Chapter 5: Proteins: Primary Structure

Proteins: Primary Structure

Matching

• A) Electrophoresis: A technique used to separate molecules based on their size and charge by applying an electric field.
• B) Hydrophobic Interaction: The tendency of nonpolar substances to aggregate in aqueous solution to minimize contact with water.
• C) Enzyme-Linked Immunosorbent Assay (ELISA): A biochemical technique used mainly to detect the presence of an antibody or an antigen in a sample.
• D) Three-Dimensional Shape: The specific conformation a protein adopts, crucial for its function, determined by its primary structure and influenced by the environment.
• E) N-Terminal Amino Acid: The amino acid residue at the beginning of a polypeptide chain, possessing a free amino group.
• F) Negative Charge: The characteristic of a molecule with more electrons than protons, causing it to move towards the anode in electrophoresis.
• G) Nucleases: Enzymes that cleave the phosphodiester bonds between nucleotide subunits of nucleic acids.
• H) Chromophore: A molecule or part of a molecule that absorbs light, giving a substance its color.
• I) Foaming: The formation of a stable froth or foam on a liquid, often due to the presence of proteins that reduce surface tension.
• J) High Level Expression: The production of a large quantity of a specific protein, often achieved through recombinant DNA technology in host organisms like bacteria.
• K) 2-Mercaptoethanol: A reducing agent used to break disulfide bonds in proteins, commonly used in SDS-PAGE.
• L) Positive Charge: The characteristic of a molecule with fewer electrons than protons, causing it to move towards the cathode in electrophoresis.
• M) Cation Exchange: A type of ion exchange chromatography where positively charged ions (cations) are exchanged between a solution and a solid support.
• N) pI (Isoelectric Point): The pH at which a molecule carries no net electrical charge.
• O) Chymotrypsin: A proteolytic enzyme that cleaves peptide bonds on the C-terminal side of aromatic amino acids such as phenylalanine, tyrosine, and tryptophan.
• P) C-Terminal Amino Acid: The amino acid residue at the end of a polypeptide chain, possessing a free carboxyl group.
• Q) Sodium Dodecyl Sulfate (SDS): An anionic detergent used to denature proteins and impart a negative charge, essential for SDS-PAGE.

Fill-in-the-Blanks

1. One of the reasons the primary structure is important for a protein is that it determines the three-dimensional shape the molecule adopts in aqueous solutions.
   • Answer: D
   • Explanation: The amino acid sequence (primary structure) dictates how a protein folds into its functional 3D structure.
2. If the cDNA for a protein has been cloned, it may be possible to obtain large quantities of the protein by high level expression in bacteria.
   • Answer: J
   • Explanation: Cloning a protein's cDNA allows for its overexpression in a host organism like bacteria to produce a large quantity of the protein.
3. To help prevent denaturation of proteins in solution, steps are taken to avoid foaming and adsorption to surfaces.
   • Answer: I
   • Explanation: Foaming and adsorption to surfaces can lead to protein denaturation, so these conditions should be avoided.
4. Molecules that contain a(n) chromophore are capable of absorbing light.
   • Answer: H
   • Explanation: Chromophores are the light-absorbing parts of molecules, crucial for spectrophotometry and other optical techniques.
5. If antibodies to the protein being assayed are available, a(n) enzyme-linked immunosorbent assay can be carried out.
   • Answer: C
   • Explanation: ELISA is a powerful technique to detect and quantify a protein using specific antibodies.
6. In general, proteins are least soluble in water when the pH is close to the pI.
   • Answer: N
   • Explanation: At the isoelectric point (pI), a protein has no net charge, leading to minimum solubility and increased aggregation.
7. Hydrophobic interaction chromatography is a method of fractionating a protein mixture according to differences in polarity.
   • Answer: B
   • Explanation: Hydrophobic interaction chromatography separates proteins based on their hydrophobic properties.
8. In order for DEAE to act as an anion exchanger, it must have a positive charge.
   • Answer: L
   • Explanation: Anion exchangers are positively charged resins that bind negatively charged molecules.
9. In cation exchange chromatography, a protein mixture must be applied to the column at a low pH so that the proteins will have a net positive charge and bind to the column.
   • Answer: M
   • Explanation: Cation exchange chromatography uses negatively charged resins, so a low pH is used to protonate proteins, giving them a positive charge.
10. In SDS-PAGE, disulfide-linked polypeptides can be separated after reacting the protein first with 2-mercaptoethanol.
    • Answer: K
    • Explanation: 2-Mercaptoethanol is used to reduce and break disulfide bonds, allowing complete separation of polypeptide chains during SDS-PAGE.
11. Either dansyl chloride or Edman's reagent can be used to identify the N-terminal amino acid of a protein.
    • Answer: E
    • Explanation: Dansyl chloride and Edman's reagent are used to label and identify the N-terminal amino acid of a protein.
12. The endoprotease chymotrypsin cleaves polypeptides on the C-terminal side of certain bulky hydrophobic amino acid residues.
    • Answer: O
    • Explanation: Chymotrypsin is an endoprotease that preferentially cleaves peptide bonds adjacent to aromatic amino acid residues.

Multiple Choice

13. Proteins are synthesized in vivo by the translation of
    • A) cDNA.
    • B) tRNA.
    • C) rRNA.
    • D) exons.
    • E) mRNA.
    • Answer: E
    • Explanation: mRNA (messenger RNA) carries the genetic code from DNA to ribosomes for protein synthesis.
14. Since there are 20 standard amino acids, the number of possible linear polypeptides of length N can be expressed as:
    • A) n × 20
    • B) 20^n
    • C) 20 × 10^n
    • D) n^{20}
    • E) n × 10^{20}
    • Answer: B
    • Explanation: For each position in a polypeptide of length n, there are 20 possible amino acids, so the total number of possible polypeptides is 20^n.
15. Natural proteins most commonly contain linear polypeptides between 100 and 1000 residues in length. One of the reasons polypeptides outside this range may be disfavored is that
    • A) larger polypeptides would likely be insoluble.
    • B) smaller polypeptides do not form stable folded structures.
    • C) smaller polypeptides typically assemble into prion-like aggregates.
    • D) amide linkages are not strong enough to keep larger polypeptides intact.
    • E) ribosomes are unable to synthesize larger polypeptides.
    • Answer: B
    • Explanation: Polypeptides shorter than 100 amino acids often lack sufficient interactions to form stable, well-defined 3D structures.
16. The vast majority of polypeptides contain between amino acid residues.
    • A) 10 and 50
    • B) 50 and 100
    • C) 100 and 1000
    • D) 1000 and 2000
    • E) 2000 and 34,000
    • Answer: C
    • Explanation: Most functional proteins fall within this size range because they require a certain number of residues to fold correctly, but excessively large proteins can be prone to errors during synthesis.
17. Which of the following has the most dramatic influence on the characteristics of an individual protein?
    • A) the amino-acid sequence
    • B) the amino-acid composition
    • C) the location of its encoding gene within the genome
    • D) the stereochemistry at the α-carbon
    • E) the sequence of tRNA molecules involved in its translation
    • Answer: A
    • Explanation: The amino acid sequence (primary structure) directly determines the protein's folding, structure, and function.
18. Which statement about insulin is correct?
    • A) Insulin is composed of two polypeptides, the A chain and the B chain.
    • B) Insulin contains an intrachain disulfide bond.
    • C) Insulin contains interchain disulfide bonds.
    • D) The A chain and the B chain of insulin are encoded by a single gene.
    • E) All of the above are correct.
    • Answer: E
    • Explanation: Insulin consists of two polypeptide chains (A and B) linked by disulfide bonds and includes an intrachain disulfide bond within the A chain. Both chains are derived from a single precursor gene.
19. A fast and common method for determining the protein concentration in column effluent is
    • A) tandem mass spectrometry.
    • B) salting in with ammonium sulfate.
    • C) drying a portion and weighing the solid.
    • D) measuring light absorption at 280 nm.
    • E) Edman degradation.
    • Answer: D
    • Explanation: Aromatic amino acids in proteins absorb light at 280 nm, allowing for quick and easy protein concentration determination.
20. The salting in of proteins can be explained by:
    • A) salt counter-ions reducing electrostatic attractions between protein molecules.
    • B) salt ions reducing the polarity of the solution.
    • C) salt ions increasing the hydrophobic interactions.
    • D) releasing hydrophobic proteins from nonpolar tissue environments.
    • E) hydration of the salt ions reducing solubility of proteins.
    • Answer: A
    • Explanation: At low salt concentrations, salt ions can shield charges on proteins, reducing aggregation and increasing solubility.
21. The quantitation of proteins due to their absorbance at ~280 nm (UV region) is due to the large absorptivity of the amino acids.
    • A) anionic
    • B) dansylated
    • C) cleaved
    • D) polar
    • E) aromatic
    • Answer: E
    • Explanation: Tryptophan, tyrosine, and phenylalanine, which are aromatic amino acids, absorb UV light at ~280 nm due to their ring structures.
22. Which of the following ‘assays’ would be most specific for a particular protein?
    • A) Bradford assay
    • B) UV absorptivity
    • C) radioimmunoassay
    • D) molar absorptivity
    • E) amino acid analysis
    • Answer: C
    • Explanation: Radioimmunoassay (RIA) uses antibodies that specifically bind to the protein of interest, making it highly specific.
23. An enzyme-linked immunosorbent assay requires
    • A) a radioactive substrate.
    • B) a radioactive standard for binding to the antibody.
    • C) aromatic amino acids.
    • D) an antibody that binds the protein of interest.
    • E) a catalytic antibody.
    • Answer: D
    • Explanation: ELISA relies on specific antibody-antigen interactions, where the antibody binds to the protein of interest.
24. ELISA is an example of a(n):
    • A) enzyme assay.
    • B) biological assay.
    • C) binding assay.
    • D) immunological assay.
    • E) none of the above
    • Answer: D
    • Explanation: ELISA utilizes antibody-antigen interactions, making it an immunological assay to detect and quantify substances.
25. You are purifying a nuclease by affinity chromatography. To determine which fractions contain the protein of interest, you test samples of all fractions for their ability to break down DNA. This is an example of
    • A) a binding assay.
    • B) a biological assay.
    • C) an enzyme assay.
    • D) an immunological assay.
    • E) none of the above
    • Answer: C
    • Explanation: Testing the ability to break down DNA is a direct assessment of the nuclease enzyme's activity.
26. A radioimmunoassay requires
    • A) an enzyme-linked antibody.
    • B) a coupled enzymatic reaction.
    • C) a radiolabeled antibody.
    • D) a catalytic antibody.
    • E) a radiolabeled standard protein that is used to compete for binding to the antibody.
    • Answer: E
    • Explanation: RIA uses a radiolabeled antigen/protein that competes with the unlabeled antigen in the sample for binding to an antibody.
27. Five graduate students prepare extracts from 5 different tissues. Each student measures the total amount of alcohol dehydrogenase and the total amount of protein in his or her extract. Which extract has the highest specific activity?
    • Total protein (mg) Total alcohol dehydrogenase activity (units)
    • A 300 60,000
    • B 200 80,000
    • C 3000 96,000
    • D 5000 100,000
    • E 1000 200,000
    • Answer: B
    • Explanation: Specific activity is total activity divided by total protein. B has the highest specific activity (80,000/200 = 400 units/mg), compared with the other options.
28. Which physical characteristic is not commonly used in protein separation?
    • A) solubility
    • B) stereochemistry
    • C) size
    • D) charge
    • E) polarity
    • Answer: B
    • Explanation: Chromatography techniques usually separate proteins based on size, charge, polarity or solubility; stereochemistry isn't usually a primary factor.
29. Adding additional salt to a protein solution can cause:
    • A) an increase in solubility called ‘salting in’.
    • B) a decrease in solubility called ‘salting out’.
    • C) protein precipitation from solution.
    • D) all of the above
    • E) none of the above
    • Answer: D
    • Explanation: Adding salt can initially increase solubility (salting in) but at higher concentrations, it decreases solubility, causing precipitation (salting out).
30. A first step in purifying a protein that was initially associated with fatty substances would be
    • A) Coomassie Brilliant Blue dye staining.
    • B) analytical ultracentrifugation.
    • C) ELISA.
    • D) Western blotting.
    • E) hydrophobic interaction chromatography.
    • Answer: E
    • Explanation: Hydrophobic interaction chromatography exploits the affinity of the protein for non-polar substances, effectively separating it from the fatty substances.
31. The acronym HPLC stands for
    • A) hydrophobic protein liquid chromatography.
    • B) high performance liquid chromatography.
    • C) hydrophilic partition liquid chromatography.
    • D) high priced liquid chromatography.
    • E) hydrostatic process liquid chromatography.
    • Answer: B
    • Explanation: HPLC (High Performance Liquid Chromatography) is a separation technique to identify, quantify, and purify the individual components of a mixture.
32. A technique that can be used to separate proteins based primarily on the presence of non-polar residues on their surface is called
    • A) ion-exchange chromatography.
    • B) gel filtration chromatography.
    • C) affinity chromatography.
    • D) gel electrophoresis.
    • E) hydrophobic interaction chromatography.
    • Answer: E
    • Explanation: Hydrophobic interaction chromatography is used to separate proteins based on their hydrophobic properties by using a column with hydrophobic ligands.
33. A technique that can be used to separate proteins based primarily on their pI is called
    • A) ion-exchange chromatography.
    • B) gel filtration chromatography.
    • C) affinity chromatography.
    • D) isoelectric focusing.
    • E) hydrophobic interaction chromatography.
    • Answer: D
    • Explanation: Isoelectric focusing separates proteins based on their isoelectric points (pI) in a pH gradient under an electric field.
34. Which of the following amino acids would be last to elute at pH 8.0 from an anion-exchange column?
    • A) lysine
    • B) alanine
    • C) glutamic acid
    • D) asparagine
    • E) glycine
    • Answer: C
    • Explanation: Anion-exchange columns bind negatively charged molecules. At pH 8.0, glutamic acid is most negatively charged, resulting in the strongest binding and latest elution.
35. Which of the following amino acids would be first to elute at pH 8.0 from an anion-exchange column?
    • A) lysine
    • B) alanine
    • C) glutamic acid
    • D) asparagine
    • E) glycine
    • Answer: A
    • Explanation: At pH 8, lysine is positively charged, so it will not bind to the anion exchange column and will elute first.
36. The pK1, pK2, and pKR of the amino acid lysine are 2.2, 9.1, and 10.5, respectively. The pK1, pK2, and pKR of the amino acid arginine are 1.8, 9.0, and 12.5, respectively. A student at SDSU wants to use ion exchange chromatography to separate lysine from arginine. What pH is likely to work best for this separation?
    • A) 1.5
    • B) 2.5
    • C) 5.5
    • D) 7.5
    • E) 10.5
    • Answer: E
    • Explanation: At pH 10.5, lysine will be partially deprotonated but still have a +1 charge, while arginine will still retain a +1 charge because the pH is below its pKR. This small difference will allow for better separation.
37. The pK1, pK2, and pKR of the amino acid histidine are 1.8, 9.3, and 6.0, respectively. The pK1, pK2, and pKR of the amino acid arginine are 1.8, 9.0, and 12.5, respectively. You have a mixture of histidine and arginine, how would you try to separate these two amino acids?
    • A) anion exchange chromatography at pH 2
    • B) anion exchange chromatography at pH 4
    • C) cation exchange chromatography at pH 2
    • D) cation exchange chromatography at pH 4
    • E) cation exchange chromatography at pH 9
    • Answer: E
    • Explanation: At pH 9, histidine would be mostly neutral (close to its pKR + pK2 / 2), while arginine would still be positively charged. Using cation exchange chromatography at this pH will lead to separation.
38. What can be done to increase the rate at which a protein of interest moves down an ion-exchange chromatography column?
    • A) reduce the ion concentration in the eluant
    • B) add a small amount of a non-ionic detergents to the eluant
    • C) change the pH of the eluant
    • D) add a protease inhibitor to the eluant
    • E) reduce the temperature of the eluant
    • Answer: C
    • Explanation: Changing the pH can alter the charge of the protein and column, reducing interaction or reversing it entirely, hence increasing the elution rate.
39. Hydrophobic interaction chromatography can be used to separate proteins based on differences in
    • A) ionic charge.
    • B) solubility.
    • C) size.
    • D) polarity.
    • E) binding specificity.
    • Answer: D
    • Explanation: Hydrophobic interaction chromatography uses a hydrophobic stationary phase to bind less polar substances and separates them from more polar substances.
40. You are trying to separate five proteins, which are listed below, by gel filtration chromatography. Which of the proteins will elute first from the column?
    • A) cytochrome c (12 kDa)
    • B) RNA polymerase (99 kDa)
    • C) glutamine synthetase (621 kDa)
    • D) interferon-g (34 kDa)
    • E) hemoglobin (62 kDa)
    • Answer: C
    • Explanation: Gel filtration separates proteins based on size; larger proteins elute first because they can't enter pores in the column matrix, so they take a shorter path.
41. SDS-PAGE separates proteins primarily due to differences in
    • A) isoelectric point.
    • B) mass.
    • C) polarity.
    • D) solubility.
    • E) amino acid sequence.
    • Answer: B
    • Explanation: SDS-PAGE (sodium dodecyl sulfate polyacrylamide gel electrophoresis) separates proteins based on their size/mass after denaturation and coating with SDS.
42. Which of these techniques is used to separate proteins mainly based on mass?
    • A) polyacrylamide gel electrophoresis (in the absence of SDS)
    • B) SDS-PAGE
    • C) isoelectric focusing
    • D) immunoblotting
    • E) Western blotting
    • Answer: B
    • Explanation: SDS-PAGE separates proteins based on mass because SDS unfolds and coats all proteins with a uniform negative charge, so their migration is primarily determined by size.
43. Which of these techniques uses antibodies to detect very small amounts of specific proteins following separation by SDS-PAGE.
    • A) immunoblotting
    • B) silverstaining
    • C) Coomassie Brilliant Blue staining
    • D) ELISA
    • E) RIA
    • Answer: A
    • Explanation: Immunoblotting (Western blotting) uses antibodies to detect specific proteins after separation by SDS-PAGE, providing high specificity and sensitivity.
44. Disulfide bonds can be cleaved using
    • A) iodoacetate.
    • B) dansyl chloride.
    • C) 2-mercaptoethanol (b-ME).
    • D) trypsin.
    • E) phenylisothiocyanate.
    • Answer: C
    • Explanation: 2-Mercaptoethanol (β-ME) is a reducing agent that breaks disulfide bonds by reducing the disulfide bridges.
45. Which of these reagents is commonly used to determine the number of polypeptides in a protein?
    • A) iodoacetate
    • B) dansyl chloride
    • C) 2-mercaptoethanol (b-ME)
    • D) cyanogen bromide
    • E) DEAE
    • Answer: B
    • Explanation: Dansyl chloride labels the N-terminal amino acid of each polypeptide chain, which assists in determining the number of polypeptide chains present in the protein.
46. Enzymes that hydrolyze the internal peptide bonds (not the peptide bonds of the terminal amino acids) of a protein are classified as
    • A) oxidoreductases.
    • B) lyases.
    • C) endopeptidases.
    • D) nucleases.
    • E) exopeptidases.
    • Answer: C
    • Explanation: Endopeptidases cleave peptide bonds within the polypeptide chain, not at the terminal amino acids.
47. Which of the following substances cannot be used to cleave peptide bonds in polypeptides?
    • A) trypsin
    • B) cyanogen bromide
    • C) endopeptidases
    • D) 2-mercaptoethanol
    • E) pepsin
    • Answer: D
    • Explanation: 2-Mercaptoethanol breaks disulfide bonds, not peptide bonds; trypsin, cyanogen bromide, endopeptidases, and pepsin cleave peptide bonds.
48. Which of these are commonly used to cleave peptide bonds in polypeptides?
    • A) 2-mercaptoethanol (b-ME)
    • B) dansyl chloride
    • C) iodoacetate
    • D) sodium dodecyl sulfate
    • E) trypsin
    • Answer: E
    • Explanation: Trypsin is a protease that cleaves peptide bonds, while the other options modify proteins or disrupt their structure but do not cleave peptide bonds.
49. The peptide Leu─Cys─Arg─Ser─Gln─Met is subjected to Edman degradation. In the first cycle the peptide first reacts with phenylisothiocyanate under basic conditions. The product of this reaction is incubated with anhydrous trifluoroacetic acid and subsequently with an aqueous acid. What are the products generated in the first cycle.
    • A) PTH─Leu, PTH─Cys, PTH─Arg, PTH─Ser, PTH─Gln, and PTH─Met
    • B) PTH─Leu─Cys─Arg─Ser─Gln─Met
    • C) PTH─Met and Leu─Cys─Arg─Ser─Gln─Met
    • D) PTH─Leu─Cys and PTH─Arg─Ser─Gln─Met
    • E) PTH─Leu and Cys─Arg─Ser─Gln─Met
    • Answer: E
    • Explanation: Edman degradation removes one amino acid at a time from the N-terminus, so the products are PTH-Leu and the remaining peptide sequence.
50. Edman degradation can be used to
    • A) identify the N-terminal amino acid of a polypeptide.
    • B) identify the C-terminal amino acid of a polypeptide.
    • C) separate the subunits of a multi-subunit protein.
    • D) cleave a protein at specific sites.
    • E) cleave disulfide bonds within a protein so that the individual polypeptides can be separated.
    • Answer: A
    • Explanation: Edman degradation is a method for sequencing peptides by removing and identifying one amino acid at a time from the N-terminus.
51. Although a protein’s primary sequence can be inferred from the nucleotide sequence, modifications such as can be determined most easily by tandem mass spectrometry followed by protein database searching.
    • A) phosphorylation
    • B) disulfide crosslinks
    • C) glycosylation
    • D) acetylation
    • E) all of the above
    • Answer: E
    • Explanation: Tandem mass spectrometry is highly effective for identifying post-translational modifications, including phosphorylation, disulfide crosslinks, glycosylation and acetylation.
52. The positive charge on proteins in electrospray ionization mass spectrometry is the result of
    • A) protons fired at the gas-phase protein molecules.
    • B) protonated side chains of Asp and Glu residues.
    • C) protonated side chains of Arg and Lys residues.
    • D) a high pH.
    • E) electrons fired at the gas-phase protein molecules.
    • Answer: C
    • Explanation: Electrospray ionization (ESI) involves adding protons to basic residues, such as arginine and lysine, giving the protein a positive charge.
53. has emerged as a technique for protein sequencing.
    • A) NMR spectroscopy
    • B) Mass spectrometry
    • C) Gel electrophoresis
    • D) Phylogenetic analysis
    • E) Limited proteolysis
    • Answer: B
    • Explanation: Mass spectrometry has become a dominant technique for determining protein sequences due to its speed, sensitivity, and accuracy.
54. Protein sequences are customarily ‘reconstructed’ from sequenced fragments because
    • A) protein purification invariably results in the fragmentation of the protein of interest.
    • B) proteins are naturally and inevitably cleaved by proteolytic enzymes.
    • C) proteins are composed of multiple subunits.
    • D) large polypeptides cannot be directly sequenced.
    • E) all of the above
    • Answer: D
    • Explanation: Large polypeptides are typically digested into smaller fragments for sequencing, and the sequence is reconstructed based on overlapping fragments.
55. You have purified a new peptide hormone. To determine its amino acid sequence you have digested the polypeptide with trypsin and in a separate reaction you have cleaved the polypeptide with cyanogen bromide.
    • Cleavage with trypsin yielded 5 peptides that were sequenced by Edman degradation as shown in the following.
    • Ser─Leu
    • Asp─Val─Arg
    • Val─Met─Glu─Lys
    • Ser─Gln─Met─His─Lys
    • Ile─Phe─Met─Leu─Cys─Arg
    • Cleavage with cyanogen bromide yielded 4 peptides that were sequenced by Edman degradation:
    • His─Lys─Ser─Leu
    • Asp─Val─Arg─Val─Met
    • Glu─Lys─Ile─Phe─Met
    • Leu─Cys─Arg─Ser─Gln─Met
    • Determine the identity of the N-terminal amino acid after reconstructing the intact protein.
    • A) Asp
    • B) Ser
    • C) His
    • D) Glu
    • E) Ile
    • Answer: A
    • Explanation: Comparing the sequences, it's clear that "Asp-Val-Arg" is the N-terminal end because it is at the beginning of the cyanogen bromide fragment,