Good one .

function tf = isPangram(s)
s1='QWERTYUIOPASDFGHJKLZXCVBNM';
L=strlength(s);
cnt=0;
s2=upper(s);
for i=1:26
if isempty(findstr(s1(i),s2))==false
cnt=cnt+1;
continue;
end
end
if cnt==26
tf=true;
else
tf=false;
end
end

nice one-liner :)

Funny system they have for measuring length - this sure looks like the shortest.

. . . punctuation marks

This will fail if the sentence contains enough non-alphabetical letters!

or if it's just 'ABCDEFGHIJKLMNOPQRSTUVWXYZ';

Or if it is 'aAbBcCdDeEfFgGhHiIjJkKlLmMnNoOpPqQrRsS'

i think my solution is correct but the system seems to get it wrong

This would not work if punctuation was included in the string, or if spaces were not included in the string, or if capital and lower case letters were duplicated, but letters were still missing.

For example:
'ABCDEFabcdefghijklmnopqrstuvwxy'

this would not work if the string was 'szv'

there is a false positive on 'The quick brown fox jumps over a lazy do!'