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The incorrect statement among the following is :
Option: 1  α-D-glucose and β-D-glucose are anomers.
Option: 2 α-D-glucose and β-D-glucose are enantiomers.
Option: 3 Cellulose is a straight chain polysaccharide made up of only β-D-glucose units.
Option: 4 The penta acetate of glucose does not react with hydroxyl amine.  
 

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Posted by

vishal kumar

Match the following:
Species Deficiency disease
(i) Riboflavin    (a) Beriberi
(ii) Thiamine  (b) Scurvy
(iii) Pyridoxine (c) Cheilosis
(iv) Ascorbic acid (d) Convulsions
 
Option: 1 (i)-(a),\: (ii)-(b),\: (iii)-(c),\: (iv)-(d)
Option: 2 (i)-(c),\: (ii)-(a),\: (iii)-(d),\: (iv)-(b)
Option: 3 (i)-(c),\: (ii)-(d),\: (iii)-(a),\: (iv)-(b)
Option: 4 (i)-(d),\: (ii)-(b),\: (iii)-(a),\: (iv)-(c)

Vitamin  A     -  Night blindness , Xeropthalmia

Vitamin  (Thiamine) B_{1}  -  Beriberi

Vitamin   (Riboflavin) B_{2}  - Cheilosis

Vitamin   (Niacin) B_{3}  -  Pellagra

Vitamin  (Pyridoxine)  B_{6}  - Convulsions , Anaemia

Vitamin  B_{12}  -  Pernicious anaemia

Vitamin C  (Ascorbic acid)  -  Scurvy

Vitamin   D - Rickets ( in chilldrens )

                       Osteomalacia ( in adults )

Vitamin   E - Increased RBCs fragility , muscular weakness          

Vitamin K - Poor blood clotting

-

Hence, the correct match is 

         (i) Riboflavin              -         (c) Cheilosis
        (ii) Thiamine              -          (a) Beriberi
        (iii) Pyridoxine           -          (d) Convulsions
        (iv) Ascorbic acid       -          (b) Scurvy

Therefore, Option(2) is correct.

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Posted by

Kuldeep Maurya

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Two monomers in maltose are :
Option: 1 \alpha -D-glucose\; and\; \alpha -D-glucose
Option: 2 \alpha -D-glucose\: \: and\: \: \alpha -D-galactose
Option: 3 \alpha -D-glucose \: \: and\: \: \beta -D-glucose
Option: 4 \alpha -D-glucose\: \: and\: \: \alpha -D-fructose

As we have learnt,

Maltose is a disaccharaide of \alpha- D- Glucose. Two units of \alpha- D- Glucose are connected by the C_1-C_4 linkage to form a molecule of Maltose.

Therefore, Option(1) is correct.

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Posted by

vishal kumar

Which of the following statements is correct?
Option: 1 Gluconic acid can form cyclic (acetal/hemiacetal) structure
Option: 2 Gluconic acid is a dicarboxylic acid
Option: 3 Gluconic acid is obtained by oxidation of glucose with HNO_{3}
Option: 4 Gluconic acid is a partial oxidation product of glucose

Gluconic acid is obtained by partial oxidation of glucose by mild oxidising agents like Tollen's reagent or Fehling solution or Bromine water.

Gluconic acid can not form cyclic hemiacetal or acetal.

It is to be noted that Glucose reacts with strong oxidising agents to form Saccharic acid.

Therefore, Option(4) is correct.

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Posted by

Ritika Jonwal

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Which one of the following statements is not true ?  
Option: 1 Lactose contains \alpha -\text{glycosidic} linkage between C_{1} of galactose and \text{C}_{1} of galactose and \text{C}_{4} of glucose.    
Option: 2 Lactose is a reducing sugar and it gives Fehling's test.
Option: 3 Lactose (\text{C}_{11}\text{H}_{22}\text{O}_{11}) is a disaccharide and it contains 8 hydroxyl groups.
Option: 4 On acid hydrolysis, lactose gives one molecule of D(+)-\text {glucose} and one moecule of D(+)-\text {galctose}.

 

Lactose contains \beta -\text{glycosidic} linkage not \alpha -\text{glycosidic} between C_{1} of galactose and \text{C}_{1} of galactose and \text{C}_{4} of glucose.

So, Statement 1 is incorrect but the rest statements are correct.

Therefore, the correct option is (1).

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Posted by

Kuldeep Maurya

The number of chiral carbon(s) present in peptide, \textup{Ile-Arg-Pro}, is ______.
 

We know 

Where R is given below-

Now, Structure of \textup{Ile-Arg-Pro} will be - 

The number of chiral carbon(s) present in peptide, \textup{Ile-Arg-Pro}, is 4.

Ans  = 4

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Posted by

Kuldeep Maurya

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Which of the following is not an essential amino acid ?
Option: 1 Tyrosine
Option: 2 Leucine
Option: 3 Valine
Option: 4 Lysine

Tyrosine is not an essential amino acid.

See the list below.

Therefore, the correct option is (1).

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Posted by

Kuldeep Maurya

The number of chiral carbons present in sucrose is _________ .
Option: 1 9
Option: 2 9
Option: 3 9
Option: 4 9
Option: 5 -
Option: 6 -
Option: 7 -
Option: 8 -
Option: 9 -
Option: 10 -
Option: 11 -
Option: 12 -
Option: 13 -
Option: 14 -
Option: 15 -
Option: 16 -

Thus, the total number of chiral carbons present in sucrose is 9.

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Posted by

Kuldeep Maurya

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What are the functional groups present in the structure of maltose?
Option: 1 One ketal and one hemiketal
Option: 2 Two acetals
Option: 3 One acetal and one hemiacetal
Option: 4 One acetal and one ketal

So, One acetal and one hemiacetal functional groups are present in the structure of maltose.

Therefore, the correct option is (3).

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Posted by

Kuldeep Maurya

Which of the following will react with \mathrm{CHCl}_{3}+\text { alc. } \mathrm{KOH} ?
Option: 1 Adenine and Proline
Option: 2 Thymine and Proline
Option: 3 Adenine and Lysine
Option: 4 Adenine and Thymine

\mathrm{CHCl}_{3}+\text { alc. } \mathrm{KOH} reacts with those compounds which have \mathrm{-NH_2} group.

Only Lysine and Adenine have \mathrm{-NH_2} group.

Therefore, the correct option is (3).

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Posted by

Kuldeep Maurya

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