BIO 361 Lecture Notes - Lecture 1: Beta Sheet, Protein Purification, Proline
29 Jun 2018
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Discussion Clarification Week 1, Lectures 1-5
Lecture 1: Entropy of the System
1. You have been studying a reaction that generates a number of glucose molecules from the
hydrolysis of a single high molecular weight glucan polymer. The reaction can proceed in either
water or ice, but you have observed that when the reaction occurs in ice, the ice appears to melt
as the reaction proceeds. Why does this occur?
Any given reaction of a system can be modeled by the equation for Gibbs Free Energy, ∆G = ∆H
- T∆S. In this equation, enthalpy (H) and entropy (S) are related via temperature, and the sign on
∆G indicates whether the reaction is spontaneous (∆G < 0) or nonspontaneous (∆G > 0).
However, in this question, we are interested in the phase change the “surroundings” (i.e., the ice)
undergo as the system of interest (i.e., the glucan polymer) undergoes its hydrolysis reaction.
Therefore, we are specifically looking for a source of heat to drive the melting process of ice
water. At constant pressures, enthalpy is equivalent to heat, and hence a positive ∆H
(endothermic) indicates a gain of heat into the system from the surroundings, and a negative ∆H
(exothermic) is a loss of heat from the system into the surroundings. Because we know that the
surroundings must be experiencing a gain of heat, we know that the reaction of the system must
be exothermic in order to drive the phase transition.
Lecture 2: pH / Buffers
2. Hydrochloric acid dissociates fully into protons (or hydronium ions) and chloride ions when it
is added to water. HEPES is a zwitterionic salt that exists in both acid and base forms with a pK
close to 7.0. What is the result of the addition of 1mM HCl to a 50mM solution of HEPES
previously adjusted to pH 7?
A. results in an increase in pH as the chloride ions act as a base.
B. results in a modest increase in pH as the ratio of the basic to the acidic forms of HEPES is
increased because the HCl ties up in the acidic form of HEPES.
C. results in only a modest decrease in pH because most of the protons from HCl are neutralized
by the basic form of HEPES.
D. results in only a modest decrease in pH because addition of 1mM HCl to water alone would
decrease the pH only from 7 to 6.8.
E. has no effect on pH because HEPES is zwitterionic but HCl is fully dissociated, so it can no
longer react with HEPES.
Because HCl ionizes completely in solution into H+ and Cl-, it is classified as a strong acid. Its Ka
value is therefore significantly higher than 1, as Ka = [H+][A-] / [HA]. In contrast, water at 25 ºC
has a Kw of 10-14, indicating that it is a weak acid as its K is less than 1. Therefore, water is more
basic than HCl, and after HCl dissociates completely into conjugate acid H+ and its conjugate
base Cl-, H2O will accept essentially all of these lone protons in solution, forming H3O+ and
significantly lowering the solution’s pH (pH = -log[H3O+]; in most biochemical reactions that
take place in aqueous solutions, the classic definition of pH, -log [H+], becomes equivalent to -
log [H3O+]. However, in the presence of a buffer, these pH changes can be mitigated. Buffers
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exist as approximately equal amounts of their protonated and deprotonated forms specifically
when the pH of a solution is equal to the buffer’s pK, as seen in the Henderson-Hasselbalch
equation, pH = pK + log([A-]/[HA]) (equal amounts of an acid and its conjugate base would
cancel the log term). Since HEPES has a pK of about 7.0, and is present in a solution adjusted to
pH 7 prior to the addition of HCl, it exists in approximately equal amounts of its acid
(protonated)and base (deprotonated) forms. Therefore, once HCl is added, the base form of
HEPES will neutralize many of the dissociated protons released from HCl. Fewer of the protons
would then be left to be accepted by H2O to form H3O+, maintaining a pH close to, yet somewhat
lower than, 7, yielding answer Choice C.
Lecture 3: Ionic Interactions
3. Ionic interactions between amino acids have been postulated to stabilize some proteins. Such
an interaction could involve:
A. The N-terminal amino group of one polypeptide chain in a tetrameric protein interacting with
the C-terminal carboxyl group in another polypeptide chain.
B. The side chains of serine and cysteine interacting in the interior of a protein.
C. The ionized phenolic side chains of tyrosines interacting with the side chains of
phenylalanines.
D. The ionized side chains of glutamic acid and aspartic acid residues interacting with each other
in an otherwise hydrophobic environment.
E. The amino and carboxyl groups that form one peptide bond with the carboxyl and amino
groups that form another peptide bond.
It seems a good deal of students, from the discussions, thought that Choice A (the correct
answer) described peptide bond formation rather than an ionic interaction. The N-terminal of a
polypeptide chain will not form a peptide bond with the C-terminal of another polypeptide
chain. Under normal conditions, peptide bond formation is catalyzed through the enzymatic
activity of the ribosome during translation. This process incorporates single amino acids into a
growing polypeptide chain, and cannot join two polypeptides together. Without an enzyme that
can catalyze this reaction, the two ends of a polypeptide chains can only engage in ionic
interactions with each other.
An ionic interaction or bond can be described as the electrostatic interaction between a cation (a
positively charged ion), and an anion (a negatively charged ion). In proteins, ionic interactions
can occur between a basic and acidic residue side chain intramolecularly or intermolecularly.
Additionally, because amino acids are dipolar ions, they can be classified as zwitterions at
physiological pH (~7.4). As the amino end of a protein typically has a pKa of about 9.4, at
physiological pH it will be protonated and positive. Because the carboxyl end has a pKa of
approximately 2.2, it will be deprotonated and negative at physiological pH. Therefore, the only
answer choice with a potential for ionic interaction is Choice A. It should be noted that the
amino and carboxyl groups in a protein that are not “free” (i.e., the ones involved in peptide
bonds) are not basic or acidic at near neutral pHs, respectively: amide bonds do not undergo
protonation or deprotonation under physiological conditions.
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Document Summary
Lecture 1: entropy of the system: you have been studying a reaction that generates a number of glucose molecules from the hydrolysis of a single high molecular weight glucan polymer. The reaction can proceed in either water or ice, but you have observed that when the reaction occurs in ice, the ice appears to melt as the reaction proceeds. Any given reaction of a system can be modeled by the equation for gibbs free energy, g = h. In this equation, enthalpy (h) and entropy (s) are related via temperature, and the sign on. G indicates whether the reaction is spontaneous ( g < 0) or nonspontaneous ( g > 0). However, in this question, we are interested in the phase change the surroundings (i. e. , the ice) undergo as the system of interest (i. e. , the glucan polymer) undergoes its hydrolysis reaction. Therefore, we are specifically looking for a source of heat to drive the melting process of ice water.
