Identify the start and stop codons in the mrna sequence 5'-uauccauggcacuuuaaac-3'. what is the resulting amino acid sequence? a. start codon: uau; stop codon: aaa; protein: tyr-pro-trp-his-phe-lys b. start codon: uau; stop codon: uaa; protein: met-val-pro-stop c. start codon: caa; stop codon: uau; protein: gln-ile-phe-thr-val-leu d. start codon: aug; stop codon: uaa; protein: met-ala-leu-stop

D. Start codon: AUG; stop codon: UAA; protein: Met-Ala-Leu-stop

Explanation:

Start codon is the first codon in mRNA which is translated to an amino acid. AUG is the most common start codon and corresponds to methionine in eukaryotes and modified methionine in bacteria. Stop codon ends the translation process. There is no corresponding amino acid to it and the newly formed mRNA exits the translation machinery. There are three types of stop codons: UAA, UGA and UAG. GCA codon codes for Alanine and CUU codon codes for Leucine. Hence in the given sequence AUG is the start codon, UAA is the stop codon and the resulting amino acid sequence is Met-Ala-Leu-stop.

A) the sequence of bases would halt the production of any proteins. 

The  answer is 1 amu.