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Question

Arrange the following compounds in descending order on the basis of number of carbon atoms present in them.
(a) Palmitic acid
(b) Serine
(c) Ribose
(d) Arachidonic acid
(e) Glucose

A
d,c,a,b,e
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B
b,c,e,a,d
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C
d,a,e,c,b
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D
c,b,a,d,e
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Solution

The correct option is C d,a,e,c,b
The given compounds belong to the different category of biomolecules. Among these arachidonic acid and palmitic acid both are unsaturated and saturated fatty acids respectively. Arachidonic acid is 20-carbon unsaturated fatty acid with 4 double bonds and its molecular formula is C20H40O32. Palmitic acid is a 16-C saturated fatty acid with molecular formula C16H32O2. Glucose and ribose are monosaccharides with 6 and 5 carbon atoms respectively. Glucose (C6H12O6) is a hexose sugar while ribose (C5H10O5) is a pentose sugar. Serine is an amino acid with 3 carbon atoms.
Arachidonic acid- 20 C
Palmitic acid - 16 C
Glucose - 6 C
Ribose - 5 C
Serine - 3 C
Thus, the correct answer is option C.

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