If you haven't solved question 14 yet, pause the video now and try it on your own. Muscle contraction involves remember that those ATPs are used to allow the myosin heads to release from the actin filaments during muscle contraction and again, ATP is a form of stored chemical energy and muscle contraction will result in motion which is a form of kinetic energy.
Biological oxidation-reduction reactions always involve the transfer of electrons. While A through D are all possibilities, certainly oxygen frequently participates in redox reactions. It's not a necessity though. Likewise, water will sometimes be formed from redox reactions and specifically the, reaction at the end of electron transport will form water. But again not a necessity. And likewise, there's a lot of redox that goes on in the mitochondria. This is all stuff we're going to learn about in the next unit. And lastly, while hydrogens can be transferred, again, it's not a necessity. The only thing that is a necessity for a redox reaction is the transfer of electrons. The others are possibilities but again, they're not absolutes.
Looking at question 16, it says, "The standard reduction potentials for the following half-reactions are..." It gives us some data. Then it asks, "If you mixed succinate, fumarate, FAD, and FADH2 altogether at one molar concentrations at pH7 and basically that 1 molar concentration at pH7 is just reaffirming that we're using these conditions, the prime not conditions. If we mix it all together in the presence of succinate dehydrogenase, which of the following will happen?" Okay. So something you need to know to understand this question is what succinate dehydrogenase does. Succinate dehydrogenase takes succinate and forms fumarate. That's succinate dehydrogenase. Again, this is a reaction we're going to talk about when we get to the next unit. So, this question is saying alright, if we put all the ingredients in, given the, standard reduction potentials, so if you put all the ingredients in at these standard conditions, right? These prime not conditions. What's going to happen? So this is not saying what's going to happen at cellular conditions. That's an important difference to make, right? These conditions are not cellular conditions. So, this is the reaction that occurs in your cells but that might not necessarily be what's happening here. What we have to do is we have to look at these reduction potentials to decide what's going to happen. And the way to do this is you want to think about, you want to see which, direction gives you the greatest positive value, right? The biggest value for E, that's going to be the direction things are going to move.
So here we're given reduction potentials, right? This is the reduction of fumarate and the reduction of FAD. So what we want to know is at these conditions, how do we get the biggest value for E prime not, right? And we need to realize is that if you reduce 1 like if we're going to reduce fumarate for example, then we're going to need to oxidize FAD. So because you know, looking at this reaction up here, you see that FADH2 and fumarate are on the same side of the equation but in our half reactions right here, notice how they're on opposite sides of the equation. That means we're going to have to flip one of these values to find our answer. So we could say okay, well let's reduce FAD and let's actually oxidize fumarate. Well that means that this value here is going to become negative because this value is for the reduction of fumarate. The oxidation would be the opposite. So the oxidation of fumarate would give us negative 0.031 volts. So, if we were to oxidize if we were to oxidize this, meaning we're going to go from succinate to fumarate, right? If we reverse our half reaction here, this half reaction, we're actually going to be taking it in reverse, we get this negative value. And this half reaction with FAD is going to proceed in the direction we see here and then so we're going to keep this negative value here. So our total for e is going to be -0.031 plus -0.219 volts, right? We're going to get that's going to give us a negative number.
Now, what would happen if we did the opposite? Let's actually go ahead and do the opposite here, okay? So let's say that instead we're actually going to take we're going to take this reaction, the second half reaction in the opposite direction, right? So we're going to instead of reducing FAD, we're going to oxidize FADH2. Well, that is going to lead us to have a reduction potential of positive 0.219 volts, right? That means that if we proceed So looking at our equation up top here, if we're actually going to proceed in this direction, that's going to give us a positive value for E prime naught, which is what we want. We want the greatest positive value because that's going to be the direction things are going to proceed. Meaning that actually what we are going to have is not the reaction that normally occurs in cells but actually the opposite. We're going to go from fumarate to succinate and we're going to go from FADH2 to FAD. And that means that, that means that our answer is that in fact, fumarate will be reduced and FADH2 would become oxidized. So just to recap, basically because of the numbers we're given, so because of the conditions we're doing this reaction and it's actually more favorable for this reaction that normally occurs the way it's written here to actually occur in the opposite direction, meaning that instead of FAD getting reduced, we're going to oxidize FADH2 and instead of, instead of acting or going from succinate to fumarate, we're going to go from fumarate to succinate meaning we're going to reduce fumarate. So the answer here is B.
Alright. Moving on to question 17. The hydrolysis of phosphoenol has a delta G of about negative 62 kilojoules per mole. Remember that's that's super favorable reaction at the end of glycolysis. The greatest contributing factor to this is dynamite. No, I'm kidding. It's tautomerization. Remember that as soon as that reaction is done, pyruvate actually comes out of that reaction in the enol form, right? So PEP goes to pyruvate but it's pyruvate in the enol form and it's actually pyruvate would much rather be in the keto form. So, there is a tautomerization there. So there is a tautomerization that occurs and it is, pyruvate moving from the enol form to the keto form.
Let's move on to question 18. So the reverse reaction of phosphoglucoisomerase has a Keq of 1.97. What is the delta G for delta G prime not for this reaction using, 2.5 kilojoules per mole for RT. So, first thing we need to do is figure out what equation we're going to need for this reaction. The equation we're going to use is delta G prime not equals negative RTln of KEQ. Now, there's there's an equation very similar to this one. That's actually what we're going to use in part b. So hopefully you use the correct equations in the correct places here. So, we are being asked what is, what is our delta G for the reverse reaction? And notice that we're given the Keq for the reverse reaction. So this is this is really just a simple plug it in. Right? We're told RT is going to be 2.5 kilojoules per per mole. So we negative 2.5 kilojoules per mole times. Let me do it this way actually times ln of 1.97. And if you plug that all in, what you get is -1.7 kilojoules per mole. And you simply just plug that in on your calculator. Use the, LN button.
Looking at part B for this problem. It says, if the cellular concentration of the substrate of this reverse reaction is 1.2 millimolar and the product is 0.6 millimolar, now now we want the plain old delta G. Meaning, we're going to have to use this equation that relates those 2. So delta G equals delta G prime not plus RTln q. And q, just as a refresher, q is the ratio of products to reactants like during the process of the reaction. So, here we're given the values we need for that ratio. And again, this is just another simple plugin. So this is all equal to to and I'm just gonna hop out here so I have more room to write. So this is equal to we found, delta delta the previous equation, 1.7 kilojoules per mole plus rt. We We again know from the previous equation. It's 2.5 kilojoules per mole and that is times ln of q. And remember, here q is going to be our, products over reactants, so that's 0.6 millimolar over 1.2 millimolar which is really, right, this is really just 0.5, right? Or 1 half. So plug that all in. Solve it out and you should get negative 3.4 kilojoules per mole. All right. So hopefully you knew, which equations to use where. And after that, these are really just simple plugin problems. Nothing too tricky here other than knowing how to find Q right here. And remember, Q is products over reactants. Say concentration products over concentration reactants. Let's flip the page.
