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sequencelabinfo - Chemistry 365 Biochemistry Individual Lab...

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Unformatted text preview: Chemistry 365 Biochemistry Individual Lab Unit #7 Sequence Determination Techniques for Proteins and Nucleic Acids References: Lehninger, A.L.; Nelson, D.L.; Cox, M.M. Principles of Biochemistry, 2nd ed.; Worth Publishers: New York, 1993. Berg, J.; Tymoczko, J.; Stryer. L. Biochemistry, 5th ed.; W. H. Freeman: New York, 2002. Introduction: I. Determining Protein Amino Acid Sequence Once the protein of interest has been extracted and purified, and its molar mass determined, the next step is to completely hydrolyze the protein (6 N HCl at 110oC for 24 hours) and determine its amino acid composition. Amino acids in t he h ydrolysate a re separated and identified by Ion E xchange C hromatography, S ulfonated P olystyrene R esin ion exchange Elution Profile of Amino Acids chromatography. In most cases, the next step is to identify the first a mino a cid i n t he sequence, o r i n o ther words, the protein's amino terminal. The amino terminal is typically reacted with a labelling reagent, such as dabsyl chloride, that forms a bond stable to hydrolysis. The peptide is hydrolyzed and the labeled amino acid identified. pH 3 .25 0.2 M citrate pH 4 .25 0.2 M citrate E lution V olume pH 5 .28 0.35 M c itrate E KS A M F L E g l u lys s er ala m e t p h e l e u g l u H3 C N N H3 C N S O2 Cl dabsyl chloride H3 C N H3 C O N N dabsyl-amino acid (fluorescent) RH S O N H C O O Peptides up to 50 residues long can be sequenced by a cyclic procedure where the amino terminal is labelled, cleaved, identified and the process repeated on the shortened chain. This procedure, the Edman degradation method, has been a utomated. T his m achine, c alled a sequenator, performs the reactions, separations and identifications as well as recording all results. O EDMAN DEGRADATION 1 2 3 4 O 2 N 4 2 3 4 4 Label O O 5 3 5 4 N C S H O N C C C ser ala met phe leu glu H O NH2 O Release 4 4 H 5 5 Release and Identify 3 C H Label 3 NH2 C N Release and Identify 2 ser ala met phe leu glu H O H 3 C C phenylisothiocyanate Label 2 O N C S 5 Release and Identify 1 C H2N C 3 H H 5 Label 1 C 5 O C H O C H C Repeat N C H N O C C ser ala met phe leu glu H N H S Identify NH2 Endopeptidases and Sequencing by Fragment Overlap EKS A M F L E trypsin hydrolyzes peptide bonds on C side of lys, arg glu lys ser ala met phe leu glu glu lys ser ala met phe leu glu EK resulting peptides S AMF L E chymotrypsin hydrolyzes peptide bonds on C side of phe, trp, tyr resulting peptides Deduce sequence by comparing fragments must be glu lys ser ala met phe leu glu EKS A M F L E glu lys ser ala met phe leu glu glu lys ser ala met phe glu lys leu glu ser ala met phe leu glu trypsin glu lys ser ala met phe chymotrypsin leu glu Proteins longer than 50 or so residues must be sequenced in an additional manner. The protein chain is broken into smaller fragments by site specific chemical reagents or endopeptidases such as trypsin and chymotrypsin. The resulting segments are separated and sequenced. The correct ordering of these sequences is made possible by repeating the procedure with chemical reagents or enzymes that cleave specifically at different sites than previously. This second analysis should yield a set of different fragments, which when compared to the first set and "overlapped", will allow the deduction of the complete protein amino acid sequence. Recently, the development of recombinant gene technology and DNA sequencing methods make it possible to sequence proteins via an alternative strategy. If the gene for a given protein can be isolated, and the nucleic acid sequence determined, the protein amino acid sequence can be deduced through the genetic code. II. Determining Nucleic Acid Nucleotide Sequence The development of a technique by Frederick Sanger has made it relatively easy to sequence large DNA molecule fragments. The method depends on the abilities to: 1) find two appropriate DNA primers that border the target DNA fragment 2) synthesize the complementary strand to the target utilizing the primers, DNA polymerase and deoxynucleotides DNA SEQUENCING by the SANGER (dideoxy) METHOD P P DNA Polymerase P P P P P P 5' P P P OH OH OH P PRIMER P dATP ddATP dGTP A T T T G A TEMPLATE C G A A C A T C G (Target) HO 3' P P P P P P P P 5' 3) generate four sets of la belled complementary DNA fragments, with each set generated by a specific reaction including a dideoxynucleotide, and PRIMER CTTGAGCTG A PRIMER ddATP CTTG A PRIMER PRIMER CTTGAG C PRIMER C PRIMER CTTGAGCT G PRIMER CTTGAGC T PRIMER ddGTP CT T PRIMER TEMPLATE 3' GAACTCGACT CTT G PRIMER ddCTP CTTGA G PRIMER OH 5' CT + d CTP, d GTP, d TTP, d ATP Cycle ddTTP Sequencing 4) separate labeled complementary DNA fragments, by electrophoresis, that differ in length by only one nucleotide PRIMER CTTGAGCTG A PRIMER CTTGAGCT G PRIMER CTTGAGC T PRIMER CTTGAG C PRIMER PRIMER CTTGA G CTTG A PRIMER CTT G PRIMER CT T PRIMER CT C PRIMER ELECTROPHORESIS A 10 9 8 7 6 5 4 3 2 1 C G T - + The four sets of labelled fragments are electrophoretically separated side-by-side, and the pattern of bands produced directly yields the nucleotide sequence. For example above, the labelled complementary DNA fragment, pCTTGAGCTGA, can be produced from the target pTCAGCTCAAG by four different d ideoxynucleotide reactions, each yielding a dideoxynucleotide at the 3' end of fragments that all begin at the same starting point. The "dideoxynucleotide A" electrophoresis lane will show fragments 5 and 10 nucleotides long; the "dideoxynucleotide C" lane will have fragments 1 and 7 nucleotides long; the "dideoxynucleotide G" lane will have fragments 4, 6 and 9 nucleotides long; and the " dideoxynucleotide T" lane will have fragments 2, 3 and 8 nucleotides long. The final result is that there will be one labelled complementary DNA fragment for each length up to the total number of bases, and each of these fragments ends with the base at that number position in the sequence of the labelled complementary DNA fragment. The sequence of the original target DNA fragment may now be deduced from complementary fragment using base pairing rules. Chem 365 Biochemistry Individual Lab #7 25 points # Amino Terminal Trypsin Digest Spring 2004 Chymotrypsin Digest DNA STRAND 1 E (Y)(ACIKW)(RGGE) (RGGEAW)(CIKY) 5’CTCGTCAATCATAGC3’ 2 H (S)(EK)(AHIRTWY) (WHI)(AY)(KERST) 5’CTCTAGACAGTAGCA3’ 3 T (RT)(EWY)(EFIER) (RIY)(WE)(ERTEF) 5’AGACGTGGCCTCCAT3’ 4 F (FR)(EKE)(FGITY) (F)(T)(FIG)(EKERY) 5’CCGTTCGGTAGACTC3’ 5 G (RG)(YAKS)(FARY) (F)(RA)(SYK)(YRAG) 5’TTCCAGAGGGGGTGC3’ 6 S (VAY)(WASCRGY) (V)(CY)(YRAG)(WAS) 5’GTCTATCGGGATTAA3’ 7 H (RAH)(EKE)(GREY) (KEGER)(YAREH) 5’AGCGTTGTCGTACAT3’ 8 T (FS)(RHT)(WEENR) (S)(FERN)(WHERT) 5’CGACAGGTAGCATGC3’ 9 C (ACK)(S)(WREEFA) (FA)(SER)(WACKE) 5’CGACGGGATATGGCA3’ 10 D (AKY)(LEK)(REDE) (LEAKK)(REDEY) 5’CTCGTTGATCATAGC3’ 11 D (Y)(SYK)(RAINDWF) (FN)(IYR)(KYS)(WAD) 5’GTAATCATTAGCGCA3’ 12 S (I)(KS)(IRW)(ERTY) (EIR)(KSW)(TIRY) 5’CGAATCGATGCATGC3’ 13 C (KWMI)(CEEFFAR) (F)(K)(ACF)(EEMRIW) 5’CGAGATGCCCTCTAC3’ 14 L (E)(AK)(AHLNRWWY) (HY)(NW)(ALW)(AEKR) 5’ATCTGCTAGAGCTGA3’ 15 I (S)(EEK)(CRY)(IKN) (IKNY)(CEEKRS) 5’CGCCGATAGAGCTGA3’ 16 S (IREF)(CRS)(DERW) (DIRF)(CERSW)(ER) 5’ATGGCACGCCTCCGG3’ 17 S (CRY)(EEK)(ADRSW) (RDY)(ASW)(CEEKR) 5’CTTATTGATTCGATC3’ 18 P (K)(AR)(EE)(EKY)(ARP) (AEEKR)(AEKPRY) 5’CGTGCAAGCCTCTCG3’ 19 A (AKS)(ERY)(EFIW) (E)(FI)(ERW)(AKSY) 5’CGAAGCGTTTCGCAT3’ Assignment: 1) Assign the positions of as many amino acids as possible in the peptide sequence based on the experimental results given for your problem number. Briefly explain your reasoning for each amino acid placement. 2) Draw the final electrophoresis pattern from Sanger's method using your DNA strand as the template . 3) Write down the mRNA base sequence corresponding to your DNA template strand. 4) Write down the amino acid sequence (one letter codes) of the peptide that your DNA segment encodes. 5) Determine the number and lengths of fragments obtained from each, separate restriction endonuclease digest on your DNA strand below. Make a drawing showing where the digestion fragments would appear in the two lanes of the separation gel after electrophoresis. Indicate the anode and cathode ends of the gel slab. Assume the gel slab can separate fragments from 1 to 100 nucleotide bases. a) GØGATCC 1) 2) BamHI b) GØAATTC EcoRl TGAGTCGGATCCGAATTCTGAGTCTGAGTCTGAGTCTGAGTCTGAGTCTGAGTCTGAGTCTGAGTC TGAGTCTGAGTCGAATTCTGAGTCTGAGTCTGAGTCTGAGTCTGAGTCTGAGTCTGAGTCGGATCCTGAGTCTGAGTCTGA GTCTGAGTCTGAGTCTGAGTCTGAGTC 3) TGAGTCTGAGTCTGAGTCGGATCCGAATTCTGAGTCTGAGTCTGAGTCTGAGTCTGAGTCTGAGTC 4) TGAGTCTGAGTCTGAGTCTGAGTCGGATCCGGATCCTGAGTCTGAGTCTGAGTCTGAGTCTGAGTCGAATTCTGAGTCTGA GTCTGAGTCTGAGTCTGAGTC 5) TGAGTCTGAGTCTGAGTCTGAGTCTGAGTCGAATTCGGATCCTGAGTCTGAGTCTGAGTCTGAGTCGGATCCTGAGTCTGA GTCTGAGTCTGAGTC 6) TGAGTCGAATTCTGAGTCTGAGTCTGAGTCGGATCCTGAGTCTGAGTCTGAGTC 7) TGAGTCTGAGTCGGATCCGGATCCTGAGTCTGAATTCGAGTCGGATCCTGAGTCTGAGTCTGAGTCTGAATTCGAGTCTGA GTCTGAGTCTGAGTC 8) TGAGTCTGAGTCTGAGTCGGATCCTGAGTCTGAGTCTGAGTCTGAATTCGAGTCTGAGTCTGAGTCGGATCCTGAGTCTGA GTCTGAGTCTGAATTCGAGTCTGAGTCTGAGTC 9) TGAGTCTGAGTCTGAGTCTGAGTCGGATCCTGAGTCTGAGTCTGAGTCTGAGTCTGAGTCTGAGTCGAATTCTGAGTCTGA GTCTGAGTCTGAGTCTGAGTCTGAGTC 10) TGAGTCTGAGTCTGAATTCGAGTCTGAGTCGGATCCTGAGTCTGAGTCTGAGTCTGAGTCTGAGTCGGATCCTGAGTCTGA ATTCGAGTCTGAGTCTGAGTC 11) TGAGTCTGAGTCTGAGTCTGAGTCTGAGTCTGAGTCGAATTCTGAGTCGAATTCGGATCCTGAGTCTGAGTCTGAGTCTGA GTC 12) TGAGTCGGATCCGAATTCTGAGTCTGAGTCTGAGTCGGATCCTGAGTCTGAGTCTGAGTC 13) 14) TGAGTCGGATCCGAATTCTGAGTCTGAGTCTGAGTCTGAGTCTGAGTCTGAGTCTGAGTCTGAGTC TGAGTCTGAGTCGAATTCTGAGTCTGAGTCTGAGTCTGAGTCTGAGTCTGAGTCTGAGTCGGATCCTGAGTCTGAGTCTGA GTCTGAGTCTGAGTCTGAGTCTGAGTC 15) TGAGTCTGAGTCTGAGTCGGATCCGAATTCTGAGTCTGAGTCTGAGTCTGAGTCTGAGTCTGAGTC 16) TGAGTCTGAGTCTGAGTCTGAGTCGGATCCGGATCCTGAGTCTGAGTCTGAGTCTGAGTCTGAGTCGAATTCTGAGTCTGA GTCTGAGTCTGAGTCTGAGTC 17) TGAGTCTGAGTCTGAGTCTGAGTCTGAGTCGAATTCGGATCCTGAGTCTGAGTCTGAGTCTGAGTCGGATCCTGAGTCTGA GTCTGAGTCTGAGTC 18) TGAGTCGAATTCTGAGTCTGAGTCTGAGTCGGATCCTGAGTCTGAGTCTGAGTC 19) TGAGTCTGAGTCGGATCCGGATCCTGAGTCTGAGTCTGAGTCGGATCCTGAGTCTGAATTCGAGTCTGAGTCTGAGTCTGA ATTCGAGTCTGAGTC ...
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