CHEM 3653 Exam # 1 (03/07/13)

Transcription

1 1. Using phylogeny all living organisms can be divided into the following domains: A. Bacteria, Eukarya, and Vertebrate B. Archaea and Eukarya C. Bacteria, Eukarya, and Archaea D. Eukarya and Bacteria E. Prokaryotes, Eukaryotes and Multicellular F. None of the above 2. Which cellular compartment or organelle is involved in the synthesis of proteins and lipids? A. Endoplasmic reticulum B. Lysosomes C. Nucleolus D. Peroxisomes E. Vacuoles F. All of the above 3. Consider the reaction A + B P. H 0 = 20 kj mol -1 S 0-1 = 210 J K -1 mol A. This reaction is spontaneous only below -178 C B. This reaction is spontaneous only above -178 C C. This reaction is spontaneous only below 95 C D. This reaction is spontaneous only above 95 C E. This reaction is spontaneous at all temperatures F. This reaction is never spontaneous 4. Rank the following interactions in order of increasing strength (start with the weakest interaction). A. Ionic interactions, hydrogen bonds, London dispersion forces, covalent bonds B. London dispersion forces, hydrogen bonds, ionic interactions, covalent bonds C. London dispersion forces, ionic interactions, hydrogen bonds, covalent bonds D. Covalent bonds, London dispersion forces, ionic interactions, hydrogen bonds E. Hydrogen bonds, London dispersion forces, ionic interactions, covalent bonds 5. Methanol CH 3 OH can act both as a H-bond donor and as a H-bond acceptor. What is the maximal number of H-bonds a single molecule of methanol can form with surrounding water molecules? A. 0 B. 1 C. 2 D. 3 E. 4 F. 6

2 6. Which of the following is the best explanation for the hydrophobic effect? A. It is caused by an affinity of hydrophobic groups for each other. B. It is caused by the affinity of water for hydrophobic groups. C. It is an entropic effect, caused by the desire of water molecules to increase their entropy by forming highly ordered structures around the hydrophobic groups. D. It is an entropic effect, caused by the desire of water molecules to increase their entropy by excluding hydrophobic groups, which they must otherwise surround with highly ordered structures. E. It is an entropic effect caused by the desire of hydrophobic groups to increase their entropy by associating with other hydrophobic groups. 7. If you are planning to measure the activity of an enzyme at ph 4.0, a mixture of which acids or bases would serve as the best buffer? Acid/Base pka Formic acid 3.7 Tris 8.1 Piperidine 11.2 A. HCl and Formic Acid B. HCl and Tris C. HCl and Piperidine D. Formic Acid and NaOH E. Tris and NaOH F. Piperidine and NaOH 8. What is the net charge of the dipeptide Glu-Ala at ph=4.1, given the following data for the individual amino acids: Glu: pk a1 = 2.1; pk a2 = 9.5; pk ar = 4.1 Ala: pk a1 = 2.3; pk a2 = 9.9 A. +2 B. +1 C D. 0 E F. -1 G H The following compound is: A. Cytidine triphosphate B. Guanosine triphosphate C. Deoxyguanine diphosphate D. Adenine triphosphate E. Deoxycytidine triphosphate F. Deoxyuracil triphosphate G. Uracil triphosphate H. Thymidine triphosphate J. Deoxythymidine triphosphate

3 10. A double stranded DNA fragment contains 12% adenine residues. Calculate the percentage cytosine residues. A. 12% B. 24% C. 38% D. 50% E. 76% F. Cannot be determined given the information provided 11. Genomic DNA is, resulting in the production of. A. transcribed; mrna B. translated; trna C. transcribed; protein D. translated; protein E. translated; rrna 12. Which of the following sequences is able to form hairpin structures? A. 5 -AATCAAGGTC B. 5 -AAGGAACCTT C. 5 -AATTAACCGG D. 5 -AAGGTAGGAA E. 5 -AAGGTAAGGT 13. Paralogous genes are A. genes that do not encode protein. B. genes of slowly evolving proteins. C. relics of genes that are not expressed. D. genes of rapidly evolving proteins. E. the results of gene duplication. F. genes in different species that perform the same function 14. The following amino acid is: A. Alanine B. Arginine C. Asparagine D. Glycine E. Glutamine F. Leucine G. Lysine H. Threonine I. Tyrosine

4 15. 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 all peptide bonds in a protein. F. Cleave disulfide bonds within a protein so that the individual polypeptides can be separated. 16. Identify hydroxyproline in the following list of structures A B C D E F G H 17. A 70 amino-acid long peptide forms an -helix in solution. How long is this -helix? A. 1.1 nm B. 4.7 nm C nm D nm E. 47 nm F. 105 nm 18. A chaperonin A. helps fold some proteins into their lowest energy state. B. is required for all proteins to fold properly. C. mediates the unfolding of proteins. D. is required for protein denaturation. E. counteracts the laws of thermodynamics.

5 19. Imagine that a researcher treated a protein with a high concentration of a chaotropic agent. Which of the following is the most likely result of the treatment? I. Nonpolar portions of the protein become more soluble. II. The protein begins to denature, III. The protein stability increases due to hydrophobic collapse, A. I, II, III B. I, II C. II, III D. I, III E. I F. II G. III 20. Hydrogen bonds and maximum separation of amino acid side chains make the very stable and energetically. A. α helix and β sheet, favorable B. α helix, unfavorable C. β sheet, unfavorable D. α helix, favorable E. β sheet, favorable 21. The rearrangement of T state hemoglobin to the R state A. occurs in each protein subunit independently when its heme binds oxygen. B. requires the binding of at least three oxygen molecules. C. increases the ion pairing interactions of the C-terminal amino acids. D. involves the movement of the Fe(II) into the heme plane. E. opens a central cavity for BPG binding. 22. Which of the following increases the affinity of hemoglobin for O 2? A. an increase in BPG concentration B. the formation of N-terminal carbamates C. an increase in ph D. a decrease in ph E. an increase in CO 2 concentration 23. Which of these amino acid groups would not make a good nucleophilic catalyst? A. amino B. sulfhydryl C. imidazole D. methyl E. hydroxyl F. phenyl

6 24. Which of the following amino acid residues would provide a side chain capable of increasing the hydrophobicity of a binding site? A. histidine B. lysine C. isoleucine D. arginine E. serine F. glutamine 25. Proton transfer from an acid, lowering the free energy of a reaction s transition state, is characteristic of A. Electrophilic catalysis. B. Nucleophilic catalysis. C. General base catalysis. D. General acid catalysis. E. Entropic catalysis. F. Electrostatic catalysis. 26. Zymogens are not enzymatically active because A. the active site amino acids have been mutated. B. they have not yet bound the proper cofactor. C. their environment has the wrong ph. D. they are not yet shaped such that essential proximity and orientation catalysis can occur. E. None of the above is correct. 27. The names of the encircled structural elements that labeled as A and B on the structure of DNA topoisomerase II (the Figure to the right) are, respectively:. A. -helix and -strand B. -helix and random coil C. -helix and a loop D. -strand and -turn E. -strand and random coil F. -strand loop 28. The symmetry group for the DNA topoisomerase (shown on the Figure in the previous question) is: A. C1 B. C2 C. D2 D. D4 E. H2 F. Mirror G.Tetrahedral H. Icosahedral

7 29. Determine sequence of a pentapeptide from the following information. Amino acid composition was determined as Ile, Lys, Met, Arg, Tyr. Trypsin digest released lysine (Lys) and a tetrapeptide. Treatment with Cyanogen Bromide released a single amino acid and a tetrapeptide. Brief chymotrypsin treatment produced a dipeptide and a tripeptide. The dipeptide obtained by chymotrypsin digest was boiled at low ph to yield lysine and tyrosine. Trypsin cuts after Lys and Arg: Lys-_?_-_?_-_?_-Arg CNBr cuts after Met: Lys-_?_-_?_-Met-Arg Boiling in acid breaks all peptide bonds; the dipeptide sequence is Lys-Tyr; the entire peptide: Lys-Tyr-_?_-Met-Arg Final answer: Lys-Tyr-Ile-Met-Arg 30. Consider the reaction of pyruvate reduction to lactate: Pyruvate + NADH Lactate + NAD +. Assuming that R = 8.31 J mol -1 K -1 and the following G for the half-reactions: Pyruvate Lactate; G = 35.7 kj mol -1 NAD NADH; G = 60.8 kj mol -1 (a) Calculate the standard free energy change for the complete reaction. = 35.7 kj/mol 60.8 kj/mol = kj/mol (b) Assuming the following concentrations: pyruvate, 10 mm; NAD, 1 mm; NADH, mm; calculate the equilibrium concentration of lactate. Keq = exp (-ΔG0 /RT) = = [NAD][lactate]/([NADH][pyruvate]) [lactate] = Keq x [NADH] x [pyruvate] / [NAD] [lactate] = 3 x 10-6 x 0.01 x / = 0.75 M (if T = 25C) [lactate] = 3 x 10-6 x 0.01 x / = 0.5 M (if T = 37C)

8 31. (bonus; 10 pt). For the enzyme mechanism illustrated in the picture below, calculate how much the catalyzed reaction is faster than the uncatalyzed one. Assume that the bound substrate is stabilized by two hydrogen bonds and the transition state is stabilized by three hydrogen bonds. Assume that the free energy of a single hydrogen bond is 10 kj mol -1. Enzyme and Unbound substrate Substrate-bound Enzyme Transition state The rate acceleration by enzyme is defined by the stability of the enzyme-bound transition state compared to the enzyme-bound unstrained substrate. G = G TS - G ES = 10 kj/mol V E /V N = exp( G/RT) = exp(10000/(8.31x(273+37))) = 48 (at 37 C) or: V E /V N = exp( G/RT) = exp(10000/(8.31x(273+25))) = 56 (at 25 C)