Viewing 8 reply threads
  • Author
    • #10649

      Questions are excerpted from Deborah Goldberg AP Biology:

      "Questions 5-11
      In a lab experiment, one enzyme is combined with its substrate at time 0. The product is measured in micrograms at 20-second intervals and recorded on the data table below.

      Time(s) 0 (space) 20 (space) 40 (space) 60 (space) 80 (space)100 (space) 120
      Product 0.0 (space)0.25 (space) 0.50 (space) 0.70 (space) 0.80 (space)0.85 (space) 0.85

      7. Why is there no increase in product after 100 seconds?

      8. What would happen if you added only more enzyme after 100 seconds?

      9. What would happen if you added only more substrate after 100 seconds?"

      I know that this might seem like a homework question, when in fact this is NOT a homework question – I am trying to study for a big test.

      Please formulate your answers to questions 7-9.

      I will leave a bunch of space here so that your answers will not be influenced by the actual answers that I will type out below.

      "7. There is no increase in product after 100 seconds because all the enzymes are saturated and are already catalyzing reactions as fast as they can.

      8. Assuming there is excess substrate that enzyme is not colliding with, an addition of enzyme after 100 seconds would increase the rate of reaction until the enzyme, once again, became saturated.

      9. If you add more substrate after 100 seconds, there would be no change because the enzyme is saturated and reacting with as much substrate as it can. The enzyme cannot handle any more substrate. If you added both substrate and enzyme after 100 seconds, the reaction rate would increase temporarily until the enzyme became saturated once again."

      The work I did for this problem (as I did spend much time on this set of questions) is shown below.
      I reasoned as follows: – Well the product is what was made, right? So if you look at times 100 and 120, you can clearly see that there was no increase in product (as question 7 pointed out). But that would mean that from 100 to 120, nothing was made, right (because once again the product is what you end up with, or, in other words, what you make)?????????? So it follows that (and this was my answer to question 7) there was no increase in product because, well, the reaction had finished, right????? So if you added more enzymes (question eight), no reaction would occur and you would still have the same amount of product (which was the substrate). It would also follow, then (question 9), that if you added only more substrate after 100 seconds, then the reaction would resume (as the one enzyme you had in the beginning would work again) and there would be more and more product until that reaction, too, was finished and then the product would remain the same.

      Now, having explained my reasoning, can someone please explain to me why my reasoning was faulty???????????????? I’m not sure about this, but I have a hunch – I think that my reasoning fell apart, as after this step, my reasoning from there looks pretty good – I think that I assumed the wrong definition of product. So, if I was indeed wrong in my definition of product in this context, could someone please give me the correct definition of product in this case?

    • #88030

      I had to put the word space in between the numbers because the reply mechanism messed up my formatting. 🙁

    • #88037

      There are two possibilities:

      1) if there was only 0.85 micrograms of total substrate present, the reaction could just stop after consuming all of the substrate there was. If the reaction is not easily reversed (which can happen) then nothing happens until you add more substrate, and then the reaction can run for awhile again, accumulating more product.

      2) you are in steady state; the number of molecules of product formed is balanced by an equal number of products being consumed by the reverse reaction, regenerating substrate. Under steady-state conditions, if you add more substrate, you will transiently increase the rate of product formation until you re-establish a new steady state.

      Steady state is not the same thing as saturation. You are correctly describing what saturation is, but the experimental condition may not be at saturation yet. The steady state assumption only requires that Rate(forward) = Rate(reverse) and that will be determined by the ratio of the microscopic rate constants and the equilibrium product/substrate concentrations.

    • #88040

      Thanks a lot, blcr11 !!!!!!!!!!!!!!! 🙂

      So just to clarify: So are you saying that there are two possible interpretations, but based on the answers to the questions, you deduced that only one interpretation worked? If so, selection (2) was correct, right (because 1 would not work because as the solution to question #7 states: the enzymes are still working) ?

      If so, I have one question: then why does the solution to question 7 include the word catalyzing – I mean, wouldn’t that imply that the enzymes are only breaking up the substrate???????????

    • #88045

      My best guess is that the reaction is in steady state, but I don’t know that for a fact.

      Your typical enzyme reaction is reversible. When you first mix substrate with enzyme there is no product around. The rate of product formation is proportional to the product of the initial enzyme and substrate concentrations. The enzyme concentration remains fixed since the enzyme is a catalyst and is not consumed in any way by the reaction. But as the substrate is consumed and product is made, the enzyme can now start to catalyze the reverse reaction, regenerating substrate from product, or, if you prefer, the product is consumed by the reverse reaction. At steady state, the rate of product formation is balanced by the rate of product consumption, so there is no longer any net change in product (or substrate) concentration. It is at an equilibrium point determined by the ratio of the microscopic rate constants and the amounts of enzyme and substrate/product in the system. When you add more substrate (provided you are not at saturation) you will temporarily drive the forward reaction a bit (by good old Le Chatelier’s principle), generating more product. Fairly quickly, though, the rate of forward and reverse reactions will equilibrate again at some new ratio of substrate to product concentrations—a new steady state condition ensues. The effect of adding enzyme to a system at steady state is less certain. Most likely the same thing will happen as does when you add more substrate, but it really depends on the ratio of the microscopic rate constants; it is possible for additional enzyme to give a slight boost to either the forward or the reverse reaction, or if the equilibrium constant is 1, it may have little effect at all.

      An enzyme is a catalyst. To say that the enzyme is still “catalyzing” is just to say that the enzyme is still active. At steady state, the enzyme is “catalyzing” the production and consumption of both substrate and product. To say that the enzyme is saturated is to say that there are more substrate molecules around than there are catalytic sites for them to bind to. At or above saturation levels of substrate, the reaction is no longer sensitive to further increases in substrate concentrations. This is not the same thing as steady-state. Above saturation, there is net product formation until enough substrate is consumed and the system is no longer saturated.

    • #88047

      Thanks a lot, blcr11 !!!!!!!!! 🙂

      I just need to clarify up some points:
      1) For the "catalyzing" part, I was thinking more about catabolism (as opposed to anabolism) and totally forgot about catalysts.
      2) The problem starts out with the assumption of one enzyme – "In a lab experiment, one enzyme is combined…" so it would not be possible for the forward and backward reactions to cancel out, as there is only one enzyme (doing the reactions). I am wondering, however, if it might be possible that the backward reaction will generate the product (as the product is now the substrate as the enzyme breaks it down)? But, wait, it’s not possible because problem started out with the product as 0.0, so it took the substrate to be 0.0. So if the enzyme does the reaction two times, you will start out with the original substrate and that will count as 0.0, so you would have to subtract from 0.85. However, it still might be possible because at 120, it might be possible that the enzyme has just finished doing the reaction three times, so that the substrate became the product and then it decomposed to its original, substrate form before being reacted upon again to form the product (and that was the point in time 120, so that product was still 0.85)….

      Looking back at the problem, could it have been a typo and instead of product, could they have meant the productivity rate – because the wording of the solution made it sound like productivity level (although they used product here instead of productivity level, making me question whether or not my hypothesis is correct) ????

      However, does anyone have any alternative theories or any other possible explanations of this problem?

      Any and every insight offered with regards to this problem is greatly appreciated and I thank you in advance.

    • #88064

      "so it would not be possible for the forward and backward reactions to cancel out, as there is only one enzyme (doing the reactions)"

      OK, I can’t guarantee that’s what’s going on here, but the same enzyme can and often does catalyze the forward and the reverse reactions. It is certainly possible for the forward and reverse reactions to "cancel out." That’s the meaning of "reversible" and enzyme reactions are "reversible."

      I can’t tell from your description what the units are for the product. Total moles of product is possible. So is a concentration. So is a velocity. Read the book again and see what the book says.

    • #88075

      I didn’t say that as clearly as I should have. "Reversible" means that the same enzyme catalyzes both the forward and reverse reactions. "Steady-state" means that the two reactions "cancel out."

    • #88083

      Thanks!!!!!! 🙂

      Looking back, does anyone think that this problem could have been worded better to avoid confusion and ambiguity? If so, could someone reword the question (and post it)??????

Viewing 8 reply threads
  • You must be logged in to reply to this topic.