<v Lecturer>This lesson is called, "Explaining effects of substrate concentration and temperature on enzyme rate" And is from the unit, "Biological molecules and enzymes." Hi there.

My name's Mrs. McCready, and I'm here to guide you through today's lesson.

So thank you very much for joining me, and I hope that you enjoy our lesson today.

In our lesson today, we are going to explain how and why the rate of an enzyme reaction is affected by substrate concentration and temperature.

Now, in our lesson today, we're gonna come across a good number of keywords which are shown on the screen now.

You may wish to pause the video and make a note of them, but I will introduce them to you as we come across them.

In our lesson today, we're going to first of all look at the effect of substrate concentration on rate, and then we're going to have a look at the effect of temperature on rate.

So I hope you're ready to go, I am.

Let's get started.

So an enzyme is a biological catalyst.

It speeds up the rate of a reaction.

So it does that firstly by having both the enzyme and the substrate present within the same space.

Now, the enzyme has a special particular area within it called the active site, and it's into the active site that the substrate then fits.

And when the substrate has fitted into the active site of the enzyme, it forms the enzyme substrate complex.

At this point, the chemical reaction can then be catalysed by the enzyme, and that means that the products can then be released from the enzyme and the enzyme can be reused again to catalyse the same reaction on a different substrate.

And this process goes on many, many millions of times every second in all of the cells, in all of the living organisms on earth, catalysing many, many different chemical reactions.

Now, in enzyme catalyse reactions, it is really important that the enzyme and the substrate collide together at the active site because it is at the active site where the chemical reaction is catalysed.

So if the substrate collides with the enzyme on the other side of the active site, for instance, or anywhere else on the enzyme which is not the active site, the chemical reaction will not take place.

So it is critical that when the enzyme and the substrate collide, they collide with the active site and the substrate meeting each other.

And it is only when this form of successful collision occurs that the chemical reaction can be catalysed and the substrates turned into products.

So it's really critical that not only are there enzymes present and that there are substrates present and that the other conditions are right for the chemical reaction to take place, but that when they do collide, it is a successful collision where the substrate and the active site meet.

Now, we can measure how fast the reaction occurs, so we can see how long it takes for the substrate to be turned into product.

So we can measure how much substrate is present at the start of the chemical reaction, and then every second, minute, however long thereafter, and we should see a reduction in the amount of substrate that is present and an increase in the amount of product that is present.

And we can use the amount of substrate present or the amount of product as a measure of how fast or the rate of the chemical reaction.

So the rate of the reaction is just a measure of how much change is happening per unit of time, be that every second, every minute, or longer.

So let's just check our understanding.

Which of these is a measurement of the rate of reaction? How much substrate has been used up, in grammes? How long it took for the reaction to finish, in seconds? Or how much product was formed every second, in grammes per second? I'll give you five seconds to think about it.

Okay, so did you decide that statement C, how much product is formed every second in grammes per second is a measurement of the rate of reaction? I hope so, and well done if you did.

So the rate of the reaction requires two measures, the quantity of product or substrate, and time.

Now, the concentration of the substrate will affect the rate of the reaction, so let's look at that in a little bit more detail.

So at low substrate concentrations, there are very few substrate molecules.

And because there are fewer substrate molecules present, there are fewer successful collisions.

The frequency of the successful collisions is lower because there are simply fewer substrates crashing into enzymes, and therefore because they're happening less often, the rate of the reaction will be lower.

At higher substrate concentrations, there are more substrate molecules, and therefore there is more likelihood that those successful collisions will take place, and so the frequency of those successful collisions increases.

So you can see that as substrate concentration increases, so does rate, and this is all to do with the amount of substrate that is present.

So let's look at this in a bit more detail.

So here's a graph showing what we would expect to see if we collected data on this type of experiment and then converted it into a graph where substrate concentration is plotted on the x axis, along the bottom of the graph, and the rate of reaction is on the Y axis, up the side.

And you can see that as substrate concentration increases, so does the rate of the reaction.

However, it does get to a point where it peters out and then plateaus, where despite the substrate concentration increasing further, the rate of reaction does not increase anymore.

So let's see why that is.

Well, let's look at low concentration first of all.

At low substrate concentration, there are few substrates, and so whilst there's a number of enzymes, not all of them will have their active site filled, and therefore many of them aren't actually catalysing the chemical reaction, and so the rate of the reaction is low because lots of them are effectively sitting idle, not really doing anything, waiting for the next substrate to crash into it.

So we can see though that as the concentration increases, so does the rate.

In other words, more enzymes are being occupied, their active sites are being occupied by substrate molecules, and therefore more chemical reactions are taking place.

We can see that as substrate concentration increases, so does the rate of reaction, up until what we would call the optimum.

Now, the optimum substrate concentration is the point where all of the enzyme active sites are full all of the time, and therefore the reaction rate is as fast as it possibly can be.

That's what the optimum substrate concentration means.

So there will be more substrates present within the solution, but the enzymes are fully loaded.

The active sites are working as fast as they can, and no more chemical reactions can take place per second than is currently being facilitated by the enzyme.

And it's at this point that we see that the line of rate plateaus.

That the rate of reaction does not increase anymore.

And that's because as we continue to increase the substrate concentration beyond the optimum, there are no more enzymes present to catalyse the reaction.

So the reaction cannot go any faster because the enzymes are working completely flat out.

There's no more space for them to work any faster.

So by adding extra substrate, we're simply creating a queue, just like you might see in a checkout, you know, in a supermarket where all of the checkouts are open and they're all operating as quickly as possible.

You can add more people to go through those checkouts, but they're not gonna go through any faster because all of the checkouts are working as fast as they possibly can.

Well, that's a similar idea to what's happening here.

So how do we increase the rate further? Well, we would have to open more checkouts, or in this case, we would need to add more enzyme.

And only by doing that will we be able to increase the rate further.

So let's just check our understanding of that.

Looking at the graph there, at which point on the graph is the optimum substrate concentration? Where's the optimum? Is it A, B, or C? I'll give you five seconds to think about it.

Okay, so I hope you said that the optimum is at point B on the graph, where the graph flattens out and plateaus at the maximum rate, but right at that point where that plateauing happens.

Well done indeed.

What about this then? Why can the rate of reaction not increase further when the substrate concentration is higher than the optimum? Why is that? Is that because all of the enzyme active sites are full? Is it because few of the enzyme active sites are full, or is it because all of the enzyme has been used up? Again, I'll give you five seconds to think about it.

Well, I hope you've explained this by choosing option A, that all the enzyme active sites are full and therefore the chemical reaction cannot go any faster than it already is.

Well done.

And finally, what do you think about this question? So where on the graph could the enzyme concentration be greater than the substrate concentration? Think about that carefully.

I'll give you a few seconds to decide.

Okay, so where is enzyme concentration greater than substrate concentration? Well, I hope you've identified that as point X on the graph, where the chemical reaction could go faster if were more substrates available.

Well done indeed.

So what I'd like you to do next is to firstly draw a sketch graph to show how the rate of reaction changes as the substrate concentration increases.

So remember which axis each of those labels are going to go on.

Which axis does the substrate concentration go on? Which axis does the rate of the reaction go on? And then what will that graph look like? How will rate change as the concentration increases? Once you've done that, I would like you then to explain why the substrate concentration affects the enzyme rate.

So you might want to put those explanations around the graph that you've drawn for part one, or you might want to add your own explanation underneath the graph, maybe refer to it, but keep it as a separate item.

That's up to you to decide how you're going to present that information, but do make sure your explanation is as detailed as you can make it.

So take your time, make sure you are thinking really clearly and carefully about this, and come back to me when you are ready.

Okay, let's check our answers then.

So for the sketch graph, I asked you to draw a graph to show how the rate of reaction changes as substrate concentration increases.

So we've put substrate concentration on the X axis, along the bottom, the rate of reaction on the Y axis, at the side, and we can see that there's a straight line increasing as the substrate concentration increases so does the rate until it plateaus and then stays flat all the while substrate concentration continues to increase.

So just check your graph, does it look like mine? And then I asked you to explain why substrate concentration affects enzyme rate.

So hopefully your answer includes many of the following points, that at low substrate concentrations only a few enzyme active sites are full, and the enzymes are therefore not operating at maximum capacity, and few substrate molecules are therefore being converted into product, so the reaction rate is low.

As the substrate concentration increases, so does the reaction rate, until the optimum substrate concentration is reached.

And at this point, all of the enzyme active sites are full and the reaction rate is highest and the reaction cannot go any faster.

So beyond this, at substrate concentrations greater than the optimum, there are no more enzyme active sites available, and the reaction rate cannot and therefore does not increase any further.

And if we want to increase that rate further, we must increase the amount of enzyme available.

In other words, increasing enzyme concentration in order to provide more enzymes to catalyse more substrates into products.

Now, that is a really detailed explanation, so check over your work, see how much of that you've got.

If you've got it all, that's an absolutely terrific job.

Well done indeed.

If you need to add to your work, then please do so.

Take your time to make sure you've got a really thorough explanation about why substrate concentration is affecting enzyme rate.

And that was a really good effort, so well done indeed.

Thank you.

So let's move on now to see the effect of temperature on rate.

Now, temperature will affect the rate of a reaction, but why is that? So we need to remember that temperature is a measure of energy, so therefore at low temperatures, the particles have less energy and therefore move slowly.

And they crash into each other, the frequency of collisions is lower, and therefore the frequency of successful collisions is even lower.

Remembering that the enzyme and the substrate must collide at the active site of the enzyme for it to be considered a successful collision.

Now, as we increase temperature, we are therefore increasing the amount of available energy that these particles have, and therefore those particles are going to move faster and they are moving with more energy.

So that means that each collision has more energy in itself and they are happening more often because they're moving faster.

So if you think about being in your classroom, if you were all walking around really slowly, the chances of you bumping into each other would be very, very low indeed, because you'd be able to avoid each other really easily.

But if you were all hurtling around your classroom, please don't do this by the way, but if you were all hurtling around the classroom really fast, you would probably bump into each other fairly often because it would be much harder to avoid each other.

We've got a similar idea going on here with temperature.

So as the temperature increases, the speed at which those particles are moving increases and the amount of energy that they have increases as well, so they collide each with each other more frequently and with more energy.

So the number of successful collisions is going to increase and they'll increase with more force as well.

And therefore the rate of reaction increases too.

So let's see what this might look like if we were to collect data from an experiment and plot it as a graph.

So we know that as the temperature increases towards the optimum temperature, the best temperature for the chemical reaction to take place at, the number of successful enzyme substrate collisions increases because those successful collisions happen at a higher frequency and with more energy as the temperature increases.

So we can see on the graph with temperature on the X axis and rate on the Y axis, we can see this increase in rate as the temperature increases because the number of successful enzyme substrate collisions is increasing.

They're happening with higher frequency and with more energy, and that is therefore increasing the rate of the reaction, how fast the number of substrates is turned into products.

Now, at the optimum temperature, all the enzyme substrates are full.

They're all working as fast as they possibly can, and they're colliding with each other as frequently as they can and with as much energy as they are able to at that temperature.

So let's just check our understanding.

Izzy, Jun and Alex have been talking about this, about the effect of temperature on the rate of enzyme reactions, but whose explanation is most accurate? So Izzy says, "As temperature increases, so does the number of particles." Jun says, "As it gets hotter, the substrate molecules move faster, so they collide with the active sites more frequently." And Alex says, "Increasing the temperature increases the rate of reaction because it's more concentrated." So whose explanation is most accurate? I'll give you a few seconds to decide.

Okay, so I hope you have decided that Jun's explanation is most accurate because the substrate molecules are moving faster, so they collide with the active sites more frequently.

Well done indeed.

Okay, so let's carry on with the effect of temperature on rate.

So we know that the enzyme's active site is specific for its substrate, just like a lock and a key.

So the lock is specific to the key.

The only one key will fit a lock.

And similarly with an enzyme, only one substrate will fit an enzyme's active site.

Now, this is really important because temperature will affect the shape of the active site.

So we know that temperature affects enzymes, and we also know that temperature is a measure of how much energy there is.

So as temperature increases, this causes the molecules to move faster and with more energy.

Now, this is very true for the substrate and the enzymes.

The whole of the substrate and the whole of the enzyme, as temperature increases, move around as a whole entity faster and faster.

So as temperature increases, so does the speed at which those whole items of substrate and enzyme, those whole particles, how fast they move around and how much energy they are moving around with.

But at temperatures beyond the optimum, we could do with zooming into the enzyme itself and having a look at how fast the molecules of the enzyme are moving as well.

Because as temperature increases the particles within the enzyme, so remember the enzyme is made up of lots and lots of atoms and ions, and each of these are vibrating.

As temperature increases, so does the vibration of all of the particles within the enzyme as well.

So at lower temperatures, the vibration is perfectly okay within the enzyme and the enzyme maintains its structure.

But beyond the optimum temperature, the molecules that make up the enzyme vibrate faster and faster and with more force, so they're moving back and forth over greater distances and more often, there's more energy there.

Now, this ultimately causes the bonds that are holding those enzyme molecules together to start to break apart.

And this causes a change in the shape of the enzyme.

And you can see that in the picture there, that the molecules are moving back and forth within the enzyme faster and faster and over greater distance with more force until ultimately the enzyme changes shape at an atomic level.

And that means that overall the enzyme shape changes, and we call that change denaturing.

So the enzyme wholly changes shape because the particles that make up the enzyme are moving faster than they can be contained within their shape.

And eventually what this means is that the active site has changed shape.

Now, we know that the active site is specific for the substrate, just like a lock and its key.

So if the active site changes shape, then the substrate no longer fits into the active site and the enzyme cannot catalyse that reaction anymore.

Now, this is quite a complex idea because it involves a change of focus, from considering the enzyme being this sort of blob to understanding that the enzyme is full of individual atoms and ions, each of which is capable of moving.

So do spend a bit of time understanding the scale at which we're working here, because if you understand the scale, then you'll understand this idea of what's happening to the bonds within the enzyme itself.

You'll understand that more easily and more thoroughly.

So we can see when we go back to the graph that as the temperature increases beyond the optimum, it denatures the enzyme because the chemical bonds get broken and the enzyme's active site changes shape, it denatures, and therefore the substrate no longer fits the enzyme.

So the enzyme cannot catalyse that chemical reaction anymore, and therefore the rate decreases until it gets to zero.

So let's just check our understanding.

What happens to an enzyme at temperatures above the optimum temperature? Does the enzyme die? Does it denature or does it keep working faster? I'll give you a few seconds to decide.

So hopefully you've said that the enzyme denatures.

That's a really critical key term in describing enzyme function.

Well done.

Let's consider this then.

The enzyme denatures because the active site changes shape.

Is that true or false? What do you think? I'll give you a few seconds to decide.

So hopefully you've said that's true.

Well done.

But can you justify your answer with one of the two justifications listed there on the screen? So hopefully you've said that the active site changes shape because the bonds are broken.

Remember, it is the breaking of the bonds that causes the enzyme to change shape and therefore causes it to denature.

That is a really tricky concept so well done for getting that right, and if you feel like you need to go back over that again, there's absolutely nothing wrong with that, and do spend a little bit of time making sure you've got your head around that properly.

But well done indeed.

So we can observe the effect of temperature on the rate of reaction by doing a small experiment.

So if we take potato cores and we put some of them in cold conditions, leave some of them at room temperature and put some of them in really, really hot conditions for at least five minutes, we can then test them to see what effect that temperature has had on the rate of the reaction.

So we've put the potato cores in each of these conditions and we've left them for five minutes.

Meanwhile, we're going to put five centimetres cubed of hydrogen peroxide with washing up liquid into each of three test tubes.

Then once the potato cores have been in the conditions for at least five minutes, we're then going to add one potato core per test tube.

So we're gonna put a cold potato core into tube A, room temperature potato core into tube B and a hot potato core into tube C.

And then we're going to observe the height of bubbles produced after a few minutes.

Now, the potato contains an enzyme called catalase, which breaks hydrogen peroxide down into oxygen and water, and the oxygen is released as a gas and make bubbles in the washing up liquid so the washing up liquid will foam.

So take a look at this demonstration to see that actually happen.

So let's just check our understanding.

Which test tube best represents the expected results of a reaction at the optimum temperature? I'll give you a few seconds to think about it.

So hopefully you've identified the fact that the optimum temperature is the temperature at which the reaction will happen fastest, therefore, the most oxygen will be produced and therefore the greatest column of bubbles and therefore chosen C as a result.

Well done, good job.

So what I'd like you to do now is to bring all of that information together.

Firstly, I would like you to create a flow chart to show the effect of low, optimum and high temperature on the rate of an enzyme reaction.

And I've given you an example of how you might wish to lay that out in a flow chart, but if you wish to choose a different layout, that's fine too just as long as it's nice and clear.

Then I'd like you to draw a graph to support the flow chart and link the labels A, B, and C to the various different parts on the graph to show where that is happening as far as the graph data is showing us.

So take your time, really spend some time processing the information there about how enzymes respond to changes in temperature because it is quite complicated, but once you've understood it, you'll probably never forget it.

So do spend the time really processing that information and working it through, getting a really thorough flow chart and come back to me when you are ready.

Good luck.

Okay, let's see how we've got on.

So what I asked you to do was to create a flow chart to show the effect of low, optimum and high temperatures on the rate of enzyme reaction.

So at low temperatures, you should have said that successful enzyme substrate collisions occur at a low frequency and therefore the reaction rate is low.

At the optimum temperature, all of the active sites are full and the reaction rate is as high as it can be because successful enzyme substrate collisions have a high frequency.

In other words, the enzyme is working as fast as it can.

At temperatures beyond the optimum, so at high temperatures, you should have been therefore talking about enzymes and the bonds that make up the enzyme and the fact that these enzyme bombs have broken.

This denatures the enzyme, and this is all happening because the molecules within the enzyme are vibrating really quickly, and that breaks the bonds.

Denaturing the enzyme means that the reaction cannot take place, it cannot be catalysed, and therefore the reaction rate decreases to zero.

That's quite a complicated explanation, so just check your work thoroughly.

And well done indeed.

Next, I asked you to draw a graph to support the flow chart and link labels to it to show how the various different stages of your flow chart link to the different temperatures on the graph.

So your graph should have temperature on the X axis, along the bottom, and rate of reaction up the side on the Y axis, and have a steep line increasing the rate of reaction as temperature increases up to the optimum, where then the rate of the reaction decreases ultimately to zero.

And you could have linked your low temperature explanation at point A, that increasing temperature part of the graph.

The point about the optimum is point B, and so that should be at the very top of the graph, the peak of the graph.

And then point C is about higher temperatures where the enzyme is denatured, and that should be on the downward slope at point C, around about that point.

So just check your work over thoroughly.

Well done, that was a really hard task to do, so you've done really well to persevere with that.

Good stuff, thank you.

So thank you very much for joining me in today's lesson.

It's been a really hard lesson.

There's a real great amount of challenging content to cover with this, but I hope you've got through it well, well done for persevering.

So let's just summarise what we've covered in our lesson today.

So firstly, we saw that the substrate concentration, as that increases, so does the rate of reaction.

And at the optimum substrate concentration, all of the enzyme active sites are full, and increasing the substrate concentration further does not increase the rate because the enzymes ultimately are working as fast as they possibly can.

We've seen also with temperature that as temperature increases up to the optimum, so does the rate of reaction, and this is because there are more successful enzyme substrate collisions, they're occurring more frequently and more energetically.

Remember, that is where the enzyme at the active site collides with the substrate.

Now, we've also seen that beyond the optimum, at higher temperatures, the bonds in the enzyme itself get broken because of the amount of energy that they have.

And this denatures the enzyme so that it is unable to catalyse the reaction.

And this is why the rate decreases to zero.

Now, that is hard stuff, so well done for persevering.

Thank you very much for joining me today.

You've done a really great job and I very much hope to see you again soon.

Bye.