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Question
Differentiable between the following:
Amylose and Amylopectin
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Solution 1
Amylose:
- Water soluble components
- Constitutes 15-20% of starch
Amylopectin:
- Water insoluble component
- Constitutes 80-85% starch.
Solution 2
| Amylose | Amylopectin | |
| (a) | Only C1− C4 glycosidic linkage | C1− C4 glycosidic linkage but, branching occurs by C1− C6 glycosidic linkage |
| (b) | Water-soluble component of starch (15-20%) | Water-insoluble component of starch (80-85%) |
RELATED QUESTIONS
The number of asymmetric carbon atom(s) below the figure is/are


Glucose does not give Schiff’s test because of the formation of cyclic ____________.
The symbols D and L represents ____________.
The reaction of glucose with red P + HI is called ____________.
The α-D glucose and β-D glucose differ from each other due to difference in carbon atom with respect to its ____________.
The number of chiral carbons in ß-D(+) glucose is ____________.
Write the reactions of D-glucose which can’t be explained by its open-chain structure. How can cyclic structure of glucose explain these reactions?
On the basis of which evidences D-glucose was assigned the following structure?
\[\begin{array}{cc}
\ce{CHO}\\
|\phantom{....}\\
\phantom{..}\ce{(CHOH)4}\\
|\phantom{....}\\
\phantom{..}\ce{CH2OH}
\end{array}\]
The number of asymmetric carbon atoms in the glucose molecule in open and cyclic form is ______.
Give a reason for the following observations:
Penta-acetate of glucose does not react with hydroxylamine.
