Question

A decapeptide (M.W. 796) on complete hydrolysis gives glycine (M.W. 75), alanine and phenylalanine. Glycine contributes \[47.0{\text{ }}\% \]of the total weight of the hydrolysis products. The number of glycine units present in the decapeptide is:
A.4
B.5
C.6
D.3

Answer 1

Hint: To answer this question, you should first find the percentage of glycine contribution in the total decapeptide. Using this value, you can find the value of units of glycine involved.

Complete step by step solution:
We know that the given decapeptide molecular mass = 796. A decapeptide by definition involves 10 amide linkages. These 10 amide linkages on hydrolysis give glycine, alanine, phenylalanine.
That means it needs 9 water molecules for its hydrolysis.
So, the mass of the product \[ = \;796\; + \;9\; \times \;18\; = \;958\].
In the total mass, \[47.0{\text{ }}\% \] is glycine. Glycine molecular mass for a single glycine unit = 75.
\[{\text{No of units of glycine = }}\dfrac{{450}}{{75}}\; = \;6\]

Therefore, we can conclude that the correct answer to this question is option C.

Note: Make sure that you observe the units of the given concentration.
$\% {\text{w/w}}$ is weight concentration of a solution: If a solution is labeled as \[10\% \] glucose in water by mass, it refers to that 10g of glucose is dissolved in 90 g of water resulting in 100g of solution.
$\% {\text{v/v}}$ is the volume concentration of a solution: it refers that if 50 mL of acetic acid is added to 50 mL of water, the acetic acid is labeled as \[50\% {\text{v/v}}\] .
$\% {\text{w/v}}$ is the mass concentration of a solution: if x grams/ml of solute are present in solution it means x gram of solute is dissolved in 100ml of solution.