Question

Invertase (sucrase is the other name) is the enzyme hydrolyzing sucrose into glucose and fructose. (a) Describe how you can measure the rate of the reaction catalyzed by invertase by change in optical rotation of the reaction solution. Provide the formula for calculation of the rate (in uM/min) from the change of optical rotation that your high-school educated laboratory technician can use. (b) Can you guess why the enzyme gets the name invertase? The specific rotation (see #1 for definition) of sucrose is +66.4°, fructose (equilibrium mixture) is -92°, glucose (equilibrium mixture) is +52.7°.

Answer 1

(a) C₁₂H₂₂O₁₁ + H₂O + H⁺   $\rightarrow$    C₁₆H₁₂O₆ + C₆H₁₂O₆ + H⁺

The hydrolysis of sucrose can be considered a pseudo-first-order reaction, in which the rate of change at any time is proportional to the concentration of sucrose.

The rate law for the inversion of sucrose is in the form: d[sucrose] / dt = k [sucrose]^m [H₂O]ⁿ [H⁺] ^o (1)

Let concentration of sucrose be c. Also, the concentration of water and acid in this experiment is large and essentially constant. We can then rewrite Eq. 1 to isolate the dependence on the concentration of sucrose: – d[c]/dt = k_eff [c]^m (2)

with k_eff = k [H2O]ⁿ [H⁺]^o (3)

Integrating (2) : c = c₀ e^{-keff t}

Taking logarithms, ln c/c₀ = - kt

k = (ln c₀/c) / t

Instead of measuring concentration directly, the optical rotation, α, is measured. Optical rotation is proportional to concentration of the optically active compound.

The reaction rate constant k can be calculated using the following equation:

(4)

k = (2.303/6) log[((0) - (c))/(a(t)-0(c)]

where $\alpha$(0) is the initial rotation of pure sucrose, $\alpha (\infty )$ is the rotation at the time t (infinite time is the time when the reaction has gone to completion), and $\alpha$(t) is the rotation at some specific time t.

So by calculating optical rotation and substituting in (4) rate of reaction can be calculated

(b) Sucrose is dextrorotatory (+66.4°) but after hydrolysis, the resulting mix of glucose and fructose is slightly laevorotatory because of levorotatory fructose (-92°) having a greater molar rotation than the dextrorotatory glucose (+52.7°).

As the reaction proceeds, the angle of rotation to the right becomes less and less, and finally, the light is rotated to the left.

This inversion from dextrorotatory to laevorotatory is the reason for naming the enzyme invertase.

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