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Name: __________________________ Date: _____________ NUID: Questions from the first lecture: 1. How can just a few elements give rise to all biological diversity? At what level, if any, are all biological organisms similar? Given this biochemical similarity, how is the structural and functional diversity of living things possible? Ans: Living things are composed primarily of macromolecules, polymers of simple compounds of just a few different types. The properties of these polymers are determined by their sequence of monomers and these can be combined in many different ways. Diversity is thus achieved through the nearly limitless variety of sequences that can exist when amino acids are linked to form proteins, nucleotides are linked to form nucleic acids, and monosaccharides are linked to form polysaccharides. Branching in the latter can contribute additional heterogeneity. Each type of organism constructs a unique set of macromolecules from these monomeric units, resulting in the structural and functional diversity among species. 2. What are five periodic elements most frequently seen incorporated into compounds of biological organisms? Name three occasionally seen elements too. Ans: Hydrogen, Oxygen, Nitrogen, Carbon, Phosphorus and Sulfur are the most frequently seen elements. Chloride, Sodium, Potassium, and Calcium are frequently seen, but not incorporated into compounds. The three elements could include iron, selenium, magnesium, vanadium, chromium, manganese, molybdenum, cobalt, nickel, copper, or zinc. Page 1 3. Draw the structures of the following functional groups in their un-ionized forms: (a) hydroxyl, (b) carboxyl, (c) amino, (d) phosphoryl. Ans: 4. (a) List the types of noncovalent interactions that are important in providing stability to the three-dimensional structures of macromolecules. (b) Why is it important that these interactions be noncovalent, rather than covalent, bonds? Ans: (a) Noncovalent interactions include hydrogen bonds, ionic interactions between charged groups, van der Waals interactions, and hydrophobic interactions. (b) Because noncovalent interactions are weak, they can form, break, and re-form more rapidly and with less energy input than can covalent bonds. This is important to maintain the flexibility needed in macromolecules. 5. Why is an asymmetric carbon atom called a chiral center? Ans: An asymmetric carbon has four different substituents attached, and cannot be superimposed on its mirror image—as a right hand cannot fit into a left glove. Thus, a molecule with one chiral carbon will have two stereoisomers, which may be distinguishable from one another in a biological system. Page 2 6. A chemist working in a pharmaceutical lab synthesized a new drug as a racemic mixture. Why is it important that she separate the two enantiomers and test each for its biological activity? Ans: Biomolecules such as receptors for drugs are stereospecific, so each of the two enantiomers of the drug may have very different effects on an organism. One may be beneficial, the other toxic; or one enantiomer may be ineffective and its presence could reduce the efficacy of the other enantiomer. 7. How is the genetic information encoded in DNA and how is a new copy of DNA synthesized? Ans: The genetic information is encoded in the linear sequence (order) of the four different deoxyribonucleotides in the DNA. When a new copy of DNA is needed, the two strands of the DNA unwind and each strand serves as a template on which a new strand is synthesized. 8. Name two functions of (a) proteins, (b) nucleic acids, (c) polysaccharides, (d) lipids. Ans: Many answers are possible including: (a) proteins function as enzymes, structural elements, signal carriers, transporters; (b) nucleic acids store and transmit genetic information and act as both structural and catalytic elements; (c) polysaccharides serve as energy-yielding fuel stores and cellular and extracellular structural and recognition elements; (d) lipids function as membrane components, fuel stores, and cellular signals. Page 3 Questions from the second lecture: 9. How does the electronegativity of each atom affect the polarity of a bond? Use electronegativity to explain why water is a good solvent. Ans: Electronegativity describes the tendency of an atom to attract electrons / electron density towards itself. The electronegativity of hydrogen is 2.1, while that of oxygen is 3.5. Thus, electron density is more towards the oxygen, polarizing the bond between hydrogen and oxygen. This makes water a good solvent, especially for ions, because the negative and positive dipoles of the water can partially satisfy the charges on the ions. 10. Explain the fact that ethanol (CH3CH2OH) is more soluble in water than is ethane (CH3CH3). Ans: Ethanol can form hydrogen bonds with water molecules, but ethane cannot. When ethanol dissolves, the decrease in the system's entropy that results from formation of ordered arrays of water around the CH3CH2– group is partly compensated by the favorable interactions (hydrogen bonds) of the hydroxyl group of ethanol with water molecules. Ethane cannot form such hydrogen bonds. 11. Describe how van der Waal's interactions work. What types of molecules can participate in van der Waal's interactions? Ans: Van der Waal's interactions work because of mutually attractive induced dipoles. They are highly dependent on the distance between two participating atoms. They are very weak, but any atoms can participate in them. Page 4 12. Phosphoric acid (H3PO4) has three dissociable protons, with the pKa's shown below. Which form of phosphoric acid predominates in a solution at pH 4? Explain your answer. pKa Acid H3PO4 2.14 H2PO4– 6.86 2– HPO4 12.4 Ans: At pH 4, the first dissociable proton (pKa = 2.14) has been titrated completely, and the second (pKa = 6.86) has just started to be titrated. The dominant form at pH 4 is therefore H2PO4–, the form with one dissociated proton (see Fig. 2-15). 13. Define pKa for a weak acid in the following two ways: (1) in relation to its acid dissociation constant, Ka, and (2) by reference to a titration curve for the weak acid. Ans: (1) pKa = –log Ka. (2) See Fig. 2-17, p. 59; pKa is the value of pH at the inflection point in a plot of pH vs. extent of titration of the weak acid. At the pKa, the concentration of ionized acid equals the concentration of un-ionized acid. 14. Give the general Henderson-Hasselbalch equation and sketch the plot it describes (pH against amount of NaOH added to a weak acid). On your curve, label the pKa for the weak acid and indicate the region in which the buffering capacity of the system is greatest. Ans: The inflection point, which occurs when the weak acid has been exactly one half titrated with NaOH, occurs at a pH equal to the pKa of the weak acid. The region of greatest buffering capacity (where the titration curve is flattest) occurs at pH values of pKa ±1. (See Fig. 2-17, p. 59.) Page 5 15. You have just made a solution by combining 50 mL of a 0.1 M sodium acetate solution with 150 mL of 1 M acetic acid (pKa = 4.7). What your solution's pH? Ans: pH = pKa + log [conjugatebase] = 4.7 + log (5/150) [acid] = 4.7 – 1.48 = 3.22 16. What are the structural characteristics common to all amino acids found in naturally occurring proteins? Ans: All amino acids found in naturally occurring proteins have an 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. 17. Only one of the common amino acids has no free -amino group. Name this amino acid and draw its structure. Ans: The amino acid L-proline has no free -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). 18. 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? Ans: 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.) Page 6 19. 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%. [conjugatebase] [acid] [acid] [conjugate base] pH = pKa + log pKa – pH = log antilog (pKa – pH) = [acid] [conjugate base] antilog (6.0 – 5.4) = [acid] [conjugate base] 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. Questions from the third lecture: 20. 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. Page 7 21. A Tyr-Lys-Met B Gly-Pro-Arg C Asp-Trp-Tyr D Asp-His-Glu E Leu-Val-Phe Which one of the above tripeptides: ____(a) is most negatively charged at pH 7? ____(b) will not form an alpha helix or beta sheet? ____(c) contains the largest number of nonpolar R groups? ____(d) contains sulfur? ____(e) will have the greatest light absorbance at 280 nm? Ans: (a) D; (b) B; (c) E; (d) A; (e) C 22. Any given protein is characterized by a unique amino acid sequence (primary structure) and three-dimensional (tertiary) structure. How are these related? Ans: The three-dimensional structure is determined by the amino-acid sequence. This means that the amino-acid sequence contains all of the information that is required for the polypeptide chain to fold up into a discrete three-dimensional shape. 23. When a polypeptide is in its native conformation, there are weak interactions between its R groups. However, when it is denatured there are similar interactions between the protein groups and water. What then accounts for the greater stability of the native conformation? Ans: In the unfolded polypeptide, there are ordered solvation shells of water around the protein groups. The number of water molecules involved in such ordered shells is reduced when the protein folds, resulting in higher entropy. Hence, the lower free energy of the native conformation. Page 8 24. Draw the resonance structure of a peptide bond, and explain why there is no rotation around the C—N bond. Ans: The intermediate resonance structure imparts a partial double bond characteristic to the C—N bond, thereby prohibiting rotation. (See Fig. 4-2, p. 116.) 25. Draw the hydrogen bonding typically found between two residues in an helix. Ans: Hydrogen bonds occur between every carbonyl oxygen in the polypeptide backbone and the peptide —NH of the fourth amino acid residue toward the amino terminus of the chain. (See Fig. 4-2, p. 116.) 26. Describe three important features of an -helix structure. Provide one or two sentences describing why each feature is important. Ans: The -helical structure of a polypeptide is tightly wound around a long central axis; each turn of the right-handed helix contains 3.6 residues and stretches 5.4 Å along the axis. The peptide NH is hydrogen-bonded to the carbonyl oxygen of the fourth amino acid along the sequence toward the amino terminus. The R groups of the amino acid residues protrude outward from the helical backbone. 27. Describe three of the important features of a sheet polypeptide structure. Provide one or two sentences for each feature. Ans: In the sheet structure, several extended polypeptides, or two regions of the same polypeptide, lie side by side and are stabilized by hydrogen bonding between adjacent chains. Adjacent chains may be either parallel (with a repeat distance of about 6.5 Å) or antiparallel (7 Å repeat). The R groups are often small and alternately protrude from opposite faces of the sheet. Page 9 28. Why are glycine and proline often found within a turn? Ans: A turn results in a tight 180° reversal in the direction of the polypeptide chain. Glycine is the smallest and thus most flexible amino acid, and proline can readily assume the cis configuration, which facilitates a tight turn. 29. Explain how circular dichroism spectroscopy could be used to measure the denaturation (unfolding) of a protein. Ans: Circular dichroism spectroscopy measures the amount of -helix in a given protein. As the protein denatures, the amount of -helix should decrease as the protein chain becomes disordered; this change would be detectable using CD spectrography. 30. How can changes in pH alter the conformation of a protein? Ans: Changes in pH can influence the extent to which certain amino acid side chains (or the amino and carboxyl termini) are protonated. The result is a change in net charge on the protein, which can lead to electrostatic attractions or repulsions between different regions of the protein. The final effect is a change in the protein's three-dimensional shape or even complete denaturation. Page 10 Questions from the fourth lecture: 31. Name four factors (bonds or other forces) that contribute to stabilizing the native structure of a protein, and describe one condition or reagent that interferes with each type of stabilizing force. Ans: Any of the following forces stabilize native protein structures and are disrupted by the listed conditions or reagents: (a) disulfide bonds by reducing conditions or mercaptoethanol or dithiothreitol, (b) hydrogen bonds by pH extremes, high salt or heat, (c) hydrophobic interactions of non-polar groups in aqueous solvent by detergents, urea or guanidine hydrochloride, or organic solvent, (d) ionic interactions by changes in pH or ionic strength, and (e) van der Waals interactions by any unfolding condition. 32. What important concepts regarding protein thermal denaturation can be inferred from the egg white of a boiled egg? Ans: 1) Denatured proteins often precipitate and/or aggregate. 2) Denaturation is often not a reversible process. 33. Once a protein has been denatured, how can it be renatured? If renaturation does not occur, what might be the explanation? Ans: Because a protein may be denatured through the disruption of hydrogen bonds and hydrophobic interactions by salts or organic solvents, removal of those conditions will reestablish the original aqueous environment, often permitting the protein to fold once again into its native conformation. If the protein does not renature, it may be because the denaturing treatment removed a required prosthetic group, or because the normal folding pathway requires the presence of a polypeptide chain binding protein or molecular chaperone. The normal folding pathway could also be mediated by a larger polypeptide, which is then cleaved (e.g., insulin). Denatured insulin would not refold easily. Page 11 34. Each of the following reagents or conditions will denature a protein. For each, describe in one or two sentences what the reagent/condition does to destroy native protein structure. (a) urea / guanidine hydrochloride (b) high temperature (c) detergent (d) low pH Ans: (a) Urea or guanidine hydrochloride acts primarily by disrupting hydrophobic interactions. (b) High temperature provides thermal energy greater than the strength of the weak interactions (hydrogen bonds, electrostatic interactions, hydrophobic interactions, and van der Waals forces, breaking these interactions. (c) Detergents bind to hydrophobic regions of the protein, preventing hydrophobic interactions among several hydrophobic patches on the native protein. (d) Low pH causes protonation of the side chains of Asp, Glu, and His, preventing electrostatic interactions. 35. What is the pI, and how is it determined for a protein? 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. Hypothetically, one can average all the pKa's of a proteins amino acids to determine the pI. However in practice, some of these amino acids are satisfied by the proteins tertiary or quaternary structure. Thus, to be most accurate, pI needs to be determined empirically by isoelectric focusing gel. Page 12 36. 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: pI (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. 37. A biochemist has accidentally isolated a second protein with the protein they were trying to isolate. The reason for its co-purification is unknown, as is its identity. How might the biochemist have identified its presence and what should they do to determine its identity? Ans: The biochemist most likely identified the second protein in a PAGE gel, either SDS or isoelectric focusing or both. Since the biochemist presumably knows what species they are working with, they only need a small amount of sequence to correctly identify the protein. In this circumstance, either Edman degradation or mass spectrometry may be performed. However, mass spectrometry is the most likely to be used, since it is more readily available. Page 13 38. 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. 39. Describe a reservation about the use of x-ray crystallography in determining the threedimensional structures of biological molecules. Ans: To obtain an x-ray picture of a biomolecule, the molecule must be purified and crystallized under laboratory conditions far different from those encountered by the native molecule. Biomolecules in the cell also have more flexibility and freedom of motion than can be accommodated in a rigid crystal structure. Therefore, the static picture obtained from an x-ray analysis of a crystal may not provide a complete or accurate representation of the biomolecule in vivo. Page 14 40. Name two ways to determine the precise (high-resolution) three-dimensional structure of a protein complex. Describe why you think these methods are appropriate. Ans: The protein complex could be crystallized, and its structure determined by x-ray crystallography. The pattern of diffracted x-rays yields, by Fourier transformation (black magic), the three-dimensional distribution of electron density. By matching electron density with the known sequence of amino acids in the protein, each region of electron density is identified as a single atom. This would be an appropriate choice if the complex can be crystallized, but inappropriate if not. Sometimes, the three-dimensional structure of a small protein or peptide can be determined in solution by sophisticated analysis of the NMR spectrum of the polypeptide. However, this technique is rarely appropriate for complexes, which usually exceed the complexitiy limit. A three dimensional structure can also be determined by cryo-electron tomography, which is an excellent method to use with complexes that don't crystallize. If an advanced machine is available, molecular details may be sufficient to determine high resolution structure. This is most appropriate for larger complexes, or those with individually crystallized components, but may also be used as the first approach. Homology modeling is only an appropriate answer if the above three are likely to fail, and a good structure of a highly similar protein is available. Questions from the fifth lecture: 41. For the binding of a ligand to a protein, what is the relationship between the Ka (association constant), the Kd (dissociation constant), and the affinity of the protein for the ligand? Ans: Ka = 1/Kd. The larger the Ka (and hence the smaller the Kd), the higher the affinity of the protein for the ligand. Page 15 42. Describe how you would determine the Ka (association constant) for a ligand and a protein. Ans: An experiment would be carried out in which a fixed amount of the protein is incubated with varying amounts of ligand (long enough to reach equilibrium). The fraction of protein molecules that have a molecule of ligand bound is then determined. A plot of this fraction () vs. ligand concentration [L] should yield a hyperbola. The value of [L] when = 0.5 is equal to 1/Ka. 43. What fraction of ligand binding sites are occupied () when [ligand] = Kd? Show your work. Ans: From equation (5-8) on page 156: [Ligand] = [Ligand] + K d thus, when [Ligand] = K d , the equation becomes: = Kd 1 Kd + Kd 2 44. Explain why most multicellular organisms use an iron-containing protein for oxygen binding rather than free Fe2+. Your answer should include an explanation of (a) the role of heme and (b) the role of the protein itself. Ans: (a) Binding of free Fe2+ to oxygen would result in the formation of reactive oxygen species that can damage biological structures. Heme-bound iron is less reactive in this regard. (b) Binding of oxygen to free heme can result in irreversible oxidation of the Fe2+ to Fe3+ that does not bind oxygen. The environment of the heme group in proteins helps to prevent this from occurring. Page 16 45. Explain why the structure of myoglobin makes it function well as an oxygen-storage protein, whereas the structure of hemoglobin makes it function well as an oxygentransport protein. Ans: The hyperbolic binding of oxygen to the single binding site of myoglobin results in a high affinity even at the relatively low partial pressures of O2 that occur in tissues. In contrast, the cooperative (sigmoidal) binding of O2 to the multiple binding sites of hemoglobin results in high affinity at high partial pressures such as occur in the lungs, but lower affinity in the tissues. This permits hemoglobin to bind O2 in the lungs and release it in the tissues. 46. How does BPG binding to hemoglobin decrease its affinity for oxygen? Ans: BPG binds to a cavity between the subunits. It binds preferentially to molecules in the low-affinity T state, thereby stabilizing that conformation. 47. Fetal hemoglobin binds BPG with lower affinity than adult hemoglobin. How does this property facilitate tranfers of O2 from mother to fetus? Ans: Lower affinity for BPG means that fetal hemoglobin will have less BPG bound that the mother's hemoglobin. This shifts the fetus' fractional O2 saturation curve to the left (i.e., lower p50) of the mother's. At low pO2, O2 will dissociate from the maternal hemoglobin and can be bound by the fetal hemoglobin. 48. Why is carbon monoxide (CO) toxic to aerobic organisms? Ans: It binds to heme with a higher affinity than oxygen, and thus prevents oxygen from binding to hemoglobin. Page 17 49. Describe briefly the structure of myosin. Ans: Myosin contains two copies of a large polypeptide (heavy chain) and four copies of a small polypeptide (light chain). The helix contributes significantly to the structure of the heavy chains. At their carboxyl termini, the heavy chains are wrapped around each other in a fibrous left-handed coil. At their amino termini, they each have a globular domain with which the light chains are associated. 50. What is the role of ATP and ATP hydrolysis in the cycle of actin-myosin association and disassociation that leads to muscle contraction? Ans: ATP binding to myosin results in a conformational change that causes dissociation of actin from the myosin. ATP hydrolysis results in a change of orientation of the myosin relative to the actin filament, which allows movement to the next actin subunit. This is followed initially by release of the phosphate hydrolysis product and weak binding of the myosin to this actin subunit, and, subsequently, by tight binding and release of the ADP hydrolysis product. Page 18