exam1_WHITE_BCHM461-0102_OFFICIAL_KEY_Fall2022

.pdf

School

University of Maryland, University College *

*We aren’t endorsed by this school

Course

461

Subject

Chemistry

Date

Dec 6, 2023

Type

pdf

Pages

12

Uploaded by CoachClover395

Report
BCHM 461-0102 Name: Official Key (WHITE exams) October 10, 2022 Prof. Julin Exam I (100 points) NOTE: read the questions carefully, think about your answers, and, be sure that you answer all parts of the questions. Provide specific information that answers the question being asked. Always show your work or explain your reasoning. You may use a calculator, but only for computation. Any other use is a violation of the University’s Code of Academic Integrity . No other electronic equipment may be used - that is, turn off your cell phone ! Honor Pledge Please write the following sentence in the box, and then sign your name: “I pledge on my honor that I have not given or received any unauthorized assistance on this examination.” Some equations and formulas which might (or might not) be useful: a) Quadratic formula. For: ax 2 + bx + c = 0 x = ! b ± ( b 2 ! 4ac ) ½ 2a b) K w = [H + ][OH ! ] = 10 ! 14 c) K a = [H + ][A ! ]/[HA] pH = pK a + log([A ! ]/[HA]) d) Δ G = Δ H ! T Δ S e) Δ G = Δ G EN + RT ln{[ products ]/[ reactants ]} f) K eq = exp( ! Δ G EN /RT) g) Δ G EN = ! RTln(K eq ) K eq = [products] / [reactants] h) R = 8.3 J/K/mol = 0.0083 kJ/K/mol i) T (K) = T ( E C) + 273 j) E % Q 1 Q 2 / ε r k) fr dissoc Ka H Ka H pH pKa pH pKa . . [ ] [ ] ( ) ( ) 1 10 10 1 l) m / z = (MW + nH)/n H = 1.008 m) c = Abs / ε ( Beer’s Law ) n) ε 280 = (# Trp) × 5,500 + (# Tyr) × 1,490
1. (20 points) Given below are the names of five recipients of a Nobel Prize (in various fields). i) (4 pts) Consider only the last names (written in capital letters). Which of these last names could also be a peptide, written in single-letter code? Explain briefly and specifically why the other last names cannot be a peptide. 1) (Ei-ichi) N E G I S H I 2) (David) B A L T I M O R E 3) (Venkatraman) R A M A K R I S H N A N 4) (Doris) L E S S I N G 5) (Gabriel Garcia) M A R Q U E Z Answer: The 6 letters in the alphabet that do not stand for an amino acid are: B, J, O, U, X, Z. So, David Baltimore (Physiology or Medicine, 1975) and Gabriel Garcia Marquez (Literature, 1982) are NOT also peptides. Ei-ichi Negishi (Chemistry, 2010), Venkataraman Ramakrishnan (Chemistry, 2009), and Doris Lessing (Literature, 2007) are also peptides (whether they know it or not). ii) (16 pts) Pick ANY ONE of the last names given above that could be a peptide. Draw the complete structure of four consecutive residues from that last name, as a tetrapeptide. Your drawing must fit the following conditions: a) The amino acid that is the first letter of the name must be at the amino terminus of your peptide. b) Use the letters (amino acids) in the order (left-to-right) that they appear in the name, but DO NOT use the same amino acid twice! Skip a repeated letter and use the next one. c) Do not use A or G (too simple!). d) Draw the peptide structure in the form that would predominate in a solution at pH = 7 . e) Write the 3-letter abbreviations of the four amino acids that you have drawn (fill in the blanks, below). The name that you chose: _____________ 3-letter abbreviations: (N-term) - _____ - _____ - _____ - _____ Answer: See the textbook for amino acid structures and 3 letter abbreviations. In addition to points deducted for an incorrect structure or incorrect protonation, points were deducted for: 1) an incorrect 3-letter abbreviation, 2) using A (alanine) and/or G (glycine) in your structure, 3) the peptide did not start with the amino acid that corresponds to the first letter in the last name, 4) a letter in the name that is an amino acid was skipped and omitted from the peptide structure. 2
2. (5 points) The table given below lists information about four proteins - proteins 1 - 4 . 1) The last row in the table gives the number of protein chains in the native form of these proteins. 2) The second-to-the-last row gives the total number of amino acid residues in one chain of each protein. 3) The rest of the table lists the number of each amino acid residue in one chain of each protein. protein 1 protein 2 protein 3 protein 4 Amino acid # # # # Alanine 83 28 26 56 Arginine 44 34 12 26 Asparagine 29 22 21 3 Aspartic acid 59 25 22 19 Cysteine 4 11 3 3 Glutamic acid 36 40 25 19 Glutamine 26 21 10 6 Glycine 60 22 27 21 Histidine 14 11 11 12 Isoleucine 41 26 19 10 Leucine 65 40 26 25 Lysine 25 25 19 5 Methionine 19 13 8 7 Phenylalanine 23 20 13 6 Proline 34 23 11 14 Serine 34 25 13 17 Threonine 48 21 17 14 Tryptophan 10 0 4 3 Tyrosine 16 24 15 3 Valine 59 38 28 24 # residues: 729 469 330 293 # chains: 1 2 6 3 Suppose that you analyzed these four proteins by polyacrylamide gel electrophoresis in the presence of SDS (SDS-PAGE). Which protein would move the farthest in the gel? Explain briefly how you decided. Answer: The individual chains of a protein separate when a protein is treated with SDS (detergent) to be run on SDS-PAGE. The mobility of proteins in SDS-PAGE depends on the molecular weight of their individual chains. The molecular weights of these proteins is correlated with the number of amino acid residues in the individual chains. Protein 4 has the fewest amino acids in each chain, so protein 4 runs farthest in the gel. 3
3. (10 points) The table given below lists information about three more proteins. The table contents are the same as for the previous question. protein A protein B protein C Amino acid # # # Alanine 83 74 77 Arginine 44 21 57 Asparagine 29 25 8 Aspartic acid 59 27 23 Cysteine 4 11 6 Glutamic acid 36 41 58 Glutamine 26 14 18 Glycine 60 50 53 Histidine 14 17 15 Isoleucine 41 32 9 Leucine 65 50 76 Lysine 25 28 13 Methionine 19 9 7 Phenylalanine 23 16 33 Proline 34 30 55 Serine 34 15 24 Threonine 48 35 19 Tryptophan 10 9 22 Tyrosine 16 16 27 Valine 59 43 45 # residues: 729 563 645 # chains: 1 1 1 Estimated charges: A B C +44 +21 +57 (Arg) ! 59 ! 27 ! 23 (Asp) ! 0.4 ! 1.1 ! 0.6 (Cys) ! 36 ! 41 ! 58 (Glu) +1.4 +1.7 +1.5 (His) +25 +28 +13 (Lys) ! 25 ! 18.4 ! 10.1 :total charge Suppose that you ran a mixture of these three proteins on an ion exchange chromatography column in which the column beads are positively-charged . Which protein would elute last from the column? Explain how you decided. Answer: The protein with the greatest net negative charge will elute last from the column. Estimate the net charge on each protein by assuming that all Arg and Lys residues are positively charged and all Asp and Glu are negatively charged, at pH 7. Each His has an average charge of about +0.1 at pH 7, and each Cys has charge of ! 0.1. The Tyr residues can be ignored because they are almost completely protonated at pH = 7 and do not contribute any significant charge. See estimates of the net charges given above. Protein A elutes last from the column since it has the greatest net negative charge. 4
4. (5 points) Here are the last four proteins. The table contents are the same as for the previous two questions. protein I protein II protein III protein IV Amino acid # # # # Alanine 59 28 47 8 Arginine 36 20 16 16 Asparagine 34 13 30 6 Aspartic acid 47 16 24 15 Cysteine 12 1 14 1 Glutamic acid 42 21 35 21 Glutamine 28 16 13 7 Glycine 72 35 34 19 Histidine 24 2 18 8 Isoleucine 56 27 24 11 Leucine 64 43 49 30 Lysine 52 18 38 13 Methionine 10 8 11 9 Phenylalanine 23 8 27 13 Proline 42 11 26 18 Serine 38 14 49 17 Threonine 45 9 24 15 Tryptophan 9 7 3 2 Tyrosine 20 6 21 11 Valine 41 18 45 17 # residues: 754 321 548 257 # chains: 1 1 1 1 Suppose you have four aqueous solutions, each of which contains only one of the four proteins. The protein concentrations are the same in the four solutions. You measure the absorbance at 280 nm (A 280 ) of each solution. Which protein solution would have the greatest A 280 ? Explain briefly how you decided. Answer: The relation of protein concentration (c) to A 280 is given by Beer’s law: c = A 280 / ε 280 . The ε 280 of a protein is due to absorbance by the Trp and Tyr residues in the protein, and Trp contributes greater absorbance than does Tyr. The solution that contains Protein I would have the greatest A 280 . 5
Your preview ends here
Eager to read complete document? Join bartleby learn and gain access to the full version
  • Access to all documents
  • Unlimited textbook solutions
  • 24/7 expert homework help