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Chapter 22: Problem 88

Draw the structures of the tripeptides gly-ala-ser and ser-alagly. How many other tripeptides are possible using these three amino acids?

Short Answer

Expert verified
The structures of the tripeptides gly-ala-ser and ser-ala-gly are as follows: Gly-Ala-Ser: \[H_2N-CH_2-C(=O)-NH-CH(CH_3)-C(=O)-NH-CH(OH)CH_2-COOH\] Ser-Ala-Gly: \[H_2N-CH(OH)CH_2-C(=O)-NH-CH(CH_3)-C(=O)-NH-CH_2-COOH\] There are a total of \(3^3 = 27\) possible tripeptides using the three given amino acids (glycine, alanine, and serine).

Step by step solution

01

Identify the Structures of the Amino Acids

The three amino acids we're working with are: 1. Glycine (gly) - Its structure is H2N-CH2-COOH. 2. Alanine (ala) - Its structure is H2N-CH(CH3)-COOH. 3. Serine (ser) - Its structure is H2N-CH(OH)CH2-COOH.
02

Draw the Structure of the Tripeptide Gly-Ala-Ser

To draw the structure of the tripeptide gly-ala-ser, connect the amino acids in the following order: glycine -> alanine -> serine. The peptide bond is formed between the amino group (NH2) of one amino acid and the carboxyl group (COOH) of the next amino acid, releasing a molecule of water in the process. The structure of gly-ala-ser is: \[H_2N-CH_2-C(=O)-NH-CH(CH_3)-C(=O)-NH-CH(OH)CH_2-COOH\]
03

Draw the Structure of the Tripeptide Ser-Ala-Gly

To draw the structure of the tripeptide ser-ala-gly, connect the amino acids in the following order: serine -> alanine -> glycine. The peptide bond is formed as described in Step 2. The structure of ser-ala-gly is: \[H_2N-CH(OH)CH_2-C(=O)-NH-CH(CH_3)-C(=O)-NH-CH_2-COOH\]
04

Determine the Total Number of Possible Tripeptides Using the Given Amino Acids

To calculate the number of possible tripeptides using the three amino acids, we utilize combinatorial mathematics. Since there are three positions in a tripeptide and each position can be occupied by any one of the three amino acids, we can simply multiply the number of choices available at each position. Number of tripeptides = (number of choices for first position) x (number of choices for second position) x (number of choices for third position) Number of tripeptides = (3 choices) x (3 choices) x (3 choices) = \(3^3\) = 27 possible tripeptides

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