calculation - cd45

[ArchivalUser]

CD45,
From your conventional method,89 atp's,the number is correct.But I didnt get why would you take off 2 ATP for activation of propionyl Co-A?(When, from my knowledge on this B-Oxidation,once the fatty acid got activated by taking 2 high energy bonds and entered itochondrial matrix,its there until its completlely oxidised,and unlike the remaining acetyl co-A,even Propionyl Co-A will directly enter TCA cycleat the level of Succinyl Co-A ,there by yielding 1NADH(instead of 3NADH like Acetyl Co-A),1FADH,1GTP===yielding 6ATP ,instead of 12.)
So you need to take off,6ATP's for TCA cycle of Propionyl Co.A,(So your final calculation coincides with mine as you have taken off 6 ATP's for propionyl co-A...but I just got a little lost...From -4ATP from TCA and -2ATP for activation ...can you please clear my confusion...)
And coming to your Q, Yeah,the calcultion isn't working.I didnt know this formula before.But I felt may be I can use it for fast calculation,when the options given are not in a narrow range....if they give 88,87,89 ...may be I will go for conventional method.
I suppose from this example ,may be the formula gives a rough estimate,not exact..(though I am not sure of it..as I already said,this is first time I knew about this formula)
But thank you for this,as it would make my life easy,if wide range if opyions are given...I can just calculate roughly,mark it and move on...if I get time come back and calculate conventionally.
But if you are 100% sure the formula is exact one...then ...I dont know...bcoz both 89 and 88 are correct then...mark something pray and move on:-)o correct myself.
Hope this helps
Thanks
If some one else find my approach wrong,please correct me...I will be happy t

[ArchivalUser]

Sorry my browser had some problem words get jumbled in between,and I dont notice.
I said I will be happy t(o correct myself...this phrase went up:-)

[ArchivalUser]

just a q nautilus... how did u get 25 atp from beta oxidation?

[ArchivalUser]

Each cycle of B-Oxidation yields 1NADH and 1FADH...which yield 5ATP on Oxidative phosphorylation.
So 13-C FA,5 cycles. 5x5=25

You just need to see how many cycles of B-Oxd needed for a given number of C's in FA,what the end products are.
Eg:C-18 FA
18/2=9acetyl co-A's formed from B-oxd.
And its always 1 cycle less than no: of acetyl co-a(check it ur self manually...i just gave simple way.) So here its 8 cycles so 8x5=40ATP from B-Oxd
I acetyl co-a gives 12 ATP from TCA
So 12x9=108ATP
2 High energy bonds used initially for activation so -2ATP
total is 146 atp from 18-C FA.
If its odd C say 15-C FA; 15/2=7.5,
meaning you are left out with 1c if 7 Acetyl co-a's produced,from 7 cycles of B-oxd.
So simply correct it to 6 acetyl co-a's .now you are left with 3 carbons..i.e; propionyl co-a
So finally u got 6 Acetyl Co-a(6x12=72)
1 Propionyl Co-A(1x6=6ATP)
7x5=35ATP from B-oxd.
2ATP used up for activation.
So total yield for 15-C FA =(72+6+35-2)=111ATP


Hope I am clear..

[ArchivalUser]

wow thanks nautilus... its crystal! really appreciate it... never knew how to really calculate this!

[ArchivalUser]

Hey nautilius...could u pl. explain this:

C-13...so that leaves us with 5 acetyl COA (10C) and 1 propyonyl CoA (3C)

5 acetyl COA undergoes 4 turns and loses one ATP to get the spiral started..so

ATP from B-Oxid = 4*5 = 20 -1 = 19

ATP from TCA = 5* 12 = 60.

ATP from propionyl COA = 6

So total 19 + 60 + 6 = 84


How did u get 89??? Moreover...is the number of ATP from propionly coa 6 ??

Look at the below link

http://www.elmhurst.edu/~chm/vchembook/622overview.html

[ArchivalUser]

how does 5 acetly COA undergo 4 turns... should be 5 turns only.. 5*5= 25...

propionyl COA as nautilius explained enters TCA at succinyl COA .SO after that point only 1NADH, 1FADH2 and 1GTP are formed... so that makes 6ATP

what i dont get is the 4 ATP cd45 has subtacted... im getiing for 13 C FA= (25+ 72+ 6-2)

so im stuck!

[ArchivalUser]

ok so my above toal comes to 101ATP if i further subtract 3 ATP (for the 1 NADH not used in the TCA) then i get exactly 89 ATP... which matches the calculated formula version

but i dont know if we need to subtract that NADH ....

[ArchivalUser]

13 C-->11C-->9C-->7C-->5C-->3C+2C

5 cycles right!!!

[ArchivalUser]

Well,
If the number of carbons is 16, then by the 7th Fatty acid spiral, we get a 14 carbon CoA and an Acetyl CoA. This is what the link says:

"The fatty acid-CoA molecule is degraded into acetyl CoA molecules by a recurring cyclic sequence of four reactions. The end product of each cycle is the fatty acid shortened by 2 carbons and acetyl CoA. The 14 carbon-CoA molecule repeats steps 1-4 and finishes as 12 carbon-CoA plus acetyl CoA. The 12 carbon-CoA repeats steps 1-4 and finishes as 10 carbon-CoA plus acetyl CoA. The cycles are repeated until the last cycle begins with a 4 carbon-CoA and finishes as two acetyl CoA molecules. Since the fatty acid cycles have a definite beginning and end, it is better to refer to them as the Fatty Acid Spiral."

In order to calculate total ATP from the fatty acid spiral, you must calculate the number of turns that the spiral makes. Remember that the number of turns is found by subtracting one from the number of acetyl CoA produced. See the graphic on the left bottom.

Example with Palmitic Acid = 16 carbons = 8 acetyl groups

Number of turns of fatty acid spiral = 8-1 = 7 turns

ATP from fatty acid spiral = 7 turns and 5 per turn = 35 ATP.

This would be a good time to remember that single ATP that was needed to get the fatty acid spiral started. Therefore subtract it now.

NET ATP from Fatty Acid Spiral = 35 - 1 = 34 ATP

As a general principle, the number of fatty acid spiral turns is always one less than the number of acetyl CoA groups formed. In this example, 8 - 1 = 7 turns of the fatty acid spiral.