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Thread: Cellular Respiration

  1. #1 Cellular Respiration 
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    I am attempting to account for all the H2O used in cellular respiration as "waste products".

    C6 H12 O6 +O6 = 6CO2 + 6H2O + energy. How do we arrive at only 6H20 as the waste product?

    When you come down to the ETC, we are left with 12 H2O (technically, each of the 10 NADH and 2 FADH help form 1 water molecule?).....which does not equal 6H20 as in the initial redox equation of glucose.

    But......6H20 are ADDED during the Krebs Cycle (3 H2O times 2 turns), this would be subtracted from the 12H20 leaving 6H20, all fine and well until.....

    it turns out, that Glycolysis produces 2H2O which bumps up the ending equation to 8 H2O, not 6.

    I have accounted for the CO2 and the ATP....but the H2O slips through my hands like water!



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  3. #2  
    Forum Professor Zwirko's Avatar
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    I'm not sure, but I'd make a guess that it's because a) the electrons supplied by FADH2 (in the succinate dehydrogenase complex) don't come from the original glucose molecule, or b) the TCA cycle consumes water also.

    The FADH2 is not really a product of the TCA cycle in the normal sense, but instead is a bound component of the succinate dehydrogenase complex. It cycles back and forth between the FAD and FADH2 forms as the complex carries out the succinate to fumarate step of the TCA cycle as well as passing electrons down the electron transport chain.

    I could be wrong though - I never was great a remembering this maze of reactions.


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  4. #3 Re: Cellular Respiration 
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    I suspect you simply aren't balancing your equations correctly. Re-examine everything.

    Quote Originally Posted by praisgodrua2j
    ...but the H2O slips through my hands like water!
    I loved this. I hope you stay around. We need more bad punsters.
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  5. #4  
    Forum Cosmic Wizard i_feel_tiredsleepy's Avatar
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    First of all you got the reactions in glycolysis, the PDH, and the TCA which give you this:

    C6H1206 + 2 NAD + 2 ADP + 2 HPO4 → 2 C3H3O3 + 2NADH + 2H + 2ATP + 2 H2O

    The bond formation of ATP releases two oxygen and two hydrogen. Everything balanced so far.

    CoA-SH + Pyruvate (C3H303) + NAD → Acetyl-CoA (C2H3O S - CoA) + CO2 + NADH

    Acetyl-CoA + 3 NAD+ + Q + GDP + H2P04 + 2 H2O → CoA-SH + 3 NADH + 3 H+ + QH2 + GTP + 2 CO2

    Double the pyruvate reactions and Smack it all together, considering that free phosphate can be HPO4, or H2PO4.

    Glucose + 10 NAD + 2 Q + 2 ADP + 2 GDP + 4 Pi + 2 H2O → 10 NADH + 10 H + 2 QH2 + 2 ATP + 2 GTP + 6 CO2

    Over to the ETC, which shouldn't really be considered in balancing the equations.

    10 NADH + 10H + 2QH2 + 6O2 + some ADP + some phosphate -> 12 H2O plus some ATP

    However, since it takes approximately 3 protons going through the ATP synthase to make ATP there is always going to be enough H in the matrix to form water, and it isn't likely ever going to be the protons from the NADH or FADH2 (or the ones released from the phosphate). It also isn't likely that the actual efficiency is equivalent to the theoretical 36 ATP, because of the need to keep up a sufficient concentration of protons.

    All this leads me to say that the equation you have up there is wrong. However, it is a nice simplified version that looks balanced.
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  6. #5  
    Forum Professor Zwirko's Avatar
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    Quote Originally Posted by i_feel_tiredsleepy
    ...that looks balanced.
    Except for the fact that 6 oxygen atoms have gone AWOL.
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  7. #6  
    Forum Cosmic Wizard i_feel_tiredsleepy's Avatar
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    I imagine it's a typo and supposed to be 6O2 not 06. Either way, it's not an accurate portrayal of cellular respiration.
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  8. #7  
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    It's just the standard equation used to illustrate the complete oxidation of 1 mole of glucose to 6 moles each of carbon dioxide and water. It's taught in most elementary biology courses. Clearly it is a summary of something that is far more complex. It's taught this way for that very reason.
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  9. #8 Cellular Respiration....am I missing something? 
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    Quote Originally Posted by i_feel_tiredsleepy
    First of all you got the reactions in glycolysis, the PDH, and the TCA which give you this:

    C6H1206 + 2 NAD + 2 ADP + 2 HPO4 → 2 C3H3O3 + 2NADH + 2H + 2ATP + 2 H2O

    The bond formation of ATP releases two oxygen and two hydrogen. Everything balanced so far.

    CoA-SH + Pyruvate (C3H303) + NAD → Acetyl-CoA (C2H3O S - CoA) + CO2 + NADH

    Acetyl-CoA + 3 NAD+ + Q + GDP + H2P04 + 2 H2O → CoA-SH + 3 NADH + 3 H+ + QH2 + GTP + 2 CO2

    Double the pyruvate reactions and Smack it all together, considering that free phosphate can be HPO4, or H2PO4.

    Glucose + 10 NAD + 2 Q + 2 ADP + 2 GDP + 4 Pi + 2 H2O → 10 NADH + 10 H + 2 QH2 + 2 ATP + 2 GTP + 6 CO2

    Over to the ETC, which shouldn't really be considered in balancing the equations.

    10 NADH + 10H + 2QH2 + 6O2 + some ADP + some phosphate -> 12 H2O plus some ATP

    However, since it takes approximately 3 protons going through the ATP synthase to make ATP there is always going to be enough H in the matrix to form water, and it isn't likely ever going to be the protons from the NADH or FADH2 (or the ones released from the phosphate). It also isn't likely that the actual efficiency is equivalent to the theoretical 36 ATP, because of the need to keep up a sufficient concentration of protons.

    All this leads me to say that the equation you have up there is wrong. However, it is a nice simplified version that looks balanced.
    THANKS SO MUCH FOR YOUR INPUT, BUT.....

    I am really a little OCD about the H2O.....exactly where does it balance? I have broken down the various cycles of cellular respiration: BTW, some listed reactions on the net write the equation as C6H12O6 +6O2 +6H20 --->..... which indicates that water is supplied within some of the steps in cellular respiration. But we will deal with the traditional equation of C6H12O6+ 6O2→ 6CO2 + 6H2O+ Energy.


    Glycolysis produces 2 H2O (via reaction with enolase). .

    Then on to Pyruvate Oxidation which neither produces nor uses H2O.

    Then Krebs Cycle which uses 6 H2O (Fumaric acid-->Malic Acid.....Oxaloacetic Acid---->Citric Acid.......Ketoglutaric Acid ----->Succinic Acid....These 3H2O times 2 cycles equal 6 H2O).

    By the time its all said and done, the ETC then seems to allows the 10 NADH and 2 FADH to produce an ending net total of 12 H20. Backtracking to arrive at only 6H2O per original equation:

    From the 12 H2O, we have to add 2 from Glycolysis (=14 H2O).

    From Pyruvate Oxidation, zero (still total of 14 H2O).

    In the Krebs Cycle, we subtract the 6 H2O that are used resulting in a net total of 8 H2O.

    I know that there are two pesky H2O being used somewhere so that we can have world peace again. Dont let me have to put out an APB on them!
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  10. #9  
    Forum Cosmic Wizard i_feel_tiredsleepy's Avatar
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    It's not going to balance because some of the hydrogens and oxygens aren't coming from the glucose. That equation is just meant to illustrate in a simple manner the general process of cellular respiration, and its products.

    Also the net H2O used by two turns of the TCA is 4, because it also produces one H2O each cycle.
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  11. #10  
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    Quote Originally Posted by i_feel_tiredsleepy
    Also the net H2O used by two turns of the TCA is 4, because it also produces one H2O each cycle.
    Forgive me for being a pest, but if what you are talking about is the citric acid--->aconitic acid producing a H2O, isnt it took right back out between the aconitic acid---->isocitric acid? which would leave that particular reaction balance of H2O as zero?

    Otherwise, I am not seeing where else and H2O is produced.

    BTW.....while I am a novice.....I appreciate the love and dedication that all you teachers and professors have put into your area of expertise in order to provide the rest of us with valuable information on a much more simplistic scale. :-D
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  12. #11  
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    The TCA cycle consumes one molecule of water per turn. The conversion of fumarate to malate, by fumarase, is a hydration reaction.
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  13. #12  
    Forum Cosmic Wizard i_feel_tiredsleepy's Avatar
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    There's hydrolysis at the citrate synthase step that generates a hydroxide ion to break the bond between CoA and the citrate.
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  14. #13  
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    I'm starting to remember now why I hated learning metabolic pathways.
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