Chap 3 Problem 2 Fully protonated glycine has

Chap. 3. Problem 2. Fully protonated glycine has two dissociable protons, one on its carboxyl group (-COOH) and one on its -amino group (-NH 3+). The three forms of glycine that occur during its titration therefore are +H 3 N-CH 2 -COOH (Point I), +H 3 N-CH 2 -COO- (Point III), and H 2 N-CH 2 -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 (+H 3 N-CH 2 COOH) occurs at the beginning of the titration, i. e. , Point I. Part (b). 50% of the -COOH protons are titrated to -COO- at the p. K 1 of glycine (Point II, p. H 2. 34). At this point, half of the molecules have a net +1 charge (+H 3 N-CH 2 -COOH) and half have a 0 charge (+H 3 N-CH 2 -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 p. K 2 (Point IV, p. H 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 p. H = p. Ka. Thus, the p. H of the glycine solution is equal to the p. Ka 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 p. Hs 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, +H 3 N-CH 2 -COO-. This is Point III, which occurs at a p. H halfway between p. K 1 and p. K 2 p. I = (p. K 1 + p. K 2)/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 -NH 3+ group has been completely titrated, i. e. , Point V. Part (j). The species, +H 3 N-CH 2 -COO-, which has a net charge of zero occurs at the p. I. This species predominates at Point III, which has a p. H 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 H 2 N-CH 2 -COO-. Part (l). Glycine occurs with a 50/50 mixture of +H 3 N-CH 2 -COOH and +H 3 N-CH 2 -COO- when the -COOH group has been 50% titrated. This is at the midpoint of the first plateau in the titration curve, where p. H = p. K 1.

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 p. H regions for glycine to serve as a buffer occur at the points furthest from the two plateaus, p. H = p. K 1 and p. H = p. K 2. These regions occur at Points I, III, and V.

Part (a). The structures, equilibrium equations, and p. Kas for the three ionization reactions of histidine are shown in the diagram. e in each ionization state is (1) +2; (2) re the same as in Part a because each of histidine’s three dissociable protons. f the histidine species at each of these. Based on the net charge, histidine will Chap. 3. Problem 4.

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 k. Da polypeptides are present under both conditions, the data suggest that these two proteins are not attached to others by disulfide bonds. Because the 180 k. Da protein obtained in the first procedure disappears and a new band at 90 k. Da appears, it suggests that the 180 k. Da protein may actually consist of two 90 k. Da subunits. Thus the most likely subunit composition of this 400 k. Da protein is (160, 902, 60). The two 90 k. Da 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 p. I 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 p. I. Histones have high p. I 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 p. H, 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 (p. H 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 (Gly 2, Leu 1, Phe 1, Tyr 1)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-(Gly 2, Leu 1, Phe 1), 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 [Phe 1, Gly 2]) 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-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.