Download Set 1 Answers

BL4010
PROBLEM SET 1
FALL 2006
1. Explain the distinguishing properties of water and their chemical basis.
Water has a high dielectric constant, high specific heat, high surface tension,
and is less dense in the frozen state than the liquid state. The high surface
tension is a result of hydrogen bonding which is very strong in water (30%
covalent). High dielectric constant is due to combination of ionization of the
OH bond and phenomena of proton transfer. High specific heat is related to
both these properties. The increased density of water on freezing is due to Hbonding but also to the bent configuration of the molecule enabling packing
in a tetrahedral geometry. The hydrogen bonding network of this packed state
and the partial covalency actually sparses the water molecules into a less
compact form.
2. Explain the difference between equilibrium and steady state.
At equilibrium (chemical), no net change in concentration of products and
reactants occurs (closed system). At steady-state, a net change can occur but
this happens at a constant rate (open system).
3. State the four laws of thermodynamics.
Zeroth Law
Two systems in thermal equilibrium with a third are in thermal equilibrium with
each other.
First Law
Energy can neither be created nor destroyed. OR more accurately from its
derivation in class: The quantity of energy supplied to any isolated system in the
form of heat is equal to the work done by the system plus the change in internal
energy of the system. (equivalency of heat and energy (enthalpy) plus an
unknown).
Second Law
Spontaneous processes occur such that the overall disorder of the universe is
increased. (entropy)
Third Law
By no finite series of processes is the absolute zero attainable. (When T=0
G=H).
BL4010
PROBLEM SET 1
FALL 2006
4. A solution containing 10 mM lactic acid and 87 mM lactate has a pH = 4.80.
What is the Ka? What is the pH if the concentration of lactate equal to that of
lactic acid?
HR H+ + RpH = pKa + log([R-]/[HR])
4.8 = pKa +log(0.087/0.010)
pKa = 4.8 - log(.087/.01) = 4.8 - 0.94 = 3.86
Ka = 10-386 = 1.3 x 10-4
5. List the four main classes of biomolecules. Give one example molecule for
each class and draw the chemical structure (pH=7).
Proteins (amino acids)
H3N+
CH3 O
C
H
O ALANINE
Carbohydrates (sugars)
O
HC
HC OH
HC OH
HC OH
CH2OH
RIBOSE
Nucleic acids
O
CH3
HN
O
O
P
O
O
O
HO
O
N
OH
THYMIDINE MONOPHOSPHATE
Lipids
H3C
NH2
OH
OH
SPHINGOSINE
BL4010
PROBLEM SET 1
FALL 2006
6. Consider the ionization of lysine. Draw the structure at each ionization state.
Using the following pKa values, draw the titration curve and calculate the pI.
backbone amino pKa = 9.2
backbone carboxyl pKa = 2.3
side chain amino pKa = 10.5
NH3+
NH3+
NH3+
NH2
CH2
CH2
CH2
CH2
CH2
CH2
pKa = 2.3
CH2
CH2
CH2
H3N+
C
H
OH
H3N+
O
CH2
pKa = 9.2
pKa = 10.5
CH2
CH2
CH2
CH2
CH2
CH2
CH2
CH2
CH2
C
H
O
H2N
O
C
H
O
H2N
O
CH2
O
C
H
O
10.5
pH
9.2
2.3
0
0.5
1
1.5
2
2.5
3
Equivalents OH-
7. Draw the structure of the tripeptide I-G-E (pH=10). Identify the peptide
bonds. Label the alpha carbons, the N-terminus, and the C-terminus.
CH3
H3C
HC
H2N
CH2
CH2
H
C
H
O
N
O
O
O
CH2
H
CH2 O
N
H
N-terminus
alpha carbons
O
C-terminus
BL4010
PROBLEM SET 1
FALL 2006
8. What defines a state function? Give an example.
A state function is independent of path. In other words, it is a property of
the system that is inherent and does not depend on how the system arrived
at that state. For example, the internal energy inherent in a molecule (i.e. sum
potential energy of its bonds and collective electronic properties) is
independent of how the molecule was formed. Other state functions include
enthalpy and entropy.
9. Identify and describe the thermodynamic parameter(s) that determine(s) the
spontaneity of any given process?
Spontaneity is determined by the change in internal energy. If the energy of
the products is less than that of the reactants (i.e. the reaction goes downhill)
it is spontaneous. This is dependent on both the enthalpy (H) and the
entropy (S) according to G = H - TS.
10. Describe the probable events involved in the evolution of living cells.
Rise of organic molecules (primordial conditions - reducing atmosphere, extremes of
temperature and pH in concentrated microenvironments). Coalescence of organics
into larger molecules...eventually self-replicating molecules such as small RNA's. Rise
of lipids and membranes...imperfect barriers at first...better barrier with crossover
molecules later allowing separation of charge, control of concentration gradients and
the rise of metabolism. A minimal self-replicating structure or set of structures would
quickly proliferate. Inaccurate replication leads to radiation and evolution begins.
OR...insert your favorite theory here.
Additional problems from Voet & Voet:
3.3 In terms of thermodynamic concepts, why is it more difficult to park a car in a
small space than it is to drive it out from such a space?
Basically, this a simple matter of limited degrees of freedom. When the car is in the
open it can go forward, backward, turn. When parking in a small space there is
usually only one way in and maximal two configuration (nose in or nose out).
Entropy is reduced which is a thermodynamic penalty. When driving out..there is
usually one way out but many final configurations.