CBSE Class 12 Chemistry Biomolecules Worksheet Set C

Question. In the viscose process the solvent for cellulose consists of:
a. Ether
b. Copper sulphate
c. Sodium hydroxide
d. carbon disulphide
Answer : C, D

Question. In DNA, the complementary bases are:
a. Uracil and adenine
b. Adenine and thymine
c. guanine and cytosine
d. thymine and cytosine
Answer : B, C

Question. Which of the following compounds do not belong to lipids?
a. Fats
b. Amino acids
c. Phospholipids
d. Carbohydrates
Answer : B, D

Question. RNA is different from DNA because RNA contains:
a. thymine
b. Ribose sugar
c. Deoxyribose sugar
d. uracil
Answer : B, D

Question. Fat consists of:
a. Monohydroxy carboxylic acid
b. Monohydroxy aliphatic acid
c. saturated carboxylic acid
d. Dihydroxy aliphatic carboxylic acid
Answer : B, C

Question. Sucrose molecule is made up of:
a. A gluco pyranose
b. a fructo furanose
c. a fructo pyranose
d. A gluco furanose
Answer : A, B

Question. If an aqueous solution of glucoseis allowed to freeze than crystal of which will be separated out first?
a. Glucose
b. Water
c. Both of these
d. None of these
Answer : B

Question. Insulin is:
a. An amino acid
b. Protein
c. A carbohydrate
d. A lipid
Answer : B

Question. A nucleoside on hydrolysis gives:
a. an orthophosphoric acid
b. a homocyclic base
c. An aldopentose
d. a heterocyclic base
Answer : C, D

Question. Amino acids usually exist in the form of Zwitter ions.
This means that it consists of:
a. The basic group –NH₂ and the acidic group–COOH
b. The basic group −NH₃⁺ and the acidic group –CO₂
c. The basic group −CO₂⁻ and the acidic group NH₃⁺
d. No acidic or basic group
Answer : C

Question. Which of the following monosaccharide is a pentose?
a. Galactose
b. Glucose
c. Fructose
d. Arabinose
Answer : D

Question. For α−amino acids having the structure
R − CH − CO₂H
       |
       NH₂
Which of the following statements are true?
A. Water solubility is maximum at a pH when concentrations of anions and cations are equal
B. They give ninhydrin test
C. On reacting with nitrous acid give off N₂
a. All
b. B
c. C
d. D
Answer : B, C

Question. The ‘iodine value’ of an oil indicates:
a. Its boiling point
b. Inflammability
c. Unsaturation present in acid contents
d. Solubility of salt in oils
Answer : C

Assertion and Reason

Note: Read the Assertion (A) and Reason (R) carefully to mark the correct option out of the options given below:
a. If both assertion and reason are true and the reason is the correct explanation of the assertion.
b. If both assertion and reason are true but reason is not the correct explanation of the assertion.
c. If assertion is true but reason is false.
d. If the assertion and reason both are false.
e. If assertion is false but reason is true.

Question. Assertion: Hydrolysis of sucrose is known as inversion of cane sugar.
Reason: Sucrose is a disaccharide.
Answer : B

Question. Assertion: Sucrose is a non-reducing sugar.
Reason: It has glycosidic linkage.
Answer : A

Question. Assertion: Glycosides are hydrolyzed in acidic conditions.
Reason: Glycosides are acetals.
Answer : D

Question. Assertion: Valine is an essential amino acid.
Reason: The lack of essential amino acids in the diet causes Kwashiorkor.
Answer : B

Question. Assertion: Sucrose is a disaccharide.
Reason: Sucrose is dextro rotatory.
Answer : B

Question. Assertion: DNA molecules and RNA molecules are found in the nucleus of a cell.
Reason: On heating the enzyme do not lose their specific activity.
Answer : D

Question. Assertion: α-amino acids exist as dipolar ions or zwitter ions.
Reason: α-amino acids are the building blocks of proteins
Answer : B

Question. Assertion: Activity of an enzyme is pH-dependent.
Reason: Change in pH affects the solubility of the enzyme in water.
Answer : B

Question. Assertion: The specific rotation of a freshly prepared solution of α-glucose decreases from +112º to +52.7º while that of β glucose increase from +19º to +57.7º.
Reason: The change in specific rotation of an optically active compound with time to an equilibrium value is called mutarotation.
Answer : B

Question. Assertion: ATP molecules are energy rich molecules.
Reason: ATP consists of a purine base adenine, pentose sugar ribose and a string of three phosphate groups.
Answer : B

Comprehension Based

Paragraph-I

There are three aldohexoses, A, B and C. Compounds A and B yield the same optically active alditol when they are reduced with hydrogen and catalyst; A and B yield different phenylosazones when treated with phenyl hydrazine; B and C gives the same phenylosazone but different alditols. A, B and C, all are D sugars.

Question. The aldohexose ‘A’ is –?

Answer : A

Question. The aldohexose ‘B’ is –?

Answer : B

Question. The aldohexose ‘C’ is –?

Answer : D

Paragraph – II

An ordinary D(+) glucose has melting point 146ºC and specific rotation, [α]²⁵_D = +112°. One another form of D(+) glucose has melting point 150ºC and specific rotation [α]²⁵_D] = +18.7°. The two forms have significantly different optical rotations but when on aqueous solution of either form is allowed to stand, its rotation changes. The specific rotation of one form decreases and rotation of other increases. This change in rotation towards an equilibrium value is called mutarotation.

Question. Mutarotation is a characteristic feature of:
a. epimers
b. enantiomers
c. anomers
d. ring-chain isomers
Answer : C

Question. Two significant forms which exhibit mutarotation are assigned α and β − forms. For mannose, the mutarotation can be shown is brief as follows

Which form of mannose exists predominantly in aqueous form?
a. α- form
b. β- form
c. open chain form
d. none of these
Answer : A

Question. What percentage of β−D(+) glucopyranose is found at equilibrium in the aqueous solution?
a. 50%
b. ≈ 100%
c. 36%
d. 64% Passage
Answer : D

Match the Column

Question. Match the following:

Column I (Pair of Molecules)	Column II (Characteristics)
(A) Glucose – Fructose	1. Anomers
(B) Fructose – Mannose	2. Diastereomers
(C) Glucose – Mannose	3. Functional isomers
(D) α-DGlucopyranose -β-Dglucopyranose	4. Lobry-De-Bruyn Alberada Van Ekenstein transformation

a. A→1, 5; B→3, 4; C→2, 4; D→1, 2
b. A→3, 2; B→3, 5; C→2, 4; D→1, 2
c. A→3, 5; B→3, 4; C→2, 4; D→1, 2
d. A→3, 5; B→3, 4; C→5, 4; D→1, 2
Answer : C

Question. Match the following:

Column I (Molecules)	Column II (Characteristics)
(A) CH3.CH(OH).CHO	1. Carbohydrate
(B) HOCH2–CH(OH)CHO	2. Amino acid
(C) CH3CH(NH2)COOH	3. Positive Tollen’s test
(D) PhCH(CH2NH2)COOH	4. Ninhydrin test

a. A→2; B→1, 3; C→2, 4; D→3
b. A→3; B→1, 3; C→2, 4; D→2
c. A→3; B→1, 3; C→2, 3; D→4
d. A→4; B→2, 3; C→1, 4; D→1
Answer : B

Question. Match the following:

Column I	Column II
(A) Starch	1. Natural polymer
(B) Nylon-6	2. Synthetic polymer
(C) Peptide Bond	3. Amide linkage
(D) Maltose	4. Glycoside linkage
	5. Oligosaccharide

a. A→2, 3; B→1, 3; C→4, D→1, 5
b. A→1, 5; B→2, 4; C→2, D→4, 5
c. A→1, 4; B→1, 3; C→5, D→3, 5
d. A→1, 4; B→2, 3; C→3, D→4, 5
Answer : D

Question. Match the Column-I with Column II:

Column I	Column II
(A) Fructose	1. Protein
(B) Zwitterion	2. Inversion
(C) Peptide linkage	3. α-mino acid
(D) Hydrolysis of cane sugar	4. Carbohydroate
	5. Ketose

a. A→1, 3; B→5; C→1, 4 D→2, 4
b. A→1, 5; B→3, 5; C→1, 2; D→2, 4
c. A→4, 5; B→3; C→1, D→2, 4
d. A→4, 5; B→3, 5; C→1, 4; D-2, 4
Answer : C

Integer

Question. Maltose contains how many oxygen atoms?
Answer : 11

Question. A nanopeptide contains …….. peptide linkages:
Answer : 8

Question. A compound of mol. wt. 180 is acetylated to give a compound of mol. wt. 390. The number of amino groups in the initial compound is?
Answer : 5

Question. pH in stomach is approximately:
Answer : 2

Question. The number of essential amino acids in man is:
Answer : 10

Question. In fructose, the possible optical isomers are:
Answer : 8

Question. Oxidation of glucose is one of the most important reactions in a living cell. What is the number of ATP molecules generated in cells from one molecule of glucose?
Answer : 38

 

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