17/10/2012 § 1 Comment
On the pages of this chapter, there are these long ribbons with the letters “HL” tacked onto the very ends so that we know we’re learning HL material. Thanks, textbook, my head hadn’t noticed.
A reactant needs to gain energy in order to take part of a reaction. Activation energy is needed to break the bonds of the reactant, or the substrate. This energy is required to reach the transition state (prior to being converted into products) even if energy will be released when the substrate changes into the product. The energy released by the reaction isn’t changed by the enzyme but reactions with an enzyme have a more reduced rate of energy than reactions without an enzyme.
Induced fit VS Lock-and-key
Lock-and-key is so old school, guys. Let’s go up to that HL level we all know and love and talk INDUCED FIT. What actually happens with the proteins is that the substrate will approach the binding site and up ’til then, the binding site isn’t complete conformed to the substrate’s shape. The spa of the active site eventually changes as the substrate binds to it and eventually completely fits it. The enzyme then weakens the substrate’s bonds and reduces the activation energy and does the catalyses enzymes are so well known for. The induced fit model can explain better the way an enzyme can have a broad specificity – not just a few but a great many – of substrates it can bind to and catalyze.
There are some chemical substances that are really rude whose intentions are to reduce or completely prevent enzyme activity from occurring; they’re called inhibitors (because they inhibit, eheh). There are two types of inhibitors: competitive and non-competitive.
- competitive inhibition: here, the substrate and inhibitor are chemically similar; the inhibitor basically takes up the space provided by the active site and in occupying the active site, prevents the substrate from binding until the inhibitor decides to leave; increasing the substrate concentration would reduce the effect of this inhibitor (because it leaves anyway); example – sulfadiazine
- non-competitive inhibition: here, the substrate and inhibitor are not similar; the inhibitor is like a jamming system that alters the shape of the active site; substrate can bind but cannot convert into the product; increasing the substrate concentration would increase enzymatic activity; example – nitric oxide synthase
The metabolism refers to the chemical changes occurring in living cells. Over time, patterns in metabolism have been discovered, such as:
- the majority of chemical changes happen in small steps, which create the metabolic pathway; think of a word game that starts from one word, alters a few letters per step, and ends with a final word
- metabolic pathways involve a chain of reactions; e.g. phenylalanine uses a reaction chain to convert into fumarate and acetoacetate
- sometimes metabolic pathways are more like metabolic cycles but that’s later
More inhibition! Some molecules are beyond rude, now they’re just harsh. The product of the last reaction in the chain will inhibit the enzyme that catalyzes the first reaction. This is called an end-product inhibition, and the enzyme that inhibits the end products is an allosteric enzyme. The binding site is simply called an allosteric site while the interactions are… well… allosteric interactions.
Though this stops reactions, it can also control reactions, especially if there is an excess of the intermediates that lead to the production of the end product.
- Outline the thermal, cohesive, and solvent properties of water. (5 marks)
- Describe the significance of water to living organisms. (6 marks)
- Describe the use of carbohydrates and lipids for energy storage in animals. (5 marks)
- List three functions of lipids. (3 marks)
- Describe the significance of polar and non-polar amino acids. (5 marks)
- Outline the role of condensation and hydrolysis in the relationship between amino acids and dipeptides. (4 marks)
- Describe the structure of proteins. (9 marks)
- List four functions of proteins, giving an example of each. (4 marks)
- Distinguish between fibrous and globular proteins with reference to one example of each protein type. (6 marks)
- option i – Lactase is widely used in food processing. Explain three reasons for converting lactose to glucose and galactose during food processing (3 marks)
option ii – Simple laboratory experiments show that when the enzyme lactase is mixed with lactose, the initial rate of reaction is highest at 48°C. In food processing, lactase is used at a much lower temperature, often at 5°C. Suggest reasons for using lactase at relatively low temperatures. (2 marks)
- Outline how enzymes catalyze reactions. (7 marks)
- Explain the effect of pH on enzyme activity. (3 marks)
- Compare the induced fit model of enzyme activity with the lock and key model. (4 marks)
- Draw graphs to show the effect of enzymes on the activation energy of chemical reactions. (5 marks)
- Explain, using one named example, the effect of a competitive inhibitor on enzyme activity. (6 marks)
Guess it’s time for that test. And a lot of review.
DATA BASED QUESTIONS
Page 84 the effectiveness of enzymes
1. State which enzyme catalyses the reaction with the slowest rate in the absence of an enzyme.
2. State which enzyme catalyses its reaction at the most rapid rate.
3. Calculate the ratios between the rate of reaction with and without an enzyme for ketosteroid isomerase, nuclease and OMP decarboxylase.
- ketosteroid isomerase = 3.8 x 10^11
- nuclease = 5.6 x 10^19
- OMP decarboxylase = 1.4 x 10^24
4. Discuss which of the enzymes is the more effective catalyst.
OMP decarboxylase is the more effective catalyst because its ratio is the smallest amongst the four. Without OMP decarboxylase, it is the slowest catalyzer amongst the four but with the enzyme present, the rate is the fastest between the four enzymes. The ratio is also smaller which means that OMP decarboxylase reduces the most activation energy when it does its reaction.
5. Explain how the enzymes increase the rate of the reactions that they catalyze.
Enzymes increase the rate of reactions they catalyze by reducing the activation energy prior to turning the substrate into its product. Without an enzyme present, the activation energy remains high but with an enzyme, the overall energy level of the transition state is reduced and the rate of reaction increases.
Page 88 chapter 7 questions now showing at re: still faster than you!
Your patience has paid off, young grasshopper.
- i. Outline the effects of temperature on the activity of dissolved papain. Dissolved papain remains at a high level of digestion (100%) all the way to about 40°C. After 50°C, the levels of papain start to decrease slowly.
- ii. Explain the effects of temperature on the activity of dissolved papain. Dissolved papain starts to decrease after a certain temperature (in this case, 40°C) because the energy provided by the heat gives the papain enzyme more kinetic energy and eventually, this breaks the bonds that hold the enzyme together, denaturing it and rendering it useless for all kinds of chemical activity. This is why the activity drops after 40°C.
- i. Compare the effect of temperature on the activity of immobilized papain with the effect on dissolved papain. The graphs show that dissolved papain starts to drop in activity earlier than immobilized papain. Dissolved papain starts to drop at 40°C but immobilized papain starts to drop 10°C after that (at 50°C). Dissolved papain also decreases at a faster and steeper rate (the slope).
- ii. Suggest a reason for the difference that you have described. Immobilized papain is an example of an enzyme that remains stuck to some part of the cell. It stays in one place and does its chemical reactions there – it’s a sturdy protein compared to dissolved papain, which floats around and is more susceptible to factors that cause denaturation.
- iii. In some parts of the human body, enzymes are immobilized in membranes. Suggest one enzyme and a part of the body where it would be useful for it to be immobilized in a membrane. Maltase can be found in the human small intestine.
a) Deduce which curve shows the concentration of a substrate. A represents the products and B represents the substrates.
- i. Using the data in the graph, deduce how the activity of the enzyme changed during the experiment. First, the enzyme’s activity is normal in that it does its chemical reactions and catalyses the substrates, but gradually the activity decreases, as seen when the amount of products reaches the high and the substrates reach their low (enzymes only catalyze substrates).
- ii. Explain the change in enzyme activity during the experiment. The concentration of the substrates and products determine the enzyme’s activity. If the concentration of the substrate is greater than the product, the enzyme will do more chemical reactions because there are more substrates that need to be catalyzed. If the concentration of the product is greater than the substrates, the enzymes will know to halt chemical reactions so as not to overload the cell with too much of a certain product and this decreases (or stops) its overall chemical activity.
~ Imagine that this is an HL ribbon. ~
a) State the meaning of the term “metabolic pathway.” Metabolic pathways are chains of reactions that occur during catalytic activity to create a product.
b) Compared with concentrations during oxygen starvation, state which metabolic intermediate:
- i. increased in concentration most: PYR
- ii. decreased in concentration most: FDP
- iii. did not change in concentration: PEP
- i. The concentrations shown in Figure 25 suggest that the rate of this metabolic pathway has been greater than is needed by the heart cells. Explain how the data in the bar chart shows this. The metabolic pathway is shown to be greater than needed because from pyruvate (PYR) is large in amount at the end of the reaction. Also it’s a very high percentage (350%) which means it is accumulating.
- ii. Because rate of the pathway has been greater than necessary, the enzyme catalyzing one of the reactions in the pathway has been inhibited. Deduce which reaction this enzyme catalyses, giving reasons for your answer. The reaction that the enzyme catalyses (and that has been inhibited) is FDP because while G6P and F6P increase at a normal rate, FDP drops to a very low percentage, which shows that it’s been inhibited and that reaction isn’t allowed to be done.
4. Thousands of different enzymes have been discovered.
a) Explain the reasons for cells needing to have many different enzymes in their cytoplasm. Enzymes can both break and build things – this is proven by their ability to do hydrolysis and condensation synthesis. In the cytoplasm, many molecules (substrates) need to either become something new for the cell or need to be broken down so the cell can use it. Also, not one or a few enzymes can do all of the activities for a single cell. According to enzyme-substrate specificity (and explained by the lock-and-key concept), enzymes can only catalyze certain molecules and substrates, though typically enzymes can catalyze more than just one molecule. However, still many different enzymes are needed for the millions (endless) kinds of molecules that do exist and that need to be broken down or built – enzymes basically do everything for the cell – they are vital to the metabolism (a function of life) for the cell.
b) Explain how different chemical reactions are catalyzed by different enzymes. The lock-and-key concept explains this well in that in an enzyme, the active site’s shape and charge can only really match with one particular substrate with a shape and charge that fits with the active site. That being said, since there are many, many kinds of molecules (there are twenty main amino acids and a molecule can be built out of chains of these amino acids, so the possibilities are exponentially massive) therefore there need to be many, many enzymes that can catalyze these molecules.
c) Distinguish, with examples, between competitive and non-competitive inhibition. Competitive inhibition is when the inhibitor binds itself to the active site on the enzyme and directly keeps the substrate from binding to the site to be catalyzed. An example of a competitive inhibitor is sulfadiazine. Non-competitive inhibition is when the inhibitor binds not to the active site but somewhere on the enzyme other than the active site – it is called the allosteric site. In doing so, the interaction changes the shape of the active site and substrates won’t be able to bind there, indirectly keeping the substrates from being catalyzed. An example of a non-competitive inhibitor is nitric oxide synthase.
5. Some enzymes have commercial uses and a source of them is needed.
a) Explain the reasons for using living cells to produce enzymes rather than synthesizing them in artificial cell-free systems. Only through a cell can all of the appropriate steps be taken to produce enzymes. The structure of a protein (which is what an enzyme is) is very complex to begin with – the primary structure takes a very precise pattern that needs to be constructed from within the cell, and the DNA needed to know what exactly to make the enzyme with is located in the cell. Then the secondary and tertiary are difficult to do with an artificial system because the folding also needs to be precise for all of the bonds to work and hold the enzyme together. It’s better to let the cells make the enzymes because they already have the system to make one the right way, and it’s very difficult to artificially synthesize enzymes because the structure is so complex and small.
b) Suggest reasons for almost all enzymes being proteins rather than other biochemicals. Proteins, globular and fibrous, are more versatile in structure and can come in many different forms. There are countless combinations of amino acids, which are what make the enzymes, therefore giving an incredibly wide variety of enzymes. Also, the genes in DNA can store information that can make many different proteins or enzymes.
c) Compare the lock and key and induced-fit models of enzyme activity. The lock-and-key model shows that the enzyme’s activity is limited to only one kind of substrate. The shape and charge of the active site doesn’t change for a lock-and-key model and the substrate immediately fits. The lock-and-key model is good to explain substrate specificity but the induced-fit model better explains the process of catalyzing a substrate – how the bonds are broken. The substrate enters the active site and then the active site will alter to fit the substrate in shape and charge. This model also helps to explain how activation energy is reduced as it breaks the bonds of the substrate (or creates it in the case of condensation synthesis). The induced-fit model better explains the wide variety of substrate specificity in enzymes.