ffect by cleaving the polysaccharide chains that form the bacterial cell wall. I
ID: 276522 • Letter: F
Question
ffect by cleaving the polysaccharide chains that form the bacterial cell wall. In the absence of this rigid mechanical support the bacterial cell literally explodes due to its high internal osmotic pressure. The cell wall polysaccharide is made up of alternating sugars, N-acetylaglucosamine (NAG) and N-acetyl muramate (NAM) linked together by glycosidic bonds (see figure below). Lysozyme normally cleaves after NAM units in the chain (Between NAM and NAG), but will also cleave artificial substrates composed entirely of NAG units. When the crystal structure of lysozyme bound to a chain of three NAG units (tri-NAG) was solved, it was discovered that the binding cleft in lysozyme include six sugar binding sites, A through F that tri-NAG filled the first three of these sites. From the crystal structure it was not apparent, however, which of the five bonds between the six sugars was the one that was normally cleaved. Tri-NAG is NOT cleaved by lysozyme, although longer NAG polymers are. It was clear from modeling studies that NAM is too large to fit into site C. Where are the catalytic groups responsible for cleavage located relative to the six sugar binding sites? and why? a. Between sites A and B b. Between sites B and C c. Between sites C and D d. Between sites D and E e. Between sites E and F NAM NAG CH2OH CHOH CH OH 0 OR OR e H,o CH,?? 0%OH CHj OH acBook AiExplanation / Answer
1) peptidoglycan is a polymer of alternate NAG and NAM units.
as in the question, the minimum trisaccharide will produce when the polymer of NAG introduced.
it is also mentioned that at site C NAM will not fit. so NAM will come at B site.
the figure showed that bond between NAM and NAG is cleaved by lysozyme and disaccharide is the product.
so the catalytic group responsible for cleavage must be located between site B and C. option B
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