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. First, understand the sequence given: 5'-AUGUUACGGAAU-3'. This sequence contains 12 nucleotides.
2. For the non-overlapping triplet code, divide the sequence into groups of three nucleotides each, starting from the 5' end. This results in four groups: AUG, UUA, CGG, AAU. Each group codes for one amino acid, so there are 4 amino acids encoded.
3. For the maximally overlapping triplet code, start at the first nucleotide and shift one nucleotide at a time to form a new triplet until reaching the end of the sequence. The first triplet is AUG, the second is UGU, the third is GUU, and so on until the last triplet starting from the tenth nucleotide. This results in 10 triplets, hence encoding 10 amino acids.
4. Compare the number of amino acids encoded in both methods: 4 amino acids for the non-overlapping code and 10 amino acids for the overlapping code.
5. The correct answer is option c: 4 (non-overlapping) and 10 (overlapping).