Enzymes

So now that we're practically experts on amino acids, protein structure and even protein techniques and this video, we're going to begin our discussions on protein function by starting to talk about enzymes. So most of you guys already know from your previous courses that most enzymes are globular proteins that catalyze or speed up the rate of chemical reactions without being consumed by the reaction. And what we really mean by without being consumed is that by the end of the reaction, the enzyme takes on its original state that it had before the reaction took place. So the enzyme is not chemically or permanently altered by the end of the reaction. Now it's important to note that not all enzymes are proteins, and that's because ribose IMEs are actually enzymes that are RNA catalysts, and they are made up of RNA nucleotides instead of amino acids. But moving forward in our course, most of the enzymes that we're going to talk about are going to be made up of proteins, so that's important to keep in mind Now. Also recall from your previous courses that the reactant of enzymes are called substrates and the substrates will specifically bind to the enzyme at the enzymes active site and so down below. In our example, what you'll see is that this green structure here represents our enzyme, and the specific region of the enzyme that binds the substrate is referred to as the active site. And so, in this example, it's this yellow structure here that is representing our substrate, which is the reactant of the reaction. And so what you'll notice is that the substrate will bind to the enzyme specifically at the enzymes active site. And that's what we're seeing here. And that forms what's known as an enzyme substrate complex, which will talk more about later on in our course. But what you'll notice is that the enzyme is able to interact with the substrate in such a way that it can speed up or catalyze the chemical reaction. And it can produce the products here at a much faster rate than what the products would be produced if the enzyme were not present. And so really, that's the main take away of this video is that enzymes are able to speed up the rate of chemical reactions. Now, in our next lesson video, we'll explain exactly how enzymes speed up the rate of chemical reactions. So I'll see you guys in our next video

Hexokinase uses ATP to convert glucose into glucose-6-phosphate during glycolysis. Which option is true?

• A. Hexokinase is a fibrous protein.
• B. Hexokinase is permanently altered in the enzymatic reaction.
• C. Hexokinase is a molecular catalyst.
• D. Hexokinase & all enzymes are proteins.

So in our last lesson video, we said that enzymes catalyze or speed up the rates of chemical reactions. But there's a common misconception about enzymes that they always convert 100% or all of the substrate into product so that there's no more substrate remaining. So in this video, I just want to clear up this idea and make sure that you guys know that this is actually not true. And instead, what enzymes do is they Onley help reactions get their substrates and products to their equilibrium concentrations and we know that equilibrium does not necessarily mean zero. And so what we can say is that enzymes, really what they do is they help reactions, get thio equilibrium at a faster rate and so down below. What we have is a typical enzyme catalyzed reaction very similar to the one from our previous lesson video. And so you can see we have the substrate here in orange, the enzyme and green. It forms an enzyme substrate complex where the enzyme can convert the substrate into the product and so down below. You can see in our graph here that we have the concentration on the Y axis and the time as the reaction progresses on the X axis. And so this orange curve right here represents the concentrations of substrate over time. And then, of course, the blue curve right here represent the concentrations of product over time, and so we could go ahead and label it as the product. And so, typically, when we're studying an enzyme catalyzed reaction in a lab, we Onley add the substrate and some enzyme. But we don't add any product, and so the product concentration is zero initially. And that's exactly what we're seeing here in our graph on the time axis, where the time is zero initially. Notice that the product concentration is also zero because we don't add any product initially. But we do have a high concentration of substrate, and we also add some enzyme is well and so that allows our reaction to proceed on on Lee one direction here initially and so what you can see is that as the reaction proceeds over time, the concentrations of substrate are decreasing and the concentration of product are increasing, and that's because the substrate is being converted into product. And so here's where the common misconception comes into play. Ah, lot of students think that enzymes will convert all of the substrate into product. And so that would mean that the substrate concentration would go down to zero. And then that would also mean that the product concentration would always go as high as the initial substrate concentration. But as we can see from this graph here, this is actually not true. And at the end of an enzyme catalyzed reaction noticed that we have some substrate concentration here, and the product concentration doesn't necessarily go all the way up to the initial substrate concentration. And so really, what we can see here is that enzymes air Onley helping the reaction get thio equilibrium at a faster rate. And so equilibrium here is being represented by this dotted line here. And we know that it's at equilibrium because noticed that the substrate concentration and the product concentrations are not changing over time. And so that's why we have these horizontal lines here and so recall from our previous lesson videos on equilibrium that the reason that the product and substrate concentrations don't change is because the rate of the forward reaction is exactly equal to the rate of the reverse reaction. And so because when we're at equilibrium, we need to consider the reverse reaction. That means that it's better to sometimes consider the reverse reaction as well in an enzyme catalyzed reaction, even though we don't typically think of enzymes being able to catalyze the reaction in the reverse direction they can in many cases. And so it's important to sometimes consider these reverse arrows. And later in our course, as we move forward, we'll be able to talk even Maura, about thes equilibrium arrows that exist in a typical enzyme catalyzed reaction. But for now, all I want you guys to realize is that the substrate concentration does not necessarily go down to zero, and instead, all enzymes do as they help reactions, get Thio equilibrium faster. And so that concludes this lesson, and I'll see you guys in our next video

Which graph below corresponds with a reaction that has an equilibrium constant less than 1 (Keq < 1)?

• A.
• B.
• C.

The enzyme catalase converts hydrogen peroxide into water and oxygen gas. Which of the following is true?

• A. All enzymes utilize ATP to catalyze reactions.
• B. All hydrogen peroxide is converted to products.
• C. Catalase stops catalyzing when reaction reaches equilibrium.
• D. Catalase catalyzes forward & reverse reactions at equilibrium.

so the way that enzyme speed up chemical reactions is by lowering their energy of activation, which can be abbreviated with either of these two symbols shown here and so a recall from your previous chemistry courses that the energy of activation is literally just the energy difference between the substrates of the reaction and the transition state and the energy of activation is required to initiate a reaction. So the transition state, which can be abbreviated with this double dagger symbol shown here, is an unstable, transient entity at the local maximum peak energy point of a reaction. And so because it symbolized with a double dagger symbol, it makes a lot more sense why the energy of activation is symbolized with Delta G with a double dagger symbol, because it's the difference and energy between the substrates and the transition state so down below. In our image, we have two different scenarios. We have a scenario where there's no enzyme that's present on the left and a scenario where there is an enzyme that's present on the right, and so over here on the left, which will notice is that it's showing the same exact reaction where we're converting substrate A into products B and C, and because there's no enzyme that's present, this is the uncapped allies reaction and down below in our energy diagram, where we have free energy on the Y axis and the reaction progress on the X axis, we can see that the local maximum peak energy point of this reaction is at the very highest point of this curve, which corresponds with the transition state. So we could go ahead and put our double dagger to symbol here for the transition state, which is right at this point, the local maximum peak. And so we know that the difference in energy between the substrates or the reactions here and the transition state, which is this symbol appears so essentially the difference and energy would be symbolized by this blue box here vertically that is going to correspond with the energy of activation which corresponds to the kinetics and the kinetics is all about the speed of the reaction. So the greater this energy of activation is, the slower the reaction will take place. And so what you'll notice is that in the presence of an enzyme, you can see that our enzyme is this red structure shown here in this example that the reaction will take place at a much faster rate than if the enzyme were not present and so down below, which will notice is in the energy diagrams. The red curve represents the enzyme catalyzed reaction, and you can see that we have the same exact curved. Except we have a smaller energy of activation for the catalyzed reaction. And the transition state is much, much lower and energy. And so because the energy of activation for the enzyme catalyzed reaction is smaller, that allows the reaction to occur faster and again. That's exactly what we say. Enzymes are responsible for doing speeding up chemical reactions. And so the enzyme catalyzed energy of activation is being represented by this yellow region right here. The difference in energy between the enzyme catalyzed transition state and the reactant. And so one thing that's really important to note is that enzymes do not affect either the thermodynamic favor ability of the reaction, meaning that there's no change to the delta g of the reaction. So notice that the Delta G, which is the difference in energy between the reactant and the products does not change even in the presence of an enzyme. So notice that the Delta G is represented by this pink box, and the pink box does not change in size from the left side, where there's no enzyme that's present, and from the right side, where there is an enzyme that's present. So that's very important to keep in mind. The enzymes do not affect this thermodynamic favorability of Delta G, and the second thing that enzymes do not affect is the equilibrium constant. So there's no change to the K e que. Which recall from our previous lesson videos, is the ratio of the concentration of products over the concentration of reactant at equilibrium. And so these are two very important things to note that enzymes do not affect enzymes. Onley affect the energy of activation. The only affect the kinetics and the speed of the reaction. So that's important to note. Moving forward and in our next couple of videos will be able to get some practice utilizing thes basic concepts. So I'll see you guys there

Which of the following is the best description of what an enzyme does?

• A. It allows a chemical reaction to proceed extremely fast.
• B. It increases the rate at which a chemical reaction reaches equilibrium relative to its uncatalyzed rate.
• C. It makes a reaction thermodynamically favorable to allow it to proceed faster.

Which of the following best describes how enzymes catalyze reactions?

• A. Alter the equilibrium constant (K_eq) of a reaction.
• B. Decrease the thermodynamic free energy (ΔG) of a reaction.
• C. Change nonspontaneous reactions into spontaneous reactions.
• D. Force reactions to proceed in only one direction.
• E. Stabilizing the transition state.
• F. c & d.