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Question

Compare the columns and find out the correct combination.

Column IColumn II
(a)

Tridax

(i)Synandrous
(b)

Dolichos

(ii)Monadelphous
(c)

Ceiba

(iii)Syngenesious
(d)

Cucurbita

(iv)Polyadephous
(v)Diadelphous

A
(a)-(iv), (b)-(v), (c)-(ii), (d)-(i)
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B
(a)-(iii), (b)-(v), (c)-(iv), (d)-(i)
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C
(a)-(iii), (b)-(v), (c)-(i), (d)-(iv)
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D
(a)-(v), (b)-(iii), (c)-(iv), (d)-(ii)
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Solution

The correct option is B (a)-(iii), (b)-(v), (c)-(iv), (d)-(i)
Column I Column II

(a)
Tridax (iii) Syngenesious
(b) Dolichos (v) Diadelphous
(c) Ceiba (iv) Polyadelphous
(d)Cucurbita (i) Synandrous
(a) Tridax belongs to the family Asteraceae, it shows the syngenesious condition of the androecium, wherein the anthers are fused and the filaments are free.
(b) Dolichos belongs to the family Fabaceae, it shows diadelphous androecium because filaments are united to form two bundles (9+2).
(c) Ceiba belongs to the family Malvaceae, it shows polyadelphous condition because the united filaments form many bundles.
(d) Cucurbita belongs to the family Cucurbitaceae, the androecium is synandrous because the stamens are united in their whole length as 2+2+1 pattern.
Hence, the correct answer is '(a)- (iii), (b)- (v), (c)- (iv), (d)- (i)'.

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