14 bimolecules

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    0-Question 14.1: Glucose or sucrose are soluble in water but cyclohexane or benzene (simple sixmembered ring compounds) are insoluble in water. Explain.

    • Answer 

    A glucose molecule contains five −! groups while a sucrose molecule contains eight −! groups."hus# glucose and sucrose undergo extensive !$bonding with water.

    !ence# these are soluble in water.

    %ut cyclohexane and benzene do not contain −! groups. !ence# they cannot undergo !$bondingwith water and as a result# are insoluble in water.

    Question 14.2: &hat are the expected products of hydrolysis of lactose'

    • Answer 

    actose is composed of $* galactose and $* glucose. "hus# on hydrolysis# it gives $* galactoseand $* glucose.

    Question 14.3: !ow do you explain the absence of aldehyde group in the pentaacetate of *$glucose'

    • Answer 

    *$glucose reacts with hydroxylamine (+!,!) to form an oxime because of the presence of aldehydic (−-!) group or carbonyl carbon. "his happens as the cyclic structure of glucose formsan open chain structure in an aueous medium# which then reacts with +! ,! to give an oxime.

    %ut pentaacetate of *$glucose does not react with +! ,!. "his is because pentaacetate does notform an open chain structure.

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    Question 14.4: "he melting points and solubility in water of amino acids are generally higherthan that of the corresponding halo acids. Explain.

    • Answer 

    %oth acidic (carboxyl) as well as basic (amino) groups are present in the same molecule of aminoacids. /n aueous solutions# the carboxyl group can lose a proton and the amino group can accepta proton# thus giving rise to a dipolar ion 0nown as a zwitter ion.

    *ue to this dipolar behaviour# they have strong electrostatic interactions within them and withwater. %ut halo$acids do not exhibit such dipolar behaviour.

    1or this reason# the melting points and the solubility of amino acids in water is higher than thoseof the corresponding halo$acids.

    Question 14.5: &here does the water present in the egg go after boiling the egg'

    • Answer 

    &hen an egg is boiled# the proteins present inside the egg get denatured and coagulate. Afterboiling the egg# the water present in it is absorbed by the coagulated protein through !$bonding.

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    Question 14.6: &hy cannot vitamin - be stored in our body'

    • Answer 

    2itamin - cannot be stored in our body because it is water soluble. As a result# it is readilyexcreted in the urine.

    Question 14.7:

    &hat products would be formed when a nucleotide from *+A containing thymine is hydrolysed'

    • Answer 

    &hen a nucleotide from the *+A containing thymine is hydrolyzed# thymine $*$,$deoxyriboseand phosphoric acid are obtained as products.

    &hen 3+A is hydrolysed# there is no relationship among the uantities of different bases obtained.

    &hat does this fact suggest about the structure of 3+A'

    Answer 

    A *+A molecule is double$stranded in which the pairing of bases occurs. Adenine always pairs withthymine# while cytosine always pairs with guanine. "herefore# on hydrolysis of *+A# the uantity

    of adenine produced is eual to that of thymine and similarly# the uantity of cytosine is eual tothat of guanine.

    %ut when 3+A is hydrolyzed# there is no relationship among the uantities of the different basesobtained. !ence# 3+A is single$stranded.

    Question 14.1: &hat are monosaccharides'

    • Answer 

    4onosaccharides are carbohydrates that cannot be hydrolysed further to give simpler units of polyhydroxy aldehyde or 0etone.

    4onosaccharides are classified on the bases of number of carbon atoms and the functional grouppresent in them. 4onosaccharides containing an aldehyde group are 0nown as aldoses and thosecontaining a 0eto group are 0nown as 0etoses. 4onosaccharides are further classified as trioses#

    tetroses# pentoses# hexoses# and heptoses according to the number of carbon atoms they contain.1or example# a 0etose containing 5 carbon atoms is called 0etotriose and an aldose containing 5carbon atoms is called aldotriose.

    Question 14.2: &hat are reducing sugars'

    • Answer 

    3educing sugars are carbohydrates that reduce 1ehling6s solution and "ollen6s reagent. Allmonosaccharides and disaccharides# excluding sucrose# are reducing sugars.

    Question 14.3: &rite two main functions of carbohydrates in plants.

    • Answer 

    "wo main functions of carbohydrates in plants are7

    (i) 8olysaccharides such as starch serve as storage molecules.

    (ii) -ellulose# a polysaccharide# is used to build the cell wall.

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    Question 14.4: -lassify the following into monosaccharides and disaccharides.

    3ibose# ,$deoxyribose# maltose# galactose# fructose and lactose

    • Answer 

    4onosaccharides7

    3ibose# ,$deoxyribose# galactose# fructose

    *isaccharides7

    4altose# lactose

    Question 14.5: &hat do you understand by the term glycosidic lin0age'

    • Answer 

    Glycosidic lin0age refers to the lin0age formed between two monosaccharide units through anoxygen atom by the loss of a water molecule.

    1or example# in a sucrose molecule# two monosaccharide units# ∝$glucose and $fructose# are 9oined together by a glycosidic lin0age.

    Question 14.6: &hat is glycogen' !ow is it different from starch'

    • Answer 

    Glycogen is a carbohydrate (polysaccharide). /n animals# carbohydrates are stored as glycogen.

    :tarch is a carbohydrate consisting of two components − amylose (;< − ,=>) and amylopectin(?= − ?).

    !owever# glycogen consists of only one component whose structure is similar to amylopectin. Also#glycogen is more branched than amylopectin.

    Question 14.7:

    &hat are the hydrolysis products of (i) sucrose and (ii) lactose'

    • Answer 

    (i) n hydrolysis# sucrose gives one molecule of ∝$* glucose and one molecule of $ *$fructose.

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    (ii) "he hydrolysis of lactose gives $*$galactose and $*$glucose.

    Question 14.8: &hat is the basic structural difference between starch and cellulose'

    • Answer 

    :tarch consists of two components − amylose and amylopectin. Amylose is a long linear chain of ∝−*−(@)−glucose units 9oined by -;−- glycosidic lin0age (∝$lin0).

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    Amylopectin is a branched$chain polymer of ∝$*$glucose units# in which the chain is formed by-;−- glycosidic lin0age and the branching occurs by -;−-B glycosidic lin0age.

    n the other hand# cellulose is a straight$chain polysaccharide of $*$glucose units 9oined by-;−- glycosidic lin0age ($lin0).

    Question 14.9: &hat happens when *$glucose is treated with the following reagents'

    (i) !/ (ii) %romine water (iii) !+5

    • Answer 

    (i) &hen *$glucose is heated with !/ for a long time# n$hexane is formed.

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    (ii) &hen *$glucose is treated with %r, water# *$ gluconic acid is produced.

    (iii) n being treated with !+5# *$glucose get oxidised to give saccharic acid.

    Question 14.10: Enumerate the reactions of *$glucose which cannot be explained by its openchain structure.

    • Answer 

    (1) Aldehydes give ,# $*+8 test# :chiff6s test# and react with +a!:   to form the hydrogensulphite addition product. !owever# glucose does not undergo these reactions.

    (2) "he pentaacetate of glucose does not react with hydroxylamine. "his indicates that a free−-! group is absent from glucose.

    (3) Glucose exists in two crystalline forms − ∝ and. "he ∝$form (m.p. C ;D ) crystallises

    from a concentrated solution of glucose at 5=5 and the $form (m.p C ,5 ) crystallises from ahot and saturated aueous solution at 5F; . "his behaviour cannot be explained by the openchain structure of glucose.

    Question 14.11: &hat are essential and non$essential amino acids' Give two examples of eachtype.

    • Answer 

    Essential amino acids are reuired by the human body# but they cannot be synthesised in thebody. "hey must be ta0en through food. 1or example7 valine and leucine

    +on$essential amino acids are also reuired by the human body# but they can be synthesised inthe body. 1or example7 glycine# and alanine

    Question 14.12: *efine the following as related to proteins

    (i) 8eptide lin0age (ii) 8rimary structure (iii) *enaturation.

    • Answer 

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    (i) Peptide linkage:

    "he amide formed between −-! group of one molecule of an amino acid and −+! , group of another molecule of the amino acid by the elimination of a water molecule is called a peptidelin0age.

    (ii) Primary structure:

    "he primary structure of protein refers to the specific seuence in which various amino acids arepresent in it# i.e.# the seuence of lin0ages between amino acids in a polypeptide chain. "heseuence in which amino acids are arranged is different in each protein. A change in the seuencecreates a different protein.

    (iii) Denaturation:

    /n a biological system# a protein is found to have a uniue 5$dimensional structure and a uniuebiological activity. /n such a situation# the protein is called native protein. !owever# when thenative protein is sub9ected to physical changes such as change in temperature or chemical

    changes such as change in p!# its !$bonds are disturbed. "his disturbance unfolds the globulesand uncoils the helix. As a result# the protein loses its biological activity. "his loss of biological

    activity by the protein is called denaturation. *uring denaturation# the secondary and the tertiarystructures of the protein get destroyed# but the primary structure remains unaltered.

    ne of the examples of denaturation of proteins is the coagulation of egg white when an egg isboiled.

    Question 14.13:&hat are the common types of secondary structure of proteins'

    • Answer 

    "here are two common types of secondary structure of proteins7

    (i) ∝$helix structure

    (ii) $pleated sheet structure

    ∝- Helix structure:

    /n this structure# the −+! group of an amino acid residue forms !$bond with the group of the ad9acent turn of the right$handed screw (∝$helix).

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    -pleated sheet structure:

    "his structure is called so because it loo0s li0e the pleated folds of drapery. /n this structure# allthe peptide chains are stretched out to nearly the maximum extension and then laid side by side."hese peptide chains are held together by intermolecular hydrogen bonds.

    Question 14.14: &hat type of bonding helps in stabilising the ∝$helix structure of proteins'

    • Answer 

    "he !$bonds formed between the −+! group of each amino acid residue and

    the group of the ad9acent turns of the ∝$helix help in stabilising the helix.

    Question 14.15: *ifferentiate between globular and fibrous proteins.

    • Answer 

    Firous protein !loular protein

    1"

    /t is a fibre$li0e structure formed by thepolypeptide chain. "hese proteins are heldtogether by strong hydrogen and disulphidebonds.

    1"

    "he polypeptide chain in thisprotein is folded around itself#giving rise to a sphericalstructure.

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    2" /t is usually insoluble in water. 2" /t is usually soluble in water.

    3"

    1ibrous proteins are usually used for structuralpurposes. 1or example# 0eratin is present innails and hair collagen in tendons andmyosin in muscles.

    3"

    All enzymes are globularproteins. :ome hormones suchas insulin are also globularproteins.

    Question 14.16:

    !ow do you explain the amphoteric behaviour of amino acids'

    • Answer 

    /n aueous solution# the carboxyl group of an amino acid can lose a proton and the amino groupcan accept a proton to give a dipolar ion 0nown as zwitter ion.

    "herefore# in zwitter ionic form# the amino acid can act both as an acid and as a base.

    "hus# amino acids show amphoteric behaviour.

    Question 14.17: &hat are enzymes'

    • Answer 

    Enzymes are proteins that catalyse biological reactions. "hey are very specific in nature andcatalyse only a particular reaction for a particular substrate. Enzymes are usually named after theparticular substrate or class of substrate and some times after the particular reaction.

    1or example# the enzyme used to catalyse the hydrolysis of maltose into glucose is named as

    maltase.

    Again# the enzymes used to catalyse the oxidation of one substrate with the simultaneousreduction of another substrate are named as oxidoreductase enzymes.

    "he name of an enzyme ends with H− ase6.

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    Question 14.18:

    &hat is the effect of denaturation on the structure of proteins'

    • Answer 

    As a result of denaturation# globules get unfolded and helixes get uncoiled. :econdary and tertiarystructures of protein are destroyed# but the primary structures remain unaltered. /t can be said

    that during denaturation# secondary and tertiary$structured proteins get converted into primary$structured proteins. Also# as the secondary and tertiary structures of a protein are destroyed# theenzyme loses its activity.

    Question 14.19: !ow are vitamins classified' +ame the vitamin responsible for the coagulation of blood.

    • Answer 

    n the basis of their solubility in water or fat# vitamins are classified into two groups.

    (i) Fat-solule #itamins: 2itamins that are soluble in fat and oils# but not in water# belong to this

    group. 1or example7 2itamins A# *# E# and

    (ii)  $ater-solule #itamins:  2itamins that are soluble in water belong to this group. 1orexample7 % group vitamins (%;# %,# %B# %;,# etc.) and vitamin -

    !owever# biotin or vitamin ! is neither soluble in water nor in fat.

    2itamin is responsible for the coagulation of blood.

    Question 14.20: &hy are vitamin A and vitamin - essential to us' Give their important sources.

    • Answer 

    "he deficiency of vitamin A leads to xerophthalmia (hardening of the cornea of the eye) and nightblindness. "he deficiency of vitamin - leads to scurvy (bleeding gums).

    "he sources of vitamin A are fish liver oil# carrots# butter# and mil0. "he sources of vitamin - arecitrus fruits# amla# and green leafy vegetables.

    Question 14.21: &hat are nucleic acids' 4ention their two important functions.

    • Answer 

    +ucleic acids are biomolecules found in the nuclei of all living cells# as one of the constituents of chromosomes. "here are mainly two types of nucleic acids − deoxyribonucleic acid (*+A) andribonucleic acid (3+A). +ucleic acids are also 0nown as polynucleotides as they are long$chainpolymers of nucleotides.

    "wo main functions of nucleic acids are7

    (i) *+A is responsible for the transmission of inherent characters from one generation to the next."his process of transmission is called heredity.

    (ii) +ucleic acids (both *+A and 3+A) are responsible for protein synthesis in a cell. Even thoughthe proteins are actually synthesised by the various 3+A molecules in a cell# the message for thesynthesis of a particular protein is present in *+A.

    Question 14.22: &hat is the difference between a nucleoside and a nucleotide'

    • Answer 

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    A nucleoside is formed by the attachment of a base to position of sugar.

    +ucleoside C :ugar @ %ase

    n the other hand# all the three basic components of nucleic acids (i.e.# pentose sugar# phosphoricacid# and base) are present in a nucleotide.

    +ucleotide C :ugar @ %ase @ 8hosphoric acid

    Question 14.23: "he two strands in *+A are not identical but are complementary. Explain.

    • Answer 

    /n the helical structure of *+A# the two strands are held together by hydrogen bonds betweenspecific pairs of bases. -ytosine forms hydrogen bond with guanine# while adenine forms hydrogen

    bond with thymine. As a result# the two strands are complementary to each other

    Question 14.24: &rite the important structural and functional differences between *+A and3+A.

    • Answer 

    "he structural differences between *+A and 3+A are as follows7

    D%& '%&

    1""he sugar moiety in *+A molecules is $*$, deoxyribose.

    1""he sugar moiety in 3+A molecules is$*$ribose.

    2"*+A contains uracil (I). /t does notcontain thymine (").

    2"3+A contains thymine ("). /t does notcontain uracil (I).

    3""he helical structure of *+A is double$stranded.

    3""he helical structure of 3+A is single$stranded.

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    "he functional differences between *+A and 3+A are as follows7

    D%& '%&

    1"*+A is the chemical basis of heredity.

    1" 3+A is not responsible for heredity.

    2"8roteins are synthesised by3+A molecules in the cells.

    2"*+A molecules do not synthesise proteins# buttransfer coded message for the synthesis of proteins in the cells.

    Question 14.25: &hat are the different types of 3+A found in the cell'

    • Answer 

    (i) 4essenger 3+A (m$3+A)

    (ii) 3ibosomal 3+A (r$3+A)

    (iii) "ransfer 3+A (t$3+A)

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