Question

In the process of generating a clonal mammalian cell line, a single cell was seeded in a well of a cell culture plate. After the first 48 hours, one of the progeny cells underwent apoptosis due to a new mutation. If the doubling time of the cells is 24 hours and no more cell death occurs, the total number of cells after a total of 7 days from seeding will be . (answer in integer)

Hint:

Grow to 4 cells by 48 h, remove 1 for apoptosis (3 left), then double 3 cells five more times (120 h / 24 h) to reach 168 h.

Answer 1

Correct Answer: 96

Step 1: Understanding the Question.
A single cell is seeded at time 0. It divides every 24 hours (its doubling time), so the population doubles at \(t=24\) h and again at \(t=48\) h. At the 48 hour mark, one of the resulting progeny cells dies by apoptosis. After that, no more cell death happens, and every surviving cell keeps doubling every 24 hours. We need the total cell count 7 days (168 hours) after seeding.

Step 2: Track the population up to the apoptosis event.
\[ t=0\text{ h}: 1 \ \text{cell} \quad\rightarrow\quad t=24\text{ h}: 2 \ \text{cells} \quad\rightarrow\quad t=48\text{ h}: 4 \ \text{cells} \]
At \(t = 48\) h, one of these 4 cells undergoes apoptosis, leaving
\[ 4 - 1 = 3 \ \text{cells} \]

Step 3: Continue doubling every 24 hours from this point.
From \(t=48\) h to \(t=168\) h (7 days) is
\[ 168 - 48 = 120 \ \text{hours} \]
Since the doubling time is 24 hours, the number of doublings in this interval is
\[ n = \frac{120}{24} = 5 \]

Step 4: Apply 5 doublings to the surviving 3 cells.
\[ N(168\text{ h}) = 3 \times 2^{5} = 3 \times 32 = 96 \ \text{cells} \]

Final Answer:
\[ \boxed{96 \ \text{cells}} \]