Question

what amino acid sequence would result from the following mrna? gene a mrna sequence: 5 ccc aug ucc uac uuc cca cag gcc uaa uuu 3 pro met ser tyr phe pro gln ala pro met ser tyr phe pro gln ala phe met ser tyr phe pro gln ala phe met ser tyr phe pro gln ala

Explanation:

Step1: Identify start codon

The start codon is AUG which codes for Methionine (Met). The mRNA sequence starts with CCC but translation begins at AUG.

Step2: Group into codons

Group the mRNA sequence into sets of three nucleotides after AUG: UCC, UAC, UUC, CCA, CAG, GCC, UAA, UUU.

Step3: Use codon - amino acid table

UCC codes for Serine (Ser), UAC for Tyrosine (Tyr), UUC for Phenylalanine (Phe), CCA for Proline (Pro), CAG for Glutamine (Gln), GCC for Alanine (Ala). UAA is a stop codon and translation stops there, so UUU is not translated.

Answer:

Met Ser Tyr Phe Pro Gln Ala