BCH3701 - UNISA
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,Question 1 [20]
A research group discovers a new enzyme, happyase that catalyses the chemical reaction
HAPPY ⇌ SAD
The researchers begin to characterize the enzyme.
1.1 In the first experiment, with [E]t at 5.5 nM they find that the Vmax is 6.3 µM.s-1. Based on this
experiment what is the kcat for happyase? (Include appropriate units.) (5)
The kcat (catalytic constant) for happyase can be calculated using the formula:
kcat = Vmax / [E]t
Given that Vmax is 6.3 µM.s-1 and [E]t is 5.5 nM (which is equivalent to 5.5 × 10-9 M), we can plug in these
values to find the kcat:
kcat = (6.3 µM.s-1) / (5.5 × 10-9 M) = (6.3 × 10-6 M.s-1) / (5.5 × 10-9 M) = 1.14 × 10^3 s-1
Therefore, the kcat for happyase is 1.14 × 10^3 s-1.
1.2 In another experiment, with [E]t at 1.75 nM, and [HAPPY] at 27.5 µM, the researchers find that V0
= 275 nM.s-1. What is the measured KM of happyase for its substrate HAPPY? (Include appropriate
units.) (5)
The KM (Michaelis constant) of happyase for its substrate HAPPY can be determined using the Michaelis-
Menten equation:
V0 = (Vmax × [HAPPY]) / ([HAPPY] + KM)
Given that V0 is 275 nM.s-1, [E]t is 1.75 nM (equivalent to 1.75 × 10-9 M), and [HAPPY] is 27.5 µM
(equivalent to 27.5 × 10-6 M), we can substitute these values into the equation and solve for KM:
275 nM.s-1 = (6.3 µM.s-1 × 27.5 × 10-6 M) / (27.5 × 10-6 M + KM)
Simplifying the equation:
275 = (6.3 × 27.5) / (27.5 + KM) 275 × (27.5 + KM) = 6.3 × 27.5 KM = (6.3 × 27.5) / 275 - 27.5 KM = 0.63 M
Therefore, the measured KM of happyase for its substrate HAPPY is 0.63 M.
1.3 Given that for the enzyme-catalyzed reaction, k1 = 4 × 106 M-1 s-1, k-1 =6 × 104 s-1 and k2= 2.0 ×
103 s-1. Determine if the enzyme – substrate binding follow the equilibrium or not ? (5)
To determine if the enzyme-substrate binding follows equilibrium, we compare the rate constants (k1, k-
1, and k2) for the enzyme-catalyzed reaction.
If k1 × k2 is equal to k-1, then the enzyme-substrate binding follows equilibrium. If not, it does not follow
equilibrium.
,Given: k1 = 4 × 10^6 M-1 s-1 k-1 = 6 × 10^4 s-1 k2 = 2.0 × 10^3 s-1
k1 × k2 = (4 × 10^6 M-1 s-1) × (2.0 × 10^3 s-1) = 8 × 10^9 M-1 s-2
Since k1 × k2 is not equal to k-1 (8 × 10^9 M-1 s-2 ≠ 6 × 10^4 s-1), the enzyme-substrate binding does not
follow equilibrium.
1.4 In the conversion of P into S in the following biochemical pathway, enzymes EPQ, EQR, and ERS have
the KM values indicated under each enzyme. If all of the substrates and products are present at a
concentration of 10-4 M, which step will be rate limiting and why? (5)
EPQ EQR ERS
P⇌Q⇌R⇌S
KM = 10-2M 10-4M 10-3M
To determine the rate-limiting step in the biochemical pathway, we compare the KM values of the enzymes
involved. The step with the highest KM value will be the rate-limiting step because it indicates the weakest
substrate binding.
Given: EPQ: KM = 10-2 M EQR: KM = 10-4 M ERS: KM = 10-3 M
Comparing the KM values, we see that the step catalyzed by EQR has the lowest KM value (10-4 M),
indicating the strongest substrate binding among the three enzymes. Therefore, the step catalyzed by EQR
will not be the rate-limiting step.
The step with the highest KM value is the one catalyzed by EPQ, with a KM value of 10-2 M. This suggests
weaker substrate binding compared to the other steps. Hence, the conversion of P to Q catalyzed by EPQ
will be the rate-limiting step in this biochemical pathway.
The rate-limiting step is determined by the enzyme with the highest KM value because a higher KM value
indicates a higher substrate concentration required for achieving half-maximal velocity (Vmax/2) and, thus,
slower enzyme-substrate complex formation.
Question 2 [6]
Draw schematic curves that show the appropriate relationship between the variables:
2.1[ES] vs time
The concentration of the enzyme-substrate complex ([ES]) over time can be represented by an initial rapid
increase followed by reaching a steady-state level.
, [ES]
↑
| /
| /
| /
| /
| /
| /
| /
|/
|/
------------------> Time
2.2[P] vs time
The concentration of the product ([P]) over time can be represented by a gradual increase as the reaction
progresses.
[P]
↑
|
| ___
| / \
| / \
| / \
| / \
| / \
|/ \
|/
------------------> Time
2.3v0 vs [E]T
SCIENCERA TUTORS
FOLLOW US ON YOUTUBE @SCIENCERA
,Question 1 [20]
A research group discovers a new enzyme, happyase that catalyses the chemical reaction
HAPPY ⇌ SAD
The researchers begin to characterize the enzyme.
1.1 In the first experiment, with [E]t at 5.5 nM they find that the Vmax is 6.3 µM.s-1. Based on this
experiment what is the kcat for happyase? (Include appropriate units.) (5)
The kcat (catalytic constant) for happyase can be calculated using the formula:
kcat = Vmax / [E]t
Given that Vmax is 6.3 µM.s-1 and [E]t is 5.5 nM (which is equivalent to 5.5 × 10-9 M), we can plug in these
values to find the kcat:
kcat = (6.3 µM.s-1) / (5.5 × 10-9 M) = (6.3 × 10-6 M.s-1) / (5.5 × 10-9 M) = 1.14 × 10^3 s-1
Therefore, the kcat for happyase is 1.14 × 10^3 s-1.
1.2 In another experiment, with [E]t at 1.75 nM, and [HAPPY] at 27.5 µM, the researchers find that V0
= 275 nM.s-1. What is the measured KM of happyase for its substrate HAPPY? (Include appropriate
units.) (5)
The KM (Michaelis constant) of happyase for its substrate HAPPY can be determined using the Michaelis-
Menten equation:
V0 = (Vmax × [HAPPY]) / ([HAPPY] + KM)
Given that V0 is 275 nM.s-1, [E]t is 1.75 nM (equivalent to 1.75 × 10-9 M), and [HAPPY] is 27.5 µM
(equivalent to 27.5 × 10-6 M), we can substitute these values into the equation and solve for KM:
275 nM.s-1 = (6.3 µM.s-1 × 27.5 × 10-6 M) / (27.5 × 10-6 M + KM)
Simplifying the equation:
275 = (6.3 × 27.5) / (27.5 + KM) 275 × (27.5 + KM) = 6.3 × 27.5 KM = (6.3 × 27.5) / 275 - 27.5 KM = 0.63 M
Therefore, the measured KM of happyase for its substrate HAPPY is 0.63 M.
1.3 Given that for the enzyme-catalyzed reaction, k1 = 4 × 106 M-1 s-1, k-1 =6 × 104 s-1 and k2= 2.0 ×
103 s-1. Determine if the enzyme – substrate binding follow the equilibrium or not ? (5)
To determine if the enzyme-substrate binding follows equilibrium, we compare the rate constants (k1, k-
1, and k2) for the enzyme-catalyzed reaction.
If k1 × k2 is equal to k-1, then the enzyme-substrate binding follows equilibrium. If not, it does not follow
equilibrium.
,Given: k1 = 4 × 10^6 M-1 s-1 k-1 = 6 × 10^4 s-1 k2 = 2.0 × 10^3 s-1
k1 × k2 = (4 × 10^6 M-1 s-1) × (2.0 × 10^3 s-1) = 8 × 10^9 M-1 s-2
Since k1 × k2 is not equal to k-1 (8 × 10^9 M-1 s-2 ≠ 6 × 10^4 s-1), the enzyme-substrate binding does not
follow equilibrium.
1.4 In the conversion of P into S in the following biochemical pathway, enzymes EPQ, EQR, and ERS have
the KM values indicated under each enzyme. If all of the substrates and products are present at a
concentration of 10-4 M, which step will be rate limiting and why? (5)
EPQ EQR ERS
P⇌Q⇌R⇌S
KM = 10-2M 10-4M 10-3M
To determine the rate-limiting step in the biochemical pathway, we compare the KM values of the enzymes
involved. The step with the highest KM value will be the rate-limiting step because it indicates the weakest
substrate binding.
Given: EPQ: KM = 10-2 M EQR: KM = 10-4 M ERS: KM = 10-3 M
Comparing the KM values, we see that the step catalyzed by EQR has the lowest KM value (10-4 M),
indicating the strongest substrate binding among the three enzymes. Therefore, the step catalyzed by EQR
will not be the rate-limiting step.
The step with the highest KM value is the one catalyzed by EPQ, with a KM value of 10-2 M. This suggests
weaker substrate binding compared to the other steps. Hence, the conversion of P to Q catalyzed by EPQ
will be the rate-limiting step in this biochemical pathway.
The rate-limiting step is determined by the enzyme with the highest KM value because a higher KM value
indicates a higher substrate concentration required for achieving half-maximal velocity (Vmax/2) and, thus,
slower enzyme-substrate complex formation.
Question 2 [6]
Draw schematic curves that show the appropriate relationship between the variables:
2.1[ES] vs time
The concentration of the enzyme-substrate complex ([ES]) over time can be represented by an initial rapid
increase followed by reaching a steady-state level.
, [ES]
↑
| /
| /
| /
| /
| /
| /
| /
|/
|/
------------------> Time
2.2[P] vs time
The concentration of the product ([P]) over time can be represented by a gradual increase as the reaction
progresses.
[P]
↑
|
| ___
| / \
| / \
| / \
| / \
| / \
|/ \
|/
------------------> Time
2.3v0 vs [E]T
