Assume that a cell culture was started with five human fibroblast cells. Two cells did
not divide even once whereas the other three cells completed three rounds of cell
division. At this stage, the total number of kinetochores in all the cells put together is
Official Answer Key:
1196 to 1196
Step-by-Step Solution & Explanation
### **Step-by-Step Solution & Mathematical Derivation (GATE BT 2020 Q.58)**
**1. Governing Concept & Definition**:
- **Topic**: Cell Biology & Mitotic Chromosome Kinetochores
- In human diploid cells ($2n = 46$ chromosomes):
- In an undivided diploid cell, each chromosome has 1 centromere / 2 kinetochores (or 46 kinetochores in $G_1$, 92 in $G_2/M$).
- For cells that undergo division, each cell produces $2^k$ progeny cells.
**2. Detailed Step-by-Step Calculation**:
- **Given Initial Inoculum**: 5 human fibroblast cells.
- 2 cells did not divide: They remain 2 cells.
- 3 cells completed 3 rounds of division:
$\text{Number of progeny cells} = 3 \times 2^3 = 3 \times 8 = 24\text{ cells}$
- Total cells present $= 2 + 24 = 26\text{ cells}$.
- **Total number of kinetochores**:
- With 46 chromosomes per diploid cell, each replicated chromosome has 2 sister chromatids with 2 kinetochores per chromosome pair (or 46 kinetochores):
$26 \times 46 = 1196\text{ kinetochores}$
**3. Final Answer**:
- **Calculated Value**: **1196**
- **Official GATE Accepted Range**: **1196 to 1196**
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