ORGANIC CHEMISTRY LL >BI<
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ISBN: 9781260561609
Author: Carey
Publisher: MCG CUSTOM
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Chapter 27, Problem 27P
Interpretation Introduction
Interpretation:
The amino acid that is
Concept Introduction:
The genetic code carried by mRNA consists of triplets of
Each codon indicates a particular amino acid.
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A random-sequence polyribonucleotide produced by polynucleotide phosphorylase, with CDP and ADP in a 5:1 molar ratio stimulated the incorporation of proline, histidine, threonine, glutamine, asparagine, and lysine in a cell-free translation system in the following proportions: 100, 23.4, 20, 3.3, 3.3, and 1.0, respectively. What does this experiment reveal about the nucleotide composition of coding triplets for these six amino acids?
Three peptides were obtained from a trypsin digestion of two different polypeptides. In each case, indicate the possible sequences from the given data and tell what further experiment should be carried out in order to determine the primary structure of the polypeptide. a. polypeptide I: 1. Val-Gly-Asp-Lys 2. Leu-Glu-Pro-Ala-Arg 3. Ala-Leu-Gly-Asp b. polypeptide II: 1. Val-Leu-Gly-Glu 2. Ala-Glu-Pro-Arg 3. Ala-Met-Gly-Lys
Deduce the sequence of a heptapeptide that contains the amino acids Ala, Arg, Glu, Gly, Leu, Phe, and Ser, from the following experimental data. Edman degradation cleaves Leu from the heptapeptide, and carboxypeptidase forms Glu and a hexapeptide. Treatment of the heptapeptide with chymotrypsin forms a hexapeptide and a single amino acid. Treatment of the heptapeptide with trypsin forms a pentapeptide and a dipeptide. Partial hydrolysis forms Glu, Leu, Phe, and the tripeptides Gly–Ala–Ser and Ala–Ser–Arg.
Chapter 27 Solutions
ORGANIC CHEMISTRY LL >BI<
Ch. 27.1 - Problem 27.1 Write a structural formula for the...Ch. 27.1 - Prob. 2PCh. 27.1 - Prob. 3PCh. 27.2 - Prob. 4PCh. 27.3 - Prob. 5PCh. 27.3 - Prob. 6PCh. 27.5 - Prob. 7PCh. 27.5 - Prob. 8PCh. 27.5 - Prob. 9PCh. 27.6 - Prob. 10P
Ch. 27.7 - Prob. 11PCh. 27.9 - Prob. 12PCh. 27.12 - 27.13 Modify Figure 27.12 so that it corresponds...Ch. 27.13 - Prob. 14PCh. 27 - Prob. 15PCh. 27 - Prob. 16PCh. 27 - Prob. 17PCh. 27 - Nebularine is a toxic nucleoside isolated from a...Ch. 27 - Prob. 19PCh. 27 - The 5-nucleotide of inosine, inosinic acid...Ch. 27 - Prob. 21PCh. 27 - (a) The two most acidic hydrogens of uracil have...Ch. 27 - The phosphorylation of -D-glucopyranose by ATP...Ch. 27 - When 6-chloropurine is heated with aqueous sodium...Ch. 27 - Prob. 25PCh. 27 - Prob. 26PCh. 27 - Prob. 27PCh. 27 - Prob. 28PCh. 27 - Oligonucleotide Synthesis In Section 27.6 we noted...Ch. 27 - Oligonucleotide Synthesis In Section 27.6 we noted...Ch. 27 - Oligonucleotide Synthesis In Section 27.6 we noted...Ch. 27 - Oligonucleotide Synthesis In Section 27.6 we noted...Ch. 27 - Oligonucleotide Synthesis In Section 27.6 we noted...Ch. 27 - Oligonucleotide Synthesis In Section 27.6 we noted...
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- Three peptides were obtained from a trypsin digestion of two different polypeptides. indicate the possible sequences from the given data and tell what further experiment should be carried out in order to determine the primary structure of the polypeptide. polypeptide I: 1. Val-Gly-Asp-Lys 2. Leu-Glu-Pro-Ala-Arg 3. Ala-Leu-Gly-Asparrow_forwardSomatostatin is a tetradecapeptide of the hypothalamus that inhibits the release of pituitary growth hormone. Its amino acid sequence has been determined by a combination of Edman degradations and enzymic hydrolysis experiments. On the basis of the following data, deduce the primary structure of somatostatin: 1. Edman degradation gave PTH-Ala. 2. Selective hydrolysis gave peptides having the following indicated sequences: Phe-Trp Thr-Ser-Cys Lys-Thr-Phe Thr-Phe-Thr-Ser-Cys Asn-Phe-Phe-Trp-Lys Ala-Gly-Cys-Lys-Asn-Phe 3. Somatostatin has a disulfide bridge.arrow_forwardOrnithine is an amino acid that is not used in the synthesis of proteins, but is an important intermediate in several metabloic pathways including the urea cycle and the synthesis of polyamines. It has a perfectly ordinary terminal amino group and terminal carboxyl group like any other amino acid (so use the pKas for those groups given on your amino acids handout), and a side chain with a single ionizable side group with a pKa of 10.3. If ornithine is placed in solution at pH 7.0, it has a net charge of +1. What would the net charge on this amino acid be if the pH of the solution was raised to pH 12.0? Please explain your reasoning.arrow_forward
- 3. A homopolymer of histidine (polyhistidine) adopts an alpha helix structure at pH 9 but is unfolded at a pH of 4. Explain why using chemical details.arrow_forwardA peptide has the following amino acid composition: 2 Met, 2 Phe, 2 Glu, 1 Arg, 1 Lys, 1 Val, 1 Leu, 1 Gly, 1 Ser Reaction of the intact peptide with dansyl chloride followed by acid hydrolysis creates a derivative of Met. A specific cleavage of the intact peptide produces fragments with the following sequences: Fragment A: Glu-Gly-Lys-Phe Fragment B: Met-Ser-Leu-Arg Fragment C: Met-Val-Glu-Phe What information do this result give about the sequence of the peptide? Explain how you arrived on your answer. a) The sequence is: Met-Val-Glu-Phe-Glu-Gly-Lys-Phe-Met-Ser-Leu-Arg b) The sequence is: Met-Ser-Leu-Arg-Met-Val-Glu-Phe-Glu-Gly-Lys-Phe c) The sequence is: Met-Val-Glu-Phe-Met-Ser-Leu-Arg-Glu-Gly-Lys-Phe d) The sequence is: Met-Ser-Leu-Arg-Glu-Gly-Lys-Phe-Met Val-Glu-Phearrow_forwardReport the isoelectric point of the peptide, Gly-Asp-Lys-Ile?arrow_forward
- Chemical Connections 20D states that the antigen in the red blood cells of a person with B-type blood is a galactose unit. Show schematically how the antibody of a person with A-type blood would aggregate the red blood cells of a B-type person if such a transfusion were made by mistake.arrow_forwardA sample of an unknown peptide was divided into two aliquots. One aliquot was treated with trypsin; the other one was treated with cyanogen bromide. Given the following sequences (N-terminal to C-terminal) of the resulting fragments, deduce the sequence of the original peptide. Trypsin treatment Asn-Thr-Trp-Met-Ile-Lys Gly-Tyr-Met-Gln-Phe Val-Leu-Gly-Met-Ser-Arg Cyanogen bromide treatment Gln-Phe Val-Leu-Gly-Met Ile-Lys-Gly-Tyr-Metarrow_forwardDetermine the amino acid sequence of a polypeptide from the following data:Complete hydrolysis of the peptide yields Arg, 2 Gly, Ile, 3 Leu, 2 Lys, 2 Met, 2 Phe, Pro, Ser, 2 Tyr, and Val.Treatment with Edman’s reagent releases PTH-Gly.Carboxypeptidase A releases Phe. Treatment with cyanogen bromide yields the following three peptides:1. Gly-Leu-Tyr-Phe-Lys-Ser-Met 2. Gly-Leu-Tyr-Lys-Val-Ile-Arg-Met 3. Leu-Pro-Phearrow_forward
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