Primary wall

A primary wall layer is one whose cellulosic microfibrils were laid down while the cell was still (capable of) growing. Once deposited and the cell has stopped growing, a primary wall layer will not acquire more cellulose, although in certain cell types it later becomes impregnated with for example lignin or cutin. Such a wall layer is still 'primary', even if lignified.

oh a-D-Galacturonic acid (a-GalA)

oh a-L-Arabinofuranose (a-Ara)

ch2oh ch2oh cooh

\joh y \joh y \joh \°h y ho ^ f ho ^ ho ^ ho ^ * oh oh

oh ho

oh a-L-Arabinopyranose (a-Ara p)

oh a-D-Glucuronic acid (a-GlcA)

cooh hoN°LV

D-lyxo-5-Hexosulose uronic acid (Hua)

T"^ V°H yi \°h y oh y^^^ oh cooh cooh cooh ho / oh oh oh a-L-Aceric acid (a-AceA)

[2-Keto]-3-deoxy-a-D-manno-octulosonic acid (a -Kdo)

[2-keto]-3-Deoxy-ß-D-lyxo -heptulosaric acid (ß-Dha)

o ch3 oh oh

Methyl a-D-galacturonate [ester] (MeGalA)

4-O-Methyl- a-D-glucuronic acid [ether]


galacturonic acid [ester] a-L-arabinose [ester]


Figure 1.1 Monosaccharide building blocks (shown as Haworth formulae) of plant cell wall polysaccharides. The figure shows all the known sugar residues of plant cell wall polysaccharides and a selection of their esters and ethers. Top row, major components of pectins; 2nd row, major components of hemicelluloses; 3rd row, minor sugars of various origins; 4th row, mainly or only known from RG-II; bottom row, a selection of sugars with non-carbohydrate substituents. Sugars with five and six C atoms are called pentoses and hexoses respectively; Rha and Fuc are deoxyhexoses.

The monosaccharides are shown as hemiacetal or hemiketal rings. However, within a polysaccharide, each sugar (except one, the reducing terminus) is present as an acetal or ketal residue, the term 'residue' implying that it is 'what remains' after losing the -OH group (shown in blue) from the anomeric carbon (the anomeric carbon is the one with single-bonds to two oxygens; it is here drawn as the right-hand extremity of the hexagon or pentagon). In a sugar residue of a polysaccharide, this particular -OH group has departed (in the form of H2O), 'taking with it' one oxygen-linked H atom from the next sugar along the polysaccharide chain. The one sugar of the polysaccharide that is not strictly a residue is the reducing terminus, so called because it has not lost its anomeric -OH group and in aqueous solution can therefore equilibrate with the straight-chain form, which possesses an oxo group (C=O, which has reducing properties).

All but two of the sugars shown are aldoses (i.e. the anomeric carbon has only one additional C atom attached to it), but Kdo and Dha are ketoses (the anomeric C is attached to two other carbons). (Hua has two anomeric carbons (C-1 and C-5) and is both an aldose and a ketose.) In aqueous solution, each illustrated hemiacetal and hemiketal equilibrates with a small percentage of a straight-chain form possessing an oxo group (an aldehyde or ketone, in aldoses and ketoses respectively) - hence the slightly redundant term 'keto' in the names of Kdo and Dha.

Each named sugar could theoretically occur as two isomeric forms (enantiomers, designated d- and l-), distinguished by the orientation of the C-O bond of the penultimate C atom. Galactose is the only wall residue known to occur as both d- and l-enantiomer. Note that d- and l-Gal differ in orientation of the C-O bond at all four non-anomeric, chiral centres (= carbons 2, 3, 4 and 5; the difference at C-5 is indicated by the placement of the -CH2OH group).

The linkage between a sugar residue and the next building-block along a polysaccharide chain can be in either of two isomeric forms (anomers, designated a- and P-) defined by the orientation of the bond between the anomeric C atom and the oxygen atom (shown in blue) that bridges the two sugars: if this C-O bond has the same orientation as that of the penultimate C atom, then the residue is a-; if opposite, P-. This means that, in these Haworth formulae, the -OH of the anomeric carbon points down in a-d- and P-l-sugars, and up in P-d- and a-l-sugars.

The sugar ring can be 6- membered (pyranose; -p) or 5- membered (furanose; - f). Api and AceA must be -f because of the absence of an oxygen on a C-5, and MeGlcA can only be -p. Ara occurs in both forms. All the others could theoretically occur in either form, but in practice occur only in the -p form illustrated.

Each sugar residue is attached, via its anomeric carbon, to an -OH group on the following sugar unit in the polysaccharide chain. Usually, there are several such -OH groups to choose from (e.g., in the case of Glcp, on carbons 2, 3, 4 or 6: the linkage is designated (1 ^2), (1 ^3), (1^4) or (1^6), accordingly). However, a given sugar unit (either a residue or the reducing terminus) can and often does have more than one sugar residue attached to it.

Once it has become part of a polysaccharide chain, a given sugar residue is ' locked' in one of the four possible ring forms (a-p, P-p, a-f, or P-f). These ring forms have a huge impact on the polysaccharide, as is obvious from the enormous differences in physical, chemical and biological properties between amylose and cellulose (which are a-p and P-p, respectively, but otherwise identical).

Although illustrated here in unionized form, the free carboxy groups (-COOH, shown in red) would often be negatively charged (-COO-) under physiological conditions of pH. Relatively hydrophobic (non-polar) groups are shown in green.

Abbreviations: The diagrams show (in parentheses) the shorthand used throughout this chapter. Thus, unless otherwise stated in the text, the ring-form (-p or -f) and enantiomer (d- or l-) are assumed to be as illustrated here; for example, ' P-Gal' implies P-d-Galp unless specified as l-Gal. Other abbreviations used (not illustrated): MeXyl, 2-O-methyl-a-d-Xylp (ether); MeFuc, 2-O-methyl-a-l-Fucp (ether); MeRha, 3- O-methyl-a-l-Rhap (ether); MeGal, 3-O-methyl-d-Galp (ether); 5AcAra, 5-O-acetyl-l-Araf (ester); 6AcGal, 6-O-acetyl-d-Galp (ester); 6AcGlc, 6-O-acetyl-d-Glcp (ester); AUA, a 4,5-unsaturated, 4-deoxy derivative of GalA or GlcA.

Was this article helpful?

0 0

Post a comment