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LS1a Fall 2014
Problem Set #2
Due Monday 10/6 at 6 pm in the drop boxes on the Science Center 2nd Floor
Note: Adequate space is given for each answer. Questions that require a brief explanation
should be answered only in the provided space.
I. Basic Concept Questions
1. (12 points) Fill out the table below using the filled-in boxes as guides. Draw the
structures of the indicated amino acids (using standard drawings) at pH = 7 and include
the one-letter and three-letter codes where appropriate, as well as a description of their
side chain properties (i.e., polar, nonpolar, positive at pH 7.0, or negative at pH 7.0).
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2. (7 points) Shown below is the structure of the amino acid arginine (drawn at pH 7).
a. (3 points) On the structure above, identify this amino acid’s side chain, amine, and
carboxylate.
b. (4 points) Shown below is a tripeptide. The resonance structure of a peptide bond
causes the nitrogen atom of the peptide bond to adopt a trigonal planar molecular
geometry (just like in the week #4 section handout and in the example of ethylene
on pages 17-19 of the notes for lecture 7).
Consider the peptide bond between amino acids 1 and 2 (amino acid 1 has “R1” as
its R-group, and amino acid 2 has “R2” as its R-group) and put a box around each of
the six atoms that are “coplanar” (i.e., those that must reside in the same plane) due
to the resonance form of the peptide bond between amino acids 1 and 2. Similarly,
draw circles around each of the six atoms that are coplanar due to the peptide bond
linking amino acids 2 and 3.
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3. (22 points) The complex series of reactions through which organisms build amino acids
often involve other amino acids as intermediates. A number of these reactions are
outlined below. Each species is identified by a number. Multi-step processes are shown
with multiple arrows ().
a. (14 points) Indicate whether each numbered species is one of the 20 naturally
occurring amino acids that is found in proteins (yes or no). If the answer is “yes,”
write its name.
Amino Acid?
Name (if Yes)
Amino Acid? Name (if Yes)
1.
2.
3.
4.
5.
6.
7.
Yes
Yes
Yes
Yes
Yes
Yes
Yes
No
No
No
No
No
No
No
_______________
_Threonine_
_______________
_______________
____Lysine___
_______________
_______________
8.
9.
10.
11.
12.
13.
14.
Yes
Yes
Yes
Yes
Yes
Yes
Yes
No
No
No
No
No
No
No
Aspartic Acid__
_Glutamic Acid_
_Glutamine___ _
_Asparagine_ __
___________________
___________________
___Arginine______
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b. (8 points) The molecule shown below can undergo one of two processes, each
leading to a different amino acid (after additional steps). The transition state from
each reaction is shown below. In the space provided, draw the product of each
transition state and give the name of the amino acid that is the most likely end
product of each transition state by taking into account amino acid side chain
structures.
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4. (8 points) As we learned in lecture, keratin, the protein in hair, contains disulfide bonds
between cysteine side chains that must be broken when hair stylists give perms. Shown
below is the arrow pushing mechanism describing how thioacetic acid (CH3COSH;
shown in bold) is used to accomplish this.
a. (4 points) In the space provide above, draw the products of the arrow pushing
mechanism.
b. (4 points) In the space below, draw the transition state for this reaction.
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5. (11 points) The reaction of the cysteine side chain with a chemical reagent called Nethylmaleimide (“NEM”) is shown below.
a. (5 points) Onto the structures of the reactants, draw the arrow pushing mechanism
that leads from the reactants to the products via the transition state that is shown.
b. (6 points) Consider two unfolded samples of a protein whose folded structure is
stabilized by disulfide bonds (for example: keratin). One of the two unfolded protein
samples is treated with NEM and the other unfolded protein sample is left
untreated. If both samples are then allowed to fold, which protein sample (the
“NEM-treated” or “untreated”) would feature a larger ratio of folded to unfolded
protein at equilibrium? Briefly explain your answer.
NEM treatment prevents cysteine residues from being able to form disulfide
bonds. You can consider the reaction between cysteines to form a disulfide
bond and the reaction of a cysteine with NEM to be two coupled reactions: a
cysteine can EITHER react with NEM to form a covalent bond OR two cysteines
can react with each other to form a covalent disulfide bond. Disulfide bonds
can only form between cysteines with available –SH (“sulfhydryl”) groups at
the ends.
Given a protein folding reaction, the “reactants” are the unfolded protein and
the “products” are the folded protein. Since disulfide bonds stabilize the
folded structure, the folded protein is lower in energy when the disulfide
bonds are made than when they are not made. Without the disulfide bonds,
ΔG°folding is therefore less favorable (since the folded protein “product” is now
higher in energy without the disulfide bonds). Since ΔG°folding is less negative,
Keq for folding will be smaller for the NEM treated sample. Since Keq is larger
for the sample without NEM treatment, the ratio of folded to unfolded protein
will be higher for the untreated sample.
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II. Applied Concept Questions
6. (17 points) Later this semester we will be using a technique called chromatography to
separate proteins in order to purify a protein called the Green Fluorescent Protein (or
“GFP”). One type of chromatography involves separating peptides based on their net
charges. One way of doing this is to pass a mixture of peptides through a column that is
lined with positive charges. As the mixture flows through the column, peptides with
negative charges interact with the positively-charged column and pass through the
column more slowly. The greater the negative charge of the peptide, the slower it will
move through a positively-charged column. Because a positively-charged column
separates different peptides based on the differences between their net charges, we first
need to choose an appropriate pH to maximize the difference between the peptides' net
charges.
A mixture of two peptides, NH2-C-D-R-Y-COOH (peptide 1) and NH2-K-H-E-W-COOH
(peptide 2), are both dissolved in three different buffered solutions, A, B, and C, shown
below.
Solution A:
pH 1.0
Solution B:
pH 5.0
Solution C:
pH 13.0
a. (10 points) In the space below, draw both peptides as they would predominately
exist in solution A. Refer to the lecture notes for pKa values of the amino acid side
chains. The pKa of the carboxyl terminus is ~ 2.3, and the pKa of the free amino
terminus is ~ 9.8. In your diagram, draw all amino acids in the trans conformation.
You do not need to show stereochemistry.
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b. (4 points) Calculate the net charges for each peptide in each of the solutions (A, B,
and C). In which solution (A, B, or C) would there be the greatest difference in net
charge between the two peptides?
Solution A: Peptide 1, +2; Peptide 2, +3
Solution B: Peptide 1, 0; Peptide 2, +1
Solution C: Peptide 1, -4; Peptide 2, -2
Solution C gives the greatest difference in net charge.
c. (3 points) Which solution (A, B, or C) would allow for the best separation of the two
peptides by a positively charged column? Which peptide would spend more time
bound to the column? Briefly explain your reasoning.
Solution C because it generates the greatest possible difference in net charge
that is possible between the two peptides. Peptide 1 is more negative and
would spend more time bound to the column than the less negative peptide.
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7. (8 points) Three-dimensional representations of two tripeptides are shown below.
These two images were generated using PyMOL, the software used in lab to visualize 3D molecules. [Note: green is carbon; blue is nitrogen; red is oxygen; yellow is sulfur;
hydrogen atoms are not shown. If you print this question out in black and white, it will
probably help to at least look at the color picture on the computer monitor while
considering this question.]
a.
(4 points) Write the amino acid sequence of each peptide starting from the Nterminus. Label the N- and C-termini in your answer.
Peptide 1: N-terminus – Glu – Trp – Pro – C-terminus
Peptide 2: N-terminus – Met – Pro – Thr – C-terminus
b.
(4 points) Circle all α-carbons present in the structures shown above. Draw an
arrow pointing to any cis peptide bond(s) that are present.
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8. (5 points) A schematic diagram of a 372 amino acid protein is shown below. Two
stretches of amino acids that interact with each other in the folded protein to form a βsheet are highlighted:
a. (3 points) The diagram below indicates two amino acids, glycine 181 and tyrosine
164. The numbers indicate the amino acid’s position in the protein’s primary
sequence (where “1” is the amino acid at the N-terminus and “372” is the amino acid
at the C-terminus of this particular protein). Based on these numbers, fill in the
blank next to each of the two indicated amino acids to show each amino acid’s
number in the primary sequence. Briefly explain how you arrived at your answer.
First one must identify the N-/C-termini; on the top strand, the N terminus is
to the left and on the bottom strand, the N-terminus is to the right. The Ile is
two positions closer to the N-terminus than Tyr 164, so it must be in position
number 162. The Ala is three positions further from the N-terminus than Gly
181, so it must be in position number 184.
b. (2 points) In the figure β-sheet shown above, draw dashed lines (
) to indicate
the hydrogen bonds that stabilize this segment of secondary structure.
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9. (10 points) Below is an image of two proteins (one green and the other cyan) binding to
one another via their β-sheets. The colors are important for this question, so you may
want to do this problem looking at a color print out or online through the course
website.
Protein A
Protein B
a. (4 points) The image to the left shows the
same two proteins in the same orientation,
but hides much of the two proteins and
focuses the regions of through which these
two proteins interact. The image shows only
the polypeptide backbone (i.e., the side
chains are hidden), and intermolecular
interactions are shown in magenta.
Use your understanding of the
interactions that stabilize secondary
structure to briefly explain the types of
interactions proteins exploit to bind one
another by aligning the edges of their βsheets. In particular, what makes the
exposed edges of β-sheets productive
places for proteins to interact?
Protein B
Protein A
β-strands adopt an extended shape in which the polypeptide backbone
carbonyls and amines of successive residues point in opposite directions.
These amide carbonyls and amides serve as hydrogen bond acceptors and
donors with adjacent β -strands. β-sheets are held together by the many
hydrogen bonds that occur between adjacent β-strands.
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While the strands within the sheets make hydrogen bonds to adjacent strands
on both of sides, the β-strands at the edge of a sheet are only adjacent to one βstrand, leaving the potential hydrogen bonds that could be made to another βstrand unsatisfied.
β-strands at an exposed edge of a sheet are therefore capable of forming
hydrogen bonds with another β-strand from another protein that is stretched
out in the same conformation (with its amines and carbonyls in the
appropriate spatial arrangement) because it is also a β-strand.
The structure of β-strands optimally orients each carbonyl and amine group
to hydrogen bond with an adjacent strand, such that a β-strand at the edge of a
β-sheet in one protein is optimally oriented to form as many hydrogen bonds
as possible with a β-strand at the edge of a β-sheet from a different protein.
b. (6 points) In addition to interacting using the edges of their β-sheets, the proteins
also interact through several side chains. One example, in which lysine, the 84th
amino acid on the cyan protein, interacts with glutamate, the 31st amino acid on the
green protein, as shown below:
K84
E31
To determine the importance of this interaction between lysine and glutamate,
scientists conducted three experiments in which they changed which amino acids
are present at these same positions.
In experiment 1, scientists replaced lysine with glutamate at amino acid position 84
of the cyan protein. In experiment 2 the lysine at position 84 of the cyan protein and
the glutamate at position 31 of the green protein were both swapped for alanines. In
experiment 3, the lysine of the cyan protein is replaced by a glutamate and the
glutamate on the green protein is replaced with a lysine. These experiments, along
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with the ΔG°binding describing the favorability of the two proteins binding one
another, are summarized in the table below:
Experiment
Control
1
2
3
Protein
Unaltered amino acid sequence
K84E
K84A, E31A
K84E, E31K
ΔG°binding
-7 kJ/mol
+3 kJ/mol
-1 kJ/mol
-7 kJ/mol
This chart uses a formalism when describing changes to a protein’s amino acid
sequence, in which the first letter indicates what the amino acid started out as, the
adjacent number details where in the primary sequence the amino acid occurs, and
the final letter indicates which amino acid the original one has been changed to:
K84E
Original amino acid
Final amino acid
Position in primary structure
This example of “K84E” translates to “amino acid 84, which is normally lysine, has
been changed to glutamate.”
Using the information in the chart above, briefly explain why each experiment
makes the protein binding interaction more or less favorable.
In the control experiment with wild-type proteins, the positively-charged
lysine makes an ionic bond with the negatively-charged glutamate.
In experiment 1, lysine is replaced by a negatively-charged glutamate, which
will be repelled by the glutamate on the green protein, making the binding
reaction less favorable.
In experiment 2, by exchanging both lysine and glutamate for alanines, the
favorable ionic bond is lost, so binding is weaker than between the two
unaltered proteins. However, since no repelling charges are introduced at the
protein interface as in experiment 1, the binding interaction between the two
proteins in experiment 2 is still more favorable than in experiment 1.
In experiment 3, the two amino acids are swapped between the two proteins:
the cyan protein now has glutamate instead of lysine, and the green protein
has lysine instead of glutamate, such that the interaction between these two
side chains is equivalent to the original, unaltered scenario.
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