By now you would have gone through your CT scripts and i'm sure that some of you would have expected to do better. Congrats to those who have achieved an A! For the rest, as this is your first major exam, take this test as a chance to learn your mistakes. Its just the beginning, so keep trying and you will get better at answering the questions. Here are some learning points for you to note other than what has been said in class.
1. Kinetic energy:
Molecules in a liquid/gas would be constantly moving in random directions, and the energy associated with this movement would be the kinetic energy (KE) of the molecules. Temperature is a measure of that KE. The higher the temp of a mixture, the higher the KE of the particles in the mixture. Heat a mixture up, and the KE of the particles increases, hence temperature increases. Do note this involves all the particles in the mixture.
Hence in enzyme experiments, when a mixture is heated to a higher temperature, both the enzyme and the substrate molecules would have more kinetic energy. One common mistake would be to mention that only "the enzyme molecules have more kinetic energy", which is not true. Another common mistake would be to use the term "heat energy" instead of kinetic energy. Finally, please always include the 5 points in the flow chart that I have emphasized on when describing temperature changes. Memorize those phrases well.
2. Describing graphs:
When you are asked to explain the graph: Firstly you must include in your answer a description of the shape and what it means, before launching into your explanation.
When describing the shape of the graph:
There is a difference between a positive gradient (ie. values are increasing, like AC in this graph) and increasing gradient (or graph becomes steeper). Remember, to obtain the gradient we usually draw a tangent at a particular point on the graph. Hence if we were to take the gradient of AB and compare it with the gradient of BC, we would find that the gradient of BC is actually gentler compared to AB. So to describe AC, we can say that the curve has a positive gradient, but we cannot say that there is an increasing gradient.
To describe CD, we can just state that now the curve has a negative gradient. Finally, at point C, at optimum temperature, the graph has a gradient = 0 (not maximum gradient).
Also, remember to quote data for the various parts of the graph!
3. Reading graphs:
One common mistake for graph reading would be not using dotted lines to read values off the curve you have drawn. Even though it may be slightly tedious to draw dotted lines down to the x and y axes, there are a few reasons for doing this:
(a) It makes it easier for the marker to see how you obtained the value. And shows the marker that you read off the graph to obtain the value. (Btw, please do not use the ratio method to calculate the predicted rate of reactions at certain temperatures. You must read off the curve you drew.)
(b) It helps you to read off the curve accurately. Those who did not draw the dotted lines have a tendency to have readings that are a little off, as your eye tends to wander as you try to read the value on the axis.
4. For the qn on whether a mixture kept at 65 degrees Celsius will show a change in rate of reaction when the temp is changed back to optimum:
Most of you managed to reason that if enzymes were kept at 65C, it would cause a certain percentage of enzymes to denature, which was an irreversible process. Hence when the temperature is changed back to 40C (optimum temp for the enzyme), it would mean only that percentage of enzyme is functional, and rate of reaction remains the same as when it was at 65C. This was the answer we were expecting and was the answer in the marking scheme.
However, 2 combined students managed to reason at a higher level and said that the reaction rate would drop lower than the measured rate at 65C. Why? Because not only will be decreased percentage of enzymes working, lowering the mixture from 65 to 40C means that the particles in the mixture have lower kinetic energy. Hence, less functional enzyme + less KE = lower reaction rate. Kudos to those 2 students for being able to think this up!
5. When enzyme or substrate concentration is increased:
Do explain that rate of reaction increases due to increased collisions between enzyme and substrate. This is due to the increased number of enzymes/substrate, not due to increased kinetic energy of the particles!
6. The essay:
Many of you managed to pinpoint the procedures for extracting amylase, however, most of you forgot to explain the rationale of the procedures. Also, many of you are not sure of the proper procedures for running the various food tests and I get steps for the lipid test when you are trying to test for protein! Please memorize food tests procedures as they are important for both your practical exam and theory!
Some problems with the design of the experiment to show the chemical nature of amylase:
(a) Bear in mind that the seed extract would contain anything inside the cells of the seed that is soluble in water. Hence, aside from amylase, other proteins would have been dissolved in water. So if you were to run the Biuret test to show that the extract contains proteins, you are only proving that the seed has protein, and not that amylase is a protein.
(b) You cannot boil the seed to extract amylase, then later use the seed extract for testing as the amylase and all other enzymes would have been denatured!
Problems for the procedures to show action of amylase:
(a) Amylase breaks down starch into maltose, not glucose! Maltose is a complex sugar, a disaccharide.
(b) We cannot use the Benedict's test to show that amylase in the seed extract has digested starch into maltose. As previously mentioned, the seed extract will contain all water soluble substances, which may include sugars. So a positive test just indicates that there are reducing sugars in the seed extract, not that amylase has acted on starch.
(c) We usually add iodine solution at the end of the experiment to show that all the starch has been digested (give it an hour for this step, not 3 minutes!), and that iodine solution remains brown (ie, no more starch left). Do not plan an experiment where iodine solution is added at the start, as when it is added at the start, it changes to a blue black colour due to presence of starch. However, as starch is digested, the colour does not change back to brown. Its just an odd observation, hence we always plan experiments where we add iodine solution at the end. (However, in this exam, we do not penalize you on this point as you have not been taught this.)
Do bear all these points in mind in the future, and hang in there!
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