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Review 2: Biosignaling, Glycolysis, Gluconeogenesis, & PP-Pathway

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if you haven't solved question 14 yet, paused the video now and try it on your own muscle. Contraction involves the conversion of chemical energy in the form of ATP to kinetic energy in the form of movement. Uh, remember that those ATP's are used. Thio allow the Miocene heads to release from the acting 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. Or let's move on to question 15. Biological oxidation reduction reactions always involved e the transfer of electrons, while a through D are all possibilities, Uh, 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 the mitochondria. This is all stuff we're gonna learn about the next unit. And lastly, while hydrogen is can be transferred again, it's not a necessity. The Onley thing that is a necessity for Redox reaction is transfer of electrons, the others air possibilities, But again, they're not absolutes looking at questions. 16. It says the standard reduction potentials for the following half reactions are gives us some data. Then it asks if you mixed second, a few Marie F A. D and F A. D. H two altogether at one Mueller Concentrations at Ph. Seven. And basically that one Mueller concentration at Ph. Seven is just reaffirming that we're using these conditions. The prime not conditions if we mix it all together in the presence of sucking eight d hydrogen is which of the falling will happen. Okay, so something you need to know to understand this question is what sucking it di hydrogen is. Does sucking it di hydrogen ISS takes second eight and forms FEMA rate. That's sickening de oops, de hydrogen ease Again, This is a reaction we're gonna talk about when we get to the next unit. So this question is saying all right if we put oh and sorry in the process, uh, f a D h two is formed, so you have f a d and you form f a. D h two. Okay, so this question is basically saying if you put all the ingredients in, given the standard reduction potentials. So if you put all the ingredients in at at these standard conditions right, these prime not conditions. Um what? What's going to happen? So this is not saying what's gonna happen. It's cellular conditions. That's 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, um, 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 they're gonna move. So here were given reduction potentials, right? This is the reduction of humor. Rate is this number and the reduction of F a d is this value. So what we want to know is, uh, at thes conditions, how do we get the biggest value for e prime? Not right. And we need to realize is that if you if you reduce one like if we're going to reduce fume rate, for example. Then we're gonna need thio oxidize f a d uh, so basic, Because looking looking at this reaction up here, you see that f a D h two and fume rate 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 gonna have to flip one of these values to find our answer. So we could say Okay, well, let's let's reduce f a d and let's actually oxidize FEMA rate. Well, that means that this value here is going to become negative because where this is a, this value is for the reduction of humor rate. The oxidation would be the opposite. So the oxidation of humor, it would give us negative 0.31 volts. So if if we were thio oxidize if we were thio oxidize this meaning we're going to go from sucking eight to FEMA, right? Right. We reverse our half reaction here. This half reaction, we're actually gonna be taking it in reverse. We get this negative value right, And this half reaction with f a D is gonna proceed in the direction we see here. And then So we're gonna keep this negative value here. So our total for E is going to be negative. 031 plus negative. 0.219 volts. Right. We're gonna get that's going to give us a negative number. Now, what would happen if we did the opposite? It's actually go ahead and do the opposite here. Okay, so let's say that instead we're actually gonna take We're going to take this reaction, the second half reaction in the opposite direction. Right? So we're going Thio instead of reducing f a. D. We're gonna oxidize f a d h two. Well, that is going to lead us to have a reduction potential of positive too point to 19 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, not which is what we want. We want the greatest positive value because that's going to be the direction things they're going to proceed, meaning that actually what we're going to have is not the reaction that normally occurs and sells, but actually the opposite. We're gonna go from few Marie Thio, Soeken eight. And we're going to go from F A D h two thio f a d. And that means that, uh that means that our answer is that in fact, humor eight will be reduced and F a d h two would become oxidized. So just to recap, basically, because of the numbers were given. So because of the conditions were doing this reaction and it's actually more favorable for this reaction that normally occurs the way it's written here, Toe actually occur in the in the opposite direction, meaning that instead of f a d getting reduced, we're gonna oxidize f a d h two and instead of instead of acting, are going from second to fume rate. We're gonna go from FEMA rate to circulate meeting. We're gonna gonna reduce FEMA rate. So the answer here is be all right. Moving on to your question 17. The hydraulics ISS of Foss Final Piru Bit has a delta G of about negative 60 to kill a Jules per mole. Remember that? That's super favorable reaction at the end of like Hollis iss. The greatest contributing factor to this is dynamite. No, I'm kidding. It's too taameri ization. Remember that as soon as that reaction is done, our pyro of eight, our pyro of it actually comes out of that reaction in the in all form. Right? So help goes to Piru of it. It's Peruvian, the Enel form. And it's actually, uh, Pirou Bit would much rather be in the key toe form so that there is a taut amortization there. So there is a tot amore ization that occurs and it is Piru bit moving from the in all form Thio, the key toe form. All right, let's move on. Thio question 18. So the reverse reaction phosphate glucose I som race has a cake of 1.97 What is the Delta G for Delta G Prime? Not for this reaction using 2.5 kg per mole for rt. So first thing you need to do is figure out what equation we're gonna need for this reaction. The equation we're gonna use is Delta G crime not equals negative. R t l n of k e que Now there's, uh there's an equation. Very similar to this one. That's actually what we're going to use in part B. So hopefully you used the correct equations in the correct place is here. So we are being asked what is what is our delta g for the reverse reaction and notice that we're given the cake for the reverse reaction. So this is this is really just a simple plug it in, right? Uh, we're told rt is gonna be 2.5. Eligible promotes. We have negative 2.5 kill a jewels per mole times. Let me do it this way. I actually times Ln of 197 And if you plug that all in, what you get is negative 1.7 killer jewels per mole, and you simply just plug that in on your calculator. Used the Ellen button. All right. Looking at be part B for this problem. It says if the cellular concentration of the substrate of this reverse reaction is 1.2 millimeter and the product is 0.6 millimeter, what is the delta g of the reverse reactions to Remember before we found the Delta G at these conditions. Now you want the plain old Delta G, meaning we're gonna have to use this equation that relates those two. So Delta G equals Delta G pretty not. Plus rt Ellen Que and cue. Just a za refresher. Que is, um, the the ratio of products to react INTs like during the process of the reaction. So here were given the values we need for that ratio. And again, this is just another simple plugged in. So this is all equal to you, and I'm just gonna hop out here, so I have more room. Right. So this is equal to We found, uh, Delta, Delta G in the previous equation, 1.7 killed jewels per mole plus rt. We again? No, from the previous equation, it's 2.5 killer Joel's Permal. And that is times Ellen of Q. And remember here, Hugh is going thio be our products. Overreacting. So that 06 million Mueller over 12 million Mueller, which is really right. This is really just 0.5 or one half. So plug that all in Solve it out and you should get negative three 0. killer jewels. Her more. All right. So hopefully you knew which equations to use where and after that, these air really just simple plug in plug in problems. Nothing too tricky here other than knowing how to find Q. Right here. And remember. Q. I'm just gonna write it over here. Q. Is products over react? INTs uh, state concentration products over concentration react INTs. All right, let's flip the page.

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