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Solutions to Problelms
Chapter 5
2. Relationship between the titration curve and
the acid-base properties of glycine
(a) Glycine
is
present
predominantly
as
the
species
H3N-CH2-COOH?
Solution: (I)
(b) The average net charge of glycine is +1/2
Solution : (II)
(c) Half of the amino groups are ionized
Solution: (IV)
(d) The pH is equal to the pKa of the carboxyl group
Solution: (II)
(e) The pH is equal to the pKa of the protonated amino group.
Solution: (IV)
(f) Glycine has its maximum buffering capacity
Solution: (II and IV)
(g) The average net charge of glycine is zero
Solution: III
(h) The carboxyl group has been completely titrated ( first
equivalence point)
Solution: (III)
(i)
Glycine is completely titrated ( second equivalence
point)
Solution: (V)
(j) The preominant species is +H3N-CH2-COOSolution: III
(k) The average net charge of glycine is –1
Solution: (V)
(l) Glycine is present predominantly as a 50:50 mixture of
+
H3N-CH2-COO-
Solution: (II)
(m) This is the isoelectric point
Solution: (III)
(n) This is the end of the titration
Solution: (V)
(o) There are the worst pH region for buffering power
Solution: (I) (III) and (V)
9. The number of tryptophon residues in bovine serum
albumin
(a)
Since a molecule of water has been eliminated from the
tryptophan ( it is a residue), the molecular weight of
tryptophan is :
Mtrp = 204-18 = 186
So the the minimum molecule weight of BSA is :
MBSA = 186 / 0.58% = 32069
(b)
Since the real molecular weight of BSA is 70000, the
ration between these two is :
This means that there are two tryptophan residues in a
molecule of BSA.
12. The isoelectric point of histones
Solution: Based on the pI of hintones, we could estimate that
the amino acid residues present at large quantities may be
Lys, Arg and His. Negatively charged phosphate groups in DNA
interact with positive charged side groups in histones.
16. Structure of a peptide antibiotic from Bacillus
brevis
Orn → Leu →
Phe → Pro → Val
↑
↓
Val ← Pro ←
Phe ←
Leu
← Orn
Explanations:
(a)
There are equal amount of Leu, Orn, Phe, Pro and Val
in this peptide.
(b)
The molecular weight of this peptide is:
MW = 2x (131+132+165+115+117)-10x18 = 1140
Approximate to the molecular weight estimation.
(c)
Since this peptide is a circle and no free c terminal
existed
the
enzyme
carboxypeptidase
cannot
hydrolysis it.
(d)
Since this peptide is a circle and no other free amino
group existed except for the Orn, the only amino acid
residue which could react with FDNB is Orn and give
out the only derivative.
(e)
This could be explained by the sequence.
Chapter 6
6. Amino acid sequence and protein structure
(a) Bends are most likely at residues 7 and 19; Pro residues
in the cis configuration accommodate turns well.
(b)The Cys residues at positions 13 and 24 can form disulfide
bonds.
© External surface; polar and charged residues (Asp, Gln,
Lys); interior: nonpolar and aliphatic residues (ala, Ile);
Thr, though polar, has a hydropathy index near zero and thus
can be found either on the external surface or interior of
the protein.
7. Bacterirhodopsin in purple membrane proteins
(a) Since the move distance of a amino acid residue in a α
helix is 5.4/3.6 = 1.5 nm,the minimum number of amino acid
residues necessary for one segment ofαhelix to traverse
the membrane completely is : 45/1.5 = 30
(b) The minimum number of amino acid residues involved in
αhelix conformation is : 7 x 30 =210
Based on the average amino acid residue weight 110, the
molecular weight of these residues involved inαhelix is :
210 x 110 = 23100
So the fraction of the bacteriorhodopsin protein that is
involved in memebrane-spanning helices is :
23100/26000 = 89%
8. Pathogenic action of Bacteria that cause gas
gangrene
Solutions: The bacterial enzyme (a collagenase) can destroy
the conne barrier of the host, allowing the bacterium to
invade the host. Bacteria do not contain collagen.
Chapter 7
3.Affinity on oxygen in myoglobin and hemoglobin
Solution: Myoglobin: Since pH, CO2 partial pressure in the
lungs and BPG concentration have no influence to myoglobin
oxygen affinity, changes on these have no effect.
Hemoglobin:
(a)
A drop in the pH of blood plasma from 7.4 to 7.2
Increasing concentration of H+ ( decrease of pH) lowers
the oxygen affinity of hemoglobin and tending to release the
oxygen from the hemoglobin molecules.
(b)
A decrease in the partial pressure of CO2 in the lungs
from 6kPa to 2kPa
Decrease in the partial pressure of CO2 in the lungs will
increase the oxygen affinity of hemoglobin and tend to
release the oxygen from the hemoglobin molecules.
(c)
An increase in the BPG level from 5mM to 8mM
BPG’s
binding to the central cavity of hemoglobin
will lower its oxygen affinity.With the increase in the BPG
level, the oxygen affinity of hemoglobin will decrease.
7. Reversible (but tight) binding to an antibody
From the definition ofθ, we could get:
θ= [L]/([L]+Kd)
Here, [L] is the concentration of antigen, and Kd = 5x10-8
(a)
When θ= 0.2, [L]= (0.2x5x10-8)/(1-0.2)= 1.25x10-8(M)
(b)
When θ= 0.5, [L]= (0.5x5x10-8)/(1-0.5)= 5x10-8(M)
(c)
When θ= 0.6, [L]= (0.6x5x10-8)/(1-0.6)= 7.5x10-8(M)
(d)
When θ= 0.8, [L]= (0.8x5x10-8)/(1-0.8)= 2x10-8(M)
9. The immune system and vaccines
Solution:Many pathogens, including HIV, have evolved
mechanisms by which they can repeatedly alter the surface
proteins to which the immune system components initially
bind. Thus the host organism regularly faces new antigens
and requires time to mount an immune response to them. As
the immune system responds to one variant, new variants
created.
10. How we become a “stiff”
Solution: Binding of ATP to myosin triggers dissociation
of myosin from the actin thin filament. In the absence
of ATP, actin and myosin bind tightly to each other.
Chapter 8
4.Protection of enzyme against denaturation
by heat
Solution: The enzyme-substrate xomplex is more stable than
the enzyme alone.
5. Requirements of active sites in enzyme
(a)
The interval between Arg145 and Glu270 is 270-145=125
The distance between the two amino acid in a α helix
is 0.15nm.
So the distance between Arg145 and Glu270 is:
123 x 0.15 = 18.8nm = 188 Å
(b)
Three-dimensional folding of the enzyme brings these
amino acid residues into close proximity.
7.Relation
between
reaction
substrate
concentration:
velocity
and
Michaelis-Menten
equation
Solution: Based on Michaelis-Menten equation
(a) Km = 0.005M
V=1/4 Vmax
[S]=1/3 Km = 1.7x10-3(M)
(b)According to Michaelis-menten equation,
V/Vmax = [S]/(Km+[S])
[S]= 1/2 Km, V= (1/2 Km)/ (1/2 Km+ Km)= 0.33
[S]= 2 Km, V= (2Km)/ (2Km+ Km)= 0.67
[S]= 10 Km, V= (10 Km)/ (10 Km+ Km)= 0.91
8.Estimation of Vmax and Km by inspection
Based on the enzyme-catalyzed reaction data obtained, we
could
see
that
with
the
increasing
of
substrate
concentration, the velocity of the reaction is approximate
to 140uM/min. In this way we could estimate the Vmax of this
reaction is 140uM/min. So the Km of this reaction is:
Km = (Vmax/V – 1)[S]
Using one set of data, e.g. V=28uM/min, [S]= 2.5x10-6M
Then, Km = (140/28 –1)x 2.5x10-6 = 1x10-5 (M)
9.Properties of an enzyme of prostaglandin synthesis
(a)
When [S]=1.0mM, V = 32.2mM/min
[S]=1.5mM, V = 36.9mM/min
V = Vmax[S]/ (S+Km)
Vmax = 51.55mM/min Km = 0.598
(b)
Using the data in the first and third columns of the table,
from the double-reciprocal plots, we could find the Vmax
doesn’t change. So the inhibion is competitive.
11.The Eadie-Hofstaee equation
Solution; Since a competitive inhibitor only decrease the
Km of the reaction and Km= -slope.
In this plot,the slope of curve B and curve C decreased while
that of curve A increased. So the only decrease in Km is curve
A, which means only curve A shows the enzyme activity when
a competitive inhibitor is added to the reaction mixture.
12.The turnover number of carbonic anhydrase
MWenzyme = 3x104
the molar of the enzyme is
10ug/(3x104)= 3.33x10-10
MWCO2 = 44
the molar of the CO2 is
0. 30g/(44)= 6.82x10-3
Kcat = 6.82x10-3 / 3.33x10-10 = 2x107 (min-1)
16. Inhibition of Carbonic Anhydrase by acetazolamide
Solution: From the plot we could find the Vmax decreased with
the adding of acetazolamide. Since a competitive inhibitor
won’t affect the Vax of a reaction acetazolamide is a
noncompetitive inhibitor.
17.pH optimum of lysozyme
Solution: pKaGlu = 5.9
pKaAsp = 4.5
At pH 5.2, Glu35 is protonated and Asp is deprotonated.
Since Glu35 and Asp52
are two amino acid residues
essential for catalysis, their protonated or deprotonated
status will affect the activity of the enzyme. When pH is
raised to higher than 5.9, the two amino acid residues will
all be deprotonated and be protonated when pH is decreased
to lower than pH4.5. Both deprotonated or protonated status
of these two amino acid residues will great decrease the
activity of lysozyme.