Subtopic: Fundamentals of Biomolecules, enzyme kinetics
Question Statement:
A protein has three identical sites arranged at the vertices of an equilateral
triangle. If one site is filled with a dye (donor), the measured quantum yield (
D )
is 0.5. Filling one site with a donor dye and a second site with an acceptor dye
results in
D of 0.25. The measured
D of one site filled with donor and the other
two sites filled with acceptor dye (rounded off to three decimal places)
is ___________.
Official Answer Key:
0.165 to 0.168
Step-by-Step Solution & Explanation
### **Step-by-Step Solution & Mathematical Derivation (GATE BT 2023 Q.57)**
**1. Governing Concept & Definition**:
- **Topic**: Biophysical Techniques & FRET De-excitation Kinetics
- Fluorescence quantum yield of donor alone:
$\Phi_D = \frac{k_r}{k_r + k_{nr}}$
- When one acceptor dye is present at distance $R$:
$\Phi_{DA} = \frac{k_r}{k_r + k_{nr} + k_T}$
- When two identical acceptor dyes are present at identical distances (equilateral triangle geometry):
$\Phi_{DA2} = \frac{k_r}{k_r + k_{nr} + 2 k_T}$
**2. Detailed Step-by-Step Calculation**:
- **Step 1: From donor alone ($\Phi_D = 0.5$)**:
$\frac{k_r}{k_r + k_{nr}} = 0.5 \implies k_r + k_{nr} = 2 k_r \implies k_{nr} = k_r$
Let $k_r = 1 \implies k_r + k_{nr} = 2$.
- **Step 2: From donor + one acceptor ($\Phi_{DA} = 0.25$)**:
$\frac{1}{2 + k_T} = 0.25 = \frac{1}{4} \implies 2 + k_T = 4 \implies k_T = 2$
- **Step 3: Donor + two identical acceptors**:
- Due to equilateral triangle symmetry, both acceptors are equidistant from the donor, each providing transfer rate $k_T = 2$:
$\Phi_{DA2} = \frac{1}{2 + 2 k_T} = \frac{1}{2 + 2(2)} = \frac{1}{2 + 4} = \frac{1}{6} \approx 0.1667 \approx 0.167$
**3. Final Answer**:
- **Calculated Value**: **0.167**
- **Official GATE Accepted Range**: **0.165 to 0.168**
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