Boswellia serrata

Boswellia serrata is a tall deciduous tree growing over 15 m in height.
It has a smooth, peeling bark secreting white gum-resin. 
It is the tree that produces Indian frankincense.
The habit of Boswellia serrata has a light, spreading crown and somewhat drooping branches.

The bark of Boswellia serrata is fairly thick, smooth, exfoliating in thin, papery flakes. 
The bark is light green, silvery white with dark blotches. 
The stem is brittle and exudates transparent yellow-green gum-resin, with a strong smell, from wounds in the bark. 
The branchlets are pubescent.

The leaves of Boswellia serrata are compound, they are pinnate with an alternate arrangement. 
The leaves are about 20 cm long and 5 cm wide. 
There are 10 to 15 pairs of leaflets per compound leaf. 
The leaflets are oblong in shape, 2.5 to 5 cm long and 1 cm wide. 
The leaflet base is oblique, the leaflet margin is entire-crenate and the leaflet apex acute. 
The rachis of the leaf is tomentose and red. 
The petiole is less than 1 cm long.    

The flowers of Boswellia serrata grow in axillary panicles about 10 to 15 cm long. 
The flowers are small, less than 1 cm across and they are bisexual. 
They are white in colour. 
Each flower has 5 to 7 ovate petals, a bell-shaped red calyx-tube, short and pubescent and 10 to 15 brown stamens. 
The flowers are fragrant.

The fruits of Boswellia serrata are trigonous ovoid drupes. 
They are 1 to 3 cm across. 
Each drupe is splitting along 3 valves which contain the pyrenes (small hard nutlets). 
Each pyrene is one seeded.    

The seeds of Boswellia serrata are brown and around 3-5 mm long. 
They are contained in pyrenes.