bio/essential-cell-bio/03.markdown @ f76b2070bb71

Chapter 3
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Questions
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### Q 3-1

It would probably not happen in a single step, because sugars have the form
(CH₂O)ₙ.  The smallest sugars have n=3, and so multiple steps are required to
build the sugar from individual H₂O and CO₂ molecules.

Heat must be generated because sugars are more complex molecules than the
components.  As order increases, the overall system must gain disorder in the
form of heat.

### Q 3-2

A. Yes
B. No
C. No
D. Yes
E. No

### Q 3-3

A. ΔG depends on the starting proportion of heads and tails.  If there are
   unequal amounts, there will be a tendency for the box to end up closer to
   1:1.  The more unequal the starting proportions the larger magnitude ΔG will
   be.  ΔG⁰ is zero, because H → T and T → H are equally likely.

B. The temperature is how vigorously you shake the box.  The activation energy
   is the minimum force needed to possibly flip a coin.  The enzyme would be
   something that lowered the minimum force needed to flip a coin (e.g. it might
   make the coins lighter).

### Q 3-4

Enzymes would reduce the activation energy required at each step, but wouldn't
change the starting or ending values at all.

### Q 3-5

It's probably limited by the concentration of the reactants.

The area under the curve for the catalyzed portion should be 10,000,000 times
the area under the uncatalyzed portion.

### Q 3-6

Enzymes don't catalyze a reaction in a single direction, they catalyze it it
*either* direction depending on the relative concentrations of the
reactants/products.

### Q 3-7

A. Rocks: ATP
   Debris: ADP + Pi
   Bucket Bottom: X
   Bucket Top: Y

B. Rocks hitting the ground without machine:     ATP →      ADP + Pi + lots of heat
   Rocks hitting the ground WITH machine:    X + ATP →  Y + ADP + Pi + some heat

### Q 3-8

ΔG depends not only on ΔG⁰ but also on the concentration of the reactants and
products in the cell:

    ΔG = ΔG⁰ + RT ln([X]/[Y])
    ΔG = ΔG⁰ + RT ln([ADP]*[Pi]/[ATP])

If the proportion `[ADP]*[Pi]/[ATP]` is less than 1 (i.e. there's more ATP in
the cell than ADP/Pi) then `ln([ADP]*[Pi]/[ATP])` will be negative and will
increase the amount of free energy liberated when the reaction happens.

A range is given because the exact concentration of ATP, ADP, and Pi in the cell
is not always constant.

### Q 3-9

A. No, this is cellular respiration.
B. Yes, this is the second stage of photosynthesis.
C. No, once you have the nucleoside triphosphates you're all set.
D. Yes, cellular respiration is used to "charge up" the carriers like ADP into ATP.
E. Yes, this is an example of D.

### Q 3-10

A. No, they might slow to a crawl but they won't stop completely.
B. No, the "high energy" here is referring to how much energy is release from
   giving up the electron, not how much energy the electron itself has.
C. Yes, roughly (109 is roughly 2 * 54)
D. If it's oxidized it's lost some of its electrons, so… not sure.
E. Yes, e.g. CoA.
F. No, this is true in the presence of oxygen but not necessarily true everywhere.
G. What?
H. No, constants are constant.  But the equilibrium constant of the chain of
   reactions *as a whole* will be shifted when compared to X → Y.

### Q 3-11

    4.2 kJ/mole * 2 = 12.6 kJ/mole

According to the table, the ratio of X to Y will be a little larger than 10².

Doubling the number of hydrogen bonds would double the energy, and result in
a ratio a little more than 10⁴.

### Q 3-12

### Q 3-13

True:

    ΔG = ΔG⁰ + RT ln([X]/[Y])

If X is more prevalent than Y, then x/y will be more than 1 and the log will be
positive.  If Y is more prevalent than X then x/y will be between 0 and 1 and
the log will be negative.  RT is always a positive constant.