Q.49 Aqueous two–phase extraction is used to recover α–amylase from a solution. A
polypropylene glycol–dextran mixture is added and the solution separates into
upper and lower phases. The partition coefficient is 4.0 and the ratio of upper to
lower phase volume is 5.0. The enzyme recovery or yield (in percentage, rounded
off to the nearest integer) is ___________.

Aqueous Two-Phase Extraction (ATPS) α-Amylase Recovery

In a polypropylene glycol–dextran aqueous two-phase extraction system, the partition
coefficient (K = 4.0) and upper-to-lower phase volume ratio (r = 5.0) give a
95% recovery of α-amylase in the upper phase.

Key Concept

Aqueous two-phase extraction partitions α-amylase between:

• Upper phase: Polypropylene glycol (less dense)
• Lower phase: Dextran (more dense)

The partition coefficient (K) defines how enzyme distributes:

K = Cupper / Clower = 4.0

Step-by-Step Calculation

Let lower phase volume VL = V.
Then upper phase volume VU = rV = 5V.

If lower concentration is CL then:

CU = K × CL = 4 CL

Total enzyme:

Total = VUCU + VLCL
       = 5V(4CL) + V(CL)
       = 20VCL + VCL
       = 21VCL

Enzyme in upper phase:

Upper = 20VCL

Recovery (yield):

20VCL / 21VCL × 100 = 95.24% ≈ 95%

Verification Formula

The standard ATPS recovery equation:

Y = (Kr / (1 + Kr)) × 100

Substitute:

Y = (4 × 5) / (1 + 4 × 5) × 100 = 20/21 × 100 = 95%

Final Answer

α-Amylase recovery in the upper phase = 95%

Introduction

Aqueous two-phase extraction (ATPS) is widely used to separate sensitive biomolecules such as α-amylase.
Using polymer-based phase separation (polypropylene glycol–dextran), engineers can recover enzymes efficiently
without denaturing heat or organic solvents.

Understanding ATPS

Two immiscible aqueous phases are formed:

• Upper phase: Polypropylene glycol (PPG)
• Lower phase: Dextran

The enzyme distributes based on affinity, described by:

K = Cupper / Clower = 4.0

Volume ratio is r =5.0, meaning the upper phase is five times larger than the lower.

Yield Calculation Summary

Given K and r:

Y = Kr / (1 + Kr) × 100

Plug in:

Y = 95%

Relevance in Exams

This question is common in:

• GATE Biotechnology
• MTech entrance tests
• Bioprocess engineering courses

Common errors include ignoring phase volume difference or using K incorrectly.

Industrial Application

ATPS enables:

• Cost-effective enzyme recovery
• Scalable downstream processing
• Retention of enzyme activity

Final Result: 95% α-amylase recovery