Saturday 30 January 2021

Chapter 6 Molecular Basis of Inheritance







Hello students, in the previous chapter, you have come to know, how the characters, pass on to their children. You are now familiar with the word GENE.


Now in this chapter, “Molecular Basis of Inheritance” we will come to know about the chemical or molecular nature of GENES, which is the main genetic material in the world of GENETICS.


Genetic material is that substance which not only controls the inheritance of traits from one generation to the next but is also able to express its effect through the formation and functioning of the traits.



A molecule that can act as a genetic material must fulfil the following criteria:

1. It must be able to replicate,

2. Store information,

3. Express information, and

4. Allow variation by mutation.


Scientists worked hard to establish that hereditary material is DNA in most

of organisms.


Meischer discovered NUCLEIN. He used this term for DNA with associated proteins.


Mendel too had given the principles of inheritance at same time and used the term ‘factor’ for genetic material.


By 1926, scientists had reached molecular level of research to establish the identity of genetic material.


The famous scientists Frederick Griffith carried out series of experiment with Streptococcus pneumoniae bacteria of S and R strain.


Mice injected with S strain developed pneumonia whereas the mice injected with R strain did not.


Then he killed S strain bacteria with heat and injected. Nothing happened, mice did not die.


Now he injected heat killed S strain and live R strain bacteria. The mice



This experiment led to thought believing DNA as genetic material.


Many scientists worked to determine the biochemical nature of ‘Transforming Principle’ in Griffith’s experiment.


Many scientists thought that genetic material was a protein.


They extracted PROTEINS, DNA and RNA from heat killed S cells.They discovered that Proteases and Rnase didn't affect transformation but Dnase did inhibit transformation.

elt was concluded that DNA is the hereditary material. Many scientists were

not convinced with this conclusion and many continued their experim




(i) Multiple Choice Questions:


1) Which of the following is false about any molecule to be considered as genetic material?

(a) replicable

(b) inneritable

(c)non mutable

(d) all


2) Frederick Griffith carried out his experiment with:

(a) Viruses

(b) Bacteria

(c) Bacteriophages

(d) none



3) How many strains of bacteria were used in Griffith's experiment?






4) Meischer discovered:

(a) Nuclein

(b) Chromatin

(c) Histones

(d) None


5) On the culture plate, the rough colonies produced by baceria of:

(a) M strain

(b) S strain

(c) R strain

(d) Z strain


(ii) True/ False:

1) The mice injected with live S strain bacteria died from pneumonia.

2) Heat killed S strain bacteria when injected alone, the mice died.

3) Genetic material snould be replicable.



(iii) Fill ups:

1) DNA is ............. acid.

2) .............. Strain is virulent Strain of streptococcus bacteria



A) Multiple Choice Questions:


1) (c) non mutable It should be able to mutate so that it may show


2) (b) Bacteria Griffith conducted a series of experiments using Streptococcus

pneumoniae bacteria.

3) (a) Griffith used two strains of pneumococcus (Streptococcus pneumoniae)

bacteria which infect mice — a type III-S (smooth) which was virulent, and a type II-

R (rough) strain which was non-virulent.

4) (a) Nuclein

5) (c) R strain The R (rough) strain, lacks the coat and produces colonies that

look rough and irregular.


B) True/ False:

1) True — The mice died due to virulent nature of S strain bacteria.

2) False - The mice remained alive after injecting heat killed S strain.

3) True - Genetic material should have the ability to replicate so that it can make

its copy.


C) Fill Ups:

1) DNA is deoxyribose nucleic acid.

2) S Strain is considered as virulent strain as it is responsible to cause disease.



1) Which biomolecules were considered to be possible genetic material?

2) Enlist various characteristics of genetic material.

3) Differentiate between S strain and R strains of Streptococcus bacteria.



1) Write the salient features of Griffith's experiment.





Dear students, as in previous topic we have learnt about Griffith's experiment

of Transformation, on mice.


TRANSFORMATION is defined as the change in genetic material of an organism by the genes from outside as from remains of its dead relatives.Griffth’s experiment however could not specify the Transforming Chemical.Today we will study about experiments conducted by various scientists to prove that DNA is the hereditary or genetic material.


A) AVERY & OTHERS EXPERIMENT:Oswald Avery, Colin MacLeod and Maclyn McCarty (1933-44) purified biochemicals (protein, DNA, RNA, etc.) from the heat killed S cells. They

discovered that DNA alone from S - bacteria caused R - bacteria to become transformed.

They also discovered that transformation was not affected by protein-digesting

enzymes (proteases), and RNA-digesting enzymes (RNases).

Hence, it was concluded that the transforming substance was not a protein or RNA. Digestion with DNase inhibited transformation. This suggested that the DNA was the cause of transformation. Thus, it was concluded that DNA is the hereditary material.


Their findings were as under:

i) DNA alone from S- bacteria caused R- bacteria to be transformed.

ii) They found that Proteases (protein digested enzymes) and RNAase (RNA

digesting enzymes) did not affect transformation.

iii) Digestion with DNAase did inhibit transformation.Thus they finally concluded that DNA is the hereditary material.



In 1952, Hershey and Chase conducted their blenders experiment on T:phage which attacks the bacterium Escherichia coli.


1. The phage particles were prepared by using radio isotopes of *S and *P .


2. These two radioactive phage preparations (one with radioactive proteins and

another with radioactive DNA) were allowed to infect the culture of E.coli.


3. The protein coats were separated from the bacterial cell walls by shaking and

centrifugation. the heavier infected bacterial cells during centrifugation pelleted to the bottom.


4. The supernatant had the lighter phage particles and other components that

failed to infect bacteria.


5. It was observed that bacteriophages with radioactive DNA gave rise to radioactive pellets *2P in DNA.


6. However in the phage particles with radioactive protein (with °°S) the bacterial pellets have almost nil radioactivity, indicating that proteins have failed to migrate into the bacterial cell.


So, it can be safely concluded that during infection by bacteriophage T, it

was DNA which entered the bacteria. It was followed by an eclipse period during which phage DNA replicates numerous times within the bacterial cell.


Lysozyme (an enzyme) brings about the lysis of host cells and releases the newly formed bacteriophages.The above experiment clearly suggests that it is phage DNA and not protein which contains the genetic information for the production of new bacteriophages.



(i) Multiple Choice Questions:


1.) The initial work which formed the base to establish DNA as genetic material was done by:

a.) Watson and Crick

b.) George Mendel

c.) Frederick Griffith

d.) Beadle and Tatum


2.) Which microorganism was involved in the study of DNA as a genetic material by Griffith?

a.) S. pneumoniae

b.) Neurospora

c.) L. odoratus

d.) L. bacillus


3.) In Griffith's experiment, the conversion of R-type to S-type of bacteria

when mixed with heat killed S-type is called:

a.) Mutation

b.) Transduction

c.) Transfection

d.) Transformation


4.) The Blenders experiment to prove DNA as genetic material was conducted by:

a.) Beadle and Tatum

b.) Frederick Griffith

c.) Hershey and Chase

d.) Hofmeister and Waldayer


5.) Which organism was allowed to infect the culture of E.coli in Blender’s experiment?

a.) Bacteriophage

b.) Bacteria

c.) Algae

d.) Plasmodium vivax


(ii) Fill Ups:

1. | The chemicals used by Hershey and chase in their experiment,were and.

2. The protein coats were separated from the bacterial cell walls by and ;


(iii) True/False:

1. Hershey and Chase worked with Viruses that infect bacteria in their experiment.

2. DNA alone from R- Bacteria, caused S-Bacteria to be transformed.





(i) Multiple Choice Questions:

1. (c) Griffith performed transformation experiment to prove that there is some genetic material which is responsible for carrying characteristic functions

2. (a), Griffith used two strains of streptococcus bacteria, S and R strain

3. (d), Transformation

4. (c), Hershey and Chase conducted blender’s experiment which is also

known as Transduction.

5. (a) Bacteriophages are the viruses which infect bacteria.


(ii) Fill Ups:

1. Radioactive Sulphur and Phosphorus

2. Shaking and centrifugation


(iii) True/False:

1. True

2. False: DNA alone from S- Bacteria caused R-Bacteria to be transformed.



1. Name the scientists who proved Griffith's experiment also DNA is the

genetic material.

2. What type of radioactive components were used in Hershey and chase experiment?

3. Define transduction.



1.) Discuss the contributions of Avery and Co-workers in the study of DNA as

genetic material.

2.) How Hershey and Chase experimentally proved that DNA is hereditary material?







DNA is a group of molecules that is responsible for carrying and transmitting the hereditary materials or the genetic instructions from parents to offspring.

It is an organic compound that has a unique molecular structure.It is found in all prokaryotic cells and eukaryotic cells.



The DNA molecule is composed of units called nucleotides and each nucleotide is composed of three different components, such as sugar, phosphate groups and nitrogen bases. When this unit lacks

phosphate group it is called nucleoside. The basic building blocks of DNA are nucleotides, which are composed of a sugar group, a phosphate group, and a nitrogen base.



DNA is found in the cells of all the living organisms except in some plant

viruses. In bacteriophages and viruses, there is a central core of DNA which

is enclosed in a protein coat. In bacteria, and in mitochondria and plastids of

eukaryotic cells, DNA is circular and lies naked in the cytoplasm.



Chemical analyses have shown that DNA is composed of three different types of molecules.

1. Phosphoric acid (H3PO,4) has three reactive (-OH) groups of which two are involved in forming the sugar phosphate backbone of DNA.


2. Pentose sugar’ DNA contains 2’-deoxy-D-ribose (or simply deoxyribose) which is the reason for the name deoxyribose nucleic acid.


3. Organic bases: The organic bases

are heterocyclic compounds containing nitrogen in their rings; hence they are also called nitrogenous bases. DNA ordinarily contains four different bases called Adenine (A),Guanine (G), Thymine (T) and Cytosine (C).


CHARGAFF’s RULE According to this rule, the DNA molecule should have an equal ratio of Pyrimidines (T and C) and Purines (A and G). There is always equality in quantity between the bases A and T and between the bases G and C. (Ais adenine, T is thymine, G is guanine, and C is cytosine.)




A=T and G=C


The number of purines and pyrimidines in the DNA exist in the ratio 1:1.


It provides the basis of base pairing.


Withthe help of thisrule, one can determine the presence of a base in the DNA and also determine the strand length.


Base ratio A+T/G+C is specific for a species but remains constant.


IMPORTANCE OF CHARGAFF RULE:-Chargaff's rules are important because they point to a kind of “grammar of biology”. a set of hidden rules that gover the structure of DNA. This grammar ought to reveal itself as patterns in DNA that are invariant across all species.




(a) Multiple choice questions:


(1) Which of the following options, are the pyrimidine bases found in DNA?

(a) uracil and thymine

(b) thymine and cytosine

(c) adenine and thymine

(d) cytosine and uracil.


(2) A nucleoside differs from a nucleotide, in lacking the:

(a) base

(b) sugar

(c) hydroxl group

(d) phosphate group


(3) The two strands in a DNA double is joined by:

(a) covalet bond

(b) hydrogen bond

(c) ionic bond

(d) both a and b


(4) DNA was first discovered by:

(a) J.D.watson

(b) Francis Crick

(c) Friedrich miescher

(d) H.G.khurana


(5) According to chargaff's rule, which one is correct?

(a) A+T=G+C

(b) AtC=G+T

(c) A+G=T+C

(d) both a and c.


(b) Fill in the blanks:

(1) nitrogenous base is not present in DNA.

(2) New strand of DNA are found only in the direction.

(3) A nucleotide consists of a a and a nitrogen base.


(c) True /false:

(1) Adenine always pairs with cytosine.

(2) sugars and phosphates make up the backbone of DNA.

(3) sugar in DNA is called ribose.



(a) MCQs:

(1) b--- thymine and cytosine are pyrimidine bases found in DNA.

(2) d---- Phosphate group

(3) b----hydrogen bonds are responsible for base pairing formation in DNA.

(4) c----friedrich miescher

(5) c ---A+G=T+C Acc. to chargaff rule any species of any organism should have a 1:1 ratio of Purine and Pyrimidine bases.


(bi Fill Ups:

(1) Uracil -—-uracil nitrogenous base is present in RNA.

(2) 5'-3'-—- new DNA is made by polymerase which require template and

synthesize DNA in 5’3’ direction.

(3) Sugar, phosphate--—a nucleotide consist of sugar, phosphate and nitrogen base.


ic) True/ false:

(1) False: (always pair with J acc. to Chargaff Rule.

(2) True

(3) False; sugar in DNA is called deoxyribose.



(1) Which bases are purines and pyrimidines?

(2) write the chemical composition of DNA.

(3) Why is base pairing rule important?



(1) If a double stranded DNA has 20% thymine calculate the percentage of

guanine in the DNA.

(2) Explain the structure of D.N.A.





Students, in our previous topic we studied about that DNA is universal genetic

material in all life forms. In our today’s topic we will discuss its structure as

explained by Watson and crick. They proposed 3-Dimensional model of DNA.

It was only in 1953 that James Watson and Francis Crick, based on X-ray diffraction data produced by Maurice Wilkins and Rosalind Franklin, proposed a very simple but famous double helix model for the structure of DNA.


The three-dimensional structure of DNA, consists of two long helical strands that

are coiled around a common axis to form a double helix.


Each DNA molecule is comprised of two biopolymer strands coiling around each



Each strand has a 5’end (with a phosphate group) and a 3'end (with a hydroxyl group).


The strands are antiparallel, meaning that one strand runs in a 5'to 3'direction,

while the other strand runs in a 3'to 5'direction.


The diameter of the double helix is 2nm and the double-helical structure repeats

at an interval of 3.4nm which corresponds to ten base pairs.


The two strands are held together by hydrogen bonds and are complementary

to each other.


The two DNA strands are called polynucleotides, as they are made of simpler monomer units called nucleotides. Basically, the DNA is composed of



The deoxyribonucleotides are linked together by 3- 5‘phosphodiester bonds.


The nitrogenous bases that compose the

deoxyribonucleotides include adenine, cytosine, thymine, and guanine.


The structure of DNA -DNA is a double helix structure because it looks like a twisted ladder.


The sides of the ladder are made of alternating sugar (deoxyribose) and

phosphate molecules while the steps of the ladder are made up of a pair of

nitrogen bases.


Asa result of the double-helical nature of DNA, the molecule has two asymmetric grooves. One groove is smaller than the other.


This asymmetry is a result of the geometrical configuration of the bonds between the phosphate,sugar, and base groups that forces the base groups

to attach at 120-degree angles instead of 180 degrees.


The larger groove is called the major groove, occurs when the backbones are far apart; while the smaller one is called the minor groove, and occurs when they are close together.


Since the major and minor grooves expose the edges of the bases, the grooves can be used to tell the base sequence of a specific DNA molecule.


The possibility for such recognition is critical since proteins must be able to

recognize specific DNA sequences on which to bind in order for the proper functions of the body and cell to be carried out.


The Watson and Crick model shows that DNA is a double helix with sugar-phosphate backbone on the outside and paired bases on the inside.



(i) Multiple Choice Questions:


1. The two strands in DNA are coiled to each other:

(a) Parallel

(b) anti parallel

(c) both a) and b)

(d) none of these


2. The nucleic acids are polymer of:

(a) amino acids

(b) amides

(c) nucleotides

(d) phosphates


3. Which of the following combination of base pair is absent in DNA?

(a) A-T

(b) A-U

(c) C-G

(d) T-A



4. In a polynucleotide chain, nucleotide are held through

(a) peptide linkage

(b) glycosidic linkage

(c) phosphodiester linkage

(d) hydrogen bond


5. A nucleotide is formed by the combination of

(a) phosphate and sugar

(b) sugar phosphate

(c) base-sugar- phosphate

(d) none of these


(ii) True/ False:

1. 8base pairs are present in one full tum of DNA-helix.

2. Base pairing pattern of DNA is A-U and G-C

3. Two strands of DNA molecule are held together by phosphodiester bond.


(iii) Fill ups:

1. The DNA molecule takes a complete turn after ..........0.00..... base


2. Anucleoside combines form ucleotide.





A) Multiple Choice Questions:

1. (b) anti parallel It is because two chains are parallel but their 5’->3

directions are opposite

2. (c) nucleotides A nucleotide consists of a base, a sugar molecule and a

phosphate molecule.

3. (b) A-U Because Uracil is present in RNA

4. (c) phosphodiester linkage

5. (c) base-sugar-phosphate


B) True/ False:

1) False - 10 base pairs are present

2) False - the base pair pattern is A-T and G-C

3) False - The hydrogen bonds between complementary nucleotides keep

two. strands of DNA helix together.


C) Fill Ups:

1) 10 base pairs

2) Phosphate group



1. Distinguish between nucleosides and nucleotides.

2. The two strands of DNA molecule are anti-parallel. What do you understand by


3. What are the four pairs of DNA bases that form in the double helix?



1. Describe Watson and Crick model of DNA with the help of well labelled






DNA, the genetic information carrier molecule of the cell is a long polymer

of nucleotides and can adopt different types of structural conformations

the various types of conformations that the DNA can adopt depend upon different factors such as:-


Hydration level


Salt concentration


DNA sequence


Quantity and direction of super-coiling


Presence of chemically modified bases


Presence of polyamines in solution









A-DNA:A DNA is a rare type of structural conformation, which can adopt under

dehydrating condition




A-DNA has shorter and more compact structure organisation


A-DNA was discovered by Rosalind Franklin


A-DNA is formed from B DNA under dehydrating condition


A-DNA is much wider and flatter than B-DNA


A-DNA is a right handed helix


The helix diameter of A-DNA is 26A


A-DNA containing 11.6 base per turn


A-DNA has narrow and deep major groves


The minor groves of A-DNA are wide and shallow



B-DNA is the most common type of DNA present in the cell it is also called

biological DNA .B-DNA was discovered by James Watson and Francis Crick.



Majority of the DNA in a cell is B DNA.


B-DNA is a right-handed helix.


The helical diameter of B-DNA is 20 A each tum in the B-DNA consists of 10 base pairs.


The B-DNA has a solid Central core.


The major grove of B-DNA is wide and deep the minor grove of B-DNA is narrow and deep.




C DNA has 9.33 base pairs which are negatively tilted from the axis of the

helix it's helical diameter is 19 A.



D-DNA has 8 base pairs which are negatively tilted from the axis of the helix.



Z DNA is a left-handed double helical confirmation of DNA in which the

double helix winds to the left in zigzag pattern, it was discovered by Andres

Wang and Alexander Rich.



The helical diameter of z DNA is 18 A, with 12 base pair.






Q1: The helical diameter of B-DNA is:

a) 30 A°

b) 21 A°

c)32 A°

d)20 A°


Q2: Major groove of B DNA is:

a) Wide and deep

b) Shallow and deep

c) Narrow and wide

d) Narrow and deep


Q3: Name the three DNA that occur under dehydration conditions:

a) A-DNA

b) B-DNA

c) C-DNA

d) Z-DNA


Q4: Left-handed DNA is:

a) A-DNA

b) B-DNA

c) Z-DNA

d) C-DNA


Q5: The helical diameter of Z- DNA is:

a) 12 A°

b) 30 A°

c) 28 A°

d) 18 A°



a) Z-DNA was discovered by.........

b) A-DNA contains........ base pair per turn

c) B-DNA is called........ DNA



a) Helical diameter of a DNA is 26 A’.

b) Each turn in B-DNA consists of 12 base pair.

c) B-DNA was discovered by James Watson and Francis Crick.




1) (d) 20A

2) (a) wide and deep

3) (a) A-DNA- as it can adopt under dehydrating conditions.

4) (d) Z-DNA- as its double helix winds in a zig-zag pattern

5) (d) 18A


(ii) FILL UPS: -

1) Andres Wang and Alexander Rich

2) 11.6 base pairs

3) Biological DNA- as it is the most common type of DNA found in the cell








1) Define DNA a molecule

2) Name different forms of DNA

3) Give main features of B-DNA



1) Give difference between A-DNA, B- DNA and Z-DNA





The DNA or deoxyribonucleic acid is not only the largest negatively charged

macromolecule but also represents genetic material of organism and molecular basics of heredity. The most important property of DNA is super coiling.In Prokaryotic cells, DNA is circular and embedded in the cytoplasm and is often called nucleoid. It is not bounded by nuclear membrane and is without histone proteins. It is termed as naked DNA.Many Prokaryotes also possess extra chromosomal small circular DNA

segments called Plasmids.


In Eukaryotic cells, DNA is linear and mainly confined to the nucleus as the

component of chromosomes. It is termed as nuclear DNA. It is associated with histone proteins to form chromatin fibres.


A small quantity of DNA is also present in the mitochondria and plastids which is termed as extra nuclear or organellar DNA. It is circular like prokaryotic DNA.



The length of the DNA is around 3 meters that need to be accommodates within the nucleus which is only a few micrometres in diameter.


In order to fit- in, the DNA molecules into the nucleus, DNA,needs to be packed into an extremely compressed and compact structure called chromatin.


DNA is condensed into a complex structure with histone and non-histone proteins. Histones are rich in lysine and arginine and are positively charged that provides structural support to a chromosome.


Histone proteins give a more compact shape to chromosomes. Positive charge of histones allows them to associate with negatively charged DNA.


There are five types of histones named H1, H2A, H2B, H3 and H4.


Histones form a unit of eight molecules called histone octamer at a centre

of a nucleosome core particle. The negatively charged DNA is wrapped

around the positively charged histone octamer to form a structure called



Linker DNA is double-stranded DNA 38-53 bp long inbetween two nucleosome cores that, in association with histone H1, holds the cores together. Linker DNA is seen as the string in the "beads and string model"


Nucleosome contribute thread like coloured bodies in the nucleus called chromatin. The nucleosomes in chromatin are seen like beads on string.


Chromatin exists in two forms Euchromatin and Heterochromatin.


Euchromatin is lightly stained and loosely packed chromatin whereas

Heterochromatin is darkly stained and more densely packed chromatin.






1. DNA is not present in:

(a) Nucleus

(b) Chloroplast

(c) Ribosomes

(d) Mitochondria


2. DNA fragments are:

(a) Negatively charged

(b) Either positively or negatively charged

(c) Positively charged

(d) Neutral


3. A nucleosome is best described as:

(a) One fully packaged DNA molecule

(b) One strand of DNA double helix

(c) Nucleosomes coiled around each other

(d) ADNA strand wrapped around histone proteins


4. What are the set of positively charged basic proteins called as?

(a) Histidine

(b) DNA

(c) RNA

(d) Histones


5. When the negatively charged DNA combines with positively charged histone octamer, which of the following is formed?

(a) Nucleus

(b) Nucleoid

(c) Nucleosome

(d) none





1. Histones are rich in and amino acids.

2. The positively charged basic proteins are called .



1. Histones are negatively charged.

2. The length of DNA is near about 3m.



MCQ’s >

1. (c) Ribosomes are present in cytoplasm as well as matrix of Mitochondria and Chloroplast. They do not contain any DNA.


2. (a) DNA is Negatively charged because of the presence of one negatively charged oxygen in phosphate group.


3. (c) ADNA strand wrapped around histone proteins.


4. (d) Histones are the basic proteins because they are rich in basic amino acids i.e. Arginine and lysine.


5. (c) Nucleosome



1. Lysine, Arginine

2. Histones



1. False Histones are the basic proteins because they are rich in basic amino acids i.e. Arginine and lysine. These are positively charged.

2. True



1. Whatis linker DNA and what is its function?

2. What type of appearance is given by nucleosome chain under electron




1. What are the reasons of stabilization of Histone-DNA interaction?





In molecular biology, DNA replication is the biological process of producing two identical replicas of DNA from one original DNA molecule.DNA replication occurs in all living organisms acting as the most essential part for biological inheritance.

DNA Replication is Semi Conservative:


DNA replication is a semi conservation process because when a new double

stranded DNA molecule is formed, one strand will be form original template molecule and one strand will be newly synthesised. That is half strand of DNA is conserved.


Meselson and Stahl Experiment to Prove that DNA Replication is Semiconservative in Nature:Matthew Meselson and Franklin Stahl performed the following experiment in 1958:


AIM : To prove the mode of DNA replication is semi conservative


EXPERIMENTAL MATERIAL: E. coli, lsotopes of nitrogen N-15, N-14


PRINCIPLE : Density gradient centrifugation

1. E. Coli grown in ‘*N medium:They grew E. coli ( Escherichia coli) in a medium containing '°NH.C!l ('°N is the heavy isotope of nitrogen)as the only nitrogen source for many generations. The result was that 'N was incorporated into newly

synthesised DNA.


This heavy DNA molecule could be distinguished from the normal DNA by centrifugation in a cesium chloride (CsCl) medium or densitygradient.


Being '°N not a radioiosotopic isotope, it can be separated from '4N only based on densities.


2. Transformation of N cell to N medium.

Then they transferred the cells into a medium with normal NH.Cl and took

samples at various definite time intervals as the cell multiplied and extracted the


DNA that remained as double-stranded helices. The various samples were

separated independently on CsCl gradients to measure the densitiesof DNA.


3. Replication.

After that the DNA was extracted from the culture, one generation after the

transfer from '°N to '4N medium , that is after 20 minutes the cells of E.coli

divides and had a hybrid or intermediate density. DNA extracted from the

culture after another generation that is after 40 minutes, 2"? generation

composed of equal amount this hybrid DNA and of light DNA.




Meselson and Stah! concluded that DNA replication is not conservative but



Each strand in DNA molecule serves as a template for synthesis of new

complementary strand.


Each strand is composed of one parent strand and one newly synthesized








1) Semiconservative mode of replication of DNA was proved by:

(a)Hershey and Chase

(b) Griffith

(c) Watson and Crick

(d) Meselson and Stahl


2) Semiconservative replication of DNA was first demonstrated in:

(a)Drosophila melanogaster

(b) Escherichia coli

(c) Streptococcus pneumonia

(d) Salmonella typhi


3) Which process was used by Meselson and Stahl for their experiment:


(b) Chromatography

(c) Density gradient centrifugation

(d) Buoyant density centrifugation




4) Which isotope out of the following was used:

(a) N

(b) O'8


(d ) all of these


5) Which medium is used for centrifugation in the experiment:

(a) cesium chloride

(b) sodium chloride

(c) ammonium chloride

(d) potassium chloride


(ii) FILL UPS :

1) The Meselson and Stahl experiment is based on the principle

2) This experiment proves that DNA replication is

3) The semiconservative replication of DNA was proved in year



1) E. coli divides in 30 minutes

2) Ammonium chloride (NHsCL) is the only source of nitrogen in the experiment





1) (d) Meselson and stahl they proved the semiconservative mode of replication .


2) (b) Escherichia coli as in E.coli replication occurs with high accuracy


3) (c) Density gradient centrifugation — This type of centrifugation separates

different particles based on their different densities.


4) (a) N'* — They use heavy isotope.

5) (a) Cesium chloride — It is a density gradient.



1) Density gradient centrifugation

2) Semi conservative

3) 1958



1) False — E-coli divides in 20 minutes

2) True



1) What are the requirements of Meselson and Stahl experiment?

2) Why did Meselson and Stahl use nitrogen?

3) What is semi conservative DNA replication? By whom was is experimentally proved?



1) Describe Meselson and Stahl’s experiment to show that DNA replication is semi conservative.





Students, in previous topic, we have studied about the properties of genetic

material. Also in the Meselson and Stahl’s experiment we have come to know

that replication of DNA is emiconservative. In our today’s topic we will discuss

DNA replication process in detail.


DNA replication is a process by which DNA make copy of itself during cell-

division. It is a semi conservative process in which one strand of the DNA is

parental and other one is newly synthesized. It is an important process that takes place in the dividing cells. The cell undergoes DNA replication during the S-phase of the cell cycle.



1. DNA replication is semi conservative.

2. It is bidirectional process.

3. It proceeds from specific point called origin of replication.

4. It proceeds from 5’ to 3’ direction.

5. It is multi enzymatic process.

6. It occurs with high degree of fidelity.


When DNA replicates, various types of enzymes are involved. These enzymes

have the ability to quicken the reaction by building or break down the items that

they act upon. Below listed are the enzymes involved in DNA replication:


|. Helicase: It opens the DNA helix by breaking hydrogen bonds between the

nitrogenous bases.


Il. Topoisomerase: It helps to relieve the stress on DNA when unwinding by

causing breaks and then resealing the DNA.


Ill. DNA polymerase I: It undergoes exonuclease activity by removing RNA primer

and replaces it with newly synthesized DNA.


IV. DNA polymerase Il: It repairs DNA.


V. DNA polymerase Ill: It is the main enzyme that adds nucleotides in the 5-3’



VI. RNA primase: It synthesize RNA primers needed to start the replication.


Vil. Ligase: It seals the gaps between the Okazaki Fragments to create one

continuous DNA strand.


VII. Single strand binding proteins (SSB): It binds to single stranded DNA to avoid

DNA rewinding back.


DNA replication involves following steps: -


1. ACTIVATION OF DEOXYRIBONUCLEOTIDES - The four nucleotides of DNA i.e., AMP, GMP, CMP & TMP are floating in nucleus. They are

activated by ATP to form  deoxyribonucleosides triphosphates called,



2. RECOGNITION OF INITIATING POINTS From a particular point unwinding of DNA molecule start with the help of specific initiator proteins.


3. UNWINDING OF DNA MOLECULE — DNA double helix unwind into single strands by the breakdown of weak hydrogen bonds. Unwinding of Helix is helped by enzyme helicase. The separation of two single strands of DNA

creates Y-shaped replication fork.


4. FORMATION OF RNA PRIMER - The DNA directed RNA polymerase form the RNA primer.


5. FORMATION OF NEW DNA CHAINS The enzymes DNA polymerase can polymerize the nucleotides only in 5’ 3’ direction. DNA polymerase is responsible for the template direct condensation of deoxyribonucleotide



One of the strands is oriented in the 3’ > 5 direction is called leading strands. The other strand oriented in the 5’ — 3' direction called lagging strands. As the results of their different orientation, the two strands are replicated differently.


A short piece of RNA called a PRIMER comes along and binds to the end of the leading strand. The primer act as the starting point for the DNA synthesis.


DNA polymerase binds to the leading strand and then walks along it,adding new complimentary nucleotide basis (A, C, G & T) to the strands of DNA in the 5’—.3’ direction.


This sort of replication is called continuous.


lagging strand:Small pieces of DNA called Okazaki fragments are than added to the lagging strand in 5’ — 3’ direction.


This type of replication is called Discontinues as the Okazaki fragments,

formed in parts, will need to be joined up.


6. REMOVAL OF RNA PRIMER —- Once the small pieces of Okazaki fragments have been formed RNA primer is removed from 5’ end.


7, PROOF READING- Proof Reading is done to make sure there is no mistake in the new DNA sequence.Finally, an enzyme called DNA ligase seals up the sequence of DNA in the

two continuous and discontinuous double strands.




(i) Multiple Choice Questions:


1. What is the nature of DNA replications?

(a) Conservative

(b) Non-Conservative

(c) Semi-Conservative

(d) None


2. The Unwinding of Helix is helped by enzymes:

(a) Helicase


(c) Polymerase

(d) Ligase


3. Which of the following is true about DNA polymerase?

(a) It can synthesize DNA in the 5’ — 3’ direction

(b) ) It can synthesize DNA in the 3’ — 5’ direction

(c) It can synthesize mRNA in the 3’ — 5’ direction

(d) It can synthesize MRNA in the 5’ — 3’ direction


4. Okazaki fragments are:

(a) Large

(b) Small

(c) Both type

(d) None


5. Which of the following reaction is required for proof reading during

DNA replication by DNA polymerase?

(a) 5° — 3' Exonuclease activity (b) 3’ — 5’ Exonuclease activity

(c) 3’ -+ 5 Endonuclease activity (d) 5° — 3’ Endonuclease activity


(ii) True/ False:

1. Unwinding of Helix is helped by Enzyme helicase.

2. The four nucleotides of DNA are ATP, GTP, CTP & TTP.


(iii) Fill ups:

1. RNAprimer is removed at ..............0...... end.

2. DNA directed ...................... polymerase forms the RNA primer.






A) Multiple Choice Questions:


1. (c) Semi-Conservative — Each DNA strand serve as a template for the synthesis of a new strand producing two new DNA molecule, each with one new strand and one old strand.


2. (A) Helicase - A DNA helicase is an enzyme that unwind the DNA double helix

by breaking the hydrogen bounds between the complimentary basis.


3. (A) It can synthesize DNA in the 5’ — 3’ direction — DNA pole can synthesize only anew DNA strandin 5’ — 3 direction only.


4. (B) Small- Okazaki fragments are small sequence of DNA nucleotides (Approximately 150 to 200 base pair in eukaryotes)


5. (B) 3’ — 5 Exonuclease activity- This removes the mis- paired nucleotides

and polymerase begins again activities known as proof reading.


B) True/ False:

1) True

2) False - The four nucleotide of DNA are AMP, GMP, CMP & TMP.


C) Fill Ups:

1) 5

2) RNA



1. What are Okazaki fragments?

2. What do you mean by leading strands?



1. Explain the process of DNA replication?






Students, in our previous topic we have learnt that DNA, or deoxyribonucleic

acid, is the hereditary material in humans and almost all other organisms. Here in

today’s topic, we will study about another important nucleic acid which is responsible for converting the genetic information of DNA into Proteins.


RNA or Ribose Nucleic Acid is the other nucleic acid present in all biological

cells. It is principally involved in the synthesis of protein, carrying the messenger instructions from DNA which itself contains the genetic instruction required for the development and maintenance of life.






RNA is unbranched single- stranded polymer of ribonucleotides.


Each nucleotide unit is composed of three molecules: a phosphate group, a 5-carbon ribose sugar, and a nitrogen containing base.


The bases in RNA are Adenine(A), Guanine(G), Uracil(U) and Cytosine(C).


The various components are linked up as in DNA i.e. Adenine bonds with Uracil and Cytosine bonds with Guanine.




There are three types of RNA in every cell.


The three types of RNA are transcribed from different regions of DNA template.


The types of RNA are classified into three types:

1. mRNA or messenger RNA

2. tRNA or transfer RNA

3. rRNA or ribosomal





The DNA that controls protein synthesis is located in the chromosomes within the nucleus.


The ribosome, on which the protein synthesis actually occurs are placed in

the cytoplasm.


Therefore, some sort of agency must exist to carry instructions from the DNA to ribosomes. The agency does exist in the form mRNA.


The mRNA carries the message (information) from DNA about the sequence of particular amino acid to be joined to form a polypeptide, hence its name, messenger RNA.


Itis also called information RNA or template RNA.



The tRNA has many varieties.


Each variety carries a specific amino acid from the amino acid pool to the mRNA on the ribosomes to form a polypeptide, hence, its name.


The tRNA form about 15% of the total RNA of a cell.


Its molecule is the smallest of all the RNA types.


A tRNA molecule, as proposed by R.W, Holley in 1965, has the form of a clover leaf that results from self-folding and base pairing creating paired stems and unpaired loops.


Ithas four regions:

1. Amino acid carrier end

2. Anticodon end

3. DArm or Enzyme Site

4. TyC Am or Ribosome Site



The rRNA molecule is greatly coiled.


In combination with protein, it forms the small and large sub-units of the ribosomes, hence its name.


It forms about 80% of the total RNA of a cell.


A eukaryotic ribosome is 80S.


Its large 60S subunit consists of 28S, 58S and 5S rRNAs and over 45 different basic proteins.


Its small 40S subunit comprises 18S RNA and about 33 different basic proteins.


A prokaryotic ribosome is 70S.


Its large 50S subunit consists of 23S and 5S rRNAs and about 34 different basic proteins.


Its small 30S subunits comprise 16S rRNA and about 21 different basic proteins.


The 3' end of 18S rRNA (16S rRNA in prokaryotes) has a binding site for the

mRNA Cap.


The 5S rRNA has a binding site for t RNA.



There are two more minor types of RNAs

1. Small nuclear RNA or snRNA

2. Small Cytoplasmic RNA or scRNA

SnNRNA Role- It helps in processing of rRNA and mRNA in the nucleus.

ScRNA Role-lt helps in binding the ribosomes to endoplasmic reticulum (ER) in the cytoplasm.


All the three kinds of RNA play an important role in PROTEIN SYNTHESIS.

Messenger RNA (mRNA), carries the genetic information copied from DNA.


Transfer RNA (tRNA) is the key to deciphering the code words in MRNA.Each type of amino acids has its own type of tRNA, which binds it and carries it to the growing end of a polypeptide chain if the next code word on MRNA calls for it.


Ribosomal RNA (rRNA) associates with a set of proteins to form



RNA serves as genetic material in some viruses.

Example: Tobacco Mosaic Virus (TMV)


RNA carries out a broad range of functions, from translating genetic

information into the molecular machines and structures of the cell to

regulating the activity of genes during development, cellular differentiation,

and changing environments.






1. How many types of main RNA present in eukaryotes?

a) 1

b) 2

c) 3

d) 4



2. Which RNA carries information from nucleus to cytoplasm?

a) rRNA

b) mRNA

c) tRNA

d) ScRNA


3. Who proposed the clover leaf shape of tRNA?

a) T.H morgan

b) George John Mendel

c) R.W Holley

d) Watson & crick


4. RNA acts as genetic material in which types of virus?

a) Potato blight

b) Salmonela typhi

c) Corona

d) Tobbaco mosaic virus (TMV)


5. What is the percentage of rRNA in total RNA?

a) 15%

b) 4%

c) 80%

d) 70%


(ii) FILL Ups:

1. RNAserveras _________ Material in Some Virus (TMV)

2. ThemRNA has______ SHAME Of Molecule.


(iii) TRUE / FALSE:

1. mRNAis also known as template RNA.

2. R.W. Holley in 1965 told that tRNA has clover leaf form.


I. MCQs:

1. (c) There are three types of RNA: mRNA, tRNA, rRNA

2. (b) Messenger RNA (mRNA) decodes the information of DNA by Transcription process in nucleus.

3. (c) R.W. Holley

4. (d) Tobbaco Mosaic virus (TMV) contains single stranded RNA as genetic material.

5. (c) rRNA constitutes approx. 80% of the total RNA in a cell.


Il. FILL Ups:

1. Genetic

2. Linear



1. True

2. True



1Q. Write the different types of RNA?

2Q. Write the difference between MRNA & tRNA?

3Q. Write the role & importance of MRNA?

4Q. Describe TMV?

5Q. Write the structure & role of RNA?



Q1. Describe the various types of RNA in details?




Dear students, as in previous topic we have learnt about DNA replication.DNA replication is the process by which a double-stranded DNA molecule is

copied to produce two identical DNA molecules.Today we will study about transcription.The process of copying or transferring genetic information from one strand of the DNA into MRNA is termed as transcription.



This process occurs in nucleus in the presence of an enzyme, RNA polymerase.

In other words,The process of synthesis of mRNA from DNA or flow of genetic information from DNA to mRNA is known as transcription.


Where, Why and When transcription takes place?

We know, DNA is present within the nucleus and protein synthesis occurs in cytoplasm.


Total raw material for synthesis of protein is present in cytoplasm


The whole biological information about the type of protein, which is going to form, is present in DNA.


Being larger in size DNA can not come outside nucleus through nuclear pores.


That is why DNA transfers its information to RNA (messenger RNA or mRNA).


This process occurs in G1 and G2 phase of interphase of cell cycle.Process of transcription:

General Information:

DNA is double stranded.


Biological information is present in one strand and it is Known as sense or

coding strand. This strand lies in the sequence of 5’ — 3’.


Its complementary strand is known as antisense strand and lies in sequence of 3’ — 5’. It acts as template for the formation of RNA.


RNA formed on template strand will be in the sequence of 5’ — 3’.


In the nucleotide of RNA, nitrogenous base URACIL is present instead of nitrogenous base THYMINE as compared to nucleotide of DNA.


RNA polymerase enzymes are required for this process.


RNA polymerase is only of one type in prokaryotes.


It has two extra factors i.e. sigma (o) factor (helps in initiation of

transcription) and rho (p) factor (helps in completion or termination of transcription).


RNA polymerase in eukaryotes is of three types

RNA polymerase | (forms ribosomal or rRNA)


RNA polymerase || (forms messenger or MRNA)


RNA polymerase III (forms transfer or tRNA)


So, it is clear that for transcription RNA polymerase II is required. This RNA

polymerase unwinds the two templates of the DNA.


From the figure given below, it is very clear that new mRNA is formed and

nucleotide sequence of newly formed MRNA is exactly same as of the nucleotide sequence of sense strand (which contains biological information) with a difference of presence of uracil in spite of thymine.


Requirements for transcription:

(1) DNA template

(ii) |The enzyme like RNA polymerase.

(iii) All four types of ribonucleosides triphosphates (ATP, CTP, GTP and UTP), are required because these ribose nucleotides join to form RNA chain.


Transcription unit:

Transcription unit is that segment of DNA which is participating in transcription.


This unit in case of prokaryotes is polycistronic (having information for many

cistrons or proteins).


In case of eukaryotes it is mono-cistronic (having information for one cistrons or proteins).


Transcription unit of DNA molecule comprises of three regions-

(i) A promoter

(ii) The structural gene

(iii)A terminator


The promoter is small sequence of DNA which provides binding site for RNA polymerase and is located towards the 5’ end of the structural gene. It is also called the upstream end.


The structural gene is the gene to be transcribed.


The two strands of the DNA in the structural gene have opposite polarity, strand has 3’ — 5’ polarity and other strand has 5’ — 3’ polarity.


Strand having 3’ — 5’ polarity acts as a template and referred as template

strand or anticoding strand.


Strand with 5’ — 3’ polarity acts as a coding strand or sense strand.


All the biological information is present on the coding or sense strand.


The terminator is small sequence of DNA which provides binding site for

terminator factor and is located towards the 3’ end of the structural gene and

also called as downstream end.


The promoter part is A (adenine), T (thymine) rich base area and also known

as TATA box.


It was discovered by Pribnow and also called Pribnow box.


The terminator part is having palindromic sequence or poly A rich base



Stages of transcription: There are three stages of transcription:

(i) Initiation

(ii) | Elongation

(iii) Termination



The RNA polymerase enzyme binds to the promoter region with the help of sigma (o) factor in case of prokaryotes and with transcription factors in eukaryotes.


Sigma (0) factor or Transcription factor and RNA polymerase together,

forming a transcription initiation complex.



RNA polymerase moves along the DNA and causes unwinding and splitting of DNA strands.


The area where DNA strand unwinds itself, become bulgy and known as transcription bubble.


This area gives site for the formation of RNA.


Now RNA polymerase will start moving from the promoter to the structural gene and will start forming RNA in the direction of 5’ — 3’ over the template strand which have the sequence of 3’ — 5’.


Now the ribonucleotides start coming over the template strand and start pairing with the bases on the template DNA strand with the help of hydrogen bonds.


The RNA polymerase helps in unwinding of DNA strand and in polymerization of ribonucleotides by forming phosphodiester bond, to form long chain of RNA in the 5’ -3’ sequence.


RNA polymerase utilizes ribonucleotide triphosphates (ATP, GTP, CTP and

UTP) for the formation of RNA by the formation of phosphodiester bonds.



When RNA polymerase enzyme reaches the terminator area, then the rho (po) factor of RNA polymerase enzyme separates the RNA polymerase enzyme and newly formed mRNA from the terminator area in prokaryotes.


RNA synthesized from 5’ end to 3’ end, antiparallel to template strand.


Now the RNA formed will be processed further.


Many non-coding areas (introns) are also formed on new MRNA.


Now non coding areas (introns) from newly formed RNA are removed by

enzymes called nucleases (splicing).


Coding areas (exons) are re-joined by ligase enzyme (union) to form a long

chain of mature MRNA.


At 5’ end a 7-methyl guanosine cap is added.


At 3' end a poly adenosine (poly A) tail is added to form a complete mRNA.


Let us know what we have learnt!


1. Multiple choice questions


(i) The RNA formed after transcription is:

a) mRNA

b) rRNA

c) tRNA

d) All the above


(ii) The process of transcription occurs in what phase of interphase of cell cycle.

a) G; phase

b) G2 phase

c) S phase

d) Gi and G2 phase


(iii) Where the process of protein synthesis occurs in a cell?

a) in nucleus

b) in cytoplasm

c) inside ribosomes

d) All the above


(iv) What is the correct sequence of the stages of transcription?

a) Initiation, Elongation, Termination

b) Termination, Elongation, Initiation

c) Elongation, Initiation, Termination

d) Initiation, Termination, Elongation


(v) Which RNA polymerase is required in translation?

a) RNA polymerase |

b) RNA polymerase II

c) RNA polymerase III

d) All the above


2. Fill in the blanks

(1) In transcription, elongation stage isfollowedby

(2) The area where DNA strand unwinds itself is known as ,

(3) TATA box of promoter in transcription unit is also known as ;


3. True / False

(1) In prokaryotes, rho (p) factor of RNA polymerase enzyme helps in initiation in


(2) The promoter site is located towards the 3’ end of the structural gene.

(3) DNA can not come out from nucleus to synthesise the proteins.




(i) (a) MRNA

(ii) (d) G; and G2 phase

(iit) (b) in cytoplasm

(iv) (a) Initiation, Elongation, Termination

(v) (b) RNA polymerase II


(1) Termination

(2) Transcription bubble (because this area of DNA strand becomes bulgy due to splitting of two strands of DNA and a bubble-like structure appears there)

(3) Pribnow box because It was discovered by Pribnow.


(1) False, because for transcription the initiation factor is sigma.

(2) False, the promoter site is located towards the 5’ end of the structural gene.

(3) True, because of its larger size, DNA can not come outside the nucleus.



(1) Why transcription is required?

(2) Name the three types of RNA polymerase. What function they perform?

(3) What do you mean by polycistronic unit of transcription?



(1) What do you understand by  transcription. Discuss with the help of steps or stage involved in transcription.





Students, as discussed in previous topics that Central Dogma is the process

in which DNA is decoded by RNA to makes proteins. The process by which DNA is

copied to RNA is called transcription, and that by which RNA is used to produce

proteins is called translation. In our today’s topic we will discuss Translation in detail.Translation is the process of translating the sequence of MRNA (messenger RNA)into a sequence of amino acids. This translation takes place during protein



This process involves the transport of amino acids to the ribosome, where they are

assembled in the polypeptide chain. After which they will assemble into proteins

somewhere in the cytoplasm. This process is accomplished by ENA and occurs in several stages.



The key components required for translation are:


Ribosomes: The ribosome is a complex organelle, present in the cytoplasm,

which serves as the site of action for protein synthesis. It provides the

enzymes needed for peptide bond formation.


mRNA: mRNA is used to convey information from DNA to the ribosome. It is a single strand molecule, complimentary to the DNA template, and is generated through transcription. Strands of MRNA are made up of codons,each of which signifies a particular amino acid to be added to the polypeptide in a certain order.


RNA: This is a single strand of RNA composed of approximately 80ribonucleotides. Specific t RNAs binds to sequence on the MRNA template and

add the corresponding amino acid to the polypeptide chain. Therefore, t RNAs

are the molecules that actually “translate” the language of m RNA into the language of Proteins.


RNAs need to interact with three factors:


They must be recognized by the correct Aminoacyl Synthetase.


They must be recognized by ribosomes.


They must bind to the correct sequence in MRNA.



Aminoacyl tRNA synthetase: Through the process of t RNA “charging,”each t RNA molecule is linked to its correct amino acid by a group of enzymes called Aminoacyl t RNA synthetases.


At least one type of Aminoacyl t RNA synthetase exists for each of the 20

amino acids.



Protein biosynthesis involves following major steps:


1. Activation of amino acids: Here, when an Amino Acid (AA) and Adenosine Triphosphate (ATP), is mediated by Aminoacyl ynthetases, enzyme in the presence of Mg? the amino acid- enzyme -AMP complex is formed. The

complex remains temporarily associated with the enzyme. The amino acid-AMP-enzyme complex is called an activated Amino Acid.


2. Charging of tRNA: The enzyme complex formed above is reacted withthe specific t RNA. As a result, an Amino Acid is transferred into t RNA and the enzyme and AMP are liberated. The reaction is catalyzed by the same Aminoacyl-t RNA Synthetase enzyme. The resulting tRNA-amino acid complex is called a charged tRNA.


3. Initiation of polypeptide chain: Now the charged t RNA shifts to the ribosome. The large and small are the two subunits of the ribosome. The MRNA binds to the small subunit and the charged t RNA bind to the larger subunit to complete the initiation complex.


The Aminoacyl t RNA binding site (A site) and Peptidyl Site (P site) are the two

sites of binding in large sub-units.


The mRNA chain has at its 5’ end an “initiator” or “start” codon (AUG) that signals 

the start of polypeptide formation or Translation.


This start codon lies close to the P site of the ribosome.


The amino acid Formyl-Methionine (methionine in eukaryotes) initiates the

process. It is carried by t RNA having UAC anticodon which bonds to AUG initiator

codon of MRNA by hydrogen bonds.Initiation factors (IF 1, IF 2 and IF 3) and GTP promote the initiation process.


4. Elongation:The first codon on MRNA binds with the anticodon of the methionyl t RNA complex in the P site.

The other Aminoacyl t RNA complex with the appropriate amino acids thus enters the ribosome and attaches to A site.The peptide bond is formed between the first and second amino acids, when the anticodon of new t RNA binds to the second codon in the m RNA in the presence of an enzyme, Peptidyl Transferase.Then, the translocation takes place i.e. when the first amino acid and the tRNA bond broken and both get released out from the complex from E site.The second t RNA from the A-site is pulled to the P site along with the m RNA.


5. Termination:Two conditions are necessary for termination of protein synthesis.

One is the presence of a stop codon that signals the chain elongation to terminate,

and the other is the presence of release factors (RF) which recognise the chain

terminating signal.There are three terminating codons, UAA, UGA and UAG for which t RNAs do not exist. Termination of polypeptide chain is signaled by one of these codons in the MRNA. Release of the Peptidyl t RNA from the ribosome is promoted by three


specific release factors, RF 1, RF2 and RF3.RF1 recognizes triplets UAA and UAG, while RF2 recognizes UAA and UGA. The

third factor RF3 does not possess any release activity of its own, but it stimulates the binding of RF1 and RF2 with the ribosome.






1. Translation is the process of polymerisation of -

(a) amino acids

(b) sugars

(c) fats

(d) energy units


2. Which of the following is responsible for the initiation of RNA polymerase activity?

a) initiation site

b) promoter region

c) sigma factor

d) rho factor


3). Translation occurs ina .....

(a) nucleus.

(b) cytoplasm

(c) ribosome

(d) lysosome


4). Which is energy rich molecule required for the initiation of translation....

(a) ATP

(b) GTP

(c) CTP

(d) AMP


5). Which of the following step of translation does not consume a high

energy phosphate bond.

(a) Translocation

(b) Amino acid activation

(c) Peptidyl transferase reaction

(d) Aminoacy] tRNA binding to A-site


B) FILL Ups:

1) A translational unit in mRNA is the sequence of RNA that is flanked by the

start codon, ........., and the codes for a polypeptide.

2) The cellular structure responsible for synthesizing roteinsis............



1. Translation takes place before transcription.

2. E.coli bacteria can synthesize all of the amino acids required for protein





1. (b) amino acids _ Translation is ail about protein synthesis. Polymer of

Amino acids joined together by peptide bond to form polypeptides.


2. (c) sigma factor is a protein needed for initiation of RNA polymerase



3. (c) Ribosome — is the structural unit in cell where protein synthesis takes place


4. (b) GTP — is required for translation initiation, for elongation and for



5. (c) Peptidyl transferase reaction involves formation of peptide bond.



(a) stop codon

(b) Ribosome



(a) FALSE; According to Central Dogma first step is DNA replication then

transcription and the last step is translation.

(b) TRUE



Q1) What is the function of mRNA?

Q2) Differentiate between transcription and translation?

Q3) What do you mean by translocation?



Q1) Describe the process of Translation?





The DNA is a blue print of life. It carries vital information of life but how information get translated into protein is still a mystery which has kept the

scientist waiting for a long time. This mystery was unfolded when the GENETIC

CODE was understood by the scientist in the field of genetics. James Watson and Crick proposed double helix modal for structure of DNA and got Nobel Prize.


During Replication and Transcription, a nucleic acid was copied to form another nucleic acid. The process of translation requires transfer of genetic information from a polymer of nucleotides to synthesise a polymer of amino acids. Change in nucleic acids were responsible for change in amino acids in proteins. It was very challenging for scientists to determine the biochemical

nature of genetic material and the structure of DNA. They found it exciting

deciphering of genetic code. This was a complex process. Many scientists

were involved from different disciplines.


George Gamow, a physicists, who came with the idea that there are only 4 nucleotide bases and they code for 20 amino acids.


It means the code should be a combination of bases.


The combination of 4° i.e. (4 x 4 x 4) would generate 64 codons, many more codons than required.


It was hard task to provide proof that the codon was a TRIPLET.


Indian origin scientist Har Gobind Khorana developed chemical method that was helpful in synthesising RNA molecules with defined combination of bases.( homopolymers and copolymers)


A Homopolymer is made up of only one type of monomers and a Copolymer is made of two or more monomers.


Marshall Nirenberg and Heinrich J. Matthaei were the first to discover the nature of codon.


At first polyuracil RNA sequence UUU coded for Phenylalanine.


The Severo Ochoa enzyme was discovered by scientist Severo Ochoa.


It is Polynucleotide Phosphorylase (PNPase) which is bifunctional enzyme with phosphorolytic 3° to 5’ exoribonuclease activity and 3’ terminal oligonucleotide polymerase activity.


The function of this enzyme is to dismantie the long RNA chain that start from 3' end and move toward 5’ end.


Finally the checker board for genetic code was prepared.


GENETIC CODE:Genetic code is the sequence of nucleotides that codes for the corresponding amino acid sequence of proteins.

1) Itis TRIPLET. Out of 64 codons, 61 codons code for amino acids and 03

codons do not code for any amino acid.


2) One codon codes for only one amino acid, hence it is unambigous and specific.


3) Some amino acids are coded by more than one codon. Hence the code is degenrate.


4) The codon is read in MRNA in contiguous fashion there are no punctuations.


5) The code is nearly universal; for example. From bacteria to human UUU

would code for phenylalanine (phe). Some expections to this rule have been

found in mitochondrial codons and in some Protozoans.


6) AUG has dual function it codes for methionine and it also act as initiator code.


7) 61 Sense codon and 03 Nonsense codons are there.UAA, UAG, UGA are STOP CODONS or NON- SENSE CODONS.



The phenomena of change occurring in DNA sequence caused either by external

factors or internal factors like smoking and UV rays is termed as mutation.

Variation caused by change in building block and single base pair of DNA is termed

as POINT MUTATION example is SICKLE CELL ANAEMIA When reading frame of genetic code is altered by insertion or deletion of one or two bases leads to frame shift mutation.



tRNA is also called SRNA soluble RNA, or activator RNA.has a role as adapter

molecule. It has an anticodon loop having bases complimentary to the code. It

has amino acid accepter. The secondary structure of tRNA looks like CLOVER

LEAF. In actual structure tRNA looks like inverted





1. Which of the following is not a feature of genetic code?

a) Triplet

b) degenerate

c) Ambiguous

d) Non overlapping


2. The codon is a .

a) Singlet

b) Triplet

c) Doublet

d) Quadruplet


3. Who was instrumental in finding chemical method of determining DNA

a) Severo Ochoa

b) George Gamow

c) Har Gobind Khorana

d) Marshall Nirenberg


4. Sickle cell anemia is an example of

a) point mutation

b) both of these

c) silent mutation

d) none of these


5. The secondary stucture of t RNA looks like

a) clover leaf like structure

b) flower like structure

c) inverted L shape structure

d) inverted T shape structure



1. The codon AUG code for amino acid methionine also serves as___.

2. tRNA has an anticodon loop that has bases the code.



1) Some amino acids are coded by more than one codon, so the genetic code

is degenerative.

2) AUG codes for lysine.

3) Severo Ochoa enzyme is polynucleotide phosphorylase.



1. c) Ambiguous The genetic code is unambiguous and specific.

2. b) Triplet genetic code is triplet

3. c) Har Gobind Khorana

4. a) point mutation

5. a) clover leaf like structure



1) Initiator codon

2) Complementary



1) True

2) False AUG codes for Methionine and also acts as initiator codon

3) True



1) Discuss role of Har Gobind Khorana in defining Genetic code.

2) How tRNA is the adapter molecule?

3) What are frame shift mutations?



1) Write the salient features of Genetic Code.





In your previous assignment, you have studied about the genetic code. Now,you know that the genetic code is the set of rules used by living cells to translate information encoded within the genetic material (DNA or mRNA) into proteins.All the genes present in the chromosome are not expressed simultaneously.

The cell permits the expression of only a few genes at a time, thus maintaining its

economy. So, in this assignment, we will discuss about the regulation of gene

expression.Gene expression is the mechanism at the molecular level by which a gene is

able to express itself in the phenotype of an organism.



From a number of studies on the metabolism of the bacterium Escherichia Coli,two French scientists, Jacob and Monod proposed a model of gene regulation,known as the Operon model.

Operon is a co-ordinated group of genes such as Structural Gene, Operator Gene, Promoter Gene and Regulator Gene which function together and regulate a metabolic pathway as a unit e.g.: Lac operon.

Lac operon is an inducer operon. A segment of DNA made up of 3- structural

genes, operator gene, promoter gene and regulator gene.


a) Transcribes mRNA for polypeptide synthesis.

b) It is the gene which receives the product of the regulator gene. It allows the functioning of the operon when it is not covered by the biochemical produced by the regulator gene.

c) Provides the attachment site for the RNA polymerase

d) It synthesises a biochemical or regulator protein which can act positively as activator and negatively as repressor. It controls the

activity of the operator gene.

e) and or co-repressor (from outside) are also found.



The lac refers to lactose. In E. Coli, breakdown of lactose requires three

enzymes. These enzymes are synthesised together in a co-ordinated manner

by functional unit of DNA, i.e., fac operon. Since the addition of lactose itself

stimulates the production of the required enzymes, thus it is called inducible

operon system.

The lac operon consists of the following:

a. Three Structural Genes: Three structural genes are:

1. Lac z: The z gene codes for B-galactosidase which is primarily responsible

for the hydrolysis of the disaccharide, lactose into its monomeric units

galactose and glucose.


2. Lac y: The y gene codes for the permease, which increases the permeability of the cell to B-galactosidase.


3. Lac a: The a gene codes for the transacetylase which can transfer acetyl

group to B-galactosidase.


b. Operator Gene: It interacts with a protein molecule or regulator molecule, which

prevents the transcription of structural genes.


c. Promoter Gene: The gene possesses site for RNA polymerase attachment


d. Regulator gene (i): The gene codes for a protein known as the repressor protein, it is synthesised all the time from the i-gene,that is why it is constitutive gene which is functional always.


SWITCHING OFF THE OPERON:The operon is switched off when the repressor protein produced by regulatory or inhibitor gene binds to operator gene. RNA polymerase gets blocked, so there would be no transcription.



Regulation of lac operon by repressor is referred to as negative control or

regulation. If lactose is provided in the growth medium of the bacteria, the lactose

is transferred into the cells through the action of Permease. A very low level of lac

operon has to be present in the cell all the time; otherwise lactose cannot enter the

cells. In the presence of an inducer, such as lactose or allolactose, the repressor is

inactivated by interaction with inducer. This allows RNA polymerase access to the

promoter and transcription proceeds.

Lac operon is under control of positive regulation as well.



(1) Multiple Choice Questions


1. Lac operon was proposed by:

(a) Hershey and Chase

(b) Meselson and Stahl

(c) Alec Jeffreys

(d) Francois Jacob and Jacques Monod


2. The equivalent of structural gene is :

(a) Operon

(b) Recon

(c) Muton

(d) Cistron


3. Which of the following is required as inducer in the Lac operon?

(a) Glucose

(b) Lactose

(c) Galactose

(d) Lactose and Galactose


4. The Lac operon consists of :

(a) Four regulatory genes only

(b) One regulatory gene and 3 structural genes

(c) Two regulatory genes

(d) 3 regulatory and 3 structural genes


5. In lac operon, the gene Z codes for which enzyme ?

(a) Transacetylase

(b) B-galactosidase

(c) Permease

(d) Repressor protein



1. In bacteria, the regulation of gene expression is usually affected through .


2. Lac operon in bacteria codes for genes responsible for the metabolism of .


3. Promoter gene provides attachment site for



1. Operator gene is the site or the gene for binding RNA polymerase in an


2. Lac operon gets switched on in E.Coli when lactose is present and it binds to the repressor.


(1) Multiple choice Questions:

1. (d) Francois Jacob and Jacques Monod


2. (d) Cistron. Structural genes are continuous stretch of DNA that regulates a particular genetic trait and Cistron also denotes a coding sequence or segment of DNA encoding a polypeptide.


3. (b) Lactose. Lactose inactivates repressor.


4. (b) One regulatory genes and three structural genes. The Lac ‘i’ gene is

regulatory, while the three structural genes are Lac z, Lac y and Lac a.


5. (b) B-galactosidase. B-galactosidase hydrolyzes lactose to galactose and glucose.


(Il) Fill Ups :

1. Operons. Gene regulation is a process in which a cell determines which gene it will express and when


2. Lactose. Lac operon has three structural genes required for the degradation of lactose.


3. RNA polymerase. Promoter gene marks the site at which transcription of mRNA starts and where RNA polymerase enzyme binds


(I) True / False

1. False. Operator gene is the segment of DNA to which a repressor binds.


2. True. Lactose binds itself to active repressor leading to change in its structure. Operon gets transcribed and enzymes are produced.



1. Study the figure given below and answer the questions that follow:

a. Name the enzyme transcribed by Z gene

b. How does the repressor molecule get inactivated

2. Name the genes that constitute an operon

3. What is the role of Lactose in lac operon



1. Define an operon. Explain an inducible operon.





Students, as you know, the history of the human race has been filled with

curiosity and discovery about our abilities and limitations. With the rapid growth

of scientific knowledge and experimental methods, humans have begun to unravel and challenge another mystery, the discovery of the entire genetic make-up of the human body. This endeavor, the Human Genome Project

(HGP), has created hopes and expectations about better health care.


HUMAN GENOME PROJECT was called a mega project. Human genome is said to have approximately 3x10°bp and if one cost of sequencing required is US $3 per bp then approximately cost of project will be 9 billion US dollars. The HGP was initiated in 1990 under the leadership of American geneticist Francis Collins, with support from the U.S. Department of Energy (DOE) and

the National Institutes of Health (NIH). The effort was soon joined by scientists from around the world.The Human Genome Project is an ambitious research effort aimed at

deciphering the chemical makeup of the entire human genetic code (i.e.,the genome).



1. Identify all the approximately 20,000-25,000genes in human DNA.

2. Determine the sequences of the 3 billion chemical base pairs that make up human DNA.

3. Share this information in databases.

4. Improve tools for data analysis.

5. Transfer related technologies to other sectors.

6. Address the ethical, legal, and social issues (ELSI)that may arise from the




The method involves two major approaches.

1. EE. This method is focussed on identifying

all the genes that are expressed as RNA.


2. . This method took the blind approach of simply sequencing the whole set of

 genomes that contained all the coding and

noncoding sequence, and later assigning different regions in the sequence with




The commonly used vectors were


[BBM (Bacterial artificial chromosomes) It is based on natural, extra-chromosomal plasmid of E. coli. These vectors can accommodate upto 300-350 kb of foreign DNA and are also being used in genome sequencing project.


[HM (Yeast artificial chromosomes) are used to clone DNA fragments of more than 1Mb in size. However, While YAC can be used for cloning and sequencing, BACs are much more stable.


FREDERICK SANGER developed a principle by which fragments were sequenced. Sanger is also credited for developing method for determination of amino acid sequences in proteins. These sequences were then arranged

based on some overlapping regions present in them. This required generations of overlapping fragments for sequencing. Alignment of these sequences was humanly not possible. Therefore, specialised computer-based programmes were developed. These sequences were subsequently annotated and were assigned to each other. The sequence of chromosome 1 was completed only

in May 2008.



1. The human genome contains 3164.7 million nucleotide bases.

2. The average gene consists of 3000 bases.

3. Total number of genes is estimated at 30,000.

4. The functions are unknown for over 50% of the discovered genes.

5. Less than 2% of genome codes for proteins.

6. Repeated sequences make up very large portion of human genome.

7. Chromosomes 1 has most genes (2968) and the Y has the fewest (231).

8. Scientists have identified about 1.4 million locations where single base DNA

differences (SNPs-single nucleotide polymorphism) occurs in humans.

9. The largest gene in humans is Dystrophin.


This information promises to revolutionise the process of finding chromosomal

locations for disease associated sequences and tracing human history.





1. Identification of human genes and their functions.

2. Understanding of polygenic disorders e.g. cancer, hypertension, & diabetes.

3. Improvement in gene therapy.

4. Improve diagnosis of diseases.

5. Development of pharmacogenesis.

6. Improved knowledge on mutations.

7. Development of biotechnology.



 We still do not know about the following: -

Gene number, exact locations and functions.


Gene regulation, DNA sequence organization


Chromosomal structure and organization and so on.





1. The Human Genome Project was initiated by:

a) NIH and DOE

b) NIH and EBI

c) NIH and DDBJ

d) DOE and DDBJ


2. Which of the following vectors were used in HGP?

a) Plasmid and cosmid

b) A Phage and M13 vector

c) Phagmid and shuttle vectors

d) BAC and YAC


3. According to HGP human genome consists of:

a) 100000 genes

b) 50000 genes

c) 30000 genes

d) 20000 genes


4. YAC means:

a) Yellow artificial chromosomes

b) Yeast artificial chromosomes

c) Yeast acquired chromosomes

d) Yeast artificial chromatids


5. The largest gene in humans is:

a) Titin

b) Dystrophin

c) Insulin

d) Phosphofructokinase



1. The human genome contains ............ million nucleotide bases.

2. The functions are unknown for over .....% of the discovered genes.



1. EXPRESSED SEQUENCE TAGS method is focussed on identifying all the

genes that are expressed as Proteins.

2. SNP means Single Nucleotide Polymorphism.




1. a (NIH and DOE) HGP was initiated by U.S. Department of Energy (DOE) and the National Institutes of Health (NIH).


2. d (BAC and YAC) Both BAC and YAC were used in HGP. While YAC can be used for cloning and sequencing, BACs are much more stable.


3. c (30000 genes)


4. b (Yeast artificial chromosomes) used to clone DNA fragments of more than 1Mb in size.


5. b (Dystrophin)



1. 3164.7

2. 50



1. FALSE EXPRESSED SEQUENCE TAGS method is focussed on identifying all the genes that are expressed as RNA.




1. Why is human genome project called a mega project?

2. What are the benefits of the human genome project?



1. Write salient features of human genome project?






DNA fingerprinting, also called DNA typing, DNA profiling, genetic fingerprinting, genotyping, or identity testing, in genetics, method ofisolating and identifying variable elements within the base-pair sequence

of DNA (deoxyribonucleic acid). It is a technique of determining nucleotide

sequences of certain areas of DNA which are unique to each individual. Each person has unique DNA fingerprint.



ALEC JEFFREY'S (1984) invented the DNA finger printing technique at Leicester University, United Kingdom.


Dr V. k. Kashyap and Dr Lal Ji Singh started DNA fingerprinting technology,in India at CCMB (Centre for cell and molecular biology) Hyderabad.



Human genome possesses numerous small noncoding but inheritable sequences of bases which are repeated many times.


The area with same sequence of bases repeated several times is called REPETITIVE DNA.


They can be separated as satellite from the bulk DNA during density gradient

centrifugation and hence called SATELLITE DNA.


SATELLITE DNA is of two types




Microsatellite also Known as Simple sequence repeats (SSRs) or short tandem repeats (STRs) are repeating sequence of 2-6 base pairs of DNA.


Mini satellite also known as (VNTR) is a section of DNA that consist of short series of bases of 10-16 base pairs.


Their analysis is useful in genetic and Biology research, forensic and DNA



An inheritable mutation occurring in a population at high frequency is referred to as DNA POLYMORPHISM.


Repeated nucleotides sequence in a non-coding DNA of an Individual is called VARIABLE NUMBER OF TANDEM REPEATS (VNTR).



1. ISOLATION OF DNA: The DNA is extracted from the nuclei of white blood cells, blood, semen, hair follicle cells.


2. DNA AMPLIFICATION: The extracted DNA is amplified through PCR (Polymerase chain reaction) It produces numerous copies of DNA.


3. FRAGMENTATION: The DNA molecule are first broken with the help of enzyme RESTRICTION ENDONUCLEASE (called Chemical knife) that cuts them into fragments.


4. SEPARATION OF VNTRs CHOPPED: DNA is exposed to electrophoresis over agarose polymer gel. It separates DNA fragments. The separated VNTRS can be recognized by staining them Acridine dye that become fluorescent

when illuminated with UV Radiations.


5. SINGLE STRANDARD DNA: VNTRS are treated with alkaline chemicals to split them in to single stranded DNAs.


6. SOUTHERN BLOTTING: The separated VNTR single stranded sequences are transferred to nylon membrane placed over a gel. The procedure is called SOUTHERN BLOTTING.


7. DNA PROBES: They are small radioactive single stranded DNA segments of known sequences of nitrogen bases.


8. HYBRIDISATION: Nylon sheet is immersed in a bath to which DNA probes are added. The probes get attached to single stranded VNTRs_ having complementary nucleotide sequence. Rest of probes are worked away.


9. EXPOSURE TO X-RAY FILMS: The nylon membrane containing the radioactive DNA probes and VNTRs is exposed to X-Ray film. The hybridized radioactive VNTRs appear as dark bands. The film gives us DNA prints. The

prints are compared and relationship found.



1. INDENTIFICATION: DNA fingerprinting is sure method of identification of criminals involved in various type of crimes including rape and murder.


2. PATERNITY- MATERNITY DISPUTES: The method can provide reliable information as to real biological father, mother on off spring.


3. HUMAN LINEAGE: It provides information to human lineage and relationship with apes.


4. MIGRATIONS: DNA finger printing can  identify racial groups, origin, historical migration and invasions.


5. DNA fingerprinting is used to identify and protect the commercial varieties of crops and livestock.




(i) Multiple Choice Questions:


1. DNA finger-printing was developed by:

(a) Francis Crick

(b) Knorana

(c) Alec Jeffery

(d) James Watson


2. The technique to distinguish the individuals based on their DNA print

patterns is called:

(a) DNA fingerprinting

(b) DNA profiling

(c) Molecular fingerprinting

(d) All of these


3. Which tissue samples are used for DNA fingerprinting?

(a) Hair

(b) Skin

(c) Blood

(d) Any of above


4. Which one is used in DNA Finger-printing?

(a) Western Blotting

(b) Flow Cytometry

(c) Northern Blotting

(d) Southern Blotting


5. DNA Fingerprinting is applied in:

(a) Identification of criminals

(b) Paternity disputes

(c) Identify and protect Commercial varieties of plants

(d) All of these


(ii) Fill ups:-

6. Polymorphism arises due to .

7. DNA molecules are broken with the help of an enzyme known as

8. Mini-satellites are tt.


(iii) True/ False:

9. The DNA finger pattern of child is 100% similar to the father’s DNA print.

10. The sequences of satellite DNA do not code for proteins.

11. DNA can be used as a useful tool in the forensics application by the presence of lysozymes in it.


A) Multiple Choice Questions:

1. (c) ALEC JEFFREY, he developed this technique in 1984


2. (d) All of the above DNA fingerprinting, also called DNA typing, DNA profiling, genetic fingerprinting, genotyping, or identity testing is the method of isolating and identifying variable elements within the base-pair sequence of DNA (deoxyribonucleic acid).


3. (d) All of the above DNA Sample can be obtained from any tissue and can be

amplified by PCR.


4. (d) Southern Blotting It is the process of separation of VNTRs from agrosegel to nylon membrane.


5. (d) All of these


B) Fill-Ups:

6) Mutation — Polymorphism arises due to mutation. It can also be known as the

discontinuous variation that is seen among the individuals of species.


7) Restriction Endonuclease - It is also called as chemical knife that cuts DNA

molecule in to fragments.


8) Minisatellites - They are the SHORT NON CODING REPITITIVE SEQUENCES present throughout the chromosomes.


C) True/False:

9) False; The DNA fingerprint pattern of a child is 50% bands similar to the father and

rest 50% to the mother.


10) True ; The small peaks formed by the DNA when kept in the density gradient

centrifugation are called as SATELLITE DNA. These DNA sequences do not code for proteins. On the contrary, they form a large portion in the genes of humans


11) False; DNA can be used as a useful tool in the forensics application by showing

some degree of polymorphism with hair follicles.



1. Whatis the principle of DNA fingerprinting?

2. What do you mean by satellite DNA?





1. What is DNA fingerprinting? Explain the different steps used in DNA fingerprinting with a suitable diagram.




Hello students, we have discussed all the topics of Molecular basis of inheritance. Let us work on the NCERT questions of the chapter. Now it is time to test our understanding of the content and we should be able to answer the related questions.



DNA and RNA are the two types of nucleic acids found in organisms.


DNA is a long polymer of  deoxyribonucleotides.


In RNA every nucleotide residue has an additional -OH group present at 2'position of ribose.


Double helix of DNA is made of two nucleotide chains.



Euchromatin is transcriptionally active chromatin, whereas Heterochromatin

is inactive


The human genome contains 3164.7 million nucleotide bases.“Polymorphism variation of genetic level arises, due to mutation.




Question 1Group the following as nitrogenous bases and nucleosides:Adenine, Cytidine, Thymine, Guanosine, Uracil and Cytosine.

Answer 1Adenine, Uracil, Cytosine, Thymine are nitrogenous bases and Cytidine

Guanosine are nucleosides.


Question. 2If a double stranded DNA has 20 percent of CYTOSINE, calculate the percentage of ADENINE in the DNA.

Answer 2According to Chargaff’s rule —


More over AMOUNT of ADENINE =AMOUNT of THYAMINE.And AMOUNT OF GUANINE =AMOUNT OF CYTOSINE.Since percentage of CYTOSINE is 20%. Percentage of GUANINE is also 20%.Now we are left with 60%. This 60% will be half Adenine and half Thymine i.e.30% Adenine and 30% Thymine.


Question.3If the sequence of one strand of DNA is written as follows:


Write down the sequence of COMPLEMENTARY STRAND in 5'—3' direction.

Answer 3


The two strands of DNA have antiparallel polarity.


Question.4If the sequence of the coding strand in a transcription unit is written as

follows: 5' -ATGCATGCATGCATGCATGCATGCATGC-3' Write down the sequence of mRNA.

Answer 4

The sequence of m RNA shall be :



Question.5Which property of DNA double helix led Watson and Crick to hypothesis semi-conservative mode of DNA replication? Explain.

Answer 5The semi conservative mode of DNA replication hypothesised by Watson

Crick was based upon the property that during replication the Two strands would separate and act as a template for the synthesis of new complimentary

 strands. After the completion of replication, each DNA molecule would have one parental and one newly synthesised strand. This scheme was termed as

“Semi conservative DNA replication.


Question 6.Depending upon the chemical nature of the template (DNA or RNA)and the nature of nucleic acids synthesised from it (DNA or RNA), listthe types of nucleic acid polymerases.

Answer 6Process of replication needs a set of enzymes. Depending upon the chemical

nature of the template( DNA or RNA) and the nature of the nucleic acids synthesised from it, there are four different types of nucleic acid polymerases:


(i) DNA-dependent DNA polymerase - Uses DNA template to catalyse the polymerisation of deoxynucleotides.


(ii) DNA-dependent RNA polymerase- Catalyses transcription of all types of RNAs in bacteria.


(iii) In eukaryotes there are three types of DNA- dependent RNA polymerase:

(a) RNA polymerase I- Transcribes rRNAS

(b) RNA polymerase Il- Transcribes precursor of MRNA i.e. hnDNA

(c) RNA polymerase Ill-Transcribes tRNA, srRNA and snRNAS. .


(iv) Reverse transcriptase for the synthesis of complementary DNA over RNA template.


Question.7How did Hershey and Chase differentiate between DNA and protein in their experiment while proving that DNA is the genetic material?

Answer 7 Harshey and Chase (1952) worked on viruses which infect bacteria called

bacteriophages .They concluded with their experiment that bacteria which were infected with viruses that had radioactive DNA were found to be radioactive.It gave the idea that DNA acted as hereditary material which was

transmitted from bacteriophage to  bacteria.Bacteria which was infected with

bacteriophages (VIRUSES) that has radioactive proteins. They were not found to be radioactive. It indicates that proteins are not transported to bacteria.


Question.8Differentiate between the followings:

(A) Repetitive DNA and Satellite DNA

(B) MRNA and tRNA

(C) Template strand and Coding strand


Answer 8. (A)Repetitive DNA refers to the sequences of DNA when a small stretch of DNA

is repeated many times. They code for proteins.Satellite DNA refers to those repetitive DNA sequence which do not code for

any protein but form a large protein of the gene.


(B)mRNA It is the straight chain molecule. Triplets of Nitrogenous bases (in the

form of codons) code for the specific amino acids and decides the sequence of

amino acid in a polypeptide. It has highly methylated 5’ cap Poly A (a long chain of adenine nucleotides) 3° tail.tRNA This molecule becomes a looped and assumes a clover leaf like structure. It has 4 important binding sites. At one end is a sequence of three bases called anticodon, opposite to it is amino acid arm, another arm is ribosome binding site and finally enzyme binding site.


(C)Template strand this is the strand of DNA with 3° > 5 polarity that functions

as template for transcriptions.Coding strand this is the strand of DNA with 5 > 3° polarity that does not code for region of RNA


Question.9Enlist two essential roles of ribosome during translation.

Answer 9) Role of ribosomes

1) Ribosomes usually forms linear or helical groups during the active protein

synthesis called polyribosomes or polysomes .


2) Ribosomes acts as a catalyst for formation of peptide bond by an enzyme peptidyl transferase .



Question.10In the medium where E. coli was growing, lactose was added, which induced the lac operon. Then, why does lac operon shut down some time after addition of lactose in the medium?

Answer.10The addition of lactose induced the lack operon. As lactose formed inducer —

repression complex, the operator gene got switched on. As the lactose were digested by the action of enzymes repressor became free and operator gene gets switched off.


Question.11Explain (in one or two lines) the function of the followings:

(A) Promoter

(B) tRNA

(C) Exons

Answer 11(A)Promoter :It is the region of DNA where RNA polymerase attaches prior to the

process of transcription


(B)tRNA :It works as adaptor molecules for carrying amino acids to the mRNA template during protein synthesis .lt bears anticodon and recognises the specific codon on MRNA .


(C)Exons: The region of a gene, which become part of m RNA and code for different regions of the protein are referred to exons


Question.12Why the Human Genome project is called a mega project?

Answer 12.Human Genome Project is called a

Mega Project due to following reasons:


1. HGP took approximately 13 years

to accomplish in 2006.


2. Human genome is set to bear 3x10° bp. The cost of the product can be imagined that if the cost of sequencing a base pair is $3

sequencing of {3x10}° base pair cost comes out to be 9 billion US dollars.


3. If such sequences are stored in books, with every page having 1000 letters and each book is of 1000 pages, total number of books formed will be 3300 to store information present in single human cell thus for

storing this data high speed computers are required.


4. Because of this project, many new areas have opened in genetics.Depending on the high magnitude and requirements of the project, HGP has been called mega project.


Question.13What is DNA fingerprinting? Mention its application.

Answer 13 DNA fingerprinting, also called DNA typing, DNA profiling, genetic

fingerprinting, genotyping, or identity testing, in genetics, method of isolating

and identifying variable elements within the base-pair sequence of DNA (deoxyribonucleic acid).DNA fingerprinting technique was developed by Alec Jeffery (1985.86) at

Leicester University, United Kingdom. Inheritance of DNA is very stable, every

person has specific pattern of sequence which shows a combination of DNA of both mother and father.Applications of DNA fingerprinting ;


1) Paternity disputes can be solved by DNA fingerprinting


2) It can solve the problems of evolution


3) It can be used to study the breeding patterns of animals facing the danger

of extinction


4) It is useful in restoring health of the patients suffering from the leukemia

(blood cancer )


5) It is very useful in the detection of crime and legal persuits


Question.14 Briefly describe the following:

(a) Transcription

(b) Polymorphism

(c) Translation

(d) Bioinformatics

Answer 14(a)Transcription

It is one of the fundamental processes that happens to our genome. It's the process of turning DNA into RNA. And you may have heard about the central dogma, which is DNA, to RNA, to protein. Well, transcription refers to that first part of going from DNA to RNA. And we transcribe DNA to RNA in specific

places. The most popular places are those things that code for these protein-

encoding genes. But there are a whole host of other RNAs that get transcribed, like transfer RNAs and ribosomal RNAs, that do other functions that are genomic as well.


(b) Polymorphism As in DNA sequence is the basis of genetic mapping of human genome as

well as of DNA fingerprinting, it is essential that we understand what DNA

polymorphism means in simple terms. A DNA Polymorphism is a sequence difference compared to a reference standard that is present in a least 1-2% of a population. Polymorphism (variation at genetic level) arises due to mutations. New mutations may arise in an individual either in somatic cells or in the germ cells.


(c) Translation is the process in which ribosomes in the cytoplasm or endoplasmic reticulum synthesize proteins after the process of transcription of DNA to RNA in the cell's nucleus. The entire process is called gene expression. In translation, messenger RNA (mRNA) is decoded in a ribosome,

outside the nucleus, to produce a specific amino acid chain, or polypeptide.

The polypeptide later folds into an active protein and performs its functions in

the cell. The ribosome facilitates decoding by inducing the binding of complementary tRNA anticodon sequences to mRNA codons. The tRNAs

carry specific amino acids that are chained together into a polypeptide as the

mRNA passes through and is "read” by the ribosome.


(d) Bioinformatics is an interdisciplinary field that develops methods and software tools for understanding biological data, in particular when the data sets are large and complex. As an interdisciplinary field of science,bioinformatics combines biology, computer science, information engineering,

mathematics and statistics to analyse and interpret the biological data.Bioinformatics has been used for in silico analyses of biological queries using mathematical and statistical techniques.






Dear students, we have got in depth knowledge of Genetic material in this chapter. We have got knowledge about strucure of DNA and why DNA is suitable to be genetic material. We have gone through DNA replication,

Transcription and Translation. We read about Human Genome Project and DNA Fingerprinting. We will revise the chapter by learning important Definitions and Differences related to the content we have studied.


IMPORTANT DEFINITIONS NUCLEOTIDE DNA and RNAi are polymers of deoxyribonucleotides and

ribonucleotides. A nucleotide has three components —

1) A nitrogenous base

2) A deoxyribose pentose sugar (DNA) OR ribose pentose sugar (RNA)

3) A phosphate group

There are two types of nitrogenous bases-

Purines ( Adenine and Guanine)Pyrimidines ( Cytosine , Uracil, Thymine)We will be able to learn about the differences between Purines and Pyrimidines. Both have a pyrimidine ring ( similar to Benzene ring ) . But  purines have pyrimidine ring fused to an imidazole ring.


When a nirogenous bases are linked to pentose sugar through N- glycosidic bond, there is formation of a Nucleoside. The phosphate group is linked through a phosphoester linkage to Nucleoside to make it a Nucleotide.




Replication is when DNA synthesis takes place.Transcription is when genetic information is copied from DNA to RNA.


Translation is process of polymerisation of amino acids to form polypeptide inturn

leading to protein synthesis. The order and sequence of amino acids are defined by

the sequence of bases in the MRNA.


Francis Crick proposed central dogma in molecular biology. It states that the

genetic information flows in a specific sequence —

DNA ---—----—- mRNA --—--------------—- Protein



When the negatively charged DNA is wrapped around the positively charged

histone octamer , it becomes a Nucleosome.



Okazaki fragments are short sequences of DNA nucleotides with approimately 150-200 base pairs which are synthesised on the lagging replication. The ligase enzyme joins the Okazaki fragments to make one strand during DNA replication.


GENETIC CODE:Genetic code is required for TRANSLATION.Genetic code is the set of rules ( triplet of base pairs) used to translate

information encoded within genetic material to specify particular amino acids to make proteins.






1) In a nucleotide nitrogenous base is linked to pentose sugar through —

(a) N-glycosidic bond

(b) phosphoester bond

(c) hydrogen bond

(d) hydroxyl bond


2) Replication refers to —

(a) Protein synthesis

(b) RNA synthesis

(c) DNA synthesis

(d) none of these


3) Genetic code is required in the process of

(a) Replication

(b) Translation

(c) Transcription

(d) All of these


4) The process of copying genetic information from one strand of DNA into RNA is termed as-

(a) Translation

(b) Transcription

(c) Replication

(d) none of these


5) The negatively charged DNA is wrapped around positively charged histone octamer to form —

(a) Nucleotide

(b) Nucleus

(c) Nucleoid

(d) Nucleosome



6) Translation is the last step of Central Dogma.

7) Histone octamer is negatively charged.

8) Genetic code is triplet.



9) Central dogma was proposed by......

10) Adenine and Guanine are.......






1) (a) N- glycosidic bond

2) (c) DNA synthesis

3) (b) Translation

4) (c) Transcription

5) (b) Nucleosome



6) True

7) False

8) True



9) Francis Crick

10) Purines



11) How nucleoside is different from nucleotide?

12) Define Genetic Code.

13) Define Nucleosome.



14) Differentiate between —

(a) Replication and Transcription

(b) Transcription and Translation.

Chapter 6 Molecular Basis of Inheritance