Introduction

The lac operon of Escherichia coli is a classical example of transcriptional regulation
in prokaryotes. The regulatory state of the operon depends on the intracellular concentration of the
lac repressor, its binding affinity (Kd), and the presence or absence of
lactose.

In this article, we solve a numerical problem to determine whether the lac operon is repressed, inducible,
or constitutively expressed.

Given Data

• Volume of an E. coli cell = 10⁻¹² cm³ = 10⁻¹⁵ L
• Number of lac repressor molecules = 60
• Binding affinity (Kd):
  • Without lactose = 10⁻⁹ M
  • With lactose = 10⁻⁸ M

Step 1: Calculation of Lac Repressor Concentration

Step 2: Comparison with Binding Affinity (Kd)

Without Lactose

Repressor concentration (10⁻⁷ M) is much greater than Kd (10⁻⁹ M).

➤ Strong binding of repressor to operator
➤ lac operon is repressed

With Lactose

Repressor concentration (10⁻⁷ M) is still higher than Kd (10⁻⁸ M), but binding affinity is reduced.

➤ Repression is relieved
➤ lac operon becomes inducible

Correct Answer

(A) Repressed and can only be induced with lactose

Explanation of All Options

(A) Repressed and can only be induced with lactose –
Correct. The repressor binds strongly without lactose and dissociates upon lactose binding.

(B) Repressed and cannot be induced with lactose –
Incorrect. Lactose decreases repressor affinity, allowing transcription.

(C) Not repressed –
Incorrect. Repressor concentration is much higher than Kd, ensuring repression.

(D) Expressed only when glucose and lactose are present –
Incorrect. Glucose causes catabolite repression; maximum expression occurs when glucose is absent.