TPR FL 1 C/P

[avalonisland888]

1) I'm not quite understanding the explanation for this question. They talk about how the Cl- ion can react with the Ag+ and how this can ultimately create more I- ions as NaCl concentrations increase while keeping the Ag+ concentrations low. I assumed that since there is no common ion effect, there shouldn't be any change at all.

2) Why does the NH+ to N in raclopride-H+ not count as oxidation if we're losing the H?

3) Why are ATPases and GTPases hydrolases instead of oxidoreductases? The way I understand it is GTP/ATP gets reduced to ADP/GDP. I looked it up online and it does that ADP is the reduced form of ATP, but am I just interpreting this incorrectly since it's more of a hydrolysis and not a reduction? I think I'm getting confused with the enzyme categories because let's say for succinyl CoA synthetase. It's a ligase and not a hydrolase despite producing GTP. Is the difference because GTP/ATPases' substrate IS the ATP/GTP whereas for the synthetases and dehydrogenases, their substrate is something else but they PRODUCE the energy?

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[Instance_Variable]

1) If you add NaCl to the solution, you will form Na+ and Cl-. Then, Cl- can react with the dissolved Ag+ to form AgCl. AgCl is quite insouble in water, so some of that will precipitate out of solution. Now, go back to the original equation with silver iodide: AgI <-> Ag+ + I-. According to Le Chatelier's Principle, removing products will cause the reaction to shift to the right. Since the Cl- is forming AgCl, Ag is being removed from the right side of the equation. The question states that AgI precipitate is present, so as Ag+ is removed, more of this will be allowed to dissolve, thus raising the I- concentration.

2) Oxidation/reduction has to do with the loss/gain of electrons, respectively. Notice how when you lose an H, the number of electrons remains the same - the electrons that were once in the N-H bond are now a lone pair on the nitrogen atom (which is why the nitrogen atom no longer has a charge). There was no net gain or loss of electrons. The lone pair isn't drawn though, so you have to be careful there and notice that since the nitrogen is no longer charged, there must be a lone pair there.

3) Hydrolases cleave bonds with the addition of water. So any time you have a reaction taking place where water goes in and you get two things in the products, then chances are it's a hydrolysis reaction and a hydrolase enzyme is involved. An oxidoreductase, on the other hand, catalyzes oxidation/reduction reactions between molecules. These reactions involve things such as NADH or NADPH and a direct transfer of electrons between molecules. Even though ATP and GTP are reduced in ATP/GTP hydrolysis, you should consider this as a byproduct of the more important thing happening: the enzyme is catalyzing the cleavage of a phosphoanhydride bond (by using water). There is no transfer of electrons occurring in ATP hydrolysis, but a bond is being broken and water is consumed to break that bond. Thus, it's a hydrolase, not an oxidoreductase.

 

[avalonisland888]

1) If you add NaCl to the solution, you will form Na+ and Cl-. Then, Cl- can react with the dissolved Ag+ to form AgCl. AgCl is quite insouble in water, so some of that will precipitate out of solution. Now, go back to the original equation with silver iodide: AgI <-> Ag+ + I-. According to Le Chatelier's Principle, removing products will cause the reaction to shift to the right. Since the Cl- is forming AgCl, Ag is being removed from the right side of the equation. The question states that AgI precipitate is present, so as Ag+ is removed, more of this will be allowed to dissolve, thus raising the I- concentration.

2) Oxidation/reduction has to do with the loss/gain of electrons, respectively. Notice how when you lose an H, the number of electrons remains the same - the electrons that were once in the N-H bond are now a lone pair on the nitrogen atom (which is why the nitrogen atom no longer has a charge). There was no net gain or loss of electrons. The lone pair isn't drawn though, so you have to be careful there and notice that since the nitrogen is no longer charged, there must be a lone pair there.

3) Hydrolases cleave bonds with the addition of water. So any time you have a reaction taking place where water goes in and you get two things in the products, then chances are it's a hydrolysis reaction and a hydrolase enzyme is involved. An oxidoreductase, on the other hand, catalyzes oxidation/reduction reactions between molecules. These reactions involve things such as NADH or NADPH and a direct transfer of electrons between molecules. Even though ATP and GTP are reduced in ATP/GTP hydrolysis, you should consider this as a byproduct of the more important thing happening: the enzyme is catalyzing the cleavage of a phosphoanhydride bond (by using water). There is no transfer of electrons occurring in ATP hydrolysis, but a bond is being broken and water is consumed to break that bond. Thus, it's a hydrolase, not an oxidoreductase.

I understand Q1 and Q2 now - thank you! For Q3, I think I still need clarification cause now I'm getting confused between lyase and hydrolase. I understand that lyase does not require energy and hydrolysis and breaks bonds just like hydrolases. But when I look at citrate synthase in the Krebs Cycle, it's apparently a lyase but why is this a case if the reaction it catalyzes is a condensation and then hydrolysis?

 

[avalonisland888]

I understand Q1 and Q2 now - thank you! For Q3, I think I still need clarification cause now I'm getting confused between lyase and hydrolase. I understand that lyase does not require energy and hydrolysis and breaks bonds just like hydrolases. But when I look at citrate synthase in the Krebs Cycle, it's apparently a lyase but why is this a case if the reaction it catalyzes is a condensation and then hydrolysis?

The same for some other enzymes like fumarase where it catalyzes a hydrolysis reaction but is still considered a lyase?