Download Features of the DNA Double Helix - E

NEHRU ARTS AND SCIENCE COLLEGE
DEPARTMENT OF MICROBIOLOGY WITH NANOTECHNOLOGY
E-LEARNING
CLASS
: II B.Sc.
SUBJECT
: ALLIED BIOCHEMISTRY I
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UNIT-I
Carbohydrates: Monosaccharides-Definition, classification, structure and properties.
Disaccharides-Definition, types, structure and biological importance. Polysaccharides-types
and properties.
Part – A
1. The monomer of a polysaccharide is called a(n) _______________
Ans : sugars
2. The polysaccharide _______________ is a component of cell walls in plants.
Ans : Cellulose
3. The substance _______________ is also known as animal starch.
Ans : glycogen
4. Starch is a
(A) Polysaccharide (B) Monosaccharide (C) Disaccharide (D) None of these
Ans: Polysaccharides
5. The most important epimer of glucose is
(A) Galactose (B) Fructose (C) Arabinose (D) Xylose
Ans: Galactose
6. A pentose is a sugar with_____ carbons
Ans: Five
7. What Is a Carbohydrate?
Carbohydrates are a common class of simple organic compouds. A carbohydrate is an
aldehyde or a ketone that has additional hydroxyl groups.
8. All of the following are storage carbohydrates except
(a) starch (b) glycogen (c) amylopectin (d). amylose (e) cellulose
Ans : Cellulose
Part – B
3. Describe the functions of carbohydrates
Carbohydrates serve several biochemical functions:
 Monosaccharides are a fuel for celular metabolism.
 Monosaccharides are used in several biosynthesis reactions.
 Monosaccharides may be converted into space-saving polysaccharides, such as
glyocogen and starch. These molecules provide stored energy for plant and animal
cells.
 Carbohydrates are used to form structural elements, such as chitin in animals and
cellulose in plants.
 Carbohydrates and modified carbohydrates are important for an organism's
fertilization, development, blood clotting and immune system function.
Part - C
1. Write a note on types and structure of carbohydrates
Carbohydrates are the main energy source for the human body. Chemically,
carbohydrates are organic molecules in which carbon, hydrogen, and oxygen bond together in
the ratio: Cx(H2O)y,
Definition: Carbohydrates are defined as poly hydroxy alcohol with aldehyde or keto
derivatives. Carbohydartes are mainly classified in to
 Monosaccharides
 Disaccharides
 Polysaccharides
Monosaccharides
These are the only sugars that can be absorbed and utilized by the body.
Disaccharides and polysaccharides must be ultimately broken down into monosaccharides in
the digestive process known as hydrolysis. Only then can they be utilized by the body. Three
monosaccharides are particularly important in the study of nutritional science: glucose,
fructose and galactose.
Glucose (also known as dextrose or grape sugar)
This monosaccharide is the most important carbohydrate in human nutrition because it is
the one that the body fuses directly to supply its energy needs. Glucose is the carbohydrate
found in the bloodstream, and it provides an immediate source of energy for the body's cells
and tissues. Glucose is also formed when stored body carbohydrate (glycogen) is broken
down for use.
Fructose (also known as levulose or fruit sugar)
Fructose, a monosaccharide, is very similar to another monosaccharide, galactose.
These two simple sugars share the same chemical formula; however, the arrangements of
their chemical groups along the chemical chain differ. Fructose is the sweetest of all the
sugars and is found in fruits, vegetables and the nectar of flowers, as well as in the
unwholesome (to humans) sweeteners, molasses and honey. In humans, fructose is produced
during the hydrolysis of the disaccharide, sucrose.
Galactose
Galactose differs from the other simple sugars, glucose and fructose, in that it does
not occur free in nature. It is produced in the body in the digestion of lactose, a disaccharide.
Classification of Monosaccharides
A sugar that contains an aldehydo group is called aldose; one containing a keto group
is called a ketose. A sugar with five carbons is called a pentose; one with six carbons is
called a hexose. Structures of some simple sugars are shown here.
2. Write a note on types and structure of disaccharides
A disaccharide is formed by linking together two monosaccharides. Disaccharides can
be classified into two types. They are
 reducing and
 non-reducing
If the functional group is present in bonding with another sugar unit it is called as reducing di
saccharide
Sucrose
The disaccharide, sucrose, consists of one molecule of each of two monosaccharides—
glucose and fructose. Sucrose is found in fruits and vegetables and is particularly plentiful in
sugar beets (roots) and sugarcane (a grass).
Maltose (also known as malt sugar)
Maltose is a disaccharide of glucose and glucose.
Maltose occurs in the body as an intermediate product of starch digestion. (Starch is a
polysaccharide.) When maltose is hydrolyzed, it yields two molecules of glucose.
Lactose (also known as milk sugar)
Lactose is a disaccharide of galactose and glucose. This disaccharide is found only in
milk.
3. Give an account on types, structure and properties of polysaccharides
A polysaccharide contains many monosaccharides linked together. Like the
disaccharides, the polysaccharides cannot be directly utilized by the body. They must first be
broken down into monosaccharides, the only sugar form the body can use.
Polysaccharides contain up to 60,000 simple carbohydrate molecules. These
carbohydrate molecules are arranged in long chains in either a straight or in a branched
structure. There are four polysaccharides that are important in the study of nutritional science:
starch, dextrin, glycogen and cellulose.
Cellulose
Cellulose is, therefore, the most abundant naturally-occurring organic substance. It is
characterized by its insolubility, its chemical inertness and its physical rigidity.
This polysaccharide can be digested only by herbivores such as cows, sheep, horses,
etc., as these animals have bacteria in their rumens (stomachs) whose enzyme systems break
down cellulose molecules. Humans do not have the enzyme needed to digest cellulose, so it is
passed through the digestive tract unchanged.
In cellulose, the glucose residues are held together by b-1,4’ linkages. The resulting
structure is relatively linear.
Starch
Starch is abundant in the plant world and is found in granular form in the cells of
plants. Most starches are a mix of two different molecular structures, amylose and
amylopectin. The former has a linear structure and the latter has a branched or bushy
structure.
The jelling characteristics of starches are considered to result from the amylose
present, while amylopectin is considered to be responsible for the gummy and cohesive
properties of the paste.
Glycogen
Glycogen is the reserve carbohydrate in humans. It is to animals as starch is to plants.
Glycogen is very similar to amylopectin, having a high molecular weight and branched-chain
structures made up of thousands of glucose molecules. The main difference between
glycogen and amylopectin is that glycogen has more and shorter branches, resulting in a more
compact, bushlike molecule with greater solubility and lower viscosity (less stickiness or
gumminess).
Glycogen is stored primarily in the liver and muscles of animals. About two-thirds of total
body glycogen is stored in the muscles and about one-third is stored in the liver.
UNIT-II
Lipids:- Definition, Classification and properties of lipids. Types of fatty acids saturated, unsaturated and essential fatty acids. Classification and significance of
lipoproteins and phospholipids. Importance of steroids, structure and biological
significance of cholesterol.
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PART- A
1. Essential fatty acid:
(A)Linoleic acid (B) Linolenic acid Arachidonic acid (D) All these
Ans: (D) All these
2. An example of a saturated fatty acid is
(A) Palmitic acid (B) Oleic acid (C) Linoleic acid (D) Erucic acid
Ans: C) Linoleic acid
3. The enzymes of β-oxidation are found in
(A) Mitochondria (B) Cytosol (C) Golgi apparatus (D) Nucleus
Ans: (B) Cytosol
4. A bland diet is recommended in
(A) Peptic ulcer (B) Atherosclerosis (C) Diabetes (D) Liver disease
Ans: (A) Peptic ulcer
5. Ketone bodies are synthesized in
(A)Adipose tissue (B) Liver (C) Muscles (D) Brain
Ans: (A)Adipose tissue
PART- B
1. Write a note on Biomedical Importance of lipid bioc
They have the common property of being
(1) relatively insoluble in water and
(2) soluble in nonpolar solvents such as ether and chloroform.
They are important dietary constituents not only because of their high energy value
but also because of the fat-soluble vitamins and the essential fatty acids contained in the fat of
natural foods. Fat is stored in adipose tissue, it serves as a thermal insulator in the
subcutaneous tissues and around certain organs. Nonpolar lipids act as electrical insulators,
allowing rapid propagation of depolarization waves along myelinated nerves. Combinations
of lipid and protein (lipoproteins) are important cellular constituents, occurring both in the
cell membrane and in the mitochondria, and serving also as the means of transporting lipids
in the blood. Knowledge of lipid biochemistry is necessary in understanding many important
biomedical areas, eg, obesity, diabetes mellitus, atherosclerosis, and the role of various
polyunsaturated fatty acids in nutrition and health.
PART- C
1. Define lipid and explain about the classification of lipids
Definition
The lipids are a heterogeneous group of compounds, including fats, oils, steroids,
waxes, and related compounds, which are related more by their physical than by their
chemical properties.
Lipids Classification
Lipids are classified in to
1. Simple lipids: Esters of fatty acids with various alcohols.
a. Fats: Esters of fatty acids with glycerol. Oils are fats in the liquid state.
b. Waxes: Esters of fatty acids with higher molecular weight monohydric alcohols.
2. Complex lipids: Esters of fatty acids containing groups in addition to an alcohol and a
fatty acid.
a. Phospholipids: Lipids containing, in addition to fatty acids and an alcohol, a
phosphoric acid residue. They frequently have nitrogen containing bases and other
substituents,
(eg) in glycerophospholipids the alcohol is glycerol and in sphingophospholipids
the alcohol is sphingosine.
b. Glycolipids (glycosphingolipids): Lipids containing a fatty acid, sphingosine, and
carbohydrate.
c. Other complex lipids: Lipids such as sulfolipids and aminolipids. Lipoproteins
may also be placed in this category.
3. Precursor and derived lipids: These include fatty acids, glycerol, steroids, other
alcohols, fatty aldehydes, and ketone bodies (Chapter 22), hydrocarbons, lipid-soluble
vitamins, and hormones.
Because they are uncharged, acylglycerols (glycerides), cholesterol, and cholesteryl esters
are termed neutral lipids.
UNIT-III
Amino acids: Classification of amino acids, essential amino acids, reactions of amino
and carboxyl groups of amino acids. Proteins: Definition, classification and function of
Proteins, structural levels of organization (Preliminary treatment). Denaturation and
isoelectric point of Proteins.
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PART – A
1. Define – Amino acid
Amino acids are organic acids containing an amine group. The most common amino acids
are α-amino acids and the most common α amino acids are the L-α-amino acids.
2. Sulphur containing amino acid is
(A)Methionine (B) Leucine (C) Valine (D) Asparagine
Ans: A)Methionine
3. All proteins contain the
(A) Same 20 amino acids (B) Different amino acids
(C) 300 Amino acids occurring in nature (D) Only a few amino acids
Ans: (A) Same 20 amino acids
4. An aromatic amino acid is
(A)Lysine (B) Tyrosine (C) Taurine (D) Arginine
Ans: (B) Tyrosine
5. An amino acid not found in proteins is
(A)β-Alanine (B) Proline (C)Lysine (D) Histidine
Ans: (B) Proline
6. The protein present in hair is
(A) Keratin (B) Elastin (C) Myosin (D) Tropocollagen
Ans: (A) Keratin
7. A disulphide bond can be formed between
(A)Two methionine residues (B)Two cysteine residues
(C) methionine and a cysteine residue (D)All of these
Ans: (D)All of these
8. A Zwitterion is
(A)Positive ion (B) Negative ion (C) Both (A) and (B) (D) None of these
Ans: (C) Both (A) and (B)
9. Which among the following is a basic amino acid?
(A)Aspargine (B) Arginine (C) Proline (D) Alanine
Ans: (D) Alanine
10. The useful reagent for detection of amino acids is
(A) Molisch reagent (B) Dichlorophenol Indophenol (C) Ninhydrin (D) Biuret
Ans: (C) Ninhydrin
PART - B
1. Write a note on essential amino acids
All tissues have some capability for synthesis of the non-essential amino acids, amino
acid remodeling, and conversion of non-amino acid carbon skeletons into amino acids and
other derivatives that contain nitrogen. In times of dietary surplus, the potentially toxic
nitrogen of amino acids is eliminated via transaminations, deamination, and urea formation.
Essential vs. Nonessential Amino Acids
Nonessential
Essential
Alanine
Arginine*
Asparagine
Histidine
Aspartate
Isoleucine
Cysteine
Leucine
Glutamate
Lysine
Glutamine
Methionine*
Glycine
Phenylalanine*
Proline
Threonine
Serine
Tyrptophan
Tyrosine
Valine
The amino acids arginine, methionine and phenylalanine are considered essential
for reasons not directly related to lack of synthesis. Arginine is synthesized by mammalian
cells but at a rate that is insufficient to meet the growth needs of the body and the majority
that is synthesized is cleaved to form urea. Methionine is required in large amounts to
produce cysteine if the latter amino acid is not adequately supplied in the diet. Similarly,
phenyalanine is needed in large amounts to form tyrosine if the latter is not adequately
supplied in the diet.
2. Give an account on functions of protein
 Antibodies - are specialized proteins involved in defending the body from antigens
(foreign invaders).
 Contractile Proteins - are responsible for movement. Examples include actin and
myosin. These proteins are involved in muscle contraction and movement.
 Enzymes - are proteins that facilitate biochemical reactions. They are often referred
to as catalysts because they speed up chemical reactions. Examples include the
enzymes lactase and pepsin. Lactase breaks down the sugar lactose found in milk.
Pepsin is a digestive enzyme that works in the stomach to break down proteins in
food.
 Hormonal Proteins - are messenger proteins which help to coordinate certain bodily
activities. Examples include insulin, oxytocin, and somatotropin. Insulin regulates
glucose metabolism by controlling the blood-sugar concentration. Oxytocin stimulates
contractions in females during childbirth. Somatotropin is a growth hormone that
stimulates protein production in muscle cells.
 Structural Proteins - are fibrous and stringy and provide support. Examples include
keratin, collagen, and elastin. Keratins strengthen protective coverings such as hair,


quills, feathers, horns, and beaks. Collagens and elastin provide support for
connective tissues such as tendons and ligaments.
Storage Proteins - store amino acids. Examples include ovalbumin and casein.
Ovalbumin is found in egg whites and casein is a milk-based protein.
Transport Proteins - are carrier proteins which move molecules from one place to
another around the body. Examples include hemoglobin and cytochromes.
Hemoglobin transports oxygen through the blood. Cytochromes operate in the
electron transport chain as electron carrier proteins.
3. Discuss about denaturation of Proteins
Introduction
Denaturation of proteins involves the disruption and possible destruction of both the
secondary and tertiary structures. Since denaturation reactions are not strong enough to break
the peptide bonds, the primary structure (sequence of amino acids) remains the same after a
denaturation process. Denaturation disrupts the normal alpha-helix and beta sheets in a
protein and uncoils it into a random shape.
Denaturation occurs because the bonding interactions responsible for the secondary
structure (hydrogen bonds to amides) and tertiary structure are disrupted. In tertiary structure
there are four types of bonding interactions between "side chains" including:
 hydrogen bonding
 salt bridges
 disulfide bonds
 non-polar hydrophobic interactions which may be disrupted.
Therefore, a variety of reagents and conditions can cause denaturation. The most common
observation in the denaturation process is the precipitation or coagulation of the protein.
Heat
Heat can be used to disrupt hydrogen bonds and non-polar hydrophobic interactions.
This occurs because heat increases the kinetic energy and causes the molecules to vibrate so
rapidly and violently that the bonds are disrupted. The proteins in eggs denature and
coagulate during cooking. Other foods are cooked to denature the proteins to make it easier
for enzymes to digest them. Medical supplies and instruments are sterilized by heating to
denature proteins in bacteria and thus destroy the bacteria.
4. Write a note on isoelectric point
(i) When ionized form of amino acid is placed in an electric field it will migrate towards
the opposite electrode.
(ii) Depending upon the pH of the medium following three thing may happen.
(a) In acidic medium, the cation move towards cathode.
(b) In basic medium, the anion move towards anode.
(c) The Zwitter ion does not move towards any of the electrodes.
(iii) At a certain pH (i.e. H+ concentration), the amino acid molecules show no tendency
to migrate towards any of the electrodes and exists as a neutral dipolar ion, when
placed in electric field is known as isoelectric point.
(iv) All amino acids do not have the same isoelectric point & it depends upon the nature of
R – linked to α- carbon atom.
PART - C
1. Discuss about the classification of Amino Acids
Definition
Amino acids are organic acids containing an amine group. The most common amino
acids are α-amino acids and the most common α amino acids are the L-α-amino acids.
It is important to understand the following about amino acid structure:
 Only 20 L-α-amino acids are used to make proteins.
 Side groups (labelled "R" in Figure 5.4) are what distinguish the α amino acids
from each other.
 Amino acids can exist as zwitterions - substances containing equal numbers of
positive and negative charge - due to their carboxyl and amine groups, which can
benegatively and positively charged, respectively.
2. Explain about the classification of proteins
Definition
A protein is a complex, high molecular weight organic compound that consists of amino
acids joined by peptide bonds.
Proteins are essential to the structure and function of all living cells and viruses. Many
proteins are enzymes or subunits of enzymes. Proteins are one of the classes of biomacromolecules, alongside polysaccharides and nucleic acids, that make up the primary
constituents of living things.
Classification of proteins
Proteins are classified as
1. Simple Proteins
2. Conjugated Proteins
3. Derived Protein
1. Simple Proteins
a. Albumins: blood (serumbumin); milk (lactalbumin); egg white (ovolbumin); lentils
(legumelin); kidney beans (phaseolin); wheat (leucosin). Globular protein; soluble in
water and dilute salt solution; precipitated by saturation with ammonium sulfate
solution; coagulated by heat; found in plant and animal tissues.
b. Globulins: blood (serum globulins); muscle (myosin); potato (tuberin); Brazil nuts
(excelsin); hemp (edestin); lentils (legumin). Globular protein; sparingly soluble in
water; soluble in neutral solutions; precipitated by dilute ammonium sulfate and
coagulated by heat; distributed in both plant and animal tissues.
c. Glutelins: wheat (glutenin); rice (oryzenin). Insoluble in water and dilute salt
solutions; soluble in dilute acids; found in grains and cereals.
d. Prolamines: wheat and rye (gliadin); corn (zein); rye (secaline); barley (hordein).
Insoluble in water and absolute alcohol; soluble in 70% alcohol; high in amide
nitrogen and proline; occurs in grain seeds.
e. Protamines: sturgeon (sturine); mackerel (scombrine); salmon (salmine); herring
(clapeine). Soluble in water; not coagulated by heat; strongly basic; high in arginine;
associate with DNA; occurs in sperm cells.
f. Histones: Thymus gland; pancreas; nucleoproteins (nucleohistone). Soluble in water,
salt solutions, and dilute acids; insoluble in ammonium hydroxide; yields large
amounts of lysine and arginine; combined with nucleic acids within cells.
g. Scleroproteins: Connective tissues and hard tissues. Fibrous protein; insoluble in all
solvents and resistant to digestion.
• Collagen: connective tissues, bones, cartilage, and gelatin. Resistant to
digestive enzymes but altered to digest gelatin by boiling water, acid, or alkali;
high in hydroxylrpline.
• Elastin: Ligaments, tendons, and arteries. Similar to collagen but cannot be
converted to gelatin.
• Keratin: Hair, nails, hooves, horns, and feathers. Partially resistant to
digestive enzymes; contains large amounts of sulfur, as cystine.
2.Conjugated Proteins
h. Nucleoproteins: cytoplasm of cells (ribonucleoprotein); nucleas of chromosomes
(deoxyribonucleoprotein) viruses, and bacteriophages. Contains nucleic acids,
nitrogen, and phosphorus. Present in chromosomes and in all living forms as a
combination of protein with either RNA or DNA.
i. Mucoprotein: saliva (mucin); egg white (ovomucoid). Proteins combined with amino
sugars, sugar acids, and sulfates.
j. Glycoprotein:
bone
(osseomucoid);
tendons
(tendomucoid);
carilage
(chondromucoid). Containing more than 4% hexosamine, mucoproteins; if less than
4%, then glycoproteins.
k. Phosphoproteins: milk (casein); egg yolk (ovovitellin). Phosphoric acid joined in
ester linkage to protein.
l. Chromoproteins: hemoglobin; myoglobin; flavoproteins; respiratory pigments;
cytochromes. Protein compounds with such nonprotein pigments as heme; colored
proteins.
m. Lipoproteins: serum lipoprotein; brain, nerve tissues, milk, and eggs. Water-soluble
protein conjugated with lipids; found dispersed widely in all cells and all living forms.
n. Metallo proteins: ferritin; carbonic anhydrase; ceruloplasmin. Proteins combined
with metallic atoms that are not parts of a nonprotein prosthetic group.
3.Derived Proteins
o. Proteans: edestan (from elastin) and myosan (from myosin). Results from short
action of acids or enzymes; insolvent in water.
p. Proteases: intermediate products of protein digestion. Soluble in water; uncoagulated
by heat; and precipitated by saturated ammonium sulfate; result from partial digestion
of protein by pepsin or trypsin.
q. Peptones: intermediate products of protein digestion. Same properties as proteases
except that they cannot be salted out; of smaller molecular weight that proteases.
r. Peptides: intermediate products of protein digestion. Two or more amino acids joined
by a peptide linkage; hydrolyzed to individual amino acids.
3. Describe the structural level of proteins with suitable diagram
Polypeptide:
-Co-NH-groups joined with peptide bonds.
Structure of proteins:
Primary structure:
Linear arrangement of polypeptide.
Sequence of amino acids determines protein specificity.
E.g. Haemoglobin.
His – Val – Leu – Leu – Thr – Pro – Glu – Glu – Lys
His – Val – Leu – Leu – Thr – Pro – Val – Glu – Lys
Secondary structure:
Folding and twisting of the polypeptide chain.
Hydrogen bond formation between the neighboring amino acid of amide and
carboxyl group.

α – Helical structure.

β- Pleated structure or β – conformation.

Random coiled structure.
1. α – Helical structure:
3.6 amino acid per turn.
5.4A0 per helix.
Hydrogen bond formation between the neighboring amino acid of amide and
carboxyl group within a polypeptide chain.
2. β- Pleated structure or β – conformation:
Hydrogen bond formation between the neighboring amino acid of amide and
carboxyl group with adjacent polypeptide chain.
Parallel and Anti parallel.
3. Random coiled structure:
No relationship between the polypeptide chains.
Hydrogen bond formed irregularly.
Tertiary structure:
Hydrogen, Ionic, Hydrophobic, and Disulphide bond.
Folding of secondary structure leads to three dimensional structure.
Globular protein.
Quaternary structure:
Association of several poly peptide chains leads to formation of tetrahedral
structure.
UNIT-IV
Nucleic acids: Components of DNA and RNA. Double helical structure of DNA.
Structure and types of RNA. Denaturation and renaturation of DNA. Genetic code.
Protein synthesis (an outline)
___________________________________________________________________________
PART – A
1. Nucleoside consists of
(A) Nitrogenous base
(B) Purine or pyrimidine base + sugar
(C) Purine or pyrimidine base + phosphorous (D) Purine + pyrimidine base + sugar +
phosphorous
Ans: (B) Purine or pyrimidine base + sugar
2. A nucleotide consists of
(A) A nitrogenous base like choline (B) Purine + pyrimidine base + sugar + phosphorous
(C) Purine or pyrimidine base + sugar (D) Purine or pyrimidine base + phosphorous
Ans: (B) Purine + pyrimidine base + sugar + phosphorous
3. RNA does not contain
(A) Uracil (B) Adenine (C) Thymine (D) Ribose
Ans: (C) Thymine
4. The sugar moiety present in RNA is
(A)Ribulose (B) Arabinose (C)Ribose (D) Deoxyribose
Ans: (C) Ribose
5. The structure of tRNA appears like a
(A)Helix (B) Hair pin (C) Clover leaf (D) Coil
Ans: (C) Clover leaf
6. DNA rich in G-C pairs have
(A)1 Hydrogen bond (B) 2 Hydrogen bonds (C) 3 Hydrogen bonds (D) 4 Hydrogen bonds
Ans: (C) 3 Hydrogen bonds
PART – B
1. Describe the component of DNA with structure
Nucleotides are molecules that, when joined together, make up the structural units of
RNA and DNA. In addition, nucleotides play central roles in metabolism. A nucleotide is
composed of a nucleobase (nitrogenous base), a five-carbon sugar (either ribose or 2'deoxyribose), and one to three phosphate groups. Together, the nucleobase and sugar
comprise a nucleoside.
Structure of nitrogenous base
Structure of sugar molecules
2. Illustrate the structure of ribosomal RNA (rRNA)
In prokaryotes, the ribosomal RNA (rRNA) has three types: 23S, 5S, and 16S. In
mammals, four types of rRNA have been found : 28S, 5.8S, 5S and 18S. The unit "S"
stands for Svedberg, which is a measure of the sedimentation rate. After rRNA molecules
are produced in the nucleus, they are transported to the cytoplasm, where they combine with
tens of specific proteins to form a ribosome. In prokaryotes, the size of a ribosome is 70S,
consisting of two subunits: 50S and 30S. The size of a mammalian ribosome is 80S,
comprising a 60S and a 40S subunit. Proteins in the larger subunit are designated as L1, L2,
L3, etc. (L = large). In the smaller subunit, proteins are denoted by S1, S2, S3, etc.
PART – C
1. DNA is double helical structure in nature - justify.
Components of DNA
DNA is a polymer. The monomer units of DNA are nucleotides, and the polymer is
known as a "polynucleotide." Each nucleotide consists of a 5-carbon sugar (deoxyribose), a
nitrogen containing base attached to the sugar, and a phosphate group. There are four
different types of nucleotides found in DNA, differing only in the nitrogenous base. The four
nucleotides are given one letter abbreviations as shorthand for the four bases.
 A is for adenine
 G is for guanine
 C is for cytosine
 T is for thymine
Purine Bases
Adenine and guanine are purines. Purines are the larger of the two types of bases found in
DNA.
Prymidine Bases
Cytosine and thymine are purines.
Base Pairs
Within the DNA double helix, A forms 2 hydrogen bonds with T on the opposite strand,
and G forms 3 hyrdorgen bonds with C on the opposite strand.
Features of the DNA Double Helix
 Two DNA strands form a helical spiral, winding around a helix axis in a right-handed
spiral
 The two polynucleotide chains run in opposite directions
 The sugar-phosphate backbones of the two DNA strands wind around the helix axis
like the railing of a sprial staircase
 The bases of the individual nucleotides are on the inside of the helix, stacked on top
of each other like the steps of a spiral staircase.
2. Illustrate the structure and types and RNA
The major role of RNA is to participate in protein synthesis, which requires three classes
of RNA:
 messenger RNA (mRNA)
 transfer RNA (tRNA)
 ribosomal RNA (rRNA)
messenger RNA (mRNA)
mRNA is transcribed from DNA, carrying information for protein synthesis. Three
consecutive nucleotides in mRNA encode an amino acid or a stop signal for protein synthesis
(see Genetic Code). The trinucleotide is know as a codon.
Structure of transfer RNA (tRNA)
The major role of tRNA is to translate mRNA sequence into amino acid sequence.
A tRNA molecule consists of 70-80 nucleotides. Its secondary and tertiary structures are
shown in Figures 3-C-2 and 3-C-3, respectively. Some nucleotides in tRNA have been
modified, such as dihydrouridine (D), pseudouridine (Y), and inosine (I). In
dihydrouridine, a hydrogen atom is added to each C5 and C6 of uracil. In pseudouridine,
the ribose is attached to C5, instead of the normal N1. Inosine plays an important role in
codon recognition. In addition to these modifications, a few nucleosides are methylated.
3. Give an account on genetic code and its properties.
Genetic code
The genetic code consists of 64 triplets of nucleotides. These triplets are called
codons. With three exceptions, each codon encodes for one of the 20 amino acids used in the
synthesis of proteins. That produces some redundancy in the code: most of the amino acids
being encoded by more than one codon.
One codon, AUG serves two related functions:


it signals the start of translation
it codes for the incorporation of the amino acid methionine (Met) into the growing
polypeptide chain
The genetic code can be expressed as either RNA codons or DNA codons. RNA
codons occur in messenger RNA (mRNA) and are the codons that are actually "read" during
the synthesis of polypeptides (the process called translation). But each mRNA molecule
acquires its sequence of nucleotides by transcription from the corresponding gene. Because
DNA sequencing has become so rapid and because most genes are now being discovered at
the level of DNA before they are discovered as mRNA or as a protein product, it is extremely
useful to have a table of codons expressed as DNA.
Properties of the genetic code
1. The code is read in non overlapping groups of three mRNA nucleotides. Each group
is called a codon.
2. There are no spaces or commas separating neighboring codons. This is like having a
sentence in English consisting entirely of 3 letter words where there are no spaces
between the words. This property is especially important in understand the effects of
mutations on proteins.
3. The genetic code is redundant. There are 64 possible codons but only 20 amino acids.
4. There is a start codon corresponding to the amino acid methionine. When translation
begins the first amino acid is always methionine. After translation this amino acid is
removed as part of editing the protein. Note though that once translation has started,
methionine can occur in the protein.
5. There are three non coding stop or nonsense codons. These tell the machinery of
translation that the end of the protein has been reached.
6. Not all amino acids have an equal number of codons coding for it. Observe that
tryptophan has one codon while arginine has six codons!
7. The code is almost universal. The table shown above shows the standard genetic code
shared by most organisms on the planet. However, certain bacteria, mitochondria and
protista have minor variations in their codes. The near universality of the code
suggests that the code arose very early in the evolution of life.
UNIT-V
Enzymes: Classification of enzymes with examples, coenzymes and cofactors
(structures not needed). Active site: Lock and Key model, Induced fit hypothesis.
Factors affecting enzyme activity. Types of inhibition of enzyme action. Chemical and
industrial applications of enzymes.
___________________________________________________________________________
PART – A
1. Enzymes, which are produced in inactive form in the living cells, are called
a. Papain (B) Lysozymes (C) Apoenzymes (D) Proenzymes
Ans: Proenzymes
2. Fischer’s ‘lock and key’ model of the enzyme action implies that
a. The active site is complementary in shape to that of substance only after
interaction.
b. The active site is complementary in shape to that of substance
c. Substrates change conformation prior to active site interaction
d. The active site is flexible and adjusts to substrate
Ans: (B) The active site is complementary in shape to that of substance
3. In enzyme kinetics Vmax reflects
a. The amount of an active enzyme (B) Substrate concentration
b. Half the substrate concentration (D) Enzyme substrate complex
Ans: (B) Substrate concentration
4. The enzymes of the citric acid cycle are located in
a. Mitochondrial matrix (B) Extra mitochondrial soluble fraction of the cell
b. Nucleus (D) Endoplasmic reticulum
Ans: (A) Mitochondrial matrix
5. Pyruvate dehydrogenase a multienzyme complex is required for the production of
a. Acetyl-CoA (B) Lactate (C) Phosphoenolpyruvate (D) Enolpyruvate
Ans: (A) Acetyl-CoA
6. An example of enzyme inhibition is
(A) Reversible inhibition (B) Irreversible inhibition (C) Allosteric inhibition (D) All
of these
Ans: (D) All of these
7. A cofactor required for the conversion of acetyl-CoA to malonyl-CoA in
extramitochondrial fatty acid synthesis is
Biotin (B) FMN (C) NAD (D) NADP
Ans: (D) NADP
PART – B
1. Explain about the models proposed for binding of substrate with enzyme.
Introduction
The basic mechanism by which enzymes catalyze chemical reactions begins with the
binding of the substrate (or substrates) to the active site on the enzyme. The active site is the
specific region of the enzyme which combines with the substrate. The binding of the
substrate to the enzyme causes changes in the distribution of electrons in the chemical bonds
of the substrate and ultimately causes the reactions that lead to the formation of products. The
products are released from the enzyme surface to regenerate the enzyme for another reaction
cycle.
The active site has a unique geometric shape that is complementary to the geometric
shape of a substrate molecule, similar to the fit of puzzle pieces. This means that enzymes
specifically react with only one or a very few similar compounds.
Lock and Key Theory
The specific action of an enzyme with a single substrate can be explained using a Lock
and Key analogy first postulated in 1894 by Emil Fischer. In this analogy, the lock is the
enzyme and the key is the substrate. Only the correctly sized key (substrate) fits into the key
hole (active site) of the lock (enzyme).
Induced Fit Theory
Not all experimental evidence can be adequately explained by using the so-called rigid
enzyme model assumed by the lock and key theory. For this reason, a modification called the
induced-fit theory has been proposed.
The induced-fit theory assumes that the substrate plays a role in determining the final
shape of the enzyme and that the enzyme is partially flexible. This explains why certain
compounds can bind to the enzyme but do not react because the enzyme has been distorted
too much. Other molecules may be too small to induce the proper alignment and therefore
cannot react. Only the proper substrate is capable of inducing the proper alignment of the
active site.
PART – C
1.Give an account on classification of enzymes
Enzymes can be classified by the kind of chemical reaction catalyzed.
I. Addition or removal of water
A. Hydrolases - these include esterases, carbohydrases, nucleases, deaminases,
amidases, and proteases
B. Hydrases such as fumarase, enolase, aconitase and carbonic anhydrase
II. Transfer of electrons
A. Oxidases
B. Dehydrogenases
III. Transfer of a radical
A. Transglycosidases - of monosaccharides
B. Transphosphorylases and phosphomutases - of a phosphate group
C. Transaminases - of amino group
D. Transmethylases - of a methyl group
E. Transacetylases - of an acetyl group
IV. Splitting or forming a C-C bond
A. Desmolases
V. Changing geometry or structure of a molecule
A. Isomerases
VI. Joining two molecules through hydrolysis of pyrophosphate bond in ATP or
other tri-phosphate
A. Ligases
2. Analyze the factors affective enzyme activity
Enzymes are globular proteins which act as biological catalysts. This means that they
speed up the rate of reaction by lowering the activation energy, that is the energy required to
break bonds. Enzymes are a complex tertiary and sometimes quaternary shape and catalyse
reactions by forming a complex (known as the enzyme substrate complex) at a specific region
of the enzyme called the active site.
Enzyme + substrate
enzyme substrate complex
product
Factors affecting enzyme activity
1. Temperature
Enzymes have an optimum temperature – this is the temperature at which they work
most rapidly. Below the optimum temperature, increasing temperature will increase the rate
of the reaction. This is because temperature increases the kinetic energy of the system,
effectively increasing the number of collisions between the substrate and the enzyme’s active
site.
Temperatures above the optimum will lead to denaturation. This occurs because the
hydrogen bonds and disulphide bridges which maintain the shape of the active site are
broken. Thus, enzyme substrate complexes can no longer be formed.
2. pH
The effect of a change in pH on enzyme activity is shown in Fig. As with temperature,
each enzyme has an optimum pH. If pH increases or decreases much beyond this optimum,
the ionisation of groups at the active site and on the substrate may change, effectively
slowing or preventing the formation of the enzyme substrate complex. At extreme pH, the
bonds which maintain the tertiary structure – hence the active site – are disrupted and the
enzyme is irreversibly denatured.
3. Enzyme concentration
The effect of enzyme concentration on the rate of reaction is shown in Fig. At low
enzyme concentrations there are more substrate molecules than there are available active
sites. Increasing the number of active sites by increasing the concentration of the enzyme,
therefore, effectively increases the rate of the reaction. Eventually, at point x, increasing the
enzyme concentration has no effect on the rate of reaction. This is because it is now the
number of substrate molecules which has become the limiting factor.
4. Substrate concentration
At low substrate concentration the reaction proceeds slowly. This is because there are
not enough substrate molecules to occupy all of the active sites on the enzyme. As substrate
concentration increases, the rate increases because there are more enzyme substrate
complexes formed. At point x, however, increasing the substrate concentration will have no
further effect on the rate of reaction. This is because all of the enzyme’s active sites are now
occupied by substrate molecules – increasing the substrate concentration further will have no
effect, because no more enzyme substrate complexes can form.
3. Explain about the mechanism of enzyme inhibition by inhibitors
Introduction
Enzyme inhibitors are molecules that interact in some way with the enzyme to prevent
it from working in the normal manner.
The enzymologists recognize several classes of enzyme inhibition. These are:
1. Reversible inhibitors
There are two types of reversible inhibitor:
1. competitive reversible inhibitor
2. non-competitive reversible inhibitor
3. Uncompetitive inhibition
2. Irreversible inhibitors
Competitive inhibition
A competitive inhibitor is any compound which closely resembles the chemical
structure and molecular geometry of the substrate. The inhibitor competes for the same active
site as the substrate molecule. The inhibitor may interact with the enzyme at the active site,
but no reaction takes place. The inhibitor is "stuck" on the enzyme and prevents any substrate
molecules from reacting with the enzyme. Competitive inhibition
A typical plot for competitive inhibition looks as follows:
is usually reversible if sufficient substrate molecules are available to ultimately displace the
inhibitor. Therefore, the amount of enzyme inhibition depends upon the inhibitor
concentration, substrate concentration, and the relative affinities of the inhibitor and substrate
for the active site.
Non competitive Inhibitors
A noncompetitive inhibitor is a substance that interacts with the enyzme, but usually
not at the active site. The noncompetitive inhibitor reacts either remote from or very close to
the active site. The net effect of a non competitive inhibitor is to change the shape of the
enzyme and thus the active site, so that the substrate can no longer interact with the enzyme
to give a reaction. Non competitive inhibitors are usually reversible, but are not influenced by
concentrations of the substrate as is the case for a reversible competive inhibitor. See the
graphic on the left.
A typical set of plts for such inhibition is shown below.
Uncompetitive inhibition
An uncompetitive inhibitor binds at a site distinct from the substrate active site and,
unlike a competitive inhibitor, binds only to the ES complex. In uncompetitive inhibition the
inhibitor binds only to the enzyme-substrate complex. This type of enzyme inhibition is
relatively rare. It gives a third type of plot as is shown below.
Summary of Enzyme Inhibition Modes
Irreversible Inhibitors
Irreversible Inhibitors form strong covalent bonds with an enzyme. These inhibitors
may act at, near, or remote from the active site. Consequently, they may not be displaced by
the addition of excess substrate. In any case, the basic structure of the enzyme is modified to
the degree that it ceases to work.
Since many enzymes contain sulfhydral (-SH), alcohol, or acid groups as part of their
active sites, any chemical which can react with them acts as an irreversible inhibitor. Heavy
metals such as Ag+, Hg2+, Pb2+ have strong affinities for -SH groups.
Nerve gases such as diisopropylfluorophosphate (DFP) inhibit the active site of
acetylcholine esterase by reacting with the hydroxyl group of serine to make an ester.
4. Write a note on Industrial applications of enzymes
1. Proteases - Four kinds of proteases
Applications
2. Amylase
3. Cellulose
 Detergent application
 Textile industry (Stonewashing of denim or biopolishig)
 Deinking of newspaper
4. Lipase
 Detergent application
 Food industry
 Organic synthesis
 Pulp processing
 Synthesis of cocoa butter
 Lipases are used in degradation, synthesis, or purification of stereospecific esters :
Important in pharmaceutical industry