Download 1. a. (6 points) HEPES (N-2-hydroxyethylpiperazine-N`-2

1. a. (6 points) HEPES (N-2-hydroxyethylpiperazine-N’-2-ethanesulfonic acid) has a
pK value of 7.47, which is in an acceptable range for use as a blood buffering
system. (The optimal blood pH is 7.4.) Assuming this were nontoxic and the
body were able to produce this buffer, why is it still an insufficient substitute for
the bicarbonate blood buffering system that our body uses? What is the
advantage of the bicarbonate system?
Bicarbonate not only serves as a blood buffering system, which HEPES could regulate,
but also facilitates CO2 expiration. Most CO2 in the blood is carried in the form of
bicarbonate. Carbonic anhydrase catalyzes this slow reaction as follows:
CO2 + H2O ↔ H+ + HCO3H+ generated from highly active muscles is taken up by Hb, thus causing Hb to release O2
and simultaneously causing further conversion of CO2 to bicarbonate (LeChatlier’s
principle). Bicarbonate can then travel through the blood to the lungs. In the blood
vessels of the lungs, Hb binds O2, causing the release of the proton. Addition of H+,
drives the formation of CO2 (again by LeChatlier’s principle), which is released from the
lungs.
b. (6 points) In a hospital laboratory, a 10.0 mL sample of gastric juices, obtained
several hours after a meal, was titrated with 0.1 M NaOH to neutrality; 7.2 mL of
NaOH was required. The stomach contained no ingested food or drink, thus
assume that no buffers were present. What was the pH of the gastric juice?
Multiplying volume (ml) by molar concentration (mol/L) gives the number of moles in
the volume added or present. If x is the concentration of gastric HCl (mol/L):
(10 mL) x = (7.2 mL)(0.1 mol/L)
x = 0.072 M gastric HCl
Since by definition, pH = - log[H+], and since HCl is a strong acid:
pH = -log(7.2 x 10-2) = 1.1
c. (3 points) Calculate the number of pentapeptides that can contain one residue
of each of the following: Ala, Gly, His, Lys, and Val.
The first residue can be one of 5 amino acids, the second can be one of 4 amino acids,
etc.
N = 5 x 4 x 3 x 2 x 1 = 120
2. a. (6 points) Draw the titration curve of the amino acid histidine.
Midpoint 3
9.33
pH
6.04
Midpoint 2
1.80
Midpoint 1
Equivalents OHi. (6 points) Label each midpoint on the graph and draw the chemical
structures present at that point.
At midpoint 1, fully protonated His and His with a negatively charged carboxyl group
would be present in equal amounts. At midpoint 2, His species with the protonated
amino group, protonated imidazole ring & negatively charged carboxyl group would be
present in equal amounts with the His species containing the protonated amino group,
negatively charged carboxyl group & the neutral imidazole ring (the side chain has
become deprotonated. At midpoint 3, the latter would be present in equal amounts as the
fully deprotonated His species - negatively charged carboxyl group and neutral amino
group and imidazole ring.
ii. (3 points) On the graph, mark the pH ranges at which histidine could
act as a buffer.
His could act as a buffer at pH values that are one unit above and below each of its pKa
values:
0.8 < pH < 2.8, 5.04 < pH < 7.04, 8.33 < pH < 10.33
b. (6 points) A peptide isolated from the brain has the sequence
Glu-His-Trp-Ser-Tyr-Gly-Leu-Arg-Pro-Gly
Determine the net charge on the molecule at pH 3. What is the net charge at pH
5.5? At pH 8? At pH 11? Estimate the pI for this peptide. (Use pKa values for
side chains and terminal amino and carboxyl group as given in Table 4-1.)
When pH > pKa, ionizing groups lose their protons. The pKa values of importance here
are those of the amino-terminal (2.3) and carboxyl-terminal (9.6) and those of the R
groups of Glu (4.3), His (6.0), Tyr (10.1), and Arg (12.5).
pH
3.0
5.5
8.0
11
+
H3N-Glu-His-Trp-Ser-Tyr-Gly-Leu-Arg-Pro-Gly-COO+1 0 +1
0
+1
-1
+1 -1 +1
0
+1
-1
+1 -1 0
0
+1
-1
0 -1 0
-1
+1
-1
Net Charge
+2
+1
0
-2
To calculate the pI:
Avearge the pKa values for the two side chains that ionize near pH 8, the amino-terminal
α-amino group of Glu and the His imidazole group:
pI = (9.6 + 6.0) = 7.8
2
c. (4 points) Some years ago two drug companies marketed a drug under the trade
names Dexedrine and Benzedrine. The structure of the drug is shown below.
The physical properties (C, H, and N analysis, melting point, solubility, etc.) of
Dexedrine and Benzedrine were identical. The recommended oral dosage of
Dexedrine (which is still available) was 5 mg/d, but the recommended dosage of
Benzedrine was significantly higher. Apparently it required considerably more
Benzedrine than Dexedrine to yield the same physiological response. Provide a
possible simple answer to this apparent contradiction.
Only one of the two enantiomers of the drug (which has a chiral center) is physiologically
active, due to interaction with a stereospecific receptor site. Dexadrine, as manufactured,
consists of only the single enantiomer (D-amphetamine) recognized by the receptor site.
In contrast, Benzedrine is a racemic mixture (equal amounts of the D and L isomers).
Thus, a much larger dose of Benzedrine is required to obtain the same effect.
3. a. A biochemist discovers and purifies a new enzyme, generating the purification
table below:
Procedure
Total Protein
Activity
(mg)
(units)
1. Crude extract
20,000
4,000,000
2. Precipitation (salt)
5,000
3,000,000
3. Precipitation (pH)
4,000
1,000,000
4. Ion exchange chromatography
200
800,000
5. Affinity chromatography
50
750,000
6. Size-exclusion chromatography
45
675,000
i. (6 points) From the information given in the table, calculate the specific
activity (units/mg) of the enzyme solution after each purification
procedure.
From the percentage recovery of activity (units), we can calculate percentage yield and
the specific activity (units/mg).
Procedure
1.
2.
3.
4.
5.
6.
Protein
(mg)
20,000
5,000
4,000
200
50
45
Activity
(units)
4,000,000
3,000,000
1,000,000
800,000
750,000
675,000
% Yield
(100)
75
25
20
19
17
Specific activity Purification
(units/mg)
factor
200
(1.0)
600
x 3.0
250
x 1.25
4,000
x 20
15,000
x 75
15,000
x 75
ii. (4 points) Which of the purification procedures used for this enzyme is
most effective (i.e., gives the greatest increase in purity)?
Step 4, ion-exchange chromatography, gives the greatest increase in specific activity,
which is an index of purity and degree of increase in purification.
iii. (4 points) Which of the purification procedures is least effective?
Step 3, pH precipitation, in which two-thirds of the total activity from the previous step
were lost.
iv. (3 points) Is there any indication in this table that the enzyme is now
pure? What else could be done to estimate the purity of the enzyme
preparation?
Yes, the specific activity was constant after step 5, affinity chromatography. SDSpolyacrylamide gel electrophoresis is an excellent, standard way of checking
homogeneity and purity. Mass spectrometry is another possibility.
b. (3 points) You place concentrated solutions of a mix of the following three
proteins, myoglobin, serum albumin, and fibrinogen, in water inside a dialysis
tubing. You then place the dialysis tubing inside a large beaker containing a
solution of ammonium sulfate at an ionic strength of 6. Describe what you think
would happen to the proteins inside the dialysis tubing.
Fibrinogen would essentially be totally precipitated; the serum albumin would partially
precipitate, while the myoglobin would all remain in solution. See Figure 6-2.
4. A group of peptides that influence nerve transmission in certain parts of the brain
has been isolated from normal brain tissue. These peptides are known as opioids,
because they bind to specific receptors that bind opiate drugs, such as morphine and
naloxone. Opioids thus mimic some of the properties of opiates. Some researchers
consider these peptides to be the brain’s own pain killers. Using the information below,
determine the amino acid sequence of the opioid leucine enkaphalin. Explain how your
structure is consistent with each piece of information.
a. (4 points) Complete hydrolysis by 1 M HCl at 110 ºC followed by amino acid
analysis indicated by the presence of Gly, Leu, Phe, and Tyr, in a 2:1:1:1 molar
ratio.
The empirical composition is (2Gly, Leu, Phe, Tyr)n.
b. (4 points) Treatment of the peptide with 1-fluoro-2,4-dinitrobenzene followed
by complete hydrolysis and chromatography indicated the presence of the 2,4dinitrophenyl derivative of tyrosine. No free tyrosine could be found.
Tyr is the amino-terminal residue, so the sequence is Tyr-(2Gly, Leu, Phe).
c. (4 points) Complete digestion of the peptide with pepsin followed by
chromatography yielded a dipeptide containing Phe and Leu, plus a tripeptide
containing Tyr and Gly in a 1:2 ratio.
Referring to Table 7-2, we see that pepsin cleaves on the amino-side of aromatic residues
(Phe, Tyr, Trp) and Leu. Since we know Tyr is amino-terminal, we can deduce that the
sequence of the first part of the peptide fragment must be Tyr-Gly-Gly. Since the
dipeptide consists of two amino acids that can both be cleaved by pepsin (Phe and Leu),
we must conclude that the peptidase was inefficient at cleaving the bond between the
two, and therefore we are unable to determine in which order these will appear. Thus the
sequence can be one of two possibilities:
Tyr-Gly-Gly-Leu-Phe or Tyr-Gly-Gly-Phe-Leu
It is possible that n > 1 (i.e., that this sequence repeats). A good experimentalist would
test this by a determination of Mr).
5. a. (6 points) The membrane protein bacteriorhodopsin is involved in proton pumping
and contains an aspartic residue at the 96 position (D96) of the protein. D96 is an
internal amino acid residue surrounded by a leucine barrel with two
phenylalanines (F42 and F219) covering the barrel. It was found that the pKa
value of D96 was 11.4 while it remained in this closed state. The limited solvent
accessibility is probably caused by F42, which shields D96 from the cytoplasmic
surface. Structural changes, such as slight kinking or tilting of one or more helices
during protein activity, may allow water to diffuse into this part of the protein.
(Zscherp et al. Proc Natl Acad Sci; 1999; 96(10): 5498–5503.) What effect do
you think this structural change will have on the pKa value of D96? Please
provide a concise explanation for this.
As discussed in Lecture 2 (see the chart on slide 20), the pKa value of an amino acid is
subject to change based on the dielectric constant of its microenvironment. As its
surroundings become increasingly hydrophobic/nonpolar, the dielectric constant is
lowered and the pKa value of the ionizable side chain is raised, hence it becomes less
acidic. The structural information provided suggests that the reason for the exceptionally
high pKa is the very hydrophobic vicinity of D96. When the protein undergoes a
structural change and D96 is subsequently exposed to the aqueous solution of the
cytoplasm, the dielectric constant of the microenvironment surrounding the amino acid
would be raised; hence the pKa value of the ionizable side chain would lower.
(According to the authors, time-resolved ATR/FT-IR experiments at different pH values
allowed in situ determination of the change in pKa. The pKa of D96 appears to be at
7.1 ± 0.2 during the lifetime of the N intermediate.)
b. (6 points) Under physiological conditions, polylysine assumes a random coil
conformation. Under what conditions might it form an α helix?
At physiological pH, the positively charged lysine side chains repel each other.
Increasing the pH above the pKa of the side chain amino group (~ pKa 10.5) would
neutralize the side chains and allow an α helix to form.
6. a. (4 points) Explain the specific problem that would arise in oxygen transport
through the body if blood pH were too high.
See Fig. 10-6. Basic conditions shift the oxygen-binding curve of Hb to the left.
Therefore, at low pO2 (venous pressure), YO2 is higher than normal, i.e. Hb binds oxygen
with a greater affinity. Under these conditions, Hb would be less likely to release O2 to
the tissues.
(4 points) What if blood pH were too low? (4 points)
Acidic conditions shift the oxygen-binding curve of Hb to the right. Therefore, at low
pO2 (venous pressure), YO2 is lower than normal, i.e. Hb binds oxygen with a lower
affinity. Under these conditions, Hb would bind less O2 from the lungs and retain less O2
before reaching the tissues.
b. Studies of oxygen transport in pregnant mammals have shown that the O2saturation curves of fetal and maternal blood are markedly different when
measured under the same conditions. Fetal erythrocytes contain a structural
variant of hemoglobin, hemoglobin F, consisting of two α and two γ subunits
(α2γ2), whereas maternal erythrocytes contain the usual hemoglobin A (α2β2).
i. (4 points) Which hemoglobin has a higher affinity for oxygen under
physiological conditions, hemoglobin A or hemoglobin F? Explain how
you came to this conclusion.
Hemoglobin F; its pO2 (the pressure of oxygen required for half-saturation) is lower than
that of HbA, thus HbF binds oxygen better than does HbA at low oxygen levels. An
alternative way of looking at this is that at pO2 = 4 kPa, HbF is 58% saturated, whereas
HbA is only 33% saturated. Both comparisons indicate that HbF binds oxygen more
efficiently than HbA under physiological conditions.
ii. (4 points) What is the physiological significance of the different oxygen
affinities? Explain briefly.
The higher oxygen affinity (lower pO2 for half-saturation) of HbF means that oxygen
passes (or is pulled) from adult (maternal) blood to fetal blood in the placenta.