BIOMOLECULE

BIOMOLECULE

Write the sequence that related biomolecules to living organism.

Ans:- Living bodies are built up with biomolecules. The sequence that relates biomolecules to living organisms is as follows.

Biomolecules→Organelles→Cells→Tissues→Organs→Living organism.

What are endergonic and exergonic reactions ? Explain these reactions in a cell.

Ans:- Cells need energy fpr several types of activities. Molecules like those of glucose undergo complex and controlled oxidation by means of biocatalysts known as enzymes and liberate energy. Endergonic reactions are those in which Gibb’s energy    change G > 0. While exergonic reactions are those in which Gibb’s energy G<0. Endergonic reactions as such appears as forbidden reactions, they are made to take place by coupling it with suitable exergonic reactions.

Eg: Some metabolic processes in human body are with G > 0, but they take place simultaneously with highly exergonic reactions.

Write brief note on photosynthesis.

Ans:- Plants absorb CO2 from air and H2O from soil makes them combine to form carbohydrates in the presence of sunlight and chlorophyll.

Photosynthesis involves two types of reactions. 1) Light reactions which require energy, take place in the presence of light. 2) Dark reactions which do not require energy, take place in the absence of light. In dark reactions ATP undergoes hydrolysis and carries the dark reaction with liberated energy from its hydrolysis.

Depending on the nature of plants and the type of reaction, the glucose formed initially converts into disaccharides, polysaccharides like starch, cellulose or proteins or oils.

Explain the classification of carbohydrates.

Ans:- Carbohydrates are classified in different ways basing on the characteristic selected.

First type classification : The basis is the term saccharide. Saccharides are classified on the basis of the products during their hydrolysis.

The carbohydrates which contain up to six carbon atoms and cannot be hydrolysed into simpler carbohydrates are called monosaccharides. Eg: Glucose, fructose.

The carbohydrates which give two monosaccharides on hydrolysis are called disaccharides.

The two monosaccharide products may be same are different. Eg : Sucrose gives glucose and fructose, maltose gives two glucose molecules.

The carbohydrates which give 3-10 molecules of monosaccharides on hydrolysis are called oligosaccharides.

The carbohydrates which give more than 10 monosaccharides on hydrolysis are called polysaccharides Eg : Starch, cellulose.

The monosaccharides are further classified into polyhydroxy aldehydes and polyhydroxy ketones depending on the functional group. The polyhydroxy aldehydes are called aldoses and the polyhydroxy ketones are called ketoses. Depending upon the number of carbon atoms present in an aldose or ketose they again grouped.

 

No. of carbon General Aldose Ketose
atoms Term (Aldehyde group) (Ketone group)
3 Triose Aldotriose Ketotriose
4 Tetrose Aldotetrose Ketotetrose
5 Pentose Aldopentose Ketopentose
6 Hexose Aldohexose Ketohexose
7 Heptose Aldoheptose Ketoheptose

 

Second type classification: This is based on taste. The mono, di and oligosaccharides which are generally crystalline, water soluble are sweet in taste are called sugars. Polysaccharides are generally amorphous, water insoluble and tasteless. These are called non-sugars.

Third type classification: The saccharides which can reduce Fehling solution, Tollen’s reagent etc., are called reducing sugars. The saccharides which cannot reduce fehling solution, Tollen’s reagent etc., are called non-reducing sugars.

Explain the structure of glucose on the basis of its properties.

Ans:-

  1. The molecular formula of glucose is experimentally found to be C6H12O6.
  2. Acylation of glucose with acetic anhydride gives pentaacetyl derivative. This indicates that glucose contain five ‘OH’ groups.
  3. Glucose reacts with hydroxyl amine to form oxime, HCN to form cyanohydrin. These reactions indicate that glucose contain a carbonyl group.
  4. Glucose reduces Tollen’s reagent and Fehling’s solution. Glucose is oxidised to gluconic acid in these reactions. Bromine water or alkaline solution of iodine oxidises glucose to gluconic acid. These reactions indicate that carbonyl group of glucose is aldehyde group.
  5. Oxidation of glucose with nitric acid gives saccharic acid. This indicates that glucose contains a primary alcoholic group – CH2
  6. Glucose on reduction with HI gives n-Hexane. This suggests that.in glucose.all the six carbon atoms in a straight chain. ,
  7. With concentrated solution of sodium hydroxide glucose first turns yellow, then brown and finally resinifies.
  8. With dilute solution of NaOH glucose undergoes reversible isomerisation to give a mixture of D – glucose, D – mannose and D – fructose. This reaction is known as Lobry – de Bruyn – Van Ekenstein rearrangement.

Basing on the above properties the following open chain structure was proposed to glucose by Baeyer.

This open chain structure explains the properties of glucose as follows.

Though the above chain structure explains the most of the properties of glucose but it could not explain the following properties.

  1. Glucose does not react with sodium bisulphite and ammonia though it contains – CHO group.
  2. Glucose does not give the Schiff’s test.
  3. Pentaacetate of glucose does not react with hydroxyl amine.
  4. Glucose forms osazone with phenylhydrazine.
  5. Glucose shows mutarotation.

The spontaneous change in specific rotation of an optically active compound is called mutarota­tion. – glucose crystallised from water at 30°C have m.p. 146°C and its specific optical rotation is +111°.  α- orβ- glucose crystallised from hot saturated solution at above 98°C have m.p. 150°C and its specific optical rotation is +19.2°. α- orβ- Either  glucoses dissolved in water and allowed to stand, the specific rotation of solution changes and slowly reaches to a constant value +52.5°C.

These observations confirm that glucose exists in isomers. This is confirmed by the reaction of methanol in the presence of dry HCl gas to give methyl  D- glucoside and methyl glucoside.

In the cyclic structure the C1 carbon becomes assymetric with OH group may be either on right side or on left side resulting in two isomers. This C1 carbon is called anomeric carbon and the isomer formed due to the difference in the configuration at anomeric carbon are called anomers.

The six membered cyclic structure was similar to that pyran which is a cyclic six membered compound with one oxygen and 5 carbon atoms in the ring. So it is called pyranose structure and represented by Haworth formula as follows.

In the cyclic structure the aldehyde group is not free. This explains why glucose shows mutarotation, and the existence of  isomers of glucose.

Explain the pyranose ring structure of glucose molecule.

Ans:- Though the chain structure explains the most of the properties of glucose but it could not explain the following properties.

  1. Glucose does not react with sodium bisulphite and ammonia though it contains – CHO group,
  2. Glucose does not give the Schiff’s test,
  3. Pentaacetate of glucose does not react with hydroxyl amine,
  4. Glucose forms osazone with phenyl hydrazine,
  5. Glucose shows mutarotation.

 

The spontaneous change in specific rotation of an optically active compound is called mutarota­tion. a – glucose crystallised from water at 30°C have melting point 146°C and its specific optical rotation is + 111°. p-glucose crystallised from hot saturated solution at above 98°C have melting point 150°C and its specific optical rotation is + 19.2°. α- orβ- Either  glucoses dissolved in water and allowed to stand, the specific rotation of solution changes and slowly reaches to a constant value + 52.5°C.

These observations confirm that glucose exists in α-,β-  isomers. This is confirmed by the reaction of methanol in the presence of dry HCl gas to give methyl α-β- glucoside and methy! P – D – glucoside.

In the cyclic structure the C1 carbon becomes assymetric with OH group may be either on right side or on left side resulting in two isomers. This C1 carbon is called anomeric carbon and the isomer formed due to the difference in the configuration at anomeric carbon are called anomers.

The six membered cyclic structure was similar to that pyran which is a cyclic six membered compound with one oxygen and 5 carbon atoms in the ring. So it is called pyranose structure and represented by Haworth formula as

In the cyclic structure the aldehyde group is not free. This explains why glucose shows mutarotation, and the existence of α, β- isomers of glucose.

Explain the D,L notation of monosaccharide stereoisomers.

Ans:- A. Glyceraldehyde exists in two optically active isomers (+) – glyceraldehyde and (-) – glyceraldehyde. While writing the formulae of glyceraldehyde Emil Fischer assumed that the OH group is on right side for (+) glyceraldehyde while OH group is on left side for (-) glyceraldehyde. While writing the absolute configurations of optical isomers.

Any compound or isomer that can be prepared from D – (+) – glyceraldehyde or can be converted into D – (+) – glyceraldehyde belongs to D – series similarly any compound or isomer that can be prepared from L – (-) – glyceraldehyde or can be converted into L – (-) – glyceraldehyde belongs to L – series. The D – and L – prefixes tell about the configuration of the highest numbered stereogenic (assymetric) centre.

Write molecular structure of any two disaccharides.

Ans:-

Write a polysaccharide.

Ans:- Carbohydrates containing large number of monosaccharide units joined through glycosidic linkage as called polysaccharides.

Eg: Starch, Cellulose.

Starch:

It is a white amorphous powder almost insoluble in cold but relatively more soluble in hot water.

Starch solution gives blue colour with iodine in cold but on heating blue colour disappears. Starch on complete hydrolysis gives blue colour.

Starch cannot reduce Tollen’s reagent or Fehling’s solution. Starch does not form osazone. These faces indicate that the -OH groups of glucose units at C – 1 are in glycosidic form.

Starch contains two polysaccharide units called 10-20 % amylase and 80 – 90 % amylopectin.

Amylase is water soluble and gives blue colour with iodine. In its molecule D – glucose units are joined by a- glucocidic linkages between C- l of one glucose unit and

C- 4 of the next. Its molecular mass varies from 10,000 to 50,000.

Amylopectin is a branched chain polysacctharide. It is water insoluble. It does not give blue colour with iodine. Its molecule contains 25 – 30 D – glucose units combined by a-D-glycosidic link between C – 1 of one glucose unit and C – 4 of the next. The branches arise due to 1, 6 linkages.

What are essential and non – essential amino acids ? Give examples.

Ans:- The amino acids which cannot be synthesized in the body but must be supplied to the body through the diet are called essential amino acids.

The amino acids which are synthesized in the body are known as non – essential amino acids.

Eg:   Glycine H2N – CH2 – COOH

Write a brief note on the structure of proteins at four different levels.

Ans:- The structure of proteins and their shapes are studied at four different levels as i) Primary ii) Secondary     iii) Tertiary       iv) Quaternary structures.

i) Primary Structure: A protein is formed by combining different amino acid molecules in a specific sequence. The sequence of different amino acid in a protein gives the Primary structure. Any change in the sequence of aminbacid produces a different protein.

ii) Secondary structure: The structure which a protein molecule assumes as a result of bonding between two or more polypeptide chains is known as Secondary structure. The secondary structure is also produced through formation of disulphide (-S – S-) linkages.

The segments of the protein back bone fold either a-helix (3-pleated sheet or coil conformation).

iii) Tertiary structure: It is the three dimensional arrangement of all the atoms in the protein. The amino acid residues lying at the top and bottom of the helix are held together by hydrogen bond and thus gives rise to tertiary structure. Further folding of secondary structure occurs in fibrous and globular shapes.

iv) Quaternary structure: Proteins having more than one peptide chain are known as oligomers. These individual chains are called sub units. The arrangement of subunits gives the quarternary structure.

 

What are peptides? Give an example.

Ans:- Amino acids contain both carboxylic acid group (- COOH) and amine group (- NH2). The amine group of one amino acid reacts with the carboxyl group of another amino acid molecule to form an amide bond. Amide linkages between amino acids are known as Peptide linkages or peptide bonds. The product formed is called Peptide, Depending on the number two, three four and many amino acid molecules linked they are called di, tri, tetra and polypeptides respectively.

Explain the denaturation of protein.

Ans:- Breaking the highly organised tertiary structure of a protein is called denaturation. During the denaturation of the native protein, the biological activity of protein is lost but the primary structure does not change. Denaturation may be reversible or irreversible. The coagulation of egg white on boiling is an irreversible denaturation. Renaturation is the reverse of denaturation.

Denaturation can be carried out by changing the pH or by heating or by adding reagents like urea or by adding detergents.

Explain enzymes.

Ans:- Enzymes are naturally occurring simple or conjugate proteins with some exceptions. An enzyme molecule may contain a non – protein component called prosthetic group. The prosthetic group that is covalently bonded with the enzyme component is called cofactor.The prosthetic groups attached at the time of reaction are called coenzymes. Enzymes act as specific catalysts in biological reactions. Enzyme catalysed reactions involvethe formation of complex (ES) between enzyme and substrate, conversion of the complex to an enzyme – intermediate complex (EI), further conversion to a complex between enzyme and product (EP). dissociation of the enzyme product complex leaving the free enzyme. The enzymes are deactivated during the reactions and they should be replaced by synthesis in the body.

What are ribose and deoxyribose ? Give their structures.

Ans:- The ribose and deoxyribose are’carbohydrates containing 5 carbons. They are .pentoses. Deoxyribose differs from ribose in not having an -OH on C – 2.

What are nucleosides? Give examples.

Ans:- The pentose sugar combined with a base is called nucleoside

Nucleosides are named as adenosine, guanosine, citidine, thymidine and uridine when they contain adenine, guanine, cytosine, thymine and uracil respectively.

If the pentose sugar is deoxyribose, the word ‘deoxy’ is added before the name of nucleoside.

Write the structure of purine and pyrimidine bases of DNA and RNA.   Pyramidine bases

Ans:-

What are nucleosides and nucletodies ? Write the structures of nucleosides and nucleotides with suitable examples.

Nucleoside:

Ans:-The pentose sugar combined with a base is called nucleoside.

Base + Sugar→Nucleoside

Nucleosides are named as adenosine, guanine,   citidine, thymidine and uridine when they contain adenine, guanine, cytosine, thymine and uracil respectively.

Nucleotide:

A nucleotide is a phosphate ester of a nucleoside. A nucleotide contains a pentose sugar, a purine or pyramidine base and one to three phosphate groups. The phosphate group in a nucleotide is attached either to 51 or to 31  OH group.

Explain the structures of DNA and RNA.

Ans:- Nucleic acids are long chain biopolymers of nucleotides with a polyphosphate ester chain. Nucleic acids are two types, i) Ribonucleic acid ( RNA ) and (ii) Deoxyribonucleic acid ( DNA ).

DNA contains deoxyribose, a pentose sugar, phosphoric acid and purine bases adenine (A), guanine (G) or pyramidine bases thymine (T) and cytosine (C). RNA contains ribose, a pentose sugar, phosphoric acid and bases that present in DNA except thymine and in the place of thymine it contains uracil.

When a pentose sugar combines with a base it is called nucleoside.

Base + Sugar à Nucleoside

When a nucleoside combines with a phosphate group it is called  nucleotide.

Base + Sugar + Phosphate à Nucleotide.

When several nucleotides are combined through phosphate ester group  nucleic acids are formed. The nucleic acid chain can be shown as follows.

While nucleic acids are formed the nucleotides combine at 3 of one pentose molecule to 5 of another pentose molecule through phosphate ester group.

In the formation of deoxyribonucleic acid (DNA) the amount of adenine is equal to that of thymine (A = T) and the amount of cytosine is equal to that of guanine (C = G). The total amounts of purine bases are equal to the total amount of pyramidine bases (A + G = C + T). The ratio of AT/ GC varies from species to species.

X-ray diffraction studies indicate that DNA has double helical structure, composed of two right handed polynucleotide chains(strands) around same central axis and run antiparallel.Their5′ → 3′ phosphate diester linkages run in opposite directions. The bases are placed in an order helix, perpendicular to helical axis. The two strands are held together by hydrogen bonds between bases. Adenine “forms two hydrogen bonds with thymine and guanine forms three hydro­gen bonds with cytosine.

The diameter of the double helix is 2nm and the double helical structure repeats at an interval of 3.4 nm (one complete turn).

Explain heredity and genetic code.

Ans:- Nucleic acids control heredity. DNA preserve the information and uses in the following manner.

  1. DNA molecules can duplicate themselves so that they can synthesise other DNA molecules identical with the original. This process is known as replication
  2. A single strand of DNA acts as a template on which a molecule of RNA is synthesised in a specific manner. This process is called as Transcription
  3. RNA molecules directs the synthesis of specific proteins which are characteristic of each type of organism. This process is called translation

 

The base sequence in DNA controls the sequence of amino acids in the protein and therefore the hereditary. The hereditary message is written in a language of only 4 letters, in a code with each word of 3 letters called codon, standing for a particular amino acid.

Explain the transcription of nucleic acids.

Ans:- The synthesis of RNA from a DNA is called transcription. The double helix of DNA partially uncoils and on one of the two strands, a chain of RNA is formed. Here a ribose sugar is incorporated against a deoxyribose sugar of the parent DNA strand and a uracil is added opposite each adenine of DNA. The newly formed chain of RNA is complementary to a segment of the DNA chain.

What is translation of nucleic acid, DNA?

Ans:- Translation is the process by which the genetic message in DNA, that has been passed to mRNA is decoded and used to build proteins. A protein is synthesised from N-terminal end to its C-terminal end by the sequence of bases along   m  RNA strand in the

5→3 direction. A sequence of three bases, called codon specifies a particular amino acid, that is to be incorporated into a protein. The amino acid specified by each three base sequence is called genetic code.

A difference of simple base is the DNA molecule or a single error in the reading of the code can cause a change in the amino acid sequence which leads to mutation. Every t-RNA molecule has an amino acid attachment site having three complimentary nucleotides for recognition of the triplets in mRNA (anticodon).

Discuss the classification, structure and functions of lipids.

Ans:- Lipids are naturally occurring compounds related to fatty acids and include fats, oils, waxes and other related compounds.

Lipids are classified into three types.

  1. i) Simple or homolipids.
  2. ii) Compound or heterolipids.

iii) Derived lipids.

  1. Simple lipids: These are long chain fatty acid esters of trihydric alcohol i.e., glycerol. These include neutral fats and waxes. Liquids are known as oils and solids are known as fats. Structure is

CH2 OOCR1

|

CHOOCR2

|

CH2 OOCR3

R1, R2 and R3 may be same type or different type of alkyl groups of saturated or unsaturated fatty acids. If R1, R2 and R3 are different it is called mixed fat.

Waxes are the esters of long chain fatty acids with long chain monohydric alcohol. The fatty acid contain C-14 to C-36 • The alcohol may range between C-16 and C-36.

Eg : Bee wax is an ester of palmitic acid ( C15 H31 COOH ) and myricyl alcohol

( C30H 61OH ).

  1. Compound or heterolipids: These contain addition groups which include phospholipids or phophosides, glycolipids and terpenes.
  2. i) Phospholipids : These are made by fatty acid, glycerol or other alcohol, nitrogenous base and phosphoric acid. Its structure.

CH2O – fatty acid

 |

CHO – fatty acid

|

CH2O – O – P-base.

( P = phosphoric acid )

These are further classified into

  1. a) Glycerophosphatides: These contain glycerol, fatty acid, phosphoric acid and a base-like choline, serine etc.
  2. b) Phosphoinosides: In these, the cyclic hexahydric alcohol (inositol) replaces the base.
  3. c) Phosphosphringosides: In these, the glycerol is replaced by complex amino alcohol
  4. ii) Glycolipids: These do not contain phosphoric acid. These contain sphingol, carbohydrate and fatty acid.

iii) Terpenes: These are polymers of isoprene. The side chains of vitamin A, E and K and carotene belong to this group.

  1. Derived fats : These are hydrolysis products of some simple and compound lipids. The products include glycerol, fatty acids, sphingosime, steroids, terpenes and carotenoids

Functions of liquids :

  1. They are energy sources in the body.
  2. They are structural components of cell membrane.
  3. They are heat insulators.
  4. They are emulsifiers used as detergents to emulsify fats for transport of connective tissues.
  5. Vitamin carriers of A, D, E.
  6. They are enzyme, activators.

What are hormones ? How are they classified? Give two functions of any four hormones.

Ans:- Hormones are molecules of carbon compounds that transfer biological information from one group of cells to distant tissues of organs.

Classification of hormones :

Hormones are classified into three groups on the basis of their site of production.

1)  Steroid hormones: These are produced by the adrenal cortex, testis and ovary.

2)  Protein hormones: These are produced by pancreas, parathyroid, pituitary and the gastrointes­tinal mucosa.

3)  Amino acid derivatives: These are produced by thyroid and adrenal medulla.

The classification basing on structure.

Functions of hormones:

  1. Testosterone: This is responsible for the development of male secondary sexual characteristics like deep voice, facial hair, sturdy physical structure.
  2. Estradiol: It is responsible for the development of secondary female sex characteristics’ like breast development, shrill voice and long hair.   It also controls the menstrual cycle.
  3. Progesterone: is useful for preparing the uterus for the implantation of the fertilized egg. It is also useful as birth control agent.
  4. Insulin: It is responsible for the entry of glucose and other sugars into the living cell.  It helps in the decrease of glucose in the blood.

What are plant hormones ? What are functions of plant hormones ?

Ans:- Plant hormones are carbon compounds   produced by plants.   They regulate, .growth and physiological functions at a site remote from the place of secretion. Plant hormones are also called as phytohoraiones. There are no specific glands for the secretion of plant hormones.  Plant hormones are 1) Auxins 2) Gibberlins 3) Cytokinins 4) Ethylene 5) Traumatic acid 6) Absicic acid 7) Morphachin.

  1. Auxins promote growth along longitudinal axis.   These are cyclic compounds and pentene derivatives..
  2. Gibberlins cause bolting ( shoot elongation ) and flowering. These are derivatives of gibbane.
  3. Cytokinins stimulate cell division, promote cell elongation, induces flowering in short day plants These are purine derivatives.
  4. Ethylene accelerates the colouring of the harvested fruits. Induces rooting and flowering. It is an unsaturated hydrocarbon.
  5. Traumatic acid induces cell division. This is an open chain dicarboxylic acid with one double bond.
  6. Absicic acid induces prototropism. It is a sesquiterpene.
  7. Morphactin inhibits mitosis. Formation of branches is promoted. These are derivatives of fluorine – 9 – carboxylic acid.

What are vitamins? How are they classified? Give the sources and deficiencies caused hy A, D, C, B, K.

Ans:-Vitamins are organic, organometallic or metalloorganic compounds which are essential in small amounts in the metabolic processes of most living systems. Their absence in the human body causes deficiency diseases or disorders.

Classification: Vitamins are classified into two groups depending on their solubility.

  1. Fat soluble vitamins: Vitamins A, D, E and K are fat soluble vitamins.
  2. Water soluble vitamins: Vitamins C and B-complex are water soluble vitamins.

Sources and diseases due to deficiency:

S

Vitamin Source Deficiency disease
A Fish oil, liver, rice Night blindness, Redness in eyes,
(Retinol) polishes, kidney. degeneration of lachrymal glands
D Fish oils, butter, milk, egg Rickets in children osteomalacia in adults.
(Calciferol).    
C Green leafy vegetables, Scurvy, delay in wound healing.
{Ascorbic acid) citrus fruits.  
K Green leafy vegetables Blood coagulation is prevented
(Antihaemorrhagic) intestinal flora in continuous bleeding
B1 Cereals, outer brain ayers, yeast, Beri Beri
(Thiamin) 1 milk, green vegetables.  
B2 Yeasts vegetables, milk, Dark red tongue, dermatitis, chelosis
(Riboflavin) egg white, liver and kidney. fissuring at corners of mouth and lips.
B3 Present in ail foodstuffs. Burning feet
(Pentothenic acid)    
B5 Meat, yeast, milk, green Pellagra dermatitis, diarrhoea.
(Nicotinic acid or Niacin) leafy vegetables.  
B6 Cereals, grams, yeast, Dermatitis convulsion
(Pyridoxine) egg yolk, meat.  
B7 Yeast; liver kidney Dermatitis, Blood cholesterol
(Biotin). milk. increases, Loss of hair, Paralysis.
B9 Spinach leaf intestinal Anaemia, inflammation of tongue,
(Folic acid) bacteria. gastro intestinal disorder.
B12 Liver of ox, pig, fish etc., Pernicious anaemia
(Cyanocobalamine)   hyperglycemia.

 

Explain cyclic structures of glucose.

Glucose occurs as six membered cyclic ring known as pyranose structure. In cyclic form C – 1 is assymetric and hence glucose exists in two stereoisomeric forms, the a – glucopyranose and p -glucopyranose. The glucose with OH group on right side is a – form and that with OH on left side is (3 – form. Such pair of stereoiosmers which differ is configuration only at C -1 are called anomers.

Explain the importance of carbohydrates

1)   Carbohydrates are essential for the life of both plants and animals.

2)   Honey has been used for a long time as instant source of energy.

3)   Carbohydrates are used as storage molecules.

Ex :   Starch in plants, glycogen in animals.

4)   Cell walls of bacteria and plants are made up of cellulose.

5)   D-ribose and 2 – deoxy – D – ribose are present in nucleic acids.

6)   Antibiotics such as streptomycin, kenamycin, neomycin and gentamycins are            carbohydrate antibiotics.

 

Discuss the structure of fructose.

Fructose has molecular formula O2H12O6 ft contains ketonic group at carbon number 2.1t is laevorotatonjj and belongs to D – series, hence written as D –  H – fructose..

Fructose exists in two cyclic furanose structures, a – D –(-) fructosefuranose and ÷β – D -(-) – fructo-furanose. The Haworth structures of fructose are as given below.

Explain DNA finger printing.

The sequence of bases in DNA is unique for a person and this sequence cannot be altered by any treatment. The information regarding this sequence is called DNA finger printing. DNA finger printing is used for

1)   identification of criminals

2)   determining the paternity

3)   identification of dead bodies.

4)   identification of racial groups.

 

What are lipids ? How are they classified ?

Lipids are naturally occurring compounds related to fatty acids and include fats, oils, waxes and other related compounds. Lipids are classified into three types,

  1. i) Simple or homolipids ii) Compound or heterolipids iii)  Derived lipids.

Name one example each for different types of lipids.

Lipids are three types.

  1. i) Simple or Homolipids : These include neutral fats .and waxes.
  2. a) Neutral fat: Triglycerides or triacyl glycerols.

CH2OOCR1

I

CHOOCR2

I

CH2OOCR3

R1, R2 and R3 may be same or different alkyl groups of saturated or unsaturated fatty acid.

  1. b) Waxes : These are esters of long chain fatty acid. Eg : Bee wax
  2. ii) Compound or Heterolipids: These contain additional groups, which include phospholipid or phosphotids and glycolipids.

iii) Derived fats : These are the hydrolysis products of simple and compound lipids. The products include glycerol, fatty acids, sphingosine, steroids, terpenes, and carotenoids.

Give any four functions of lipids.

  1. 1. In the body the fats serve as an efficient source of energy.
  2. Phospholipids serve as structural materials of cells and tissues such as cell membrane.
  3. Lipids are essential for the absorption of fat soluble vitamins such as A, D, E and K. 4. They act as heat insulators.
  4. They act as enzyme activators.

What are vitamins ? How are they classified ?

Vitamins are organic, organometallic or metalloorganic compounds which are essential in small

amounts in the metabolic processes of most living systems. Their absence in the human body causes deficiency diseases or disorders. Vitamins are two types.

  1. Fat soluble vitamins : Vitamins A, D, E and K are fat soluble vitamins.
  2. Water soluble vitamins : Vitamins C and B-complex. are water soluble vitamins.

 

Give the sources and deficiency diseases of a) Vitamin A b)Vitamin D c)Vitamin C     d)Vitamin E    e)Vitamin K

  1. Vitamin A is present in fish oils, liver, rice polish, kidney. Deficiency of vitamin A causes night blindness, redness in eyes, degeneration of lachrymal glands.
  2. Vitamin D is available in fish oils, butter, milk and egg. Its deficiency causes rickets in children, osteomalacia in adults.
  3. Vitamin C is present in green leafy vegetables and citrus fruits. Its deficiency causes scurvy and in wound healing.
  4. Vitamin E is present wheat germ oil, vegetable oils, egg yolk- and green leafy vegetables. Its  deficiency causes sterhty nutritional nuclear dystrophy,- neurosis of heart muscles.
  5. Vitamin K is present in green leafy vegetables and intestinal flora. Its deficiency causes the prevention of blood coagulation during continuous bleeding.

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