Question

Study the given data and answer the questions that follow.

A sample of an enzyme called lactase was isolated from the intestinal lining of a calf. Assays were undertaken to evaluate the activity of the enzyme sample.

The substrate of lactase is the disaccharide lactose. Lactase breaks a lactose molecule in two, producing a glucose molecule and a galactose molecule. Two assays were carried out.

Assay 1

Lactose concentration (% w/v)	15	15	15	15	15	15
Concentration of enzyme sample(% v/v)	0	5	10	15	20	25
Reaction rate ($\mu$ mole glucose $se{c}^{-1}m{L}^{-1}$)	0	25	50	75	100	125

Assay 2

Lactose concentration (% w/v)	0	5	15	20	25	30
Concentration of enzyme sample(% v/v)	5	5	5	5	5	5
Reaction rate ($\mu$ mole glucose $se{c}^{-1}m{L}^{-1}$)	0	15	25	35	40	40

Which of the following assays would you expect to have the highest reaction rate?

• A. Lactose concentration (% w.v) - 15
  Concentration of enzyme sample (%v/v) - 5
• B. Lactose concentration (% w.v) - 30
  Concentration of enzyme sample (%v/v) - 5
• C. Lactose concentration (% w.v) - 15
  Concentration of enzyme sample (%v/v) - 25
• D. Lactose concentration (% w.v) - 30
  Concentration of enzyme sample (%v/v) - 25

Answer 1

The correct option is C Lactose concentration (% w.v) - 15

Concentration of enzyme sample (%v/v) - 25

The reaction rate is proportional to the amount of enzyme present at a lactose concentration of 15% w/v. The rate of reaction is proportional with the substrate concentration only to a limit, after which any increase in substrate concentration makes no difference in rate of reaction, the rate of reaction then remains constant.

Thus, the maximum reaction rate is at Lactose concentration - 15% w/v and enzyme sample concentration- 25% v/v.

So, the correct option is (C).