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sprint123
Forum Guru Topics: 129 Posts: 849 |
What is the net energy yield in high eneregy po4 bonds from each molecule of lauric acid?(12 carbon molecule) A.93 B.95 C.97 D.99 E.101 Please explain
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Clozapine
| Forum Senior  Topics: 16 Posts: 160
Consider oxidation of FA w/ even # of C atoms I remember this way ... if its an 16 C atom FA u`ll get (16/2 -1) i.e. (8-1)=7 FADH2 =7X2 ATP =14 ATP ; (16/2-1) =7 NADH =7X3=21 ATP and 16/2 Acetyl CoA = 8 Acetyl CoA =8X12= 96 ATPS Total = 14+21+96 = 131 ATPS For ur Q , it will be (12/2-1)X2 + (12/2-1)X3 + (12/2) X 12 =5X2 + 5X3 + 6X12 = 25+72 = 97 ATPS ATP = net energy yield in high eneregy po4 bonds hope it helps
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Clozapine
| Forum Senior Topics: 16 Posts: 160
For exam purpose take half of the # of C atom and multiply with 12 and add (half of the # of C atom -1)X 5 ... that should give u the result.
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sany
| Forum Senior Topics: 7 Posts: 117
Hello I need your advice.Does this type of calculation really high yield for the exam?I am seriously studying but I usually don't give attention to this type of question.What is the experience of seniors? Thank you Sany
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Ancylostoma
| Forum Guru Topics: 43 Posts: 632
It not a hard question, she clearly explained how to do it. Memorize what clozapine said and you will be fine. Maybe 10 mins of your time .
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sany
| Forum Senior Topics: 7 Posts: 117
Hello Ankylostoma Thank you.You are indirectly telling me that such things should not be overlooked!! Sany
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sprint123
| Forum Guru Topics: 129 Posts: 849
Thanks a lot clozapine...That was a nice way to rememeber. I think we have to substract 2 atp's for the activation of fatty acid ..since the answer given was 95.. Clozapine,Can you tell is there a similar way for other cycles which are important for the exam.It would be very useful.Thanks again
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Clozapine
| Forum Senior Topics: 16 Posts: 160
I`m glad I could help Sprint, but I`m a lil confused about the answer .... for activation of FA by FA-CoA synthase we need only 1 ATP, as far as I remember  I guess we are counting the ATP required to convert OAA to Acetyl CoA after it`s been transported out of mitochon to cyosol ....  95, agreed Well, the only other pathway with ATP issues  I guess is aerobic and anerobic glycolysis ... for anerobic its only 2 substrate level net ATP production ... (NADH used up by LDH) ... it`s the 2nd option of body for glycolysis, so yields only 2 ATPS ( thats my way  )And for aerobic its 12 from TCA cycle once again so from 1 mol. of Glucose (12x2) =24 from TCA cycle plus net 2 substrate level ATP, 2 cytosolic NADH (2X2 or 2X3) depending on the shuttle and another 2 NADH produced by pyruvate DH .... so in total 24+ 2 + (3X2) + (2 X 3 or 2) = 36 to 38 ATPs For FA synthesis just reverse the oxidation .... U`ll need 'half the number of carbon atom Acetyl coA' ; half the number of carbon atom minus 1 ATPs and since we don use FADH2`s here, so, half the number of carbon atom minus 1 time 2 NADH .... in other words replace the FADH2s by NADHs e.g. for making palmitate (16c) we`ll need 8 Acetyl CoA+ 7 ATP + 7 NADH + 7 NADH for rest of the pathways, just trying to memorize all the regulatory enz so that I can name them off the top of my head GL
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Ancylostoma
| Forum Guru Topics: 43 Posts: 632
Its a kaplan q bank question, I'm glad you brought it up.. I totally forgot about the activation by fattyacyl coa synthase. Anyway, i found the answer and it says that activation requirs 2 atp. Doesnt really explain. Edited by Ancylostoma on Feb 05, 2007 - 11:39 PM
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Ancylostoma
| Forum Guru Topics: 43 Posts: 632
By the way, I found it in lippincotts illustrated reviews, it does take 2 atp for activation, even though it doesnt mention it in kaplan. I just made a note next to it in my book.
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vijay vegesna
| Forum Newbie Topics: 1 Posts: 17
Answer is 95. it has 5 breaks in b-oxidation generating 5 FADH2 and 5NADH2. A total of 5x5=25 ATP in B-Oxidation. each acetyl co a enters the krebs cycle and generates 12 ATP so a total of 72 ATP. 2 ATP are used in activation of the fatty acid. so a net of 25+72-2=95
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