How many amino acids would be encoded in the sequence 5′ AUGUUACGGAAU 3′ by a non-overlapping and maximally overlapping code?

1 answer:

5 0

<span>In non-overlapping code there would be four amino acids, because there are just four triplets in the code, of which none is the stop codone. For maximally overlapping code there would be ten amino acids encoded, the four starting from the first letter: AUG, UUA, CGG, AAU, then three starting from the second letter: UGU, UAC, GGA, and the three starting from the third letter: GUU, ACG, GAA.</span>

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