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Today we are going to come up with new task, that is about “Question 8 (1 point) the graph of enzyme activity (Vmax) vs pH shows an increase to a maximum at pH 7. There are 2 points of inflection in the bell shaped curve, on…”. We have found the solution to this question, along with interesting information related to it, thanks to its novelty and interest. This will help you to develop a more research-oriented mindset, and make it easier for you to answer these types of questions. This is where we will focus to gain this useful and new knowledge!
Question: ” Question 8 (1 point) the graph of enzyme activity (Vmax) vs pH shows an increase to a maximum at pH 7. There are 2 points of inflection in the bell shaped curve, on…”
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Question 8 (1 point) The graph of enzyme activity (Vmax) vs pH shows an increase to a maximum at pH 7. There are 2 points of inflection in the bell shaped curve, one at pH= 4.1 and another at pH= 9.
Which of the the following is always true?
a. An ionizable group with a pK of about 4 must be protonated from the enzyme to be most active.
b. lysine is a required residue in catalysis.
c. An ionizable group with a pK of about 9 must be protonated from the enzyme to be most active.
d. An ionizable group with a pK of about 7 must be deprotonated from the enzyme to be most active.
e. none of the above.
Question 9 (1 point) If AG <0 then K > 1
This is a three-part problem. The first is what happens to that function. As s approaches infinite, this is our best function. These are the variables. It says in the corner that A is Constance and K is Constance. These are just numbers. As this approach to infinity, it will be all that is required. Really? Look at the leading terms for the numerator as well as the denominator. Again, s is our variable. On the top, there is only one term which is the leading term at the bottom. Kay is a constant here so it can fade to the background as it is not our leading term. The variable with the highest power is the leading term. This is the only one that has all the variables. This is the one. Now we have only two terms. One on top and one at the bottom. If you are looking at behavior, and if it can be reduced, that’s great. Now, let’s look at the bottom. There is no way to cancel outs, so you will only have this value of a. This is our overall limiting response value or limiting value in this problem. That’s why a is called so when s approaches infinite. You simply end up with an overall limited reaction rate. As your overall result. Ok, now for part B. It suggests that s equal to come. Let’s see how the overall results look. Instead of thinking s equals two k, I suggest that we think K equals s. In other words, rewrite the expression This function with K bit value beings. Observe how the course and the name writer will be the same when this happens. Then, enter the nominator. You’ll get yes, plus ass. This is basically what you end up with. Notice how s is at the top and bottom. Which you can cancel out. These will be removed and replaced with one. This will give you exactly half your limiting value. The living value is, naturally, Okay. It reading up having is now our function. They’ve also given us some different values to use in this function. It states, “OK, find the S value that creates a function value that is 75% less than the limiting value with the limiting value.” We wanted to allow our member’s function value to be 75% as a result of the example of the limiting value of 0.1. What exactly is this value? Take your 0.1. Now, you’re a value. 75% of it is multiplied by 750.75. All that’s going to do is move the decimal one place left. We’ll end with 1075 four-hour functions values as our best value. Okay, so this value, we’re going let be 0.75, and our value will be listed as this value. Red, we’ll have it set at 0.1 and then we’ll go. Let Kay equal 1.25. This will allow us to solve for us as the only variable we have left in this equation. So how can we solve for snow? It’s the best vehicle. We didn’t just want to get rid of the year. You can immediately multiply your fraction. This denominator equals 1.25 + s. Can you multiply both sides of this equation? It would cancel out completely over there, as it is on the top, but it would still be over here. Let’s divide both sides by the entire denominator, which is one plus 25 or 1.25 additional processes. Our 0.75 will be left on the right, multiplied with this whole expression of 1.2 fine plus the s. This denominator will cancel, as we have already said on the left-hand side. This was the purpose of multiplying it. We’ll end up with this 0.01 multiplied by the s. What can we do? We can give this 0.7 point 075 between these terms and it will allow us to do more than before. This will allow us to freely move some terms from one side to the other as we wish. This means that if you divide the 0.75 between these terms, you get 0.9 375 for your first term. After multiplying the 0.75 times at 1.25, you will have 0.75 multiplied with s. You get exactly what you had before. It’s fine to use the 0.1. Now that you have this distributed through, move this term to one side to join the other term involving s. This will allow all the SS to be together. It’ll make it simpler for him to solve the problem. Now, subtract 0.75 uh s from the 0.1 ass instead and you’ll get the same 0.9375 left. You’ll find a 0.0-5 s left on this page. This is because the 0.1 was 0.75 less than the 0.1. So that’s it. Ss is right here. Simply divide the sides by 0.25. This will give you as value of three points 75.
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