Midterm 1 Last, First

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1 Midterm 1 BIS 105 Prof. T. Murphy April 23, 2014 There should be 6 pages in this exam. Exam instructions (1) Please write your name on the top of every page of the exam (2) Show all work for full credit (3) If you need more space, continue on the back of the page (label please) (4) You may use a calculator for numerical calculations, not for recalling facts or equations. (5) Note the helpful information on the last page Please read and sign the statements below: Authorization for public distribution I authorize the return of my exam for pickup in a class bin. I do not authorize the public return of my exam. (If you choose this option or fail to choose either option, you will pick up your exam from the instructor during office hours.) Signature Date Honor code statement My signature below affirms that I wrote this exam in the spirit of the honor system of UCD. I neither received nor furnished any help during the exam. Signature

2 1. The standard state ΔG o for hydrolysis of glucose-1-p, (glucose-1-p+ H 2 O glucose + Pi) is kj/mol. a. (8) Calculate ΔG in kj/mol for the hydrolysis reaction in 10 mm glucose-1-p, 5 mm glucose, and 10 mm Pi. Assume 25 o C; R = J/ o K-mol ΔG = -21,000 + (8.315)(298) ln ((0.005)(0.01))/0.01) = kj/mol 6 pts for formula; 2 pts for correct answer units b. (4) Under the conditions above, could this reaction be used to synthesize ATP (ADP + Pi ATP + H 2 O), assuming appropriate enzymes and that ΔG for ATP synthesis = +50 kj/mol? Explain. In order to proceed, the ΔG for the coupled reaction must be negative. ΔG for the coupled reaction would be = kj/mol. At these concentrations the reaction would not proceed (go backwards: ATP hydrolysis would form glucose-1-p). (correct direction, 1 pt; correct reasoning, 3 pts) 2. What is the ph of a 0.1 M solution of: a. (4) HCl ph = -log(0.01) = 1 b. (4) KOH ph = 14 - poh = 14-1 = (8) Given 0.1 M solutions of monobasic and dibasic Na phosphate (NaH 2 PO 4 and Na 2 HPO 4 ) and the three pka s of H 3 PO 4 (pka 1 = 2.15, pka 2 = 7.2, pka 3 = 12.4), describe the calculation and the preparation of 1 liter of 0.1 M buffer at ph 6.9. By H-H: ph = pka + log ([HPO 4 ( -2 )]/[H 2 PO 4 ( - )]) [HPO 4 ( -2 )]/[H 2 PO 4 ( - )] = exp 10 ( ) = exp 10 (-0.3) = 0.5 [HPO 4 ( -2 )] = 0.5*[H 2 PO 4 ( - )] ml of HPO 4 ( -2 ) + ml of H 2 PO 4 ( - ) = 1000 To 667 ml of NaH 2 PO 4, add 333 ml of Na 2 HPO 4 4 pts for correct formula, with or without data insertion 2 pts for relative amts of components; 2 pts for actual mls

3 4. (12) Draw the chemical structure of the tripeptide, N-tyrosine-glutamine-arginine-C, as it would appear at ph 7. Include letters for all atoms (C, N, O, H) and show ionic charges (+,-). 3 pts each for the correct side chains; 1 pt each for the correct acid/base form of the N- terminal and the C-terminal; 1 pt for correct peptide bonds. O H C N _ O H C N _ 5. (12) In the square below, draw an appropriate titration curve for 0.1 M histidine (4 pts). Label the axes, including units, if appropriate; identify the part of histidine that corresponds to each inflection point of your curve (2+3 pts). Be sure to note the positions of the inflection points on the x axis (3 pts) : N-term ph 7 6.0: side chain 1.8: C-term OH - (M)

4 6. (12 pts) The picture below shows a diagram of a structure in a protein. Name the structure. Describe where in a protein you might expect to find this particular one, and explain your reasoning. (4) This represents an α- helix, looking down the axis. (2+2) Amino acids on the left (val, ile, leu, trp) are in the interior of the protein, because sidechains are hydrophobic. (2+2) The others, from asp clockwise to arg, are facing solvent (exterior), because side chains are hydrophylic. 7. (10) The table below shows three steps in the purification of an enzyme. Complete the table, giving reasonable units and values in each open box. Steps Protein (mg) Enzyme (units) Specific Activity Fold purification % Yield Initial Starting Material Ammonium sulfate precipitation Gel filtration ph pt each

5 BIS105 Midterm 1 Last, First 8. (10 pts) The picture below shows one step in the hydrolysis of a protein by chymotrypsin. Explain (1) the importance of the formation of a "low-barrier hydrogen bond" in the reaction and (2) why this enzyme does not work at a ph below 6. (1) (5 pts) The formation of the low-barrier-hydrogen bond allows to move to the right, forming a bond between the histidine N and the -H from serine, and releasing the electrons from the serine O so they can attack the substrate C=O. (2) (5 pts) Below ph 6, the histidine ring is doubly protonated (and + charged), and thus it cannot accept the serine -H in order to release the serine O electrons. 9. (12 pts) The KM of triose phosphate isomerase for its substrate, glyceraldehyde-3-p, is 18 µm (1.8 x 10-5 M). When [glyceraldehyde-3-p] = 30 µm (3 x 10-5 M), the rate of the reaction, V, was 82.5 µmol/ml-sec. Using the Michaelis-Menten equation, calculate Vmax for this enzyme. V = Vmax[S]/(Km + [S]) 82.5 = Vmax*(30)/( ) Vmax = 82.5*( )/30 Vmax = 132 µmol/ml-sec 4 pts for equation; 4 pts for set-up (may be combined); 4 pts for correct answer. 10. (4) Provide the one-letter codes for the following amino acid sequence: phe-ile-asn-ile-ser-his-glu-asp-tyr-ile-pro-pro-glu-glu F I N I S H E D Y I P P E E

6 R = J/mol-K K = o C Amino acid α-cooh pka α-nh 3 + pka R group pka Alanine Arginine Asparagine Aspartic acid Cysteine Glutamic acid Glutamine Glycine Histidine Isoleucine Leucine Lysine Methionine Phenylalanine Proline Serine ~13 Threonine ~13 Tryptophan Tyrosine Valine Question Possible Points Total 100