PowerPoint PresentationSlide 2Slide 3Slide 4Slide 5Slide 6Slide 7Slide 8Slide 9Slide 10Slide 11Slide 12Slide 13Slide 14Slide 15Slide 16Slide 17Slide 18Slide 19Slide 20Slide 21Slide 22Slide 23Slide 24Slide 25Slide 26Slide 27Slide 28Slide 29Slide 30Slide 31Slide 32Slide 33Slide 34Slide 35Slide 36Slide 37Slide 38Slide 39Slide 40Slide 41Slide 42Slide 43EXAM 1 Thursday, Feb 12 MCB 450On material covered inLectures 1 through 7 REVIEW FOR EXAM 1 For room assignments, please seecourse web site R1-11. Questions like homework questions in difficulty and format (all M-C)Information & Tips4. Some Qs may require calculations in order to identify correct answer option.You will need a scientific calculator for the examNo graphing or programmable calculators allowed 6. If you get stuck on a Q or are not sure of the answer, move on and do the onesyou can - then go back to the ones that caused problems.R1-23. A copy of last semester's exam has been posted7. Look at the "Learning Objectives" slides at the ends of each lecture2. The exam will have ~ 35-40 questions5. You will be give a table of pKas: Use those in your calculations,unless the question gives you another pKa value1. Amino acids: structures, properties, titration2. Isoelectric point of a peptide3. Buffer questions4. Ion exchange chromatography5. CarbohydratesThings that have come upR1-3Simple:Amide:Carboxyl group:Grouping amino acids by relationships & similaritiesAsp, Asn, Glu, GlnSerCys, MetPhe, Tyr, TrpLeuLys, Arg, His THE R-GROUPSON THESE 14 ALL START WITH CH2R1-4Hydroxyl group:Contains S:Aromatic:R1-5Branched chain hydrophobic:Basic:Imino acid:R1-6VYY-shaped side chain:V-shaped side chain:L-shapedside chainGrouping by structural relationships (of R-group)Simple: Gly, AlaBranched chain hydrophobic: Leu, Ile, ValImino acid: ProAromatic: Phe, Tyr, TrpAmide: Asn, GlnHydroxyl group: Ser, Thr, TyrCarboxyl group: Asp, GluBasic: Arg, Lys, HisContain S: CysMetNON-POLARPOLAR,UNCHARGEDPOLAR, UNCHARGEDNON-POLAR-vely CHARGED @ pH 7+vely CHARGED @ pH 7(except Y-OH,W indole N)Things to know about amino acidsR1-7KNOW THE3-LETTERABBREVIATIONSIonizable side chainsAsp, Glu carboxyl groupArg guanidiniumLys aminoHis imidazoleTyr hydroxylCys sulfhydryl Things to know about amino acidsR1-8Conservative changes in amino acids indifferent proteinsAcidic side chain D, EBasic: K, R, (H)Aromatic: F, Y*, W* *some polarityHydrophobic: I, L, V, M, F (A)Hydroxyl group: S, TAmide: N, QConsidered "similar" for construction of asequence alignment:R1-9KNOW THESINGLE-LETTERABBREVIATIONSCovalent modifications of side chainsO-linked sugars: Ser, ThrN-linked sugars: AsnPhosphorylation: Ser, Thr, TyrAcetylation: LysMethylation: ArgThings to know about amino acidsNHOCH3CR1-10H+H+fully protonated:net charge: +1 0 -1Titration of glycineISOELECTRIC POINT net charge = 0Removal of COOH proton essentially complete,Removal of NH3+ protonhas just begunTHE pI IS THE AVERAGEOF pKa1 and pKa2! 9.0 2.1 5.55R1-11Titrations of free amino acidsGeneralization:Amino acids with an R-group that does not ionize….…..have similar titration curves to Gly(just the -NH2 and -COOH titrate)R1-12net charge:+1 0 -1 -2H+H+H+Titration of glutamic acidR1-13 pKR=4.1 pK2=9.0 pK1=2.1 pKas governing formationof the +1 and -1 formsgovern formation ofuncharged glutamate, andpI =2.1 + 4.12= 3.11. Need to identify the amino acids with ionizable groups:-amino, -carboxyl, side chain2. Need to be able to find their pKas in the table provided3. Need to know what charge the ionizable group will have whenprotonated or deprotonated:protonated deprotonated-amino: NH3+(+) NH2 (0)-carboxyl, Asp/Glu carboxyl: COOH (0) COO- (-)His imidazole: N+H (+) N (0)Lys -amino: NH3+(+) NH2(0)Tyr side-chain OH: OH (0) O-(-)Arg guanidino N+H2(+) NH2(-)R1-14Determining the isoelectric point of a simple peptide-11. straightforward if the pKa's of the ionizable groups are well-separatedDetermining the isoelectric point of a simple peptide2. identify the two ionization steps that govern formation of the form ofthe peptide with no net charge (i.e. the neutral form with 0 charge)3. = ionization steps in which…the +1 charged form of the peptide is converted to the 0 charge form, and…the 0 charge form is converted to the –1 charged form.4. the pI is the average of the pKas that govern these two ionization steps5. a simple way to do this is to start with the form of the peptide with all its ionizablegroups protonated (i.e. most +vely-charged form):6. then see what happens as the pH is raised: the ionizable groups will become deprotonated in the order of ascending pKaR1-15What's the pI of Met-Asp-Gly-His?NOHNOHNOHMet & Gly side chains don't ionize, so ignoreAsppKa 4.1HispKa 6.0H3N+COOHCOOH N+HAsppKa 4.1HispKa 6.0terminal-aminopKa 8.0terminal-carboxylpKa 3.1start with the form of the peptide with all its ionizable groups protonated(i.e. most +vely-charged form):R1-16pI = (4.1 + 6.0)/2 = 5.05H3N+COOHH3N+COO-H3N+COO-H3N+COO-H2NCOOHCOOHCOO-COO-COO-COO-N+HN+HN+HNN+2+10-1-2pKa 3.1pKa 6.0pKa 8.0pKa 4.1Net chargeTHESE TWO IONIZATIONSTEPS GOVERN FORMATIONOF THE ZERO-CHARGE(= ISOELECTRIC) FORMOF THE PEPTIDE:What's the pI of Met-Asp-Gly-His?R1-17EquationsR1-18pH = -log10 [H+] [H+] [A-] [HA]Ka =pKa = -log10KapH = pKa + log10 [A-] [HA]pKa = pH - log10 [A-] [HA]a) What is the concentration of a lactic acid buffer (pKa = 3.9) that contains 2.701 M of CH3CH(OH)COOH and 0.901 M of CH3CH(OH)COO-? Give your answer using 2 decimal points .[Buffer] = [HA] + [A-]= 2.701 M + 0.901MANSWER: Final concentration of lactic acid buffer = 3.60 M b) What is the pH? Give your answer using 1 decimal point.pH = pKa + log10 [A-]/[HA]= 3.9 + log10 (0.901 M/2.701 M) = 3.9 + log10 (0.34)= 3.9 + (-0.5)ANSWER: pH = 3.4 H.H.R1-19ProblemQ. A molecule has pKa values of 7, 8, and 9. At pH equals 7.0, estimate roughly what percentage of each of the three groups is inthe deprotonated state. Use 2 decimal points.pH = pKa + log10 ([A-]/[HA])pKa 7 group:7.0 – 7.0 = log10 ([A-]/[HA]) = 0Antilog 0 = ([A-]/[HA]) = 1 (ratio: 1 part A- and 1 part HA), therefore A- is 1 part out of 2 parts total (1/2 x 100)Answer: 50.00 %pKa 8 group:7.0 – 8.0 = log10 ([A-]/[HA]) = -1Antilog -1 = ([A-]/[HA]) = 0.1 (ratio: 1 part A- and 10 parts HA) therefore A- is 1 part out of 11 parts total (1/11 x 100)Answer: 9.09 %pKa 9 group:7.0 – 9.0 = log10 ([A-]/[HA]) = -2Antilog -2 =
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