Determine 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-Phe
Treatment with trypsin yields the following four peptides:
1. Gly-Leu-Tyr-Phe-Lys
2. Ser-Met-Gly-Leu-Tyr-Lys
3. Val-Ile-Arg 4. Met-Leu-Pro-Phe

Solution :

1. It is given that when Edman's reagent is used, it releases PTH-Glycin. The degradation of Edmans is used for the amino acids to sequencing of the peptide/protein and is also removes N terminal amino acid and thus gives us the PTH amino acid product.

Therefore, the first amino acid is Glycine.

2. The Carboxypeptidase A releases the C terminal amino acid, and therefore the last amino acid is Phe.

3. The Cyanogen Bromide helps to chop after the Med residue, therefore the sequence is 1 - 2 - 3 or 2 - 1 - 3.

4. The trysin chops after the basic amino acid - Arg and the Lys unless followed by Proline and so the sequence of the peptide is 1 - 2 - 3 4.

answer: the value of enthalpy change will be -152.80 kj.

solution:

pressure = 1atm , temperature = = 298 k

given that pressure and temperature are same in both gases , the number of moles of both gases can be determined by the ideal gas equation.

according to reaction 3 moles no are reacting with 1 mol of ,then 0.49 mol of no will react with moles of ozone that is 0.16 moles of ozone. since, no is limiting reagent and is an excessive reagent.

enthalpy change for the reaction between the given volumes of the gases is :

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