The Question is from neet-botany cell-cycle-and-cell-division mitosis, and it was created by the author D VINOD

Question

A researcher is studying mitosis in a novel diploid organism with 2n = 16 chromosomes.  During metaphase, a drug is applied that inhibits the separation of sister chromatids but allows the cell to progress through anaphase and telophase. At the end of this aberrant cell division, what is the most likely chromosomal composition of the resulting daughter cells?

D VINOD · Accepted Answer

## CORE INFORMATION
✓ **Answer**: C
## CONCEPT FRAMEWORK
 **Primary Concept**: Mitosis and chromosome segregation
 **Related Topics**: Cell cycle, cytokinesis, effects of drugs on cell division
 **Prerequisites**: Understanding of mitosis phases, chromosome structure (sister chromatids), and ploidy.
## SOLUTION PATHWAY
**Given Information**:
  * Diploid organism (2n = 16 chromosomes)
  * Drug inhibits sister chromatid separation during metaphase
  * Cell completes anaphase and telophase
**Method & Steps**:
 1. **Normal Mitosis**: In a normal mitotic division, sister chromatids separate during anaphase, resulting in two daughter cells each with 16 chromosomes (2n).
 2. **Effect of the Drug**: The drug prevents sister chromatids from separating.  Therefore, each chromosome will consist of two sister chromatids that migrate together to one pole during anaphase.
 3. **Resulting Daughter Cell**: Since sister chromatids don't separate, both sets of duplicated chromosomes end up in one daughter cell.  This results in doubling the chromosome number. 
  Reasoning: The cell completes anaphase and telophase, meaning that cytokinesis occurs, leading to the formation of a single cell.
  Formula/Rule: None directly applicable, but understanding of chromosome behavior in mitosis is crucial.
 4. **Final Chromosome Number**:  The resulting cell will have 32 chromosomes (2 x 16 = 32), representing a tetraploid (4n) cell.
  Result: One tetraploid cell (4n=32)
  Verification: The total number of chromosomes is doubled as sister chromatids remain together.
 ## OPTION ANALYSIS
A :x: Two diploid cells (2n=16) each: This would occur if sister chromatids separated normally.
B :x: One diploid cell (2n=16) and one cell with no chromosomes:  This is not possible given that the cell completes anaphase and telophase.
 C ✓: One tetraploid cell (4n=32): This is the correct result due to the failure of sister chromatid separation.
D :x: Two cells with 8 chromosomes each: This would imply that only half the chromosomes segregated to each daughter cell, which is not consistent with the drug's effect and the completion of anaphase and telophase.
 ## LEARNING AIDS
 **Common Mistakes**:
  1.Assuming normal sister chromatid separation despite the drug's effect: Carefully read and understand the effect of the drug on the cell division process.
  2.Forgetting that cytokinesis still occurs: Even with aberrant chromosome segregation, the cell will usually still undergo cytokinesis.
 **Quick Tips**:
  1.Visualize the process: Draw diagrams of the chromosomes during metaphase and anaphase to better understand the effect of the drug.
  2.Remember the definition of ploidy:  A diploid cell has two sets of chromosomes (2n), while a tetraploid cell has four sets (4n).

D VINOD · Answer

Two diploid cells (2n=16) each

D VINOD · Answer

One diploid cell (2n=16) and one cell with no chromosomes

D VINOD · Answer

The correct Option is "One tetraploid cell (4n=32)"

D VINOD · Answer

Two cells with 8 chromosomes each

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