Question:
Which among the followings is an example of omega three fatty acid?
Which among the followings is an example of omega three fatty acid?
Show Hint
Remember the classic designations:
- Oleic acid: \( 18:1 \) (omega-9)
- Linoleic acid: \( 18:2 \) (omega-6)
- \( \alpha \)-Linolenic acid: \( 18:3 \) (omega-3)
- Oleic acid: \( 18:1 \) (omega-9)
- Linoleic acid: \( 18:2 \) (omega-6)
- \( \alpha \)-Linolenic acid: \( 18:3 \) (omega-3)
Updated On: Jul 21, 2026
- Oleic acid
- Linoleic acid
- $\alpha$-Linolenic acid
- Arachidonic acid
Show Solution
Verified By Collegedunia
The Correct Option is C
Solution and Explanation
Step 1: Understanding the Concept:
Unsaturated fatty acids are classified based on the position of the first double bond from the methyl end (omega, \( \omega \), or \( n \) end) of the carbon chain.
An omega-3 (\( \omega \)-3) fatty acid has its first double bond at the third carbon-carbon bond from the methyl end.
Step 2: Detailed Explanation:
Let us analyze the structure of each fatty acid.
1. Oleic acid:
An 18-carbon monounsaturated fatty acid (\( 18:1 \)).
The double bond is located at the 9th carbon from the methyl end, making it an omega-9 (\( \omega \)-9) fatty acid.
2. Linoleic acid:
An 18-carbon polyunsaturated fatty acid with two double bonds (\( 18:2 \)).
The first double bond is at the 6th carbon from the methyl end, making it an omega-6 (\( \omega \)-6) fatty acid.
3. \( \alpha \)-Linolenic acid (ALA):
An 18-carbon polyunsaturated fatty acid with three double bonds (\( 18:3 \)).
The first double bond is at the 3rd carbon from the methyl end, classifying it as an omega-3 (\( \omega \)-3) fatty acid.
4. Arachidonic acid:
A 20-carbon polyunsaturated fatty acid with four double bonds (\( 20:4 \)).
The first double bond is at the 6th carbon from the methyl end, making it an omega-6 (\( \omega \)-6) fatty acid.
Thus, \( \alpha \)-Linolenic acid is an omega-3 fatty acid.
Step 3: Final Answer:
The omega-3 fatty acid is \( \alpha \)-Linolenic acid, corresponding to option 3.
Unsaturated fatty acids are classified based on the position of the first double bond from the methyl end (omega, \( \omega \), or \( n \) end) of the carbon chain.
An omega-3 (\( \omega \)-3) fatty acid has its first double bond at the third carbon-carbon bond from the methyl end.
Step 2: Detailed Explanation:
Let us analyze the structure of each fatty acid.
1. Oleic acid:
An 18-carbon monounsaturated fatty acid (\( 18:1 \)).
The double bond is located at the 9th carbon from the methyl end, making it an omega-9 (\( \omega \)-9) fatty acid.
2. Linoleic acid:
An 18-carbon polyunsaturated fatty acid with two double bonds (\( 18:2 \)).
The first double bond is at the 6th carbon from the methyl end, making it an omega-6 (\( \omega \)-6) fatty acid.
3. \( \alpha \)-Linolenic acid (ALA):
An 18-carbon polyunsaturated fatty acid with three double bonds (\( 18:3 \)).
The first double bond is at the 3rd carbon from the methyl end, classifying it as an omega-3 (\( \omega \)-3) fatty acid.
4. Arachidonic acid:
A 20-carbon polyunsaturated fatty acid with four double bonds (\( 20:4 \)).
The first double bond is at the 6th carbon from the methyl end, making it an omega-6 (\( \omega \)-6) fatty acid.
Thus, \( \alpha \)-Linolenic acid is an omega-3 fatty acid.
Step 3: Final Answer:
The omega-3 fatty acid is \( \alpha \)-Linolenic acid, corresponding to option 3.
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