The peptide Phe-Leu-Ile-Met-Ser-Leu has six amino acids, each encodable by multiple codons as specified. The number of unique DNA sequences equals the product of codon options per position, since codons are independent. Calculation: 2 (Phe) × 6 (Leu) × 3 (Ile) × 1 (Met) × 4 (Ser) × 6 (Leu) = 864.

Step-by-Step Solution

Multiply codon counts sequentially for the sequence:

• Position 1 (Phe): 2 options.

• Position 2 (Leu): 6 options → 2 × 6 = 12.

• Position 3 (Ile): 3 options → 12 × 3 = 36.

• Position 4 (Met): 1 option → 36 × 1 = 36.

• Position 5 (Ser): 4 options → 36 × 4 = 144.

• Position 6 (Leu): 6 options → 144 × 6 = 864.​

This assumes standard genetic code degeneracy, where DNA sequences reflect the template strand but codon counts match mRNA equivalents given.​

The genetic code’s degeneracy allows multiple codons per amino acid, key for questions like “number of unique DNA sequences encoding Phe-Leu-Ile-Met-Ser-Leu peptide.” In competitive exams such as GATE Biotechnology or IIT JAM, this tests understanding of codon tables where Phe uses 2 codons (UUU, UUC), Leu 6 (UUA, UUG, CUU, CUC, CUA, CUG), Ile 3 (AUU, AUC, AUA), Met 1 (AUG), and Ser 4 (UCU, UCC, UCA, UCG).​

For the peptide sequence Phe-Leu-Ile-Met-Ser-Leu, unique DNA sequences arise from multiplying codon possibilities at each position, ignoring start/stop codons as it’s a fragment. No options are provided, but common MCQ traps include addition (wrong), forgetting repeat Leu, or using RNA vs DNA mismatch—always multiply for independent positions.​

Codon Breakdown Table

This 864 result appears in prior GATE 2020 papers, emphasizing multiplication over summation for degeneracy calculations.​