GATE BT 2020 • Q11
MCQ • 1 Mark
Section 2: General Biotechnology
Biomolecules, enzyme kinetics
Subtopic: Fundamentals of Biomolecules, enzyme kinetics
Question Statement:
Protein P becomes functional upon phosphorylation of a serine residue. Replacing
this serine with
will result in a phosphomimic mutant of P.
Options:
-
(A) alanine
-
(B) aspartic acid
-
(C) phenylalanine
-
(D) lysine
Step-by-Step Solution & Explanation
### **Official Solution & Concept Analysis (GATE BT 2020 Q.11)**
**1. Governing Concept**:
- **Topic**: Biochemistry & Phosphomimetic Mutations
- Protein phosphorylation adds a bulky, negatively charged phosphate group ($\text{PO}_4^{2-}$) to the hydroxyl group of serine, threonine, or tyrosine residues.
- To mimic this negative electrostatic charge constitutively in genetic engineering (**phosphomimetic mutation**), the target residue is substituted with a negatively charged acidic amino acid:
- **Aspartic acid (Asp, D)** or **Glutamic acid (Glu, E)**, which carry negative carboxylate charges ($\text{COO}^-$) at physiological pH.
- In contrast, substitution with **Alanine (Ala, A)** abolishes phosphorylation (**phospho-dead / phospho-null** mutant).
**2. Option Analysis**:
- **Option (A) [alanine] is INCORRECT**: Creates a phospho-dead/unphosphorylatable mutant.
- **Option (B) [aspartic acid] is CORRECT**: Negatively charged side-chain mimics the phosphorylated state.
- **Option (C) [phenylalanine] is INCORRECT**: Bulky non-polar aromatic residue.
- **Option (D) [lysine] is INCORRECT**: Positively charged basic residue ($\text{NH}_3^+$).
**3. Final Answer**:
- **Correct Choice**: **B**
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