1CHEM205 Problem Set 1.Amino Acids, Peptides, and Protein Structure.Question 1 (6 points). A peptide has the sequenceasp-glu-his-trp-ser-gly-leu-arg-gly-lysWhat is the net charge of the molecule at pH 3, 8, and 11?Generally for negatively charged ions (R-COO-, R-S-, R-O-) the dissociationequilibrium is:R-COOH = RCOO- + H+So the acid dissociation constant is:Ka = [RCOO-][H+]/[RCOOH]We know [H+] from the pH, we know Ka from the side-chain pKa,so if we set theionic species [RCOO-] = xxacidic = (Ka/[H+])/(1 + Ka/[H+])For positively charged ions (R-NH3+, R-NH-C(NH2)=NH2+, imH+) setting the ionicspecies (eg R-NH3+) = xxbasic = 1/(1 + Ka/[H+])xpHN-ter(9.0)asp(3.9)glu(4.3)his(6.0)ser(13)arg(12.5)lys(10.5)C-ter(2.0)total3+1.000-0.112-0.048+0.9990.000+1.000+1.000-0.909+2.9308+0.909-1.000-1.000+0.0100.000+1.000+0.997-1.000-0.08411+0.010-1.000-1.0000.000-0.010+0.969+0.240-1.000-1.7912Question 2 (4 points). What is the pI of the peptide in Question 1? You can checkyour answer at http://www.embl-heidelberg.de/cgi/pi-wrapper.plH3N+(9)-asp(3.9)-glu(4.3)-his(6.0)-trp-ser(13)-gly-leu-arg(12.5)-gly-lys(10.5)-CO2-(2.0)At low pH (<1) the peptide is as positively charged as it can possible be:H3N+-asp-glu-his+-trp-ser-gly-leu-arg+-gly-lys+-CO2H total charge ~ 4+To get to a total charge of zero we need to remove protons by increasing the pH.So at pH 6:H3N+-asp--glu--his+0.5-trp-ser-gly-leu-arg+-gly-lys+-CO2-total charge ~ 0.5This is an estimate: the actual charge (calculated in the same way as for q1) is+0.527. We need to remove more protons to get to zero, so let’s go to the nexthighest pKa, which is the N-terminus (9)The total charge is about –0.5 (it is exactly –0.530).So we estimate the pI to be halfway between these values: pI = (6.0+9.0)/2 = 7.5.Check this: using the same procedure as q1, at pH 7.5 the total charge = 0.000,which is pretty good!3Question 3 (10 points). A toxin is isolated from a scorpion. Upon treatment withdithiothreitol the mass spectrum of the protein shows one peak at an averagemass of 6891 Da. Treatment of the protein with glycyl endopeptidase followed byEdman sequencing yields nine fragments. Glycyl endopeptidase cuts at the C-terminal side of glycine. Two of the fragments are too large to sequencecompletely, as indicated by the ellipsis (…)RECG, KEG, SSG, IKKG, CKLSCFIRPS…, LPNWVKVWDRA…, YCG,YCAWPACYCYG, YLMDHEGTreatment of the same protein with trypsin followed by Edman sequencing yieldsseven fragments and the amino acids K and C. Two of the fragments are toolarge to sequence completely, as indicated by the ellipsis (…)ECGIK, ATNK, GSSGYCAWPA…, EGYLMDHEGC…, PSGYCGR, LSCFIR,VWDRTreatment of the protein with cyanogen bromide gives two fragments. Edmansequencing of one of the fragments yields KEGYL-hsl (hsl is homoserinelactone). The second fragment is too large to sequence entirely, but the first tenN-terminal amino acids of this fragment are DHEGCKLSCF...What is the complete amino acid sequence of the scorpion toxin?There is only one M, because CNBr cleavage gives only 2 fragments. The N-terminal fragment will have a C-terminal homoserine lactone, so we start withKEGYL -Glycyl endopeotidase cuts after G, so we should have a fragment YL… in theglycyl endopeptidase digest. Trypsin cuts after R and K so there should be afragment EGYL… in the trypsin digest. The longest fragment is EGYLMDHEGC..from the trypsin digest. So we now haveKEGYLMDHEGC- but we don’t know what comes after the CGlycyl endopeptidase cuts between the G and the C, so there should be afragment that starts with C in the glycylendopeptidase digest. And there is:CKLSCFIRPS…, which gives usKEGYLMDHEGCKLSCFIRPS…Trypsin cuts between R and P, so we’re looking for a fragment in the trypsindigest that starts with PS: it’s PSGYCGR, which givesKEGYLMDHEGCKLSCFIRPSGYCGRThe complete sequence is KEGYLMDHEGCKLSCFIRPSGYCGRECGIKKGS-SGYCAWPACYCYGLPNWVKVWDRATNKC which weighs 6891 Da.4Question 4 (10 points). The following sequence is a stable, monomeric a-helix atpH 7:AEAAAKEAAAKEAAAKEAAAKADraw this sequence on the helical wheel diagram below (start top center), andindicate the favorable electrostatic interactions that stabilize the structure.Favorable electrostatic interactions occur between same-color amino acid sidechains (E- and K+)Draw the same sequence on a helical wheel diagram as a 310 helix. Do youexpect this sequence to adopt a stable 310 helix?AAAAAAEKAAA E AE A KA K AK A E12AAKAAEAAE A A K A E AK A A E A K A1Now we’re looking for K-Einteractions vertically above andbelow each other. The onlyinteractions we have in the 310helix are separated by A, sothey’re too far apart to besignificant. We wouldn’t expectthe 310 helical arrangement ofthis sequence to be stable.5Question 5 (10 points). The following amino acid sequence adopts an a-helicalcoiled-coil motif. Use helical wheel diagrams to predict whether the coiled-coil isparallel or antiparallel.RMKQLEDKLEELESKLYHLENKLARLEKAntiparallel arrangement showing the leu/met residues forming the hydrophobicinterface. In this orientation the Ks are electrostatically repulsive, as are the Es.Parallel arrangement showing the leu/met residues forming the hydrophobicinterface. In this orientation the Ks and Es are electrostatically attractive.KKKRKEYAEEEELMR H E QK N S D1L L LL L LRKKKEEEELLR H E QA Y E KK N S DL L LM L LNCKKKRKEYAEEEELMR H E QK N S D1L L LL L LEEEERKKKLMK E Y AR H E QK N S DL L LL L LNN6Question 6 (12 points). Consider the following amino acid sequence:AELQAKSAIAHELQAKSAIAHAWhen treated with ATP in the presence of a kinase, the serine side-chains arephosphorylated as follows:At pH 5 the phosphorylated sequence is an a-helix. Use a helical wheel diagramto indicate the stabilizing interactions of the phosphorylated sequence.The ser now has a PO42- group, making it look more like a glu with 2 –vecharges. There’s an electrostatic attraction to the nearby his, which is 90%protonated at pH5.At pH 8, the non-phosphorylated sequence exhibits predominantly b-structure.Use a zig-zag illustration like that in Figure 6.11 to indicate whether you expectthe structure to form parallel or antiparallel b-sheets.Amino acids in the blue circles are below the plane of the backbone. Amino acidsin black circles are above the plane of the backbone:NHHNOHOONHHNOOOP O-O-OATP, kinaseADPAAAALASKLAA E IE I KA H QH Q