During normal embryonic development of the mice paw, elimination of cells from the inter-digital space is due to ________.
- (A)
apoptosis
- (B)
meiosis
- (C)
mutagenesis
- (D)
necrosis
During normal embryonic development of the mice paw, elimination of cells from the inter-digital space is due to ________.
apoptosis
meiosis
mutagenesis
necrosis
ANSWER
GATE BT 2024 Q.19: The correct answer is (A) apoptosis.
Official answer: ASource: Official answer key, GATE 2024 BT (IISc Bengaluru)
Identify this as a developmental biology fact-recall question on limb/paw morphogenesis.
(A) Apoptosis: programmed, regulated cell death is the established mechanism that sculpts individual digits by eliminating the interdigital webbing during normal embryonic limb development — correct.
(B) Meiosis: a nuclear division process restricted to gamete formation; it has no role in somatic tissue remodeling or cell elimination — incorrect.
(C) Mutagenesis: the induction of mutations in DNA; it is not a mechanism of cell elimination or tissue sculpting — incorrect.
(D) Necrosis: unregulated, pathological cell death arising from injury or damage, accompanied by inflammation; it is not the clean, regulated process used in normal development — incorrect.
Conclusion: only apoptosis correctly describes the normal, regulated developmental process that removes interdigital cells.
KEY IDEA
Digit individualization during limb development occurs by programmed cell death (apoptosis) of the interdigital mesenchyme/webbing, a classic, well-characterized example of developmental apoptosis (regulated by BMP signaling in the interdigital regions).
WATCH OUT FOR
Confusing apoptosis (programmed, regulated cell death used in developmental sculpting) with necrosis (unregulated, injury-induced cell death causing inflammation) — necrosis is never the mechanism of normal developmental tissue remodeling.
From the GATE 2024 BT paper. Every question of this paper · Download the full paper · All 15 Cell biology questions, with notes