C H A P T E R
5
MOLECULAR BASIS OF
INHERITANCE
Learning Objectives
Students will be able to learn about
1. Structure of DNA and its significance.
2. Formation of proteins via DNA through mRNA.
3. Gene Expression and control.
4. Genome projects and its advantages.
List of Topics
Topic-1: Nucleic Acid –
DNA and RNA �������������������������������������������
Topic-2 : Genetic Code, Translation, Lac Operon,
HGP and DNA Fingerprinting���������������
TOPIC-1 Nucleic Acid – DNA and RNA
Concepts Covered
 Nucleic acids, packaging of DNA helix,  Nucleosome,  Experiments to show DNA as a genetic material,
 RNA, process of protein synthesis.
Revision Notes
Genetic Material
¾¾ Nucleic Acids
DNA and RNA are the two types of nucleic acids.
DNA is the genetic material in all organisms except
some viruses.
RNA is the genetic material in some viruses.
RNA mostly functions as messenger.
¾¾ Structure of Polynucleotide Chain
Polynucleotides are the polymers of nucleotides.
DNA and RNA are examples of polynucleotides.
A nucleotide has 3 components :
(i) A nitrogenous base
(ii) A pentose sugar (ribose in RNA and deoxyribose
in DNA)
(iii) A phosphate group
Nitrogen bases are of 2 types :
(a) Purines : It includes Adenine (A) and Guanine
(G).
(b) Pyrimidines : It includes Cytosine (C), Thymine
(T) and Uracil (U). Thymine (5-methyl Uracil)
present only in DNA and Uracil only in RNA ( In
place of thymine).
A nitrogenous base is linked to the pentose sugar
through an N-glycosidic linkage to form nucleoside.
Nucleosides in
RNA
Nucleosides in
DNA
Adenosine
Deoxyadenosine
Guanosine
Deoxyguanosine
Cytidine
Deoxycytidine
Uridine
Deoxythymidine
Nitrogen base + sugar + phosphate group =
Nucleotide (deoxyribonucleotide). In RNA, every
nucleoside residue has an additional –OH group
present at 2’-position in the ribose (nucleoside=Ribose
sugar+ Base pair) phosphate group is absent in
nucleoside.
2 nucleotides are linked through
3’ → 5’ phosphodiester bond to form
dinucleotide.
When series of nucleotides are linked
together, it forms polynucleotide.
¾¾ Structure of DNA
Johann Friedrich Miescher (1869): Identified DNA and
named it as ‘Nuclein’.
James Watson & Francis Crick proposed the double
helix model of DNA. It was based on the X-ray
diffraction data produced by Maurice Wilkins &
Rosalind Franklin.
DNA is made of two polynucleotide chains coiled in a
right-handed fashion. Its backbone is formed of sugar
and phosphates. The bases project inside.
The two chains have anti-parallel polarity i.e., one
chain has the polarity 5’ → 3’ and the other has 3’ → 5’.
Nitrogen bases of opposite chains are held together by
hydrogen bonds forming base pairs (bp).
There are two hydrogen bonds between A and T (A =
T) and three H-bonds between C and G (C ≡ G).
Purine comes opposite to a pyrimidine. This generates
a uniform distance between the two strands.
¾¾ Erwin Chargaff’s Rule
Purines and pyrimidines are always in equal amounts
i.e., A + G = T + C.
In DNA, the proportion of A is equal to T and the
proportion of G is equal to C i.e., A = T and G = C.
Molecular Basis of Inheritance
Q. Calculate the number of beaded structures
(nucleosomes) present in the nucleus of diploid
eukaryotic cell which possess 2.2 × 106 bp.
Sol. One nucleosome has 200 bp.
The number of beaded structures (nucleosomes)
present in the nucleus of diploid eukaryotic cell
which possess 2.2 × 106 bp.
\
2.2 ´10 6
= 1.1 × 104 or 11 × 103 nucleosomes
200
¾
­
¾
¾
¾
¾ The Search for Genetic Material
Griffith’s Experiment - Transforming Principle
Griffith (1928) used mice and a bacterial strain,
Streptococcus pneumoniae.
Streptococcus pneumoniae has two strains :
(a) Smooth (S) strain (Virulent): Has polysaccharide
mucous coat. Causes pneumonia.
(b) Rough (R) strain (Non-virulent): No mucous
coat. Does not cause pneumonia.
¾ Experiment
S-strain → Inject into mice → Mice die
R-strain → Inject into mice → Mice live
S-strain (Hk) → Inject into mice → Mice live
S-strain (Hk) + R-strain (live) → Inject into mice
→ Mice die
He concluded that there exists some ‘transforming
principle’, that is transferred from heat-killed S-strain
to R-strain. It enabled R-strain to synthesise smooth
polysaccharide coat and become virulent. This must be
due to the transfer of genetic material.
¾ Biochemical Characterisation of Transforming
Principle
Oswald Avery, Colin MacLeod & Maclyn McCarty in
1944 worked to determine the biochemical nature of
‘transforming principle’ in Griffith’s experiment.
They purified biochemicals (proteins, DNA, RNA, etc.)
from heat-killed S cells using suitable enzymes.
They discovered that:
(a) Digestion of protein and RNA (using Proteases
and RNases) did not affect transformation. So, the
transforming substance was not a protein or RNA.
(b) Digestion of DNA with DNase inhibited
transformation. It means that DNA caused the
transformation of R cells to S cells i.e., DNA was
the transforming substance.
¾ The Genetic Material is DNA
The fact that DNA is the genetic material also came
from the experiments of Alfred Hershey and Martha
Chase (1952).
They worked with viruses that infect bacteria and are
called bacteriophages.
¾ Hershey-Chase Experiment—Blender Experiment
Hershey and Chase made two preparations of
bacteriophage – In one, proteins were labelled with
35
S by putting in a medium containing radioactive
sulphur (35S). In the second, DNA was labelled with
32
P by putting in a medium containing radioactive
Phosphorous (32P).
¾
¾
¾
¾ Packaging of DNA Helix
In prokaryotes (e.g., E. coli), the DNA molecule is
held with some positively charged non-histone basic
proteins like negatively charged polyamines and form
‘nucleoid’.
In eukaryotes, it involves a number of molecules. Histones, Histone octamer, Nucleosome, Chromatin.
Two types of chromatin are:
(a) Euchromatin: Loosely packed and transcriptionally
active chromatin and is light-stained.
(b) Heterochromatin: Densely packed and inactive
region of chromatin and stains dark.
In eukaryotes, there is a set of positively charged basic
proteins called histones.
Histone proteins are rich in positively charged basic
amino acid residues lysine and arginine.
There are five types of histone proteins-H1, H2A, H2B,
H3 and H4.
Two molecules each of H2A, H2B, H3 and H4 organize
to form a unit of eight molecules called as histone
octamer.
Negatively charged DNA is wrapped around positively
charged histone octamer to form a structure called a
nucleosome.
Nucleosomes are connected with the help of linker
DNA on which H1 Histone is present.
¾ Nucleosome
A typical nucleosome contains 200 bp of DNA helix.
Therefore, the total number of nucleosomes in human
= 6.6 × 109 bp/200 bp = 3.3 × 107.
Nucleosomes constitute the repeated unit to form
chromatin.
Chromatin is the thread-like stained bodies.
Nucleosomes in chromatin appears as “beadson-string” when it is viewed under the electron
microscope.
Chromatin is packaged to form a solenoid or a zig-zag
structure.
Further supercoiling constitute a looped structure
called chromatin fibre.
These chromatin fibres further coil and condense at the
metaphase stage of cell division to form chromosomes.
Chromatin is packaged → solenoid → chromatin
fibres → coiled and condensed at metaphase stage →
chromosomes.
Higher level packaging of chromatin requires nonhistone chromosomal (NHC) proteins.
Example-1
The base ratio A + T/G + C may vary from species to
species but constant for a given species.
Length of DNA = number of base pairs × distance
between two adjacent base pairs.
f 174 (a bacteriophage) has 5386 nucleotides.
Bacteriophage lambda has 48502 base pairs (bp).
E. coli has 4.6 × 106 bp.
Haploid content of human DNA = 3.3 × 109 bp.
Number of base pairs in human = 6.6 × 109
Length of DNA in humans = 6.6 × 109 bp × 0.34 × 10–9
m/bp = 2.2 m
Length of DNA in E. coli =1.36 mm (1.36 × 10–3 m).
\ The number of base pairs = 1.36 × 10–3 m/0.34 × 10–9
m/bp = 4 × 106 bp.
BIOLOGY, Class-XII
Presence of thymine
Presence of uracil
Absence of 2’-OH
Presence of 2’-OH
¾
The two DNA strands are complementary. On heating,
they separate. When appropriate conditions are
provided they come together. (In Griffith’s experiment,
when the bacteria were heat-killed, some properties of
DNA did not destroy).
¾ RNA World
RNA is a single-stranded structure but it is often folded
back upon itself forming helices. Nitrogenous bases are
like those of DNA except that there is uracil in place of
thymine.
RNA was the first regulatory chemical and genetic
material in early life forms.
¾
¾
¾
¾
¾
Single-stranded
Double-stranded
­
Reasons for mutability
(high reactivity) of
RNA
­
Reasons for stability
(less reactivity) of
DNA
It acts as genetic material and biocatalyst.
Essential life processes (metabolism, translation,
splicing, etc) evolved around RNA.
DNA has evolved from RNA with chemical
modifications that made it more stable.
¾ Central Dogma of Molecular Biology
It was proposed by Francis Crick (1958). It states that
the genetic information flows unidirectionally from
DNA → RNA → Protein.
¾ Reverse Transcription: H. Temin and Baltimore in 1978
gave the concept of reverse flow of genetic information i.e., the formation of DNA from RNA. This is called
Reverse Central Dogma or Teminism or reverse transcription. This takes place in some of the viruses in the
presence of an enzyme called reverse transcriptase.
¾ Types of RNA
RNA is of 3 types –mRNA, tRNA and rRNA.
mRNA constitutes 2–5% of the total cellular RNA,
tRNA is about 15% and rRNA is about 70–80%.
mRNA (messenger RNA): Provides a template for
translation (protein synthesis) and is transcribed from
DNA.
rRNA (ribosomal RNA): Structural and catalytic role
during translation. e.g., 23S rRNA in bacteria acts as
ribozyme.
It is the component of ribosome and is the most stable
type of RNA.
tRNA (transfer RNA or sRNA or soluble RNA or
adaptor RNA): Brings amino acids for protein synthesis
and reads the genetic code.
tRNA is the smallest amongst all the RNA and is made
up of 70–80 nucleotides only.
¾ DNA Replication
[Board 2019]
Replication is the copying of DNA from parental DNA.
Watson & Crick proposed a semi-conservative mode
of replication.
It suggests that the parental DNA strands act as a
template for the synthesis of new complementary
strands. After the completion of replication, each
DNA molecule would have one parental and one new
strand.
¾ Experimental Proof
Mathew Meselson & Franklin Stahl (1958)
experimentally proved semi-conservative mode.
Meselson & Stahl’s Experiment:
They cultured E. coli in a medium
containing N15H4Cl (N15: heavy
isotope of N). N15 was incorporated
into both strands of bacterial DNA
and the DNA became heavier.
Another preparation containing N
salts labelled with N14 was also made.
N14 was also incorporated in both strands of DNA and
became lighter.
These two types of DNA can be separated by
centrifugation in a CsCl density gradient.
They took E. coli cells from the N15 medium and
transferred them to the N14 medium.
After one generation (i.e., after 20 minutes), they
isolated and centrifuged the DNA. Its density was
intermediate (hybrid) between 15N DNA and 14N DNA.
¾
¾
These preparations were used separately to infect
E. coli.
After infection, the E. coli cells were gently agitated
in a blender to separate the phage particles from the
bacteria.
Then the culture was centrifuged. Heavier bacterial
cells were formed as a pellet at the bottom. Lighter
viral components outside the bacterial cells remained
in the supernatant.
They found that,
(a) Supernatant contains viral protein labelled with
35
S, i.e., the viral protein had not entered the
bacterial cells.
(b) The bacterial pellet contains radioactive 32P. This
shows that viral DNA labelled with 32P had
entered the bacterial cells. This proves that DNA
is the genetic material.
¾ Properties of Genetic Material
A molecule that can act as a genetic material must fulfil
the following criteria:
(a) Be able to generate its replica by the process of
replication.
(b) Chemically and structurally be stable.
(c) Allow slow changes, the mutations that are
required for evolution.
(d) It should be able to store genetic information
which can be inherited.
(e) Be able to express itself as ‘Mendelian Characters’.
¾ DNA is a better Genetic Material than RNA due to the
following reasons :
DNA is chemically less reactive and structurally more
stable. It can to undergo repair.
Due to the unstable nature of RNA, RNA viruses (e.g.,
Qb Bacteriophage, Tobacco Mosaic Virus, etc.) mutate
and evolve faster.
For the storage of genetic information, DNA is better
due to its stability. But for the transmission of genetic
information, RNA is better.
RNA can directly code for protein synthesis, hence can
easily express the characters. DNA is dependent on
RNA for protein synthesis.
Molecular Basis of Inheritance
¾
¾
¾
It consists of 3 regions :
(a) A promoter (Transcription start site) : Binding
site for RNA polymerase.
(b) Structural gene : The region between promoter
and terminator where transcription takes place.
(c) A terminator : The site where transcription stops.
The DNA- dependent RNA polymerase catalyses the
polymerization only in 5’→3’direction.
3’→5’ acts as the template strand. 5’→3’ acts as the
coding strand.
3’–ATGCATGCATGCATGCATGCATGC–5’
template
strand.
5’–TACGTACGTACGTACGTACGTACG–3’
coding
strand.
¾ Transcription Unit and the Gene
Gene: Gene is Functional unit of inheritance. It is the
DNA sequence coding for RNA molecule.
Cistron: A segment of DNA coding for a polypeptide.
Structural gene in a transcription unit is of two types :
(a) Monocistronic structural genes
(split genes): It is seen in
eukaryotes. Here, the coding
sequences (expressed sequences
or exons) are interrupted by
introns (intervening sequences).
(b) Polycistronic structural genes:
It is seen in prokaryotes. Here,
there are no split genes.
Exons and Introns: In eukaryotes, the monocistronic
structural genes have interrupted coding sequences
i.e., the genes in eukaryotes are split. The coding
sequences or expressed sequences are called as exons.
Exons are said to be those sequences that appear in
mature or processed RNA. The exons are interrupted
by introns. Introns or intervening sequences do not
appear in mature or processed RNA.
¾ Steps of transcription in prokaryotes
Initiation: Here, the enzyme RNA
polymerase binds at the promoter
site of DNA. This causes the local
unwinding of the DNA double helix.
An initiation factor (s factor) present
in RNA polymerase initiates the RNA
synthesis.
Elongation: The RNA chain is synthesised in the 5’3’ direction. In this process, activated ribonucleoside
triphosphates (ATP, GTP, UTP & CTP) are added. This
is complementary to the base sequence in the DNA
template.
Termination: A termination factor (r factor) binds to
the RNA polymerase and terminates the transcription.
In bacteria (Prokaryotes), transcription and translation
can be coupled (Translation can begin before mRNA
is fully transcribed) because mRNA requires no
processing to become active.
Transcription and translation take place in the same
compartment (no separation of cytosol and nucleus).
¾ In eukaryotes, there are 2 additional complexities:
(a) There are three RNA polymerases :
RNA polymerase I: Transcribes rRNAs (28S,
18S & 5.8S).
¾
¾
¾
This showed that in the newly formed DNA, one strand
is old (N15 type) and one strand is new (N14 type). This
confirms the semi-conservative mode of replication.
After II generations (i.e., after 40 minutes), there were
equal amounts of hybrid DNA and light DNA.
Taylor et. al (1958) performed similar experiments on
Vicia faba (faba beans) using radioactive thymidine to
detect distribution of newly synthesised DNA in the
chromosomes. It proved that the DNA in chromosomes
also replicate semi-conservatively.
¾ The Machinery and Enzymes for Replication
DNA replication starts at a point called origin (ori).
A unit of replication with one origin is called a replicon.
During replication, the two strands unwind and
separate by breaking H-bonds in the presence of an
enzyme, Helicase.
Unwinding of the DNA molecule at a point forms a ‘Y’shaped structure called replication fork.
The separated strands act as templates for the synthesis
of new strands.
DNA replicates in the 5’→3’ direction.
Deoxyribonucleoside triphosphates (dATP, dGTP,
dCTP & dTTP) act as substrate and also provide energy
for polymerisation.
Firstly, a small RNA primer is synthesised in presence
of an enzyme, primase.
In the presence of an enzyme, DNA dependent DNA
polymerase, many nucleotides join with one another to
primer strand and form a polynucleotide chain (new
strand).
The DNA polymerase forms one new strand (leading
strand) on a continuous stretch in the 3’→5’ direction
(Continuous synthesis).
The other new strand is formed in small stretches
(Okazaki fragments) in the 5’→3’ direction
(Discontinuous synthesis).
The Okazaki fragments are then joined together to
form a new strand by an enzyme, DNA ligase. This
new strand is called lagging strand.
If a wrong base is introduced in the new strand, DNA
polymerase can do proofreading.
E. coli completes replication within 38 minutes i.e.,
2000 bp per second.
In eukaryotes, the replication of DNA takes place at the
S-phase of the cell cycle. Failure in cell division after
DNA replication results in polyploidy.
¾ Transcription
[Board 2020]
It is the process of copying genetic information from
one strand of the DNA into RNA.
Here, adenine pairs with uracil instead of thymine.
Both strands are not copied during transcription,
because
(a) The code for protein is different in both strands.
This complicates the translation.
(b) If two RNA molecules are produced simultaneously they would be complementary to each other,
hence form a double-stranded RNA. This prevents
translation.
¾ Transcription Unit
It is the segment of DNA between the sites of initiation
and termination of transcription.
BIOLOGY, Class-XII
-
Capping: Here, a nucleotide methyl guanosine
triphosphate (cap) is added to the 5’ end of
hnRNA.
Tailing (Polyadenylation): Here, adenylate
residues (200–300) are added at 3’ end. It is the
fully processed hnRNA, now called mRNA.
-
RNA polymerase II: Transcribes the hete­
rogeneous nuclear RNA (hnRNA). It is the
precursor of mRNA.
RNA polymerase III: Transcribes tRNA, 5S
rRNA and snRNAs (small nuclear RNAs).
(b) The primary transcripts (hnRNA): They contain
both the exons and introns and are non functional.
Hence introns have to be removed. For this, it
undergoes the following processes :
Splicing: From hnRNA, introns are removed
(by the spliceosome) and exons are spliced
(joined) together.
[Board, 2015]
MNEMONICS
1. Concept: Erwin Chargaff’s Rule
Mnemonics: AayaTha; ChalaGya
Interpretations: A- Adenine = T- Thymine G-Guanine
= C- Cytosine
2. Concept: Central Dogma of Molecular Biology
Mnemonics: Doctors Recovered Patients
Interpretations: DNA RNA → Protein
Important Diagrams :
Fig 5.1 : A polynucleotide Chain of DNA
Fig 5.2 : Double Stranded polynucleotide chain
DNA
H1 histone
Histone
octamer
Core of histone
molecules
Fig 5.3 : Central Dogma
Fig 5.4 : Nucleosome
Molecular Basis of Inheritance
Bacteriophage
Radioactive
(35S) labelled
protein capsule
Radioactive (32P)
labelled DNA
GC
AT
AT
GC
TA
1. Infection
CG
AT
GC
TA
2. Blending
3. Centrifugation
No Radioactive (35S)
detected in cells
+
Radioactive [35S]
detected in
supernatant
5’
No Radioactive (32P)
detected in cells
+
No radioactivity
detected in
supernatant
14
CG
A
T
C
A
C
G
GC
TA
AT
CG
AT
CG
AT
GC
AT
GC
A
T
GC
TA
TA
GC
Generation II
N-DNA
N-DNA
14
N-DNA 40 min
15
20 min
N-DNA
14
Gravitational
force
N-DNA
14
15 15
N N
Heavy
14 15
N N
Hybrid
N14N 14N15N
Light Hybrid
14
Separation of DNA
by centrifugation
Fig. 5.7 : Meselson and Stahl’s Experiment
Fig 5.8 : Replicating Fork
3’
Fig 5.6 : Watson Crick model of
Semi-conservative DNA replication
Generation I
N-DNA
G
TA
GC
Fig 5.5 : The Hershey and Chase Experiment
15
T
A
T
Fig 5.9 : Schematic structure of a transcription unit
BIOLOGY, Class-XII
5'
3'
Initiation
5'
s
Promoter
Sigma factor
RNA polymerase
DNA helix
5'
3'
Elongation
3'
5'
RNA Terminator
3'
Termination
3'
5'
5'
r
RNA
3'
Rho factor
RNA polymerase
Fig 5.10 : Process of Transcription in Bacteria
5'
3'
Cappingcap
m
5' Gppp
3' mRNA
Intron
Exon
RNA splicing
Polyadenylation
3'
Poly a tail
m
Gppp
3'
m
Gppp
m
Gppp
Messenger RNA (mRNA)
Fig 5.11 : Process of Transcription in Eukaryotes
SUBJECTIVE TYPE QUESTIONS
Very Short Answer Type Questions
(1 mark each)
1. What happen’s when in a bacterium RNApolymerase binds to the promoter on a transcription
unit during transcription?
K [Delhi Set-1, 2020]
Concept Applied Transcription
Ans. When RNA-polymerase binds to the promoter on a
transcription unit during transcription it initiates the
process.
2. Name one amino acid, which is coded by only one
codon.
Ap [Comptt, Set-1, 2018]
3. A region of a coding DNA strand has the following
nucleotide sequence :
– ATGC –
What shall be the nucleotide sequence in
(i) sister DNA segment it replicates, and
(ii) m-RNA polynucleotide it transcribes ?
K [Foreign Set-1, 2, 3, 2017]
This Question is for practice and its solution is given at the end of the chapter.
Concept Applied Replication
Ans. (i) - TACG - (ii) - AUGC- ½+½
[Marking Scheme, 2017]
4. Mention one difference to distinguish an exon from
an intron.
U [Foreign Set-1, 2016]
Concept Applied Transcription
Ans. Exon : Coded/expressed sequence of nucleotides in
mRNA.
Intron : Intervening sequence of nucleotides not
appearing in processed mRNA. 1
[Marking Scheme, 2016]
Molecular Basis of Inheritance
5. Retroviruses have no DNA. However the DNA of
the infected host cell does possess viral DNA. How
is it possible?
Ap [Outside Delhi Set-I, 2015]
Ans. RNA is the genetic material in retrovirus. This RNA
forms DNA by the process of reverse transcription
with the help of the enzyme called reversetranscriptase.
1
functional only after processing by splicing.
During splicing, the introns are removed and exons
are joined. hnRNA also undergoes two additional
processes called capping and tailing. During
capping, the unusual nucleotide methyl guanosine
triphosphate (mGPPP) is added to the 5' end of
hnRNA. In tailing, about 200–300 adenylate residues
are added at the 3' end of mRNA. Now, this is the
fully processed hnRNA which is called mRNA. It
is now functional and is ready for translation. The
splicing of hnRNA occurs in the nucleus.
2
Short Answer Type Questions-I
(2 marks each)
Short Answer Type Questions-II
(3 marks each)
1. What are 'SNPs'? Where are they located in a human
cell? State any two ways the discovery of SNPs can
be of importance to humans.
R + Ap [Delhi Set-1, 2020]
Ans. SNPs are a Single nucleotide polymorphism. It is the
variation in the genome of the organisms within a
particular species because of changes in the sequence
of a single nucleotide.
SNPs are located within the chromosome in certain
discrete locations.
Importance of SNPs :
(i) Used to identify genetic regions associated with
certain genetic disorders.
(ii) Used in forensic sciences.
1+1+1
2. (a) Identify the polarity of x to x’ in the diagram
below and mention how many more amino acids
are expected to be added to this polypeptide
chain.
1. Carefully examine structures A and B of pentose
sugar given below. Which one of the two is more
reactive? Give reasons.
A [SQP, 2020-21]
Ans. A is more reactive
½
2'–OH group present in the pentose sugar
½
makes it more labile, catalytic and easily
degradable.
½ + ½ = 1
[CBSE Marking Scheme, 2020]
2. Although a prokaryotic cell has no defined nucleus,
yet DNA is not scattered throughout the cell.
Explain.
U [CBSE, Outside Delhi/Delhi, 2018]
Bor
Ans. DNA is a negatively charged, positively charged
protein, holds it in place, in large loops (in a region
termed as nucleoid).
½×4
[CBSE Marking Scheme, 2018]
AG
G
3. Describe the structure of a nucleosome.
U [Foreign, Set-II, 2017]
Brr
Concept Applied Packaging of DNA
Concept Applied Transcription
Ans. hnRNA is the primary transcript. It is non functional.
It contains both the coding region-exons and noncoding regions called introns in RNA. It is called
heterogeneous nuclear RNA or hnRNA. It gets
CUCUUGGGUCCGCAGUUUAA
x
Ans. (i) A typical nucleosome contains 200 bp of DNA
helix.
(ii) Negatively charged DNA is wrapped around
positively charged histone octamer.
(iii) It constitutes the repeating unit to form
chromatin.
(iv) The chromatin appears as ’’beads on string’’.
(v) The chromatin is packed to form a solenoid
structure and further supercoiling constitute a
looped structure called chromatin fibre.
(vi) Higher level packaging of chromatin requires
non-histone chromosomal (NHC) proteins.
4. Why does hnRNA need to undergo splicing? Where
does splicing occur in the cell?
U [CBSE, Delhi Set-1 & 3 Comptt. 2016]
x'
(b) Mention the codon and anticodon for alanine.
(c) Why are some untranslated sequences of bases
seen in mRNA coding for a polypeptide? Where
exactly are they present on mRNA?
K [CBSE Solve 2021] [3]
Concept Applied Translation
Ans. (a) x to x’ is 5' → 3' [½]
No more amino acids will be added [½]
(b) Codon is GCA [½]
Anticodon is CGU [½]
(c) The untranslated regions are required for an
efficient translation process. [½]
They are present before the initiation codon at the
5’ – end and after the stop/termination codon, at the
3’ – end. [½]
BIOLOGY, Class-XII
5. (a) State the hypothesis that S.L. Miller tried to
prove in the laboratory with the help of the set
up given above.
(b) Name the organic compound observed by him
in the liquid water at the end of his experiment.
(c) A scientist simulated a similar set up and added
CH4, NH3 and water vapour at 800°C. Mention
the important component that is missing in his
experiment?
K [SQP 2023–24]
3. Explain the mechanism of DNA replication with
the help of a replication fork. What role does the
enzyme DNA-ligase play in a DNA replication
fork?
U [Delhi Set-1, 2019]
Concept Applied DNA Replication
(a) The process of DNA replication begins at a point
called the origin of replication (ori), to form a
replication fork.
Ans:(a) Chemical evolution: First form of life
originated from pre-existing non-living
organic molecules.
(b) Amino acids
(c) H2
[Marking Scheme SQP 2023–24]
Long Answer Type Questions Ans. RNA polymerase II. 1
Capping- unusual nucleotide (methyl guanosine
triphosphate) is added to the 5’ end of the hnRNA,
1
Tailing-adenylate residues are added at 3’ end in a
template-independent manner 1
[Marking Scheme, 2019]
Concept Applied DNA is a Genetic material
Ans.(a)
Alfred Hershey and Martha Chase carried out
their experiments to prove that DNA is the
genetic material and not the protein.
(b) (i) They used radioactive phosphorus (32P) and
radioactive sulphur (35S).
They grew some viruses on a medium that
contained radioactive phosphorus and some
others on medium that contained radioactive
sulphur. Viruses grown in the presence of
radioactive phosphorus contained radioactive
DNA but not radioactive protein because DNA
contains phosphorus but protein does not.
Similarly, viruses grown on radioactive sulphur
contained radioactive protein but not radioactive
DNA because DNA does not contain sulphur.
(ii) Blender : To separate the viral protein coats that
are still attached to the surface of bacteria.
Centrifuge : To separate lighter supernatant
(containing viral protein coats) from the denser
residue (containing bacteria).
(iii) Observations :
(a) Bacteria that were infected with viruses having
radioactive DNA were found to contain
radioactive DNA.
(b) Bacteria that were infected with viruses having
radioactive protein coat were not found to
contain radioactivity.
Conclusion : DNA is the genetic material that
is passed from virus to bacteria and not the
protein.
1+2+2
(b) The separated strands act as templates for the
synthesis of new strands.
(c) DNA replicates in the 5’ → 3’ direction.
(d) dNTPs (Deoxyribonucleotide triphosphate) act as
substrate and also provide energy for the polymerization of nucleotides.
(e) DNA polymerase is an enzyme that assembles a
new DNA strand that is complementary to the
template strand.
(f) DNA polymerase continues to move along the
template strand and add new nucleotides to the
growing or complementary strand until the entire
genome is replicated.
(g) The DNA polymerase forms one new strand
(leading strand) in a continuous stretch in the
5’ → 3’ direction (continuous synthesis).
(h) The other new strand is formed in small stretches
(Okazaki fragments) in the 5’ → 3’ direction
(discontinuous synthesis).
(i) The Okazaki fragments are then joined together to
form a new strand by an enzyme, DNA ligase. This
new strand is called the lagging strand.
The function of DNA ligase is to join two
nucleotides. During the DNA replication process,
it joins Okazaki fragments together to form the
complete DNA strand .
3
4. Name the enzyme that transcribes hnRNA in
eukaryotes. Explain the steps that the hnRNA
undergoes before it is processed into mRNA.
U [Outside Delhi Set-2, 2019]
(5 marks each)
1. (a) State the reasons for which Hershey and Chase
carried out their experiments.
(b) Answer the following questions based on the
experiments of Hershey and Chase :
(i) Name the different radioactive isotopes they
used, and explain how they used them.
(ii) Why did they need to agitate and spin their
culture?
(iii) Write their observations and the conclusions
they arrived at.
U [Outside Delhi Set-3, 2019]
Molecular Basis of Inheritance
(a) State the ‘Central dogma’ as proposed by Francis Crick. Are there any exceptions to it? Support your answer
with a reason and an example.
(b) Explain how the biochemical characterisation (nature) of “Transforming Principle” was determined, which
was not defined from Griffith’s experiments.
Ap [Delhi 2018]
2.
Topper's Answer, 2018
BIOLOGY, Class-XII
4.
5. In the mid-twentieth century, scientists were still
unsure as to whether DNA or protein was the
genetic material of the cell. It was known that some
viruses consisted solely of DNA and a protein
coat and could transfer their genetic material into
hosts. In 1952, Alfred Hershey and Martha Chase
conducted a series of experiments.
(a) State the reasons for which Hershey and Chase
carried out their experiments.
(b) Answer the following questions based on the
experiments of Hershey and Chase:
(i) Name the different radioactive isotopes they
used, and explain how they used them.
(ii) Why did they need to agitate and spin their
culture?
(iii) Write their observations they arrived at.
[SQP 2023 – 24]
OR
(i) Describe the series of experiments of F. Griffith.
Comment on the significance of the results
obtained.
(ii) State the contribution of Macleod, McCarty and
Avery.
U + A [Outside Delhi Set-2, 2016]
OR
(i) Explain with the help of Griffith’s experiment how
the search for genetic material was conducted and
what was the conclusion drawn?
(ii) How did Macleod, McCarty and Avery establish
the biochemical nature of the so called “genetic
material” identified by Griffith in his experiment.
U + Ap [Delhi Set-2, Comptt. 2016]
Concept Applied Transcription + Translation
Ans: 5’--ATG ACC GTA TTT TCT GTA GTG CCC GTA
CTT CAG GCA TAA—3’= Coding strand
(a) 3’- TAC TGG CAT AAA AGA CAT CAC GGG
CAT GAA GTC CGT ATT---5’= Template
strand1
5’---AUG ACC GUA UUU UCU GUA GUG CCC
GUA CUU CAG GCA UAA---3’ == mRNA
strand1
(b) i. In a bacterium
5’---AUG ACC GUA UUU UCU GUA GUG CCC
GUA CUU CAG GCA UAA---3’ 1
ii. In humans
5’---mGpppAUG ACC UUU UCU GUG CCC CUU
CAG GCA UAA- Poly A tail--3’
1
(c) Nine amino acids in the polypeptide because
UAA is stop/terminator codon and does not
code for any amino acid.
[Marking Scheme SQP 2023–24]
(ii) Sigma factor associates with RNA polymerase
to initiate transcription, Rho factor gets
associated with RNA polymerase to terminate
transcription.
½+½
(iii) RNA polymerase I - Transcribes -RNAs
½
RNA polymerase III - Transcribes tRNA /
5srRNA / 5SnRNA
½
[Marking Scheme, 2016]
3. Given below is a stretch of DNA showing the
coding strand of a structural gene of a transcription
unit?
5’--ATG ACC GTA TTT TCT GTA GTG CCC GTA
CTT CAG GCA TAA—3’
(a) Write the corresponding template strand and the
mRNA strand that will be transcribed, along with
its polarity.
(b) If GUA of the transcribed mRNA is an intron,
depict the sequence involved in the formation of
mRNA /the mature processed hnRNA strand.
(i) In a bacterium
(ii) In humans
(c) Upon translation, how many amino acids will the
resulting polypeptide have? Justify.
U + K [SQP 2023–24]
Ans. (i) A-Template strand 1
B-Coding strand
1
Template strand has polarity 3’→ 5’
Coding strand has polarity 5’ → 3’
On the basis of polarity with respect to
promoter.
½+½
U + Ap [Delhi Set-3, 2014; Delhi Set-2, Comptt. 2015]
OR
Describe Frederick Griffith’s experiment on
Streptococcus pneumoniae. Discuss the conclusion
U + A [Outside Delhi Set-2, 2014]
he arrived at.
(i) Identify strands ‘A’ and ‘B’ in the diagram of
transcription unit given above and write the basis
on which you identified them.
(ii) S
tate the functions of Sigma factor and Rho factor
in the transcription process in a bacterium.
(iii) W
rite the functions of RNA polymerase-I and
RNA polymerase-III in eukaryotes.
Ap [Foreign Set-1, 2016]
OR
(i) Describe the various steps of Griffith’s experiment
that led to the conclusion of the ‘Transforming
principle’.
(ii) How did the chemical nature of the ‘Transforming
principle’ get established ?
Ans. (a) Alfred Hershey and Martha Chase carried out
their experiments to prove that DNA is the
genetic material and not the protein.
(b) (i) They used radioactive phosphorus (32P) and
radioactive sulphur (35S).
They grew some viruses on a medium that contained
radioactive phosphorus and some others on a
medium that contained radioactive sulphur. Viruses
Molecular Basis of Inheritance
s
production of proteins. Bacterial transcription
occurs in the cytoplasm alongside translation.
The process of transcription is completed in three
steps : Initiation, elongation and termination.
(a) Initiation : The enzyme binds at the promoter
site of DNA and initiates the process of
transcription. It causes the local unwinding
of the DNA double helix. In initiation, sigma
factor ( ) present in RNA polymerase initiates
the RNA synthesis.
(b) Elongation : The RNA chain is synthesized
in the 5’→ 3’ direction. RNA polymerase
uses nucleoside triphosphate as substrate
and polymerisation occurs according to
complementarity.
(c) Termination : Termination occurs when the
termination factor (rho) alters the specificity of
RNA polymerase to terminate the transcription.
As the RNA polymerase proceeds to perform
elongation, a short stretch of RNA remains
bound to the enzyme. As the enzyme reaches
6. Describe the packaging of DNA helix in a
the termination region, this nascent RNA falls
prokaryotic cell and an eukaryotic nucleus.
off and transcription is terminated.
3
K [Foreign Set-1, 2016]
(ii) The precursor of mRNA i.e., hnRNA contains
7. (i) Describe the process of transcription in bacteria.
both introns and exons. Introns are removed
(ii) Explain the processing the hnRNA needs to
and exons are joined by a process called
undergo before becoming functional mRNA in
splicing. The remaining mRNA is processed in
eukaryotes.
K [Outside Delhi Set-1, 2016]
two ways :
OR
(a) Capping: Here, an unusual nucleotide called
Explain the process of transcription in a prokaryote.
methyl guanosine triphosphate (cap) is added
to the 5’ end of hnRNA.
U [Outside Delhi Comptt. 2017, Set-2]
(b) Tailing: Here, adenylate residues (200–300) are
added at the 3’ end of hnRNA in a template
Concept Applied Synthesis of RNA
independent manner.
When hnRNA is fully processed, it is known as
Ans. (i) Bacterial transcription is the process in which
mRNA, which is transported out of the nucleus
messenger RNA transcripts of genetic material
to get translated.
2
in bacteria are produced to be translated for the
8. (a) Explain the process of DNA replication with the help of schematic diagram.
(b) In which phase of the cell cycle does replication occur in eukaryotes? What would happen if cell-division is
not followed after DNA replication?
K [Delhi Comptt. Set-1, 2015]
grown in the presence of radioactive phosphorus
contained radioactive DNA but not radioactive
protein because DNA contains phosphorus but
protein does not. Similarly, viruses grown on
radioactive sulphur contained radioactive protein
but not radioactive DNA because DNA does not
contain sulphur.
(ii) Blender: To separate the viral protein coats that are
still attached to the surface of bacteria.
Centrifuge: To separate lighter supernatant (containing
viral protein coats) from the denser residue (containing
bacteria). 1
(iii) Observations:
• Bacteria that were infected with viruses having
radioactive DNA were found to contain radioactive
DNA.
• Bacteria that were infected with viruses having
radioactive protein coat were not found to contain
radioactivity. [Marking Scheme SQP 2023 – 24]
– Replication of DNA begins at ori site to form a replication fork.½
– DNA dependant DNA polymerase forms a new strand in 5’ → 3’ direction.½
– Role of DNA ligase is to join discontinuously synthesised fragments.½
Ans. (a)
(b) S-phase.
Polyploidy
This Question is for practice and its solution is given at the end of the chapter.
2½
½
[Marking Scheme, 2014] ½
BIOLOGY, Class-XII
TOPIC-2 Genetic Code, Translation, Lac Operon, HGP and DNA Fingerprinting
Concepts Covered
 Genetic code  Translation  Gene expression  Lac Operon  HGP  Rice Genome Project
 Steps and application of DNA Fingerprinting
Revision Notes
[Board, 2019]
The relationships between genes and
DNA are best understood by mutation
studies.
Effects of large deletions and
rearrangements in a segment of DNA
may result in loss or gain of a gene and so a function.
A classical example of point mutation is a change of
single base pair in the gene for beta globin chain of
haemoglobin that results in the change of amino acid
¾
¾
¾ The Adaptor Molecule – tRNA
The tRNA is a molecule that has about 60% of its part
double-stranded and the rest remains single stranded
which has unpaired bases.
The tRNA has
(a) An anticodon (NODOC) loop that has bases
complementary to the CODON with which it gets
attached in mRNA.
(b) An amino acid acceptor end to which amino acid
binds. This end or site lies at the 3' end & CCA–
OH group. The 5' end bears G.
(c) T Y C loop : This is the site for attaching with
the ribosome. This has some unusual bases like Y
(pseudouridine) and ribothymidine.
(d) DHU-Loop : It is the binding site for the enzyme
aminoacyl synthetase. It is the largest loop and
has Dihydrouridine.
(e) Extra arm : It is a variable side arm lying between
T Y C and anticodon loop.
tRNA is called an adaptor molecule because it picks up
amino acids from the cytoplasm and transfers them to
ribosomes during protein synthesis.
For initiation, there is another tRNA called initiator
tRNA.
There are no tRNAs for stop codons.
2-D structure of tRNA looks like a clover-leaf according
to Robert Holly (1965). The 3-D structure looks like
inverted ‘L’ according to Klug (1974).
¾ Translation – Protein Synthesis
It takes place in ribosomes. It includes 4 steps :
1. Charging of tRNA (aminoacylation of tRNA)
• Formation of a peptide bond requires energy
obtained from ATP.
• For this, amino acids are activated (amino acid
+ ATP) and linked to their cognate tRNA in the
presence of aminoacyl tRNA synthetase. So, the
tRNA becomes charged.
2. Initiation
• It begins at the 5’-end of mRNA in the presence of
an initiation factor.
• The mRNA binds to the small subunit of the
ribosome. Now the large subunit binds to the small
subunit to complete the initiation complex.
¾
¾
¾
¾
¾ Genetic Code: It is the sequence of nucleotides in
mRNA that contains information for protein synthesis
(translation).
¾ 20 amino acids are involved in translation.
George Gamow: Suggested that for coding 20 amino
acids, the code should be made up of 3 consecutive
nucleotides.
Har Gobind Khorana: Developed the chemical
method in synthesising RNA molecules with
defined combinations of bases (homopolymers and
copolymers).
Marshall Nirenberg : Developed a cell-free system for
protein synthesis.
Severo Ochoa (polynucleotide phosphorylase) enzyme
is used to polymerise RNA with defined sequences in a
template-independent manner.
¾ Salient Features of Genetic Code
The genetic code is a triplet code (three-letter code)
where three adjacent nitrogen bases code for a single
amino acid.
61 codons code for amino acids. 3 codons (UAA, UAG
and UGA) do not code for any amino acids. They
function as stop codons (Termination codons or nonsense codons).
Genetic code is universal e.g., From bacteria to human
UUU codes for Phenylalanine. Some exceptions
are found in mitochondrial codons and in some
protozoans.
No punctuations between adjacent codons (comma
less code). The codon is read in mRNA in a continuous
fashion.
Genetic code is non-overlapping.
A single amino acid is represented by many codons
(except AUG for methionine and UGG for tryptophan).
Such codons are called degenerate codons.
Genetic code is unambiguous and specific. i.e., one
codon specifies for only one amino acid.
The codon is read in the 5' ® 3' direction.
AUG has dual functions. It codes for Methionine (met)
and also acts as an initiator codon. In eukaryotes,
methionine is the first amino acid and formyl
methionine is the first amino acid in prokaryotes.
¾ Mutations and Genetic Code
residue glutamate to valine. It results into a diseased
condition called sickle cell anaemia.
Insertion or deletion of one or two bases changes the
reading frame from the point of insertion or deletion.
When there is shifting of the reading frame due to
insertion or deletion of the nucleotide, such mutation
is known as frameshift mutation.
This forms the genetic basis of proof that the codon is a
triplet and is read in a continuous manner.
Molecular Basis of Inheritance
MNEMONICS
¾
¾
¾
¾
¾
¾
b
¾
Concept: Translation process (It include 4 steps)
Mnemonics: Come In Evening Time
Interpretations: Charging of tRNA, Initiation, Elongation,
Termination
time.
However, there are housekeeping genes that are
always expressed in the cell.
The metabolic, physiological and environmental
conditions regulate the expression of genes. e.g.,
In E. coli, the enzyme beta-galactosidase hydrolyses
lactose into galactose and glucose. In the absence of
lactose, the synthesis of beta-galactosidase stops.
The development and differentiation of an embryo
into an adult the result of the regulation of several set
of genes.
Operon Concept : This is a regulatory system that is
observed in bacteria where a group of gene control a
metabolic pathway.
“Each metabolic reaction is controlled by a set of genes”.
All the genes regulating a metabolic reaction constitute
an Operon e.g., lac operon, trp operon, ara operon, his
operon, val operon etc.
When a substrate is added to growth medium of
bacteria, a set of genes is switched on to metabolize it.
This is called induction.
When a metabolite (product) is added, the genes to
produce it are turned off. This is called repression.
The Lac Operon
Lac operon in E. coli : The operon controlling lactose
metabolism. It consists of a regulator gene, 3-structural
genes, an operator gene, promoter gene, a repressor
and an inducer.
(a) A regulatory or inhibitor gene : Codes for the
repressor.
(b) 3 structural genes :
(i) z gene : Codes for -galactosidase (hydrolyze
lactose to galactose and glucose).
(ii) y gene : Codes for permease (increase
permeability of the cell to lactose).
(iii) a gene : Codes for a transacetylase.
The genes present in the operon function together in
the same or related metabolic pathway. There is an
operator region for each operon.
If there is no lactose (inducer), lac operon remains
switched off. In the absence of inducer, repressor
gene is active. The regulator gene synthesizes mRNA
to produce the repressor protein, this protein binds
to the operator genes and blocks RNA polymerase
movement. So, the structural genes are not expressed.
In the absence of glucose, If lactose is provided in the
growth medium, the lactose is transported into the E.
coli cells by the action of permease. Lactose (inducer)
binds with repressor protein.
So, repressor protein cannot bind to operator gene.
The operator gene becomes free and induces the
RNA polymerase to bind with promoter gene then
transcription starts. Regulation of lac operon by
repressor is called negative regulation.
Transport of mRNA from the nucleus to the cytoplasm.
Translational level.
Importance of regulation of gene expression:
Gene regulation is the process to switch off or switch
on the genes as per the requirement of the organism.
Gene regulation is required so that there is no waste
of energy in expressing the genes not required at the
¾
¾
¾
• Large subunit has 2 binding sites for tRNAaminoacyl tRNA binding site (A site) and peptidyl
site (P site).
• Initiation codon for methionine is AUG. So,
methionyl tRNA complex would have UAC at the
anticodon site.
3. Elongation
• At the P-site the first codon of mRNA binds with
anticodon of methionyl tRNA complex.
• Another aminoacyl tRNA complex with an
appropriate amino acid enters the ribosome and
attaches to A site.
• Its anticodon binds to the second codon on the
mRNA and a peptide bond is formed between first
and second amino acids in presence of an enzyme,
peptidyl transferase.
• The uncharged tRNA moves from the P site to the E
site and the peptidyl-tRNA moves to the P site. This
is called a translocation.
• Then 3rd codon comes into A site and a suitable
tRNA with 3rd amino acid binds at the A site. This
process is repeated.
• A group of ribosomes associated with a single mRNA
for translation is called a polyribosome (polysomes).
• A ribozyme is a ribonucleic acid (RNA) enzyme
that catalyses a chemical reaction. The ribozyme
catalyses specific reactions in a similar way to that
of protein synthesis. Also called catalytic RNA,
ribozyme are found in ribosome where they join
amino acids together to form protein chains.
4. Termination
• When aminoacyl tRNA reaches the termination
codon like UAA, UAG & UGA, the termination of
translation occurs. The polypeptide and tRNA are
released from the ribosomes.
• The ribosome dissociates into large and small
subunits at the end of protein synthesis.
An mRNA has additional sequences that are not
translated (untranslated regions or UTR). UTRs
are present at both 5’-end (before start codon) and
3’-end (after stop codon). They are required for an
efficient translation process.
Regulation of Gene Expression
Gene expression results in the formation of a
polypeptide. In eukaryotes, the regulation includes the
following levels :
Transcriptional level (formation of primary transcript).
Processing level (regulation of splicing).
Human Genome Project (HGP) [Board, 2017, 2023]
The entire DNA in the haploid set of chromosome of
an organism is called a Genome.
In human genome, DNA is packed in 23 chromosomes.
Human Genome Project (1990–2003) is the first effort in
identifying the sequence of nucleotides and mapping
of all the genes in the human genome.
BIOLOGY, Class-XII
¾
¾
¾
­
­
¾
¾
About 1.4 million locations where single-base
DNA differences (SNPs- Single nucleotide
polymorphism or ‘snips’) occur in humans.
¾ Rice Genome Project
Rice is one of the most largely consumed foods in India.
Also, the population is increasing with a rapid pace,
so, to meet this requirement, Rice genome project has
been launched to increase the production of rice. Rice
has the smallest genome of 430Mb nucleotides located
on chromosome 12.
¾ Rice Genome : It is a joint project of National Institute of Aerobiological Sciences (NIAS), forestry and
fisheries (STAFF), Ministry of Agriculture, Forestry
and Fisheries (NAFF), Society for Techno-innovation of
Agriculture genome research program.
Arabidopsis is an experiment plant of rice genome
because it has fast life cycle and can be easily grown.
It has smaller genome and high diversity and helps in
enhancing the molecular products.
Need for sequencing rice genome:
To know the functioning of genes by accurate gene
sequencing.
It is important for agronomic traits which requires
mapping of genomic sequences.
Improvement of other cereals will become easier.
¾ DNA Fingerprinting (DNA profiling)
[Board, 2020]
It is the technique to compare the DNA fragments of
two individuals.
Developed by Alec Jeffreys (1985). He is considered
as the father of DNA fingerprinting. Lalji Singh is the
Father of Indian DNA fingerprinting.
¾ Basis of DNA Fingerprinting
[Board, 2017]
DNA carries some non-coding sequences called
repetitive sequence [Variable Number of Tandem
Repeats (VNTR)].
Number of repeats is specific. It varies from person to
person and is specific to a person.
The size of VNTR varies from 0.1 to 20 kb.
Repetitive DNA is separated from bulk genomic DNA as
different peaks during density gradient centrifugation.
The bulk DNA forms a major peak and the other small
peaks are called satellite DNA.
Satellite DNA is classified into many categories (microsatellites, mini-satellites, etc.) based on the base
composition (A-T rich or G-C rich), length of segment
and number of repetitive units.
An inheritable mutation observed in a population at
high frequency is called DNA polymorphism (variation
at genetic level).
Polymorphism is higher in noncoding DNA sequence. This is because
mutations in these sequences may
not have any immediate effect on an
individual’s reproductive ability.
These
mutations
accumulate
generation after generation and
cause polymorphism. For evolution &
speciation, polymorphisms play an important role.
¾
¾
¾
¾
¾
¾ Goals of HGP
(a) To identify all the estimated genes in human DNA.
(b) To determine the sequences of the 3 billion
chemical base pairs that make up human DNA.
(c) To store this information in databases.
(d) To improve tools for data analysis.
(e) To transfer related technologies developed during
the project of society to other sectors of society.
(f) To address the Ethical, Legal and Social Issues
(ELSI) that may arise from the project.
¾ HGP was Closely Associated with Bioinformatics
The application of computer science and information
technology to the field of biology and medicine helps
in analysing DNA sequence data.
¾ Methodologies of HGP
There are two major approaches
namely,
ESTs
and
sequence
annotation.
Expressed Sequence Tags (ESTs) :
Focused on identifying all the genes
that are expressed as RNA and
sequencing the same.
Sequence annotation : Sequencing whole set of the
genome containing all the coding & non-coding regions
and later assigning functions to different regions.
¾ Procedure :
Isolate total DNA from a cell → Convert into random
fragments of smaller size → Clone in suitable host (e.g.,
BAC – bacterial artificial chromosomes & YAC – yeast
artificial chromosomes) for amplification through
PCR (polymerase chain reaction) → Fragments are
sequenced using Automated DNA sequencers (using
Frederick Sanger method) → Sequences are arranged
based of the overlapping regions → Alignment of
sequences using computer-based programs → Genetic
and physical maps on the genome were generated
using the information on polymorphism of restriction
endonuclease recognition sites and some repetitive
DNA sequences (micro-satellites).
¾ Salient Features of Human Genome
(a) Human genome contains 3164.7 million
nucleotide bases pairs.
(b) Total number of genes = about 25,000.
(c) Average gene consists of 3000 bases, but sizes
vary. The largest known human gene (dystrophin
on X-chromosome) contains 2.4 million bases.
(d) 99.9% of nucleotide bases are identical in all
people. It is 0.1% which makes each of us unique.
(e) Functions of over 50% of discovered genes are
unknown.
(f) Chromosome I has the most genes (2968) and Y
has the fewest (231).
(g) Less than 2% of the genome codes for proteins.
(h) Repeated sequences make up a very large portion
of the human genome. Repetitive sequences are
stretches of DNA sequences that are repeated
many times. They have no direct coding functions
but they shed light on chromosome structure,
dynamics and evolution.
(i)
Human genome contains about 3 × 10 bp.
¾ Steps of DNA Fingerprinting (Southern Blotting
Technique)
(a) Isolate DNA (from any cells like blood stains,
semen stains or hair roots).
¾
9
Molecular Basis of Inheritance
the DNA fragment on the membrane to form a
hybridized DNA.
(h) The filter paper is washed to remove unbound
probe.
(i) The hybridised DNA is photographed on to an
X-ray film by autoradiography. The image (in the
form of dark & light bands) obtained is called a
DNA fingerprint. This gives the characteristic
pattern of an individual's DNA.
¾
¾ Applications of DNA Fingerprinting are :
Forensic tool to solve paternity, rape, murder, etc.
For the diagnosis of genetic diseases.
To determine the phylogenetic status of animals.
(b) Make copies (amplification) of DNA by
Polymerase Chain Reaction (PCR) if the amount
of isolated DNA is small.
(c) Digest DNA by restriction endonucleases.
(d) Separate DNA fragments by gel electrophoresis
over agarose polymer gel.
(e) Treat with alkali solution (NaOH) to denature
DNA bonds so as to split them into singlestranded DNAs in the gel.
(f) Transfer (blotting) single-stranded DNA fragments
to synthetic membranes such as nitrocellulose or
nylon, and then baked in a vacuum oven at 80°C
for 3-5 hours (to fix the DNA fragment on the
membrane).
(g) Nitrocellulose filter membrane is placed in a
solution containing a radioactive labelled singlestranded DNA probe. The DNA probes are small
radioactive synthetic DNA segments of known
sequences of nitrogen bases. These DNA probe
binds with the complementary sequences of
KEY-TERM
VNTR: Variable Number of Tandem Repeats
KEY FACT
¾
¾ D
NA finger printing is based upon principle of
polymorphism in DNA sequence.
IMPORTANT DIAGRAMS :
Tyr
Bor
AG
G
Ser
tRNA
Brr
5'
U C A Anticodon
A G U Codon
AUG
U C A mRNA
CUCUUGGGUCCGCAGUUUAA
3'
5'
Fig 5.12: tRNA- the adapter molecule
P
I
P O
3'
Fig 5.13: Translation
Z
Y A In absence of inducer
Repressor binds to the operator
Repressor mRNA region (o) and prevents RNA polymerase
from transcribing the operon
Repressor
P
I
P O
Z
Y A In presence of inducer
Transcription
Repressor mRNA
lac mRNA
Translation
Permease
-Glactosidse
Transacetylase
(Inactive repressor)
Inducer
Fig 5.14: The lac Operon
BIOLOGY, Class-XII
Example-2
Q. When does lac operon get switched off ?
Sol. The lac operon comprises of one regulatory gene or inhibitor gene (i), are promoter gene, one operator gene
and three structural genes. Regulator gene codes for a protein known as repressor protein, it is synthesised all
the time from the i-gene.
The operon gets switched off when repressor protein produced by regulatory or inhibitor gene binds to
operation gene. RNA polymerase gets blocked, so there is no transcription.
Repressor protein + Operator gene → Switched off
SUBJECTIVE TYPE QUESTIONS
(1 mark each)
1. Name different components which are part of
transcription unit in DNA. [OEB]
K
Regulation of Genen
Expression
Ans. The segment of DNA that takes part in transcription is
called transcription unit. It has three components (i)
a promoter, (ii) the structural gene, (iii) a terminator.
1
2. Mention two applications of DNA-polymorphism.
K [Foreign Set-1, 2016]
3. Which property of genetic code is utilised in Wobble
hypothesis?
U
Concept Applied Genetic code
Ans. Degeneracy of genetic code is utilised in wobble
hypothesis.
1
4. Name the site, where during transcription, RNA
polymerase binds to the DNA.
[OEB]
K
Short Answer Type Questions-I (2 marks each)
1.State the roles of AUG codon at 5' end and UAG at
3' end of a certain m-RNA during translation.
Ans.
Concept Applied Translation
Concept Applied
3. Following are the features of genetic codes. What
does each one indicate ? Stop codon, Unambiguous
codon, Degenerate codon, Universal codon.
U [Outside Delhi Set-1, 2016]
Detailed Answer :
The features of genetic codes are :
(i) Stop codon : Termination codons or stop
codons are UAA, UAG and UGA. They do not
code for any amino acids. They represent the
termination of translation.
(ii) Unambiguous codon : The genetic code is
specific and non-ambiguous i.e., one codon
specifies only one amino acid.
(iii) Degenerate codon : This indicates that a single
amino acid is represented by more than one
codons.
(iv) Universal codon : This indicates that one codon
codes for the same amino acid in all species. From
bacteria to human, UUU codes for phenyl alanine.
½×4=2
4. Differentiate between the features of genetic code
given below :
(a) Unambiguous and Universal
(b) Degenerate and Initiator
U [Outside Delhi, 2017, Set-1, 2, 3]
Very Short Answer Type Questions
(a)
K [Delhi Comptt. - 2017, Set-1, 2, 3]
Ans. AUG codon at 5' end = Start codon (for translation)/
codes for methionine.
1
UAG codon at 3' end = Stop codon (for translation)/terminate polypeptide chain.
1
[Marking Scheme, 2017]
2. Explain when is a genetic code said to be
(a) Degenerate
(b) Universal
Ap [Delhi Set-1, 2020]
OR
Unambiguous
Universal
One codon codes Genetic code or codons
for only one amino are (nearly) same for
acid
all organisms or from
bacteria to human
Degenerate
Initiator
(b)
More than one Start codon or AUG
codon code for the
same amino acid.
Topper's Answer, 2017
This Question is for practice and its solution is given at the end of the chapter.
Molecular Basis of Inheritance
5. What is aminoacylation ? State its significance.
K [Outside Delhi Set-II, 2016]
Concept Applied Translation
Short Answer Type Questions-II
(3 marks each)
Ans.(a) (i) Point mutation/ single base substitution½
(ii) Point mutation/ single base deletion½
(b) (i) 4 amino acids1
(ii) 4 amino acids
1
[Marking Scheme, 2020]
2. (a) Study the table given below and identify (i), (ii),
(iii) and (iv)
Amino acid
Phe
Val
This Question is for practice and its solution is given at the end of the chapter.
CAC
Codon in mRNA
(i)
(ii)
Anticodon in tRNA
(iii)
(iv)
A T G C A T G C A T G C
5’
“RNA molecule”
U [Delhi Set-1, 2019]
Concept Applied Transcription
Ans. Transcription is catalyzed by DNA dependent
RNA polymerase. As RNA have uracil at the place
of thymine, the given sequence is coding strand of
DNA and not the RNA strand. However, considering
this coding sequence of DNA, for the given RNA the
transcription unit will be :
[Template strand]
3’ T A C G T A C G T A C G 5’
5’ A T G C A T G C A T G C 3’
[Coding strand]
↓
mRNA
5’ A U G C A U G C A U G C 3’
3
4. A criminal blew himself up in a local market
when he was chased by cops. His face was beyond
recognition. Suggest and describe a modern
technique that can help establish his identity.
Ap [Delhi, 2017, Set-2]
OR
A number of passengers were severely burnt
beyond recognition during a train accident. Name
and describe a modern technique that can help
establish their identity.
Ap [Delhi, 2017, Set-1]
OR
During a fire in an auditorium, a large number of
assembled guests got burnt beyond recognition.
1. A small stretch of DNA strand that codes for a
polypeptide is shown below :
3' ..........CAT CAT AGA TGA AAC ........... 5'
(a) Which type of mutation could have occurred in
each type resulting in the following mistakes
during replication of the above original sequence?
(i) 3`… … … …CAT CAT AGA TGA ATC… …
…5`
(ii) 3`… … … …CAT ATA GAT GAA AC… … … 5`
(b) How many amino acids will be translated from
each of the above strands (i) and (ii) ?
Ap [SQP, 2020-21]
AAA
(b)A polypeptide consists of 14 different amino acids.
(i) How many base pairs must be there in
the processed mRNA that codes for this
polypeptide?
(ii) How many different types of tRNA are needed
for the synthesis of this polypeptide?
K [SQP 2020]
3. Construct and label a transcription unit from
which the RNA segment given below has been
transcribed. Write the complete name of the enzyme
that transcribed this RNA.
Ans. Aminoacylation is the process of adding an activated
amino acid to the acceptor arm of a transfer RNA.
It is an essential step for the synthesis of protein as
it activates the amino acids (amino acid + ATP) and
helps in linking them to their cognate tRNA in the
presence of an enzyme aminoacyl tRNA synthetase.
1+1=2
6. Study the figure below and answer the following
questions:
(a) Name the process depicted in the figure. Define
it.
(b) Name the subunits of Ribosome.
U [SQP 2023 – 24]
Ans: (a) This process is known as translation. Translation
involves decoding a messenger RNA (mRNA) to be
translated into amino acids. Amino acids are linked
together to build a polypeptide.
(b) Ribosomes consist of two major components: the
small ribosomal subunit, which reads the RNA, and
the large subunit, which joins amino acids to form a
polypeptide chain.
DNA code in gene
Suggest and describe a modern technique that can
help hand over the dead to their relatives.
Ap [Delhi, 2017, Set-3]
BIOLOGY, Class-XII
Concept Applied DNA Fingerprinting
DNA with fragments ranging from 200 base pairs to
2500 base pairs was electrophoresed on agarose gel
with the following arrangement.
U [SQP, 2022-23]
Ans. DNA fingerprinting is the technique of determination
of nucleotide sequence of certain areas of DNA,
which are unique to each individual.
Step/Procedure in DNA Fingerprinting :
(i) Extraction of DNA — using high speed
refrigerated centrifuge.
(ii) Amplification — many copies are made using
PCR.
(a) What result will be obtained on staining with
ethidium bromide? Explain with reason.
(b) The above set-up was modified and a band
with 250 base pairs was obtained at X.
(iii) Restriction Digestion → using restriction
(iv) Separation of DNA fragments → using
enzymes DNA is cut into fragments.
electrophoresis agarose polymer gel.
(v) Southern
→
Blotting
Separated
DNA
sequences are transferred to nitrocellulose or
nylon membranes.
(vi) Hybridisation
→
The
nylon
membranes
exposed to radioactive probes.
(vii) Autoradiography → The dark bands develop at
the probe site.
(viii) Matching the banding pattern so obtained with
that of the relative.
5. (a) How does mutation occur ?
What change(s) were made to the previous
design to obtain a band at X? Why did the
band appear at the position X?
shift mutation.
Ans.
(a)
No bands will be obtained as/All DNA will be
U [Outside Delhi, Set-1, 2019]
seen in the well only;
[½]
Ans. (a) Loss (deletion) or gain (insertion / duplication
DNA fragments being negatively charged will
/addition) or change in position of DNA
not move towards negative end/ cathode. DNA
segments /chromosome
1
being negatively charged will remain stationed
(b) mutation due to change in a single base pair
at the positive end/ anode end of the agar block;
of DNA is point mutation,
1
[1]
Insertion or deletion of one or two bases
(b) (i) Position of the positive terminal/ end/ anode
changes the reading frame from the point of
and the negative terminal/ end/ cathode was
insertion or deletion.
1
inter-changed.
[½]
(ii) The fragment with least base pairs will get
[Marking Scheme, 2019]
separated faster and move faster toward the a
6. Carefully observe the given picture. A mixture of
node end.
7.
(a) Explain VNTR and describe its role in DNA fingerprinting.
(b) List any two applications of DNA fingerprinting technique.
Ap [Outside Delhi/Delhi, 2018]
(b) Differentiate between point mutation and frame
Topper's Answer, 2018
Molecular Basis of Inheritance
(a) List the two methodologies which were involved in human genome project. Mention how they were used.
(b) Expand ‘YAC‘ and mention what was it used for.
C [Delhi 2017]
8.
Concept Applied Human Genome Project
Topper's Answer, 2017
Ans.
Ans: (a)
The four prime goals of Human Genome Project were:
To determine the complete nucleotide sequence of
the human genome.
9. Human Genome Project (HGP) was a mega project
launched in the year 1990 with some important
goals.
(a) Enlist any four prime goals of HGP.
(b) Name any one common non-human animal model
organism which has also been sequenced thereafter.
U [Delhi Set-1, 2023]
3
To identify all the genes present in the human
genome.
To study the function of each gene in the human
genome.
To develop new technologies and tools for
analyzing and understanding genomic data.
(b) One common non-human animal model
organism that has also been sequenced is the
mouse (Mus musculus).
[Marking Scheme, Delhi, 2023]
BIOLOGY, Class-XII
(5 marks each)
1. (a) Name the type of DNA that forms the basis of
DNA fingerprinting and mention two features
of this DNA.
(b) Write the steps carried out in the process of
DNA fingerprinting technique and mention its
application.
K [Outside Delhi Set-2, 2020]
Concept Applied DNA Fingerprinting
Ans. (a) The variation between individuals in the
lengths of their DNA satellites forms the basis
of DNA fingerprinting.
Features of DNA satellites :
(i) They are divided into minisatellites and
microsatellites whose characteristic makes them
simple for identification between two samples
as the DNA is polymorphic. It was called as
Variable Number Tandem Repeats (VNTR).
(ii) They are also inheritable from parents to
offspring and can be used for paternity testing.
(b) Steps of DNA fingerprinting are :
Isolation of DNA (From any cells like blood
stains, semen stains or hair roots).
Make copies (amplification) of DNA by
polymerase chain reaction (PCR).
Digestion of DNA by restriction endonucleases.
Separation of DNA fragments by gel
electrophoresis.
Transferring (blotting) of separated DNA
fragments to synthetic membranes, such as
nitrocellulose or nylon and then baked in a
vacuum oven at 80°C for 3–5 hours (to fix the
DNA fragment on the membrane).
Double stranded DNA made single stranded.
Hybridisation using labelled VNTR probe.
Detection of hybridized DNA fragments by
autoradiography. After hybridisation with
VNTR probe the autoradiogram gives many
bands of different sizes. These bands give a
characteristic pattern for an individual DNA.
It differs from individual to individual.
The image (in the form of dark & light bands)
obtained is called DNA fingerprint.
The DNA from a single cell is enough to
perform DNA fingerprinting.
Application of DNA fingerprinting
It is used in forensic science to identify
potential crime suspects.
It is used to establish paternity and family
relationships.
It is used to identify and protect the commercial
varieties of crops and livestock.
It is used to find out evolutionary history of an
organism and trace out the linkages between
various groups of organisms.
(Any two)
2+2+1
2. (a) List any four major goals of Human Genome
project.
This Question is for practice and its solution is given at the end of the chapter.
(b) Write any four ways the knowledge from HGP is
of significance for humans.
(c) Explain BAC and mention its importance.
K [Delhi Set-1, 2020]
3. The lac operon is a polycistronic gene that helps a
bacterial cell in metabolising lactose. It consists of
an inducer (i) gene that represses the transcription of
lac genes under certain environmental conditions.
(a) Why is the lac gene called polycistronic?
(b) What would happen if there was a mutation
blocking the translation of:
(i) gene z
(ii) gene y
(c) What happens to the expression of the lac operon
when the growth medium is provided with:
(i) both glucose and lactose
(ii) only galactose
U [CFPQ]
Ans:
(a) It has a single promoter for multiple connected
genes.
OR
A single mRNA is transcribed to be translated to
multiple proteins.
(b) 1 mark each for the following:
(i) Lactose would not be able to enter/permeate
into the bacterial cell.
(ii) Lactose would enter the cell but not be broken
down into glucose and galactose.
(c) 1 mark each for the following:
(i) Glucose is the preferred carbon source is
consumed first while lactose induces the
lac operon producing small levels of the lac
proteins.
(ii) In the absence of lactose, the repressor protein
will continue binding to the operator of the lac
operon preventing transcription of its genes.
4. Summarize the process by which the sequence
of DNA bases in Human Genome Project was
determined using the method developed by
Frederick Sanger. Name a free living nonpathogenic nematode whose DNA has been
completely sequenced.
U [SQP 2020-21]
Long Answer Type Questions
Concept Applied Human Genome Project
Ans. Frederick Sanger developed a method to sequence
fragments, using automated DNA sequencers.
On the basis of overlapping regions on DNA
fragments, these sequences are arranged.
For alignment of these sequences, computer-based
programs were used.
Finally, by using the information on polymorphism
of restriction endonuclease recognition sites and
certain repetitive DNA sequences, the genetic
and physical maps of the genome were generated
Caenorhabditis elegans.4
5. (a) Write the contributions of the following
scientists in deciphering the genetic code.
Molecular Basis of Inheritance
Ans. Mechanism of translation :
(a) Charging of tRNA (Aminoacylation of tRNA):
Here, amino acids are activated (amino acid +
ATP) and linked to their cognate tRNA in the
presence of aminoacyl tRNA synthetase. This
process is commonly known as charging of tRNA
or aminoacylation of tRNA. If two such charged
tRNAs are brought close enough, the formation of
peptide bonds between them would be favoured
energetically. This is an essential step as only
activated amino acids are carried to the site of
protein synthesis by their respective tRNA.
(b) Translation is initiated by formation of an initiation
complex consisting of 30S ribosomal subunit,
formyl-methionyl (fMet) tRNA, and mRNA. It
begins at the 5’-end of mRNA in the presence
of an initiation factor. The mRNA binds to the
small subunit of ribosome. AUG is recognised
by the initiator tRNA. The initiation codon for
methionine is AUG. So methionyl tRNA complex
would have UAC at the Anticodon site. Now the
large subunit (50S) binds to the small subunit to
complete the initiation complex. Large subunit
(70S) has two binding sites to which tRNA-carrying
amino acids can bind. One is called aminoacyl
tRNA binding site (A site) and the other is called
peptidyl site (P site). There is also a third site
called the exit or E site where tRNAs are released.
(c) An mRNA has additional sequences that are not
translated (untranslated regions or UTR). UTRs
are present at both 5’-end (before start codon) and
3’-end (after stop codon). They are required for an
efficient translation process.
[Marking Scheme SQP 2023 – 24]
7. (i) What do ‘Y’ and ‘B’ stand for in ‘YAC’ and ‘BAC’
used in Human Genome Project (HGP). Mention
their role in the project.
(ii)Write the percentage of the total human genome
that codes for proteins and the percentage of
discovered genes whose functions are known as
observed during HGP.
(iii) Expand ‘SNPs’ identified by scientists in HGP.
K [Outside Delhi Set-1, 2016]
Ans. (i) YAC (Yeast Artificial Chromosomes) and BAC
(Bacterial Artificial chromosomes) are cloning
vectors. They are used in Human genome
project for cloning or amplification of human
DNA fragments.
(ii) Total number of genes (coding for protein) in
the human genome is 30,000 which is less than
2% of the total genome and almost 50% of the
discovered genes have unknown functions.
(iii) SNPs
stand
for
Single
Nucleotide
Polymorphism.
George Gamow; Har Gobind Khorana; Marshall
Nirenberg; Severo Ochoa.
(b) State the importance of a Genetic code in protein
synthesis K [Delhi, Set-1, 2019]
Ans. (a) George Gamow : He suggested that in order to
code for all the 20 amino acids, the code should
be made up of three nucleotides. This is because
a permutation combination of 43 (4 × 4 × 4)
would generate 64 codons; generating many
more codons than required. So, the codon was
proposed to be a triplet.
Har Gobind Khorana : He developed a chemical
method to synthesise RNA molecules with
defined combinations of bases (homopolymers
and co-polymers).
Marshall Nirenberg : He developed a cell-free
system for protein synthesis which helped the
code to be deciphered.
Severo Ochoa : He discovered an enzyme
(polynucleotide phosphorylase), which helped
in the synthesis of RNA with defined sequences
in a template-independent manner (enzymatic
synthesis of RNA).
(b) The genetic code consists of the sequence of
nitrogenous bases in the DNA.
Translation of the nitrogenous base code to an
amino acid sequence in a protein is the basis for
protein synthesis.
Thus, correct translation is important for
protein biosynthesis.
4+1
6. (a) Explain the process of amino-acylation of tRNA
and its role in the process of translation.
(b) How does the initiation of the translation process
occur in prokaryotes? Explain.
(c) Where are the untranslated regions located on
m-RNA and why?
[SQP 2023 – 24]
OR
Explain the mechanism of translation that occurs in
the ribosomes in a prokaryote.
U [Outside Delhi 2019, Set-1, 2]
OBJECTIVE TYPE QUESTIONS
(1 mark each)
[A] Multiple Choice Questions
1. Taylor and colleagues performed experiments
on ............. using radioactive ........... to prove that
the DNA in chromosomes replicate semi-conser
vatively.
[Term-I 2021] U
This Question is for practice and its solution is given at the end of the chapter.
Select the correct option for the blanks.
(A) Vicia faba, Uridine
(B) E. coli, Uridine
(C) Vicia faba, Thymidine
(D) E coli, Thymidine
BIOLOGY, Class-XII
Ans. Option (C) is correct.
Explanation: In 1958, Taylor and colleagues
performed experiments on Vicia faba using
radioactive thymidine to prove that the DNA in
chromosomes replicate semi-conservatively.
(i)
(ii)
(iii)
(iv)
Select the option which is incorrectly representing
the experiment.
(A) (i) and (iii)
(B) (ii) and (iii)
(C) (iii) and (iv)
(D) (ii) and (iv)
Ans. Option (C) is correct.
Explanation : Griffith (1928) used mice and
a bacterial strain, Streptococcus pneumoniae.
Streptococcus pneumoniae has two strains :
(a) Smooth (S) strain (Virulent) : Has polysaccharide
mucous coat. Causes pneumonia.
(b) Rough (R) strain (Non-virulent) : No mucous
coat. Does not cause pneumonia.
Experiment
• S-strain → Inject into mice → Mice die
• R-strain → Inject into mice → Mice live
• S-strain (Hk) → Inject into mice → Mice live
• S-strain (Hk) + R-strain (live) → Inject into mice
→ Mice die
He concluded that there exists some ‘transforming
principle’, that is transferred from heat-killed
S-strain to R-strain. It enabled R-strain to synthesize
smooth polysaccharide coat and become virulent.
This must be due to the transfer of genetic material.
Ans. Option (B) is correct.
Explanation: The genetic code is degenerate,
meaning that multiple codons can code for the same
amino acid. This redundancy in the genetic code
makes it challenging to determine the exact mRNA
sequence solely based on the protein sequence.
6. Which one of the following diagram correctly
represents DNA replication in eukaryotes?
[SQP 2023 – 24]
2. Histone proteins that help in forming the
nucleosomes in the nucleus are rich in basic amino
acids such as K [Delhi Term-I 2021]
(A) Arginine & tyrosine
(B) Lysine & histidine
(C) Arginine & Lysine
(D) Histidine & Tryptophan
3.The reactive hydroxyl group in the nucleotide of RNA
is [Delhi Term-I 2021]
K
(A) 5' OH
(B) 4' OH
(C) 2' OH
(D) 3' OH
4. Study the given diagrammatic representation of
Griffith's experiment to demonstrate transfor­
mation in bacteria. [Delhi Term-I 2021] U
5. Arun thinks that identifying the exact mRNA
sequence from the protein sequence is difficult. Is
he correct and why?
(a) No, as the genetic code is universal.
(b) Yes, as the genetic code is degenerate.
(c) No, as the mRNA is translated into a protein
sequence.
(d) Yes, as the mRNA contains introns which are
non-coding sequences.
K [APQ 2023 – 24]
(A)
(B)
(C)
(D)
Ans. Option (A) is correct.
Explanation : DNA replication take place in the 5'
to 3' direction because DNA polymerase acts on
the 3'-OH of the existing strand for adding free
nucleotides.
5'
3'
Template DNA
(patental strands)
A
B
Continuous
synthesis
3'
5'
Discontinuous
synthesis
C
3'
5'
7. Which of the following statements about
Untranslated regions is/are true?
I. present on rRNA
II. present on mRNA at 3’ position only
III. present on mRNA at 5’ position only
IV. present on mRNA at both 3’ and 5’position
V. not required in translation process.
Molecular Basis of Inheritance
Ans. Option (A) is correct.
Explanation : As mRNA is formed from template
strand hence, the sequence of mRNA is
complementary to template strand.
10. A DNA molecule is 160 base pairs long. If it has
20% adenine, how many cytosine bases are present
in this DNA molecule? [SQP 2023–24] K
(A) 48
(B) 64
(C) 96
(D) 192
Ans. Option (C) is correct.
Explanation : No. of bases in 160 base pairs
= 160 × 2 = 320 bases.
Given, Adenine (A) = 20%
= 20/100 × 320 = 64 bases
As, Adenine (A) = Thymine (T) therefore;
T = 64 bases
A + T = 64 +64 = 128 bases
Total bases of Cytosine (C) and Guanine (G)
This Question is for practice and its solution is given at the end of the chapter.
Ans. Option(A) is correct.
Explanation: DNA, B. H1 histone, C. Histone
octamer
A
B
C
12. Given below is a sequence of bases in mRNA of a
bacterial cell. Identify the amino acid that would
be incorporated at codon position 3 and codon
position 5 during the process of its translation. 1 3'
AUCAGGUUUGUGAUGGUACGA 5 '
K [Delhi Set-1, 2023]
(A) Phenylalanine, Methionine.
(B) Cysteine, Glycine
(C) Alanine, Proline
(D) Serine, Valine
K
Ans. Option (C) is correct.
Explanation : In the presence of an inducer, such
as lactose or allolactose, the repressor is inactivated
by interaction with the inducer. As lactose binds
itself to active repressor and changes its structure,
the repressor fails to bind to the operator. The
RNA polymerase starts transcription of operon by
binding to the promoter site of the promoter and
transcription proceeds.
9. A template strand in a bacterial DNA has the
given base sequence: 5' – AGGTTTAACG – 3' What
would be the RNA sequence transcribed from this
template strand ?
[Term-I 2021]
(A) 5' – CGUUAAACCU – 3'
(B) 5' – TCCAAATTGC – 3'
(C) 5' – AGGUUUUUCG – 3'
(D) 5' – AGGTTTAACG – 3'
= 320 – 128 = 192 bases
As, C = G hence, the number of cytosine bases
present in this DNA molecule is
= 192/2 = 96 bases
11. Identify A, B and C in the given diagram.
[APQ 2023 – 24]
(A) A. DNA, B. H1 histone, C. Histone octamer
(B) A. Histone octamer, B. DNA, C.H1 histone
(C) A. DNA, B. Histone octamer, C. H1 histone
(D) A. Histone octamer, B. H1 histone, C. DNA
Ans. Option (A) is correct.
Explanation: Codons UUU and AUG represent
the amino acids phenylalanine and methionine,
respectively. Therefore, during the process of
translation, the amino acids phenylalanine and
methionine would be merged at codon positions 3
and 5, accordingly.
13. Identify the correct pair of codon with its
corresponding pair of amino acids
[Term-I 2021]
K
(A) UAA: Leucine
(B) UGA: Serine
(C) AUG: Histidine
(D) UUU: Phenylalanine
14. ‘A codon is a Triplet of bases’ was suggested by:
(A) Marshall Nirenberg
(B) Har Gobind Khorana
(C) George Gamow
(D) Francis Crick K [SQP 2023–24]
Ans. Option (D) is correct.
Explanation: These UTRs are positioned at both
the 3' and 5' ends of mRNA. While UTRs play
regulatory roles in gene expression, helping control
processes like mRNA stability and localisation,
they are not directly involved in the translation
of proteins. However, UTRs are crucial for the
efficient translation process, contributing to the
regulation of protein synthesis and influencing
the interactions between mRNA and ribosomes.
In summary, UTRs on mRNA are essential for the
fine-tuning of gene expression, but their primary
role is not in the direct translation of proteins.
8. In the presence of allolactose, the lac repressor in
the operon of E.coli [Term-I 2021]
(A) binds to the operator
(B) binds to the promoter
(C) cannot bind to the operator
(D) binds to the regulator
VI. required for efficient translation process.
(A) I only
(B) II and V
(C) III and VI
(D) IV and VI
K [APQ 2023 – 24]
Ans. Option (B) is correct.
Explanation: The concept that "a codon is a
triplet of bases" was a significant breakthrough in
deciphering the genetic code. This insight, crucial to
understanding how DNA information is translated
into proteins, was independently contributed to by
BIOLOGY, Class-XII
[B] Assertion & Reason
Directions : In the following questions, a statement
of assertion (A) is followed by a statement of
reason (R). Mark the correct choice as :
(A) Both assertion (A) and reason (R) are true and reason
(R) is the correct explanation of assertion (A).
(B) Both assertion (A) and reason (R) are true but reason
(R) is not the correct explanation of assertion (A).
(C) Assertion (A) is true but reason (R) is false.
(D) Assertion (A) is false but reason (R) is true.
1. Assertion: In the process of transcription, template
strand with polarity 3¢ ® 5¢ plays a major role.
Reason: DNA dependent RNA polymerase
catalyses the polymerisation in only one direction,
that is 5¢ ® 3¢.
K [APQ 2023 – 24]
Ans. Option (A) is correct.
Explanation: the template strand guides
transcription in the 3¢ ® 5¢ direction, while RNA
synthesis occurs in the 5¢ ® 3¢ direction, aligning
with the process of transcription.
2. Assertion (A) : Primary transcripts in eukaryotes
are non-functional.
Reason (R) : Methyl guanosine triphosphate is
attached to 5’ – end of hnRNA.
U [SQP 2020-21]
Ans. Option (B) is correct.
This Question is for practice and its solution is given at the end of the chapter.
Ans: Option (D) is correct.
Explanation: Meselson and Stahl performed the
experiments on the DNA and proved that DNA
is semi-conservation or disruptive type. The
experiment used the radioactively labelled N14. This
labelled nitrogen atom helps to detect and analyses
the replication pattern. The new strand is formed
by using one of the old strands as a template.
Explanation: Primary transcripts contains both
introns and exon, in which introns are non- coding
parts. At 5' end of hnRNA, a cap is formed by
modification of GTP into 7-methyl guanosine in
the process called capping.
3. Assertion (A): In Griffith’s experiment, the dead
R strain bacteria was capable of causing the
transformation of the live S-strain bacteria.
Reason (R): The S-strain is non-virulent strain.
[OEB]
K
4. Assertion (A): Ribosomal RNA is synthesised in the
nucleus of the cell.
Reason (R): It is translated with the enzyme RNA
polymerase III.
[SQP 2023 – 24]
Ans. Option (C) is correct.
Explanation: Nucleolus is the site of synthesis of
rRNA and synthesis of ribosomes. It is therefore
called as ribosomal factory. RNA polymerase I is an
enzyme that transcribes ribosomal RNA.
5. Assertion (A): Genetic codes are commaless.
Reason (R): Genetic codes are overlapping.
[SQP 2023–24] U
Ans. Option (C) is correct.
Explanation : A commaless genetic code means
that no punctuations are needed between any two
words. The genetic code is non-overlapping. In
actual practice six bases code for not more than two
amino acids.
6. Assertion (A): The nucleosome is a repeating unit
of a structure in nucleus called chromatin.
Reason (R): The negatively charged DNA is
wrapped around the positively charged histone
octamer to form a structure called nucleosome.
[SQP 2023 – 24]
U
Ans. Option (B) is correct.
Explanation: The length of DNA in a human
diploid cell is around 2.2 metres. It is greater than
the dimension of a typical nucleus. In order to
fit them in the nucleus, DNA is wrapped around
histone octamer to form a nucleosome. The
nucleosome in chromatin gives a ‘beads’ on string
appearance.
7. Assertion (A): RNA polymerases are able to catalyse
all three steps of translation.
Reason (R): RNA polymerases contain an initiation
factor and a termination factor in them.
[SQP 2023–24]
U
Ans. Option (D) is correct.
Explanation: DNA-dependent RNA polymerase
is the “only enzyme” that has the “capability” to
catalyse initiation, elongation, and termination
in the process of transcription in prokaryotes.
Initiation factor sigma helps in the process of
initiation while the termination factor Rho help in
the termination of the process.
8. Assertion (A): Teminism is unidirectional flow of
information.
Reason(R) : Teminism was reported in 1978 by H.
Temen and D. Baltimore.
[OEB] U
scientists such as Har Gobind Khorana. Khorana,
along with Marshall Nirenberg and Robert W. Holley,
played a key role in identifying that a sequence
of three nucleotide bases, known as a codon,
corresponds to a specific amino acid in a protein.
15. What is the smallest part of a DNA molecule that
can be changed by a point mutation?
(A) Oligonucleotide
(B) Codon
(C) Gene
(D) Nucleotide
K [SQP 2023–24]
Ans. Option (D) is correct.
Explanation : Nucleotide: It is the smallest unit of
DNA which consists of nucleoside and phosphate
groups. It is the monomeric unit of nucleic acids
such as DNA and RNA. It can be changed by a
point mutation
16. Meselson and Stahl’s experiment proved:
[SQP 2023–24]
(A) Transduction.
(B) Transformation.
(C) DNA is the genetic material.
(D) Disruptive DNA replication.
Molecular Basis of Inheritance
Ans. Option (D) is correct.
Explanation: Teminism is a theoretical concept. In
1978, H.Temin and Baltimore independently discovered Teminism. It is an exception to central dogma
of molecular biology. Teminism theory explains
that RNA can act as template for the DNA formation, that is, DNA can be synthesised from RNA. As
teminism is bidirectional flow of information, it is
popularly known as reverse transcription.
COMPETENCY BASED QUESTIONS
(1 mark each)
(C) Both (i) and (ii) are correct
[C] Case Based MCQs
I. Read the following passage and answer any four
questions given below:
The lac operon consists of a regulation gene and
three structural gene. The lactose acts as inducer. In
the presence of an Inducer such as lactose, the repressor is in inactivated during the interaction. This
allows RNA polymerase access to the promoter and
transcription proceeds. The repressor is synthesized which in turn binds with the operator region
of the operon and prevents RNA polymerase from
transcribing the operon.
[OEB] A
1. When the process of Lac operon is blocked by a
repressor, it represents :
(A) positive regulation
(B) negative regulation
(C) sometimes positive sometimes negative
(D) both positive and negative regulation
Ans. Option (B) is correct.
Explanation: The lac operon regulation can be in
both negative and positive ways. It is a negative control system because expression is typically blocked
by an active repressor (the lac repressor) that turns
off transcription. And when CAP (catabolite gene
activating protein) binds upstream of this operator
region near the promoter and transcription increases, this is an example of a positive system.
2. Identify the correct sequence of the structural
genes in the lac operon.
(A) lacA-lacZ-lacY
(B) lacZ-lacA-lacY
(C) lacZ-lacY-lacA
(D) lacA-lacY-lacZ
Ans. Option (C) is correct.
Explanation: The lac operon consists of 3 structural
genes, and a promoter, a terminator, regulator, and
an operator. The three structural genes are: lacZ,
lacY, and lacA.
3. Which of the following statement is true in
reference to the lac operon process in E.coli?
(i) Galactosidase is the only enzyme produced in
large quantities when lac operon is turned on
(ii) The messenger RNA in lac operon is a
polycistronic mRNA
(A) Only (i) is correct
(B) Only (ii) is correct
(D) None of them are correct
Ans. Option (B) is correct.
Explanation : The messenger RNA produced by
transcription carries information for the synthesis
of all three proteins found in all three structural
genes. Hence, it is a polycistronic messenger RNA.
4. What provides binding site to RNA polymerase?
(A) Exons
(B) Promoter
(C) Inducer
(D) Repressor
Ans. Option (B) is correct.
Explanation : Promoter helps in starting the process
of transcription and provides a binding site to RNA
polymerase.
5. The lac operon of E. coli contains genes involved
in lactose metabolism. It's expressed only
when lactose is____________(1) and glucose is
_________(2).
(A) 1: Present, 2: Absent
(B) 1: Absent, 2: Present
(C) 1: More, 2: Less
(D) 1: Repressed, 2: Promoted
Ans. Option (A) is correct.
Explanation: The lac operon of E. coli contains genes
involved in lactose metabolism. It's expressed only
when lactose is present and glucose is absent.
[D] Case Based Subjective Questions
I. Look at the lac operon in a bacterium
BIOLOGY, Class-XII
Predict how these mutations would affect the
function of the operon in the presence and absence
of lactose(inducer): Refer fig. 5.14
U [CFPQ]
1. Mutation of the regulatory gene (i); mutated
repressor won’t bind to lactose but will bind to the
operator.
1 mark
Ans: The mutated repressor binds to the operator on the
DNA, and would continuously repress the operon;
enzymes for lactose utilisation would not be made,
whether or not lactose was present.
2. Mutation of the operator (o); repressor will not bind
to the operator.
1 mark
Ans: The lac genes would continue to be transcribed
and the enzymes made, whether or not lactose was
present
3. Mutation of the promoter (p); RNA polymerase will
not attach to the promoter. 2 marks
Ans. RNA polymerase would not be able to transcribe the
genes and no proteins would be made, whether or
not lactose was present.
II. Shown below is a nucleotide sequence and the
genetic code.
5' - ATGCGTAGACTCGTA - 3' [APQ 2023 – 24]
1. Identify the protein sequence formed by this
sequence.
1 mark
Ans. TYR ALA SER GLU HIS
2. The first guanine base in the nucleotide sequence
changes to cytosine. Identify the type of mutation
caused by this change.
1 mark
Ans: point mutation
3. Will the mutated sequence form an mRNA and
protein? Justify.
2 marks
Ans: - mRNA will be formed
- protein will not be formed
- The first codon is a stop codon due to which
translation will not happen
SOLUTIONS FOR PRACTICE QUESTIONS (TOPIC-1)
VSATQ
Ans. 2. Methionine / Tryptophan
1
[Marking Scheme, 2018]
LATQ
Ans. 6. Prokaryotes: Negatively charged DNA is held
with positively charged proteins in nucleoid, DNA in
nucleoid is organised in large loops held by protein.
½×4=2
Eukaryotes: In nucleus, the negatively charged
DNA is wrapped around positively charged
histone octamer to form nucleosome, nucleosomes
are repeated to constitute chromatin at higher
level, additional set of non-histone chromosomal
protein gets associated with chromatin. ½ × 6 = 3
[Marking Scheme, 2016]
SOLUTIONS FOR PRACTICE QUESTIONS (TOPIC-2)
VSATQ
Ans. 2: Genetic mapping & DNA-finger printing.
1
[Marking Scheme, 2016]
Ans. 4: The DNA site at which RNA polymerase binds, is
called promoter.
1
SATQ-II
Ans. 2: (a)
Amino acid
Ans. 2: (a) Genetic code is said to be degenerate when
two or more codons specify a particular amino acid.
(b)Genetic code is said to be universal when its
coded information specifies the same amino acid
across different species.
1+1
SATQ-I
Phe
Val
DNA code in gene
AAA
CAC
Codon in mRNA
(i) UUU
(ii) GUG
Anticodon in tRNA
(iii) AAA (iv) CAC
1
(b) (i) A polypeptide containing 14 different amino
acid = 14 × 3 = 42 base pairs.
1
(ii) 14 different types of RNA are needed for the
synthesis of polypeptide.
1
Molecular Basis of Inheritance
It has paved the way for personalised medicine in
future based on one’s own genome.
It has shed a lot of light on human evolution and
phylogenetics.
LATQ
Goals of HGP
To identify all the estimated genes in human DNA
To determine the sequences of the 3 billion chemical
base pairs that makes up human DNA.
To store this information in databases.
To improve tools for data analysis.
To transfer related technologies to other sectors.
To address the ethical, legal and social issues (ELSI)
that may arise from the project.
(b) Advantages of the Human Genome Project are:
It has led to a better understanding of human
biology and genetics in general.
¾
(a)
¾
¾
¾
¾
¾
Ans. 2:
(c) BAC stands for Bacterial artificial chromosome. It is
an artificially constructed vector containing the origin
of replication and a selectable marker for identification.
It is capable of carrying large DNA fragments and can
replicate easily inside a bacterial cell. It is used in the
human genome project for cloning large chunks of the
fragmented human genome with ease.2+2+1
SOLUTIONS FOR PRACTICE QUESTIONS (MCQ)
Explanation: UUU code for Phenylalanine(phe).
UAA and UGA are stop codon while, AUG codes
for the Methionine(met) .
MCQS
Ans. 2. Option (C) is correct.
Explanation: Histones are rich in the basic amino
residues lysine and arginines. Both of these amino
acid residues carry positive charges in their side
chains
Ans. 3. Option (C) is correct.
Explanation: 2''OH group present in RNA(in every
nucleotide) makes it more reactive than DNA.
Ans. 13. Option (D) is correct.
A&R
Ans. 3. Option (D) is correct.
Explanation: In Griffith's experiment, some
'transforming principle', transferred from heat
-killed S strain had enabled the R strain to synthesis
a smooth polysaccharide coat. Due to the transfer of
the genetic material, the R-strain became virulent.
REFLECTION
1.
2.
According to Chargaff's rule, Purines and Pyrimidines are always in equal amounts i.e., A + G = T +
C. Suppose, if a double stranded DNA has 20 percent cytosine, were you able to calculate the percent
of Adenine in DNA.
As you have studied in this chapter that insertion or deletion of one or two bases is termed as Point
mutation, so, can you recall the term given to the insertion or deletion of the nucleotide due to which
there is shifting of the reading frame?
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