The Student Room Group
Reply 1
:confused:
you might have to explain yourself better

:worm:
Reply 3
it's the "follows the Q10" bit that's confusing me... and u need to be a bit more specific!!!!
Reply 4
I think i know what you mean...so here goes....

In most enzyme-controlled reactions, the rate of reaction more or less doubles with a 10°C rise in temperature. This relationship can be described mathematically by calculating the temperature coefficient or Q10.

Example:

Calculating Q10

Catalase breaks down hydrogen peroxide to produce water and oxygen. A mixture of catalase and hydrogen peroxide gave off, say 24cm3 of oxygen in 1 minute at 20°C. An identical mixture gave off 50cm3 of oxygen in 1 minute at 30°C. So,

Q10 = Rate at 30°C/Rate at 20°C
= 50/24
= 2.1

Most enzyme-controlled reactions have a Q10 of approximately 2.

With yeast this would be the same concept. Im guessing your measuring the rate of respiration in the breakdown of yeast by the enzyme zymase (contained within the yeast). Your Q10 value would depend on the investigation you carry out, as it differs from experiment to experiment (taking into account errors in practical work, anomalies in results, etc.). But as a rule of thumb, the rate of reaction doubles with a 10°C rise in temperature and enzyme-controlled reactions, like the yeast one, have a Q10 of approximately 2.

Hope this answers your question! :smile:
theicon
But as a rule of thumb, the rate of reaction doubles with a 10°C rise in temperature and enzyme-controlled reactions, like the yeast one, have a Q10 of approximately 2.


until you reach the optimum temperature I guess.
Reply 6
theicon
I think i know what you mean...so here goes....

In most enzyme-controlled reactions, the rate of reaction more or less doubles with a 10°C rise in temperature. This relationship can be described mathematically by calculating the temperature coefficient or Q10.

Example:

Calculating Q10

Catalase breaks down hydrogen peroxide to produce water and oxygen. A mixture of catalase and hydrogen peroxide gave off, say 24cm3 of oxygen in 1 minute at 20°C. An identical mixture gave off 50cm3 of oxygen in 1 minute at 30°C. So,

Q10 = Rate at 30°C/Rate at 20°C
= 50/24
= 2.1

Most enzyme-controlled reactions have a Q10 of approximately 2.

With yeast this would be the same concept. Im guessing your measuring the rate of respiration in the breakdown of yeast by the enzyme zymase (contained within the yeast). Your Q10 value would depend on the investigation you carry out, as it differs from experiment to experiment (taking into account errors in practical work, anomalies in results, etc.). But as a rule of thumb, the rate of reaction doubles with a 10°C rise in temperature and enzyme-controlled reactions, like the yeast one, have a Q10 of approximately 2.

Hope this answers your question! :smile:


thanks for that. so in my investigation for yeast, if i was to use the temperatures 20, 30,40,50,60 would that be fine??
Reply 7
G_S
thanks for that. so in my investigation for yeast, if i was to use the temperatures 20, 30,40,50,60 would that be fine??


Yep that would be fine. Most enzyme-controlled reactions have an optimum temp. of around 40 and denature around 60 and you could also test the Q10 theory as well. So yeah, uve pretty much covered the different aspects of the enzyme using that range of temperatures. Although, you could also try 0 degrees, to see what effect lowering the temp (below room temp) would have on the enzyme. We already know what would happen, but just more to write about i guess!! :wink: