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Ch. 3 Amino Acids, Peptides, & Proteins (Essay Questions)
Terms in this set (39)
What are the structural characteristics common to all amino acids found in naturally occurring proteins?
All amino acids found in naturally occurring proteins have an a carbon to which are attached a carboxylic acid, an amine, a hydrogen, and a variable side chain. All the amino acids are also in the l configuration.
Only one of the common amino acids has no free a-amino group. Name this amino acid and draw its structure.
The amino acid l-proline has no free a-amino group, but rather has an imino group formed by cyclization of the R-group aliphatic chain with the amino group (see Fig. 3-5, p. 79).
Briefly describe the five major groupings of amino acids.
Amino acids may be categorized by the chemistry of their R groups: (1) nonpolar aliphatics; (2) polar, uncharged; (3) aromatic; (4) positively charged; (5) negatively charged. (See Fig. 3-5, p. 79.)
A B C D E
Tyr-Lys-Met Gly-Pro-Arg Asp-Trp-Tyr Asp-His-Glu Leu-Val-Phe
Which one of the above tripeptides:
____(a) is most negatively charged at pH 7?
____(b) will yield DNP-tyrosine when reacted with l-fluoro-2,4-dinitrobenzene and hydrolyzed in acid?
____(c) contains the largest number of nonpolar R groups?
____(d) contains sulfur?
____(e) will have the greatest light absorbance at 280 nm?
(a) D; (b) A; (c) E; (d) A; (e) C
Draw the structures of the amino acids phenylalanine and aspartate in the ionization state you would expect at pH 7.0. Why is aspartate very soluble in water, whereas phenylalanine is much less soluble?
Aspartate has a polar (hydrophilic) side chain, which forms hydrogen bonds with water. In contrast, phenylalanine has a nonpolar (hydrophobic) side chain. (See Fig. 3-5, p. 79 for structures.)
Name two uncommon amino acids that occur in proteins. By what route do they get into proteins?
Some examples are 4-hydroxyproline, 5-hydroxylysine, g-carboxyglutamate, N-methyllysine, desmosine, and selenocysteine. Uncommon amino acids in proteins (other than selenocysteine) usually result from chemical modifications of standard amino acid R groups after a protein has been synthesized.
Why do amino acids, when dissolved in water, become zwitterions?
Near pH = 7, the carboxylic acid group (—COOH) will dissociate to become a negatively charged —COO- group, and the —NH2 amino group will attract a proton to become a positively charged —NH3+ group.
As more OH- equivalents (base) are added to an amino acid solution, what titration reaction will occur around pH = 9.5?
Ans: Around pH = 9.5, the —NH3+ group will be titrated according to the reaction: —NH3+ + OH- ® —NH2 + H2O.
In the amino acid glycine, what effect does the positively charged —NH3+ group have on the pKa of an amino acid's —COOH group?
Ans: The positively charged amino group stabilizes the negatively charged ionized form of the carboxyl group, —COO-, and repels the departing H+ thereby promoting deprotonation. The effect is to lower the pKa of the carboxyl group. (See Fig. 3-11, p. 80.)
How does the shape of a titration curve confirm the fact that the pH region of greatest buffering power for an amino acid solution is around its pK's?
Ans: In a certain range around the pKa's of an amino acid, the titration curve levels off. This indicates that for a solution with pH » pK, any given addition of base or acid equivalents will result in the smallest change in pH—which is the definition of a buffer.
Leucine has two dissociable protons: one with a pKa of 2.3, the other with a pKa of 9.7. Sketch a properly labeled titration curve for leucine titrated with NaOH; indicate where the pH = pK and the region(s) in which buffering occurs.
Ans: See the titration curve for glycine in Fig. 3-10, p. 79.
What is the pI, and how is it determined for amino acids that have nonionizable R groups?
Ans: The pI is the isoelectric point. It occurs at a characteristic pH when a molecule has an equal number of positive and negative charges, or no net charge. For amino acids with nonionizable R groups, pI is the arithmetic mean of a molecule's two pKa values:
pI = 1/2 (pK1 + pK2)
The amino acid histidine has a side chain for which the pKa is 6.0. Calculate what fraction of the histidine side chains will carry a positive charge at pH 5.4. Be sure to show your work.
pH = pKa + log
pKa - pH = log
antilog (pKa - pH) =
antilog (6.0 - 5.4) =
4 = [acid]/[conjugate base], or
4[conjugate base] = [acid]
Therefore, at pH 5.4, 4/5 (80%) of the histidine will be in the protonated form.
The amino acid histidine has three ionizable groups, with pKa values of 1.8, 6.0, and 9.2. (a) Which pKa corresponds to the histidine side chain? (b) In a solution at pH 5.4, what percentage of the histidine side chains will carry a positive charge?
Ans: (a) 6.0; (b) 80%. (See the previous problem for expanded solution to this problem.)
What is the uniquely important acid-base characteristic of the histidine R group?
Ans: Only the imidazole ring of the histidine R group has a pKa near physiological pH (pKa = 6.0), which suggests that histidine may provide buffering power in intercellular and intracellular fluids.
How can a polypeptide have only one free amino group and one free carboxyl group?
Ans: This is possible only if the peptide has no side chains with carboxyl or amino groups. Then, with the exception of the single amino-terminal amino acid and the single carboxyl-terminal amino acid, all the other a-amino and carboxyl groups are covalently condensed into peptide bonds.
Hydrolysis of peptide bonds is an exergonic reaction. Why, then, are peptide bonds quite stable?
Ans: Peptide bonds are stable because hydrolysis of peptide (or amide) bonds has a high activation energy and as a result occurs very slowly.
Draw the structure of Gly-Ala-Glu in the ionic form that predominates at pH 7.
Ans: The peptide must have an amino-terminal Gly residue, a carboxyl-terminal Glu residue, and ionized amino and carboxyl groups.
The artificial sweetener NutraSweet®, also called aspartame, is a simple dipeptide, aspartylphenylalanine methyl ester, on which the free carboxyl of the dipeptide is esterified to methyl alcohol. Draw the structure of aspartame, showing the ionizable groups in the form they have at pH 7. (The ionizable group in the side chain of aspartate has a pKa of 3.96.)
Ans: See the structure on p. 83.
If the average molecular weight of the 20 standard amino acids is 138, why do biochemists divide a protein's molecular weight by 110 to estimate its number of amino acid residues?
Ans: For each peptide bond formed, a molecule of water is lost, bringing the average molecular weight down to 120. To reflect the preponderance of low-molecular-weight amino acids, the average molecular weight is lowered further to 110.
Lys residues make up 10.5% of the weight of ribonuclease. The ribonuclease molecule contains 10 Lys residues. Calculate the molecular weight of ribonuclease.
Ans: From the structure of lysine, we can calculate its molecular weight (146); when it condenses (loses H2O, Mr = 18) to form a peptide bond, the resulting residue contributes 146 - 18 = 128 to the protein's molecular weight. If 10 Lys residues contribute 10.5% of the protein's molecular weight, each Lys residue is 1.05%. To calculate the total molecular weight, divide 128 by 1.05% (0.0105); the result is 12,190. (The actual value is 13,700.)
Why do smaller molecules elute after large molecules when a mixture of proteins is passed through a size-exclusion (gel filtration) column?
Ans: The column matrix is composed of cross-linked polymers with pores of selected sizes. Smaller molecules can enter pores in the polymer beads from which larger molecules would be excluded. Smaller molecules therefore have a larger three-dimensional space in which to diffuse, making their path through the column longer. Larger molecules migrate faster because they pass directly through the column, unhindered by the bead pores.
For each of these methods of separating proteins, describe the principle of the method, and tell what property of proteins allows their separation by this technique.
(a) ion-exchange chromatography
(b) size-exclusion (gel filtration) chromatography
(c) affinity chromatography
Ans: (a) Ion-exchange chromatography separates proteins on the basis of their charges. (b) Size-exclusion or gel filtration chromatography separates on the basis of size. (c) Affinity chromatography separates proteins with specific, high affinity for some ligand (attached to an inert support) from other proteins with no such affinity. (See Fig. 3-17, p. 87.)
A biochemist is attempting to separate a DNA-binding protein (protein X) from other proteins in a solution. Only three other proteins (A, B, and C) are present. The proteins have the following properties:
(isoelectric Size Bind to
point) Mr DNA?
protein A 7.4 82,000 yes
protein B 3.8 21,500 yes
protein C 7.9 23,000 no
protein X 7.8 22,000 yes
What type of protein separation techniques might she use to separate:
(a) protein X from protein A?
(b) protein X from protein B?
(c) protein X from protein C?
Ans: (a) Size-exclusion (gel filtration) chromatography to separate on the basis of size; (b) ion-exchange chromatography or isoelectric focusing to separate on the basis of charge; (c) specific affinity chromatography, using immobilized DNA.
What factors would make it difficult to interpret the results of a gel electrophoresis of proteins in the absence of sodium dodecyl sulfate (SDS)?
Ans: Without SDS, protein migration through a gel would be influenced by the protein's intrinsic net charge—which could be positive or negative—and its unique three-dimensional shape, in addition to its molecular weight. Thus, it would be difficult to ascertain the difference between proteins based upon a comparison of their mobilities in gel electrophoresis.
How can isoelectric focusing be used in conjunction with SDS gel electrophoresis?
Ans: Isoelectric focusing can separate proteins of the same molecular weight on the basis of differing isoelectric points. SDS gel electrophoresis can then separate proteins with the same isoelectric points on the basis of differing molecular weights. When combined in two-dimensional electrophoresis, a great resolution of large numbers of proteins can be achieved.
You are given a solution containing an enzyme that converts B into A. Describe what you would do to determine the specific activity of this enzyme solution.
Ans: First, add a known volume of the enzyme solution (say, 0.01 mL) to a solution of its substrate B and measure the initial rate at which product A is formed, expressed as mmol/mL of enzyme solution/min. Then measure the total protein concentration, expressed as mg/mL. Finally, divide the enzyme activity (mmol/min/mL) by the protein concentration (mg/mL); the quotient is the specific activity.
As a protein is purified, both the amount of total protein and the activity of the purified protein decrease. Why, then, does the specific activity of the purified protein increase?
Ans: Specific activity is the units of enzyme activity (mmol of product/min) divided by the amount of protein (mg). As the protein is purified, some of it is lost in each step, resulting in a drop in activity. However, other contaminating proteins are lost to a much greater extent. Therefore, with each purification step, the purified protein constitutes a greater proportion of the total, resulting in an increase in specific activity. (See also Table 3-5, p. 88.)
Define the primary structure of a protein.
Ans: The primary structure of a protein is its unique sequence of amino acids and any disulfide bridges present in the native structure, that is, its covalent bond structure.
In one or two sentences, describe the usefulness of each of the following reagents or reactions in the analysis of protein structure:
(a) Edman reagent (phenylisothiocyanate)
(c) reducing agent (dithiothreitol or b-mercaptoethanol)
Ans: (a) Used in determination of the amino acid sequence of a peptide, starting at its amino terminus; (b) used to produce specific peptide fragments from a polypeptide; (c) used to break disulfide bonds or "bridges" or to keep them from forming and to keep Cys residues in their reduced form.
A polypeptide is hydrolyzed, and it is determined that there are 3 Lys residues and 2 Arg residues (as well as other residues). How many peptide fragments can be expected when the native polypeptide is incubated with the proteolytic enzyme trypsin?
Ans: Six fragments would be expected, unless the carboxyl-terminal residue is Lys or Arg; in which case there would be five.
The following reagents are often used in protein chemistry. Match the reagent with the purpose for which it is best suited. Some answers may be used more than once or not at all; more than one reagent may be suitable for a given purpose.
(a) CNBr (cyanogen bromide) (e) performic acid
(b) Edman reagent (phenylisothiocyanate) (f) chymotrypsin
(c) Sanger reagent (FDNB) (g) trypsin
(d) reducing agent (dithiothreitol) (h) iodoacetamide
___ hydrolysis of peptide bonds on the carboxyl side of Lys and Arg
___ cleavage of peptide bonds on the carboxyl side of Met
___ breakage of disulfide (—S—S—) bonds
___ modification of sulfhydryl groups of Cys
___ determination of the amino acid sequence of a peptide
___ determining the amino-terminal amino acid in a polypeptide
Ans: g; a; d and e; h; b; c
Conjugated proteins contain chemical substituents in addition to amino acids. List three classes of conjugated proteins and identify the type of prosthetic group associated with each one.
Ans: Any of the following are acceptable answers:
Lipoproteins, with lipid groups
Glycoproteins, with carbohydrate groups
Phosphoproteins, with phosphoryl groups
Hemoproteins, with heme groups
Flavoproteins, with flavin nucleotide groups
Metalloproteins, with metal ions (zinc, iron, calcium, etc.)
Provide a brief definition for a polymorphic protein.
Ans: A polymorphic protein is one whose amino acid sequence varies among the human population. The variants have little or no differences in the function or activity of the protein.
A biochemist wishes to determine the sequence of a protein that contains 123 amino acid residues. After breaking all of the disulfide bonds, the protein is treated with cyanogen bromide (CNBr), and it is determined that that this treatment breaks up the protein into seven conveniently sized peptides, which are separated from each other. It is your turn to take over. Outline the steps you would take to determine, unambiguously, the sequence of amino acid residues in the original protein.
Ans: (1) Use Edman degradation to determine the sequence of each peptide. (2) Create a second set of peptides by treatment of the protein with a specific protease (e.g., trypsin), and determine the sequence of each of these. (3) Place the peptides in order by their overlaps. (4) Finally, by a similar analysis of the original protein without first breaking disulfide bonds, determine the number and location of —S—S— bridges.
You are trying to determine the sequence of a protein that you know is pure. Give the most likely explanation for each of the following experimental observations. You may use a simple diagram for your answer.
(a) The Sanger reagent (FDNB, fluorodinitrobenzene) identifies Ala and Leu as amino-terminal residues, in roughly equal amounts.
(b) Your protein has an apparent Mr of 80,000, as determined by SDS-polyacrylamide gel electrophoresis. After treatment of the protein with performic acid, the same technique reveals two proteins of Mr 35,000 and 45,000.
(c) Size-exclusion chromatography (gel filtration) experiments indicate the native protein has an apparent Mr of 160,000.
Ans: (a) The protein has some multiple of two subunits, with Ala and Leu as the amino-terminal residues. (b) The protein has two subunits (Mr 35,000 and 45,000), joined by one or more disulfide bonds. (c) The native protein (Mr 160,000) has two Mr 35,000 subunits and two Mr 40,000 subunits.
Describe two major differences between chemical synthesis of polypeptides and synthesis of polypeptides in the living cell.
Ans: There are many such differences; here are a few. (1) Chemical synthesis proceeds from carboxyl terminus to amino terminus; in the living cell, the process starts at the amino terminus and ends at the carboxyl terminus. In the living cell, synthesis occurs under physiological conditions; chemical synthesis does not. Chemical synthesis is only capable of synthesizing short polypeptides; cells can produce proteins of several thousand amino acids.
Distinguish between homologs, paralogs, and orthologs as classes of related proteins.
Ans: Homologs are any members of a particular protein family, paralogs are two homologs present in the same species, and orthologs are are two homologs present in different species.
How are "signature sequences" useful in analyzing groups of functionally related proteins?
Ans: Such sequences are often present in one taxonomic group or shared by closely related taxonomic groups but are absent in evolutionarily more distant groups. They thus aid in constructing more elaborate evolutionary trees based on protein sequences.
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