One of the possibilities considered about the genetic code was that the code was overlapping, meaning that a single base could be part of up to three codons. How many amino acids would be encoded in the sequence 5′-AUGUUACGGAAU-3′ by a non-overlapping and a maximally overlapping triplet code? a. 4 (non-overlapping) and 16 (overlapping) b. 4 and 12 c. 4 and 10 d. 12 and 4

Verified step by step guidance

1. Identify the sequence given: 5′-AUGUUACGGAAU-3′. This sequence contains 12 nucleotides.
2. For the non-overlapping triplet code: Divide the total number of nucleotides by 3 (since each codon is composed of 3 nucleotides) to find out how many codons can be formed and thus how many amino acids can be encoded.
3. For the maximally overlapping triplet code: Start from the first nucleotide and form a codon, then move one nucleotide to the right and form the next codon, and continue this process until you reach the end of the sequence. Calculate how many codons can be formed in this manner.
4. Compare the number of codons formed in both non-overlapping and overlapping scenarios to determine how many amino acids each scenario encodes.
5. Match the calculated numbers of amino acids encoded in both scenarios with the options provided to find the correct answer.