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:
pH = 11
Explanation:
The concentration of OH⁻ in <em>pure water</em> is 10⁻⁷ M. If a substance increases OH⁻ concentration by 10⁴, the new concentration will be:
[OH⁻] = 10⁴ x 10⁻⁷ M = 10⁻³M
We can calculate pOH using it's definition:
pOH = -log[OH⁻] = -log (10⁻³) = 3
Then, we can find out pH using the following relation:
pH + pOH = 14
pH = 14 - pOH = 14 -3 = 11
Since pH = 11 is higher than 7, we can confirm that the substance is a base.
Answer:
∴ Fractional distillation is the technique used to separate the fraction
Explanation:
A: Refinery gas
B: Gasoline fraction
C: Naphtha
D: Kerosene
E: Diesel Oil
F: Fuel oil fraction
G: Lubricating fraction
H: Bitumen
