You perform a colorimetric enzyme assay to determine the act

You perform a colorimetric enzyme assay to determine the activity of invertase in a bioreactor (total volume = 1L) used to produce inverted sugar. You prepare your standard curve by mixing known monosaccharide dilutions (3ml) with the DNS reagent (2ml). The standard curve follows the following linear function: Absorbance (540nm) = 1.82 times monosaccharide concentration (M) You mix 1.5ml of the enzyme solution from the bioreactor with 1.5ml of substrate solution and let them react for exactly 10 min, after which you stop the reaction by adding 2ml of DNS reagent. You then add 20ml of distilled water and read the absorbance at 540nm. You obtain an absorbance value of 0.395. Determine the reaction rate (mal monosaccharide formed/min) in the bioreactor.

Solution

Invert sugar refers to the carbohydrate sucrose , which is a combination of both glucose anf fructose. Therefore, one mole of sucrose will split to give rise to one mole of glucose and one mole of fructose.

Now, let us determine the dilution ratio of the substrate in the reaction mixture -

Amount of substrate = 1.5mL
Amount of enzyme = 1.5mL
Amount of DNS reagent = 2mL
Amount of D/W = 20 mL

Total amount of mixture = 25mL

Ratio of Substrate (which contains the sugar) : Total amount = (1.5 : 25)

Dilution ratio = 25 / 1.5 = 16.67

Now, from given information,

Absorbance (540nm) = 1.82 * Monosaccaride concentration (M)

Therefore, Monosaccaride concentration (M) = (Absorbance (540nm) ) / 1.82

= 0.395 / 1.82

= 0.22 M

Therefore, Monosaccaride concentration (M) = 0.22M

However, we have to multiply it by the dilution factor, which is 16.67

Therefore,

Monosaccaride concentration (M) = 0.22 * 16.67 = 3.67M

That means, in 10 minutes, 3.67M of monosacarides are formed.

Therefore, in one minute,

Monosaccaride concentration (M) = ( 3.67 / 10 ) = 0.367 M

Rate of reaction = 0.367 Moles/ minute