The correct answer is (2) a > b > c.
Introduction
Catalytic efficiency (kcat/Km) is a key measure in enzyme kinetics, representing how rapidly and effectively an enzyme converts substrate to product. It incorporates both substrate affinity and turnover rate, making it crucial for comparing enzymes or variants. This article explains how to analyze the catalytic efficiency of three proteins based on normalized activity data, guiding students and researchers through the logic and mathematics of enzyme kinetic comparisons.
Concept: Normalizing Enzyme Activity
When enzyme concentrations differ, comparing raw activities alone is misleading. Normalized activity (activity per mg protein) reveals true catalytic efficiency independent of enzyme quantity:
Catalytic Efficiency=kcatKmMeasured Activity÷Enzyme Concentration
Data from Plot
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Protein a: Activity curve with 1 mg/ml protein
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Protein b: Activity curve with 4 mg/ml protein
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Protein c: Activity curve with 2 mg/ml protein
From the plot:
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b has highest raw activity, but with 4-fold more protein.
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a has intermediate activity, with least protein.
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c has lowest curve, with intermediate protein.
Stepwise Calculation of Relative Efficiency
1. Calculate “normalized” activity
Assume final plateau values for activity (visual estimation from graph):
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Aa∼15 units at 1 mg/ml
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Ab∼28 units at 4 mg/ml
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Ac∼10 units at 2 mg/ml
Divide plateau activity by protein concentration:
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a: 15/1=15 units/mg
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b: 28/4=7 units/mg
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c: 10/2=5 units/mg
2. Compare efficiencies
Ranking by normalized activity (proxy for kcat/Km):
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a (15 units/mg) > b (7 units/mg) > c (5 units/mg)
Thus, a > b > c.
Importance of Catalytic Efficiency kcat/Km
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High efficiency means rapid substrate conversion even at low concentrations.
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Used in enzyme selection for industrial, diagnostic, and pharmacological applications.
Summary Table
| Protein | Activity (units) | Protein (mg/ml) | Activity/mg | Relative Efficiency |
|---|---|---|---|---|
| a | 15 | 1 | 15 | Highest |
| b | 28 | 4 | 7 | Medium |
| c | 10 | 2 | 5 | Lowest |
Conclusion
Based on normalized activity (activity per mg protein), protein a has the highest catalytic efficiency, followed by protein b, and then protein c. Thus, the correct statement is a > b > c.
36 Comments
Varsha Tatla
September 13, 2025Done sir
Varsha Tatla
September 13, 2025A>b>c
Khushi Vaishnav
September 13, 2025a>b>c
Aakansha sharma Sharma
September 13, 2025The correct answer is (2) a > b > c.
Mohd juber Ali
September 14, 2025A >B>c
A 20/1=20
B 30/4 = 7.5
C 10/2 =5
Rishita
September 14, 2025Option 2
Aafreen Khan
September 14, 2025Option 2nd is correct
a > b > c
Dharmpal Swami
September 14, 2025Protein a has the high catalytic activity
So- a>b>C are correct
Soniya Shekhawat
September 14, 2025Specific activity=no. Of enzyme units (unit of catalystic activity)/ protein in mg/ml so a>b>c is correct option.
Kanica Sunwalka
September 14, 2025option 2 is correct
Roopal Sharma
September 14, 2025Protein a has high catalytic activity.
Santosh Saini
September 14, 2025A=20, B=30, C=10, Kcat/km so A=20/1=20, B=30/4=7.5,C=10/2=5 20unit/Mg > 7.5unit/Mg > 5unit/Mg protein A has the highest catalytic efficiency followed by B and then C , thus A >B>C
Santosh Saini
September 14, 2025A=20, B=30, C=10, Kcat/km so A=20/1=20, B=30/4=7.5,C=10/2=5 , 20unit/Mg > 7.5unit/Mg > 5unit/Mg , protein A has the highest catalytic efficiency followed by B and then C , thus A >B>C
Santosh Saini
September 14, 2025A=20, B=30, C=10, Kcat/km so A=20/1=20, B=30/4=7.5,C=10/2=5 , 20unit/Mg > 7.5unit/Mg > 5unit/Mg , protein A has the highest catalytic efficiency followed by B and then C , thus A >B>C
Kirti Agarwal
September 14, 2025Opt 2
Neha Yadav
September 14, 2025A>B>C
Sakshi Kanwar
September 14, 2025According to estimate value of graphcatalytic value a = 15 , b= 28 , c= 10
So we get a = 15/1 = 15
b= 28/4 = 7
C= 10/2 = 5
a>b>c
Vanshika Sharma
September 14, 2025Opt B
Konika Naval
September 14, 2025a>b>c
Sakshi yadav
September 14, 2025A=15/1=15,B=28/4=7, C=10/2=5 a>b>c
HIMANI FAUJDAR
September 14, 2025Ans protein a has the highest catalytic efficiency, followed by protein b, and then protein c.
a=15/1=15
b=28/4=7
c=10/2=5
So,a>b>c is correct answer.
Priya dhakad
September 14, 2025a=15/1=15
b=28/4=7
c=10/2=5
protein a has the highest catalytic efficiency, followed by protein b, and then protein c.
Anurag Giri
September 15, 2025The correct answer is (2) a > b > c.
Asha Gurzzar
September 15, 2025Correct answer is 2option
Nilofar Khan
September 15, 2025Correct answer is b
a= 15/1= 15
b= 30/4= 7.5
C=10/2= 5
So a> b>c
Payal Gaur
September 15, 2025A>B>C
A. 15/1=15
B. 30/4=7.5
C. 10/2=5
Tanvi Panwar
September 15, 2025Normalised activity = Measured activity/enzyme concentration.
a = 20/1= 20
b= 30/4= 7.5
c= 10/2= 5
so, a>b>c.
Palak Sharma
September 16, 2025from graph, we can estimate specific activity= enzyme activity/ protein conc
a>b>c
Avni
September 16, 2025protein a has the highest catalytic efficiency
a > b > c.
Mansukh Kapoor
September 16, 2025The correct answer is option 2nd
Priti khandal
September 17, 20252is right
Muskan Yadav
September 18, 2025Protein A has the highest catalytic efficiency, followed by protein B, and then protein C. Thus, the correct statement is a > b > c.
Deepika sheoran
September 18, 2025Option A
A>B>C
anjani sharma
September 19, 2025Answer 2
Protein A has the highest catalytic efficiency followed by B and then C , so answer would be A >B>C
Minal Sethi
September 19, 2025a>b>c
option 2
Anjali Kushwaha
September 24, 2025A>B>C