biology enzyme questions help!!

hi people i need help in some questions about enzymes that i cannot work out it would be most appreciative if you could help me thanks in advance;

1. substance rate of rection
liver 3
ground liver and sand 5
sand 0
cold boiled liver 0
manganese dioxide 4
cold boiled manganese dioxide 4

a) explain how tube c acts as a control ( what is a control )?
b) use the result to describe the difference between an enzyme such as catalase and a chemical catylist such as manganese dioxide.( please correct me if i am wrong but i think the answer is because the temperature does not affect the chemical catylist and does affect the biological catylist?)

2. a cylinder was placed in container with 100cm3 of hydrogen peroxide. the graph shows the percentage loss in mass over the container over the following 15 minutes.
basically the graph shows a line and it slowly curves when coming down

2a) explain why there is loss in mass over this peroid
b)describe how the curve would have differed if the potato cylinder had been cut into disks before being added to hydrogen peroxide. ( could it be that the graph line would curve more down in a shorter amount of time because of less surafce area to volume ratio?)
c) explain your answer to (B) in terms of enzyme-substrate complex.

Reply 1

thelibertines

6

I've done the exact same sheet in bio a couple of weeks ago!

1.

a) It shows that the sand in test tube had no effect on the rate of reaction, only the ground liver did.

b) Yes, you're correct. The results suggest that a chemical catalyst such as manganese dioxide is not denatured by temp unlike an enzyme like the catalase.

2)

a) Hydrogen peroxide reacts with catalase to give off oxygen. The drop in mass is the oxygen given off.

b) Yes

c) The reaction would be quicker due to increased amount of potato, and therefore a larger amount of catalase. This increases chance of collision between enzyme and substrate hence more enzyme-substrate complexes are formed.

Reply 2

5*frogsplash

OP

thanks buddy

Reply 3

Cirsium

19

Ok sorry for any mistakes but i am skim reading!

a) A control is an experiment you set up to prove that whatever you are measuring does not happen automatically... So e.g. this reaction needs a catalyst

b) I think it probably is to do with surface area because by otherwise you wudnt grind the liver...

2a) Because the peroxide is broken down to water + o2 if memory serves so some mass will be lost as O2 is released into the atmosphere

b)describe how the curve would have differed if the potato cylinder had been cut into disks before being added to hydrogen peroxide. ( could it be that the graph line would curve more down in a shorter amount of time because of less surafce area to volume ratio?) Yup - because that wud increase rate

Reply 4

Ben.S.

11

5*frogsplash

hi people i need help in some questions about enzymes that i cannot work out it would be most appreciative if you could help me thanks in advance;

1. substance rate of rection
liver 3
ground liver and sand 5
sand 0
cold boiled liver 0
manganese dioxide 4
cold boiled manganese dioxide 4

a) explain how tube c acts as a control ( what is a control )?
b) use the result to describe the difference between an enzyme such as catalase and a chemical catylist such as manganese dioxide.( please correct me if i am wrong but i think the answer is because the temperature does not affect the chemical catylist and does affect the biological catylist?)

2. a cylinder was placed in container with 100cm3 of hydrogen peroxide. the graph shows the percentage loss in mass over the container over the following 15 minutes.
basically the graph shows a line and it slowly curves when coming down

2a) explain why there is loss in mass over this peroid
b)describe how the curve would have differed if the potato cylinder had been cut into disks before being added to hydrogen peroxide. ( could it be that the graph line would curve more down in a shorter amount of time because of less surafce area to volume ratio?)
c) explain your answer to (B) in terms of enzyme-substrate complex.

1.
a) Control - allows you to ascertain backgound rates (so that you can subtract them from your other experimentally observed rates in order to get true values). This experiment doesn't contain any catalyst, so you wouldn't expect there to be an appreciable reaction velocity (or rate, if you want to be a chemist).

b) You are correct.

2.
a) Loss in mass as H202 converted to water and oxygen (which escapes after evolution).

b) Initial gradient would be steeper (more negative), but the line would reach the same value as before (albeit in a shorter amount of time). This would be because there is a greater SA/V ratio (larger surface area divided by the same volume as previously).

c) More enzyme exposed to sustrate results in more collisions between enzyme and substrate, which causes a faster rate of ES complex formation which results in a faster rate of catalysis and, hence, mass loss. The number of substrate molecules does not change so the final mass after reaction will not change.

Ben

Reply 5

thelibertines

6

No probs!

Just remember on the graph one, that the line will level off quicker but at the same % of mass. Therefore the two lines should meet towards the end of the graph.

biology enzyme questions help!!

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