Question 22.5: The structure of an mRNA segment obtained from a DNA templat......
The structure of an mRNA segment obtained from a DNA template strand is
mRNA 3′ AUU – CCG – UAC – GAC 5′
What polypeptide amino acid sequence will be synthesized using this mRNA?
The directionality of an mRNA segment obtained from template DNA is 3^{\prime}-to-5^{\prime} because the two segments must be antiparallel to each other (Section 22.5). The codons in an mRNA must be read in the 5^{\prime}-to-3^{\prime} direction to correctly use genetic code relationships to determine the sequence of amino acids in the peptide. Rewriting the given mRNA with reversed directionality (5^{\prime}-to-3^{\prime} direction) gives
mRNA 5′ CAG – CAU – GCC – UUA 3′
Note that in reversing the directionality from 3^{\prime}-to-5^{\prime} to 5^{\prime}-to-3^{\prime} the sequence of bases in a codon is also reversed; for example, GAC becomes CAG.
Using the genetic code relationships between codon and amino acid (Table 22.2) shows that this mRNA codon sequence codes for the amino acid sequence
Gln–His–Ala–Leu
