CHAPTER-14 BIOMOLECULES

Key Points

1.Carbohydrates- These are optically active polyhydroxy  aldehydes or ketones due to presence of chiral `C’ or the compounds which produce these on hydrolysis except dihydroxy acetone is not optically active.

2. Classification-

(i) Monosaccharide’s – Those carbohydrates which cannot get hydrolysed e.g. glucose ,fructose, galactose etc.

(ii) Oligosaccharides- Those carbohydrates which give to or more monosaccharide’s on hydrolysis e.g. sucrose on hydrolysis gives glucose and fructose. Raffinose on hydrolysis gives glucose, fructose and galactose.

(iii) Polysaccharides- Those carbohydrates which on hydrolysis give large number of monosaccharide’s hydrolysis.eg  starch, cellulose, glycogen.

3. Sugar-

(i)Reducing Sugars- Those which reduce Fehling’s or Tollen’s reagent. They have free aldehydic groups, eg , glucose, fructose , galactose

(ii)Non Reducing Sugars- Those which do not reduce Fehling’s or Tollen’s reagent. They do not have free functional group ,e.g.,  sucrose

4. Glucose- It is a monosaccharide’s with molecular formula C₆H₁₂O₆

 (ii) Cyclic Structure Of Glucose: the straight chain is unable to explain the following reactions.

 (a) It does not give the 2, 4-DNP test, Schiff’s Test and does not form the hydrogensulphite product with NaHSO₃ .

(b) The pentacetate of glucose does not react with NH₂OH, indicating the absence of free aldehydic group.

 (iii) Glucose exist in 2 different crystalline forms α and β forms. These are called anomers. They differ in optical rotation, they also differ in melting point.

Anomers are isomers which have a different configuration across  C-1 (first chiral ‘C’ atom).

 7. Glycosidic Linkage: The linkage between two monosaccharide units through oxygen is called the     glycosidic linkage.

8. Proteins: These are micro molecules made up of amino acids joined via a peptide link ( -(-CONH-)- is the peptide linkage). These are required for growth and development of the body.

9. Amino Acids: These contain an amino (-NH₂) and an acidic (-COOH) group and are therefore amphoteric in nature. In solution they exist in the form of zwitter ion.

10. Classification       

Fibrous Protein	Globular Protein
(i) Polypeptide chains run parallel or anti-parallel and held together by hydrogen and disulphide bonds.	(i) Chains of Polypeptide coil around  to give a spherical shape.
(ii) Generally insoluble in water. e.g. Keratin, collagen,myosin,fibroin.	(ii) Usually soluble in water. e.g., insulin, thyroglobin,albumin, haemoglobin and fibrinogen gets converted into fibrous protein fibroin on clotting of blood.

11. Structure And Shape of Protein

Primary Strutcure	Secondary Structure
The specific sequence of amino acids in the polypeptide chain. Change in amino acids sequence changes the protein. They have covalent bonds.	It is the shape in which the long polypeptide chain can exist. It is of two types : α- helix and β- pleated. These structures arise due to regular folding of the backbone of the polypeptide chain due to H-bonding between the C=o and –NH- groups of the peptide bond.

Tertiary Structure	Quaternary Structure
Represents overall folding of the polypeptide chain. It gives rise to the fibrous or globular molecular shapes. Forces stabilizing the 2^o and 3^ostructures are hydrogen bonds, disulphide linkages, van der waal’s and electrostatic forces of attraction.	Protein can be composed of two or more polypeptide chains called sub units. The spatial arrangement of these sub units with respect to each other quaternary structure of the protein.

12. Denaturaion  of Protein: The protein in native state, when subjected to a physical change like temperature, pH etc undergoes uncoiling and looses it’s biological activity. The 2^o and 3^o structures are destroyed, only  1^o structure is retained.

Renaturation of Protein:

Some proteins regain their biological activity by reversible process it is called Renaturation of Proteins. In such a cases, when temperature in pH of a denatured proteins is brought back to conditions in which the native protein is stable, secondary and tertiary structures of proteins are restored to which leads to recovery of biological activity.

13. Enzymes: These are biocatalyst and generally globular proteins e.g., invertase,zymase,phenyl, alaninehydroxylase, urease etc.

14. Vitamins: They are organic compounds required in the diet in small amounts to perform specific biological functions for maintenance of optimum growth and health of the organism. They are classified as follows

(i) Fat Soluble Vitamins: Vitamin A,D,E and K. They are stored in liver and adipose tissues.

(ii) Water Soluble Vitamins: B group vitamins and vitamin C.  They need to supplied regularly in diet as they are excreted in urine and cannot be stored(except vitamin B₁₂) in our body.

Their deficiency causes diseases.

Biotin (Vit H) is however neither fat nor water soluble. It’s deficiency leads to loss of hair.

15. Nucleic Acids: These are biomolecules which are long chain polymers of nucleotides. They are:

(i) Deoxyribonucleic acid (DNA)

(ii) Ribonucleic acid (RNA)

They are responsible for protein synthesis and transfer of genetic characteristics to offspring’s.

16. Composition of Nucleic Acid:

They are made up of pentose sugar(β-D-2-deoxyribose in DNA and β-D-ribose in RNA), phosphoric acid and a  nitrogen containing heterocyclic compound(base).

DNA- Bases present are Adenine(A), Thymine(T),Guanine(G) and Cytosine(C).

RNA- contains Adenine(A), Guanine(G), Cytosine(C) and Uracil(U).

17. Nucleoside: The unit formed by the attachment of a base to n1’-position of sugar (Base+Sugar).

18. Nucleotide: Nucleoside and phosphoric  acid at 5’-position. Nucleotides are bonded by phosphodiester linkages between 5’ and 3’ carbon atoms of pentose sugar (Base+ Sugar+ Phosphoric Acid).

19.DNA : has a double helical structure with AT and GC linked together through 2 and 3 hydrogen bonds respectively. It is responsible for transfer of genetic characteristics.

20. RNA: is of three types- messenger RNA(m-RNA), ribosomal RNA(r-RNA) and transfer RNA (t-RNA). RNA helps in protein synthesis.

21. Biological Functions of Nuclei Acid: DNA is chemical basis of hereditary and have the coded message for proteins to be synthesized in the cell. RNA carry out the protein synthesis in the cell.

Questions

Q1 – Which functional groups are present in monosaccharides?

Ans -  ―OH and ―CHO or ―OH and >CO

Q2 – Name an aldopentose, aldohexose and ketohexose.

Ans – Ribose,glucose and fructose respectively.

Q3 – What is animal starch?

Ans – Glycogen.

Q4 – Which types of bonds are present in a protein molecule?

Ans – Peptide bonds, hydrogen bonds, sulphide bonds, ionic bonds etc.

Q5 – Which is more stable? and why

            α-helix ,  β-helix

Ans – α-helix is right handed coil and is more stable due to intermolecular H bonding between first and fourth amino acid.

Q6 – The sequence of bases in one strand of DNA is TACGGACA. What is the sequence of bases of complementary strand of DNA.

Ans – ATGCCTGT.

Q7 – Name the vitamin whose deficiency causes rickets?

Ans – Vitamin D.

Q8 – Name the purines present in DNA.

Ans – Adenine and guanine.

Q9 – Give an example of (a)water soluble (b)fat soluble vitamins.

Ans – (a)Vitamin C  (b)Vitamin D.

Q10 – Name a protein which is insoluble in water.

Ans – Keratin.

SAI Type Questions – (2 Marks)

Q1 – Write the components of starch and  what  is the difference between them?

Ans – Amylose which is linear polymer of α-glucose and amytopectin which is branched polymer of α-

              glucose. Amylose is water soluble where as amylopectine is water insoluble.

Q2 – What are anomers?

Ans – Monosaccharides which differ only in the orientation of the ―OH group at C-1.e.g, α-glucose and β-

         glucose.

Q3 – Where does the water present in the egg go after boiling the egg?

Ans – On boiling during denaturation process water makes inter molecular H- bond with denatured protein

          molecule.

Q4 – What do you understand by glycosidic linkage?

         Ans – During condensation of two monosaccharides, a water molecule given out and two

         monosaccharides get linked together by an oxide or ethereal linkage (―O―) called as glycosidic

        linkage.

Q5 – What are essential and non essential amino acid? Give two examples of each type.

       Ans – Essential amino acids are those which are not produced in our body.e.g.,valine,leucine.

       Non-essential amino acids are those which are produced by our body.e.g.glycine and alanine.

Q6 – How do you explain the amphoteric behavior of amino acids?

     Ans – Amino acids have both acidic as well as basic group and they react both with acids as well as

        bases,therefore they are amphoteric in nature.

Q7 – What is the structural difference between a nucleoside and a nucleotide?

         Ans -  Nucleoside = sugar + base

         Nucleotide = sugar + base + phosphoric acid

Q8– Define (a)Enzymes (b)Antibody

Ans – (a)Enzymes – they are biological catalyst which catalyse biochemical reactions.e.g.,

       C₁₂H₂₂0₁₁ + H₂0     ---------->   C₆H₁₂0₆ + C₆H₁₂0₆

          ^{sucrose  glucose fructose}

This reaction is catalysed by the enzyme invertase.

(b)Antibody – they are chemical substances which destroy antigens that cause infections.e.g.,vaccination for typhoid produces antibodies in our body to prevent typhoid.

Q9 – What is invert sugar?

Ans – An equimolar aqueous solution of glucose and fructose is called invert sugar.

                                         SAII Type Questions – (3 Marks)

Q1 – Give three differences between DNA and RNA.

Ans –

DNA	RNA
1. it has deoxyribose as sugar. 2. it contains thymine along with adenine, cytosine and guanine as bases. 3. it is responsible for maintaining heredity traits from generation to generation.	1.it contains ribose as sugar. 2.it contains uracil in place of thymine with other bases. 3. it is responsible for protein   synthesis.

Q2 – Difference between globular protein and fibrous protein.

Ans –

Globular Protein	Fibrous Protein
1. they form α-helix structure. 2. they are water soluble. 3. they involve intra molecular H bonding.	1. they have β-pleated structure. 2. they are water insoluble. 3. they have strong intermolecular forces of attraction.

Q3 – Give reactions with support cyclic structure of glucose.

Ans – (a)Glucose does not give 2,4-DNP test, Schiff`s test and sodium hydrogen suphide test.

          (b)The pentaacetate of glucose does not react with NH₂OH indicating absence of free ―CHO group.

            (c)Glucose exists in two crystalline form α and β.

Q4 – Define with example

(a)Isoelectric point (b)Mutarotation (c)Transcription

Ans –

  (a)Isoelctric point – the pH at which there is no net migration of any ion towards electrode.e.g,amino acids  have isoelectric point at pH = 5.5-6.3

(b)Mutarotation -  it is spontaneous change in optical rotation when an optically active substance is dissolved in water.e.g, α-glucose when dissolved in water changes its optical rotation from 111^o to 52.5 ^o.

(c)Transcription – it is process by which m―RNA is generated from DNA.

Q5 – What happens when glucose reacts with

         (a)HI             (b) HNO₃            (c)Br₂ water  (d) NH₂OH   (e) Acetic anhydride

Ans –

(a) C₆H₁₂0₆  + HI ------> n-hexane C₆H₁₄

(b) C₆H₁₂0₆ + HNO₃ -------> saccharic acid / Glucric acid

(c) C₆H₁₂0₆ +Br₂ water --------> gluconic acid

(d)  C₆H₁₂0₆ +Br₂ water --------> gluconic acid

(e) C₆H₁₂0₆ + NH₂OH -------->      Oxime

(f) C₆H₁₂0₆ + Acetic anhydride --------> glucose penta acetate

Q6. Mention structural differences between amylopectin and cellulose.

Ans.

Amylopectin	Cellulose
1. It is linear polymer of α-glucose. 2. It consists of branched chains of  α-glucose.	1. It is linear polymer of  β-glucose. 2. In cellulose, the chains are arranged to form bundles and held together by hydrogen bond between glucose and adjacent strands.

  Q7. What deficiency diseases are caused due to lack of vitamins B₁ ,  B₆  and K in human diet.

Ans.

Vitamins	Deficiency Disease
B1	Beri beri (loss of appetite)
B6	Convulsions
K	Increased blood clotting time

Q8. Glucose or Sucrose are soluble in water but cyclohexane and benzene are insoluble in water. Explain.

Ans. Glucose contains five-OH groups and Sucrose contains eight-OH groups, because of this they form intermolecular hydrogen bonding, so they are soluble in water. But benzene and cyclohexane does not contain –OH groups, hence does not form intermolecular hydrogen bonding, so they are not soluble in water.

HOTS Questions

Q1. Write the formula of Zwitter ion for Glycine.

Ans.

            R-CH-COO^-                                                                                    CH₂-COO^-

     |                                                      |

                NH₃⁺                                                                                               NH₃⁺

            General Formula                          Zwitter ion of glycine

Q2. Which proteins possess α-Helix structure?

Ans. Keratin and myosin possess α-Helix structure.

Q3. What is the native state of protein?

Ans. The energetically most stable shape of the protein at normal pH and temperature is called native state.

Q4. Fresh tomatoes are a better source of Vitamin C that which have been stored for some time. Explain.

Ans. Vitamin C is destroyed on prolonged exposure to air due to its oxidation.

Q5. Why are carbohydrates generally active?

Ans. It is due to the presence of Chiral Carbon atoms in their molecules.

Q6. What type of linkages hold together monomers in DNA?

Ans. Monomers in DNA are linked by phosphate linkages.

Q7. Why is cellulose not digested in human body?

Ans. It is due to the fact that human beings do not have enzyme to digest cellulose.

Q8. Name the enzyme that is used to dissolve blood clots?

Ans. Streptokinase.

Q9. Name two diseases caused due to deficiency of enzymes.

Ans. Albinism and phenylketonuria.

Q1. Give reasons for the following-

(i) On electrolysis in acidic solution amino acids migrate towards cathode, while in alkaline solution these migrate towards anode.

(ii) The monoamino monocarboxylic acids have two pK_a values.

Ans. (i) In acidic solution, the carboxylate anion accept a proton and gets converted into carboxylic group resulting in the formation of positive ion.

In presence of a base the NH₃⁺ ion changes to —NH₂ group by losing a proton and this gives a negative ion.

This means that in acidic medium, the amino acid migrates towards the cathode while in alkaline solution it migrates towards anode on electrolysis.

(ii) In aqueous solution, monoamino monocarboxylic amino acid behave like salt at isoelectric point. At a pH lower than isoelectric point (i.e. in acidic medium) it act as cation and

and at a pH higher than isoelectric point,it acts as anion.        

Q2. Which forces are responsible for the stability of α-helix? Why is it named as 3.6₁₃ helix?

Ans. Hydorgen bonds between –  N-H and —C=O groups of peptide bonds give stability to the structure.

It is known as 3.6₁₃ helix, since each turn of helix has approximately 3.6 amino acid residue and a 13 member ring is formed by hydrogen bonding.

Q3. Write about the following protein synthesis-

       (i) Name the location where the protein synthesis occurs?

Ans. Protein synthesis occurs at the ribosome in cytoplasm.

(ii) How do 64 codones  code for only 20 amino acids?

Ans. The 64 codones for 20 amino acids; more than one codon can code for same amino acids, e.g., CUU and CUU both can code leucine. Proline is encoded by CCU, CCA, CCG, and CCC.

Q4. What is replication of DNA.

Ans.  The process by which a DNA molecule produces two identical copies of itself is called replication of   DNA

Q5. Answer the following queries about proteins-

(i) How are proteins related to amino acids?

Ans. Proteins consist of large number of amino acids linked to each other by peptide linkage, having 3- dimensional structure. Thus, proteins are biopolymers of amino acids.

(ii) When is protein said to be denatured?

Ans. When nature proteins are subjected to the action of heat, acids or alkalies, they are coagulated or precipitated. The protein in this state is said to be denatured. During denaturation process the water soluble form of globular protein change to water  insoluble fibrous protein.

Q6. Consider D and L- glucose. Are these enantiomers?

Yes , these two fischer projections are called enatiomers.

Q7. Glycine exists as a Zwitter ion but 0-and p-amino benzoic acids do not. Explain.

Ans. The lone pair of N-atom in O- and p-aminobenzoic acid is involved in resonance. The lone pair of N-atom in transferred towards benzene ring. This decreases the acidic character of –NH₂ group. Therefore these groups do not transfer and accept H⁺ ions, respectively.

Q8. The melting points and solubility in water of amino acids are generally higher than that of the corresponding halo acids. Explain.

Ans. The amino acids exists as zwitter ion (H₃N—CHR—COO^-). They have salt like structure. There are strong dipole-dipole and electrostatic attractions. Therefore, amino acids have high melting points. Amino acids strongly interact with water molecules and are soluble in it. The halo-acids do not have salt like structure and have low melting points. Halo-acids do not interact as strongly with water molecules as do amino acids. Therefore, solubility of amino acids in water is more than those of halo-acids.

Q9. A tripeptide on complete hydolysis gives glycine, alanine and phynylalanine using three letter symbols write down the possible sequence of tripeptide.