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Einstein Classes, Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road

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BIO-MOLECULES

C1A CARBOHYDRATES :

Carbohydrates have formula Cx(H

2O)

y. These can be OPTICALLY ACTIVE POLYHYDROXY

ALDEHYDES OR KETONES. Carbohydrates are also known as Saccharides. They are used in Food,cloth and for shelter.

C1B CLASSIFICATION :

(i) MONOSACCHARIDES : Cannot be hydrolysed further to simpler compounds.

e.g. : Glucose, Fructose, Galactose.

(ii) OLIGOSACCARIDES : Carbohydrates which yield few (2-10) of monosaccharides on hydrolysis.

e.g. :

)Galactose,Fructose,ecosGlu(

Hydrolysis

)Fructose(6126

)ecosGlu(6126

Hydrolysis

)Sucrose(112212

Hydrolysis

ridesMonosoccha3)rafinose(idesTrisacchar

OHCOHCOHC

ridesMonosaceha2ideDiasacchar

* Monosaccharides and Oligosaccarides are sweet sugar, crystaline solid and soluble in water.

(iii) POLYSACHRIDES : High molecular mass carbodydrates which yield many molecules of monosaccha-rides on hydrolysis. e.g. : Starch & Cellulose, both having general formula (C

6H

10O

5)n

(C6H

10O

5)n + nH

2O

H nC

6H

12O

6.

* They are insoluble in water, they are amorphous solid and tasteless.

* These are NON SUGARS.

REDUCING AND NON-REDUCING SUGAR :

(i) Carbohydrates with and groups reducible by FEHLING’S solution and TOLLEN’S

solution are Reducing sugars All Monosaccharides.

(ii) Dicsacharides

* If reducing group in monosaccharides bonded i.e. Aldehyde & Ketone group, then they are non-reducing.

Practice Problems :

1. What are reducing sugars ?

[Answers : (1) Carbohydrates which react with Fehling’s solution to give a red ppt. of Cu2O or with

Tollen’s reagent to give metallic metallic Ag are called reducing sugars. All monosaccharides (bothaldoses and ketoses) and disaccharides except sucrose and reducing sugars. Thus, D–(+)–glucose,D–(+)–galactose, D–(–)–fructose, D–(+)–maltose and D–(+)–lactose are all reducing sugars.Monosaccharides containing keto group also reduce Fehling solution because they exist asequilibrium mixture of aldehydric and keto forms. Sucrose is not reducing because the link betweenglucose and fructose is through aldehydic group]

C2 MONOSACCHARIDES :

All carbohydrates are either mono-saccharides or get converted to mono-saccharide on hydrolysis. Glucoseand Fructose are specific examples of aldohexose & ketohexose respectively.

e.g. : Glucose and Fructose Fruit Sugar..

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Preparation of Glucose :

1. From Sucrose (cane sugar)

Fructose6126

ecosGlu6126

H2

Sucrose112212 OHCOHCOHOHC

Fructose reduces Felhing’s & tollen’s reagent in alk. medium though it does not contain –CHO group

2. From Starch

ecosGlu

6126pressurehigh,K393

H2n5106 OHCnOHnOHC

Structure of D-Glucose & L-Glucose

Stereochemistry of all sugar has been determined by relating it to that of D– or L–Glyceraldehyde

Practice Problems :

1. What are the expected products of hydrolysis of lactose ?

2. What are monosaccharides ?

3. Write two main functions of carbohydrates is plants.

4. Classify the following into monosaccharides and disaccharides.

Ribose, 2-deoxyribose, maltose, galactose, fructose and lactose.

[Answers : (1) Galactose)(D

6126Glucose)(D

6126

OH

Lactaseor2Lactose

112212 OHCOHCOHOHC3

(2) Monosaccharides are car--

bohydrates which cannot be hydrolysed to smaller molecules. Their general formula is (CH2O)

n,

where n = 3 – 7. Those which contain an aldehyde group (–CHO) are called aldoses and those whichcontain a keto (C = O) group are called Ketoses (3) (i) The polysaccharide cellulose acts as the chiefstructural material of the plant cell walls. (ii) The polysaccharide starch is the major reserve foodmaterial in the plants. It is stored in seeds and acts as the reserve food material for the tiny plant tillit is capable of making its own food by photosynthesis (4) Monosaccharides : Ribose, 2-deoxyribose,galactose and fructose. Disaccharides : Maltose and lactose]

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C3A PROPERTIES OF GLUCOSE :

Practice Problems :

1. Glucose and sucrose are soluble in water but cyclohexane or benzene (simple six membered ringcompounds) are insoluble in water. Explain.

2. How do you explain the absence of aldehyde group in the pentaacetate of D-glucose ?

3. What happens when D-glucose is treated with the following reagents ?

(i) HI (ii) Bromine water (iii) HNO3

4. Enumerate the reactions of D-glucose which cannot be explained by its open chain structure.

[Answers : (1) Glucose containing five and sucrose containing eight –OH groups from H-bonds withwater. Due to this extensive intermolecular H-bonding, glucose and sucrose are soluble in waterinspite of the fact that molecular masses (180 g mol–1 and 342 g mol–1, respectively) are high. Incontrast, benzene (molecular mass = 78 g mol–1) and cyclohexane (molecular mass = 84 g mol–1) have

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low molecular mass. Even then they are insoluble in water. It is because these compounds do notcontain –OH groups and hence do not form H-bonds with water (2) when glucose is treated withacetic anhydride, the OH group at C-1, along with the four other OH groups at C-2, C-3, C-4 andC-6 form a pentaacetate. Pentaacetate of glucose does not contain a free OH group at C-1. Thereforeit cannot get hydrolysed in aqueous solution to produce the open chain aldehydic form and hence itdoes not react with NH

2OH to form glucose oxime (4) Reactions of glucose which cannot be

explained by open chain structure are : (i) Despite having the aldehydic group, glucose does not give2, 4 Dinitrophenylhydrazine (2, 4-DNP) test and Schiff’s test. It also does not form hydrogensulphiteaddition compound with NaHSO

3. (ii) The pentaacetate of glucose does not react with hydroxy-

lamine showing the absence of aldehydic group. (iii) Glucose exists in two different forms -glucoseand -glucose. -glucose has melting point 419 K while -glucose has melting point 423 K. -form ofglucose is obtained by crystallisation from concentrated solution of glucose at 303 K while and-form is obtained by crystallisation from hot and saturated aqueous solution at 371 K.]

C3B CYCLIC STRUCTURE OF GLUCOSE :

MUTOROTATION :

When 2 forms D glucose are dissolved in H

2O and allowed to stand in water, then specific rotation changes

slowly to +52.50.

This spt. change in specific rotation of an optically active glucose when dissolved in water is Mutarotation.Also shown by Fructose.

– D glucose – D – glucose

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C4 DISACCHARIDES :

disaccharides OH2 2 moles of monosaccharides (same or either different)

OH

Sucrose112212

2OHC glucose + fructose

OH

Lactose112212

2OHC glucose + galactose

maltose OH2 glucose + glucose

C12

H22

O11

The disaccharides may be reducing or non-reducing depending upon

between 2 monosaccharides units.

(i) If the glycoside linkage involves the carbonyl fuctional group of both monosaccharides units, the resultingdisaccharides is non-reducing e.g. Sucrose.

(ii) If one of carbonyl functional group of any monosacchride units is free, then resulting disaccharide wouldbe reducing e.g. Maltose & Lactose.

SUCROSE : (Sugarcane, beet root) property colourless, water soluble and sweet substance.

Fructose6126

ecosGlu6126

HCl

Lactorse2112212 OHCOHCOHOHC

Dextrorotatory but after hydrolysis gives dextrorotatory glucose + leavo fructose. (mixture is Laevorotary.(partial reacmisation)) as leavo rotation is more than dextorotation of glucose.

Change in rotation INVERSION

Mixture is INVERT SUGAR.

Fructose6126

ecosGlu6126

Invertase2

)sucrose(112212 OHCOHCOHOHC

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+ given to optically active compound which rotates plane of polarized light

clockwise or left

(i) Starch occur in form of granules, which vary in size and shape depending on their plant source

(ii) Starch is amorphous powder, insoluble in cold water

(iii) Starch solution gives blue colour in water disappear on heating

(iv) On hydrolysis it break down to molecules of variable size completely

(v) It indicates the presence of glucosidic linkage

(vi) Starch is a mixture of two polysaccharide

Amylose Amylopectin

(i) water soluble & give blue colour with I2

Branched chain polysaccharide insolublein water, which does not give blue colourwith I

2.

(ii) having D-Glucose units joined together It is composed of 25-30 D-glucose units by -Glucosidiclinkage involving C

1joined by -D-glycosidic linkage between

and C4 of another glucose C-1 of one glucose unit and C

4 of next

glucose unit (similar to amylase)

(iii) It can have 100-3000 D-Glucose However these chains are connected witheach other by 1,6-linkages.

Starch is major food material. It is hydrolysed by enzyme amylase present in saliva. The end product isglucose which is essential nutrient.

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Practice Problems :

1. What do you understand by the term glycosidic linkage ?

2. What is glycogen ? How is it different from starch ?

3. What are the hydrolysis product of (i) sucrose and (ii) lactose ?

4. What is the basic structural difference between starch and cellulose ?

[Answers : (1) The oxygen linkage through which two monosaccharides are joined together by theloss of a water molecule to form a molecule of disaccharide is called the glycosidic linkage.

(2) Starch is not a single compound but is a mixture of two

component – a water soluble components called amylose (15-20%) and water insoluble componentamylopectin (80-85%). Amylose is a linear polymer of -D-glucose. Both glycogen and amylopectinare branched polymer of -D-glucose. Glycogen is more highly branched than amylopectin.

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Amylopecting chains consists of 20-25 glucose units while glycogen chains consist of 10-14 glucose

units (3) Fructose)(D

6126Glucose)(D

6126

OH

Invertaseor2Sucrose

112212 OHCOHCOHOHC3

alactoseG)(D6126

Glucose)(D6126

OH

actaseLor2Lactose

112212 OHCOHCOHOHC3

(4) Starch consists of amylose and

amylopectin. Amylose is a linear polymer of -D-glucose while cellulose is a linear polymer of-D-glucose. In amylose, C-1 of one glucose unit is connected to C-4 of the other through-glycosidic linkage as shown under carbohydrates in “Chapter At A Glance”. However, incellulose, C-1 of one glucose unit is connected to C-4 of the other through -glcosidic linkage.Amylopectin, on the other hand, has highly branched structure]

C5 Lipid :

Lipid are naturally occuring compound related to fatty acids and it include Fats, Oils, Waxes.

In body they serve as source of energy, They are store in adipose tissues. They are lyophobic in nature andsoluble in organic solvent. Phospholipid is important consistuent of cell membrane

(A) Simple Lipid : They are triglycerides i.e. esters of glycerol with long chain fatty acids.

(i) Fatty acids have even numbers of carbons may be saturated or unsaturated.

Unsaturated/Saturated Fatty Acids

(ii) The three fatty acid in triglycerides may be same or different.

(iii) Fats are glycerides of saturated fatty acids e.g. tripalmition and tristearin.

(iv) Oils contain unsaturated fatty acids e.g. triolein, -oleo--palmito-–-stearin is an example of mixedtriglyceride.

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The presence of double bond with less stable cis stereochemistry in unsaturated fatty acid. In oleic acid it isat C-9. In linoleic acid C

17H

31COOH at C9 & C12. In linoleic acid C

17H

29COOH at C9 & C12. In solid state

the molecules of saturated fatty acids fit closely due to zig-zag tetrahedral structure. The cis unsaturatedfatty acid chains have a bend at the double bond & do not fit clearly resulting in lowering of melting pointof the fat. Application of lipid : Fats are important food reserves of animals and plant cells. Phospholipidare structural component of cell memberane and are also used as detergent to emulsify fat for transportwithin the body.

Proteins :

Proteins are high polymers. They are polyamides which contain C, H, N, O and S. Proteins are derived from-amino carboxylic acid monomers. A simple protein may contain hundreds or even thousands of theseamino acid units. In living organism, twenty -amino acids occur which combine to form different proteinmolecules.

(i) Amino Acid : A simple -amino acid can be represented as .

Due to the transfer of proton from carboxy to the amino group, -amino acids exist as dipolar ions orzwitter ions. Amino acids contain an acidic (carboxyl group) and a basic (amino group) within the samemolecule. In aqueous solution, the carboxyl group loses a proton while the amino group accepts it. As aresult, a dipolar or zwitter ion is formed. In zwitter ionic form, -amino acids show amphoteric behaviouras they react with both acids and bases. In the acidic medium, COO– ion of the zwitter ion accepts a proton

to form the cation (I) while in the basic medium, 3HN

ion loses a proton to form the anion (II).

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The amino acids are of two types :

(i) Non-essential amino acids

(ii) Essential amino acids

-Amino acids which are needed for health and growth of human beings but are not synthesised by thehuman body are called essential amino acids. For example, valine, leucine, arginine, etc. On the other hand,-amino acids which are needed for health and growth of human beings and are synthesised by the humanbody are called non-essential amino acids. For example, glycine, alaline, glutamine, etc.

Polypeptide Formation :

Peptide bond : Proteins are condensation polymers of -amino acids in which the same or different-amino acids are connected by peptide bonds. Peptide bond is an amide linkage formed between –COOHgroup of one -amino acid and NH

2 group of the other -amino acid by loss of a molecule of water. For

example :

Structure of Proteins : Proteins have three structures :

(i) Primary structure

(ii) Secondary structure

(a) -helix structure

(b) -pleated structure

(iii) Tertiary structure

(a) Fibrous structure

(b) Glubular structure

Primary structure : Proteins may contain one or more polypeptide chains. Each polypeptide chain has alarge number of -amino acids which are linked to one another in a specific manner. The specific sequencein which the various -amino acids present in a protein are linked to one another is called its primarystructure.

Secondary structure : The fixed conformation which the polypeptide chains assume as a result ofhydrogen bonding is called the secondary structure of the proteins. The two types of secondary structuresare : -helix and -pleated sheet structure

The -helix structure of proteins is stabilised by intramolecular H-bonding between C = O of one aminoacid residue and the N – H of the fourth amino acid residue in the chain.

Tertiary structure : It implies the three dimensional structure of proteins.

Globular proteins : This structure results when the chains of polypeptides coil around to give sphericalshape. These proteins are usually soluble in water. Insulin and albumin are the common examples ofglobular proteins.

Fibrous proteins : When the polypeptide chains run parallel and are held together by hydrogen bond anddisulphide bond, then fibre-like structure is obtained. Such proteins are called fibre proteins. Fibre proteinsare insoluble in water. Some examples of fibrous proteins are keratin and myosin.

Denaturation of protein : Each protein in the biological system has a unique three-dimensional structureand has specific biological activity. This is called native form of a protein. When a protein in its native formis subjected to physical changes such as change in temperature, pH etc., hydrogen bonds are broken.Consequently, unfolding of protein molecule occurs and the protein loses its biological activity. This loss ofbiological activity is called denaturation. During denaturation, 20 and 30 structures of proteins are destroyedbut 10 structure remains intact. An example of denaturation of proteins is the coagulation of albumin presentin the white of an egg.

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Practice Problems :

1. The melting points and solubility in water of amino acids are generally higher than that of thecorresponding halo acids. Explain.

2. Where does the water present in the egg go after boiling the egg ?

3. Why cannot vitamin C be stored in our body ?

4. What products would be formed when a nucleotide from DNA containing thymine is hydrolysed ?

5. When RNA is hydrolysed, there is no relationship among the quantities of different bases obtained.What does this fact suggest about the structure of RNA ?

6. What are essential and non-essential amino acids ? Give two examples of each type.

7. Define the following as related to proteins (i) Peptide linkage (ii) Primary structure(iii) Denaturation.

8. What are the common types of secondary structure of proteins ?

9. What type of bonding helps in stabilizing the -helix structure of proteins ?

10. Differentiate between globular and fibrous proteins.

11. How do you explain the amphoteric behaviour of amino acids ?

12. What is the effect of denaturation on the structure of proteins ?

[Answers : (1) The amono acids exist as zwitter ions,

COOCHRNH3 . Thus they acquire salt

like structure. Due to this dipolar salt like character they have strong dipole-dipole attractions. There-fore, their melting points are higher than haloacids which do not have salt like character. Again dueto salt like character, they interact strongly with H

2O. Polar solutes dissolve in polar solvents. As a

result, solubility in water of amino acids is higher than that of the corresponding haloacids. (3) Beingwater soluble, vitamin C is readily excreted in urine and hence cannot be stored in the body (4)Thymine, 2-deoxy-D-ribose and phosphoric acid (5) This can be explained in terms of single standstructure of RNA. A DNA molecules has two strands in which the four complementary bases paireach other, i.e., cytosine (C) always pairs with guanine (G) while thymine (T) always pairs withadenine (T). Therefore, when a DNA molecule is hydrolysed, the molar amounts of cytosine is alwaysequal to that of guanine and that of adenine is always equal to that of thymine. In RNA, there is norelationship between the quantities of four bases (C, G, A and U) obtained, that is the base-pairingprinciple is not followed. Therefore, unlike DNA, RNA has a single strand (2) When the egg is boiled,the proteins first undergo denaturation and then coagulation, and the water present in the egg getsabsorbed in the coagulated proteins (8) The fixed conformation which the polypeptide chainsassume as a result of hydrogen bonding is called the secondary structure of the proteins. The twotypes of secondary structures are : -helix and -pleated sheet structure (9) The -helix structure ofproteins is stabilised by intramolecular H-bonding between C = O of one amino acid residue and theN – H of the fourth amino acid residue in the chain (12) During denaturation, secondary and tertiarystructures of proteins are destroyed but primary structure remains unchanged. As a result ofdenaturation, the globular proteins (soluble in H

2O) are converted into fibrous proteins (insoluble in

H2O) and their biological activity in lost]

C6 Enzymes :

Enzymes are biological catalysts. Each biological system requires a different enzymes. thus, in contrast toconventional catalysts, enzymes are very specific and efficient in their action. They are required in onlysmall quantity and work at optimum temperature (310 K) and pH (7.4) under one atmospheric pressure.Chemically, they are globular proteins. However, some enzymes are also associated with some non-proteincomponent called the cofactor for their activity.

Cofactors are of two types :

(a) Inorganic ions such as Zn2+, Mg2+, Mn2+, Fe2+, Cu2+, Co2+, etc.

(b) Organic molecules : These are also of two types :

(i) Coenzymes : These are usually derived from vitamins such as thiamine, riboflavin, etc.

(ii) Prosthetic group : These are also derived from vitamins such as biotin but are tightly held to the

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protein molecule by covalent bonds. They can be separated only be careful hydrolysis.

Practice Problems :

1. What are enzymes ?

C7 Nucleic Acid :

Nucleic acids are biomolecules which are found in the nuclei of all living cells in form of nucleoproteins orchromosomes (proteins containing nucleic acids as the prosthetic group).

Nucleic acids are of two types :

(i) deoxyribonucleic acid (DNA) and

(ii) ribonucleic acid (RNA).

Important functions of nucleic acids :

(i) DNA is responsible for transmission of hereditary effects from one generation to another. This is due to theunique property of replication during cell division and two identical DNA strands are transferred to thedaughter cells.

(ii) DNA and RNA are responsible for protein synthesis needed for the growth and maintenance of our body.Actually the proteins are synthesised by various RNA molecules (r-RNA, m-RNA and t-RNA) in the cellbut the message for the synthesis of a particular protein is given by DNA molecules.

DNA : It occurs in the nucleus of the cell. The two strands in DNA molecule are held together by hydrogenbonds between purine base of one stand and pyrimidine base of the other and vice versa. Because ofdifferent sizes and geometries of the bases, the only possible pairing in DNA are G (guanine) and C(cytosine) through three H-bonds, i.e, (C G) and T (thymine) through two H-bonds (i.e., A = T). Due tothis base-pairing principle, the sequence of bases in one strand automatically fixes the sequence of bases inthe other strand. Thus, the two strands are complimentary and not identical

RNA : It occurs in the cytoplasm of the cell. The sugar present in RNA is D(-)-ribose. It has asingle-stranded -helix structure. In RNA, purine base are adenine (A) and guanine (G) while pyrimidinebases are cytosine (C) and uracil (U). RNA usually does not undergo replication. It controls the synthesis ofproteins.

Practice Problems :

1. What are nucleic acids ? Mention their two important functions.

2. What is the difference between a nucleoside and a nucleotide ?

3. The two strands in DNA are not identical but are complementary. Explain.

4. Write the important structural and functional differences between DNA and RNA.

5. What are the different types of RNA found in the cell ?

[Answers : (2) A nucleoside is formed when 1-position of a pyrimidine (cytosine, thiamine or uracil)or 9-position of a purine (guanine or adenine) base is attached to C-1 of sugar (ribose anddeoxyribose) by a -linkage. Thus, in general, nucleosides may be represented as : Sugar-Base]. Anucleotide contains all the three basic components of nucleic acids, i.e., a phosphoric acid group, apentose sugar and a nitrogeneous base. These are obtained by esterification of C

5-OH group of the

pentose sugar by phosphoric acid (4) DNA : (i) The sugar present in DNA is 2-deoxy-D-(-)ribose.(ii) DNA contains cytosine and thymine as pyrimidine bases. (iii) DNA has double stranded -helixstructure, their molecular mass may vary from 6 × 106 – 16 × 106 u. (iv) DNA has unique property ofreplication. (v) DNA controls the transmission of hereditary effects. RNA : (i) The sugar present inRNA is D-(-) ribose. (ii) RNA contains cytosine and uracil as pyrimidine bases. (iii) RNA has singlestranded -helix structure, their molecular mass ranging from 20,000 to 40,000 u. (iv) RNA usuallydoes not replicate. (v) RNA controls the synthesis of proteins (5) There are three types of RNAs :(i) Ribosomal RNA (rRNA) (ii) Messenger RNA (mRNA) (iii) Transfer RNA (tRNA)]

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C8 Vitamins :

They are the chemical substances which are needed in small amounts for the growth and health of humanbeings. Vitamins cannot be synthesised in the body (except Vitamin D) and hence must be supplied in thefood. Their deficiency can cause one or the other disease. Vitamin D may be produced in the skin by theirradation of sterols present in oils and fats.

Vitamins are classified into two groups depending upon their solubility :

(i) Water soluble vitamins : Vitamin B-complex (B1, B

2, B

5, i.e., nicotinic acid, B

6, B

12, pantothenic acid and

folic acid) and vitamin C.

(ii) Fat soluble vitamins : Vitamins A, D, E and K. They are soluble in liver and adipose (fat storing tissues).However, biotin, i.e., vitamin H is neither soluble in water nor in fat.

Practice Problems :

1. How are vitamins classified ? Name the vitamin responsible for the coagulation of blood ?

2. Why are vitamin A and vitamin C essential to us ? Give their important sources.

[Answers : (2) Vitamin A is essential as its deficiency causes xerophthalmia (hardening of cornea ofeye) and night blindness. Sources : Fish liver oil, carrots, butter and milk. Vitamin C is essential asits deficiency causes scurvy (bleeding gums) and pyorrhoea (loosening and bleeding of teeth).

OBJECTIVE EXERCISE

1. Cane sugar on hydrolysis gives

(a) glucose only

(b) glucose and maltose

(c) glucose and fructose

(d) glucose and lactose

2. Which of the following monosaccharide is apentose

(a) glucose (b) fructose

(c) arabinose (d) galactose

3. On hydrolysis, proteins give

(a) nucleotides (b) nucleosides

(c) amides (d) amino acids

4. The change in optical rotation with time of freshlyprepared solutions of sugar, is known as

(a) specific rotation (b) inversion

(c) rotatory motion (d) mutarotation

5. Glucose when heated with CH3OH in presence of

dry HCl gas, - and -methyl glucosides are formed.This is because it contains

(a) an aldehydic group

(b) –CH2OH group

(c) a ring structure

(d) five hydroxyl groups

6. The number of asymmetric carbon atoms in theglucose molecule is

(a) 1 (b) 2

(c) 4 (d) 6

7. Ascorbic acid is

(a) a vitamin (b) an enzyme

(c) a protein (d) an amino acid

8. Main structural unit of protein is

(a) ester linkage (b) ether linkage

(c) peptide linkage (d) all the above

9. Glucose cannot be classified as

(a) a hexose

(b) an oligosaccharide

(c) a carbohydrate

(d) an aldose

10. Point out the wrong statement about proteins

(a) they are nitrogenous organic compoundof high molecular masses

(b) on hydrolysis by enzymes, they giveamino acids

(c) many of them are enzymes

(d) they do not contain polypeptide linkages

11. On heating glucose with Fehling’s solution we get aprecipitate whose colour is

(a) yellow (b) red

(c) black (d) white

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12. A certain compound gives negative test withninhydrin and positive test with Benedict’ssolution, the compound is

(a) a protein

(b) a lipid

(c) a monosaccharide

(d) an amino acid

13. The organic compound of high physiologicalimportance which are essential in small amountsfor the well-being of all human beings are

(a) proteins (b) vitamins

(c) mineral salts (d) enzymes

14. The reactions with sugars are carried out inneutral or acid medium and not in alkali mediumbecause in alkaline medium sugar undergo

(a) decomposition (b) racemization

(c) inversion (d) rearrangement

15. The pH value of a solution in which a polar aminoacid does not migrate under the influence ofelectric field called

(a) iso-electric point

(b) iso-electronic point

(c) neutralisation point

(d) none

16. The digestion of fat in intestines is aided by

(a) diffusion (b) peptization

(c) emulsification (d) neutralisation

17. Which of the following gives maximum energy inmetalic process

(a) proteins (b) fats

(c) carbohydrates (d) vitamins

18. Enzyme Trypsin converts

(a) amino acids into proteins

(b) glucose into glycogen

(c) starch into sugar

(d) proteins into amino acids

19. Biuret test is used for the detection of

(a) sugars (b) fats

(c) proteins (d) saturated oils

20. Proteins give purple colour with

(a) Benedict’s reagent

(b) iodine solution

(c) ninhydrin reagent

(d) biuret

21. Which of the following gives maximum energy inmetabolic process

(a) carbohydrates (b) fats

(c) proteins (d) vitamins

22. Which of the following statements about proteinsis not true

(a) amino acid residue join together to makea protein molecule

(b) proteins are polymers with formula(C

6H

10O

5)

n

(c) eggs are richer in proteins

(d) pulses are good source of proteins

23. Secondary structure of protein refers to

(a) three dimensional structure, specially thebond between amino acids residue thatare distant from each other in thepolypeptide chain.

(b) regular folding patterns of thepolypeptide chain

(c) mainly denatured proteins and structuresof prosthetic groups

(d) linear sequence of amino acid residues inthe polypeptide chain

24. The function of enzymes in the living system is to

(a) transport oxygen

(b) provide immunity

(c) catalyse biochemical reactions

(d) provide energy

25. The digestion of fats in the intestines is aided by

(a) diffusion (b) peptization

(c) protection (d) emulsification

26. Of the following statements about enzymes, whichones are true

(i) enzymes lack in nucleophillic groups

(ii) enzymes are highly specific both inbinding chiral substrates and incatalysing their reactions

(iii) enzymes catalyse chemical reactions bylowering the activation energy

(iv) Pepsin in proteolytic enzyme

(a) (i) (b) (i) and (iv)

(c) (i) and (iii) (d) (ii), (iii) & (iv)

27. Which statement is not correct for enzyme

(a) it acts as a biocatalyst

(b) it can catalyse any chemical reaction

(c) its aqueous solution in colloidal

(d) its catalytic efficiency is temperaturedependent

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28. The vitamin which is water soluble and antioxidantis

(a) Vitamin C (b) Vitamin B

(c) Vitamin E (d) Vitamin D

29. Which one of the following contains cobalt

(a) Chlorophyll (b) Haemoglobin

(c) Vitamin B12

(d) Vitamin C

30. Which one of the following vitamins contains ametal atom

(a) Vitamin A (b) Vitamin B2

(c) Vitamin B6

(d) Vitamin B12

31. Beri-beri is caused due to

(a) Vitamin A (b) Vitamin B

(c) Vitamin C (d) Vitamin D

32. A vitamin which plays a vital role in thecoagulating property of blood is

(a) Vitamin A (b) Vitamin D

(c) Vitamin E (d) Vitamin K

33. Which of the following statements about ribose inincorrect

(a) it is an aldehyde sugar

(b) it is polyhydroxy compound

(c) it has six carbon atoms

(d) it exhibits optical activity

34. -D-glucose and -D-glucose differ from each otherdue to the difference in one of the carbons withrespect to its

(a) configuration

(b) number of —OH groups

(c) conformation

(d) size of hemiacetal ring

35. The number of asymmetric carbon atoms infructose are

(a) 2 (b) 3

(c) 4 (d) 5

36. Which of the following is the structure of D-xylose

(a)

(b)

(c)

(d)

37. Sugars are characterised by the preparation ofosazone derivatives. Which sugars have identicalosazones

(a) Glucose and fructose

(b) Glucose and arabinose

(c) Glucose and lactose

(d) Glucose and maltose

38. Which one of the following is laevorotatory

(a) Glucose (b) Fructose

(c) Sucrose (d) None of these

39. An example of non-reducing sugar is

(a) cane sugar (b) fructose

(c) lactose (d) cellobiose

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40. Which one of the following is the reagent used toidentify glucose

(a) neutral FeCl3

(b) CHCl3 and alc. KOH

(c) ammonical AgNO3

(d) C2H

5ONa

41. Starch on hydrolysis by a dilute inorganic mineralacid gives

(a) Glucose (b) Fructose

(c) Sucrose (d) Maltose

42. Which one of the following chars, when warmedwith conc. H

2SO

4.

(a) Fats (b) Carbohydrates

(c) Proteins (d) Hydrocarbon

43. On hydrolysis of starch, we finally get :

(a) glucose

(b) fructose

(c) glucose and fructose both

(d) sucrose

44. Molisch test is used for the detection of

(a) fats (b) carbohydrates

(c) alkyl halide (d) alkaloid

ANSWERS OBJECTIVE

1. c

2. c

3. d

4. d

5. c

6. c

7. a

8. c

9. b

10. d

11. b

12. c

13. b

14. d

15. a

16. c

17. b

18. d

19. c

20. c

21. b

22. b

23. b

24. c

25. b

26. d

27. b

28. a

29. c

30. d

31. b

32. d

33. c

34. a

35. b

36. c

37. b

38. b

39. a

40. c

41. d

42. b

43. a

44. b

bio-molecules - einstein classeseinsteinclasses.com/bio-molecules.pdf · and fructose are specific...

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