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Aerobic Respiration

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    • Thread Starter

    Hi :-)
    Okay, so most of the biology aqa course I'm okay with. But there is one blank spot I have that regardless of what method I try, I simply cannot fill.

    Aerobic respiration.

    Glycolysis, the Link Reaction, Krebs and the ETC just refuse to stick in my head. And the stages are impossible to stick into a song. So can someone help me out?
    I've been looking everywhere, but I can't find a simple step-by-step for each. By 'simple', I don't mean dumb it down. Just literally step by step without arrows flying everywhere.

    If someone could write out a step by step or direct me to an easy to look at step by step, I would be super grateful!!

    Either way, thanks for reading until the end, if you did, and good luck to everyone with their exams this year!

    Step by step. Woah ok. I'm speaking for AQA, I don't know if your exam board is different. I'll try my best to be concise and informative. I'll find the nicest diagrams.
    [Occurs In the Cell Cytoplasm]
    The Preparatory Phase
    1. Glucose is phosphorylated twice, i.e. we add two Phosphate groups to glucose via two ATP. The ATP hydrolyses to ADP and Pi.
    This forms phosphorylated glucose.

    2. Phosphorylated glucose breaks down into two three carbon compounds called Triose Phosphate.

    The Pay-Off Phase
    3. Triose Phosphate is oxidised by NAD+ to Pyruvate in a dehydrogenation reaction, i.e. it loses hydrogen to NAD+ to form NADH or reduced NAD.

    4. In the process, two phosphate groups are removed from triose phosphate in a dephosphorylation reaction which are used to form two ATP.
    You might think, how can two phosphate groups be removed if Triose Phosphate contains one phosphate group?
    That's because you actually add another phosphate group while oxidising it.
    5. One Triose Phosphate = 2 ATP
    Two Triose Phosphate Produced, ∴ 2 x 2 = 4 ATP produced.
    Prepratory Phase (i.e. costs) = 2 ATP
    ∴ ATP gained is 4 - 2 = 2 ATP
    NADH gained = 2

    [Occurs In the Mitochondrial Matrix]
    1. Pyruvate is actively transported into the mitochondria via a carrier protein.

    2. Pyruvate is oxidised by NAD+in a dehydrogenation reaction, i.e. it loses hydrogen to NAD+ to form NADH or reduced NAD.

    3. Pyruvate then undergoes a decarboxylation reaction, i.e. it loses CO2. This forms a two carbon compound called Acetate.
    Steps 2 and 3 are catalysed by enzymes.
    Unimaginatively, their names are Pyruvate Decarboxylase and Pyruvate Dehydrogenase respectively.
    4. Acetate combines with a coenzyme called Coenzyme A or CoA. This forms a molecule called Acetyl-CoA
    A coenzyme is something that assists the function of an enzyme, usually by transferring the substrate to the enzyme or between enzymes. In this case, CoA is transferring Acetate to the next enzyme in the next process.
    5. One Pyruvate = 1 NADH
    One Pyruvate = 1 CO2Two Pyruvate Produced in Glycolysis∴ NADH gained = 2 CO2 produced = 2

    THE KREBS CYCLE[Occurs In the Mitochondrial Matrix]
    1. Acetyl-CoA enters the Krebs Cycle. It combines with a four carbon compound to form a six carbon compound called Citrate.
    Remember, 2C from Acetate + 4C from this compound = 6C aka Citrate.
    The four carbon compound is called oxaloacetate.
    This is also why the Krebs Cycle is formally called the Citric Acid Cycle.
    2. When Citrate is formed, Coenzyme A (CoA) is removed and goes back to the Link Reaction.

    3. Through a series of redox reactions, Citrate is oxidised back to the four carbon compound. This is detailed in the next steps.

    4. Citrate is oxidised by NAD+in a dehydrogenation reaction to form NADH or reduced NAD. This happens three times.

    5. Citrate is also oxidised by FAD+in a dehydrogenation reaction to form FADH2or reduced FAD. This happens once.

    6. Citrate also undergoes two decarboxylation reactions, i.e. it loses two CO2.
    Why two? Well if we're trying to get Citrate (6C) back to our 4C compound, then we need to lose two carbon atoms.
    This is done by the loss of two carbon dioxide molecules.
    i.e. 6C - 2C = 4C.
    7. 1 ATP is also produced via substrate level phosphorylation, i.e. the addition of phosphate to another molecule directly from a phosphorylated substrate.
    We've seen that in glycolysis, when Triose Phosphate adds Phosphate to ADP.
    If yo're interested, Succinyl-CoA is dephosphorylated to Succinate, the phosphate released combines with ADP.
    8. One Acetyl-CoA = 1 ATP
    One Acetyl-CoA = 3 NADH
    One Acetyl-CoA = 1 FADH2
    One Acetyl-CoA = 2 CO2
    One Pyruvate = One Acetyl-CoA
    Two pyruvate produced in glycolysis ∴ two Acetyl-CoA

    ∴ ATP gained is 2 x 1 = 2NADH gained is 2 x 3 = 6
    FADH2 gained is 2 x 1 = 2
    CO2 produced is 2 x 2 = 4

    THE ELECTRON TRANSPORT CHAIN (ETC)[Occurs At the Inner Mitochondrial Membrane]
    1. The Electron Transport Chain is a series of four proteins, called complexes,that span the whole of the inner mitochondrial membrane.

    2. NADH is oxidised at the first protein and loses the Hydrogen Ion, H+, (which is just a proton) and two electrons to form NAD+. This returns to the Glycolysis Pathway, the Link Reaction and the Krebs Cycle.

    3. FADH2 is oxidised at the second protein and loses the two Hydrogen Ions, 2H+,(which are just protons) and two electrons to form FAD+. This returns to the Krebs Cycle.

    4. The electrons are carried along the Electron Transport Chain by an Electron Carrier. As they move from one protein to the next, the electrons drop in energy level. This releases energy.
    The electron carrier is called Q. Because it undergoes several redox reactions, it has many states and thus many names. We just call it Q.
    5. The energy of the electrons is used to pump the protons from the matrix to the intermembrane space i.e. the space between the inner and outer mitochondrial membranes.

    6. There is a high concentration of protons in the intermembrane space.
    There is a low concentration of protons in the matrix.
    The membrane is impermeable to protons, i.e. they cannot passively diffuse
    Thus, there is a concentration gradient, referred to as the proton gradient.

    7. Protons diffuse down their concentration gradient via a protein called ATP Synthase. It uses the energy of the protons to catalyse the synthesis of ATP by combining ADP and Pi. This process is called chemiosmosis.
    Three protons = 1 ATP
    More or less, sometimes it can be 4.
    8. At the fourth and final protein, the electrons are donated to Oxygen. Protons are also donated.
    Altogether, Oxygen is reduced to Water via the addition of protons and electrons at the final protein.
    i.e. O2 + 4e- + 4H+ ---> 2H2O

    Oxygen is called the Final/Terminal Electron Acceptor.

    9. This is called Oxidative Phosphorylation as it uses Oxygen to phosphorylate ADP to form ATP.

    10. Research has shown that the energy of NADH = 2.5 ATP
    and that the energy of FADH2 = 1.5 ATP.
    Collate all the NADH and FADH2 in all the previous processes.
    Glycolysis = 2 NADH
    Link Reaction = 2 NADH
    Krebs Cycle = 6 NADH
    ∴ Total NADH = 10

    Krebs Cycle = 2 FADH2
    Total FADH2 = 2

    ∴ ATP produced in ETC is (2.5 x 10) + (1.5 x 2) = 28

    Oxygen used = 6
    NADH and FADH2 both release 2 electrons.
    Oxygen accepts 4 electrons.
    10 NADH = 20 electrons
    2 FADH2 = 4 electrons.
    ∴ Total electrons in ETC = 24
    ∴ Number of Oxygen to accept electrons is 24/4 = 6
    Add up all the ATP from each process.
    2 ATP in Glycolysis.
    2 ATP in Krebs Cycle.
    28 ATP in ETC.

    ∴ TOTAL ATP is 28 + 2 + 2 = 32 ATP
    The number varies slightly for reasons outside the specification. It's do with how the NADH from Glycolysis is transported to the ETC.
    I hope this is condensed. If it looks too scary or something's confusing still, just say it.
    • Thread Starter

    Thank you so much!! This is really helpful, you're a lifesaver :-) I'll be printing this out to learn it by heart, thank you again!! :-)))

    (I hope I posted the reply right, I'm such a noob with these sorts of things :') )

    (Original post by cherrybrains)
    Thank you so much!! This is really helpful, you're a lifesaver :-) I'll be printing this out to learn it by heart, thank you again!! :-)))

    (I hope I posted the reply right, I'm such a noob with these sorts of things :' )
    It's all good. I got a notification in the thread.
    Awesome, hope you understand and good luck with your studies
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