Catalysed Revision

1 C 6 B 11 C 16 D
2 C 7 B 12 D 17 D
3 A 8 A 13 A 18 A
4 B 9 D 14 D 19 A
5 D 10 B 15 D 20 B



Qn5
- As long as lysine increases, the amount of feedback inhibition will increase as well to restore the condition i.e. ensure a consistent amount of lysine present.
- Most students will put an increase in enzyme 2 for an immediate increase without realizing the above.
- However, when enzyme 4 is decreased, even if there is increased inhibition due to lysine, we are decreasing the total amount of inhibition, and thus the overall amount of lysine will increase.

Qn6
A - not all is hydrolysed; see 50oC.
B - we see a plateau and the potential to produce more sugar at other temps.
C - see amt of sugar at t=0
D - not enough information to make that deduction; what if I change the temp to 55oC? Maybe the slope will be steeper?

Qn8
I apologize 2 3B. I did not notice the Y-axis. I doubt it is fdbk inhibition because a non-comp inhibitor was already added at the start.
Truly, there is no answer. It was decided that the question is faulty but I conceded A (official ans) because of the characteristic shape of graph makes it the best answer. Yet it is not perfect.





Qn9
Good Question
Common Error: B. KE is determined by temperature alone. Higher temp means greater KE. It is not greatest at the optimum temp.
D – Denaturation for pH is the disruption of ionic bonds.

Qn10
In this situation, we realized that the amount of unreacted substrates reaches a plateau after some time. If we examine other options, with enough time, the amount of substrates should reached zero. However for end-product inhibition, if the amount of product is high enough, we can imagine all the enzymes being inhibited (note: you only need a small amount of enzymes to carry out a reaction)

Qn11
Good Question
C or D we deliberate. As highlighted in class before – look at the shape. One is more symmetrical.

Qn12

*additional comments:
This is one question that really makes us scratch our heads so let me try to explain why it is unlikely to be B ...(after much debate and thought because CB question can be weird or incoherent)

Ok.I would explain it this way: "Equal quantities of starch is added to equal quantities of the various solution". Therefore, if all the volumes are constant (and they should), in the case of solution 1, there will be 50% saliva and 50% d.HCl compared to solution 5 100% saliva. If there is 50% less saliva, we can expect a slow rate of reaction in solution 1 (disregard d.HCl 1st) - so solution one will never reach 3 in the first place but be more than 3. Similarly for d.HCl, since there is only 50%, the reading should be more than 4. But since the colorimeter reading is 4 for solution 1, it implies that there is still enzymatic activity and thus not all enzymes are denatured or are at least fully denatured.
NOTE: CB is not concerned about degree of denaturation. it is either denatured or not.


We compare solution 5 and 2 for the answer which is the best possible one.

Qn14
The rest is too unspecific.

Qn16
Issue is similar to 9.

Qn18
For all graph questions, pls examine the axes labels. Some would put B as the answer but over time, the conc of product should not decrease but at least stay constant as in A.

Qn19
You probably need more time to digest this. Look at the table carefully.
B – see Tube 1
C – see Tube 4 (there is still rxn)
D – see Tube 1 & 2