6. Alpha-Keratin is alpha helical so that its polypeptide chain can stretch before covalent bonds are broken. Silk however is made up of mostly beta sheets which have limited flexibiity.

8. No proline cannot fit in the crowded central region of the
collagen helix.

Voet and Voet, Chapter 9

1. Hyperventilation to remove excess CO2 would raise the pH of the blood due to the Bohr Effect.

H + HCO3 ------ H2O + CO2

Raising the pH would increase the affinity of O2 for hemoglobin thus
having an opposite effect to what was hoped to be achieved.

6. Depletion of BPD shifts hemoglobins fractional saturation curve
so that the amount of O2 that hemoglobin can deliver is reduced.
Thus week old blood is not as effective.

Voet and Voet, Chapter 13

3. Ks = k-1/k1 = 4 x 10 -4 M

Km = k-1 +k2/k1 = 4 x 10 -4M

4. Remember to plot the data as inverse of substrate concentration and velocity values.

a. Km = -1/Km of X intercept

1/0.3 = 3.3 mM

Vmax = 1/Vm of Y intercept

1/0.1 = 10 µM/sec

Inhibitor 2: same Vmax, different Km implies competitive inhibition

Vmax (app) = Vmax = 10 µM/sec

Km (app) = -1/0.13 = 7.7 mM

Km (app) = Km(1 +I/Ki)

solve using Km (app) and Km to find Ki = 7.5 mM

Inhibitor 3: same Km, different Vmax implies non-competitive inhibition

Vmax = 10 µM/sec

Vmax (app) = 1/0.3 = 3.3 µM/sec

Using Vmax/Vmax (app) = 1 I/Ki to find Ki = 5 mM

b. To solve for the fraction of enzyme molecules that have bound substrate we need to remember [ES] = [Et] [S]/ Km + [S]

such that [ES]/ [Et] = [S]/ Km + [S]

thus for no inhibitor [ES]/ [Et] = 5mM/ 3.3mM + 5mM = 0.6

For the competitive inhibitor [ES]/ [Et] = [S]/ Km
(1 I/Ki) + [S] = 0.36

For the non-competitive inhibitor [ES]/[Et] = 0.2

6. Calculate Vmax at 0.2M substrate using v = (Vmax) (S) / Km
+ (S) and you will find Vmax = 0.4 x 10-6, now using that Vmax calculate
v at 0.02 M substrate.