please help 1. On a blank page you need to diagram the oxidation of the fatty acid 16:2 (49,12) to acetyl-CoA via B-oxidation. The diagram needs to include the following:a. B-oxidation:i. A diagram of the 1st, the 4th, and the 5th round of the pathway showing thestructures of the fatty acids produced by each step in each round of the pathway.Please also list the names of the enzymes used as well as the coenzymes (coenzymestructures are not needed).ii. Showing the correct stoichiometry (per 16:2(49,12)) of the following metabolites.Note this stoichiometry includes ALL of the needed rounds of B-oxidation to convert16:2(49,12) acetyl-CoA. It does not include the CAC1. FAD/FADH₂2. NAD+/NADH3. Acetyl-CoA

Here we are considering the fatty acid . This means that the fatty acid has 16 carbons and 2 double bonds.The number of rounds of beta oxidation required to completely breakdown a fatty acid with even number of carbon atoms is given by the equation below.So, 7 rounds of beta-oxidation are required here.The fatty acid is converted to its acyl-CoA form, before it can undergo beta-oxidation i.e. the carboxyl end of the fatty acid is attached to a CoA molecule via a thioester linkage. Each round of beta-oxidation is indicated via a BLUE arrow. (Neglect in the breakage in Round 4 BLUE arrow which occurred as the diagram was split into 2 imagea).The enzymes of concern are (in the order they first appear in the pathway);Acyl-CoA dehydrogenaseEnoyl-CoA hydratase-hydroxyacyl-CoA dehydrogenase-ketothiolaseCis- enoyl-CoA isomerase2,4-dienoyl-CoA reductaseEnoyl-CoA isomerase(ii) Rounds 1, 2, 3, 6 and 7 are normal rounds of beta-oxidation. In each of these rounds, 1 NADH and 1 FADH2 are produced. Rounds 4 and 5 are rounds of beta-oxidation that deal with the 2 inherent double bonds that were in the fatty acid. Round 4 yields 1 NADH. Round 5 yields 1 NADH and 1 FADH2, but note that 1 NADPH is consumed in this round.So, total NADH produced is 7.Total FADH2 produced is 6.1 acetyl-CoA is released at the end of each round, except the last round where 2 acetyl-Co-A are released. So, a total of 8 acetyl-CoA are released.FAD/FADH2 : 6NAD+/NADH : 7Acetyl-CoA : 8Note : NADPH and NADH are of similar energy. So, some consider subtracting the 1 NADPH that is consumed during the metabolism from the 7 NADH produced. If we consider this, then NAD/NADH becomes 6.