Chap. 3. Problem 2.
Fully protonated glycine has two dissociable protons, one on its carboxyl group (-COOH) and one on its -amino group (-NH3+).
The three forms of glycine that occur during its titration
therefore are +H3N-CH2-COOH (Point I), +H3N-CH2-COO- (Point
III), and H2N-CH2-COO- (Point V). Using this information, all
parts to the question can be answered (next three slides).
Chap. 3. Problem 2. (continued)
Part (a). The fully protonated species of glycine (+H3N-CH2COOH) occurs at the beginning of the titration, i.e., Point I.
Part (b). 50% of the -COOH protons are titrated to -COO- at
the pK1 of glycine (Point II, pH 2.34). At this point, half of the
molecules have a net +1 charge (+H3N-CH2-COOH) and half have
a 0 charge (+H3N-CH2-COO-). Thus the average net charge of
glycine is +1/2 at Point II.
Part (c). Half of the -amino groups of glycine molecules are
ionized at its pK2 (Point IV, pH 9.6).
Part (d). From the HH equation, a 1/1 ratio of the conjugate
base/conjugate acid forms of a weak acid occur when pH = pKa.
Thus, the pH of the glycine solution is equal to the pKa of the carboxyl group at Point II where, 50% of the -COOH groups have
been titrated.
Part (e). Point IV. See the explanation in the answer to Part (d).
Part (f). The optimum buffering pHs for a solution of glycine
occur at the two plateaus in the titration curve (Points II & IV).
This is where the concentrations of the conjugate base and
conjugate acid forms undergoing titration are the highest.
Chap. 3. Problem 2. (continued)
Part (g). The average net charge of glycine is zero at its
isoelectric point. This occurs after the -COOH group of the
amino acid has been titrated and the most prevalent species in
solution has the structure, +H3N-CH2-COO-. This is Point III,
which occurs at a pH halfway between pK1 and pK2
pI = (pK1 + pK2)/2 = (2.34 + 9.60)/2 = 5.97.
Part (h). The first equivalence point in the titration occurs when
the -COOH group has been completely titrated, i.e., Point III.
Part (i). The second equivalence point in the titration occurs when
the -NH3+ group has been completely titrated, i.e., Point V.
Part (j). The species, +H3N-CH2-COO-, which has a net charge of
zero occurs at the pI. This species predominates at Point III,
which has a pH of 5.97.
Part (k). The average net charge of glycine occurs at Point V,
where both dissociable groups have been completely titrated and
the structure of the molecule is H2N-CH2-COO-.
Part (l). Glycine occurs with a 50/50 mixture of +H3N-CH2-COOH
and +H3N-CH2-COO- when the -COOH group has been 50%
titrated. This is at the midpoint of the first plateau in the
titration curve, where pH = pK1.
Chap. 3. Problem 2. (continued)
Part (m). The isoelectric point occurs at Point III. (See answers
(g) and (j).
Part (n). Glycine has two dissociable groups, so the end of the
titration occurs when 2.0 equivalents of strong base have been
added. This corresponds to Point V on the curve.
Part (o). The worst pH regions for glycine to serve as a buffer
occur at the points furthest from the two plateaus, pH = pK1 and
pH = pK2. These regions occur at Points I, III, and V.
Chap. 3. Problem 4.
Part (a). The structures,
equilibrium equations, and pKas
for the three ionization reactions
of histidine are shown in the
diagram.
The net charge on histidine in each ionization state is (1) +2; (2)
+1; (3) 0; and (4) -1.
Part (b). The structures are the same as in Part a because each
pH listed flanks the pKas of histidine’s three dissociable protons.
Part (c). The net charge of the histidine species at each of these
pHs is shown in the figure. Based on the net charge, histidine will
migrate towards the (1) cathode; (2) cathode; (3) not migrate;
and (4) anode when placed in an electric field.
Chap. 3. Problem 8.
The average Mr of an amino acid in a protein is 128. After
correcting for the loss of water (Mr 18) that occurs in peptide
bond formation the average Mr becomes 110. Based on this value,
the size of the protein is approximately
682 aa x 110 aa-1 = 75,020 = 75,000 = Mr
Chap. 3. Problem 10.
Both types of electrophoresis used in this question separate
polypeptides on the basis of molecular weight. The second
procedure also breaks intra- and intermolecular disulfide bonds
that may be present. Because the 160 and 60 kDa polypeptides
are present under both conditions, the data suggest that these
two proteins are not attached to others by disulfide bonds.
Because the 180 kDa protein obtained in the first procedure
disappears and a new band at 90 kDa appears, it suggests that
the 180 kDa protein may actually consist of two 90 kDa subunits.
Thus the most likely subunit composition of this 400 kDa protein
is (160, 902, 60). The two 90 kDa subunits are linked by an
intermolecular disulfide bond(s).
Chap. 3. Problem 13.
Like amino acids, peptides and proteins have isoelectric points at
which the sum of all charged residues on the protein is zero. If a
protein has a relatively large number of basic residues (His, Arg,
Lys) then the pI of the protein will be high. Conversely, if it has
a relatively large number of acidic residues (Asp, Glu), then the
protein will have a low pI.
Histones have high pI values because they have large numbers of
His, Arg, and Lys residues. Because the side-chains of these
residues are partially (His) or fully (Arg & Lys) positively charged
at neutral pH, histones bind strongly to the negatively charged
phosphate groups of DNA via ionic interactions.
Chap. 3. Problem 15.
Part (a). The specific activity (S.A.) of
an enzyme sample is defined as the
number of units of activity divided by
the total mg of protein in the sample
(U/mg). Using this formula, the S.A.
obtained after each step of the
purification listed in the table is Step
1) 200; 2) 600; 3) 250; 4) 4,000; 5)
15,000; and 6) 15,000 U/mg.
Part (b). The most effective step in
the purification is the one that results
in the greatest increase in S.A. of the
enzyme. This corresponds to Step 4
(ion-exchange chromatography) wherein
the S.A. increased by 16-fold
(4,000/250).
Part (c). By a similar logic to that used
in Part (b), Step 3 (pH precipitation) is
the least effective step of the
procedure. Two-thirds of the total
activity of the protein actually was lost
in this step.
Part (d). The enzyme is pure after
Step 6 based on the fact that the
S.A. did not increase between Steps 5
and 6. SDS PAGE could be used to
assess the final purity of the enzyme.
Chap. 3. Problem 18.
Part (a). The amino acid composition
of the peptide is obtained using this
procedure. Thus we can write the
empirical composition of the peptide
as (Gly2, Leu1, Phe1, Tyr1)n, where n
indicates that these residues
theoretically could be present in two
or more copies.
Part (b). FDNB is used to determine
the N-terminal amino acid of a
peptide or polypeptide. Since no
other Tyr residue was found following
complete hydrolysis of the FDNBlabeled peptide, this indicates that n
= 1. At this stage of the analyses,
the sequence of the peptide can be
written as Tyr-(Gly2, Leu1, Phe1),
where the last four residues could
occur in any order.
Part (c). The enzyme chymotrypsin cleaves polypeptides on the carboxyl
side of aromatic residues (Phe, Trp, and Tyr). Since three products (Tyr,
Leu, and the tripeptide [Phe1, Gly2]) were obtained by enzymatic cleavage,
the data indicate that Phe must be internal to the peptide sequence and
not at the C-terminus. Thus Leu must be at the C-terminus. The only
sequence consistent with the data therefore is Tyr-Gly-Gly-Phe-Leu.
Chap. 3. Problem 21.
The conventions used in writing
consensus sequences in proteins
are explained in Box 3-2. Also
refer to Table 3-1 for amino
acid symbols.
Part (a). The invariant residues in
this sequence are Tyr (Y) 1, Phe
(F) 7, and Arg (R) 9.
Part (b). Positively charged amino
acids must be present at
positions 4 and 9. Lys (K) is more
common at position 4, whereas
Arg (R) is invariant at position 9.
Part (c). Acidic amino acids with
negatively charged side-chains
must be present at positions 5
and 10. Glu (E) predominates at
both of these positions.
Part (d). As indicated by the x in
the sequence, position 2 can be
any amino acid. However, Ser
(S), appears most often there.
Chap. 3. Problem 22.
Part (a). Peptide 2 has the highest
proportion of negatively charged to
positively charged side-chains (7 to
1, out of 18 total residues). Thus,
it will migrate the slowest through
an anion exchange column which has
positively charged groups bound to
the resin.
Part (b). Peptide 1 has the highest
proportion of positively charged to
negatively charged side-chains (11
to 1, out of 44 total residues).
Thus, it will migrate the slowest
through an cation exchange column
which has negatively charged
groups bound to the resin.
Part (c). A gel filtration column
separates proteins on the basis of
size. Since peptide 2 is the
smallest, it will move through a gel
filtration column at the slowest
rate.
Part (d). Peptide 3 has the
sequence G-x(2)-G-x-G-K-S which
corresponds to the logo.