A 20-bp GC-rich promoter undergoes a conformational transition between the B-form to Z-form of DNA, involving bases 6 to 13. The rest of the bases always remain in the B-form. The B- to Z-DNA transition has a half-life of 2 s, and Z- to B-DNA transition has a half-life of 6 s.
If the cognate transcription factor requires 10 continuous B-DNA base pairs for binding, for how many seconds in a 20 s window can the transcription factor remain bound to the promoter, assuming steady-state conditions?

Correct option is B

Correct Answer
(b) 5 s
Explanation
From half-life, rate constants are proportional to ln2/t½.
Thus, B→Z rate ∝ 1/2 and Z→B rate ∝ 1/6.
At steady state, fraction in B-form = (1/6) / (1/2 + 1/6) = 1/4.
Therefore, the promoter is suitable for binding 25% of the time.
In a 20 s window, this equals 0.25 × 20 = 5 s.
Information Booster
• DNA can adopt multiple conformations such as A, B, and Z forms.
• Z-DNA is favored in GC-rich regions and under torsional stress.
• Half-life reflects the time for half the population to change state.
• Steady state depends on relative forward and reverse rates, not absolute values.
Additional Knowledge
Option (a) is incorrect because it corresponds to 20% occupancy rather than the calculated 25%.
Option (c) assumes equal B and Z populations, which is not supported by the given half-lives.
Option (d) would require the promoter to be in B-form most of the time, which contradicts the faster B→Z transition.