ATAR Notes: Forum

VCE Stuff => VCE Science => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Biology => Topic started by: Cammmeron! on March 12, 2013, 07:44:37 pm

Title: Unit 3 SAC 2 Questions.
Post by: Cammmeron! on March 12, 2013, 07:44:37 pm
Hey guys, just wanted to ask some questions about the second unit 3 SAC. Can someone please tell me why we added sand to the liver before it was ground up? Was it to help ground it up further and increase the SA:V?
Also, why is it necessary for cells to quickly break down hydrogen peroxide and how can I conclude the activity was because of enzymes, and lastly, in what tissue was catalase found? Isn't it in liver tissue?
Title: Re: Unit 3 SAC 2 Questions.
Post by: Bad Student on March 12, 2013, 08:17:35 pm
You need ta quickly break down da H2O2 b-coz its like super toxic and it oxidises all the stuff in de cell like the DNA and the proteins and stuff.

U can conclude that the activity waz b-coz of da enzymes b-coz ur control experiment, which u conducted with exactly the same conditions apart from da absence of da enzyme, didn't show any activity.

And ur rite 'bout catalase being in da liver cell but I fink it can also be found in nearly every other cell in da body.
Title: Re: Unit 3 SAC 2 Questions.
Post by: Snorlax on March 12, 2013, 08:42:28 pm
I has question ov my own.
So, in relation to the rate of reaction, is a competitive inhibitor better off (because an enzyme can counter it by increasing substrate concentration) or non-competitive (because the active site goes back to its normal state after the inhibitor moves on)?
I thought competitive inhibitors "permanently excluded the normal substrate"
And non-competitive inhibitors allowed cells to maintain a balance of conc. of products.
How does this work?
Title: Re: Unit 3 SAC 2 Questions.
Post by: Cammmeron! on March 12, 2013, 09:34:57 pm
So why do I need to add sand to the liver while I'm grinding it up?