00:01
In this problem, we are asked to draw the condensed structural formula for the activated form of erichitic acid and indicate the alpha and beta carbons in erichidal co -a.
00:10
We are then asked how many cycles of beta oxidation are needed, how many acetylcoa molecules are produced, and then to complete a table describing the total atp yield from complete beta oxidation.
00:22
Starting with part a, we are asked to draw the condensed structural formula for the activated form of erichitic acid.
00:28
In fatty acid activation, we find our fatty acyl co -a through a combination of our starting fatty acid, which is erichitic acid here, and co -enzyme a.
00:40
We'll see that what i've written here in red on co -enzyme a forms a bond with this carbon on our fatty acid to yield the fatty acyl -coa.
00:51
So the fatty acyl -coa product looks like this.
00:56
And you'll notice that the contribution from co -enzyme a is shown here.
01:00
Again in red.
01:02
Now we can determine our alpha and beta carbons.
01:05
We know that our alpha carbon is the carbon atom next to the ketone group, so it is this carbon here.
01:11
And we also know that our beta carbon is next to our alpha carbon, so it will be located here.
01:18
Next we are asked how many cycles of beta oxidation are needed.
01:22
We know that for every cycle of this process, two carbon units are cleaved off.
01:27
We have 20 carbon units total and 18 that we wish to cleave located here.
01:33
So we know that we need nine cycles total if we are cleaving two carbon units per cycle...