Wenhao's Biology 3/4 Ques Thread!!!

[pi]

^Good post

*pH - In order to maintain a constant level of pH, amino acids in the enzyme will donate or accept hydrogen ions, so that the pH level is stabilised. However, this changes the intermolecular bonds that form between the secondary and tertiary structures (as you've pretty much changed the chemical structure of the amino acid), and hence the protein will denature. The curve for pH affecting enzyme activity resembles a bell-curve because the pH scale is a logarithmic scale, which means that it goes up in factors of 10 (pH 6 is 10 times more acidic, and has 10x more hydrogen ions than pH 7), and so the drop in enzyme activity is not as dramatic as temperature.

Just be careful with this one, I got marked down in my SAC because I didn't say that it is only the R-group of the amino acid (or the N-terminus or C-terminus) that can actually donate/accept protons, just something to keep in mind

[LOLs99]

Just describing how each factor actually affects enzyme activity:

*Heat - By adding heat, you overcome/break the weak intermolecular bonds (such as hydrophobic interactions and hydrogen bonding) that hold the shape of the secondary and tertiary structures in the enzyme, and hence the enzyme will denature and enzyme activity will sharply drop after the critical point (the temperature where enzymes begin to denature)
*pH - In order to maintain a constant level of pH, amino acids in the enzyme will donate or accept hydrogen ions, so that the pH level is stabilised. However, this changes the intermolecular bonds that form between the secondary and tertiary structures (as you've pretty much changed the chemical structure of the amino acid), and hence the protein will denature. The curve for pH affecting enzyme activity resembles a bell-curve because the pH scale is a logarithmic scale, which means that it goes up in factors of 10 (pH 6 is 10 times more acidic, and has 10x more hydrogen ions than pH 7), and so the drop in enzyme activity is not as dramatic as temperature.
*Co-enzymes - these basically help the enzyme to accommodate the substrate better, and make it easier to fit. Hence, enzyme activity will increase as more substrates will easily bind to the active site of the enzyme.
*Inhibitors - These are divided into two categories, non-competitive and competitive, which bind completely to an enzyme's active site, preventing the substrate from binding to the enzyme. If you have a high concentration of competitive inhibitors, enzyme activity will decrease as less substrate is able to bind to the active sites of enzymes. However, if the concentration of substrate is greater than the concentration of competitive inhibitors, enzyme activity will decrease less as the substrate becomes more 'competitive', and there is a lesser likelihood of an inhibitor binding to an active site. Non-competitive inhibitors almost permanently bind to enzymes, halting reactions between substrates and enzymes completely.
-Concentration of enzyme - Obviously, as you increase the concentrations of enzyme, there will be a great increase in the rate of reaction - more enzymes are bumping and binding into substrates.
-Concentration of substrate - Initially, increasing the substrate concentration will increase the rate of reaction as there is a greater chance of substrates binding to active sites, but it slowly decreases as the enzyme concentration becomes 'saturated', which is when the enzymes are reacting at full capacity and cannot increase the rate of reaction.

I love your explaination! Great! Do you mind if I print this one?

Okay, so which factors do you guys think it's the best(safest) to use, so I dont get deducted for no reason..
Do we have to draw Induced Fit Model and Lock-and-Key Model?...How do I draw both of them, it kinda looks the same when drawing(theory different tho)..

[LOLs99]

^Good post

*pH - In order to maintain a constant level of pH, amino acids in the enzyme will donate or accept hydrogen ions, so that the pH level is stabilised. However, this changes the intermolecular bonds that form between the secondary and tertiary structures (as you've pretty much changed the chemical structure of the amino acid), and hence the protein will denature. The curve for pH affecting enzyme activity resembles a bell-curve because the pH scale is a logarithmic scale, which means that it goes up in factors of 10 (pH 6 is 10 times more acidic, and has 10x more hydrogen ions than pH 7), and so the drop in enzyme activity is not as dramatic as temperature.

Just be careful with this one, I got marked down in my SAC because I didn't say that it is only the R-group of the amino acid (or the N-terminus or C-terminus) that can actually donate/accept protons, just something to keep in mind

Hang On! I dont get you about that(as for what kind of questions do I have to aware of ).Are you talking about the R group of amino acid?Mr Mac told me it's the safest to say it's a variable group though, so is he right or both is fine?

[pi]

Both are fine.

[LOLs99]

Just be careful with this one, I got marked down in my SAC because I didn't say that it is only the R-group of the amino acid (or the N-terminus or C-terminus) that can actually donate/accept protons, just something to keep in mind

Hmm, okay.. so can you briefly tell me the one above because I think no one ever taught me that.

[Shenz0r]

Just describing how each factor actually affects enzyme activity:

*Heat - By adding heat, you overcome/break the weak intermolecular bonds (such as hydrophobic interactions and hydrogen bonding) that hold the shape of the secondary and tertiary structures in the enzyme, and hence the enzyme will denature and enzyme activity will sharply drop after the critical point (the temperature where enzymes begin to denature)
*pH - In order to maintain a constant level of pH, amino acids in the enzyme will donate or accept hydrogen ions, so that the pH level is stabilised. However, this changes the intermolecular bonds that form between the secondary and tertiary structures (as you've pretty much changed the chemical structure of the amino acid), and hence the protein will denature. The curve for pH affecting enzyme activity resembles a bell-curve because the pH scale is a logarithmic scale, which means that it goes up in factors of 10 (pH 6 is 10 times more acidic, and has 10x more hydrogen ions than pH 7), and so the drop in enzyme activity is not as dramatic as temperature.
*Co-enzymes - these basically help the enzyme to accommodate the substrate better, and make it easier to fit. Hence, enzyme activity will increase as more substrates will easily bind to the active site of the enzyme.
*Inhibitors - These are divided into two categories, non-competitive and competitive, which bind completely to an enzyme's active site, preventing the substrate from binding to the enzyme. If you have a high concentration of competitive inhibitors, enzyme activity will decrease as less substrate is able to bind to the active sites of enzymes. However, if the concentration of substrate is greater than the concentration of competitive inhibitors, enzyme activity will decrease less as the substrate becomes more 'competitive', and there is a lesser likelihood of an inhibitor binding to an active site. Non-competitive inhibitors almost permanently bind to enzymes, halting reactions between substrates and enzymes completely.
-Concentration of enzyme - Obviously, as you increase the concentrations of enzyme, there will be a great increase in the rate of reaction - more enzymes are bumping and binding into substrates.
-Concentration of substrate - Initially, increasing the substrate concentration will increase the rate of reaction as there is a greater chance of substrates binding to active sites, but it slowly decreases as the enzyme concentration becomes 'saturated', which is when the enzymes are reacting at full capacity and cannot increase the rate of reaction.

I love your explaination! Great! Do you mind if I print this one?

Okay, so which factors do you guys think it's the best(safest) to use, so I dont get deducted for no reason..
Do we have to draw Induced Fit Model and Lock-and-Key Model?...How do I draw both of them, it kinda looks the same when drawing(theory different tho)..

Yeah go ahead, I don't mind.

You may have to draw both models, remember that the induced fit model proposes that the active site of an enzyme changes shape in order to accommodate the substrate, while the lock-and-key model proposes that the substrate and active site bind together in a perfect fit with the structure remaining constant.

So when you draw the induced fit-diagram, you would draw the active site and then show how the active site changes shape before it binds with the substrate, while in lock-and-key model you wouldn't change the shape of the active site.

[LOLs99]

Thanks mate!! So that means I might have to draw  2-3 diagrams(procedures) for each model? And one important thing, do I have to name it enzyme-substrate complex when the substrate attaches to the enzymes?

Btw do I say the substrate attaches to enzymes or attaches to active site?Sometimes Im really fuss about this!

[Shenz0r]

Draw 3 diagrams - when the enzyme and the substrate are not in contact, then when there is a enzyme-substrate complex (label it), then when the products are released (in an induced fit model, the enzyme would change shape after reacting with the substrate)

It attaches to the active site of the enzyme, to be more specific.

[LOLs99]

So basically draw a bit different for the enzyme  but same substrate shape( induced.fit.model) or should I be more specific like label the enzymes changing its shape?

It attaches to the active site of the enzyme, to be more specific.

That's what I'm searching for. Ty  very much!!!

[LOLs99]

Hey again guys,

1. Im just wondering how do you set up an experimental design question?
    I havent done this before and kinda get a bit lost when this kind of questions come up.
2. For Krebs Cycle, you dont need to use up oxygen for this stage but do you need oxygen in presence in order to go through this stage?

That's all I have atm..everyone welcome to help me. THANK YOU!

[paulsterio]

2. For Krebs Cycle, you dont need to use up oxygen for this stage but do you need oxygen in presence in order to go through this stage?

That's right, oxygen is not reduced to water until the electron transport chain, the main aim of the kreb cycle is to transfer the energy inside pyruvate to the electron carriers NAD and FAD, reducing them to NADH and FADH

[nubs]

You do need oxygen for that cycle to commence, glycolisis is the only one that can go ahead without oxygen present

For 2nd and 3rd stages of Cellular Respiration, oxygen is needed

[paulsterio]

You need oxygen, however, oxygen is not used up inside the Kreb cycle, it is only used up in the Electron Transport Chain.

The only reason why the Kreb Cycle cannot proceed without oxygen is that because without Oxygen, there would be a build up of NADH and FADH2, because Electron Transport Fails, this inhibits the Kreb Cycle, so it no longer produces NADH and FADH2, this is as far as I know.

[LOLs99]

Thanks guys.. what about the first question because my teacher told me that there will be questions about that in exams or sacs ...Im not sure how to set up the experimental design ques.

So NADH and FADH2 are not produced in Krebs Cycle?
 NADPH is the loaded accepter electrons for photosynthesis while NADH  is the one occurring in cellular respiration, right? And am I using the right term as " loaded accepter electrons"?

[paulsterio]

They are "produced" in the Kreb Cycle, however, it's not really termed "produced"

The Kreb Cycle reduces NAD+ and FAD2+ to NADH and FADH2