Question

Given the non-template strand of DNA, draw the template strand with the 5' and 3' ends labeled. Draw the RNA molecule with the proper 5' and 3' ends labeled

Non-Template Strand: 3'- AAT GCT CGT AGC TTC GAT CGG ATC GA-5'

My answers: Template= 5'- TAT CGA GCA TCG AAG CTA GCC TAG CT-3'

RNA Molecule= 3'- AUA GCU CGU UGC UUC GAU CGG AUC GA-5'

Next it asks, how many amino acids would the RNA molecule code for? - how would I figure this out?

Answer 1

Non-Template Strand: 3'- AAT GCT CGT AGC TTC GAT CGG ATC GA-5'
Your answer, Template= 5'- TAT CGA GCA TCG AAG CTA GCC TAG CT-3'
Correct Template = 5' - TTA CGA GCA TCG AAG CTA GCC TAG CT - 3' (ANSWER)
You didn't convert the second and third base in AAT into TTA.
Hence, your subsequent RNA molecule is also incorrect:

Template Strand...........................5' - TTA CGA GCA TCG AAG CTA GCC TAG CT - 3'
Template Strand's mRNA............3' -  AAU GCU CGU AGC UC GAU CGG AUC GA - 5'
Non-Template Strand..................3' - AAT GCT CGT AGC TTC GAT CGG ATC GA - 5'
Non-Template Strand's mRNA...5' - UUA CGA GCA UCG AAG CUA GCC UAG CU - 3'

How many amino acids would the RNA molecule code for?
Technically speaking, RNA molecules code for 1 amino acide for every 3 base pairs that they contain. These 3 base pairs constitute a codon. So we can expect the 26-base pair RNA stretch to code for 8 amino acids.
However going into advanced concepts we understand that Transcription only occurs from 5' to 3' direction and starts at AUG codons called Start codon sequences. Only the Template Strand's mRNA contains an AUG codon in the 5' to 3' direction, and hence gives us when reversed:
5' - AG CUA GGC UAG CU CGA UGC UCG UAA - 3'
After regrouping into groups of three starting from AUG we get:
5' - AGC UAG GCU AGC UCG AUG CUC GUA A - 3'
Hence the RNA molecule will code for 3 amino acids, namely: MLV protein. ANSWER