This must be an error

Just as proteins have a polarity defined by the amino and carboxyl termini, oligosaccharides have a polarity defined by their reducing and nonreducing ends. The carbohydrate unit at the reducing end has a free anomeric carbon atom that has reducing activity because it can form the open-chain form, as discussed earlier (p. 325). By convention, this end of the oligosaccharide is still called the nonreducing end even when it is bound to another molecule such as a protein and thus no longer has reducing properties.

In my opinion, this sentence only makes sense if it would read reducing. Can anyone confirm?

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    $\begingroup$ Just a quick comment. You would have better to write a title that is a bit more explicit than "Error in textbook". I could not have any idea of whether I'd have the skills to answer your question. (Hopefully you use one tag that is slightly more informative). $\endgroup$ – Remi.b May 29 '14 at 18:58
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    $\begingroup$ I agree that, purely from the point of view of the language used in the last sentence, it must mean 'reducing'. $\endgroup$ – Alan Boyd May 30 '14 at 7:02

In case of oligosaccharides, only the final ends matter and not the ends of the individual monosaccharides involved in glycosidic linkage. As a consequence, most oligosaccharides have only one reducing and one non-reducing ends.
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I am not sure what the textbook means by the last line but the following two possibilities seem to be consistent with the understanding of oligosaccharide structure.

  1. There is no mistake and the end being talked about is the left end of the monosaccharide on left in the diagram . This monosaccharide has both its ends non-reducing, but only one of these ends will be called the non-reducing end even though the other end(formerly reducing end) is now involved in glycosidic linkage and is hence non-reducing. The it in the last line in this case refers to the entire monosaccharide (and not the end of concern), which has lost its reducing ability.

  2. There is a mistake. And the end being talked about is the right end of the monosaccharide on the left. It is usually not referred to as the reducing end (which is the rightmost end), but for the case of explaining the polarity of the monosaccharide, it can be referred to as the reducing end.

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    $\begingroup$ +1 for admirable attempt to give the author the benefit of the doubt. I agree with OP it's just a slip, your (2). $\endgroup$ – daniel Jun 1 '14 at 1:53

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