Question

CoA-SH АТР Mg2+ 0 AMP + 2HPO- Thiokinase Cytosol Mitochondria Active fatty acid CH, CH-CH-CH-CS-CoA | Acyl-CoA FAD dehydrogenase FADH, Η ο CH,(CH2) 14-CC-c-S-COA IB a H HO trans-Enoyl-CoA hydratase Η Η ο CH (CH4-C-C-c-s-CoA Ιβ Ια OHH NAD Hydroxyacyldehydrogenase NADH + H+ i o CH(CH2-C-CH-C-S-COA Thiolase COA-SH 5 (B-ketothiolase) CH,(CH2) 4-2-s-CoA + CH, -C-S-COA Aeyl COA Acetyl CoA Acetyl COA eel Acetyl COA Acetyl COA

1A) How many rounds/cycles of beta-oxidation would be needed to completely break down a 16-carbon fatty acid molecule?

8

6

10

7

9

1B) What type of reaction is step 2?

isomerization

hydration

oxidation-reduction reaction

hydrolysis

1C) What type of enzyme catalyzes step 3?

oxidoreductase

lyase

hydrolase

transferase

ligase

1D) How many net ATP equivalents are generated by the complete break down of a 12-Carbon fatty acid molecule?

97 ATP

95 ATP

100 ATP

146 ATP

124 ATP

117 ATP

Answer 1

Ans 1A. (d) 7

Foe even numbered saturated fatty acid, (n - 1) number of cycles of beta-oxidation are required, where 2n = number of carbon in fatty acid.

2n = 16 → n = 8, therefore

Number of cycles = 8 -1 = 7

Ans 1B. (c) Oxidation reduction reaction.

In step 2, FAD got reduced into FADH_​2, and the reactant got oxidised into the product.

Ans 1C. (c) Hydrolase.

The enzymes that cleaves bonds by addition of water.

Ans 1D. (b) 95 ATPs

For even numbered fatty acids,

Total ATP = (n - 1) * 17 + 12 - 2, where n is half the number of carbons in fatty acid

Here, n = 6

→ (6 - 1) * 17 + 12 - 2

→ 5 * 17 + 12 - 2

→ 85 + 12 - 2

→ 97 - 2 = 95 ATPs