Difficulty: Easy
Correct Answer: Log of the number of cells versus time (semilog plot).
Explanation:
Introduction:
The classic bacterial growth curve exhibits lag, exponential, stationary, and death phases. Representing data on appropriate axes is crucial for interpreting kinetics and estimating parameters such as specific growth rate.
Given Data / Assumptions:
Concept / Approach:
Plotting log(N) versus time linearizes exponential growth (N = N0 * exp(mu * t)), producing a straight line of slope mu / ln(10) on a base-10 semilog plot. This makes phase boundaries and rate estimation clearer than linear plots of N vs time.
Step-by-Step Solution:
1) Collect N(t) via plate counts or convert OD to cell numbers.2) Compute log10(N) at each time point.3) Plot log10(N) vs t; the exponential region is linear.4) Determine mu from slope: mu = slope * ln(10).
Verification / Alternative check:
Cross-validate with a linear plot of ln(N) vs time (natural log), which should also be linear in the exponential region with slope mu.
Why Other Options Are Wrong:
Common Pitfalls:
Using OD beyond linear range; failing to account for lag; mixing base-10 and natural logs inconsistently.
Final Answer:
Log of the number of cells versus time (semilog plot).
Discussion & Comments