ATAR Notes: Forum
VCE Stuff => VCE Science => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Biology => Topic started by: Kaille on March 27, 2011, 09:07:22 pm
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Hey guys
our school is prretttyyyy slowwww.
I was wondering if i am scientifically correct in saying that at high temperatures, the enzyme and substrate are supplied energy as heat and as a result move faster, causing more collisions between the two and hence an increase in the rate of reaction?
Thanks :)
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not sure about collisions, but yes do move faster....but up to certain point. after that it starts to be critical temp...where denaturation occurs!
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yeah i'm not sure about collisions either cept i can't think of a better way to describe it :S
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Collisions is correct. And yeh, make sure to mention the denaturation thing though as well. The graph tends to look something like:
(http://www.bbc.co.uk/schools/gcsebitesize/science/images/gcsechem_18part2.gif)
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can you say that an enzyme is inactive after it is denatured? cause it doesnt work no more?
and also, wen you increase enzyme concentration is there such thing as saturation point, or is that only applicable for increasing substrate concentration?
o and ye i agree. collisions is correct cause it means that there is more chance for substrate to bump into and bind with enzyme's active site, since there is more heat vibration.
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Although the inactive is kinda correct you're better of saying just denatured. And nah there isn't a saturation point for increasing enzyme concentration its only applicable for substrate concentration. Because remember saturated means fully occupied and if there are too many enzymes then the reaction proceeds until you have no more substrate to act on so its kinda like the opposite :D
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o ok. thanks for clarifying that. :)
i wrote something like this on my sac: blabla becomes denatured. As a result the enzyme is inactive and no longer carries out its catalytic functional role.
would you mark that as correct?
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I reckon you should be alright. Unless you have one mega tight ass teacher because inactive can technically mean the enzyme can function again.
i.e. you say an enzyme has become 'inactive' when the temperature is low, and can be re-activated when the temperature warms up again. However, a denatured enzyme cannot ever be 'active' again [by definition] - thus, the need for the word "denature".
Last year I had a mega tight teacher (stiglecs) who took off marks if you wrote something correct, then you said something that contradicted it, Which imo, makes you a better student but yeah.. rage. (in retrospect, meh 'twas year 11 :P)
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inactive usually refers to a) a enzyme newly created that is inactive and must be activated.
b) an enzyme that has been inactivated by a form of regulator (e.g. allosteric)
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:(
im not too happy about the sac anymore the teacher sed some people in the class failed the sac O.O
but im getting it back tmro so lets see wat happens
btw, my teacher sed that if you increase enzyme concentration IT DOES reach a point where the rate of reaction goes constant. because each enzyme can only work on one substrate at once; say you have a certain amount of substrates and increase the no . of enzymes the rate will increase at first but eventually all the substrates would be worked on and the extra enzymes would be just hanging around?
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your teacher is partially correct.
In a closed system, such as test tube, there is a limited amount of substrate, and inceasing enzyme concentration will speed up the rate of reaction until the reaction has finished.
In an open system (most sytems, standard referall in exams and tests), there is "infinite" substrate (practically) so increasing enzyme conc will increase the reaction rate indefinately. A human body is an example of an open system
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biology is so confusing, dont you think :(
so are you saying that in a closed system my teachers thing about reaching a point where the rate of reaction does not increase any further applies
but in an open system it doesnt? rate of reaction just goes up and up?
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yes that's right. Except in a closed system the rate of reaction would become 0 as there is no reaction occuring when all substrate is gone.
Your teacher is talking about a closed system, if the question/SAC didn't specify the system, or it was impossible to infer, its a bad SAC/question, and you can raise this with your teacher
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o ok thanks! where did you get this information from btw? :O
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your teacher is partially correct.
In a closed system, such as test tube, there is a limited amount of substrate, and inceasing enzyme concentration will speed up the rate of reaction until the reaction has finished.
In an open system (most sytems, standard referall in exams and tests), there is "infinite" substrate (practically) so increasing enzyme conc will increase the reaction rate indefinately. A human body is an example of an open system
It will speed up the rate of reaction until (as said) there's no more substrate for it to interact with.
And the body is not always an example of an "open" system, there are plenty of situations where low levels of substrates are used to manipulate certain reactions in the presence of overwhelming enzyme concentrations (glycolysis is a good example).