Etiket Arşivleri: Problem Set
Alcohol vapor is being absorbed from a mixture of alcohol vapor and water vapor by means of a nonvolatile solvent in which alcohol is soluble but water is not. The temp. is 97 ºC, and Pt=760 mmHg. The alcohol vapor can be considered to be diffusing through a film of alcohol-water- vapor mixture 0.1 mm thick The molal concentration of the alcohol vapor at the outside of the film is 80 %, and that on the inside is, next to the solvent is, 10 %. D Et-OH- water =0.15 cm 2 /sec at 25 ºC and 1 atm. Calculate the rate of diffusion of alcohol vapor in pounds per hour if the area of the film is 100 ft 2 .
BIOCHEMISTRY I (CHMI 2227 E) PROBLEMS and SOLUTIONS Eric R. Gauthier, Ph.D. Department of Chemistry and Biochemistry January 2007
Note: This problem set has been prepared for students taking the course Biochemistry I (CHMI 2227E), as offered at Laurentian University. It contains several problems taken from textbooks and from the author’s imagination. While the vast majority of the problems found in this book can be relatively easily solved with the help of the class notes, more difficult questions have also been included. Questions marked by a star (*) will require more work from the student. As for the questions labeled with two stars (**), they constitute a good challenge to any student interested in tackling them. After the « Problems » section, the complete, detailed solution for every question is found. For obvious reasons, we strongly encourage students to look at the solutions only as a last resource. The list of pKas and pI for the 20 natural amino acids, as well as the table of the genetic code, can be found after the “Problems” section. The following texts were consulted while writing this manual: 1) Kuchel, P. W. and Ralston, G. B. Biochimistry. Schaum Series. McGraw-Hill. 1989. 2) Lehninger, A. L., Nelson, D. L., Cox, M. M. Principles of Biochemistry . 2nd édition. Worth Publishers. 1993. 3) Mathews, C. K. et van Holde, K. E. Biochemistry . 2nd édition. Benjamin/Cummings Publishing Company, INC. 1996. 4) Rawns, J. D. Biochemistry . Editions du renouveau pédagogique. 1990. 5) Wood, W. B., Wilson, J. H., Benbow, R. M., Hood, L. E. Biochemistry. A Problems Approach . Benjamin/Cummings Publishing Company, INC. 1981. 6) Zubay, G. L., Parson, W. W., Vance, D. E. Principles of Biochemistry . Wm. C. Brown Publishers. 1995. More problems and questions can be found in these and other references.
Chapter 1: Acid-Base Equilibrium and Spectrophotometry 1.1 Acid-Base Equilibrium : What is the pH of the following solutions? a) 0.35 M hydrochloric acid b) 0.35 M acetic acid (pKa = 4.76) c) 0.035 M acetic acid. 1.2 Acid-Base Equilibrium : A weak acid, HA, has a total concentration of 0.20M and is ionized (dissociated) to 2%; a) Calculate the Ka for this acid. b) Calculate the pH for this acidic solution. 1.3 Acid-Base Equilibrium : Calculate the pH of the following mixtures: a) 1M acetic acid and 0.5M sodium acetate b) 0.3M phosphoric acid and 0.8M KH PO (pKa=2.14) 2 4 1.4 Acid-Base Equilibrium : You need to prepare a buffer solution at pH = 7.00 with KH PO and Na HPO (pKa=7.21). If 2 4 2 4 you use a 0.1M solution of KH PO , what would be the concentration of Na HPO needed? 2 4 2 4 1.5 Acid-Base Equilibrium : You need to prepare a buffer solution at pH = 7.00 with KH PO and Na HPO . What would be 2 4 2 4 the respective concentration of these substances if you wished to obtain a final phosphate concentration ([HPO -2] + [H PO -1]) of 0.3M? 4 2 4 1.6 Spectrophotometry : What is the concentration of the amino acid tyrosine (ε=1 420 L mol-1 cm-1) if you obtain an absorbance of 0.71 with a 1 cm cuvette? With a 0.1 cm cuvette? 1.7 Spectrophotometry : What would be the absorbance reading of a 37 mM solution of tyrosine? 1.8 Spectrophotometry : You wish to determine the concentration of haemoglobin in a blood sample by spectrophotometry. You first create a standard curve of the absorbance at 412 nm of several solutions of known haemoglobin concentrations. The data for the standard curve is shown below. What is the concentration (in µg/mL) of haemoglobin in your sample if the absorbance obtained at 412 nm was 0.303?
Concentration of Absorbance standard solution (412nm) (µg/ml) 0.069 1 0.113 2 0.201 4 0.377 8 0.730 16 Chapter 2: Amino acids * 2.1. Molecular mass of an amino acid. 1.812 g of a crystallized α-amino acid (pKa1: 2.4; pKa2; 9.7) has a pH of 10.4 when dissolved in 100 mL of 0.1M NaOH. Calculate the molecular mass of this amino acid. 2.2. Titration curve Calculate the pI of histidine and draw its titration curve. Indicate the position of all pKas and the pI as well as the percentages of each ionic form at the start and finish of the titration and at all pKas. The list of pKas for all 20 amino acids can be found at the end of the “Problems” section of this problem set. 2.3. Net charges of amino acids What is the net charge (+, 0, -) of the amino acids glycine, serine, aspartic acid, glutamine and arginine at: a) pH 2.01 b) pH 3.96 c) pH 5.68 d) pH 10.76 2.4. Ionic exchange chromatography A mixture of lysine, glycine, alanine, isoleucine and glutamic acid are separated by ionic exchange chromatography. What is the order of elution of these amino acids if you use gradient buffer system from pH 10 to pH 2: a) with a cation exchange resin? b) with an anion exchange resin? Which column would give the best separation? 2.5. Amino acids What amino acids can be converted into another amino acid with gentle hydrolysis, resulting in release of ammonia?
2.7. Amino acids Phosphoserine is found after enzymatic hydrolysis of casein, a milk protein. However, it does not belong to the 20 amino acids coded during protein synthesis. Give a plausible explanation. CH2-CH2-CH-COOH O NH2 -2 PO3 Phosphoserine *2.8. Ionic exchange chromatography Glycine, alanine, valine and leucine can be successfully separated by ionic exchange chromatography even though their pKas are almost identical. Explain the behaviour of these amino acids. 2.9. Peptides. A peptide is hydrolyzed and its amino acid content analyzed. Hydrolysis destroys the amino acid tryptophan, therefore the content of tryptophan can be estimated with spectrophotometry. Establish the empirical formula of the peptide with the following information. Amino acids mmol Ala 2.74 Glu 1.41 Leu 0.69 Lys 2.81 Arg 0.72 Trp 0.65 2.10. Peptides. Draw the structure of the following peptide GWYQR. Indicate the ionic form of the peptide at the following pH: a) pH 2.0 b) pH 7.0 c) pH 10.5
Chapter 3. General properties and purification of proteins 3.1.Protein Purification Why do we often use ammonium sulphate precipitation in initial purification steps of proteins? 3.2. Protein Purification DEAE cellulose columns are rarely used at pH greater than 8.5. Why? 3.3. Protein Purification 6-phosphogluconate dehydrogenase has a pI of 6. Explain why the buffer used for a chromatography on DEAE-cellulose must have a pH greater than 6 but less than 9 in order to ensure the enzyme is efficiently bound to the column. 3.4. Protein Purification. Would the enzyme, 6-phosphogluconate dehydrogenase bind to a CM-cellulose resin if the same conditions as the previous problem were used? Why? 3.5. Protein Purification. What pH would the buffer need to be in order to permit the dehydrogenase in the previous problem to bind to the CM-cellulose resin? 3.6. Protein Purification. We load a DEAE-cellulose column adjusted to a pH of 6.5 with the following mixture of proteins: ovalbumin (pI = 4.6), urease (pI = 5.0), and myoglobin (pI = 7.0). The proteins are eluted first with a buffer of weak ionic strength at a pH of 6.5, and then the same buffer containing increasing amounts of sodium chloride is used to elute the proteins. What order are the proteins eluted? 3.7. Protein Purification. An enzyme (MW 24 kDa, pI 5.5) is contaminated with two other proteins, one with a similar molecular mass and a pI of 7.0 while the other has a molecular mass of 100 kDa and a pI of 5.4. Suggest a procedure to purify the contaminated enzyme. 3.8. Protein Purification. A procedure used to purify 6-gluconate dehydrogenase from E. coli is presented below. a) Calculate (1) the specific activity, (2) the percent yield based on the initial quantity of the enzyme and (3) the degree of purification for each step (i.e. fold increase in purification). b) Indicate which step purifies the protein the most. c) Assuming the protein is pure after gel permeation chromatography (on Bio-Gel A), what percent of the initial extract contained 6-gluconate dehydrogenase?
Total protein Enzymatic activity Purification step Volume (mL) (mg) (µg/min) 1- Cellular extract 2 800 70 000 2 700 2- Ammonium sulfate 3 000 25 400 2 300 3- Heat denaturation 3 000 16 500 1 980 4- DEAE 80.00 390.00 1 680 chromatography 5- CM-cellulose 50.00 47.00 1 350 chromatography 6- Bio-Gel A 7.00 35.00 1 120 chromatography 3.9 Protein Purification. Why is SDS omitted when proteins need to undergo isoelectric focusing? 3.10. Protein Purification. A series of proteins with known molecular mass and an enzyme of unknown molecular mass are separated by chromatography on a Sephadex G-200 column. The elution volume (Ve) for each protein is indicated in the table below. Estimate the molecular mass of the unknown protein. Protein Mr V (mL) e Blue dextran 1 000 kDa 85.00 lysozyme 14 kDa 200.00 Chymotrypsinogen 25 kDa 190.00 ovalbumin 45 kDa 170.00 Serum albumin 65 kDa 150.00 aldolase 150 kDa 125.00 urease 500 kDa 90.00 ferritin 700 kDa 92.00 ovomucoid 28 kDa 160.00 unknown ? 130.00 *3.11. Protein Purification. Referring to the previous problem, give a plausible explanation for the bizarre behaviour ferritin’s elution from the sephadex column
3.12. Protein Purification. A student isolates a protein from anaerobic bacteria and analyses the protein by polyacrylamide gel electrophoresis containing SDS (PAGE-SDS). Following protein staining, a single band appears, which excites the student’s supervisor. To be certain, the supervisor suggests that the student run a second electrophoresis under native conditions (i.e. non-denaturing, or without SDS). This gel shows two bands after staining. Assuming no errors were committed during these experiments, explain the observations. 3.13. Protein Purification. A student from CHMI 2227 analyses bovine serum albumin (BSA) with a polyacrylamide gel electrophoresis (PAGE-SDS). During the experiment, the student forgets to add β- mercaptoethanol to the sample. When comparing his sample to those of his classmates he realizes that the molecular mass of his BSA sample determined by PAGE-SDS is 57 kDa, while all the other students (those that added β-mercaptoethanol) found a molecular mass of 68 kDa. Explain this difference. 3.14. Polypeptide sequencing Consider the following peptide: A-L-K-M-P-E-Y-I-S-T-D-Q-S-N-W-H-H-R Indicate the fragments generated after the following digestions : a) trypsin b) pepsin c) protease V8 d) cyanogen bromide 3.15 Polypeptide sequencing Deduce the polypeptide sequence that generated the following results: a) acid hydrolysis: (Ala , Arg , Lys , Met, Phe, Ser ); 2 2 2 b) Carboxypeptidase A digestion: Ala; c) Trypsin digestion: (Ala, Arg) (Lys, Phe, Ser) (Lys) (Ala, Met, Ser) d) cyanogen bromide treatment: (Ala, Arg, Lys2 , Met, Phe, Ser) (Ala, Ser) e) thermolysine digestion: (Ala) (Ala, Arg, Ser) (Lys2 , Met, Phe, Ser)
3.16. Polypeptide sequencing A polypeptide is reduced by β-mercaptoethanol to yield two peptide fragments with the following sequences : fragment 1: A-C-F-P-K-R-W-C-R-R-V-C fragment 2: C-Y-C-F-C The non-reduced polypeptide is digested with thermolysine and yields the following fragments : (A,C,C,V) (R,K,F,P) (R,R,C,C,W,Y) (C,C,F) Indicate the positions of disulfide bridges in the polypeptide. 3.17. Polypeptide sequencing An analysis of the polypeptide Shawi isolated from the bacteria Chretientus negativii, yields the following results : a) acid hydrolysis: (Ala , Val, Lys , Arg, Gly, Asp, Met, Pro, Trp) 4 2 b) carboxypeptidase digestion: Lys c) dinitrofluorobenzene treatment: Val d) cyanogen bromide treatment: generates two polypeptides: peptide A: (Gly, Arg, Trp, Asp, Lys, Ala); Treatment of this peptide with DNBF and carboxypeptidase yields : DNFB: Gly Carboxypeptidase: Lys peptide B: (Ala3 , Lys, Val, Met, Pro); Treatment of this peptide with DNFB and carboxypeptidase yields: DNFB: Val Carboxypeptidase: Met e) trypsine digestion: yields three peptides peptide C: (Lys, Trp, Ala); Treatment of this peptide with DNFB and carboxypeptidase yields : DNFB: Trp peptide D: (Ala3 , Val, Lys, Pro) peptide E: (Met, Asp, Gly, Arg); Treatment of this peptide with DNFB and carboxypeptidase yields : DNFB: Met
Finally, treating peptide D with thermolysine yields the following: Val Ala Ala (Ala, Lys, Pro) What is the primary structure of this peptide? Chapter 4. Three dimensional structures of proteins 4.1. 3-D Structures of proteins What amino acids among the following would you expect to find a) inside, and b) at the surface of a typical globular protein in an aqueous solution of pH 7? Glu Arg Val Phe Ileu Asn Lys Ser Thr 4.2. 3-D Structures of proteins According to the structure of urea, deduce how this compound can promote denaturation of proteins. 4.3. 3-D Structures of proteins Phenylalanine, a hydrophobic amino acid, is frequently found at the surface of natives and functional proteins. Give the most probable role of phenylalanine in this situation. *4.4. 3-D Structures of proteins Aspartic acid, a charged amino acid, is frequently found inside of native and functional proteins. Give the most probable role of phenylalanine in this situation. 4.5. 3-D Structures of proteins The following table describes the amino acid compositions of three proteins. Number of residues per molecule Amino acids protein 1 protein 2 protein 3 Polar residues Arg 12.00 4.00 7.00 Asn 9.00 6.00 5.00 Asp 14.00 5.00 9.00
Number of residues per molecule Amino acids protein 1 protein 2 protein 3 Cys 7.00 2.00 6.00 Gln 8.00 7.00 6.00 Glu 11.00 4.00 6.00 His 4.00 2.00 4.00 Lys 22.00 6.00 15.00 Ser 20.00 5.00 11.00 Thr 15.00 3.00 11.00 Trp 2.00 3.00 3.00 Tyr 7.00 7.00 6.00 Non-polar residues Ala 14.00 28.00 25.00 Gly 9.00 9.00 8.00 Ileu 5.00 16.00 9.00 Leu 3.00 19.00 7.00 Met 7.00 11.00 9.00 Phe 9.00 13.00 11.00 Pro 8.00 13.00 10.00 Val 16.00 29.00 21.00 Knowing that protein A has a rod-like form, protein B is a monomeric globular protein, and protein C is a globular protein with four identical sub-units, deduce the corresponding amino acid composition of these proteins. 4.6. 3-D Structures of proteins Indicate which secondary structure or structures (α -helix, β -pleated, random coil) will the following peptide adopt in an aqueous solution at pH 7 Ileu-Glu-Asn-Glu-Gln-Asn-Met-Ala-His-Phe-Trp-Tyr 4.7. 3-D Structures of proteins Indicate which secondary structure or structures (α -helix, β -pleated, random coil) will the following peptide adopt in an aqueous solution at pH 7 Gly-Ala-Gly-Ala-Gly-Ser-Gly-Ala-Gly-Ser-Gly-Ala 4.8. 3-D Structures of proteins Indicate which secondary structure or structures (α -helix, β -pleated, random coil) will the following peptide adopt in an aqueous solution at pH 7 Lys-Gly-Arg-Arg-Lys-Gly-Arg-Gly-Arg-Pro 4.9. 3-D Structures of proteins Indicate which secondary structure or structures (α -helix, β -pleated, random coil) will the following peptide adopt in an aqueous solution at pH 7 1 10 Gly-Pro-Glu-Ser-Ala-Tyr-Lys-Thr-Leu-Phe-Asp-Val-Pro-Asp-Asp-Glu-Asp-Gly-Gly
20 26 Ser-Ala-Gly-Ser-Ser-Gly-Ala 4.10. 3-D Structures of proteins The following table describes the amino acid composition of three proteins. Determine what structure these proteins will adopt: α-helical, β-pleated or a triple helix of collagen. protein A B C protein A B C Ala 29.40 5.00 10.70 Leu 0.50 6.90 2.40 Arg 0.50 7.20 5.00 Lys 0.30 2.30 3.40 Asp 1.30 6.00 4.50 Met – 0.50 0.80 Cys – 11.20 – Phe 0.50 2.50 1.20 Glu 1.00 12.10 7.10 Pro 0.30 7.50 12.20 Gly 44.60 8.10 33.00 Ser 12.20 10.20 4.30 His 0.20 0.70 0.40 Trp 0.20 1.20 – Hypro – – 9.40 Tyr 5.20 4.20 0.40 Ileu 0.70 2.80 0.90 Val 2.20 5.10 2.30 Chapter 5. Enzymology 5.1. Enzymatic kinetics With the following enzyme activity results determine: a) Vmax -3 b) why is the velocity v constant at [S] greater than 2 x 10 M? c) what is the free [E] at [S] = 2 x 10-2 M? [S] (mol/L) v (μmol/min) 2 x 10-1 60.00 2 x 10-2 60.00 2 x 10-3 60.00 2 x 10-4 48.00 1,5 x 10-4 45.00 1,3 x 10-5 12.00 5.2. Enzymatic kinetics The results for enzyme activity analysis can be found below. Without using a graph, determine : a) Vmax; b) Km; c) initial velocity at [S] = 1 x 10-1 M;
d) the amount of product formed during the first 5 minutes at [S] = 2 x 10-3 M. At a [S] of 2 x 10- 6 M? e) what is Km and Vmax if the free [E] is increased by a factor of 4? [S] (mol/L) v (μmol/min) 5 x 10-2 0.25 5 x 10-3 0.25 5x 10-4 0.25 5x 10-5 0.20 5 x 10-6 0.07 5 x 10-7 0.01 5.3. Enzymatic kinetics The following table describes the results from an enzymology experiment. Using a Lineweaver- Burke plot determine: a) Km; b) Vmax; [S] (mol/L) v (μmol/min) 1 x 10-3 65.00 -4 63.00 5 x 10 -4 51.00 1x 10 5x 10-5 42.00 3 x 10-5 33.00 2 x 10-5 27.00 1 x 10-5 17.00 5 x 10-6 9.50 1 x 10-6 2.20 5 x 10-7 1.10 5.4. Enzymatic kinetics We study the effect of pH on the enzymatic activity of 6-phosphogluconate dehydrogenase. This enzyme catalyzes the reaction: 6-phosphogluconate + NADP 6- phosphogluconic acid + NADPH2 NADPH absorbs light at 340 nm. The activity of the dehydrogenase is measured 2
spectrophotometrically by monitoring the absorbance (A) at 340nm, which is proportional to the concentration of NADPH . 2 4 Increase in A Increase in A at pH [S] x 10 M at pH 7.6 9.0 0.174 0.074 0.034 0.267 0.085 0.047 0.526 0.098 0.075 1.666 0.114 0.128 4.000 – 0.167 At what pH will the enzyme have more affinity for the substrate? 5.5. Enzymatic kinetics The following results describe the effect of an inhibitor on enzyme activity of an enzyme. Determine: a) Vmax in the presence and the absence of an inhibitor b) Km in the presence and the absence of an inhibitor c) Ki d) type of inhibition [S] (mol/L) Without inhibitor With inhibitor -4 v (μmol/min) [I] = 2,2 x 10 M v (μmol/min) 1 x 10-4 28.00 17.00 1,5 x 10-4 36.00 23.00 2x 10-4 43.00 29.00 5x 10-4 65.00 50.00 7,5 x 10-4 74.00 61.00 5.6. Enzymatic kinetics A biochemist studies the properties of a metabolic enzyme she has just isolated. She obtains kinetic data in the presence and in the absence of two different inhibitors (A and B). The identity of the inhibitors is unknown but we know that one of these is an substrate analog while the other is an alkylating agent.
Determine: a) Km and Vmax of the enzyme ; b) which inhibitor is the substrate analog? Which is the alkylating agent? c) Ki for both inhibitors; -4 d) what would be the Vo for this enzymatic reaction at [S] = 3 x 10 M and in the presence of the -5 inhibitor [A] = 2 x 10 M? [S] (mol/L) Without inhibitor With inhibitor A With inhibitor B v (µmol/min) [I] = 5 x 10-4 M [I] = 3,2 x 10-6 M v (µmol/min) v (µmol/min) 5 x 10-4 1.25 0.82 0.48 2,5 x 10-4 0.87 0.49 0.33 1,7 x 10-4 0.67 0.36 0.25 1,2 x 10-4 0.54 0.26 0.20 1 x 10-4 0.45 0.23 0.17 5.7. Enzymatic catalysis The effect of pH on the activity of an enzyme is demonstrated in the following graph : ) % ( y t i v i t c a e m y z n E pH How would you explain the effect of pH on enzyme activity? 5.8. Enzyme catalysis Several enzymes show a dependance on pH similar to the one shown in the previous problem. However, the optimal pH varies a great deal from one enzyme to another. What side chains would you expect to find on active sites of enzymes if the optimal pH is: a) pH 4 b) pH 11
5.9. Allosteric enzymes We study the kinetic properties of two enzymes (A and B). From the results shown below, determine if they constitue an ordinary enzyme or an allosteric enzyme. Explain the shape of the curves representing the velocity, v, in relation to the concentration of substrate, [S]. 3 v (enzyme A) v (enzyme B) [S] (x 10 M) (μmol/min) (μmol/min) 0.00 0.00 0.00 0.50 8.80 0.30 1.00 14.00 1.00 2.00 19.00 4.70 3.00 21.50 12.40 4.00 22.80 19.00 5.00 22.30 21.80 6.00 23.50 22.80 8.00 23.60 23.30 Chapter 6. Structure and properties of nucleic acids. 6.1. Nucleic acid structure. Consider the following polynucleotide: AUUACGUGGUGCACUCGGGAACAUCCCGAGUGCACCACGUAAUGGA Draw the two most stable intramolecular secondary structures this polymer can adopt *6.2. Nucleic acid structure. A solution of double stranded DNA is heated and then cooled to room temperature for two minutes. Predict, qualitatively, the variation in absorbance at 260 nm in the following conditions: a) the solution is heated to a temperature slightly above Tm before being cooled; b) the solution is heated to a temperature way above Tm before being cooled; c) suggest the structure of two polynucleotides (synthetic or natural) which will result in an absorbance profile following a cooling which is the perfect inverse of the pattern obtained in (b). 6.3. Nucleic acid structure. Explain why, RNA, and not DNA, is hydrolyzed under basic pH conditions.
6.4. Nucleic acid structure. The following results were obtained during a denaturation/renaturation experiment of a simple nucleic acid (polyA :polyU). How would you interpret these results? Solution cooled rapidly ) m n 0 6 2 ( e c n Solution cooled slowly a b r o s b A o Temperature ( C) Tm 6.5. Nucleic acid structure. IMP (inosine monophosphate) is present in chez E. coli as an intermediate of biosynthesis of purines and it is possible to incorporate IMP to DNA if the ITP (inosine triphsophate) is present in the reaction medium. However, in nature, IMP is never present in DNA. Propose an explanation. 6.6. Nucleic acid structure What are the products of the digestion of the oligoribonucleotide 5’pACGAUGCUAUC3′ by each of the following enzymes: a) pancreatic ribonuclease; b) T2 ribonuclease; c) T1 ribonuclease; 6.7. Nucleic acid structure Lets proceed to the analysis of an RNA molecule. Its global base composition is 2A, 2C, 1U, 1G. Its treatment with the serpent venom phosphodiesterase yields pC. Its hydrolysis by pancreatic ribonuclease yields 1C, a dinucleotide containing A and C, and a trinucleotide containing A, G, and U.
The action of RNase T2 yields pAp, a dinucleotide containing U and C and a trinucleotide containing A, G and C. What is the primary structure of this RNA? 6.8. Nucleic acid structure Let’s proceed to the analysis of an RNA molecule whose global base composition is 2A, 4C, 2G, 1U. Pancreatic ribonuclease treatment yields 2Cp, two dinucleotides, one containing G and C and the other containing A and U, and a trinucleotide containing A, C and G. A mixture of RNase T1 and RNase T2 yields C, Ap, pGp and two trinucleotides, one containing A and C and the second containing CG and U. The serpent venom phosphodiesterase yields pC. What is the formula of this RNA? 6.9. Nucleic acid structure What is the global charge of the trinucleotide ApGpUpC at neutral pH? 6.10. Nucleic acid structure Why does a circular double stranded DNA renature more rapidly than a linear double stranded DNA? 6.11. Nucleic acid structure Why does DNA denature in pure water, that is where the ionic strength is close to zero? 6.12. Nucleic acid structure The size of the E. coli chromosome is 4000 kpb. What length of DNA does it contain? 6.13. Nucleic acid synthesis. During an experiment similar to that performed by Meselson and Stahl, you grow bacteria for 3 14 generations (instead of 2 as in the classic experiment) in a mixture containing only N. Following DNA isolation and analysis by analytical centrifugation, what proportion of heavy DNA, hybrid DNA and light DNA will you obtain? 6.14. Nucleic acid synthesis. An isolated strand (+) of DNA (base composition: 10% of A, 20% of G, 30% of C and 40% of T) is replicated by E. coli DNA polymerase into a complimentary starnd (-). The double- stranded DNA is then used as a model for the E. coli RNA polymerase which transcribes the (-) strand. Indicate the base composition of the formed strand (in % of A, C, G of T/U).
*6.15. Nucleic acid synthesis. The time required to completely synthesise the E.coli genome is 40 minutes. However, it takes only 20 minutes for these bacteria to produce one generation. Can you explain this paradox? **6.16. Nucleic acid synthesis. You are the first scientist to successfully analyze a micro-organism found on Mars. Because this bacterium contains double-stranded DNA as genetic material, you decide to analyze using Meselson-Stahl techniques. You obtain the following results: 14 Generations after N transfer LL HL HH 0 1 2 a) how would you interpret these results? b) in order to better understand this phenomenon, you isolate the components implicated in DNA replication in this organism. You identify : – a RNA polymerase activity; – a DNA polymerase which functions only on single-stranded; – a new enzyme which can generate a product sensitive to DNAse in the presence of NADH and a product insensitive to DNase and resistant to heat. According to this information, deduce the mechanism by which this micro-organism replicates its DNA. 6.17. mRNA and transcription Differently than DNA polymerase, RNA polymerase does not proofread and edit its products. a) Why does this absence of proofreading/correction in the synthesis of RNA not threaten the cells’ viability? b) How would an enzyme using RNA as a template for DNA synthesis modify the rate of mutations for an organism? *6.18. mRNA and transcription The great majority of mRNAs have a very short half life – in the order of 3 minutes in bacteria. What caused evolution to form mRNA molecules so unstable?
6.19. mRNA and transcription If RNA polymerase lengthens RNA at a speed of 35 to 70 nucleotides per second and if each molecule of polymerase binds to 70 base pairs of DNA : a) What is the maximum speed of transcription per minute where a gene of 6000 base pairs is transcribed into RNA molecules? b) What is the maximum number of molecules of polymerase that could be found bound to this gene at any given time? 6.20. Protein coding Consider the following mRNA: AGU CUC UGU CUC CAU UUG AAG AAG GGG AAG GGG a) indicate the amino acid sequence which would be coded (read from 5’ to 3’). The table containing the genetic code can be found in the appendix. b) you obtain mutations which consist of additions or deletions of one nucleotide. If we insert G between the third and forth nucleotide, and we eliminate the 10th nucleotide from the right (it is a G), what would be the peptide sequence? 6.21. Protein coding. The amino acid sequence from part of lysozyme isolated from a wild type and a mutant bacteriophage T4 is given below: wild type: -Tyr-Lys-Ser-Pro-Ser-Leu-Asn-Ala-Ala-Lys- mutant: -Tyr-Lys-Val-His-His-Leu-Met-Ala-Ala-Lys- a) can this mutant be the result of a change in a single base pair in the DNA of phage T4? If not how was this mutant produced? b) what is the base sequence of the mRNA which codes for the five amino acids in the wild type which are different than those of the mutant type? 6.22. Protein coding. A strand of DNA has the following sequence: 5′ TCGTTTACGATCCCCATTTCGTACTCGA 3′ a) what is the sequence of its complementary strand? b) what is the base sequence of mRNA transcribed from the first strand? c) what is the coded amino acid sequence?
d) what is the coded amino acid sequence if the second T from the 3’ end of the DNA is deleted? 6.23. Genetic engineering. Give the restriction fragments obtained following digestion of the following nucleic acid with the enzyme EcoR I: 5’ATGCTCGATCGATCGAATTCTATAGCCCGGGGCTGGATCCAGGTACCAAGTTAAGCTTG3’ 3’TACGAGCTAGCTAGCTTAAGATATCGGGCCCCGACCTAGGTCCATGGTTCAATTCGAAC5’ 6.24. Genetic engineering. Give the restriction fragments obtained following digestion of the following nucleic acid with the enzyme BamHI: 5’ATGCTCGATCGATCGAATTCTATAGCCCGGGGCTGGATCCAGGTACCAAGTTAAGCTTG3’ 3’TACGAGCTAGCTAGCTTAAGATATCGGGCCCCGACCTAGGTCCATGGTTCAATTCGAAC5’ 6.25. Genetic engineering. Give the restriction fragments obtained following digestion of the following nucleic acid with the enzyme Sma I: 5’ATGCTCGATCGATCGAATTCTATAGCCCGGGGCTGGATCCAGGTACCAAGTTAAGCTTG3’ 3’TACGAGCTAGCTAGCTTAAGATATCGGGCCCCGACCTAGGTCCATGGTTCAATTCGAAC5’ 6.26. Genetic engineering. Give the restriction fragments obtained following digestion of the following nucleic acid with the enzyme KpnI and Hind III: 5’ATGCTCGATCGATCGAATTCTATAGCCCGGGGCTGGATCCAGGTACCAAGTTAAGCTTG3’ 3’TACGAGCTAGCTAGCTTAAGATATCGGGCCCCGACCTAGGTCCATGGTTCAATTCGAAC5’ 6.27. Genetic engineering. You want to map the genome of the λ bacteriophage (a double stranded linear DNA). To accomplish this, you label the genome of phage λ (total length of 48 500 bp) at the 5’ end with a radioactive phosphorous (32P). You then digest the marked genome with different restriction enzymes under conditions which will permit partial digestion of the DNA. You analyze the resulting fragments by agarose electrophoresis and then visualize the bands with autoradiography. The results are shown in the table below. a) Calculate the length of each restriction fragment obtained. b) Create the restriction map of the λ phage. DNA standard Apa I Pvu I BamH I Length (bp) Distance Distance Distance Distance migrated (cm) migrated (cm) migrated (cm) migrated (cm) 23 130 3.5 2.76 2.76 2.76 9 416 4.1 4.12 3.02 2.89
6 557 4.5 3.29 3.06 4 361 4.9 3.98 3.24 2 320 5.25 3.43 2 027 6.15 4.65 560 6.7
pKas and pI Values for Common Amino Acids pKa1 pKa2 pKR pI G 2,34 9,60 5,97 A 2,34 9,69 6,01 V 2,32 9,62 5,97 L 2,36 9,60 5,98 I 2,36 9,68 6,02 P 1,99 10,6 6,48 F 1,83 9,13 5,48 Y 2,20 9,11 10,07 5,66 W 2,83 9,39 5,89 S 2,21 9,15 13,60 5,68 T 2,63 10.43 13,60 5,87 C 1,71 10.78 8.33 5,07 M 2,28 9,21 5,74 N 2,02 8,80 5,41 Q 2,17 9,13 5,65 D 2.09 9,82 3,86 2,77 E 2,19 9,67 4,25 3,22 K 2,18 8,95 10,79 9,74 R 2,17 9,04 12,48 10,76 H 1,82 9,17 6,00 7,59
The genetic code Base at 5′ Central bases Base at 3′ ↓ U C A G ↓ Phe Ser Tyr Cys U Phe Ser Tyr Cys C U Leu Ser Stop Stop A Leu Ser Stop Trp G Leu Pro His Arg U Leu Pro His Arg C C Leu Pro Gln Arg A Leu Pro Gln Arg G Ile Thr Asn Ser U A Ile Thr Asn Ser C Ile Thr Lys Arg A Met Thr Lys Arg G Val Ala Asp Gly U Val Ala Asp Gly C G Val Ala Glu Gly A Val Ala Glu Gly G
Acid-Base Equilibrium and Spectrophotometry 1.1 Acid-base equilibrium : a) Since HCl is a strong acid, it will completely dissociate when in solution: + – HCl Æ H + Cl Stoichiometry tells us that, since the initial HCl concentration is 0.35M, the final concentration of H+ in the solution will also be 0.35M. This gives us: + pH = – log[H ] = -log 0.35 = 0.46 b) Acetic acid will also dissociate in solution: – + CH -COOH CH3-COO + H 3 However, since it is a weak acid, it will not completely dissociate, and we have to take into account the association constant (Ka) in our calculations. This constant is described as follows: pKa = – log Ka Ka = 1/10pKa = 1.74 x 10-5M We can now easily determine the H+ concentration: + – Ka = [H ] [CH3COO ] [CH3COOH] -5 + – 1.74 x 10 M = [H ] [CH3COO ] 0.35 M -5 + – + 2 1.74 x 10 M x 0.35M = [H ] [CH3COO ] = [H ] + -6 2 1/2 -3 [H ] = (6.09 x 10 M ) = 2.47 x 10 M Finally: pH = – log [H+] = – log 2,47 x 10-3 = 2.61 c) Following the same steps as in (b), we get a pH of 3.11.
1.2 Acid-base equilibrium : a) We have a weak acid. The acid-base equilibrium is: + – HA H + A We can determine Ka as follows: + – Ka = [H ] [A ] [HA] The question stipulates that this acid is only 2% ionised (or 0.02, that’s the same thing). + – This allows us to obtain the respective concentrations of species HA, H and A : + – [H ] = [A ] = 0.20M x 0.02 = 0.004M + [HA] = 0.2M – [H ] = 0.196 M Therefore : + – Ka = [H ] [A ] [HA] Ka = 0.004M x 0.004M 0.196 M -5 Ka = 8.16 x 10 M b) The pH of this solution is: pH = – log [H+] = – log 0.004M = 2.39 1.3 Acid-base equilibrium: a) This mixture is a buffer solution made of acetic acid and its conjugated base, sodium acetate: + – + CH COOH H + CH COO Na 3 3 This pH of this type of solution can be determined with the Henderson-Hasselbach equation: pH = pKa + log [Conjugated base] [Acid]
pH = 4.76 + log 0.5M = 4.46 1 M b) We have the following acid-base equilibrium : + – + H PO H + H PO K 3 4 2 4 Using the same procedure as in (a), we get: pH = pKa + log [H2PO4-] [H PO ] 3 4 pH = 2.14 + log 0.8 M 0.3 M pH = 2.57 1.4 Acid-base equilibrium : We have the following equilibrium: – + -2 H2PO4 H + HPO4 And the pKa for this equilibrium is 7.21. The Henderson-Hasselbach equation gives us: pH = pKa + log [HPO4-2] [H2PO4-] 7.00 = 7.21 + log [x] 0.1 M -0.21 = log x – log 0.1 M -0.21 + log 0.1 M = log x = -1.21 x = 10logx = 0.062 M 1.5 Acid-base equilibrium: We have the following acid-base equilibrium: – + -2 H2PO4 H + HPO4
According to the question, we have: [H2PO4-] + [HPO4-2] = 0.3M – -2 Hence: [H2PO4 ] = 0.3 M – [HPO4 ] From the Henderson-Hasselbach equation, we have: pH = pKa + log [HPO4-2] [H PO -] 2 4 7,00 = 7.21 + log [HPO4-2] [H PO -] 2 4 7,00 = 7.21 + log [HPO4-2] [H2PO4-] -0.21 = log [HPO4-2] [H PO -] 2 4 -0.21 -2 10 = [HPO4 ] [H PO -] 2 4 -2 0.616 = [HPO4 ] [H2PO4-] – -2 0.616 x [H2PO4 ] = [HPO4 ] Which is identical to : 0.616 x (0.3M – [HPO4-2]) = [HPO4-2] -2 -2 0.185 – 0.616 x [HPO4 ] = [HPO4 ] -2 0.185 = 1.616 x [HPO4 ] -2 0.114 M = [HPO4 ] And the concentration in H PO – will be : 2 4 – -2 [H2PO4 ] = 0.3M – [HPO4 ] = 0.186 M 1.6 Spectrophotometry The relationship between the absorbance and the concentration of a solution is given by the Beer-Lambert equation: A = εcl
Where: A = absorbance ε = Molar extinction coefficient (units: litres x mol-1 x cm-1) c = concentration (units : mol/l = M) l = light path (thickness of the cuvette; units: cm) We get the following: 0.71 = 1.420 L mol-1 cm-1 x c x 1 cm c = 0.71 mol cm 1420 L x 1 cm -4 c = 5 x 10 M If we use a cuvette where c=0.1 cm, we get: 0.71 = 1.420 L mol-1 cm-1 x c x 0.1 cm c = 0.71 mol cm 1420 L x 0.1 cm c = 0.005 M 1.7 Spectrophotometry With the Beer-Lambert equation, we have: A = εcl Therefore: A = 1420 L mol-1 cm-1 x (37 x 10-3M) x 1cm A = 52.54 1.8 Spectrophotometry We first have to graph the standard cuve of the absorbance as a function of the concentration of the haemoglobin standards. This graph is shown on the following page. Since the unknown has an absorbance of 0.303, we can use the standard curve to determine the corresponding haemoglobin concentration, in this case 6.31µg/mL. A more accurate value can be obtained by using the familiar equation: y = mx + b where y = value on the y axis x = value on the x axis m = slope
b = intersect on the y axis The values for m and b are easily obtained from the graph or by linear regression of the data (the latter being, by far, the best method). Therefore, we get: y = (0.0441 mLµg-1)x + 0.0246 x = (y-b)/m x = (0.303-0.0246)/0.0441mLµg-1 = 6.31 µg/mL. 0.8 0.7 0.6 e c 0.5 n a b A = 0.303 r o s 0.4 b A 0.3 0.2 6.31 µg/mL 0.1 0 0 2 4 6 8 10 12 14 16 18 Haemoglobin concentration (µg/mL)
Chapter 2: Amino acids * 2.1. Molecular mass of an amino acid. By adding NaOH, we shift the acid-base equilibrium of the amino acid towards the base: NH + – – -CH(R)-COO NH -CH(R)-COO 3 2 With the Henderson-Hasselbach equation, it is possible to obtain the proportion of the amino acid in the base and acid form after the addition of NaOH: pH = pKa + Log [base] / [acid] 10.4 = 9.7 + Log [base] / [acid] 0.7 = Log [base] / [acid] 5.011 = [base] / [acid] Since 0.01 mol of NaOH (100 mL of a 0.1M solution) was required to obtain a pH value of 10.4, this indicates that 0.01 mol of this amino acid has been converted to the basic form. Therefore: 5,011 = [base] / [acid] 5,011 = 0,01 mol /[acid] [acid] = 0,002 mol The sum of the amount of amino acid in the base and acid forms will give us the total amount of amino acid in the solution, which is 0,012 mol. All we have to do next is determine the molecular mass: 1,812 g = 0,012 mol Therefore, the molecular mass of the amion acid is: g / mol = 1,812 g /0,012 mol = 151 g / mol 2.2. Titration of amino acids. The titration of histidine involves the following equilibria: + + H N H N N N CH – CH – COOH CH – CH – COO CH – CH – COO 2 2 CH2 – CH – COO 2 + + HN NH HN NH + HN NH 3 3 HN NH3 2 pKa1 pKaR pKa2
To draw the titration curve, we need the values for the pKas and the inflexion points. For histidine, the pKas are as follows: pKa1: 1.82 pKaR: 6.00 pKa2: 9.17 By definition, the pKa is the pH where half the the amino acid lost a proton, while the other half is still protonated. The inflection points are obtained by the average value of two consecutive pKas: Inflection poin # 1: (1.82 + 6.00) / 2 = 3.91 Inflection point # 2: ( 6.00 + 9.17) / 2 = 7.59 The pI is defined as the pH where the net charge of the amino acid is 0. It is determined by calculating the average of the pKas preceding and following the inflexion point where the amion acid carries no net charge. For histidine, the pI will be 7,59 (inflection point #2). We can now easily draw the titration curve (see next page). 2.3. Net charge of amino acids. To determine the net charge of amino acids at various pH values, we first need to determine the pI for each amino acid With this information, we can deduce the charge of the amino acids : by definition, the amino acid carries no net charges with the pH equals the pI. When the pH is lower than the pI, the amino acid will carry a net positive charge. When the pH is superior to the pI, then the amino acid carries a net negative charge. Therefore, we obtain the following: pH Glycine Serine Aspartic Acid Glutamine Arginine (pI: 5.97) (pI: 5.68) (pI: 2.77) (pI: 5.65) (pI: 10.76) 2.01 + + + + + 3.96 + + – + + 5.68 + 0 – 0 + 10.76 – – – – 0
Problem 2.2. pKa2=9.17 9.0 pI = 7.59 N 7.5 100% CH – CH – COO 2 HN NH3+ pKaR=6.0 H 6.0 p + H N 3.91 100% CH2 – CH – COO HN NH3+ pKa1=1.82 1.82 0.5 1.0 1.5 2.0 2.5 3.0 – Equivalents of OH ions + H N 50% CH – CH – COOH + 2 H N HN NH3+ 50% CH – CH – COO 2 pKa1 + H N + HN NH3 50% CH – CH – COO pKaR 2 N HN NH3+ 50% CH – CH – COO 2 HN NH3+ N CH – CH – COO 50% 2 HN NH3+ pKa2 N CH – CH – COO 50% 2 HN NH 2
2.4. Ion exchange chromatography. We start the chromatography at pH 10. At this particular pH, all the amino acids in the mixture are negatively charged (the pH greater than the pI of each amino acid). a) using a cation exchange chromatography, none of the amino acids will stick to the resin and they will all be found in the eluent. b) using an anion exchange chromatography, all the amino acids will bind the resin. As we decrease the pH, the amino acids will progressively elute when the pH of the buffer becomes lower that their pI. We will collect the amino acids in this order: 1- lysine 2- isoleucine, alanine, glycine (all three will elute pretty much at the same time since their pI is similar) 3- glutamic acid. Obviously, the anion exchange resin is the best choice for this experiement. 2.5. Amino acids Only two amino acids can be converted to other amino acids and at the same time generate ammonia : asparagine and glutamine: H O NH O 2 3 O C-CH -CH-COOH C-CH -CH-COOH 2 2 H N HO 2 H N 2 H N 2 ASN ASP H O NH O 2 3 O C-CH -CH -CH-COOH C-CH -CH -CH-COOH 2 2 2 2 H N HO 2 H N H N 2 2 GLN GLU *2.6. Amino acids This observation can only be explain if we assume that serine is first inserted into casein, and that a phosphate group is then added one protein synthesis is over, producing phosphoserine. The experimental data confirm this hypothesis.
*2.7. Ion exchange chromatography. Since these four amino acids can be separated by ion exchange chromatography, but that their pI are virtually identical, another physico-chemical property must be responsible for this behaviour. A close look at the structure of glycine, alanine, valine and leucine reveals a progressive increase in the hydrophobic character of their side chain. We can deduce that these amino acids can establish hydrophobic interactions with the ion exchange resin, allowing their separation. 2.8. Peptides. Since the yield in leucine, arginine and tryptophan is similar, we can conclude that they are present in equal propotions in the polypeptide. Twice this amount was obtained as glutamate. Finally, alanine and lysine were recovered in proportions corresponding to four times the amount of arginine/leucine/tryptophan. Therefore, the empirical formula of this peptide would be: (Arg, Leu, Trp, Glu , Ala , Lys ) 2 4 4 n This experiment does not allow us to determine the amino acid sequence of this peptide. 2.9. Peptides. The structure of the peptide GWYQR (Glycyl-Tryptophanyl-Tyrosyl-Glutaminyl-Arginine) is: O O O O H N-CH -C-NH-CH-C-NH-CH-C-NH-CH-C-NH-CH-COOH 2 2 CH CH CH (CH ) 2 2 2 2 3 CH NH 2 C C NH NH O NH2 OH NH2 To determine the form of this peptide at each pH, all we have to do is note the pKa values of all the charged groups (including the side chains), and to deduce the ionization status of each of these groups (pHpKa: non-protonated): At pH 2 : Every group is protonated: O O O O + H N-CH -C-NH-CH-C-NH-CH-C-NH-CH-C-NH-CH-COOH 3 2 CH CH CH (CH ) 2 2 2 2 3 CH2 NH C C NH + 2 NH O NH2 NH OH 2
At pH 7 : The carboxyl group of arginine is ionized (pH>pKa). However, the side chain of arginine and the amino group of glycine remain protonated (pH<pKa): O O O O + H N-CH -C-NH-CH-C-NH-CH-C-NH-CH-C-NH-CH-COO 3 2 CH CH CH (CH ) 2 2 2 2 3 CH2 NH C C NH2+ NH O NH2 NH OH 2 At pH 10.5 : Here, the amino group of glycine and the side chain of tyrosine are deprotonated (pH>pKa). However, the amino group of the side chain of arginine remains protonated (pH<pKa): O O O O H N-CH -C-NH-CH-C-NH-CH-C-NH-CH-C-NH-CH-COO 2 2 CH CH CH (CH ) 2 2 2 2 3 CH2 NH + C C NH2 NH O – NH2 NH O 2 Chapter 3. General properties and purification of proteins 3.1. Protein purification. Ammonium sulphate precipitation enables the concentration of our favorite protein by the non- specific precipitation of a large proportion of the proteins of the extract.We can therefore easily obtain a partial purification of our favorite protein, which can then be purified further using other methods. 3.2. Protein purification. + The diethylamino group (-CH -CH -NH -CH -CH ) of DEAE-cellulose carries a positive charge 2 2 2 3 which is responsible for the ion-binding properties of this resin. Effectively, negatively charged amino acids/proteins will interact with the diethylamino group (via electrostatic interactions), while positively charged amino acids/proteins will be eluted. Since the diethylamino group has a pKa close to 8.5, it will be deprotonated at pH values above 8.5 and will use all ability to bind negatively charged molecules. 3.3. Protein purification. At a pI above 6, 6-phosphogluconate dehydrogenase has a net negative charge (pH>pI): it will bind the resin. At a pH value above 9, the diethylamino group of the resin is deprotonated, preventing any separation of the enzyme as a function of its charge. 3.4. Protein purification. No because CM-cellulose (CM = carboxymethyl = -CH2-COOH) is a cation-exchange resin: at a pH above 6, 6-phosphogluconate dehydrogenase is negatively charged (see problem 3.3) and will not bind the resin. 3.5. Protein purification. In order to separate 6-phosphogluconate dehydrogenase using a CM-cellulose column, we must ensure that the protein has a net positive charge. The pH of the buffer will have to be below the protein’s pI, therefore below a value of 6. 3.6. Protein purification. The biochemist has 2 choices when it comes to eluting proteins bound to ion exchange resins: use a pH gradient, or a salt gradient (usually NaCl). In the situation where a salt gradient is chosen, one would usually start with a buffer of low NaCl concentration (i.e. low ionic strength), and then progressively introduce a buffer with a greater and greater salt concentration. The Na+ – or Cl ions will elute proteins from the column by neutralizing the negative or positive charges on the proteins which interact with the resin. The end result is the elution of proteins as a function of their charge density : to be eluted from a – cation exchange resin, those proteins with less positive charges will require less Cl ions (thus, a lower NaCl concentration) to neutralize them than those proteins with more positive charges. + The same logic can be used to explain the ability of Na ions to elute proteins bound to anion exchange resins (e.g. CM-cellulose). Regarding question 3.6, we can conclude that: – myoglobin will elute first, since it will not bind to the resin (pH
contaminating protein (Mr 100 kDa). Note: Similar results would be obtained if one were to first perform the size exclusion chromatography and then the ion exchange chromatography. 3.8. Protein purification. a) Specific activity is defined as the enzymatic activity per mg protein and is an indication of the relative concentration of the enzyme in the solution (the greater the amount of enzyme in the solution relative to all other proteins, the greater the specific activity will be). All we have to do is to divide the enzymatic activity obtained at each step by the corresponding protein concentration. For example, for the heat treatment step, we get: specific activity = enzymatic activity / mg protein specific activity = 1 980 U/16 500 mg = 0,12 U/mg The percent yield is definec as the amount of enzyme (or protein) recovered at each step with reference to the amount present at the start of the purification procedure. This value is obtained by dividing the enzymatic activity at step X by the initial enzymatic activity. Again, for the heat treatment step, we get Percent yield = Enzyme activity step X / Initial enzyme activity Percent yield = (1 980 U/2 700U) x 100 = 73,3 % The degree of purification is obtained by dividing the specific activity after step X by the initial specific activity. We therefore obtain the fold increase in enzyme purification after the purification procedure. For the heat treatment step, we get: Degree of purification = Specific activity step X/ Initial specific activity Degree of purification = 0,12/0,039 = 3,07 fold. The results for each step of the purification procedure are shown in the table below: Purification step Specific activity Percent yield Degree of (U/mg) purification (fold increase) Cell extract 0,039 100% — Ammonium 0,09 85,2% 2,30 sulphate Heat treatment 0,12 73,3% 3,07
Purification step Specific activity Percent yield Degree of (U/mg) purification (fold increase) DEAE chromato. 4,31 62,2% 110,50 CM-cellulose 28,72 50% 736,40 chromato. Bio-Gel A 32,00 41,5% 820,50 b) To determine which purification step was the most effective, all one has to do is divide the value for the degree of purification of step X by the value for the preceding step. Thus, the DEAE chromatography step was the most efficient, with a 36 fold increase in enzyme purity. c) Considering that the protein is pure after molecular sieve chromatography, 35 mg of 6- gluconate dehydrogenase were obtained. This corresponds to 41,5 % of the amount of enzyme initially present in the extract (refer to percent yield). Thus, in the initial extract we had: 35 mg / 41,5% = 75,9 mg And this amount of enzyme was initially present in an extract containinfg a total of 70 000 mg proteins. Thus, in the initial extract we had: (75,9 mg / 70 000 mg) x 100 = 0,108 % of 6-gluconate dehydrogenase 3.9. Protein purification. The aim of isoelectric focusing is to separate proteins according to their pI, thus according to their charge at different pH values. Adding SDS to the protein sample would give all proteins the same charge density and would prevent their separation by this type of electrophoresis. 3.10. Protein purification. To determine the molecular mass of our unknown protein, we must first draw the graph of the elution profile of our sandards as a function of the log of their molecular mass (see graph on following page). Protein Log Mr Vel (ml) dextran blue 6.000 85.00 Lysozyme 4.146 200.00 Chymotrypsinogen 4.398 190.00 Ovalbumin 4.653 170.00 serum albumin 4.813 150.00 aldolase 5.176 125.00
Protein Log Mr Vel (ml) urease 5.699 90.00 Ferritin 5.845 92.00 ovomucoide 4.447 160.00 From this graph, we obtain a molecular mass of 139 kDa for our unknown protein. Note : ferritin and ovomucoid were excluded of the standard curve because they obviously behaved differently to the other proteins when subjected to molecular sieve chromatography. Graph, problem 3.10 Lysozyme 200 Chymotrypsinogen 180 Ovalbumin 160 e n o Ovomucoide m i t u u l Albumin l o é 140 ‘v d n e Aldolase o i m t u l u 120 l o V E Ferritin 100 Log Mr =5.14 Urease Dextran Mr = 139 kDa 80 60 4 4.2 4.4 4.6 4.8 5 5.2 5.4 5.6 5.8 6 Log masse moléculaire Log molecular mass *3.11. Protein purification. Ferritin has an iron core, giving it a greater density than other proteins of similar size and influencing its behaviour when subjected to molecular sieve chromatography. 3.12. Protein purification. In the presence of SDS, all the proteins have an identical negative charge density: this is what
allows us to use PAGE-SDS to determine the molecular mass of proteins. Therefore, two proteins of different pI but identical molecular mass will co-migrate as a single band on PAGE- SDS. However, in the absence of SDS (thus under native or non-denaturing conditions), protein migration towards the positive or negative electrodes will be driven by its net charge, in other words by the protein’s pI. In this case, two proteins of identical molecular mass but different pI will give two distinct bands upon gel staining. 3.13. Protein purification In the absence of β-mercaptoethanol, the protein’s disulfide bonds remain intact. Thus, the protein will have a more compact shape and will migrate more rapidly during PAGE-SDS than the same protein whose disulfide bonds have been reduced. 3.14. Peptide sequencing. a) Trypsin hydrolyzes the peptide bond on the carboxyl-side of the basic amino acids lysine and arginine. Therefore, every peptide fragment generated by trypsin will have Arg or Lys at their C- terminus (with the exception, of course, of the fragment corresponding to the C-terminal end of the peptide). Therefore, the peptide given in this example will give us the following fragments: A-L-K M-P-E-Y-I-S-T-D-Q-S-N-W-H-H-R b) Pepsin hydrolyzes the peptide bond on the N-terminal side of the aromatic amino acids Phe, Trp, and Tyr. Therefore, the fragments obtained after pepsin digestion will all contain Tyr, Phe or Trp at their N-terminus (with the notable exception of the fragment corresponding to the N- terminus of the initial peptide). Using the peptide shown here, we obtain the following fragments: A-L-K-M-P-E Y-I-S-T-D-Q-S-N W-H-H-R c) Protease V8 hydrolyzes the peptide bond on the C-terminal side of the acidic amino acids Asp and Glu. Therefore, every peptide fragment generated by protease V8 will have Asp or Glu at their C-terminus (with the exception, of course, of the fragment corresponding to the C-terminal end of the peptide). Therefore, the peptide given in this example will give us the following fragments: A-L-K-M-P-E Y-I-S-T-D Q-S-N-W-H-H-R d) Cyanogen bromide hydrolyzes the peptide bond on the C-terminal side of Met. Therefore, every peptide fragment generated by cyanogen bromide will have Met at their C-terminus (with the exception, of course, of the fragment corresponding to the C-terminal end of the peptide). Therefore, the peptide given in this example will give us the following fragments: A-L-K-M P-E-Y-I-S-T-D-Q-S-N-W-H-H-R
3.15. Peptide sequencing Digesting with carboxypeptidase A tells us that the C-terminal residue of the peptide is Ala. Digesting with trypsin allows us to partially order two of the four fragments (remember: trypsing generates fragments whose C-terminal end is Arg or Lys): Ala-Arg (Phe, Ser)-Lys Furthermore, since digesting with trypsin generates a free Lys residue, this indicates that this Lys is on the C-terminal side of either Arg or Lys. Trypsin digestion also indicates that the tripeptide (Ala, Met, Ser) is the C-terminus of the peptide (it doesn’t end with Arg or Lys). Furthermore, CNBr digestion allows us to determine the position of Met in this tripeptide: Met-(Ala, Ser) Since Ala is the C-terminal residu of this peptide (see digestion with carboxypeptidase A), the sequence of the last 3 amino acids of the polypeptide will be: Met-Ser-Ala Thermolysin cuts the peptide bond on the N-terminal side of hydrophobic amino acids: we can therefore deduce the position of the hydrophobic amino acids of the two fragments: Ala-(Arg, Ser) and Phe-(Lys, Lys,)-Met-Ser With this information, and considering the trypsin digestion pattern, we can conclude that the sequence of this polypeptide is: Ala-Arg-Ser-Phe-Lys-Lys-Met-Ser-Ala 3.16. Peptide sequencing Thermolysin cuts the peptide bond on the N-terminal side of hydrophobic amino acids. Digesting the two fragments after reduction of the difulphide bonds gives: fragment 1: A-C F-P-R-K W-C-R-R V-C fragment 2: C Y-C F-C Since the disulfide bonds of the peptide were intact when the peptide was digested with thermolysin, some of the peptide fragments shown above will be linked together with disulphide bonds involving Cys residues. From the fragments obtained, we can deduce the position of the disulphide bonds as follows :
S-S A-C-F-P-K-R-W-C-R-R-V-C S – S C-Y-C-F-C – S S – Note : Don’t forget that both inter-chain and intra-chain S-S- bonds can be present in the molecule. 3.17. Peptide sequencing Digestion with carboxypeptidase tells us that the C-terminus is Lys. DNFB treatment indicates that the N-terminus is Val. Trypsin digestion allows us to partially order the peptides, as follows: peptide C: Try-Ala-Lys peptide D: Val-(Ala, Ala, Ala, Pro)-Lys (remember: Val = N- terminus) peptide E: Met-(Asp, Gly)-Arg We can order the rest of the residues with the results from CNBr treatment: Met-Gly-Asp-Arg Finally treating peptide D with thermolysine allows us to order the three Ala and the Pro: Val-Ala-Ala-Ala-Pro-Lys Therefore, the sequence of the peptide is: Val-Ala-Ala-Ala-Lys-Pro-Met-Gly-Asp-Arg-Try-Ala-Lys Chapter 4. Three dimensional structures of proteins 4.1. 3-D structure of proteins Generally speaking, hydrophobic amino acids are found buried inside proteins (away from water), while polar and charged amino acids are most often found on the surface of proteins. We will then get the following distribution for the amino acids : Buried inside: Val, Phe, Ileu On the surface: Glu, Arg, Asn, Lys, Ser, Thr
FE 356 PROBLEM SET NO 2
Macroscopic Energy Balances