How do I make a cell with the following contents?

I am making a program that basically takes a string s as a single strand of DNA and returns the amino acid sequence of the longest gene it finds. Whereby, a gene is defined as a reading frame that: starts with AUG codon, ends with one of UAA,UAG, or UGA codon.

I tried making a cell of different "frames" but since they are not the same length I can't put them into an array. How do i work around this? Here's my code:

function [ptn]=Seq_transcribe2(x)
y=seq_transcribe1(x);
frames={};
frames={x(1:end) x(2:end) x(3:end) y(1:end) y(2:end)
 y(3:end)};
starts=[];
stops=[];
allorfs={};
for i=1:3:numel(frames)-2
  codon= frames([i i+1 i+2])
    if codon=='AUG'
    starts(end+1)=codon;
if strcmp(codon,'UAA') || strcmp(codon,'UAG') || strcmp(codon,'UGA')
    stops(end+1)=codon;
end
stops= find(stops>starts,1)
lengthofthisstart=stops-starts
allorfs{end+1}=frame(starts:stops-1)

[Niels]

would be helpfull if you add the error message

[Mohannad Abboushi]

Error using vertcat Dimensions of matrices being concatenated are not consistent.

Answer 1

[Guillaume]

If I understood correctly, a simple way to find all genes would be:

[genesequences, starts, stops] = regexp(x, 'AUG.*?(UAA|UAG|UGA)', 'match', 'start', 'end');

And the longest sequence is of course:

[~, longestidx] = max(stops - starts);
longestsequence = genesequences{longestidx}

[Arthur Goldsipe]

I think you need a slight change to account for the fact that all codons are 3 characters long:

    [genesequences, starts, stops] = regexp(x, 'AUG(...)*?(UAA|UAG|UGA)', 'match', 'start', 'end');

[Guillaume]

Oh yes, as I know nothing about genes and codons, I didn't know that the number of characters between the start and end codon must be a multiple of three, but I should have inferred that from the original code.

Thanks.

Answer 2

[Niels]

if i understood you right your problem is in one of the following lines:

frames={x(1:end) x(2:end) x(3:end) y(1:end) y(2:end)
allorfs{end+1}=frame(starts:stops-1)

if so, i cant replicate your problem, in cell arrays the length of the elements is irrelevant

>> a=1:3;
>> b=1:4;
>> c=1:5;
>> cell={a b c}
cell =
  1×3 cell array
    [1×3 double]    [1×4 double]    [1×5 double]
%=================================

>> a={}
  a =
    0×0 empty cell array
  >> a{end+1}=1
  a =
    cell
      [1]
  >> a{end+1}=2
  a =
    1×2 cell array
      [1]    [2]
  >> a{end+1}=[2 1]
  a =
    1×3 cell array
      [1]    [2]    [1×2 double]

[Mohannad Abboushi]

I feel like there's got to be an easier way to do this

[Guillaume]

If the following line

frames={x(1:end) x(2:end) x(3:end) y(1:end) y(2:end)
  y(3:end)};

is indeed written on two lines, then yes matlab is going to issue a concatenation error since the line return is interpreted as a vertical concatenation.

frames={x(1:end) x(2:end) x(3:end) y(1:end) y(2:end) y(3:end)};

or

frames={x(1:end) x(2:end) x(3:end) y(1:end) y(2:end) ...
  y(3:end)};

would fix the error