ONLY DO A AND C PLEASE 11. Invertase \"Inverts\" Sucrose The hydrolysis of sucro

Question

ONLY DO A AND C PLEASE

11. Invertase "Inverts" Sucrose The hydrolysis of sucrose (specific rotation +66.5) yields an equimolar mixture of D-glucose (specific rotation +52.5) and D-fructose (specific rotation -92°). (See Problem 4 for details of specific rotation.) (a) Suggest a convenient way to determine the rate of hydrolysis of sucrose by an enzyme preparation extracted from the lining of the small intestine. (b) Explain why, in the food industry, an equimolar mix- ture of D-glucose and D-fructose formed by hydrolysis of sucrose is called invert sugar. (c) The enzyme invertase (now commonly called sucrase) is allowed to act on a 10% (0.1 g/mL) solution of sucrose until hydrolysis is complete. What will be the observed optical rota- tion of the solution in a 10 cm cell? (Ignore a possible small contribution from the enzyme.)

Explanation / Answer

a) Hydrolysis of sucrose yields equimolar mixture of glucose and fructose which is called invert sugar.

C12H22O11 + H2O -Invertage------> C6H12O6 + C6H12O6

Sucrose + water -------> Glucose + Fructose

Sucrose (specific rotation 6.5o)

D-Glucose (specific rotation 52.5o)

D- Fructose (specific rotation - 92.0o)

Mixture of glucose and fructose is levorotatory with optical rotation (+52.5o – 92.0o) = – 39.5°.

Sucrase an Intestinal enzyme extracted from the lining of the small intestine converts sucrose equimolar mixture of glucose and fructose. Sucrase activity can be asserted by observing the change in optical rotation of a solution of 100% sucrose (specific rotation 6.5o) as it is converted to a 1:1 mixture of D-glucose and D-fructose.

(c)

Calculate the increase in weight of product after hydrolysis:

Molar mass of water = 18 g/mol

Molar mass of sucrose = 342g/mol

An increase in weight of the product = (18/342)(100%) = 5.26% with respect to the starting sugar.

So, a 10% sucrose solution yields a [10 + (0.053 x 10)]% = 10.5% of invert sugar.

Of this 10.5%, 5.25% (0.0525 g/ml) is D-glucose and 5.25% is D-fructose (0.0525 g/ml).

Now determine the optical rotation of each sugar in the mixture in a 10 cm (1dm) cell by using following formula

Optical activity = Observed optical rotation/Optical patch length (dm) x concentration (g/ml)

Observed optical rotation = Optical activity *Optical patch length (dm) * concentration (g/ml)

Optical rotation of glucose = (52.5o)*(1 dm)*(0.0525 g/ml) = 2.76O

Optical rotation of fructose = (- 92.0o)*(1 dm)*(0.0525 g/mL) = -4.8o

The observed optical rotation of the solution (2.76O) + ­(-4.8o ) = -2.04

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