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Chapter 14- Biomolecules Interview Questions Answers

Question 1 :

How do you explain the absence of aldehyde group in thepentaacetate of D-glucose?

Answer 1 :

D-glucose reacts with hydroxylamine (NH2OH)to form an oxime because of the presence of aldehydic (−CHO) group or carbonylcarbon. This happens as the cyclic structure of glucose forms an open chainstructure in an aqueous medium, which then reacts with NH2OHto give an oxime.

But pentaacetate of D-glucose does not reactwith NH2OH. This is because pentaacetate does not form an openchain structure.

Question 2 :

The melting points and solubility in water of amino acidsare generally higher than that of the corresponding halo acids. Explain.

Answer 2 :

Both acidic (carboxyl) as well as basic (amino) groups arepresent in the same molecule of amino acids. In aqueous solutions, the carboxylgroup can lose a proton and the amino group can accept a proton, thus givingrise to a dipolar ion known as a zwitter ion.

Due to this dipolar behaviour, they have strongelectrostatic interactions within them and with water. But halo-acids do notexhibit such dipolar behaviour.

For this reason, the melting points and the solubility ofamino acids in water is higher than those of the corresponding halo-acids.

Question 3 :

Where does the water present in the egg go after boilingthe egg?

Answer 3 :

When an egg is boiled, the proteins present inside the eggget denatured and coagulate. After boiling the egg, the water present in it isabsorbed by the coagulated protein through H-bonding.

Question 4 :

Why cannot vitamin C be stored in our body?

Answer 4 :

Vitamin C cannot be stored in our body because it is watersoluble. As a result, it is readily excreted in the urine.

Question 5 :

What products would be formed when a nucleotide from DNAcontaining thymine is hydrolysed?

Answer 5 :

When a nucleotide from the DNA containing thymine ishydrolyzed, thymine β-D-2-deoxyribose and phosphoric acid are obtained as products.

Question 6 :

When RNA is hydrolysed, there is no relationship among thequantities of different bases obtained. What does this fact suggest about thestructure of RNA?

Answer 6 :

A DNA molecule is double-stranded in which the pairing ofbases occurs. Adenine always pairs with thymine, while cytosine always pairswith guanine. Therefore, on hydrolysis of DNA, the quantity of adenine producedis equal to that of thymine and similarly, the quantity of cytosine is equal tothat of guanine.

But when RNA is hydrolyzed, there is no relationship amongthe quantities of the different bases obtained. Hence, RNA is single-stranded.

Question 7 :

What are monosaccharides?

Answer 7 :

Monosaccharides are carbohydrates that cannot behydrolysed further to give simpler units of polyhydroxy aldehyde or ketone.

Monosaccharides are classified on the bases of numberof carbon atoms and the functional group present in them. Monosaccharidescontaining an aldehyde group are known as aldoses and those containing a ketogroup are known as ketoses. Monosaccharides are further classified as trioses,tetroses, pentoses, hexoses, and heptoses according to the number of carbonatoms they contain. For example, a ketose containing 3 carbon atoms is calledketotriose and an aldose containing 3 carbon atoms is called aldotriose.

Question 8 :

What are reducing sugars?

Answer 8 :

Reducing sugars are carbohydrates that reduce Fehling’ssolution and Tollen’s reagent. All monosaccharides and disaccharides, excludingsucrose, are reducing sugars.

Question 9 :

Write two main functions of carbohydrates in plants.

Answer 9 :

Two main functions of carbohydrates in plants are:

(i) Polysaccharidessuch as starch serve as storage molecules.

(ii) Cellulose,a polysaccharide, is used to build the cell wall.

Question 10 :

Classify the following into monosaccharides anddisaccharides. Ribose, 2-deoxyribose, maltose, galactose, fructose andlactose

Answer 10 :

Monosaccharides:

Ribose, 2-deoxyribose, galactose, fructose

Disaccharides:

Maltose, lactose 


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Chapter 14- Biomolecules Contributors

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