A gene whose template strand contains the sequence 3’-TACTTGTCCGATATC-5’ is mutated to 3’-TACTTGTCCAATACC-5’.

A gene whose template strand contains the sequence 3’-TACTTGTCCGATATC-5’ is mutated to 3’-TACTTGTCCAATACC-5’. For normal and mutant genes, draw the mRNAs, and the polypeptide each encodes. What is the effect of the mutation on the amino acid sequence?

Your teacher is good. The only thing I wish your teacher had added was a phrase that this fragment is properly aligned within an exon's open reading frame.

The mRNA is complementary and antiparallel to the given template strand. You know that these are the base pairings:
DNA RNA
A .....U
T .....A
C .....G
G .....C

So for your normal
3’-TAC TTG TCC GAT ATC-5’ (template strand)
5'-AUG AAC AGG CUA UAG-3' (mRNA)

Look up the mRNA codons on a genetic code chart
http://en.wikipedia.org/wiki/Genetic_cod…
Met-Asn-Arg-Leu (carboxyl terminus)

And for your mutant
3’-TAC TTG TCC AAT ACC-5’ (template strand)
5'-AUG AAC AGG UUA UGG-3' (mRNA)
Met-Asn-Arg-Leu-Trp- (more amino acid residues follow)

You have a couple of single base pair substitutions. The first one is a "synonymous mutation" with both normal and mutant strands leading to Leucine in the polypeptide product.

The second is a biggie. The STOP codon in the normal strand is replaced by a codon that codes for Tryptophan. The polypeptide will thus not stop being translated here but could include more residues until either a stop codon is encountered, or the ribosome falls off when the RNA ends.