Jump to content

Verschaffeltia

From Wikipedia, the free encyclopedia
(Redirected from Stilt palm)

Verschaffeltia
Scientific classification Edit this classification
Kingdom: Plantae
Clade: Tracheophytes
Clade: Angiosperms
Clade: Monocots
Clade: Commelinids
Order: Arecales
Family: Arecaceae
Subfamily: Arecoideae
Tribe: Areceae
Subtribe: Verschaffeltiinae
Genus: Verschaffeltia
H.Wendl.
Species:
V. splendida
Binomial name
Verschaffeltia splendida
H.A. Wendl.

Verschaffeltia splendida ("Latanier Latte" or stilt palm[2]) is a species of flowering plant in the family Arecaceae. It is the only species in the genus Verschaffeltia.[citation needed]

It is found only in Seychelles where it is threatened by habitat loss. The name comes from the Belgian Ambroise Verschaffelt.

Description

[edit]

This species can be distinguished from all other palm species of the Seychelles, by its characteristic stilt-roots.

The slender trunk has a very hard outer covering. The leaves are initially unbroken, and those of the young plants have black spines on their stalks. They bear green-brown fruits with unique seeds.

References

[edit]
  1. ^ Ismail, S.; Huber, M.J.; Mougal, J. (2011). "Verschaffeltia splendida". IUCN Red List of Threatened Species. 2011: e.T38722A10145395. doi:10.2305/IUCN.UK.2011-2.RLTS.T38722A10145395.en. Retrieved 14 November 2021.
  2. ^ Palmpedia