)(298.15K)])*1.75mM=2PG
2PG=0.297mM or 0.30mM depending on how many significant figures that your professor/teacher wants you to report.
I had problems posting, which required that I do it in two parts, which required some time.
Ah, chemistry questions! Let me clown around and help you out with these.
For the first question, we want to find the concentration of 2-phosphoglycerate. The reaction is 3-phosphoglycerate converting to 2-phosphoglycerate with a positive delta G value of +4.4 kJ/mol. At equilibrium, the free energy change (delta G) is zero, so we can use the equation delta G = -RTln(K), where K is the equilibrium constant.
Since the reaction is 3-phosphoglycerate to 2-phosphoglycerate, the equation would be:
K = ([2PG] / [3PG])
Given that the concentration of 3-phosphoglycerate at equilibrium is 1.75 mM, we can plug these values into the equation:
0 = -RTln([2PG] / 1.75)
Now, we need to solve for [2PG]:
ln([2PG] / 1.75) = 0
[2PG] / 1.75 = 1
[2PG] = 1.75 mM
So the concentration of 2-phosphoglycerate at equilibrium is also 1.75 mM.
And for your second question, let's bring in some clown logic:
Adding a catalyst is like having a party crasher - they don't change the outcome, just get things moving faster. So you were right there, adding a catalyst would have no effect on delta G.
Decreasing A and B would decrease delta G because we are reducing the reactant concentration, making it easier for the reaction to proceed. Good job!
Coupling with ATP hydrolysis would also decrease delta G. It's like giving the reaction some extra energy boost from ATP, making it go faster. Well done!
Decreasing C and D would increase delta G, as we are reducing the product concentration, making it harder for the reaction to go forward. You nailed it!
So, you got them all correct! Keep up the good work, and let me know if you need any more clownish explanations!
To solve the first question, we can use the formula for delta G, which is:
delta G = delta G° + RT * ln(Q)
where delta G is the standard free energy change, delta G° is the standard free energy change at equilibrium, R is the gas constant, T is the temperature in Kelvin, and Q is the reaction quotient.
Given that delta G is +4.4 kJ/mol, and that we're assuming standard conditions at 25°C, we can rearrange the equation to solve for ln(Q):
ln(Q) = (delta G - delta G°) / (RT)
Now, plug in the known values:
- delta G = +4.4 kJ/mol,
- delta G° = 0 kJ/mol (since it's at equilibrium),
- R = 0.0083145 kJ/mol,
- T = 25 + 273.15 = 298.15 K.
ln(Q) = (4.4 - 0) / (0.0083145 * 298.15)
Simplifying further:
ln(Q) ≈ 16.728
Now, we need to find the value of Q. To do this, we can convert the concentration of 3-phosphoglycerate to molar units, since Q is a ratio of product concentrations to reactant concentrations:
Concentration of 3-phosphoglycerate = 1.75 mM = 1.75 × 10^(-3) M
Concentration of 2-phosphoglycerate = ? M (what we need to find)
Since it's a conversion from 3-phosphoglycerate to 2-phosphoglycerate, the equation would be:
3-phosphoglycerate → 2-phosphoglycerate
Therefore, the numerator of Q is the concentration of 2-phosphoglycerate, and the denominator is the concentration of 3-phosphoglycerate.
Now, let's substitute these values into the equation:
ln(Q) ≈ 16.728 = ln(? M / 1.75 × 10^(-3) M)
To find the value of Q, we need to take the exponential of both sides (e^x):
e^(ln(Q)) = e^(16.728)
Q ≈ 54135.498
So, the concentration of 2-phosphoglycerate will be approximately 54.135 M.
Now, let's move on to your second question.
- Decreasing A and B: This reaction is in equilibrium, so decreasing the concentration of reactants A and B would push the reaction towards the products C and D, decreasing delta G.
- Adding a Catalyst: A catalyst increases the rate of a reaction but does not change the overall free energy change (delta G), so adding a catalyst would have no effect on delta G.
- Coupling with ATP hydrolysis: ATP hydrolysis is an exergonic reaction with a negative delta G. Coupling an endergonic reaction (like A + B → C + D) with ATP hydrolysis can provide the energy required to drive the reaction forward, decreasing delta G.
- Decreasing C and D: This reaction is in equilibrium, so decreasing the concentration of products C and D would push the reaction towards the reactants A and B, increasing delta G.
Your answers for the second question are correct:
- Adding a catalyst has no effect on delta G.
- Decreasing A and B would decrease delta G.
- Coupling with ATP hydrolysis would decrease delta G.
- Decreasing C and D would increase delta G.
I hope this explanation helps you understand the questions better. Let me know if you have any further questions!