Final answer:
It would take 5 codon sequences to produce the given amino acid sequence of alanine, proline, serine, isoleucine, and a stop signal, as each amino acid and the stop signal are coded by separate three-nucleotide codons.
Step-by-step explanation:
To determine the number of codon sequences required to produce the amino acid sequence of alanine - proline - serine - isoleucine - stop, we need to understand that the genetic code uses triplets of nucleotides - codons - to specify amino acids. Since each amino acid in the sequence is coded by a three-nucleotide codon, and there is an additional codon for the stop signal, we need a separate codon for each amino acid plus one for the stop signal.
Therefore, alanine, proline, serine, and isoleucine would each require one codon. The stop signal also requires one codon. Summing these, we have 4 codons for the amino acids, plus 1 stop codon, resulting in a total of 5 codons needed to code for the sequence.
The correct answer is A. It would take 5 codon sequences to produce the amino acid sequence of alanine - proline - serine - isoleucine - stop.