Jump to content

Hemieva

From Wikipedia, the free encyclopedia
(Redirected from Suksdorfia ranunculifolia)

Hemieva
Scientific classification Edit this classification
Kingdom: Plantae
Clade: Tracheophytes
Clade: Angiosperms
Clade: Eudicots
Order: Saxifragales
Family: Saxifragaceae
Genus: Hemieva
Raf. (1837), nom. rej.
Species:
H. ranunculifolia
Binomial name
Hemieva ranunculifolia
Raf. (1837)
Synonyms[1]
  • Boykinia ranunculifolia (Hook.) Greene (1891)
  • Saxifraga ranunculifolia Hook. (1832)
  • Suksdorfia ranunculifolia (Hook.) Engl. (1930)

Hemieva ranunculifolia is a species of flowering plant in the saxifrage family known by the common name buttercup suksdorfia.[2] It is the sole species in genus Hemieva.[1] It is native to western North America from British Columbia and Alberta south to northern California. It grows in moist, rocky habitat in mountains and foothills. It is a non-rhizomatous perennial herb growing up to 40 centimeters tall. The leaves have rounded blades up to 4 centimeters wide with several large lobes edged with rounded teeth. The blades are light green, slightly fleshy, hairless in texture, and are borne on petioles up to 15 centimeters long. The inflorescence is a dense, flat-topped cluster of up to 35 flowers borne atop a mostly naked, hairy, glandular stalk. Each flower has a bell-shaped calyx of pointed sepals and five white or pink-tipped petals. The fruit is an oval brown capsule measuring 4 millimeters in length.[3]

References

[edit]
  1. ^ a b Hemieva ranunculifolia Raf. Plants of the World Online. Retrieved 11 April 2024.
  2. ^ NRCS. "Suksdorfia ranunculifolia". PLANTS Database. United States Department of Agriculture (USDA). Retrieved 4 December 2015.
  3. ^ "Suksdorfia violacea". Lady Bird Johnson Wildflower Center. Retrieved January 20, 2015.
[edit]