Blue–white screening logic: IPTG is a gratuitous inducer and X-gal is a chromogenic substrate for β-galactosidase. When α-complementation occurs (insert absent), colonies on X-gal plates appear what color?

Introduction / Context:
Blue–white screening distinguishes recombinant clones using disruption of the lacZα fragment in a suitable host. IPTG induces expression, and X-gal reveals β-galactosidase activity by producing a colored product.

Given Data / Assumptions:

• Host provides lacZΔM15 (ω fragment) and vector provides lacZα (α fragment).
• IPTG induces expression; X-gal turns blue when cleaved by functional β-gal.
• α-complementation restores enzyme activity only when the insert does not disrupt lacZα.

Concept / Approach:
Functional β-gal cleaves X-gal to produce an insoluble blue dye. If an insert disrupts lacZα, complementation fails and colonies remain white. Therefore, α-complementation (insert absent) yields blue colonies.

Step-by-Step Solution:
If lacZα intact → α-complementation with host ω fragment → active β-gal.Active β-gal + X-gal (with IPTG induction) → blue product.Hence, colonies appear blue when α-complementation occurs.

Verification / Alternative check:
Control plates with empty vector yield predominantly blue colonies; insert-containing clones screen as white.

Why Other Options Are Wrong:

• White indicates disrupted lacZα (insert present).
• Brown or no color are not typical outcomes under correct conditions.

Common Pitfalls:
Using glucose in medium (catabolite repression) suppresses color; incorrect host genotype or antibiotic levels can also confound results.

Final Answer:
Blue