Chem Biomolecules

R. K. MALIK’S NEWTON CLASSES

JEE (MAIN & ADV.), MEDICAL + BOARD, NDA, X & IX

Chapter 28

Biomolecules Solutions

SECTION - A

Objective Type Questions (One option is correct) 1.



50 C

H / H2O    Y      H N – CH COOH Compound X is X + NH3  2 2

(1) Chloroacetic acid

(2) Bromoacetic acid

(3) Both (1) & (2)

(4) Acetic acid

Sol. Answer (3) NH

 

Hydrolyses

3 X — CH2COOH     H2N — CH2 — CO O N H4       NH2 — CH2COOH

(x)

50C

X may be chloro or bromoacetic acid 2.

-helical structure of protein is stabilized by (1) Pe Peptide bond

(2) Di Dipeptide bond

(3) van der Waal’s forces

(4) Hydrogen bond

Sol. Answer (4) -helical structure of protein is stabilized by straight H-bonds between imide group (–NH–) of one amino acid and carbonyl group (–CO–) of fourth amino acid residue. 3.

When Wh en pro prote tein in is is subj subjec ecte ted d to de dena natu tura rati tion on (1)) It is hydrolyse (1 hydrolysed d to constitue constituent nt amino amino acids (2)) Elec (2 Electric tric field field has no influen influence ce on its migratio migration n (3)) Cons (3 Constitu tituent ent amino amino acids acids are separat separated ed (4)) It uncoils from an ordered and specific conformation (4 conformation into a more more random conformation conformation and precipitates precipitates from solution

Sol. Answer (4) Protein denaturation is disruption of stabilizing interchain bonds which destroy 3 dimensional form of proteins. The latter becomes non-functional.

Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1, Ph.: 0651-2562523, 9835508812, 8507613968

R. K. MALIK’S NEWTON CLASSES 82

4.


Bi o mo l e c ul e s

Solution of Assignment

Fructo Fru ctose se red reduce uces s Toll ollen’ en’s s reag reagent ent due to (1)) Pres (1 Presenc ence e of keto ketonic nic grou group p (2) Pr Pres esen ence ce of of NH4OH is Tollen’s reagent (3)) Rearrange (3 Rearrangement ment of fructose fructose into a mixture mixture of glucose, glucose, fructose fructose and mannose mannose (4) Bo Both th (2 (2)) & (3 (3))

Sol. Answer (4) When fructose is treated with dil. solution of an alkali, it undergoes reversible isomerization to form an equilibrium mixture of D-glucose, D-fructose and D-mannose. CHO

CHOH

CHO

CH2OH

1,2-enolisation H

C

OH

C

OH

HO

C

H

C

O

(CHOH)3

(CHOH)3

(CHOH)3

(CHOH)3

CH2OH

CH2 OH

CH2OH

CH2 OH

Enediol

D(+) Mannose

D(+) Gluco se

D(–) Fructose

isomerisation that It is due to isomerisation that fructose reduces Tollen’s reagent although it does not contain an–CHO group 5.

In alkaline alkaline soluti solution on D-gluco D-glucose, se, D-manno D-mannose se and D-fruct D-fructose ose are in equili equilibrium brium.. This reactio reaction n is known known as (1)) Fri (1 Fries es rea rearra rrange ngeme ment nt (2)) Lobry de Bruyn-van (2 Bruyn-van Enkel Enkelstein stein rearran rearrangeme gement nt (3)) Hof (3 Hofman mann n rear rearran rangem gement ent (4) Kol Kolbe be's 's rea react ction ion

Sol. Answer (2) 6.

D-glucose D-glucos e reacts reacts with with phenylhydra phenylhydrazine zine to to make make osazone. osazone. How many many molecules molecules of phenylhydra phenylhydrazine zine are are used used for this reaction per molecule of D-glucose? (1) One

(2) Two

(3) Three

(4) Four

Sol. Answer (3) CHO

CH = N.NHC N.NHC6H5 C6H5NHNH2 CHOH

CH = NNH NHC C6H5

(CHOH)3

(CHOH)3

(CHOH)3

CH 2OH

CH2OH

CH2OH

CHOH

C6H5NH NH2 –H2O

C=O

C6H5NHNH2

D – (+) – glucose CH = N. NH C6 H5 C = N. NHC NHC6H5 CH (OH)4 CH2 OH (Osazone)


R. K. MALIK’S NEWTON CLASSES Solution of Assignment

7.

JEE (MAIN & ADV.), MEDICAL + BOARD, NDA, X & IX Biomolecules

83

During mutarotation of -D-glucose in aqueous solution angle of optical rotation (1) Remains constant value of + 111° (2) Remains constant value of + 19.2° (3) Changes from an angle of +112° to a constant value of +52.5° (4) Changes from an angle of +19.2° to a constant value of +52.5°

Sol. Answer (4)

 – D(+ ) – glucose)

equilibrium mixture

 (D) glucose

+52.5°

+19.20

+III°

This spontaneous change in specific rotation of an optically active compound with time, to an equilibrium value is called mutarotation. 8.

Which of the following is correct about cellulose? (1) It is an important food material (2) It has only -glucosidic linkage between two D-glucose unit (3) It has only -glucosidic linkage between two D-glucose unit (4) It is sweet in taste

Sol. Answer (3) -1, 4-glycosidic linkage of cellulose CH2OH H O

9.

CH2OH O

H

H OH

H

H

OH

O H

O

H OH

H

H

OH

O H

Which of the following is incorrect about isoelectric point of amino acid? (1) At this point amino acid is present in the form of zwitter ion (2) At this point amino acid is electrically neutral (3) If pH > isoelectric point amino acid will move toward anode (4) If pH > isoelectric point amino acid will move towards cathode

Sol. Answer (4) The pH at which there is no net migration of the amino acid under the influence of an electric field is called isoelectric point. At this pH amino acid exist in the form of zwittor ion R | H N — CH — COOH



+

H (at lower pH)  (moves towards cathode) 3

R | + H N — CH — COO Zwittor ion

–

3

–

OH

(at higher pH)

R | H N — CH — COO

–

2



(moves towards anode)

10. If a native protein is subjected to physical or chemical treatment which may disrupt its higher structure without affecting primary structure then this process is called (1) Inversion of protein

(2) Denaturation of protein (3) Renaturation of protein (4) Fermentation




Biomolecules


Sol. Answer (2) It is disruption of stabilising inter chain bonds which destroys 3 dimensional form of proteins. The latter becomes non-functional. 11.

Ring structure of glucose is due to formation of hemiacetal and ring formation between (1) C1 and C5

(2) C1 and C4

(3) C1 and C3

(4) C2 and C4

Sol. Answer (1) H — — O C — | H — C — OH | HO — C — H | H — C — OH | H—C—OH



CH OH

H — OH C — | H — C — OH | HO — C — H | H — C — OH | H—C | CH OH

CH OH 6 2

H

 O

O

5 H 4

OH

H 1

OH 3

H 2

H

OH

OH

2

2

12. Acid hydrolysis of sucrose causes (1) Esterification

(2) Saponification

(3) Inversion

(4) Rosenmund reduction

Sol. Answer (3) HCl

C12H22O11  H2O     C6H12O6  C6H12O6 D  glu cos e [  ]D  52.5

[  ]D  66.5

D  fructose [  ]D  92.4

Hydrolysis of sucrose to an equimolar mixture of D (+) glucose and D (–) fructose is accompanied by a change in the sign of optical rotation from dextro rotatory to laevorotatory , the overall process is called inversion of sugar 13. Which of the following gives an optically inactive aldaric acid on oxidation with dilute nitric acid? CHO

(1)

CHO

H

OH

H

OH H

HO

OH

H

(2)

HO

H

HO

H

CH2OH

CHO

CHO

CH2OH

H

(3)

HO H

HO

OH H OH CH2OH

(4)

H

H

OH

H

OH CH2OH

Sol. Answer (3) CHO H HO H

COOH

OH H OH CH OH 2

H dil.HNO3

HO H

OH H OH COOH Optically inactive





85

Biomolecules

14. (+) Arabinose is (2R, 3S, 4S)-aldopentose which of the following is (+) – arabinose? CHO

CHO

(1)

H

OH

H

OH

(2)

H

HO

OH

H HO

H

HO

H

(3)

HO

OH

H HO

H

(4)

OH

H

CH2OH

CH2OH

CHO

CHO

H

H

OH

H

OH CH2OH

CH2OH

Sol. Answer (2) CHO H HO HO

R S S

OH H H

CH2OH

15. Peptides are composed of amino acids joined by amide bonds. which of the following statements is not correct? (1) Amide group are more resistant to hydrolysis than one similar ester groups (2) p- resonance stabilizes the amide bond (3) Stable conformations of peptides are restricted to those having planar amide groups (4) Amide groups do not participate in hydrogen bonding interaction Sol. Answer (4) Amide groups participate in H-bonding 16. A t ripe ptide is compose d e quall y o f L -vali ne, L-tyrosine and L-alanine (one molecule of each). How many isomeric tripeptides of this kind may exist? (1) 3

(2) 4

(3) 6

(4) 8

Sol. Answer (3) Six tripeptides are possible from three different amino acids 17. Which of the following is an essential amino acid? (1) Asparagine

(2) Glutamine

(3) Histidine

(4) Alanine

(3) Serine

(4) Glycine

(3) Carbohydrates

(4) Nucleic acids

(3) Aspartic acid

(4) Glycine

Sol. Answer (3) 18. Which of the following amino acid is optically inactive? (1) Lysine

(2) Glutamine

Sol. Answer (4) 19. Glycosidic linkage is present in (1) Proteins

(2) Lipids

Sol. Answer (3) 20. Which of the following is basic amino acid? (1) Lysine

(2) Valine

Sol. Answer (1)




Biomolecules


21. Which of the following disaccharide is called invert sugar? (1) Lactose

(2) Maltose

(3) Sucrose

(4) All of these

Sol. Answer (3) 22. Tertiary structures of proteins is given by (1) -helix

(2) Folding of secondary structure

(3) -pleated sheets

(4) All of these

Sol. Answer (2) 23. The helical structure of proteins is stabilised by (1) Peptide bonds

(2) Hydrogen bonds

(3) Dipeptide bonds

(4) van der Waals’ forces

Sol. Answer (2) 24. Glucose when reacted with acetic anhydride forms (1) Diacetate

(2) Hexa-acetate

(3) Pentacetate

(4) Tetra-acetate

Sol. Answer (3) 25. Complementary bases present in DNA are (1) Uracil & Adenine : Cytosine & Guanine

(2) Thymine & Adenine : Guanine & Cytosine

(3) Adenine & Thymine : Guanine & Uracil

(4) Adenine & Guanine : Thymine & Cytosine

Sol. Answer (2) 26. Choose the correct statement about isoelectric point (1) If pH > isoelectric point amino acid will move towards cathode (2) At this point amino acid is present in the form of Zwitter ion (3) If pH < isoelectric point, amino acid will move towards anode (4) At this point amino acids are basic in nature Sol. Answer (2) 27. Mutarotation in aqueous solution is shown by (1) Glycogen

(2) Sucrose

(3) Cellulose

(4) Maltose

(3) Amylose

(4) Nylon

Sol. Answer (4) 28. Branched chain structure is shown by (1) Amylopectin

(2) Cellulose

Sol. Answer (1) 29. The number of amino acids required to form a tripeptide bond are (1) Seven

(2) Two

(3) Six

(4) Four

Sol. Answer (4) 30. Glucose does not give (1) Schiff’s test

(2) Hydrogensulphite addition product with NaHSO3

(3) 2, 4 DNP test

(4) All of these

Sol. Answer (4)





Biomolecules

87

31. The two cyclic hemiacetal forms of glucose differing only in the configuration of the hydroxyl group at C-1 are called (1) Anomers

(2) Enantiomers

(3) Epimers

(4) Metamers

(3) n-hexane

(4) Gluconic

Sol. Answer (1) 32. Glucose reacts with HCN to give (1) Saccharic acid

(2) Cyanohydrin

Sol. Answer (2) 33. Cellulose is soluble in (1) Water

(2) Organic solvents

(3) Ammonical cupric hydroxide solution

(4) All of these

Sol. Answer (3) 34. Glucose does not react with (1) NH2OH

(2) C6H5NHNH2

(3) NaHSO3

(4) HCN

(3) Tertiary structure

(4) Quaternary structure

(3) Glucose + Glucose

(4) Glucose + Lactose

Sol. Answer (3) 35. Which structure of proteins involve in denaturation? (1) Primary structure

(2) Secondary structure

Sol. Answer (4) 36. Cane sugar on hydrolysis gives (1) Fructose + Glucose

(2) Fructose + Fructose

Sol. Answer (1) 37. Which is correct statement? (1) Starch is a polymer of -glucose (2) Amylose is a component of cellulose (3) Proteins are composed of only one type of amino acids (4) In cyclic structure of pyranose, there are five carbons and one oxygen atom Sol. Answer (1) 38. Enzymes in the living systems (1) Provide energy

(2) Provide immunity

(3) Transport oxygen

(4) Catalyze biological reactions

Sol. Answer (4) 39. Which of the following is not an amino acid? (1) Glycine

(2) Alanine

(3) Histidine

(4) Benzidine

Sol. Answer (4)





Biomolecules

43. The following carbohydrate is H

89

[IIT-JEE 2011]

OH H O

HO

OH

HO

H HO

H

H

(1) A ketohexose

(2)

An aldohexose

(3) An -furanose

(4) An -pyranose

Sol. Answer (2) OH O HO HO

OH

HO

 — D — Glucopyranose, which is cyclic form of an aldohexose SECTION - B

Objective Type Questions (More than one options are correct) 1.

When D-Glucose reacts with three moles of phenylhydrazine it forms glucosazone. Which of the following carbohydrates will give same osazone as that of D-Glucose? CH2OH O

(1)

HO

H

(2)

HO

H

HO

H

H

OH

H

OH

H

OH

H

OH

CH2OH

CHO

CHO

CHO

(3)

H

OH

H

OH

HO

(4)

H

H

OH

HO

H

H

OH

HO

H

H

OH

CH2OH

CHO

CH2OH

Sol. Answer (1, 2) During osazone formation, the reaction occurs only at C 1 and C2 while rest of the molecule remains intact since glucose and fructose differ from each other only in the arrangement of atoms at C 1 and C2, therefore they give same osazone. 2.

Which of the following compounds contain amide linkage? (1) Nylon - 6

(2) Acetamide

(3) Proteins

(4) Cellulose

Sol. Answer (1, 2, 3)

O || — C — NH — linkage is called amide linkage. O Nylon – 6 : — NH — (CH ) — C — 2 5

n

O

Acetamide :

— NH 2 — C — N H2 — n

O || H N — CH — C — NH — CH2 — COOH Proteins  peptide  2 | CH3 Alany lg lycine



3.


Biomolecules


What would be the correct structure of -D-Glucose? CH2OH

CH2OH

O

H H

(1)

OH

H

HO H

H

H

(2)

H

O

H

OH

OH

H

H

OH

OH

HO

OH

H

H OH

OH

HO

HO O

(3) HO

O

(4)

H H

H

HO OH

H

HO

H

HO

H H

HO

Sol. Answer (1, 3) St-I is Howarth project ion while. St-III is chair form of -D-Glucose 4.

Which of the following sugars will form osazone?

H

O

OH

O

H

O

(1)

(2)

CH2OH O H OH

(3)

H

HO H

OH

O

H

H

H

CH2OH H

OH

OH

H

H

CH2OH

(4) O

O

H

H CH2OH

OH

H

H

OH

HO

OH

OMe H

Sol. Answer (1, 2) Sugars having free hemiacetal group will form osazone. 5.

Which of the following amino acids will have +2 net charge at pH = 1? CH3 O

(1)

H2N — C — C — OH

HO O

(2)

H2N — C — C — OH

H N —H N

H (3) H2N — C — COOH (CH2— )3 COOH

H (4) H2N — C — COOH (CH2— )2 NH 2





Biomolecules

91

Sol. Answer (2, 4) In acidic medium N atom of –NH 2 group absorb H + and becomes +vely charged. +2 net charge will be present on those amino acids which are having two basic amino groups. 6.

Which of the following aldoses will give achiral product with NaBH4? CHO CHO

CHO H H

(1)

OH

HO

H

HO

H

(2)

CHO

OH

HO

H

HO

H

H

H

(3)

OH

CH2OH

H

OH

HO

(4)

H

H

OH

CH2OH

OH

HO

H

H

OH

H

OH CH2OH

CH2OH

Sol. Answer (2, 3) CHO

2

*

H (1)

CH OH OH

HO

H

HO

H

*

H NaBH

4

HO

H

HO

H

CH OH

CH OH (Chiral molecule)

2

2

CHO H

*

CH OH 2

OH

HO (2)

NaBH

H

HO

4

H

H

OH

H

OH

HO

H

HO

H

H

OH

C H OH

CH OH (Achiral)

2

2

CHO H (3)

*

CH OH 2

OH

HO

H

H

OH

NaBH

4

H

OH

HO

H

H

OH

CH OH

CH OH (Achiral)

2

2

CH O

(4)

CH 2OH

H

OH

HO

H

H H

*

H NaBH4

OH

HO

H

OH

H

OH

OH

H

OH

CH2OH

7.

OH

CH 2OH

Which of the following upon hydrolysis form only glucose? (1) Lactose

(2) Cellobiose

(3) Maltose

(4) Sucrose



Biomolecules

JEE (MAIN & ADV.), MEDICAL + BOARD, NDA, X & IX Solution of Assignment

Sol. Answer (2, 3) C12H22O11

H

 H2O     C6H12O6  C6H12O6

sucrose

glucose

fructose

H

   C6H12O6 C12H22O11  H2O  Maltose

 C6H12O6

glucose

glucose

H

   C6H12O6 C12H22O11  H2O  Lactose

 C6H12O6

glucose

galactose

dil.H2SO4

C12H22O11  H2 O   D  glucose  D  glucose under presure Cellulose

8.

Denaturation of protein can be brought about by (1) Changing concentration

(2) Changing pH

(3) Changing Temperature

(4) Changing amino acid sequence

Sol. Answer (2, 3) Denaturation of proteins can be brought about by agents like soap, detergents, acid, alcohol, heat and some disinfectants. 9.

Choose correct statements about proteins (1) Primary structure of proteins refer to amino acid sequence (2) Interaction between three polypeptide chains give rise to tertiary structure of proteins (3) Association between four polypeptide chains give quaternary structure of proteins (4) Folding of polypeptide chain due to interaction between carbonyl group and

— N — H group

of peptide

linkage give rise to secondry structure Sol. Answer (1, 4) Primary or 1° structure of Protein R

1

O

H

O

R

3

O

H N— C — C — N — C — C — N — C — C H R

H

1

R

H

2

Peptide bond

H

Peptide bond

Secondary Structure R1 H2N

C C

H H-Bond

H

C

N H

O

C

R4 C H H N

R3

O NH C O C

C H

H N

H C O

NH

O R7

H-Bond

HC–R 2

C R6

HC–R 5 C O N H

H C N O

H C R8





Biomolecules

93

10. Which of the following is/are stabilizing interaction in protein folding? (1) Hydrogen bond formation

(2) Peptide bonds

(3) Disulphide bonds

(4) Hydrophobic interaction

Sol. Answer (1, 3, 4) 11.

The correct statement(s) about the following sugars X and Y is(are)

[IIT-JEE-2009]

CH2OH O H

H

H

HO

O

HOH2C

H

O

OH

H

H

OH

H OH

HO

CH2OH

H

X

CH2OH O H

CH2OH O H

O

H

HO

OH

H

H

OH

H

H H OH

H HO

OH

H

Y

(1) X is a reducing sugar and Y is a non-reducing sugar (2) X is a non-reducing sugar and Y is a reducing sugar (3) The glucosidic linkages in X and Y are  and , respectively (4) The glucosidic linkages in X and Y are  and , respectively Sol. Answer (2, 3) No hemiacetal linkage is free in X. Therefore it is Non reducing. While Y having hemiacetal linkage would be reducing sugar.

H HO

CH2OH O H OH

H

H

OH

O

H HOH2 C O H

OH

HO H  -Glucosidic linkage

H CH2OH

H

HOH 2C O H HO

H

O H

CH2OH O H H

OH

HO

H

H OH

 -Glucosidic linkage

12. The structure of D-(+)-glucose is CHO H HO

OH H

H

OH

H

OH CH2OH




Biomolecules


The structure of L-(–)-glucose is CHO HO

(1)

CHO

H

H

H HO

OH

HO

H

HO

H

[JEE(Advanced)-2015]

(2)

H

CH2OH

OH

HO

H

HO

H

H

HO

H

HO

H

HO

H

OH

HO

CHO

CHO

(3)

H

H

OH

HO

CH2OH

(4)

H

H

CH2OH

OH CH2OH

Sol. Answer (1) CHO H

CHO

OH

HO

HO

H

H

H

OH

H

OH

HO

H

H

OH

HO

H

CH2OH

Mirror

D-(+)-glucose

CH2OH L-(–)-glucose SECTION - C

Linked Comprehension Type Questions Comprehension-I Carboxylic acids containing an amino group (—NH 2) as a substituent are called amino acids. When this amino group substitution is at -position with respect to carboxylic group, the aminoacid is called -amino acid. The aminoacids with two carboxylic acids are called acidic amino acids whereas aminoacids with two amino group are called basic amino acids. Aminoacids show amphoteric character. All chiral natural amino acids exist in Lform. 1.

Each of the following represent natural alanine (an amino acid) except

COOH

(1)

H2N

H

CH3

(2)

HOOC

H

(3)

COOH

COOH

C

C CH3

H2N

(4)

H CH3

NH2

H3C H

NH2

Sol. Answer (4) 2.

Which of the following is not an -amino acid? COOH COOH

O

(1)

C — OH

(2)

H

O OH

N

COOH

(3)

H3C

C

H

(4)

H

C

NH2 H

CH2NH2

NH2

Sol. Answer (3) NH2 group is not attached to  carbon atom


H




3.

Biomolecules

95

What would be the net charge on the given amino acid at pH = 14?

COOH

O

H2N — C — ( CH2)— C — OH 4 H (1) + 1

(2) + 2

(3) – 1

(4) – 2

Sol. Answer (4) pH = 14, (Basic medium) the structure of amino acid will be COO O | || H2N — C — (CH2 )4 — C — O | H Comprehension-II When either form of D-Glucose is dissolved in water, the solution gradually changes its optical rotation and finally attains a constant optical rotation of +52º. CHO

CH2OH O

H OH

H

HO

OH

H HO

OH H

H

OH

H

OH

H

H OH (+ 19º)

1.

H HO

CH2OH

CH2OH O H OH

H

H

OH

H OH

(+ 112º)

Which form of D-Glucose is more stable? (1) -D-Glucose

(2) -D-Glucose

(3) Open chain

(4) All forms are equally stable

Sol. Answer (2) The  and  anomers of D-glucose interconvert in aqueous solution by a process called mutarotation. Thus a solution of -D glucose and a solution of -D glucose eventually form identical equilibrium mixtures having 1

identical optical properties. This mixture contains

3

2

rd of -D glucose and

3

rd of -D glucose and few linear

and five membered rings also. 2.

What is the percentage of open chain D-Glucose in solution? (1) 36%

(2) 64%

(3) 0.5%

(4) 33.3%

Sol. Answer (3) 3.

Which of the following statements is true? (1) Because of the presence of alcoholic group in D-Glucose it exhibits mutarotation (2) Because of the presence of C == O group in Glucose it exhibits mutarotation (3) All sugars exhibit mutarotation (4) Because of free hemiacetal linkage in cyclic form, D-Glucose exhibit mutarotation




Biomolecules


Sol. Answer (4) The hemiacetal structure can be easily hydrolysed in the aqueous solution to give the open chain form of glucose. Therefore in aqueous solution either  or  anomer is converted through the open chain form into an equilibrium mixture consisting of both the anomers with a very small amount of the open chain form. Comprehension-III Aldehydes and ketones are converted to acetals by treatment with an alcohol and a trace of acid catalyst. These conditions also convert aldoses and ketoses to the acetals, call glycosides. In glycosides, an aglycone is the group bonded to the anomeric carbon atom. Example : Methanol is the aglycone in a methyl glycoside. 1.

In ethyl -D glucopyranoside, aglycone part is (1) CH3CH2—

(2) CH3—

(3) CH3CH2O—

(4) CH3OH

Sol. Answer (3)

CH2OH O

H H OH

2.

H CH3 OH, H

H

OH

(A)

OH H

OH

A is (1) Methyl -D-glucopyranoside, -glycosidic bond (2) Methyl -D-glucopyranoside, -glycosidic bond (3) Methyl--D-glucopyranoside + methyl -D glucopyranoside (4) No reaction can take place Sol. Answer (3) 3.

In sucrose, two menosaccharides unit linked by CH2OH H

O

H H OH

H

H

OH

OH

HOCH2 O

H

O HO

H OH

H

CH2OH

(1) Only -glycosidic linkage is present (2) Only -glycosidic linkage is present (3) -glycosidic linkage on glucose and -glycosidic linkage on fructose (4) -glycosidic linkage on fructose and -glycosidic linkage on glucose Sol. Answer (2)




97

SECTION - D

Assertion-Reason Type Questio 1.

STATEMENT-1 : Primary structure of proteins is determined by amino acid sequence. and

STATEMENT-2 : Intramolecular hydrogen bonding between carbonyl group and

N

H group

is responsible

for the folding of polypeptide chain. Sol. Answer (2) The respective amino acid residues are linked by peptide bonds only. 2.

STATEMENT-1 : Glucose and fructose can be differentiated by Fehling's solution. and STATEMENT-2 : Glucose is an aldose while fructose is a ketose (having keto functional group).

Sol. Answer (4) Glucose and fructose can not be differentiated by Fehling solution because in alkaline medium an equilibrium mixture of D-glucose, D-fructose, and D-mannose. 3.

STATEMENT-1 : D-Glucose and D-Mannose are C-2 epimers and STATEMENT-2 : They only have different configuration at carbon number-2.

Sol. Answer (1) Diastereomers which differ in configuration only at one C atoms 4.

STATEMENT-1 : Enzymes are protein but protein are not enzymes. and STATEMENT-2 : Enzymes are bio-catalyst and posses a stable configuration having a active site poket.

Sol. Answer (2) Fact. 5.

STATEMENT-1 : Cellulose is a polymer of glucose. and STATEMENT-2 : Reducing sugars undergo mutarotation.

Sol. Answer (2) 6.

STATEMENT-1 : Glucose gives a reddish-brown precipitate with Fehling’s solution. and STATEMENT-2 : Reaction of glucose with Fehling’s solution gives CuO and gluconic acid.

[IIT-JEE-2007]

Sol. Answer (3)



Biomolecules

JEE (MAIN & ADV.), MEDICAL + BOARD, NDA, X & IX Solution of Assignment

SECTION - E

Matrix-Match Type Questions 1.

Match the following Column-I

Column-II

(A) Glucose

(p)

Undergoes hydrolysis

(B) Fructose

(q)

React with Fehling solution

(C) Sucrose

(r)

React with Tollen’s reagent

(D) Maltose

(s)

Glycosidic linkage

Sol. Answer: A(q, r), B(q, r), C(p, q, r, s), D(p, q, r, s) (A) Glucose  can react with Fehling solution and Tollen’s reagent due to presence of –CHO group. (B) Fructose  can react with Fehling solution and Tollen’s reagent due to presence of –CHO group. (C) Sucrose  undergoes hydrolysis to give glucose and fructose.  Non reducing sugar so can not reduce Fehlng solution and Tollen’s reagent.  Glycosidic linkage is present (D) Maltose 

 undergoes hydrolysis to give glucose only  Due to free carbonyl group it is reducing hence reacts with Tollen’s reagent and Fehling solution  glycosidic linkage is present.

2.

Match column-I with column-II Column-I

Column-II

(A) Maltose

(p)

Polymer of D-Glucose

(B) Cellulose

(q)

Non-reducing sugar

(C) Amylose

(r)

Disaccharide

(D) Invertose

(s)

Reducing sugar

Sol. Answer: A(r, s), B(p, q), C(p, q), D(r, s) (A) Maltose is formed by combination of 2 molecules of glucose. It is a reducing sugar because it possesses a free aldose group. (B) Cellulose is formed by formation of 1, 4 glycosidic linkage between 6000 glucose units, it is a nonreducing sugar as it does not reduce Tollen’s regaent or Fehling’s solution. (C) Amylose is formed by formation of , 1, 4 glycosidic linkage between 200-1000 glucose units, Amylose also cannot reduce Tollen’s reagent or Fehling’s solution. (D) Invertose is formed by combination of glucose and fructose. 3.

Match the following Column I

Column II

(A) Sucrose

(p)

Diasaccharide

(B) Cellobiose

(q)

Reducing sugar

(C) Maltose

(r)

-Glycosidic linkage

(D) Fructose

(s)

-Glycosidic linkage

(t)

Shows mutarotation

Sol. Answer: A(p, r, s), B(p, q, s, t), C(p, q, r, t), D(q, t) Fact



4.


99


Column-II

(A) Glucose

(p)

React with Tollen’s reagent

(B) Fructose

(q)

React with Fehling reagent

(C) Sucrose

(r)

React with phenyl hydrazine to form osazone

(D) Maltose

(s)

It is a disachharide

(t)

It is a furanose sugar

Sol. Answer: A(p, q, r), B(p, q, r, t), C(r, s), D(p, q, r, s) 5.


Column-II

(Pair of molecules)

(Characteristic)

(A) Glucose-Fructose

(p)

Anomer

(B) Fructose-Mannose

(q)

Diastereomer

(C) Glucose-Mannose

(r)

Functional isomer

(D) -D-Glycopyronose -D-Glucopyronose

(s)

Lobry-De-Bryun Van-Ekenstein rearrangement

(t)

Both are reducing sugars

Sol. Answer: A(q, r, s, t), B(q, r, s, t), C(q, r, s, t), D(p, q, s, t) 6.


Column-II

(A) Glycine

(p)

Optically inactive

(B) Alanine

(q)

Optically active

(C) Lysine

(r)

Bear two —COOH

(D) Glutamic acid

(s)

Bear two —NH2

(t)

Cationic in highly acidic medium

Sol. Answer: A(p, t), B(q, t), C(q, s, t), D(q, r, t) SECTION - F

Integer Answer Type Questions 1.

What is the total number of acidic amino acids found in human proteins?

Sol. Answer (2) Aspartic acid and glutamic acids are acidic amino acids. 2.

Net charge available on a basic amino acid at pH = 1 would be ____.

Sol. Answer (2) Basic amino acids have two amino groups and hence net charge at pH = 1 would be = +1 3.

How many tripeptides are possible when Glycine, Alanine and Phenylalanine are allowed to form peptide bonds?

Sol. Answer (6) Fact




100 Biomolecules


O 4.

The iso-electric point of the given amino acid is, HO – C – CH – CH – CH – COOH (pK = 2) (pK = 4) + NH (pK = 9) 2

2

a

a

3

a

Sol. Answer (3) 5.

The number of chiral carbon in one molecule of -D glucose is ________.

Sol. Answer (5) 6.

The total number of basic groups in the following form of lysine is H3N

CH2

CH2

CH2

CH2

[IIT-JEE-2010]

O CH

C

H2N

O

Sol. Answer (2) H3N

CH2

CH2

CH2

CH 2

O CH

Basic site (Carboxylate ion will also behave as proton acceptor)

C

H2N

O

Basic site 7.

A decapeptide (Mol. Wt. 796) on complete hydrolysis gives glycine (Mol. Wt. 75), alanine and phenylalanine. Glycine contributes 47.0% to the total weight of the hydrolysed products. The number of glycine units present [IIT-JEE-2011] in the decapeptide is

Sol. Answer (6) A decapeptide will have (9) peptide linkage Mass of hydrolyzed product is (796 + 162) gm/mole Number of glycine molecule = 8.

0.4 7  958 75

6

The substituents R1 and R2 for nine peptides are listed in the table given below. How many of these peptides are [IIT-JEE-2012] positively charged at pH = 7.0? H3N

CH

CO

H

NH

CH

CO

R1

NH

CH

CO

NH

CH

COO

H

R2

Peptide

R1

R2

I

H

H

II

H

CH3

III

CH2 COOH

H

IV

CH2CONH2

(CH 2)4NH2

V

CH2CONH2

CH 2CONH 2

VI

(CH2)4 NH2

(CH 2)4NH2

VII

CH2 COOH

CH 2CONH 2

VIII

CH2OH

(CH 2)4NH2

IX

(CH2)4 NH2

CH 3



JEE (MAIN & ADV.), MEDICAL + BOARD, NDA, X & IX Biomolecules 101


Sol. Answer (4) When any group in R 1 and R2 is basic group then amino acid is positively charged at pH = 7.0. So, answers are peptide IV, VI, VIII and IX. 9.

When the following aldohexose exists in its D-configuration, the total number of stereoisomers in its pyranose [IIT-JEE-2012] form is CHO CH2 CHOH CHOH CHOH CH2OH

Sol. Answer (8) Given CHO CH2 CHOH CHOH CHOH

has D configuration

CH2OH

CH2OH O HO

H *

HO

*

OH C

*

are chiral carbon atoms. Hence total stereoisomers are 8.

H

*

H

10. The total number of distinct naturally occurring amino acids obtained by complete acidic hydrolysis of the [JEE(Advanced)-2014] peptide shown below is O O

O

N N

H N

CH2

H

O

N O

H N

H

H

O

N O

H

N N

CH2

O

O

Sol. Answer (1) On hydrolysis only glycine is formed as natural amino acid.




102 Biomolecules


SECTION - G

Multiple True-False Type Questions 1.

STATEMENT-1 : Glycine is an achiral amino acid. STATEMENT-2 : pH of alanine is less than 7. STATEMENT-3 : Almost all proteins are polypeptides. (1) T T T

(2)

FFF

(3)

TFT

(4)

F TF

(3)

F TF

(4)

F TT

(4)

FF F

(4)

TTF

Sol. Answer (1) Fact 2.

STATEMENT-1 : Sucrose is a reducing sugar. STATEMENT-2 : Sucrose has two glycosidic linkages. STATEMENT-3 : Sucrose shows mutarotation. (1) T T F

(2)

TFT

Sol. Answer (3) Sucrose is nonreducing sugar and it doesn’t show mutarotation. 3.

STATEMENT-1 : -D-glucose shows mutarotation. STATEMENT-2 : Sucrose on hydrolysis produces -D-glucose. STATEMENT-3 : Mannose is an aldohexose. (1) T F T

(2)

F TF

(3)

T TT

Sol. Answer (1) 4.

STATEMENT-1 : Essential amino acids are produced by body. STATEMENT-2 : Non-essential amino acids are not produced by body. STATEMENT-3 : All carbohydrates must have chiral carbon. (1) T T T

(2)

F FT

(3)

F FF

Sol. Answer (3) 5.

STATEMENT-1 : Biologically active form of protein is quaternary structure. STATEMENT-2 : -sheets are secondary structure of protein. STATEMENT-3 : Basic sequence of amino acids is defined by primary structure of protein. (1) T T T

(2)

F FF

(3)

FTT

(4)

FFT

Sol. Answer (1) SECTION - H

Aakash Challengers Questions 1.

In a paper electrophoresis amino acids and peptides can be separated by their differential migration in an electric field. To the center of a strip of paper, wet with buffer at pH = 6 is applied a mixture of the following three peptides in a single small spot : Gly-Ala, Gly-Asp and Gly-Arg. A positively charged electrode (anode) is attached to the left side of the paper and a negatively charged electrode (cathode) to the rightside. A voltage is applied across the ends of the paper for a time, after which the peptides have separated into three spots. One near the location of the original spot which peptide is in each spot? Explain.

Sol. Gly-Asp will move towards Anode while Gly-Arg will move towards cathode and Gly-Ald will remain hear spot.





2.

Starch consist of amylose and amylopectin structure of amylopectin is given as below H

CH2OH

O

H

O

H

H

HO

H

OH H

CH2OH

O

H

O

H

H

HO

OH H

O

H

CH2

O H

O

H

H

HO

OH H

O

In structure of amylopectin (1) -1,4-glycosidic bonds

(2)

-1,6-glycosidic bonds

(3) -1,4-glycosidic bonds

(4)

-1,4-glycosidic bonds and -1,6-glycosidic bonds

Sol. Answer (4) In amylo pectin, -1,4 and -1,6 linkage that provides the attachment point for another chain. 3.

Which of the strucutres 1 through 4 is methyl -D-galactopyranoside? CHO OH

H HO

H

HO

H

H

OH CH2OH (D-galactose)

CH2OH HO

(1)

CH2OH

O OCH3

HO

OH H

O

HO

(2)

H

HO

OH OCH3

HO

HO CH2OH

CH2OH

O OCH3

(3) HO

OH H

(4)

O H

HO

OH OCH3

Sol. Answer (4) OH CH OH 2 HO

H OH

OCH3

-D-Galctopyranoside




104 Biomolecules

4.


Which of the following statements would correctly describe the isoelectric point of cysteic acid an oxidation product of cysteine? H

C H3N

COOH CH2

SO3 Cysteic acid

(1) Lower than that of aspartic acid

(2)

About the same as that of aspartic acid

(3) About the same as that of Cysteine

(4)

Higher than that of lysine

Sol. Answer (1) 5.

Which two of the following compounds are reduced to the chiral alditol by NaBH 4? CHO

CHO

CHO

H

OH

H

H

OH

HO

H

HO

H

HO

H

H

HO

OH

CHO OH

H

HO

H

H

H

OH

OH

H

OH

CH2OH

CH2OH

CH2OH

(A)

(B)

(C)

CH2OH (D)

(1) A & B

(2)

B&C

(3) C & D

(4)

A&C

Sol. Answer (2) 6.

The active site of a biomolecule R is shown below

A biomolecule R

The biomolecule R binds to a tripeptide because its binding site is complementry to that of tripeptide. The shape of the tripeptide would be Note : Complimentarity in the figure is shown by shape

(1)

(2)

(3)

(4)





Sol. Answer (4) Tripeptide will exactly fit into the binding site due to exact complimentarity of shape. 7.

A mixture of three amino acids X-(pH = 3.2), Y-(pH = 5.7) and Z-(pH = 9.7) under electrophoresis at pH = 7.7, in which direction will each component of the mixture move? (1) X to anode, Y and Z to cathode (2) X to anode Y stationary, Z to cathode (3) X to cathode, Y stationary, Z to anode (4) X and Y to anode, Z to cathode

Sol. Answer (3) 8.

Convert D-erythrose to next higher aldose. CN

CH O

Sol.

H

OH

H

OH CH 2OH

HO aq. HCN

COOH H

(i) Ba (OH)2

H

OH

H

OH

( i) H2S O4

H

OH

H

OH

CH 2OH

D-erythrose

H

HO

CH 2OH



CHO HO

CO H

H

OH

H

OH

HO Na/Hg H2SO4

CH 2OH

H

H

O

OH

H CH 2OH

D-arabinose

9.

Glucose, Mannose and fructose give identical osazones. Explain.

Sol. Only C-1 and C-2 are involved in osazone formation. Hence, aldohexoses and keto hexoses, which have the same configuration at C-3, C-4 and C-5 give the same osazone. 10. -glucopyranose is oxidised by HIO 4 more rapidly than the -anomer at the 1,2 bond. Suggest a reason. Sol. -glucopyranose is cis-1, 2-diol and the -anomer is the trans 1, 2 diol. Since the former can form a cyclic ester with periodic acid and latter can not, the former is oxidised more rapidly than the latter. 11.

Calculate how much of the -anomer and -anomers are present in an equilibrium mixture with a specific rotation of +52.6º?

Sol. Answer -anomer  36% -anomer  64%




106 Biomolecules


12. Compound (A) C5H10O5, give a tetra-acetate with Ac 2O and oxidation of (A) with Br 2–H2O gives an acid, C 5H10O6. Reduction of (A) with HI and red phosphorous gives 2-methyl butane. What is the structure of (A)? Sol. Answer CHO CHOH HO–CH2–C–OH C H2O H

13. Number of possible stereoisomers of glucose and fructose. Sol. Answer Glucose – 16 Fructose – 8 






Chem Biomolecules

Recommend Documents