If your idea is sufficient, here's how to do it:

my ($name) = m/sequence (.*?) consists/i;
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$_ = join ' ', split;
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Not just to be contrary, but this is no place for the non-greedy quantifiers. What if the "unpredictable sequence" contains the string "consists"?

On the assumption that the rest of the string is less "unpredictable", you want to match everything on this line between the first "sequence" and the last "consists". Hence my suggestion:

my ($name) = m/Sequence (.*) consists/;
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(If the rest of the string may be as unpredictable, I'd suggest you write a full parser instead. Looks like there's a fair chance you won't have to though.)

Consult the docs for "non-greedy" quantifiers. /Sequence \s+ (.*?) \s+ consists/x

Simply use capturing groups like this: (You also want a non greedy operator as other posts have pointed out.)

#The regex, commented for your convinence.
$Error =~ /Sequence\s #opening anchor, don't capture
              ( # open capturing group to get target pattern
                 .+? # target pattern, non-greedy catch all
              ) # close group
              \sconsists # closing anchor, don't capture
          /x; # end regex
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After that your target name will be inside $1. If you want to place it all on one line, you can use =~ in list context and get the results as a list like this:

($Missing_Living_Being) = $Error =~ /Sequence\s(.+?)consists/;
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Just be sure that the list that you're assigning to has the same number of elements as there are capturing groups (the parts of the pattern inside parens) or else you'll run into trouble with values going to the wrong place. If you're paranoid about this you could assign to an array and shift it to get the first group, but that adds a lot of uneccessary overhead.