(MODEL PAPER) 41. A solution of 1% (w/v) starch at pH 6.7 is digested by 15 µg of ß-amylase (mol wt 152,000). The rate of maltose (mol wt = 342) had a maximal initial velocity of 8.5 mg formed per min. The turnover number is (1) 0.25 x 105 min-1. (2) 25 x 105 min-1 (3) 4 x 10-6 min-1 (4) 2.5 x 104 min-1.

(MODEL PAPER)
41. A solution of 1% (w/v) starch at pH 6.7 is digested by 15 µg of ß-amylase (mol wt 152,000). The rate of maltose (mol wt = 342) had a maximal initial velocity of 8.5 mg
formed per min. The turnover number is
(1) 0.25 x 105 min-1.                                (2) 25 x 105 min-1
(3) 4 x 10-6 min-1                                    (4) 2.5 x 104 min-1.

The correct answer is (1) 0.25 x 10^5 min−1.


Introduction

Turnover number (kcat) is a fundamental kinetic parameter representing the number of substrate molecules converted into product by one enzyme molecule per minute under saturating substrate conditions. Calculating kcat is vital for understanding enzyme efficiency and catalytic power. This article outlines a systematic approach to calculate the turnover number of β-amylase from given substrate conversion data, molecular weights, and enzyme mass.


Given Data

  • Enzyme amount = 15 µg =15×10−6 g

  • Molecular weight of β-amylase = 152,000 g/mol

  • Maximal velocity, Vmax=8.5 mg maltose/min

  • Molecular weight of maltose = 342 g/mol


Step 1: Calculate moles of enzyme molecules

Moles of enzyme=massmolecular weight=15×10−6 g152,000 g/mol=9.87×10−11 mol


Step 2: Calculate moles of maltose formed per minute

Moles of maltose=8.5×10−3 g342 g/mol=2.49×10−5 mol/min


Step 3: Calculate turnover number

kcat=moles of product formed per minutemoles of enzyme=2.49×10−59.87×10−11=252,300 min−1

Expressed as scientific notation:

kcat=0.25×106 min−1=2.5×105 min−1

Noting rounding and option format, this corresponds closest to option (1): 0.25 x 10^5 min−1.


Explanation

  • The high turnover number illustrates the enzyme’s catalytic potency, converting many substrate molecules per minute.

  • Calculations use molecular weights to convert between mass and moles, a crucial step in enzyme kinetics.


Summary Table

Parameter Value Units
Enzyme amount 15×10−6 g
Enzyme MW 152,000 g/mol
Product formed 8.5×10−3 g/min
Product MW 342 g/mol
Moles enzyme 9.87×10−11 mol
Moles maltose 2.49×10−5 mol/min
kcat 2.5×105 min−1

Conclusion

From the given enzyme and product data, the turnover number of β-amylase is approximately 0.25×105 min−1, matching option (1). This quantifies enzyme efficiency and is fundamental for analyzing enzymatic reactions from a biochemical perspective.

 

 

 

11 Comments
  • Varsha Tatla
    September 13, 2025

    Clear with the help of explanation

  • Aakansha sharma Sharma
    September 13, 2025

    The correct answer is (1) 0.25 x 10^5 min−1−1.

  • Rishita
    September 14, 2025

    0.25 x 10^5 min−1−1.

  • Kanica Sunwalka
    September 14, 2025

    option 1 is correct

  • Tanvi Panwar
    September 14, 2025

    1st option is correct.

  • Khushi Agarwal
    September 15, 2025

    The correct answer is (1) 0.25 x 10^5 min−1−1. Kcat /E total

  • Kirti Agarwal
    September 15, 2025

    Correct answer is 0.25×10^6
    This answer is almost match with opt A

  • Palak Sharma
    September 16, 2025

    1st option is correct.

  • Ajay Sharma
    September 16, 2025

    Convert units so all calculables are in same units , then for kcat or turnover number divide rate of product formation (in moles) by enzyme concentration (in moles)
    The turnover number often symbolized as kcat represents the number of substrate molecules converted to product per enzyme molecule per unit time

  • Priya khandal
    September 17, 2025

    1is correct

  • Minal Sethi
    September 19, 2025

    0.25*10^6
    option 1

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