How to determine the most likely reading frame of a DNA sequence?
This is what we classify as a homework question, but as it satisfies the criterion of the poster demonstrating an attempt to answer it, I provide the following suggestion of an answer.

I assume that as it appeared in an introductory bioinformatics module the exam question is just testing reading frames (obviously) and the punctuation of the genetic code. The meaning is not entirely clear, but as one reading frame would start with the initiation codon ATG/AUG (reverse complement of: cat - 3′)† and end with the termination codon TAA/UAA (reverse complement of: 5′ - tta) that will presumably produce “the most likely protein translation”. 

This is reading frame F4 in the output from EMBOSS Sixpack, below, in which termination codons are indicated by an asterisk.

      L  F  I  R  Q  R  H  A  R  H                                   F1
       Y  S  S  A  S  A  M  R  A  X                                  F2
        I  H  P  P  A  P  C  A  P  X                                 F3
    1 ttattcatccgccagcgccatgcgcgccat 30
      ----:----|----:----|----:----|
    1 aataagtaggcggtcgcggtacgcgcggta 30
       X  N  M  R  W  R  W  A  R  W                                  F6
      X  I  *  G  G  A  G  H  A  G                                   F5
        *  E  D  A  L  A  M  R  A  M                                 F4


The conceptual translation, reading N to C, is MARMALADE, which is obviously meant to be humorous and suggests that it is indeed the intended answer. 

Open Reading Frames

The poster asks for clarification of the difference between reading frame and open reading frame. There is a Wikipedia entry for open reading frame but I provide an explanation of my own to relate it to the example.

There are always six reading-frames for the conceptual translation of a piece of double-stranded DNA, as shown in the example. 

I would define an open reading frame as one that is not excluded from being translated by the punctuation of the genetic code. It has the theoretical potential to be translated considering only the punctuation of the code, although it may not actual be translated. It can start either with the first AUG after a termination codon (even though it cannot be certain that this is the actual AUG used) or the start of a sequenced fragment of DNA (with the assumption that an AUG is possible preceding the 5′ end of the fragment). It can end either with a termination codon or the end of the sequenced fragment (with the assumption that a termination codon will lie 5′ to the end of the fragment).

By these criteria, reading frames F1, F2, F3 and F6, above are completely open (even though the internal methionine could theoretically be an initiation codon), F5 contains the open reading frame GAHGAGG, and F4 is a complete open reading frame, perhaps excluding the termination codon (depending on your semantic definition of the precise end of an open reading frame).

† Reverse Complement

If we take a section of DNA written in the 5′-to-3′ direction — according to standard convention — and use the Watson–Crick base-pairing equivalences (A=T, G=C) to generate the complementary strand, this will be in the 3′-to-5′ direction. For the strand in the question, 

5′ - ttattcatccgccagcgccatgcgcgccat - 3′


the complementary strand is: 

3′ - aataagtaggcggtcgcggtacgcgcggta - 5′


As shown above.

For ease of translation manually — and for any computer program that manipulates sequences — one needs to reverse this to the  5′-to-3′ direction:

5′ - atggcgcgcatggcgctggcggatgaataa - 3′


This is the reverse complement. Now the starts of the three reverse reading frames are easy to read as:

ATG...
 TGG...
  GGC...