Biomolecules

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53 Terms

1

Biomolecules

  • Chemical compounds found in living organisms.

  • Responsible for their growth, maintenance and ability to reproduce.

    Examples:

  • Carbohydrates

  • Proteins

  • Nucleic acids

  • Lipids

  • Enzymes

  • Vitamins

  • Hormones

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Carbohydrates or Saccharides

  • These are the hydrates of carbon and most of them have a general formula Cₓ(H₂O)ᵧ (Old definition)

  • Polyhydroxy aldehydes or ketones or the compounds which produce carbohydrates on hydrolysis. (Modern definition) e.g.: glucose, sucrose, cellulose, starch etc.

  • Optically active compounds

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Classification of Carbohydrates

  • Based on their behaviour on hydrolysis

  • Based on reducing character

  • Based on their physical properties

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Based on Their Behavior on Hydrolysis

  1. Monosaccharides

    • Simplest carbohydrates which cannot be further hydrolysed

    • Eg: Glucose, Fructose, Galactose, Ribose etc

  2. Oligosaccharides

    • Carbohydrates which give 2-10 units of monosaccharides on hydrolysis

    • Eg: Sucrose, Maltose, Lactose

    • This can be further classified into:

      • Disaccharides:

        • Sucrose (hydrolysed into Glucose + Fructose)

        • Maltose (hydrolysed into 2 molecules of Glucose)

        • Lactose (hydrolysed into Glucose + Galactose)

        • Rafinose (hydrolysed into Glucose + Fructose + Galactose)

      • Trisaccharides

      • Tetrasaccharides

  3. Polysaccharides

    • Carbohydrates which give a large number of monosaccharide units on hydrolysis.

    • Eg: Starch, Cellulose, Glycogen, Gums etc.

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Invert Sugar

  • Cane sugar is sucrose, which on hydrolysis gives an equimolar mixture of D(+)glucose and D(-)fructose.

  • C₁₂H₂₂O₁₁ + H₂O → C₆H₁₂O₆ (D(+)Glucose, +52.5°) + C₆H₁₂O₆ (D(-)Fructose, -92.4°)

  • Sucrose is dextro rotatory but after hydrolysis gives dextro rotatory glucose and laevo rotatory fructose. Since the laevo rotation of fructose (-92.40°) is more than the dextro rotation of glucose (+52.50°), the mixture is laevo rotatory.

  • Thus, hydrolysis of sucrose brings about a change in the sign of rotation from dextro (+) to laevo (-), and the product is named as invert sugar. The process is called inversion of cane sugar.

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Based on Reducing Character

  1. Reducing Sugars

    • Carbohydrates which contain free aldehydic or ketonic group and reduce Fehling’s Solution and Tollen’s Reagent

    • Eg: All monosaccharides, disaccharides like maltose and lactose

  2. Non-Reducing Sugars

    • Carbohydrates which do not have an aldehydic or ketonic group and do not reduce Fehling’s Solution and Tollen’s Reagent

    • Eg: Sucrose, all polysaccharides

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Based on Their Physical Properties

  1. Sugars

    • Carbohydrates which are sweet in taste, crystalline and water soluble.

    • Eg: all monosaccharides and disaccharides

  2. Non-Sugars

    • Carbohydrates which have no sweet taste, not crystalline and water insoluble.

    • Eg: all polysaccharides

<ol><li><p><strong>Sugars</strong></p><ul><li><p>Carbohydrates which are sweet in taste, crystalline and water soluble.</p></li><li><p>Eg: all monosaccharides and disaccharides</p></li></ul></li><li><p><strong>Non-Sugars</strong></p><ul><li><p>Carbohydrates which have no sweet taste, not crystalline and water insoluble. </p></li><li><p>Eg: all polysaccharides</p></li></ul></li></ol><p></p>
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Monosaccharides Classification

Based on Functional Group

  1. Aldose: Monosaccharide containing an aldehyde (-CHO) group.

  2. Ketose: Monosaccharide containing a keto (>C = 0) group.

Based on the Number of Carbon Atoms

Monosaccharides containing 3 carbon atoms are called triose, 4 carbon atoms are called tetrose etc. (see image)

<p><strong>Based on Functional Group</strong></p><ol><li><p><u>Aldose:</u> Monosaccharide containing an aldehyde (-CHO) group.</p></li><li><p><u>Ketose:</u> Monosaccharide containing a keto (&gt;C = 0) group.</p></li></ol><p><strong>Based on the Number of Carbon Atoms</strong></p><p>Monosaccharides containing 3 carbon atoms are called triose, 4 carbon atoms are called tetrose etc. (see image)</p><p></p>
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Glucose

  • Glucose is an aldohexose

  • Molecular formula is C6H1206

<ul><li><p>Glucose is an aldohexose</p></li><li><p>Molecular formula is C<sub>6</sub>H<sub>12</sub>0<sub>6</sub></p></li></ul><p></p>
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Reactions of Glucose

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

When glucose reacts with different reagents, the following products are formed along with key inferences:

  • With HI:

    • Product: n-Hexane

    • Inference: All the 6 carbon atoms are linked in a straight chain.

  • With HCN:

    • Product: Cyanohydrin

    • Inference: Indicates the presence of a carbonyl group.

  • With NH₂OH:

    • Product: Oxime

    • Inference: Confirms the presence of a carbonyl group.

    • A: Pentaacetate of D-glucose in aqueous medium does not form an open chain structure and thus when it reacts with NH₂OH, it does not form oxime indicating that a free aldehyde (–CHO) group is absent.

  • With Bromine water (Br₂/H₂O):

    • Product: Gluconic acid

    • Inference: This shows that the carbonyl group is an aldehyde.

  • With Acetic anhydride:

    • Product: Glucose pentaacetate

    • Inference: Indicates the presence of five –OH groups.

  • With Conc. HNO₃:

    • Product: Saccharic acid

    • Inference: Confirms that one of the –OH groups is primary.

<p>When <strong>glucose reacts with different reagents</strong>, the following products are formed along with key inferences:</p><ul><li><p><strong>With HI:</strong></p><ul><li><p><strong>Product:</strong> <em>n-Hexane</em></p></li><li><p><strong>Inference:</strong> All the 6 carbon atoms are linked in a straight chain.</p></li></ul></li><li><p><strong>With HCN:</strong></p><ul><li><p><strong>Product:</strong> <em>Cyanohydrin</em></p></li><li><p><strong>Inference:</strong> Indicates the presence of a carbonyl group.</p></li></ul></li><li><p><strong>With NH₂OH:</strong></p><ul><li><p><strong>Product:</strong> <em>Oxime</em></p></li><li><p><strong>Inference:</strong> Confirms the presence of a carbonyl group.</p></li><li><p><strong><u>A:</u></strong> Pentaacetate of D-glucose in aqueous medium does not form an open chain structure and thus when it reacts with NH₂OH, it does not form oxime indicating that a free aldehyde (–CHO) group is absent.</p></li></ul></li><li><p><strong>With Bromine water (Br₂/H₂O):</strong></p><ul><li><p><strong>Product:</strong> <em>Gluconic acid</em></p></li><li><p><strong>Inference:</strong> This shows that the carbonyl group is an aldehyde.</p></li></ul></li><li><p><strong>With Acetic anhydride:</strong></p><ul><li><p><strong>Product:</strong> <em>Glucose pentaacetate</em></p></li><li><p><strong>Inference:</strong> Indicates the presence of five –OH groups.</p></li></ul></li><li><p><strong>With Conc. HNO₃:</strong></p><ul><li><p><strong>Product:</strong> <em>Saccharic acid</em></p></li><li><p><strong>Inference:</strong> Confirms that one of the –OH groups is primary.</p></li></ul></li></ul><p></p>
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<p>Open Chain Structure of Glucose</p>

Open Chain Structure of Glucose

But this open chain structure cannot explain the following observations:

  1. Glucose does not react with 2,4-Dinitrophenyl hydrazine, Schiff’s reagent, and with NaHSO₃.

  2. The pentaacetate of glucose does not react with hydroxylamine, indicating the absence of a free —CHO group.

  3. The existence of two different crystalline forms of glucose - (α and β form).

  • In order to explain the above, it was proposed that one of the —OH groups may add to the —CHO group and form a cyclic hemi-acetal structure. The —OH at C₅ is involved in ring formation (1,5-oxide ring).

  • Thus, the two cyclic forms exist in equilibrium with the open chain structure.

<p><strong>But this open chain structure cannot explain the following observations:</strong></p><ol><li><p>Glucose does not react with 2,4-Dinitrophenyl hydrazine, Schiff’s reagent, and with NaHSO₃.</p></li><li><p>The pentaacetate of glucose does not react with hydroxylamine, indicating the absence of a free —CHO group.</p></li><li><p>The existence of two different crystalline forms of glucose - (α and β form).</p></li></ol><ul><li><p>In order to explain the above, it was proposed that one of the —OH groups may add to the —CHO group and form a cyclic hemi-acetal structure. The —OH at C₅ is involved in ring formation (1,5-oxide ring).</p></li><li><p>Thus, the two cyclic forms exist in equilibrium with the open chain structure.</p></li></ul><p></p>
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Anomers

They are stereoisomers which differ only in the configuration at the hemiacetal or hemiketal carbon.

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Pyranose Structure of Glucose

knowt flashcard image
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Fructose

  • Fructose is an important ketohexose also with the molecular formula C₆H₁₂O₆.

  • It belongs to the D-series and is a laevorotatory compound, and it is written as D-(-)-Fructose.

  • Reactions show that fructose has a ketonic group at the second carbon.


<ul><li><p>Fructose is an important ketohexose also with the molecular formula C₆H₁₂O₆.</p></li><li><p>It belongs to the D-series and is a laevorotatory compound, and it is written as D-(-)-Fructose.</p></li><li><p>Reactions show that fructose has a ketonic group at the second carbon.</p></li></ul><div data-type="horizontalRule"><hr></div><p></p>
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<p>Gycosidic Linkage</p><p></p><p><span>Assertion (A) β-glycosidic linkage is present in maltose.<br>Reason (R) Maltose is composed of two glucose units in which C-1 of one glucose unit is linked to C-4 of another glucose unit.</span></p>

Gycosidic Linkage

Assertion (A) β-glycosidic linkage is present in maltose.
Reason (R) Maltose is composed of two glucose units in which C-1 of one glucose unit is linked to C-4 of another glucose unit.

The oxide linkage formed by the loss of a water molecule when two monosaccharides are joined together through oxygen atom is called glycosidic linkage.

A: d) A is false, R is true

since the linkage is between two alpha molecules not beta

<p>The oxide linkage formed by the loss of a water molecule when two monosaccharides are joined together through oxygen atom is called glycosidic linkage.</p><p></p><p>A: d) A is false, R is true</p><p>since the linkage is between two alpha molecules not beta</p>
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Polysaccharides

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Starch

It is a polymer of α -glucose and consists of two components — Amylose and Amylopectin

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Amylose vs Amylopectin

Write one structural differences between Amylose and Amylopectin

Amylose: (remember as amy-less so contributes lesser)

  • Water soluble component

  • Long unbranched chain polymer

  • It contains 200 – 1000 α -D-(+)- glucose units held by α - glycosidic linkages involving C1 – C4 glycosidic linkage

  • Constitutes to about 15-20% of starch

Amylopectin:

  • Water insoluble component

  • Branched chain polymer

  • It forms chain by C1 – C4 glycosidic linkage whereas branching occurs by C1 – C6 glycosidic linkage

  • It constitutes about 80-85% of starch

<p></p><p><strong>Amylose: (remember as amy-less so contributes lesser)</strong></p><ul><li><p>Water soluble component</p></li><li><p>Long unbranched chain polymer</p></li><li><p>It contains 200 – 1000 α -D-(+)- glucose units held by α - glycosidic linkages involving C1 – C4 glycosidic linkage</p></li><li><p>Constitutes to about 15-20% of starch</p></li></ul><p></p><p><strong>Amylopectin:</strong></p><ul><li><p>Water insoluble component</p></li><li><p>Branched chain polymer</p></li><li><p>It forms chain by C1 – C4 glycosidic linkage whereas branching occurs by C1 – C6 glycosidic linkage</p></li><li><p>It constitutes about 80-85% of starch</p></li></ul><p></p><p></p>
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Cellulose

  • It is a polysaccharide composed only of β -D-glucose units which are joined by glycosidic linkage between C1 of one glucose unit and C4 of the next glucose unit.

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Glycogen

  • The carbohydrates are stored in animal body as glycogen

  • Also known as animal starch because its structure is similar to Amylopectin

  • It is present in liver, muscles and brain

  • Enzymes break it down into glucose when the body needs it

  • It is also found in yeast and fungi

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Amino acids

  • Amino acids contain amino (–NH2) and carboxyl (–COOH) functional groups

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Types of Amino Acids

  • Essential Amino Acids

    • Cannot be synthesized by our body

    • Should be taken in our diet

    • Eg: Valine, Leucine

  • Non-Essential Amino Acids

    • Can be synthesized by our body

    • Does not need to be taken in our diet

    • Eg: Glycine, Alanine

    Note:

  • Except glycine, all other naturally occurring α-amino acids are optically active, since the α-carbon atom is asymmetric. These exist both in ‘D’ and ‘L’ forms.

  • Most naturally occurring amino acids have L-configuration. L-Amino acids are represented by writing the –NH₂ group on the left-hand side.

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Amphoteric Nature/Zwitter ion form of amino acids

  • Amino acids act as salts rather than simple amines or carboxylic acids

  • This is due to the presence of both basic (amino) and an acidic (carboxyl) group

  • In aqueous solutions, the carboxyl group can lose a proton and the amino group can gain a proton giving rise to a dipolar ion known as zwitter ion

  • They are neutral but contain both positive and negative charges

  • In zwitter ionic form, amino acids show amphoteric behaviour as they react both with

    acids and bases

<ul><li><p>Amino acids act as salts rather than simple amines or carboxylic acids</p></li><li><p>This is due to the presence of both basic (amino) and an acidic (carboxyl) group</p></li><li><p>In aqueous solutions, the carboxyl group can lose a proton and the amino group can gain a proton giving rise to a dipolar ion known as zwitter ion</p></li><li><p>They are neutral but contain both positive and negative charges</p></li><li><p>In zwitter ionic form, amino acids show amphoteric behaviour as they react both with</p><p>acids and bases<br></p></li></ul><p></p>
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Isoelectronic point

The pH of the solution at which the dipolar ion exists as a neutral ion and does not migrate to either electrode

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Proteins

Polymers of β-amino acids and they are connected to each other by peptide linkage

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Peptide Linkage

Peptide linkage is an amide linkage formed by condensation reaction between –COOH group of one amino acid and –NH2 group of another amino acid

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Functional vs Structural Proteins

  1. Fibrous/Structural Proteins:

    • Responsible for the structure of the body

    • Water insoluble

    • Polypeptide chains run parallel and are held together by hydrogen and di-sulphide bonds

    • Eg: Keratin (hair), Myosin (Mucles)

  1. Globular/Functional Proteins:

    • Responsible for the function of the body

    • Water soluble

    • Polypeptide chains coil around each other to give a spherical shape

    • Eg: Insulin, Albumins

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Primary Structure of Proteins

The sequence of amino acids is said to be the primary structure of a protein.

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Secondary Structure of Proteins

Refers to the shape in which a long polypeptide chain can exist

  1. α-Helix

    • A protein thread is folded in the form of a helix. This structure is maintained by H-bonds which are formed between –NH group of one amino acid and –CO group 4 amino acids away. It has only right-handed helices.

  2. β-Pleated Sheet

    • In this, the polypeptide chains lie side by side in a zig-zag manner with alternate R groups on the same side situated at fixed distances apart.

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Tertiary Structure of Proteins

It represents the overall folding of the polypeptide chain i.e., further folding of the 2° structure. It gives rise to two major molecular shapes - fibrous and globular

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Types of Bonding Which Stabilize the Structures

What type of linkage is responsible for the secondary structure of proteins?

  1. 1o

    • Peptide bonds

  2. 2o

    • Hydrogen bonds

    • Peptide linkage

  3. 3o

    • Disulphide bridge

    • H – bonding

    • Salt bridge

    • Hydrophobic interactions

    • van der Waals forces

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Quaternary Structure of Proteins

  • Some proteins are composed of two or more polypeptide chains called sub-units.

  • The spatial arrangement of these subunits with respect to each other is known as quaternary structure of proteins.

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Native Protein

Protein found in a biological system with a unique 3-D structure and biological activity

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Denaturation of Proteins

  • The loss of biological activity of proteins when a protein in its native form, is subjected to physical change like a change in temperature or chemical change like a change in pH due to which globules unfold and the helix gets uncoiled and the protein loses its biological activity

  • This is called the denaturation of protein

  • During denaturation, secondary and tertiary structures are destroyed but the primary structure remains intact

  • Example: coagulation of egg white on boiling, curdling of milk

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Enzymes

  • Biological catalysts which catalyse biochemical reactions in living beings

  • Highly selective and specific

  • Increase the speed of the reactions about 10M times

  • Extremely small quantities of enzymes (as small as 1 millionth of a mole) can increase the rate by a million times

  • Enzymes are active only at temperatures of about 37oC and pH of around 7

  • Commonly used enzymes: invertase, pepsin (protein to amino acids), amylase (starch to glucose), urease (urea to carbon dioxide and ammonia)

<ul><li><p>Biological catalysts which catalyse biochemical reactions in living beings</p></li><li><p>Highly selective and specific</p></li><li><p>Increase the speed of the reactions about 10M times</p></li><li><p>Extremely small quantities of enzymes (as small as 1 millionth of a mole) can increase the rate by a million times</p></li><li><p>Enzymes are active only at temperatures of about 37<sup>o</sup>C and pH of around 7</p></li><li><p>Commonly used enzymes: invertase, pepsin (protein to amino acids), amylase (starch to glucose), urease (urea to carbon dioxide and ammonia)</p></li></ul><p></p>
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Vitamins

Organic compounds required in the diet in small amounts to perform specific biological functions for normal maintenance of optimum growth and health of the organisms

<p>Organic compounds required in the diet in small amounts to perform specific biological functions for normal maintenance of optimum growth and health of the organisms</p>
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Classification of Vitamins

Why cannot vitamin C be stored in our body?

  1. Fat Soluble Vitamins

    • Vitamins which are soluble in fat and oils but insoluble in water

    • They are stored in the liver and adipose (fat-storing) tissues

    • Eg: Vitamin A, D, E and K

  2. Water Solutble Vitamins

    • Vitamins that are soluble in water

    • They must be supplied regularly in diet because they are readily excreted in urine and cannot be stored in our body

    • Eg: Vitamin B and C

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Vitamin A

Source:

  • Milk

  • Butter

  • Fish liver oil

  • Carrots

Deficiency Diseases:

  • Xerophthalmia (hardening of cornea of eye)

  • Night blindness

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Vitamin B₁ (Thiamine)

Source:

  • Milk

  • Yeast

  • Cereals

  • Green vegetables

Deficiency Diseases:

  • Beri Beri (loss of appetite and retarded growth)

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Vitamin B

Source:

  • Milk

  • Egg white

  • Liver

  • Kidney

Deficiency Diseases:

  • Cheilosis (fissuring at corners of mouth and lips)

  • Digestive disorders

  • Burning sensation of skin

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Vitamin B₆ (Pyrodixine)

Source:

  • Milk

  • Yeast

  • Egg yolk

  • Cereals

  • Grams

Deficiency Diseases:

  • Convulsions

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Vitamin B₁₂

Source:

  • Curd

  • Meat

  • Fish

  • Egg

Deficiency Diseases:

  • Pernicious anaemia (RBC deficient in haemoglobin)

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Vitamin C (Ascorbic Acid)

Source:

  • Citrus fruits

  • Amla

  • Green leafy vegetables

Deficiency Diseases:

  • Scurvy (bleeding gums)

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Vitamin D

Source:

  • Sunlight

  • Fish

  • Egg yolk

Deficiency Diseases:

  • Rickets (bone deformities in children)

  • Osteo-malacia (soft bones and joint pain in adults)

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Vitamin E

Source:

  • Vegetable oils like:

    • Germ oil

    • Sunflower oil

Deficiency Diseases:

  • Increased fragilitiy of RBCs and muscular weakness

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Vitamin K

Source:

  • Green leafy vegetables

Deficiency Diseases:

  • Increased blood clotting time

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Nucleoside and Nucleotide

  • Nucleoside: A unit formed by the combination of pentose sugar with the base is known as nucleoside.
    (Base is attached to the 1st position of sugar)

  • Nucleotide: When nucleoside combines with the phosphoric acid group, the unit obtained is called nucleotide.
    (Phosphoric acid attached to the 5th position of sugar)

<ul><li><p><strong>Nucleoside:</strong> A unit formed by the combination of pentose sugar with the base is known as nucleoside.<br>(Base is attached to the 1st position of sugar)</p></li><li><p><strong>Nucleotide:</strong> When nucleoside combines with the phosphoric acid group, the unit obtained is called nucleotide.<br>(Phosphoric acid attached to the 5th position of sugar)</p><p></p></li></ul><p></p>
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Phosphodiester Linkage

  • The Phosphester bond is a type of covalent bond that forms between a phosphate and a hydroxyl group (-OH)

  • In the case of nucleic acids, the phosphate group forms an ester with the hydroxyl groups on the 5th carbon of one of the nucleotides and the 3rd carbon of another nucleotide.

  • As the phosphate forms two ester bonds (double ester), hence the name phosphodiester linkage

  • This bond is formed due to the condensation reaction occurring between the hydroxyl group of two sugar groups and one phosphate group

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Nucleic Acids

  • They are long-chain polymers of nucleotides

  • Found in the nuclei of all living cells in the form of nucleoproteins or chromosomes

  • Play an important role in the transmission of hereditary characteristics and the biosynthesis of proteins

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Types of Nucleic Acids (DNA vs RNA)

Write two structural differences between DNA and RNA

  1. DNA:

    • Responsible for the transfer of hereditary characteristics from one generation to another in living organisms. This is because of the unique property of replication during cell division and the transfer of two identical DNA strands to the daughter cells.

    • Contains deoxyribose pentose sugar

    • Contains four nitrogenous bases: Adenine (A), Guanine (G), Cytosine (C) and Thymine (T)

    • Has a double-stranded helical structure

  2. RNA

    • Responsible for the synthesis of proteins

    • Contains ribose pentose sugar

    • Contains four nitrogenous bases: Adenine (A), Guanine (G), Cytosine (C) and Uracil (U)

    • Has a single-stranded helical structure

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Types of RNA

i) Ribosomal RNA (rRNA)
ii) Messenger RNA (mRNA)
iii) Transfer RNA (tRNA)

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Nitrogenous Bases Present in Nucleic Acids

  • Nucleic acid contains two types Nitrogen containing bases:

    • Purines: There are two bases derived from purines - Adenine (A) and Guanine (G)

    • Pyrimidines: There are three bases derived from pyrimidines - Thymine (T), Cytosine (C) & Uracil

  • In a DNA molecule, the nitrogenous bases are present as a complimentary base pair

  • Adenine always combines with Thyamine through a double hydrogen bond

  • Cytosine always combines with Guanine with a triple hydrogen bond to make up that double strand chain of a DNA molecule

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Hydrolysis of Nucleic Acids

  1. Generally nucleic acids (both DNA & RNA) upon hydrolysis give …
    Ans: Complete hydrolysis of nucleic acids yields a pentose sugar, phosphoric acid, and nitrogen-containing heterocyclic compounds (Nitrogenous bases).

  2. What products would be formed when a nucleotide from DNA containing thymine is hydrolyzed?
    Ans: When a nucleotide from the DNA containing thymine is hydrolyzed, thymine, β-D-2-deoxyribose, and phosphoric acid are obtained as products.

  1. When RNA is hydrolyzed, there is no relationship among the quantities of different bases obtained. What does this fact suggest about the structure of RNA?
    Ans: When RNA is hydrolyzed, there is no relationship among the quantities of different bases obtained. This fact suggests that RNA is a single-strand structure, unlike DNA, which is a double-strand structure where pairing of bases occurs (e.g., adenine pairs with thymine).

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