13. Following is the diagram of a paracentric inversion heterozygote ABCDEFG/ABFEDCG involved in recombination during meiosis I:
The consequence of this recombination will be the formation of
A. A dicentric and an acentric chromosome in meiosis I as the chiasmata gets terminated.
B. No dicentric or acentric chromosome but appearance of deletion and duplication in both the chromosome.
D. Non-viable gametes from crossover products.
Which of the above statements are correct?
(1) A and B (2) A and C
(3) A and D (4) B and C
Understanding the diagram
The figure shows a paracentric inversion heterozygote (centromere outside the loop) undergoing a single crossover within the inversion loop. Standard outcomes:
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Two chromatids remain parental (normal and inverted).
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Two recombinant chromatids are:
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Dicentric (two centromeres) with deletions/duplications.
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Acentric (no centromere) with deletions/duplications.
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These recombinants lead to non‑viable gametes, so inversion behaves as a crossover suppressor.
Evaluating each statement
A. “A dicentric and an acentric chromosome in meiosis I as the chiasmata gets terminated.” – Correct
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This is the classical cytological result of a single crossover within a paracentric inversion loop.
B. “No dicentric or acentric chromosome but appearance of deletion and duplication in both the chromosome.” – Incorrect
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This description fits pericentric inversions, not paracentric. In paracentric inversions, acentric/dicentric chromatids appear.
C. “All non‑viable gametes.” – Incorrect
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Gametes receiving non‑crossover parental chromatids are usually viable; only those with recombinant (acentric/dicentric) chromatids are non‑viable.
D. “Non‑viable gametes from crossover products.” – Correct
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Gametes that receive the dicentric or acentric recombinant chromatids contain large deletions/duplications and are inviable.
Thus, the combination of correct statements for the diagram is A and D → option (3).
