Answer:
Wild type template strand
3'TGTACTTCAAACCGATT5'
Wild type template DNA strand after transcription give following mRNA sequence
5’ACAUGAAGUUUGGCUAA3’
This mRNA has five codons; out of five, four codons (AUG, AAG, UUU and GGC)encode aminoacid and one is stop codon (UAA). Therefore, after translation this mRNA gives following amino acid sequence
Methionine-Lysine-Phenylalanine-Glycine