ATAR Notes: Forum
VCE Stuff => VCE Science => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Biology => Topic started by: jalamoushi on March 21, 2012, 08:15:18 pm
-
hi guys, i just did a prac at school for sac 2, it involved 3 different forms of liver-frozen, fresh and boiled.
we had to observe the reaction the liver had with hydrogen peroxide and detergent. we measured the rate of reaction produced by noting the height of mL's reached on the measuring cylinder due to the bubbles(oxygen) rising as a product of the reaction. the results were- boiled: 15mLs, frozen: 40 mLs and fresh: 60mLs
i have the sac for it tomorrow and i need some help clarifying key knowledge
1. the boiled enzyme, reacted, giving us results of 15ml on the measuring cylinder. I'm confused about this because i thought that enzymes denature at high temperatures and therefore would not function but here it clearly has, giving us a minor reaction, so why did it react?
2. why was detergent used?
3. did any enzyme inhibiting take place in this reaction?
-
Hey :) we just did the exact same Prac lol
Anyways 1.) the enzymes may have not fully denatured, I'm
Not 100% sure, we had the same result..
2.) the detergent is used for the bubbles :)
Someone also suggested that It is to disrupt the cell membrane (therefore releasing more enzymes) but I think for Prac purposes it's mainly the bubbles?
3.) no, no inhibitor action. Purely exploring the effect of temperature on enzyme activity
-
thank you so much!