Question

Explain the development of dicot embryo.

Answer 1

A: Oospore.
B: Two celled proembryo.
e: embryonal initial;
t: suspensor initial;
m: Embryo sac membrane.
B¹ : 4-celled I-shaped proembryo;
e₁, e₂: embryonal initial; s₁, s₂ : suspensor initial.
C: Further development of embryo.
S: Suspensor, h: Hypophysis; E: Embryonal mass
D: L. S. of ovule
Endo: Endosperm in free nuclear stage.
Anti: Antipodal tissue.
Embryo: Developing embryo
E: Embryo showing further development of embryonic octants and hypophysis.
F: L. S. of ovule. Endosperm becoming cellular.
G: Embryo; Cot: Cotyledons; Hypo: Hypocotyl; Rad: Radicle; R.c.: Rootcap;
H: Mature seed; Pl: Plumule. Endosperm has been consumed almost completely

Development of dicot embryo:

1. The zygote divides to form two-celled proembryo.
2. The larger cell towards the micropyle is called basal or suspensor initial cell and smaller cell towards chalaza is called terminal or embryonal initial cell.
3. The suspensor cell divides transversely in one plane to produce filamentous suspensor of 6-10 cells.
4. The first cell of the suspensor towards the micropylar end becomes swollen and functions as a haustorium.
5. The lowermost cell of the suspensor is known as hypophysis.
6. The suspensor helps in pushing the embryo in the endosperm.
7. The embryonal initial undergoes three successive mitotic divisions to form octant.
8. The planes of divisions are at right angles to each other.
9. The lower tier of four cells of octant gives rise to hypocotyl and radicle whereas four cells of the upper-tier form the plumule and the one or two cotyledons.
10. The hypophysis by further division gives rise to the part of the radicle and root cap.
11. Subsequently, the cells in the upper tier of the octant divide into several planes so as to become heart-shaped which then forms two lateral cotyledons and a terminal plumule.
12. Further enlargement of hypocotyl and cotyledons result in a curvature of the embryo and it appears horseshoe-shaped.