The template strand DNA sequence of a small gene (with no introns) is as follows: 3' tacaggagagtcgctact 5' Determine the mRNA sequence for this...

When a eukaryotic cell needs to send genetic material out of its nucleus to create proteins, rather than send out its precious and sensitive DNA into the cytoplasm, it will copy its DNA and make a copy called messenger RNA, or mRNA for short. The copy will look much like the opposite side of a DNA strand with Adenine (a) nucleotides to match Thymines (t) on the DNA and Cytosines (c) matching Guanines (g). There will be one small difference: instead of using Thymine on the mRNA, the cell uses Uracil (u) nucelotides to match Adenines. 

We begin with our DNA code of:

tac agg aga gtc gct act

Once transcribed, before flipping the 5' and 3' sides:

aug uuc ucu cag cga uga

Each triplet of nucleotides will then code for a different amino acid to be added to the growing protein by ribosomes. All amino acids MUST start with the Met amino acid, which tells the ribosome to begin assembling a protein and end with a STOP codon, which says the protein is finished. For this specific amino acid chain the codons are:

MetPheSerGlnArgStop

Once flipped, your amino acid chain will be:

aguagcgacucuccugua

Your answer was mostly correct, except you found your amino acids after flipping the 5' and 3' ends of the mRNA, which caused you to read some codons backwards. In that case you never would have had a Met (your first codon should have been Ser) and you would not have had a Stop codon to finish the protein chain. I hope this helped you better understand the transcription process!