Q.58 Mammalian cells in active growth phase were seeded at a density of 1×10⁵ cells/ml.
After 72 hours, 1×10⁶ cells/ml were obtained.
The population doubling time of the cells in hours is (up to two decimal places) ________

Mammalian Cell Population Doubling Time: Solved (23.10 hours)

Mammalian cells seeded at 1×10⁵ cells/mL reached 1×10⁶ cells/mL after 72 hours, giving a population doubling time of 23.10 hours. This applies exponential growth during the log phase.

Calculation Method

The doubling time equation is:

t_d = t · ln(2) / ln(N_f / N_i)

• t = 72 hours
• N_i = 1 × 10⁵
• N_f = 1 × 10⁶

Compute natural log term:

ln(10⁶ / 10⁵) = ln(10) = 2.3026

Now substitute into the formula:

t_d = 72 × 0.6931 / 2.3026 = 23.10 hours

Options Analysis

• A) 18.00 hours: Linear assumption; ignores ln
• B) 23.10 hours (Correct): Exact exponential result
• C) 28.50 hours: Uses log₁₀ instead of ln
• D) 36.00 hours: Incorrect simple halving (72/2)

Cell Growth Principles

Exponential mammalian cell growth is captured by:

N_t = N₀ × 2^t/t_d

• Typical doubling time: 20–30 hours (CHO, HEK)
• Lag and stationary phases should be excluded
• Log phase data gives reliable t_d

Exam Strategies

• 10-fold increase = ~3.32 doublings (log₂10)
• Use ln(2) ≈ 0.693 for quick checks
• This problem frequently appears in GATE Biotechnology