2. The hydrolysis of starch wan stopped when the iodine test no longer gave a bl

Question

2. The hydrolysis of starch wan stopped when the iodine test no longer gave a blue color Does this mean that the starch solution was completely hydrolyzed to ghuconet Explain. 3. Which hydrolysis of the starch is faster? On the basis of this experiment estimate what will happen to the digestion of a piece of bread (containing starch) when you chew it thoroughly? usual disaccharide, two a-D-glucose units are linked together in an 1) glycosidic linkage. Is this a reducing or nonreducing disaccharide? Explain. 4. In an unusual af1 Harcourt, Inc. 404 Experiment 38

Explanation / Answer

2) Starch contains two fractions, amylose and amylopectine. Only Amylose binds to iodine (rather I3-) to give the deep blue color.

During hydrolysis, starch (amylose and amylopectine) converts first to mono, di and trisachharides and then finally converts to glucose.

As amylose is linear and amylopectine is highly branched, amylose will be hydrolyzed faster than amylopectine.

Therefore, negative starch test confirms only disappearence of amylose, not the complete conversion to glucose. The solution may stillcontain residual amylopectine and tri, di, mono saccharides.

3) Which hydrolysis is faster? The question is not clear. I assume here we are comparing between starch to mono, di, tri saccharide conversion and from mono, di, tri to glucose formation.

Considering the change in molecular size during the two steps, I think the second step, i.e. formation of glucose is faster.

When we chew the bread, salivary glands produce alpha--amylase. This breaks down alpha (1-4) glycosidic bonds of amylose and amyopectine to form three different oligosaccharides: maltose (disaccharide), maltotriose (trisaccharide), and a group of alpha-limit dextrins (branch points from amylopectin).

These three reacts with enzyme complex called sucrase-isomaltase (or maltase) on the lumenal surface of the small intestine to produce glucose monomers.

These are then transported to the small intestine by co-transport with sodium ions.

4) Reducing sugar acts as a reducing agent due to presence of free aldehyde or ketone group.

If two monosaccharides are linked through alpha (1-1) glycosidic linkage , then there anomeric carbons are linked. That means they are stuck in the cyclic form and cannot convert to open-chain form with free aldehyde group.

Therefore, the disaccharide does not have free aldehyde and it is non-reducing.  

Disaccharides are formed from and can be classified as either reducing or nonreducing. Nonreducing disaccharides like sucrose and trehalose have glycosidic bonds between their anomeric carbons and thus

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