PCL481H1 Final: Comprehensive Notes Part I (5 pages long!)

[1] 1. (a) Solve 2 – 31 < 7. Solution: We have – 7<-3 < 7 which gives -4 0 on (-0, 1] U [2,). Next, we need to see where Vr2 – 3r +2 -1 > 0. If r2 - 3.c + 2 – 1 > 0, then V.22 – 3.0 +2 > . Observe that this is definitely true if r < 0. If r > 0, then squar- ing both sides we get 22 - 3.+2 > 1 2 > 3. V A VICO Thus, the domain of f is (-00, ). [2] (c) Find the exact value of tan(sec-14). Solution: Let y = sec-14. Then, sec y = 4. From the diagram we get that tan(sec-? 4) = tan y = 15 [2] (d) Find all r such that sin(2.x) = COS I. Solution: We have 2 sin r cos r = COS I, SO 0 = 2 sin Cos I -Cos I = cos .r (2 sin c - 1). Hence, sin(2x) = cost when cos x = 0, or when sin r = Thus, r= +ik, kez, I= +2nk, ke Z, or r= +2nk, kez [2] (e) State the precise mathematical definition of lim f(x) = 0. Solution: For every e > 0 there exists a 8 >0 such that if 0 < x-al <, then f(c) >

1) The following equilibrium constants have been determined at 298K for these two reactions that are part of the citric acid cycle (TCA or Krebs cycle).

I AcetylCoA + oxaloacetate + H2O -->

<-- CoA + citrate

II L-malate + NAD_ox -->

<-- oxaloacetate + NAD_red

Given that K_I = 6.97 x 107 and K_II = 8.87 x 10-6, calculate the standard state transformed Gibbs free energies of reaction for each of these reactions, ∆ _rG'o .

I have gotten values of ∆ _{r I}G'o

= -4474 J/mol and ∆ _{r II}G'o = 28821 J/mol

b. For reaction II, the mitochondrial matrix has the following concentrations of components under steady state condition:

[NADox] =1.0 mM [NADred] = 0.1 mM

[malate] = 3.5 mM [oxaloacetate] = 0.0003 mM

What is the solution potential of the NADred/NADox redox couple?

I am not sure how to proceed with this. I think I need to use the equation E = E⁰ –RT/abs(v)F ln K

Are these two redox couples at or near equilibrium? Explain your answer.