8. A small metabolite on average diffuses about 1 nanometre (10-9 m) in a nanosecond
(10-9 s). How far from its starting point will this metabolite reach on average in 10 s?
a. 10-6 m
b. 10-4 m
c. 10-1 m
d. 10 m
A small metabolite diffuses 1 nanometer (10⁻⁹ m) in 1 nanosecond (10⁻⁹ s), so it covers 10 meters in 10 seconds. The correct answer is option d.
Solution Explanation
Diffusion here follows a simple proportional relationship where distance scales linearly with time at constant speed. The speed equals 1 nm/ns, or 10−9 m / 10−9 s = 1 m/s. Over 10 s, distance = speed × time = 1 m/s × 10 s = 10 m.
Option Analysis
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a. 10⁻⁶ m: Equals 1 micrometer, matching diffusion over 1 microsecond, not 10 s—too short by 10 million-fold.
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b. 10⁻⁴ m: Equals 0.1 mm, corresponding to about 100 microseconds—still far too brief.
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c. 10⁻¹ m: Equals 0.1 m, as if time scaled squarely (random walk), but problem states average linear diffusion.
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d. 10 m: Matches exact calculation: 109 ns in 10 s yields 109 nm = 10 m.
In cellular biology, understanding metabolite diffusion distance proves crucial for CSIR NET Life Sciences aspirants tackling questions like: a small metabolite diffuses 1 nanometer (10⁻⁹ m) in a nanosecond (10⁻⁹ s)—how far in 10 seconds? This tests speed-time proportionality over random walk misconceptions.
Key Calculation Steps:
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Speed: 10−9 m / 10−9 s = 1 m/s.
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Distance: 1 m/s × 10 s = 10 m (option d).
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Traps: Options a-c assume slower scaling, common in Brownian motion but not this linear case.
Exam Relevance: CSIR NET emphasizes such biophysics for cell communication and enzyme kinetics. Practice reveals diffusion covers ~10 m/s effectively, vital for metabolic modeling.
