The Question is from neet-botany cell-cycle-and-cell-division meiosis, and it was created by the author B VENKAT

Question

During meiosis, a hexaploid organism (6n) undergoes a rare mutation where one pair of homologous chromosomes fails to separate during Anaphase I.  What will be the ploidy of the resulting gametes?

B VENKAT · Accepted Answer

## CORE INFORMATION
✓ **Answer**: C
## CONCEPT FRAMEWORK
 **Primary Concept**: Meiosis and Nondisjunction
 **Related Topics**: Ploidy, Chromosome segregation, Gamete formation
 **Prerequisites**: Understanding of meiosis I and II, homologous chromosomes, and ploidy.
## SOLUTION PATHWAY
**Given Information**:
  * Hexaploid organism (6n)
  * Nondisjunction during Anaphase I (one homologous chromosome pair fails to separate)
**Method & Steps**:
 1. **Normal Meiosis I:** In a normal hexaploid, meiosis I would reduce the chromosome number by half, resulting in gametes with a ploidy of 3n. 
  Reasoning:  A hexaploid has six sets of chromosomes (6n). Meiosis I separates homologous chromosomes, reducing the chromosome number to half (3n) in each daughter cell.
  Formula/Rule:  Ploidy after meiosis I = $\frac{6n}{2} = 3n$ 
 2. **Nondisjunction:**  During Anaphase I, one pair of homologous chromosomes fails to separate. This means one daughter cell receives an extra chromosome from that pair, while the other daughter cell lacks one chromosome from that pair.
  Working: Let's represent the chromosome number as (3n + 1) and (3n -1) where 3n represents the normal haploid number after meiosis I and the (+1) and (-1) represent the extra and missing chromosome respectively. 
  Key Point: This nondisjunction event happens only in one pair of homologous chromosomes.
 3. **Meiosis II:** Meiosis II proceeds normally, separating sister chromatids. This will result in four gametes with two different ploidies.
  Result: Two gametes will have (3n + 1) chromosomes, and two gametes will have (3n - 1) chromosomes.
  Verification: The total number of chromosomes remains the same (2(3n+1) + 2(3n-1) = 12n). 
 ## OPTION ANALYSIS
[A] :x: All gametes will be 3n. This would only be true if meiosis proceeded normally without nondisjunction.
[B] :x: Two gametes will be 4n and two will be 2n. This would imply a more significant error in chromosome segregation than just one pair failing to separate.
[C] ✓: Two gametes will be (3n+1) and two will be (3n-1). This accurately reflects the outcome of nondisjunction during Anaphase I.
[D] :x: One gamete will be 6n, and three will be 0n. This is highly improbable and would result from a catastrophic failure of chromosome segregation.
 ## LEARNING AIDS
 **Common Mistakes**:
  1.Assuming normal meiosis: Failing to account for the nondisjunction event.
  [How to avoid]: Carefully read the problem statement and identify any deviations from normal meiosis.
  2.Incorrectly applying ploidy changes:  Miscalculating the ploidy after nondisjunction.
  [How to avoid]:  Visualizing chromosome segregation during meiosis and carefully tracking chromosome numbers.
 **Quick Tips**:
  1.Draw a diagram:  Visualizing the chromosome segregation in meiosis helps to avoid errors.
  2.Remember the impact of nondisjunction: Nondisjunction leads to aneuploidy (abnormal chromosome number) in gametes.

B VENKAT · Answer

All gametes will be 3n

B VENKAT · Answer

Two gametes will be 4n and two will be 2n

B VENKAT · Answer

The correct Option is "Two gametes will be (3n+1) and two will be (3n-1)"

B VENKAT · Answer

One gamete will be 6n, and three will be 0n

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