Use this sequence of DNA, make complementary DNA, then mRNA and lastly the protein coded

ATG GGG CTG TTA GGA CAT GAT AAG GAC CAT.

Complementary sequence for the given DNA:

3'TACCCCGACAATCCTGTACTATTCCTGGTA5' 

OR

5'ATGGTCCTTATCATGTCCTAACAGCCCCAT3'

mRNA for given strand of DNA:

5' AUGGUCCUUAUCAUGUCCUAA CAGCCCCAU 3'

Protein or amino acid sequence for given mRNA:

Methionine-Valine-Leucine-isoleucine-Methionine-Serine

The sequence of given strand of DNA is:

5' ATG GGG CTG TTA GGA CAT GAT AAG GAC CAT 3'

Hence, 

the complementary or coding DNA sequence should be:

3' TAC CCC GAC AAT CCT GTA CTA TTC CTG GTA 5'

OR

5' ATGGTCCTTATCATGTCCTAA CAGCCCCAT 3'

mRNA is the exact copy of coding strand or non template except for uracil instead of thymine.

Hence, sequence of mRNA for given DNA strand should be:

5' AUGGUCCUUAUCAUGUCCUAA CAGCCCCAU 3'

Amino acid sequence for this mRNA should be :

Met-Val-leu-Ile-Met-Ser

Because,

AUG codes for Methionine

GUC codes for Valine

CUU codes for Leucine

AUC codes for Isoleucine

AUG codes for Methionine

UCC codes for Serine, next codon is

UAA which is a stop codon hence, translation should terminate at this point and codons beyond it can't be translated.

Hence,

the length of a protein is restricted to first six amino acids as seventh codon is stop codon or termination codon.