Problem Set 3 Answers.doc - MCB102 - Problem Set 3 This...

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MCB102 - Problem Set 3 This ungraded problem set covers the material discussed in Lectures 8-10 (and retro to lecture 4). See if you can work out the answers yourself, before looking at the answers that will be posted in a few days. Also, try the problems at the end of Chapters 5-6 in the Lehninger textbook. If you need help, talk to your GSIs or to me. 1. From the following experimental data, deduce the amino acid sequence of the corresponding polypeptide. Acid hydrolysis of the whole peptide yielded the amino acid composition: Ala x 2, Arg, Lys x 2, Met, Phe, Ser x 2. Carboxypeptidase A digestion yielded only Ala. Trypsin digestion produced fragments with the following amino acid composition: (Ala, Arg); (Lys, Phe, Ser); (lys); and (Ala, Met, Ser). CNBr digestion yielded fragments with the following amino acid composition: (Ala, Arg, Homoser, Lys x 2, Phe, Ser) and (Ala, Ser). Thermolysin digestion yielded fragments with the following amino acid composition: (Ala, Arg, Ser) and (Ala, Lys x 2, Met, Phe, Ser). The amino acid sequence of the original polypeptide is: ANS: ARSFKKMSA (i) Acid hydrolysis - Our peptide consists of 9 amino acid residues containing the following residues: A, A, R, K, K, M, F, S, and S (ii) Carboxypeptidase A digestion gives us our C-terminal amino acid residue: A So far, we have the sequence _ _ _ _ _ _ _ _ A (iii) Trypsin cuts at the carboxyl end of K and R, and we are given 4 segments with this digestion. T1 is A, R (The A residue here is not the carboxyl terminal residue since trypsin cuts at the carboxyl end of R.) The sequence for T1 is AR. T2 includes K, F, and S. We know that trypsin cuts here at the carboxyl end of K. Therefore the sequence of T2 is _ _ K. T3 is simply K. We know that is must follow a peptide fragment that has been cut by the trypsin enzyme since it's not the carboxyl terminal residue (A is the C- terminal end). T4 includes A, M, and S. We know that there are no residues in T4 cut by trypsin, and that alanine is the last residue. So the sequence of T4 is _ _ A. (iv) CNBr cuts at the C-terminal end of M. In this reaction, the reaction with the M residue gives a product of Homoserine. Wherever we see a homoser residue, we can infer that it's the M residue when sequencing a peptide. This digestion gives us two peptide fragments. C1 includes A, R, Homoser (M), K, K, F, and S. We know that all these amino acids must precede M, and that C2 (A, S) must come after it. If we look back at the trypsin experiment, we see that T4 includes A, M, and S. Overlapping C2 and T4, we get the sequence: MSA as the last three amino acids in the peptide. (v) Thermolysin digestion cuts at one place in the peptide, also. It targets very hydrophobic side chains and cuts at the amino end of the residue. A quick scan reveals F as the most hydrophobic residue and most likely candidate for digestion. We get the fragments: TL1 which includes A, R, and S.
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TL2 gives us A, K, K, M, F, and S. Since F is cut at the amino end, we can place it at
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This note was uploaded on 07/15/2008 for the course MCB 102 taught by Professor Staff during the Spring '08 term at University of California, Berkeley.

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Problem Set 3 Answers.doc - MCB102 - Problem Set 3 This...

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