CHMI 2227 EL Biochemistry I Test 1 26 January 2007 Prof : Eric R. Gauthier, Ph.D. Guidelines: 1) Duration: 55 min 2) 14 questions, on 7 pages. For 70 marks (5 marks per question). Worth 15 % of the final mark. 3) Multiple choice questions: use the answer sheet provided to you. 4) The use of the University-approved calculator is allowed. 5) The use of notes or books is strictly forbidden. 6) Good Luck!
For each of the following question, select the one, best answer. 1) What will be the ph of a solution made of 0.1M acetate and 0.3 M acetic acid (Ka = 1.76 x 10-5 M)? a) 4.28 b) 4.15 c) 5.23 d) 7.76 e) 4.24 What we have here is a buffer solution with the following acid-base equilibrium: Acetic acid acetate H The ph is easily obtained using the Henderson-Hasselbach equation: ph = pka log (conjugated base / acid) The pka value will be : -log [Ka] = -log 1.76 x 10-5 M = 4.75 Hence : ph = 4.75 log (0.1M / 0.3M) = 4.75 (-0.48) = 4.272 ~ 4.28 2) From the structure of the 5 amino acids shown below, indicate which one is glutamine? A B C D E 2
3) Using the following molecular mass values: H=1 g/mol, C=12g/mol, N=14 g/mol, O=16 g/mol, S= 32 g/mol, give the exact molecular mass of amino acid R at ph 13.5. a) 175 g/mol b) 174 g/mol c) 173 g/mol d) 172 g/mol e) 176 g/mol R is the one-letter abbreviation of arginine, the structure of which is shown below: The pka values for various ionizable groups in arginine are as follows: 1) COOH : ph 1.8 2) NH 2 (side chain): 12.5 3) NH 2 (bound to Cα) : 9 Therefore, at ph 13.5, all these groups will be deprotonated (see structure). The molecular mass is then easily calculated using the values provided for each individual atoms. COO - CH- NH 2 ( ) 3 NH C=NH NH 2 4) You are working hard to characterize a peptide composed only of the amino acid L- leucine. You hydrolyze the peptide with HCl and perform the quantification of the amino acid with ninhydrin (absorbance at 570 nm). The data obtained for the quantification, along with the results of a standard curve made with L-leucine are shown in the Table below. Determine the amount (in µg/ml) of leucine present in your sample. a) 8.01 µg/ml b) 9.32 µg/ml c) 9.98 µg/ml d) 10.51 µg/ml e) 11.01 µg/ml Leucine concentration (µg/ml) Absorbance 1 0,092 2 0,117 4 0,167 8 0,267 16 0,467 Leucine in sample 0.300 3
Here, all we have to do is use the values provided in the table to draw a standard curve of the absorbance at 570 nm as a function of the Leucine concentration: 0,5 0,45 Absorbance 570 nm 0,4 0,35 0,3 0,25 0,2 0,15 0,1 y = 0,025x 0,067 0,05 0 0 2 4 6 8 10 12 14 16 18 Leucine concentration (µg/ml) All that remains to be done is take the absorbance value obtained for the unknown sample (0.3) and use the standard curve to find the corresponding leucine concentration. 5) Starting with 1 mole of leucine, what is the exact amount (in moles) of the zwitterion and acid form of the amino acid at ph 4? a) 0.99 moles Zwitterion and 0.01 moles acid b) 0.96 moles Zwitterion and 0.04 moles acid c) 0.95 moles Zwitterion and 0.05 moles acid d) 0.98 moles Zwitterion and 0.02 moles acid e) 0.97 moles Zwitterion and 0.03 moles acid 4
What we have here is an acid-base equilibrium involving the COOH group of leucine (it s the only group in this amino acid that can be ionized at ph 4 see the list of pka values). We have the following equilibrium : COOH CH-NH 3 CH-CH 3 CH 3 pka = 2.3 COO - CH-NH 3 CH-CH 3 CH 3 Zwitterion We are given the pka (2.3) ph (4) values, and we are looking for the relative amounts of the acid (COOH) and zwitterion (aka conjugated base) of Leu. The Henderson- Hasselbach equation can therefore be used: ph = pka log ([conjugated base] / [acid]) ph- pka = log ([conjugated base] / [acid]) 4-2.3 = 1.7 = log ([conjugated base] / [acid]) ([conjugated base] / [acid]) = 50.1 ~ 50 Since we have a total amount of 1 mole of leucine in our solution, we can easily calculate the proportion of each form: 1 mole / 50 = 0.02 mole of the acid form And 1 mole 0.02 mole = 0.98 mole of the zwitterion. 6) What amino acid can be modified by a protein kinase: a) W b) F c) H d) T e) C 5
Protein kinases are enzymes which catalyze the addition of a phosphate group onto amino acids. The addition of this phosphate group always occurs on a hydroxyl group. Therefore, only Thr, Tyr or Ser can be targeted by protein kinases. 7) The absorption coefficient for the amino acid Trp is 5690 M -1 cm -1. What is the concentration of this amino acid in a solution that gives an absorbance value of 0.458 at 280nm (cuvette size = 1 cm)? a) 8.05 x 10-5 M b) 2,6 x 10 3 M c) 4.1 M d) 1.2 x 10 4 M e) Impossible to know without a standard curve. From the absorbance value and the extinction coefficient, we can easily calculate the concentration of Trp using the Beer-Lambert equation: A = εcl c = A / (εl) c = 0.458 / (5690 M -1 cm -1 x 1 cm) c = 8.05 x 10-5 M 8) What is the approximate pi (range of ph) of the peptide Ala-Tyr-His-Asp-Glu-Lys? a) 2-4 b) 4-6 c) 6-9 d) < 2 e) >9 6
We first have to find the pka values for each ionizable group of this peptide: 3 HN-CH-CO-NH-CH-CO-NH-CH-CO-NH-CH-CO-NH-CH-CO-NH-CH-COO - CH 3 ( ) 2 ( ) 4 NH OH COO - COO - NH 3 NH Ala Tyr His Asp Glu Lys Therefore : Ala has only one ionizable group (NH 3 terminus; pka 9.9) Tyr does not possess an ionizable group in this peptide His has only one ionizable group (side chain; pka 6.0) Asp has only one ionizable group (side chain; pka 3.9) Glu has only one ionizable group (side chain; pka 4.1) Lys has two ionizable groups (side chain; pka 10.5 / COO - terminus; pka 2.2) The pi is the ph value (or ph interval) where the peptide is neutral. In this case, this is only possible when the ph value is between 4 and 6 (as shown in the structure above). 9) Which of the peptide structures shown below is the correct one at ph 6? A O O 3H N-CH-C-NH-CH-C-NH-CH-COOH CH 3 ( ) 4 NH 3 COOH B O O 3 HN-CH-C-NH-CH-C-NH-CH-COO - CH ( ) 4 3 NH 2 COO - C O O 2HN-CH-C-NH-CH-C-NH-CH-COOH CH 3 ( ) 4 NH 2 COOH D O O 2HN-CH-C-NH-CH-C-NH-CH-COO - CH ( ) 4 3 NH 3 COO - 7
E O 3H N-CH-C-NH-CH-C-NH-CH-COO - O CH ( ) 4 3 NH 3 COO - This peptide is made of the amino acids Ala-Glu-Lys. The ionizable groups at play here are the following: Ala : NH 3 terminus; pka 9.9 Glu : COOH (side chain); pka = 4.1 Lys : NH 3 (side chain) (pka = 10.5) ; COOH terminus (pka = 2.2) Therefore, at ph 6, the peptide will have the structure shown in E. 10) This is the only amino acid with no stereoisomer: a) F b) G c) C d) L e) W Answer : The Cα of glycine is not chiral and, therefore, Gly will not have stereoisomers. 8
11) Which amino acid possesses a sulfur group that cannot be oxidized: a) Met b) Asp c) Cys d) Asn e) His Methionine has a sulfur atom covalently bonded to two methyl groups and, therefore, cannot be oxidized or reduced. 12) Which one of these amino acids will be neutral at ph 3, but not at ph 5: a) Lys b) Glu c) Gln d) Asn e) His At ph 3,the COOH group of the side chain of Glu is protonated (pka 4.1), while its COOH bonded to the Cα is ionized(pka 2.1). Since the pka value for the NH 3 group is equal to 9.5, it will be protonated at ph 3. However, at ph 5, the COOH group of the side chain will be ionized. Hence, Glu will be neutral at ph 3 and ionized at ph 5. 13) The bond linking two amino acids together to form a peptide is called: a) The conjugated bond b) The hydrogen bond c) The disulfide bond d) The peptide bond e) The covalent bond 9
14) In the peptide shown below, which amino acid is at the C-terminal end? a) A b) T c) L d) H e) I O 3H N-CH-C-NH-CH-C-NH-CH-COO - O CH H CH-CH 3 CH 3 CH 3 CH 3 10