4. The Michaelis-Menten combined rate constant Km, is defined for the following kinetic mechanism as k 1 k 2 E + S ES E + P k -1

Transcription

1 Fall 2000 CH 595C Exam 1 Answer Key Multiple Choice 1. One of the reasons that enzymes are such efficient catalysts is that a) the energy level of the enzyme-transition state complex is much higher than for an uncatalyzed reaction b) enzymes can lower the activation energy for the reaction c) the translational entropy of the substrate is greatly increased upon binding to the enzyme d) the enzyme typically binds the substrate much more strongly than the transition state e) enzymes can employ nucleophilic reaction mechanisms while reactions in free solution can only proceed by electrophilic mechanisms 2. For an enzyme reaction that shows typical Michaelis-Menten kinetics, doubling the concentration of enzyme will a. triple the Km b. double the Km c. halve the Km d. not alter the Km 3. If the following section of a polypeptide is folded into an α-helix, to which amino acid is the carbonyl group of alanine noncovalently bonded? ala-ser-val-asp-glu-leu-gly a. serine b. glutamic acid c. aspartic acid d. leucine 4. The Michaelis-Menten combined rate constant Km, is defined for the following kinetic mechanism as k 1 k 2 E + S ES E + P k -1 a. (k 1 + k 2 )/k -1 b. (k -1 + k 2 )/k 1 c. (k 1 + k -1 )k 2 d. k -1 /k 1 5. The highly charged 2,3-bisphosphoglycerate binds to hemoglobin a. on the exterior surface b. on the heme group c. at the Fe +2 ion d. in the interior cavity

2 6. At high altitudes, the concentration of bisphosphoglycerate (BPG) in erythrocytes is higher than at low altitudes. Which of the following statements is logically correct? a. The P 50 of Hb at high altitudes is less than the P 50 of Hb in erythrocytes at low altitudes. b. Hb adjusted to high altitudes will not show a Bohr effect. c. The P 50 of Hb increases as the concentration of BPG in erythrocytes increases. d. The P 50 of myoglobin is increased as the concentration of BPG goes up in erythrocytes adjusted to high altitudes. 7. An interesting characteristic in the allosteric enzyme, ATCase, which is not found in hemoglobin, is a. two different types of protein subunits in ATCase b. allosteric effector sites are on both subunit types c. binding of both positive and negative heterotropic effectors d. one subunit type binds only the substrate, while the other subunit type binds only heterotropic effectors. 8. Lysozyme has a ph optimum centered around ph 5.0. The active site of lysozyme contains a glutamic acid residue (pka = 5.5) and an aspartic acid residue (pka = 4.0). Which of the following statements is correct about the mechanism of lysozyme? a. The glutamic acid residue is in a more polar environment that the aspartic acid. b. During the entire catalytic mechanism, the aspartic acid residue remains unprotonated. c. During the mechanism, an oxyanion transition state forms. d. The glutamic acid residue acts as a general base catalyst. 9. The good transition state analog is one which would serve also as an effective a. competitive inhibitor b. noncompetitive inhibitor c. allosteric effector d. uncompetitive inhinbitor 10. Which of the following is correct with respect to the amino acid composition of proteins? a. Proteins with the same molecular weight have the same amino acid composition. b. Proteins contain at least one each of the twenty different standard amino acids. c. Larger proteins have a more uniform distribution of amino acids than smaller proteins. d. Proteins with different functions usually differ significantly in their amino acid composition. e. The average molecular weight of an amino acid in a protein increases with the size of the protein. 11. For amino acids with neutral R groups, at any ph below the pi of the amino acid, the population of amino acids in solution will: a. have no charged groups. b. have no net charge. c. have a net positive charge. d. have positive and negative charges in equal concentration. e. have a net negative charge.

3 12. Which of the following amino acids have two chiral centers? a. threonine b. proline c. isoleucine d. a and b e. a and c 13. Which of the following amino acids lacks an asymmetric carbon atom? a) serine b) glycine c) histidine d) glutamine e) leucine 14. Enzymes are catalysts which a) accelerate the rates of biological reactions b) inhibit the formation of unwanted metabolites c) change the equilibrium constants of reactions d) are generally non-specific in reactions they catalyze e) only work in hydrophobic surroundings 15. The partial double bond character of the peptide bond a) restricts rotation around the C-N bond b) allows free rotation around the C-N bond c) restricts rotation around the Cα-C bond d) restricts rotation around the N-Cα bond e) allows free rotation around both C-N and C-C bonds 16. The affinity of hemoglobin for oxygen is increased by a) a higher [H + ] concentration in the surroundings b) an increase in the organic phosphate level in red cells c) an increase in the CO 2 concentration in the surroundings d) an increase in the oxygen concentration in the surroundings 17. The aspartic acid residue (Asp 52 ) at the active site of lysozyme serves which of the following purposes as part of the catalytic mechanism a) general acid catalysis b) general base catalysis c) electrostatic stabilization of a carbonium ion d) covalent catalysis e) hydrophobic interaction in binding of the substrate 18. In an α helix, the R groups on the amino acid residues: a) are found on the outside of the helix spiral b) generate the hydrogen bonds that form the helix c) stack within the interior of the helix d) cause only right-handed helices to form e) alternate between the outside and the inside of the helix

4 19. The three-dimensional conformation of a protein may be strongly influenced by amino acid residues that are very far apart in sequence. This relationship is in contrast to secondary structure, where the amino residues are: a) always side by side b) generally near each other in sequence c) generally on different polypeptide strands d) generally near the polypeptide chain s amino terminus or carboxyl terminus e) restricted to only about seven of the twenty standard amino acids found in proteins 20. Experiments on denaturation and renaturation after the reduction and reoxidation of the S-Sbonds in the enzyme ribonuclease (RNase) have shown that: a) the primary sequence of RNase is sufficient to determine the formation of a specific secondary and tertiary structure b) the enzyme, dissolved in water, is thermodynamically stable relative to the mixture of amino acids whose residues are contained in RNase c) native ribonuclease does not have a unique secondary and tertiary structure d) the completely unfolded enzyme, with all S-S- bonds broken, is still enzymatically active e) the folding of denatured RNase into the native, active conformation, requires the input of energy in the form of heat 21. For an enzyme which obeys Michaelis-Menten kinetics, what is the V max value in µmol/min if v = 35 µmol/min when [S] = K m? a. 50 b. 70 c. 45 d. 95

5 Written Answers 1. In the use of NMR to determine the structure of biomolecules the structural information is obtained primarily by measuring the nuclear Overhauser effect. a) Very briefly describe what the nuclear Overhauser effect results from and to what degree the effect depends upon distance. Give the maximum practical observable distance. The nuclear Overhauser effect results from an interaction between the magnetic dipoles of the individual nuclei. The effect depends on the distance between nuclei, and is greatest for nuclei separated by less than about 3 Å, and is observable when nuclei are separated by up to about 6 Å. The nuclear Overhauser effect therefore can be used to determine the distance between protons, when the distances are less than or equal to about 6 Å. b) Solving macromolecular structures by X-ray crystallography depends upon measurement of the electron density at each point within the unit cell. In turn, those measurements depend upon the intensities of the diffracted X-rays and the phases of the diffracted X-rays. Briefly describe two common methods (MIR and MAD) that can be used to solve the phase problem. The multiple isomorphous replacement (MIR) method: a) diffraction data (intensities) are measured for a "native" crystal. b) diffraction data (intensities) are measured for a "heavy atom derivative" crystal, prepared by soaking a heavy atom such as mercury into the crystal, or by incorporating selenomethionine. c) By comparing the native and derivative diffraction intensities, it is possible to calculate the phases for the native crystal. Note: usually at least two different heavy atom derivatives are required for the MIR method to work. d) Once phases are known, the electron density within the crystal can be calculated. A model of the protein structure can then be built into a map of the electron density. The "multiwavelength anomolous dispersion" (MAD) methods: Additional information used in calculating phases can be obtained if x-ray diffraction intensities can be measured at wavelengths near the absorption edge of the heavy atom derivative. A tunable x-ray source is required (provided by a synchroton). In a synchrotron, accelerated electrons traveling near the speed of light emit intense x-rays. Using "MAD" data: a) often only a single heavy atom derivative is required to solve a structure. b) it is possible to solve structure of higher molecular weight molecules (such as the ribosome, at MW = 2,500,000). c) Contrast the use of NMR and X-ray crystallography as applied to biomacromolecules. Address in particular the issues of any limitations on the size of the macromolecules whose structure can be determined by either technique; what types of information that can be determined by one technique and not the other and whether both techniques always give the same structure for a particular macromolecule (both proteins and nucleic acids).

6 NMR: - obtain a set of 1 H- 1 H distances, used to construct a model of the structure - better at the short-range detail - can be done in solution, i.e. does not require a crystal - MW limit : ~30,000 X-RAY: - obtain electron density map, into which structure is built - better at large-scale structural features - proteins can undergo re-arrangements upon crystallizing leading to inaccurate models - MW limit : ~2.5 x 10 6

7 2. From the amino acid structures listed below, answer the following questions: a. What is the name of amino acid A? _glutamine b. Which of the above amino acids has the highest isoelectric point? _lysine (C) _ c. Which of the above amino acids is classified as a basic amino acid? lysine (C) d. Which of the above amino acids absorbs light centered at 280 nm? tyrosine (D) e. What is the name of amino acid D? tyrosine f. What is the name of amino acid B? valine g. Draw the structure of tryptophan showing the charge on the molecule as it would be at ph 7. h. Draw the structure of the dipeptide pro-his (at ph 7), indicating the peptide bond. i. Name 2 sulfur containing amino acids methionine cysteine j. What is the approximate pk of the side chain amino group of lysine? 10.5

8 3. a) What is meant by the statement stacking interactions are important in stabilizing the structures of nucleic acids? Describe the nature of these interactions and give examples for both DNA and RNA molecules. "Stacking interactions" in nucleic acids refers to the tendency of purine and pyrimidine bases to form extended stacks of planar, parallel molecules, with at least partial overlap of the bases. These stacking interactions are a form of van der Waals interactions. The stacking interactions are stabilized by hydrophobic forces. b) Which of the following stacking interactions will provide more stability : purine-purine or pyrimidine-pyrimidine? Explain why. The degree of stability in stacking is proportional to the degree of overlap of the bases. Because of the greater possibility for overlap (and thus enhanced van der Waals interactions ) with the larger purine bases, purine-purine stacking provides more stability than pyrimidine-pyrimidine stacking. c) Bound metals are often found in RNA structures. What kinds of roles can the metals play? Bound metals in RNA can play a general role of neutralizing the charge on the phosphodiester bridges. Positively-charged ions such as those of Na, K, and Mg can bind electrostatically to the RNA and allow for foldings of the RNA that would otherwise have been made difficult by a concentration of negative charges on the phosphate groups. Such binding of metals is non-specific. Metal ions can also bind in some specific metal pockets, such as those in trna. Here the metal can stabilize specific forms of folding. d) Hydrogen bonding also plays a role in nucleic acid structure stabilization. In what important ways does it differ from stacking and metal ion interactions? While stacking interactions are relatively non-specific in nature, hydrogen bonding is highly specific and highly directional. For example, the G-C and A-T hydrogen bonding pairs involves not just a general purine-pyrimidine interaction but favors specific Watson-Crick base pairing. Even where non-watson-crick pairing occurs, it is still highly specific (see V&V p.866).

9 4. a) Using the simple expression E + S ES E + P as a basis, describe the effect of competitive and uncompetitive inhibitors. Competitive: E + S ES E + P + + I I Uncompetitive: E + S ES E + P " " EI ESI b) With a brief description and diagrams show how these two types of inhibitors can be distinguished by Lineweaver-Burk plots. c) Using aspartate transcarbamoylase as an example, contrast feedback inhibition of an enzyme with simple competitive and uncompetitive inhibition. Feedback inhibition of an enzyme occurs when a product of a reaction pathway (often many steps away from the inhibited reaction) binds to the enzyme and causes inhibition, e.g., inhibition by "E": A B C D E In the case of aspartate transcarbamoylase, it is inhibited by CTP - an end product of the pathway. By contrast to simple competitive and uncompetitive inhibition where the inhibition typically binds to the active site of the enzyme, feedback inhibitors bind at sites different from the active site and sometimes even on separate regulatory subunits (e.g., ATCase). Feedback inhibitors typically show sigmoidal kinetics of inhibition and nonlinear Lineweaver-Burk plots.

10 5. Studies of oxygen transport in pregnant mammals have shown that the O 2 -saturation curves of fetal and maternal blood are markedly different when measured under the same conditions. Fetal erythrocytes contain a structural variant of hemoglobin, HbF, consisting of two α and two γ subunits (α 2 γ 2 ), whereas maternal erythrocytes contain HbA (α 2 β 2 ). (a) Which hemoglobin has a higher affinity for oxygen under physiological conditions, HbA or HbF? Explain. HbF has a higher affinity. According to the graph, the P 50 for HbF is approx. 2kPa and that of HbA about 4kPa. The P 50 represents the oxygen concentration for half-maximal saturation. Thus, in the presence of BPG, it takes approx. half the oxygen tension to saturate HbF as it does HbA. (b) What is the physiological significance of the different O 2 affinities? It makes sense that HbF has a higher oxygen affinity as oxygen has to be transferred from the blood of the mother to that of the fetus. The binding of oxygen by hemoglobins is a dynamic process with O 2 binding and releasing. When an O 2 molecule is released from HbA, it is more likely to be bound by HbF than another HbA. (c) When all the BPG is carefully removed from samples of HbA and HbF, the measured O 2 -saturation curves (and consequently the O 2 affinities) are displaced to the left. However, HbA now has a greater affinity for oxygen than does HbF. When BPG is reintroduced, the O 2 - saturation curves return to normal, as shown in the grwph. What is the effect of BPG on the O 2 affinity of hemoglobin? How can the above information be used to explain the different O 2 affinities of fetal and maternal hemoglobin? BPG when bound to hemoglobins (1 tetramer in the central cavity), it decreases the oxygen affinity of that molecule by stabilizing the deoxy form. The above information indicates that the binding of BPG to HbA has a greater effect than does BPG binding to HbF. This suggests that BPG binds more tightly to HbA than HbF. Stronger binding of BPG by HbA than HbF indicates that the residues lining the central cavity of the deoxy HbA are different than those in HbF.

11 6. You have a polypeptide that has been degraded with cyanogen bromide (cleaves on the C- terminal side of methionines) and trypsin (cleaves on the C-terminal side of lysines and arginines). The sequences of the fragments are listed below (using the one-letter codes). Is there enough information to derive the complete amino acid sequence of the polypeptide? If so list the complete sequence. If not, show how much of the sequence you can derive. In both cases show your methodology. What other treatments might be used to support such a deduced sequence, or complete the sequencing? Cyanogen bromide treatment: Trypsin treatment: (1) K (5) FMK (2) RFM (6) GMNIK (3) MLYCRGM (7) GLMR (4) NIKGLM (8) MLYCR There is sufficient information to derive a complete polypeptide sequence. In order to predict the sequence, one has to examine what overlaps there are between the peptides derived from each cleavage treatment. The overlaps are illustrated below. MLYCRGM GMNIK NIKGLM GLMR RFM FMK Thus, the peptide sequence is MLYCRGMNIKGLMRFMK. To confirm the deduced sequence, perform one round of Edman degradation to verify that the N-terminus is methionine, and perform a brief carboxypeptidase A digestion to verify that lysine is at the C-terminus.