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TRANSCRIPT
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Pre-Medical : Biology ALLENMOLECULAR BASIS OF INHERITANCE
NUCLEIC ACIDS* F. Meischer discovered nucleic acid in nucleus of pus cell and called it "nuclein". The term nucleic acid was
coined by "Altman."
* Nucleic acids are polymer of nucleotides.
Example : DNA and RNA.
Nucleotide = Nitrogen base + pentose sugar + phosphateNucleoside = Nitrogen base + pentose sugar.
A. Nitrogen base :On the basis of structure nitrogen bases are broadly of two types :–
1. Pyrimidines – Consist of one pyrimidine ring. Skeleton of ring composed of two nitrogen and four Carbonatoms. e.g. Cytosine, Thymine and Uracil.
CYTOSINE URACIL THYMINE2. Purines - Consist of two rings i.e. one pyrimidine ring (2N + 4C) and one imidazole ring (2N + 3C) e.g.
Adenine and Guanine.
ADENINE GUANINEB. Pentose Sugar (Number of carbon = 5) :–
(1) OOH
H
CH OH2
H H
OH
H
OH
1
23
4
5(2) O
OH
H
CH OH2
H H
H
H
OH
1
23
4
5
Ribose (in RNA) Deoxy ribose (in DNA)
C. Phosphate :– Acidic– Negative charged
Nitrogen base forms bond withfirst carbon of pentose sugar toform a nucleoside. Nitrogen offirst place (N1) forms bond withsugar in case of pyrimidines whilein purines nitrogen of ninthplace (N9) forms bond withsugar.Phosphate forms ester bond(covalent bond) with fifth Carbonof sugar to form a completenucleotide.
OCH2
HH
OH3'
H
H2'
N
H
N7
89
65 1
24
N H
NHH
3
P
O
OO
4'
5'
O
1'
H
N
Deoxyribose
Phosphate
Nucleoside
Nucleotide
Phosphoester bond
N–glyc
osidi
c link
age
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Pre-Medical : BiologyALLENTypes of Nucleosides and Nucleotides
1. Adenine + Ribose = Adenosine 2. Adenine + Deoxyribose = Deoxy adenosine
Adenosine + Phosphate = Adenylic acid (AMP) Deoxy adenosine + P = Deoxy adenylic acid (dAMP)
3. Guanine + Ribose = Guanosine 4. Guanine + Deoxyribose = Deoxy guanosine
Guanosine + P = Guanylic acid (GMP) Deoxy guanosine + P = Deoxy guanylic acid (dGMP)
5. Cytosine + Ribose = Cytidine 6. Cytosine + Deoxyribose = Deoxycytidine
Cytidine + P = Cytidylic acid (CMP) Deoxycytidine + P = Deoxycytidylic acid (dCMP)
7. Uracil + Ribose = Uridine 8. Thymine + Deoxyribose = Deoxy thymidine
Uridine + P = Uridylic acid (UMP) Deoxythymidine + P = Deoxythymidylic acid (dTMP)
DNA
* DNA as an acidic substance present in nucleus was first identified by Friedrich Meischer in 1869.
* DNA term was given by - Zacharis
* DNA is long polymer of deoxyribonucleotides.
* DNA is negatively charged.
* In DNA pentose sugar is deoxyribose sugar and four types of nitrogen bases A,T,G,C
* Wilkins and Franklin studied DNA molecule with the help of X-Ray crystallography.
S
S
P
P
S
S
S
P
P
P
HO
5'
3'
S
S
S
S
S
P
OH
P
P
P
P
3'
5'
A
G
A
C G
A
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Pre-Medical : Biology ALLEN* With the help of this study, Watson and Crick (1953) proposed a double helix molel for DNA. For this
model Watson, Crick and Wilkins were awarded by Noble Prize in 1962.
* One main hallmark (main point) of double helix model is complementary base pairing between purine and
pyrimidine.
* According to this model, DNA is composed of two polynucleotide chains.
* Both polynucleotide chains are complementary and antiparallel to each other.
* In both strand of DNA direction of phosphodiester bond is opposite. i.e. If direction of phosphodiester bond
in one strand is 3'-5' then it is 5'-3' in another strand.
* Both strand of DNA are held together by hydrogen bonds. These hydrogen bonds are present between nitrogen
bases of both strand.
* Adenine binds to thymine by two hydrogen bonds and cytosine binds to guanine by three hydrogen bonds.
* In a DNA molecule one purine always pairs with a pyrimidine. This generates approximately uniform distance
between the two strands of DNA.
* In DNA plane of one base pair stacks over the other in double helix. This, in addition to H-bonds, confers stability
of the helical structure of DNA.
* Chargaff's equivalency rule - In a double stranded DNA amount of purine nucleotides is equals to amount
of pyrimidine nucleotides.
Purine= Pyrimidine
[A] + [G] = [T] + [C]
[ ] [ ][ ] [ ]A G
T C
+
+= 1
* Base ratio = A TG C
++
= constant for a given species. i.e. species specific.
* In a DNA A + T > G + C Þ A – T type DNA. Base ratio of A – T type of DNA is more than one.
eg. Eukaryotic DNA
* In a DNA G + C > A + T Þ G – C type DNA. Base ratio of G – C type of DNA is less than one.
eg. Prokaryotic DNA
* Melting point of DNA depends on G – C contents.
More G – C contents means higher melting point.
Tm = Temperature of melting.
Tm of prokaryotic DNA > Tm of Eukaryotic DNA
* DNA absorbs U.V. rays of 2600Å wavelength.
* Denaturation and renaturation of DNA - If a normal DNA molecule is placed at high temperature (80-
90°C) then both strands of DNA will separate from each other due to breaking of hydrogen bonds. It is called
DNA-denaturation.
When denatured DNA molecule is placed at normal temperature then both strand of DNA attached and
recoiled to each other. It is called renaturation of DNA.
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Pre-Medical : BiologyALLENConfiguration of DNA Molecule :–
* Two strands of DNA are helically coiled like a revolving ladder. Back bone of this ladder (Reiling) is composed
of phosphates and sugars while steps (bars) are composed of pairs of nitrogen bases.
Two chains have anti-parallel polarity. It means, if one chain has the polarity 5'®3', the other has 3'®5'.
* Distance between two successive steps is 3.4 A0 . In one complete turn of DNA molecule there are such 10 steps
(10 pairs of nitrogen bases). So the length of one complete turn is 34 A0. This is called helix length.
* Diameter of DNA molecule i.e. distance between phosphates of two strands is 20A0.
* Each step of ascent is represented by a pair of bases. At each step of ascent, the strands turns 36°.
* In nucleus of eukaryotes the DNA is associated with histone protein to form nucleoprotein.
* Bond between DNA and Histone is salt linkage (Mg+2).
*f × 174 (bacteriophage) [Single stranded] 5386 Nucleotides
l bacteriophage 48502 base pair
E.coli 4.6 × 106 base pair
Human 6.6 × 109 base pair
Types of DNA :–
On the basis of direction of twisting, there are two types of DNA.
1. Right Handed DNA -
Clockwise twisting e.g. The DNA for which Watson and Crick proposed model was 'B' DNA.
DNA Helix Length No. of Distance between Diameterbase pairs two pairs
'A' 28 A0 11 pairs 2.56 A0 23 A0
'B' 34 A0 10 pairs 3.4 A0 20 A0
'C' 31 A0 9.33 pairs 3.32 A0 19 A0
'D' 24.24 A0 8 pairs 3.03 A0 19 A0
2. Left handed DNA :–
Anticlockwise twisting e.g. Z-DNA - discovered by Rich. Phosphate and sugar backbone is zig-zag.
, Helix length – 45.6 A0
Diameter – 18.4 A0
No. of base pairs – 12 (6 dimers)
Distance between 2 base - pairs – 3.75 A0
* Palindromic DNA – Wilson and Thomas
¾ ®¾C C G G T A C C G GG G C C A T G G C C
¬ ¾¾Sequence of nucleotides same from both ends.
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Pre-Medical : Biology ALLEN
PACKAGING OF DNA HELIX
The average distance between the two adjacent base pairs of 0.34 nm (0.34 × 10–9 m or 3.4 Å). Length of DNA
for a human diploid cell is 6.6 × 109 bp × 0.34 × 10–9 m/bp = 2.2 m. The length is far greater than the
dimension of a typical nucleus (approximately 10–6 m).
The number of base pairs in Escherichia coli is 4.6 × 106. The total length is 1.36 mm. The long sized DNA
accommodated in small area (about 1 mm in E. coli) only through packing or compaction. DNA is acidic due to
presence of large number of phosphate group. Compaction occurs by folding acid attachment of DNA with
basic proteins, polyamine in prokaryotes and histone in eukaryotes.
DNA packaging in Prokaryotes : DNA is found in cytoplasm in supercoiled state. The coils are maintained
by non histone basic protein like polyamines. This compact structure of DNA is called nucleoid or genophore.
DNA packaging in Eukaryotes : It is carried out with the help of lysine and ariginine rich basic proteins called
histone. The unit of compaction is nucleosome. There are five types of histone proteins H1 H2A, H2B, H3 and
H4. Four of them occur in pairs to produce histone octamer (2 copies of each – H2A, H2B, H3 and H4), called
nubody or core of nucleosome.Their positively charged ends are directed outside. The negatively charged DNA
is wrapped around the positively charged histone octamer to form a structure called nucleosome. A typical
nucleosome contains 200 bp of DNA Helix. DNA present between two adjacent nucleosome is called linker
DNA. It is attached to H1 histone protein. Length of linker DNA varies from species to species. Nucleosome
chain gives a beads on string appearance under electron microscope. The nucleosomes furthers coils to form
solenoid. It has diameter of 30 nm as found in chromatin. The beads on string structure in chromatin is pack-
aged to form chromatin fibres that are further coiled and condensed at metaphase stage of cell division to form
chromosomes. The packaging at higher level requires additional set of proteins (acidic) that collectively are
referred to as non-histone chromosomal (NHC) proteins.
Non-Histone chromosomal proteins are of three types :
(i) Structural NHC protein
(ii) Functional NHC protein e.g., DNA polymerase, RNA polymerase
(iii) Regulatory NHC protein
DNA H1 histone
Histoneoctamer
Core of histone molecules
Basic unit of DNA compaction (Nucleosome)
EM picture - "Beads-on-String
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Pre-Medical : BiologyALLEN
2 nm
10 nm
H1
Histoneoctamer
DNADNA2 nm
Nucleosme10 nm
Solenoids30 nm
Super solenoid300 nm
Chromatid700 nm
Chromosome1400 nm
1400 nm
30 n
m
300 n
m
700 n
m
Various steps in the folding and superfilding of basic chromatin components to generate an eukaryotic chromosome
In a typical nucleus, some region of chromatin are loosely packed (and stains light) and are referred to as
enchromatin. The chromatin that is more densely packed and stains dark is called as heterochromatin, specifi-
cally euchromatin is said to be transcriptionally active and heterochromatin is transcriptionally inactive.
THE SEARCH FOR GENETIC MATERIALThe experiments given below prove that DNA is the genetic material.
(I) Evidence from bacterial transformation : The transformation experiments conducted by Frederick Griffith
in 1928, are of greater importance in establishing the nature of genetic material.
He used two strains of bacterium Diplococcus or Streptococcus pneumoniae or Pneumococcus i.e., S-III and
R-II.
(i) Smooth (S) or capsulated type which have a mucous coat and produce shiny colonies. These bacteria are
virulent and cause pneumonia.
(ii) Rough (R) or non-capsulated type in which mucous coat is absent and produce rough colonies. These
bacteria are nonvirulent and do not cause pneumonia.
The experiment can be described in following four steps :
(a) Smooth type bacteria were injected into mice. The mice died as a result of pneumonia caused by
bacteria.
Live S strain ® injected into mice ® Mice died
(b) Rough type bacteria were injected into mice. The mice lived and pneumonia was not produced
Live R strain ® injected into mice ® mice lived
(c) Smooth type bacteria which normally cause disease were heat killed and then injected into the
mice. The mice lived and pneumonia was not caused.
S strain (heat killed) ® Injected into mice ® Mice lived
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Pre-Medical : Biology ALLEN(d) Rough type bacteria (living) and smooth type heat-killed bacteria (both known not to cause disease)
were injected together into mice. The mice died due to pneumonia and virulent smooth type living
bacteria could also be recovered from their bodies.
S strain (heat killed) + R strain (living) ® injected into mice ® Mice died
He concluded from fourth step of the experiment that some rough bacteria (nonvirulent) were transformed into
smooth type of bacteria (virulent). This occurred pherhaps due to absorption of some transforming substance by
rough type bacteria from heat killed smooth type bacteria. This transforming substance from smooth type
bacteria caused the synthesis of capsule which resulted in production of pneumonia and death of mice. There-
fore, transforming principle appears to control genetic characters (for example, capsule as in this case). How-
ever, the biochemical nature of genetic material was not defined from this experiments.
Virulent Non-Virulent Heat-killedVirulent
Heat-killedVirulent + Non-virulent
Died Survived Survived Died
Absent of Bacteria
Absent of Bacteria
VirulentVirulent
Bacterial transfomation experiments conducted by Griffith
Biochemical characterisation of Transofrming Principle :
Later, Avery, Macleod and McCarty (1944) repeated the experiment in vitro to identify the biochemical
nature of transforming susbtance. They proved that this substance is DNA.
Pneumococcus bacteria cause disease when capsule is present. Capsule production is under genetic
control.
In the experiments, rough type bacteria (non-capsulated and non-virulent) were grown in a culture me-
dium to which DNA extract from smooth type bacteria (capsulated and virulent) was added. Later, the
culture showed the presence of smooth type bacteria also in addition to rough type. This is possible only
if DNA of smooth type was absorbed by rough type bacteria which developed capsule and became
virulent. This process of transfer of characters of one bacterium to another by taking up DNA from
solution is caleld transformation. When DNA extract was treated with DNase (an enzyme which destroys
DNA), transformation did not occur. The transformation occurs when proteases and RNases were used.
This clearly shows that DNA is the genetic material.
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Pre-Medical : BiologyALLEN
RNon virulent DNA
extractfrom S
(Extract is non virulentbut living S is virulent)
Many cellsor R
(Non virulent)
Few cellsof S
(Virulent)
In vitro experiment of Avery and others demonstrating that DNA is genetic material
(II) Evidence from experiments with bacteriophage : T2 bacteriophage is a virus that infects bacterium
Escherichia coli and multiplies inside it . T2 phage is made up of DNA and protein coat. Thus, it is the
most suitable material to determine whether DNA or protein contains information for the production of
new virus (phage) particles. Hershey and Chase (1952) demonstrated that only DNA of the phage enters
the bacterial cell and , therefore, contains necessary genetic information for the asssembly of new phage
particle.
The functions of DNA and proteins could be found out by labelling them with radioactive tracers. DNA
contains phosphorus but not sulphur. Therefore, phage DNA was labelled with P32 by growing bacteria
infected with phages in culture medium containing 32PO4. Similarly, protein of phage contains sulphur
but no phosphorus. Thus, the phage protein coat was labelled with S35 by growing bacteria infected with
phages in another culture medium containing 35SO4. After the formation of labelled phages. Three steps
were followed, i.e., infection, blending, centrifugation.
1. Infection : Both type of labelled phages were allowed to infect normally cultured bacteria in separate
experiments.
2. Blending : These bacteria cells were agitated in a blender to break the contact between virus and
bacteria.
3. Centrifugation : The virus particles were separated from the bacteriam by spinning them in a centri-
fuge.
After the centrifugation the bacterial cells showed the presence of radioactive DNA labelled with P32
while radioactive protein labelled with S35 appeared on the outside of bacteria cells (i.e., in the medium).
Labelled DNA was also found in the next generation or phage. This clearly showed that only DNA enters
the bacterial host and not the protein. DNA, therefore, is the infective part of virus and also carries all the
genetic information. This provided the unequivocal proof that DNA is the genetic material.
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Pre-Medical : Biology ALLEN
Bacteriophage
Radioactive ( S) labelledprotein capsule
35Radioactive ( P)
labelled DNA
32
1. Infection
2. Blending
3. Centrifugation
No Radioactive ( S)detected in cells
35 Radioactive ( P)detected in cells
32
+Radioactive ( S)
detected in supernatant
35
+No Radioactivity
detected in supernatant
The Hershey-Chase experiment
Properties of Genetic mateial :
Following are the properties and functions which should be fulfilled by a substance if it is to qualify as genetic
material.
(1) The genetic material should be able to generate its own kind (replication). Both the nucleic acids (DNA
and RNA) have the ability to direct their duplications. The other molecules in the living system, such as
proteins fail to fulfill first criteria itself.
(2) It should chemically and structurally be stable. The genetic material should be stable enough not to
change with different stages of life cycle, age or with change in physiology of the organism. Stability as
one of the properties of genetic material was very evident in Griffith’s ‘transforming principle’ itself that
heat, which killed the bacteria, at least did not destroy some of the properties of genetic material. This
now can easily be explained in light of the DNA that the two strands being complementary if separated
by heating come together, when appropriate conditions are provided. Further, 2'-OH group present at
every nucleotide in RNA is a reactive group and makes RNA labile and easily degradable. RNA is also
now known to be catalytic, hence reactive. Therefore, DNA chemically is less reactive and structurally
more stable when compared to RNA. Therefore, among the two nucleic acids, the DNA is a better
genetic material. In fact, the presence of thymine at the place of uracil also confers additional stability to
DNA.
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Pre-Medical : BiologyALLEN(3) The genetic material should also be capable of undergoing mutations – Both DNA and RNA are able to
mutate. In fact, RNA being unstable, mutate at a faster rate. Consequently, viruses having RNA genome
and having shorter life span mutate and evolve faster.
(4) The genetic material should be able to transmit faithfully to the next generation, as Mendelian charac-
ters. RNA can directly code for the synthesis of proteins, hence can easily express the characters. DNA,
however, is dependent on RNA for synthesis of proteins. The protein synthesising machinery has evolved
around RNA.
The above discussion indicate that both RNA and DNA can function as genetic material, but DNA being
more stable is preferred for storage of genetic information. For the transmission of genetic information,
RNA is better.
RNA WORLD
RNA was the first genetic material. There are evidences to suggest that essential life processes, such as metabo-
lism, translation, splicing etc. evolved around RNA. RNA used to act as a genetic material as well as a catalyst,
there are some important biochemical reactions in living systems that are catalysed by RNA catalysts and not by
protein enzymes (e.g., splicing)RNA being a catalyst was reactive and hence unstable. Therefore, DNA has
evolved from RNA with chemical modifications that make it more stable. DNA being double stranded and
having complementary strand further resists changes by evolving a process of repair. RNA is adapter, structural
molecule and in some cases catalytic. Thus RNA is better material for transmission of information.
DNA REPLICATION
* While proposing the double helical structure for DNA,
Watson and Crick had immediately proposed a scheme
for replication of DNA. To quote their original statement
that is as follows:
* ‘‘It has not escaped our notice that the specific pairing
we have postulated immediately suggests a possible
copying mechanism for the genetic material’’ (Watson and
Crick, 1953).
* The scheme suggested that the two strands would sepa-
rate and act as a template for the synthesis of new
complementary strands. After the completion of replica-
tion, each DNA molecule would have one parental and
one newly synthesised strand. This scheme was termed
as semiconservative DNA replication
* D.N.A. capable of self duplication.
* All living beings have the capacity to reproduce because
of this characteristic of D.N.A.
* D.N.A replication takes place in "S - Phase" of the cell
cycle. At the time of cell division, it divides in equal parts
in the daughter cells.
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Pre-Medical : Biology ALLENSEMI CONSERVATIVE MODE OF DNA REPLICATION
Semi conservative mode of D.N.A. replication was first proposed by Watson & Crick. Later on it was
experimentally proved by Meselson & Stahl (1958) in E - Coli and Taylor in Vicia faba (1958).To
prove this method, Taylor used Radiotracer Technique in which Radioisotopes (tritiated thymidine = 1H3) were
used. Meselson and Stahl used heavy isotope of nitrogen (N15).
Matthew Meselson and Franklin Stahl performed the following experiment in 1958 :
(i) They grew E. coli in a medium containing 15NH4Cl (15N is the heavy isotope of nitrogen) as the only
nitrogen source for many generations. The result was that 15N was incorporated into newly synthesised
DNA (as well as other nitrogen containing compounds). This heavy DNA molecule could be distinguished
from the normal DNA by centrifugation in a cesium chloride (CsCl) density gradient (Please note that
15N is not a radioactive isotope, and it can be separated from 14N only based on densities).
(ii) Then they transferred the cells into a medium with normal 14NH4Cl and took samples at various definite
time intervals as the cells multiplied, and extracted the DNA that remained as double-stranded helices.
The various samples were separated independently on CsCl gradients to measure the densities of DNA.
(iii) Thus, the DNA that was extracted from the culture one generation after the transfer from 15N to 14N
medium [that is after 20 minutes; E. coli divides in 20 minutes] had a hybrid or intermediate density.
DNA extracted from the culture after another generation [that is after 40 minutes, II generation] was
composed of equal amounts of this hybrid DNA and of ‘light’ DNA.
15 N-DNA
20 min 40 min
14 N-DNA
Generation I
Generation II
14 N-DNA
15 N-DNA
14 N-DNA
14 N-DNAGravitational force
15 15N NHeavy
14 15N N 14 15N NHybrid Hybrid
14 14N NLight
Separation of DNA by Centrifugation
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Pre-Medical : BiologyALLENMECHANISM OF DNA REPLICATION
The following steps are included in DNA replication :-
(1) Unzipping (Unwinding) :–
* The separation of 2 chains of DNA is termed as unzipping. And it takes place due to the breaking of
H–bonds. The process of unzipping starts at a certain specific point which is termed as initiation point or
origin of replication. In prokaryotes there occurs only one origin of replication but in eukaryotes there
occur many origin of replication i.e. unzipping starts at many points simultaneously. At the place of origin
the enzyme responsible for unzipping (breaking the hydrogen bonds) is Helicase (= Swivelase). In the process of unzipping
Mg+2 act as cofactor.
* SSB (single stranded DNA binding protein) prevents the reformation of H-bonds.
* Topoisomerase (in prokaryotes also called as DNA gyrase) release the tension arises due to supercoiling.
Note : The process of DNA replication takes a few minutes in prokaryotes and a few hours in Eukaryotes.
(2) Formation of New Chain :–
3'5'
5'5'3'
3'Newly synthesised
strands
Discontinous synthesis
Template DNA (Parental strands)
Continous synthesis
Replicating Fork
3'5'* To start the synthesis of new chain, special type
of RNA is required which is termed as RNA Primer.The formation of RNA primer is catalysed by an
enzyme - RNA Polymerase (primase). Synthesis
of RNA–primer takes place in 5' ® 3' direction.
After the formation of new chain, this RNA is
removed. For the formation of new chain
Nucleotides are obtained from Nuceloplasm. In the
nucleoplasm, Nucleotides are present in the form
of triphosphates like dATP, dGTP, dCTP, dTTP etc.
* During replication, the 2 phosphate groups of all nucleotides are separated. In this process energy is yeilded which
is consumed in DNA replication.
* Energetically replication is a very expensive process. Daoxyribonucleoside triphosphase serve dual purposes
in addition to acting as substrates they provide energy for polymerisation.
* The formation of new chain always takes place in 5' - 3' direction. As a result of this, one chain of D.N.A.
is continuously formed and it is termed as Leading strand. The formation of second chain begins from the
centre and not from the terminal points, so this chain is discontinuous and is made up of small segments
called Okazaki Fragments. This discontinuous chain is termed as Lagging strand. Ultimately all these
segments joined together and a complete new chain is formed.
* The Okazaki fragments are joined together by an enzyme DNA Ligase.
* The formation of new chains is catalysed by an enzyme DNA Polymerase. In prokaryotes it is of 3 types:
(1) DNA - Polymerase I :- This was discovered by KORNBERG (1957). So it is also called as 'Kornberg's
enzyme'. Kornberg also synthesized DNA first of all, in the laboratory. This enzyme functions as exonuclease.
It separates RNA - primer from DNA and also fills the gap.It is also known as DNA-repair enzyme.
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Pre-Medical : Biology ALLEN(2) DNA - Polymerase II :- It is least reactive in replication process. It is also helpful in DNA-repairing
in absence of DNA-polymerase-I and DNA polymerase-III
(3) DNA - Polymerase III :- This is the main enzyme in DNA - Replication. It is most important. The
larger chains are formed by this enzyme. This is also known as Replicase.
* In the semi conservative mode of replication each daughter DNA molecule receives one chain of polynucleotides
from the mother DNA - molecule and the second chain is synthesized.
Special Point :
l All DNA polymerase I, II and III enzymes have 5'-3" polymerisation activity and 3'-5" exonuclease activity.
l DNA polymerase I also has 5'-3'' exonuclease activity.
l Any failure in cell division after DNA replication result into polyploidy.
l Difference between DNAs and DNase is that DNAs menas many DNA and DNase means DNA digestive enzymes.
BEGINNER'S BOX-11. nu body of nucleosome consists of
(1) H1 and H2A (2) H2A and H2B
(3) H3 and H4 (4) Both (2) and (3)
2. Radioactive element used to label DNA of bacteriophage in Wareing-blender experiment of Hershey and Chase
was :-
(1) S35 (2) P32 (3) N15 (4) C14
3. Bonding between deoxyribose and base in pyrimidine nucleoside molecule is :-
(1) 1'-1' glycosidic linkage (2) 1'-6' glycosidic linkage
(3) 1'-9' glycosidic linkage (4) 1'-4' glycosidic linkage
4. Tm (melting temperature) value of DNA is high when it contains
(1) A + T > G + C (2) G + C > A + T
(3) A + T = G + C (4) A + G = T + C
5. Select an incorrect statement regarding RNA molecule :-
(1) It has highly reactive 2'-OH group
(2) It shows higher rate of mutation than DNA
(3) It is genetic material in some viruses
(4) It follows Chargaff rule
6. In Meselson and Stahl's experiment, heavy isotope 15N was used in the form of
(1) 15NaNO3 (2) 15NH4Cl (3) 15KNO3 (4) 15NH4NO3
7. Assuming that 50 heavy (i.e. containing N15) DNA molecules replicated twice in a medium containing N14, we
expect
(1) 100 half heavy and half light and 150 light DNA molecules
(2) 100 half heavy and half light and 100 light DNA molecules
(3) 50 heavy and 150 light DNA molecules
(4) 50 heavy and 100 light DNA molecules
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Pre-Medical : BiologyALLEN8. The enzyme which shows polymerising activity in 5'® 3' direction is :
(1) DNA polymerase III (2) DNA polymerase II
(3) DNA polymerase I (4) All of these
9. DNA polymerase I is involved in :-
(1) Removal of RNA primer (2) Filling of gap
(3) Joining of okazaki fragments (4) Both (1) and (2)
10. DNA replication in lagging strand of most of the eukaryotic organisms is :-
(1) Conservative and continous (2) Semi-conservative but discontinous
(3) Consenrvative and semi-discontinous (4) Semi-conservative but continous
RIBO NUCLEIC ACID (RNA)
Structure of RNA is fundamentally same as DNA, but there are some differences. The differences are as follows:-
(1) In place of De-oxyribose sugar in DNA, there is present Ribose sugar in RNA.
(2) In place of nitrogen base Thymine in DNA, there is present uracil in RNA.
(3) RNA is made up of only one polynucleotide chain i.e. R.N.A. is single stranded.
Exception :–
RNA found in Reo - virus is double stranded, i.e. it has two polynucleotide chains.
Types of RNA :
1. Genetic RNA or Genomic RNA - In some viruses RNA works as genetic material and it transfers
informations from one generation to next generation.
eg. Reo virus, TMV, QB bacteriophage.
2. Non-genetic RNA - mainly of 3 types -
(1) r - RNA (2) t - RNA (3) m - RNA
* RNA functions as adapter, structural and in same cases as a catalyst (Ribozyme)
(1) Ribosomal RNA (r - RNA) :-
* This RNA is 80% of the cell's total RNA
* r RNA was discovered by Kuntze.
* It is found in ribosomes and it is produced in nucleolus.
* It is the most stable form of RNA .
* These are present in 80s type of ribosomes in Eukaryotic cells. Their subunits are 60s and 40s. In 60s
sub unit of ribosome three types of r–RNA are found – 5s, 5.8s, 28s
* 40s sub unit of ribosome has only one type of r–RNA i.e. 18s.
* So 80s ribosome has total 4 types of r–RNA.
* Prokaryotic cells have 70s type of ribosomes and its subunits are 50s and 30s.
* 50s sub unit of ribosome contains 2 types of r–RNA i.e. 5s and 23s
* 30s sub unit of ribosome has 16s type of r–RNA.
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Pre-Medical : Biology ALLEN* So 70s ribosome has total 3 types of r–RNA.
Function :–
* At the time of protein synthesis, r–RNA provides attachement site to t–RNA and m–RNA and attaches them
on the ribosome.
* It attaches t–RNA to the larger subunit of ribosome and m–RNA to smaller subunit of ribosome.
(2) Transfer – RNA (t–RNA) :–
* It is 10-15% of total RNA.
* It is synthesized in the nucleus by DNA.
* It is also known as soluble RNA (sRNA)
* It is also known as Adapter RNA.
* It is the smallest RNA (4s).
Function :– At the time of protein synthesis it acts as a carrier of amino-acids.
Structure :– The structure of t - RNA is most complicated.
CCA
5' G
3'Acceptor arm
T Y c Loop
Variable arm
Anticodon/Nodoc
Recognition Loop
DHU Loop(8-12 bases) (7 bases)
(7 bases)
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Pre-Medical : BiologyALLENHolley presented Clover leaf model of its structure. In two dimensional structure the t–RNA appears clover
leaf like but in three dimensional structure (by Kim) it appears inverted L–shaped.
The structure of tRNA looks like a clover leaf but in actual structure, the tRNA is a compact molecule which
looks like inverted 'L'.
* There are present three nucleotides in a particular sequence at 3' end of t - RNA and that sequence is CCA.
* All the 5' ends i.e. last ends are having G (guanine).
* 3' end is known as Acceptor end.
* t–RNA accepts amino acids at acceptor points. Amino acid binds to 3' end by its – COOH group.
* The molecule of t - RNA is folded and due to folding some complementary nitrogenous bases come across
with each other and form hydrogen bonds.
* There are some places where hydrogen bonds are not formed, these places are known as loop.
Loops :–
There are some abnormal nitrogenous bases in the loops, that is why hydrogen bonds are not formed.
e.g. (i) Inosine (I) (ii) Pseudouracil (Y) (iii) Dihydrouridine (DHU)
(A) T Y C Loop or Attachment loop :–
This loop connects t - RNA to the larger subunit of ribosome.
(B) Recognition Loop (Anticodon loop) :-
* This is the most specific loop of t-RNA and different types of t-RNA are different due to this loop. There
is a specific sequence of three nucleotides called Anticodon, is present at the end of this loop.
* t–RNA recognizes its place on m - RNA with the help of Anticodon.
* The anticodon of t-RNA recognises its complimentary sequence on m–RNA. This complimentary sequence
is known as codon.
(C) DHU Loop :–
* It is also known as Amino - acyl synthetase recognition loop. Amino - acyl synthetase is a specific type
of enzyme. The function of this enzyme is to activate a specific type of amino acid. after activation this enzyme
attaches the aminoacid to the 3' end of t–RNA.
* There are 20 types of enzymes for 20 types of aminoacids.
* The function of DHU loop is to recognize this specific Aminoacyl synthetase enzyme.
(3) Messenger RNA (m –RNA) :–
* The m - RNA is 1 - 5% of the cell's total RNA.
The name m-RNA was given by Jacob and Monad.
* The m - RNA is produced by genetic DNA in the nucleus.
* It is least stable RNA.
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Pre-Medical : Biology ALLEN
TRANSCRIPTION
* Formation of RNA over DNA template is called transcription. Out of two strand of DNA only one strand
participates in transcription and called ‘‘Antisense strand’’ or "Template strand".
* If both strands act as a template during transcription they would code for RNA molecule with different sequence
and If they code for proteins the sequence of aminoacid in these protein would be different and another reason
that if the two RNA molecule produced they would be complementary to each other and form a ds RNA which
prevent translation of RNA.
* A gene is defined as the functional unit of inheritance. It is difficult to literally define a gene in terms of
DNA sequence, because the DNA sequence coding for tRNA or rRNA molecule is also define a gene
(But information of protein is present on the DNA segment which code mRNA).
* The segment of DNA which contains signal for the synthesis of one polypeptide is known as ‘‘Cistron’’.
* RNA polymerase enzyme is involved in transcription. In eukaryotes there are three types of RNA polymerases.
. RNA polymerase–I for 28s rRNA, 18s rRNA, 5.8s rRNA synthesis.
. RNA polymerase–II for hn–RNA synthesis (Precursor of m-RNA)
. RNA polymerase–III for t–RNA, 5s rRNA, SnRNA synthesis.
* Prokaryotes have only one type of RNA polymerase which synthesizes all types of RNAs.
* RNA polymerase (Core enzyme) of E. Coli has five polypeptide chains b, b', a, a and w.
* s polypeptide chain is also known as s factor (sigma factor).
* Core enzyne + Sigma factor Þ RNA Polymerase
(b, b', a, a, w) (s)
A transcription unit in DNA is defined primarily by three regions in the DNA :-
(i) A promoter, (ii) The structural gene (iii) A terminator
Transcription start site
Structural geneTemplate
strandPromoter Terminator
Coding strand
3'
3'
5'
5'
TRANSCRIPTION UNIT
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Pre-Medical : BiologyALLENFollowing steps are present in transcription –
(1) INITIATION :–
* DNA has a ‘‘Promoter site’’ where RNA polymerase binds and a ‘‘Terminator site’’ where transcriptionstops.
* Sigma factor (s) recognises the promoter site of DNA.
* With the help of sigma factor RNA polymerase enzyme attached to a specific site of DNA called ‘‘Promotersite’’.
* In prokaryotes before the 10 N2 base from ‘‘Starting point’’ a sequence of 6 base pairs (TATAAT) is presenton DNA, which is called ‘‘Pribnow box’’.
* In eukaryotes before the 20 N2 base from ‘‘Starting point’’ a sequence of 7 base pairs (TATAAAA) or(TATATAT) is present on DNA which is called "TATA box or Hogness box"
* At start point RNA polymerase enzyme breaks H–bonds between two DNA strands and separates them.
* One of them strand takes part in transcription. Transcription proceeds in 5' ® 3' direction.
* Ribonucleoside triphosphate come to lie opposite complementary nitrogen bases of anti sense strand.
* These Ribonucleotides present in the form of triphosphate ATP, GTP, UTP and CTP. When they are usedin transcription, pyrophosphatase hydrolyse two phosphates from each activated nucleotide. This releases energy.This energy is used in the process of transcription.
(2) ELONGATION :–
* RNA polymerase enzyme establishes phosphodiester bond between adjacent ribonucleotides.
* Sigma factor separates and RNA polymerase moves along the anti sense strand till it reaches terminatorsite.
(3) TERMINATION :–
* When RNA polymerase enzyme reaches at terminator site, it separates from DNA templet.
* In most cases RNA polymerase enzyme can recognise the ‘Terminator site’ and stop the synthesis of RNAchain, but in prokaryotes, it recognises the terminator site with the help of Rho factor (r factor).
* Rho (r) factor is a specific protein which helps RNA polymerase enzyme to recognise the terminator site.
3' 3'
5'5'
3'
3'
5'
5'
5'5'
3'
3'
Sigma factorRNA polymerasePromoter DNA helix
TerminatorRNA
RNA polymerase
Rho factorRNA
Intitiation
Elongation
Termination
Process of Transcription in Bacteria
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Pre-Medical : Biology ALLEN
SPLIT GENEDiscovered by sharp and Roberts. They were awarded Nobel Prize in 1993. Gene which contains non functional
part along with functional part is known as split gene. Non functional part is called intron and functional
part is called exon. By transcription split gene produces a RNA which contains coding and non coding sequence
and called hn RNA (Hetero genous nuclear RNA). This hn RNA is unstable. Now 7 methyl guanonsine is
added to its 5' end, and a cap like structure is formed. It is called capping and 200-300 nucleotides of
adenylic acid are added to its 3' end, which is called poly 'A' tail, Now it becomes stable. By the process
of RNA splicing hn–RNA produces functional m-RNA that is exonic RNA. In RNA splicing non coding parts
is removed with the help of spliceosome enzyme and coding part join together with the help of RNA ligase.
Some specific proteins are also helpful in RNA - splicing called 'Small nuclear ribonucleoprotein' or 'SnRNP'or 'Snurps'. These SnRNP proteins combine with some other proteins and SnRNA to form spliceosomecomplex. This spliceosome complex uses energy of ATP to cut the RNA, releases the non-coding part and
joins the coding-part to produce functional RNA. Non coding part of hn RNA remained inside the nucleus
and not translated in to protein. Only coding part moves from nucleus to cytoplasm and gets translated into
protein.
Mostly Eukaryotic genes are example of split gene, but gene which forms histone and interferon proteinare non split gene.
Mostly prokaryotic genes are example of non split gene.
l In euckaryotes after transcription splicing process also occured.
l The split gene represent an ancient (primitive) feature of gene.
l Presence of intron is a primitive character.
l The splicing process represent the dominance of RNA world.
5 Þfunctional part Non functional part Functional part
3 AntisenseStrandof DNAExon Intron Exon
Transcription
5 3 ÞCoding part Non coding part Coding part
HnRNA(unstable)
Stabilization
Coding part Non coding part Coding part[AAA....]
Poly 'A' tail 5' end
7mG cap
by Guanyl transferase
Splicing
Ribonuclease
RNA lygase
Spliceosomecomplex
ATP
m-RNA5' end
[AAA....] Poly 'A' tail
Capping
3' end
(by Poly A Polymerase)Tailing
7mG cap
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Pre-Medical : BiologyALLEN
Capping
5'
3'
5'
3'
3'
CapmGppp5'
5'mGppp
Intron
3'
3'5'
mGppp
5'mGppp
RNA splicing
Poly A tail
Messenger RNA
Exon Polyadenylation
hnRNA
Process of Transcription in Eukaryotes
GENETIC CODE* Term Given by George Gamow.
* Discovered by Nirenberg, Matthaei and Khorana.
* The relationship between the sequence of amino acids in a polypeptide chain and nucleotide sequence of
DNA or m–RNA is called genetic code.
* There occur 20 types of amino acids which participate in protein synthesis. DNA contains information for
the synthesis of any types of polypeptide chain. In the process of transcription, information is transfered from
DNA to m–RNA in the form of complementary N2–base sequences.
* m–RNA contains code for each amino acid and it is called codon. A codon is the nucleotide sequence on
m–RNA which codes for particular amino acid ; wherease the genetic code is the sequence of nucleotides
on m–RNA molecule, which contains information for the synthesis of polypeptide chain.
Triplet Code :–
* The main problem of genetic code was to determine the exact number of nucleotide in a codon which codes
for one amino acid.
* There are four types of N2–bases in m–RNA (A, U, G, C) for 20 types of amino
acids.
* If genetic code is singlet i.e. codon is the combination of only one nitrogen
base, then only four codons are possible A, C, G and U. These are insufficient
to code for 20 types of amino acids.
. Singlet code = 41= 4 × 1 = 4 codons
A
C
G
U
Codons
Singlet Code : 4 × 1 = 4 codons
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Pre-Medical : Biology ALLEN. If genetic code is doublet (i.e. codon is the combination of two nitrogen
bases) then 16 codons are formed.
. Doublet code = 42 = 4 × 4 = 16 codons.
. 16 codons are insufficient for 20 amino acid
* Gamow (1954) pointed out the possibility of three letters code (Triplet code).
* Genetic code is triplet i.e. one codon consists of three nitrogen bases
Triplet code = 43 = 4 × 4 × 4 = 64 codons
* In this case there occurs 64 codons in dictionary of genetic code.
* 64 codons are sufficient to code 20 types of amino acids.
* The chemical method developed by Har Gobind Khorana was instrumental in synthesising RNA molecules with
defined combinations of bases (homopolymers and copolymers). Marshall Nirenberg’s cell-free system for protein
synthesis finally helped the code to be deciphered. Severo Ochoa enzyme (polynucleotide phosphorylase) was
also helpful in polymerising RNA with defined sequences in a template independent manner (enzymatic synthesis
of RNA).
Triplet codons for the various amino acids
Doublet Code : 4 × 4 = 16 codons
AA AC AG AU
CC CA CG CU
GG GA GC GU
UU UA UG UC
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Pre-Medical : BiologyALLENCharacteristics of Genetic Code :–
(i) Triplet in Nature :–
* A codon is composed of three adjacent nitrogen bases which specifies one amino acid in polypeptide chain.
For Ex. :
. In m–RNA if there are total 90 N2 – bases.
. Then this m–RNA determines 30 amino acids in polypeptide chain.
. In above example, number of nitrogen bases are 90 so codons Þ 30 and 30 codons decide 30 amino
acids in polypeptide chain.
(ii) Universality :–
The genetic code is applicable universally. The same genetic code is present in all kinds of living organism
including viruses, bacteria, unicellular and multicellular organisms.
(iii) Non – Ambiguous and specific unamfiguous :–
* Genetic code is non ambiguous i.e. one codon specifies only one amino acid and not any other.
* In this case one codon never code for two different amino acids. Exception GUG codon which codes both
valine and methionine amino acids.
(iv) Non – Overlapping :–
A nitrogen base is a constituent of only one codon.
(v) Comma less :–
* There is no punctuation (comma) between the adjacent codon i.e. each codon is immediately followed by
the next codon.
* If a nucleotide is deleted or added, the whole genetic code read differently.
* A polypeptide chain having 50 amino acids shall be specialized by a linear sequence of 150 nucleotides.
If a nucleotide is added in the middle of this sequence, the first 25 amino acids of polypeptide will be same
but next 25 amino acids will be different.
(vi) Degeneracy of Genetic code :–
* There are 64 codons for 20 types of amino acids, so most of the amino acids (except two) can be coded
by more than one codon. Single amino acid coded by more than one codon is called ‘‘Degeneracy of genetic
code’’. This incident was discovered by Baurnfield and Nirenberg.
* Only two amino acids Tryptophan and Methionine are specified by single codon.
UGG for Tryptophan
AUG for Methionine.
* All the other amino acids are specified or coded by 2 to 6 codons.
* Leucine, serine and arginine are coded or specified by 6–codons.
Leucine = CUU, CUC, CUA, CUG, UUA & UUG
Serine = UCU, UCC, UCA, UCG, AGU, AGC
Arginine = CGU, CGC, CGA, CGG, AGA, AGG
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Pre-Medical : Biology ALLEN* Degeneracy of genetic code is related to third position (3' – end of triplet codon) of codon. The third base
is described as ‘‘Wobbly base’’.
Chain Initiation and Chain Termination Codon :–
* Polypeptide chain synthesis is signalled by two initiation codons AUG or GUG.
* AUG codes methionine amino acid in eukaryotes and in prokaryotes AUG codes N–formyl methionine.
* Some times GUG also functions as start codon it codes for valine amino acid normally but when it is presentat starting position it codes for methionine amino acid.
* Out of 64 codons 3–codons are stopping or nonsense or termination codon.
Nonsense codons do not specify any amino acid.UAA (Ochre)
UAG (Amber)
üïýïþ
Non–Sense Codons or Stop codonsUGA (Opal)
* So only 61 codons are sense codons which specify 20 amino acid.
WOBBLE HYPOTHESIS* It was propounded by CRICK.
* Normally an anticodon recognises only one codon, but sometimes an anticodon recognises more than one codon.This is known as Wobbling. Wobbling normally occurs for third nucleotide of codon.
* For e.g. anticodon AAG can recognise two codons i.e. UUU and UUC, both stands for phenyl alanine.
Types of m–RNA – m–RNA is of 2 types –
(1) Monocistronic - The m - RNA in which genetic signal is present for the formation of only one polypeptidechain eg. Eukaryotes.
(2) Polycistronic :– The m–RNA, in which genetic signal is present for the formation of more than one polypeptidechains eg. Prokaryotes.
* Non sense codons are found in middle position in polycistronic m–RNA.
CENTRAL DOGMA* Central dogma was given by Crick.
* The formation (production) of m - RNA from DNA and then synthesis of protein from it, is known as CentralDogma.
* It means, it includes transcription and translation.
Reverse Transcription :–
* The formation of DNA from RNA is known as Reverse - transcription. It was discovered by Temin andBaltimore in Rous - sarcoma virus. So it is also called Teminism.
* ss–RNA of Rous–Sarcoma virus (Retro virus) produces ds–DNA in host's cell with the help of enzyme reversetranscriptase (DNA–polymerase). This DNA is called c–DNA (Complimentary DNA).
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TRANSLATION (Protein Synthesis)(1) Activation of Amino acid :–
* 20 types of amino acids participate in protein synthesis.
* Amino acid reacts with ATP to form ‘‘Amino acyl AMP enzyme complex’’ , which is also known as ‘ActivatedAmino acid’.
Amino acid + ATP Amino acyl
t RNA synthetase¾¾¾¾¾¾¾¾¾¾®- Amino acyl AMP–enzyme complex + PP
* This reaction is catalyzed by a specific ‘Amino acyl t-RNA synthetase’ enzyme.
* There is a separate ‘Amino acyl t–RNA synthetase’ enzyme for each kind of amino acid.
(2) Charging of t–RNA (Loading of t-RNA) :–
* Specific activated amino acid is recognised by its specific t–RNA.
* Now amino acid attaches to the ‘Amino acid attachment site’ of its specific t–RNA and AMP and enzymeare separated from it.
Amino acyl AMP–enzyme complex + t–RNA ® Amino acyl t–RNA complex + AMP + enzyme
* Amino acyl t–RNA complex is also called ‘Charged t–RNA’.
* Now Amino acyl t–RNA moves to the ribosome for protein synthesis.
(3) Translation :– 3 steps –
(A) Initiation of polypeptide chain :–
* In this step 30s and 50s sub units of ribosome, GTP, Mg+2, charged t–RNA, m–RNA and some initiation
factors are required.
* In prokaryotes there are three initiation factors present – IF1, IF2, IF3.
* Initiation factors are specific protein.
* GTP and initiation factors promote the initiation process.
* In prokaryotes with the help of ‘‘S D sequence’’ (Shine–Delgarno sequence) m–RNA recognises
the smaller sub unit of ribosome. A sequence of 8 N2 base is present before the 4-12 N2 base of
initiation codon on mRNA, called "SD sequence". In Smaller subunit of ribosome, a complementary
sequence of "SD sequence" is present on 16s rRNA, which is called "Anti Shine-Delgarno sequence"(ASD sequence)
* With the help of 'SD' and 'ASD sequence' mRNA recognises the smaller sub unit of ribosome.
* While in eukaryotes, smaller sub unit of ribosome is recognised by "7mG cap".
* In eukaryotes, 18s rRNA of smaller sub unit has a complementary sequence of "7mG cap".
30s sub unit + m–RNA 2IF3
Mg+¾¾¾¾® 30s m–RNA – complex
* This ‘‘30s m–RNA – complex’’ reacts with ‘Formyl methionyl t–RNA – complex’ and 30s m–RNA – formyl methionyl t–RNA – complex’’ is formed. This t–RNA attaches with codon part ofm–RNA. A GTP molecule is required.
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30s m–RNA – complex + Formyl methionyl t–RNA – complex
30s m–RNA formyl methionyl t–RNA – complex
* Now larger sub unit of ribosome (50s sub unit) joins this complex. The initiation are factor releasedand complete 70s ribosome is formed.
* In larger sub unit of ribosome there are three sites for t–RNA –
‘P’ site = Peptidyl site.
‘A’ site = Amino acyl site.
‘E’ site = Exit site
* Starting codon of m–RNA is near to ‘P’ site of ribosome, so t–RNA with formyl methionine aminoacid first attaches to ‘P’ site of ribosome and next codon of m–RNA is near to ‘A’ site of ribosome.So next new t–RNA with new amino acid always attach at ‘A’ site of ribosome but in initiation step'A' site is empty.
(B) Chain – Elongation :–
* New tRNA with new amino acid is attaches at 'A' site of ribosome.
* The link between amino acid of 'P' site of t-RNA is broken and t-RNA of P-site is discharged so –COOH of P-site A.A. becomes free.
* Now peptide bond takes place between – COOH group of P site amino acid and – NH2 group ofA-site amino acid.
* 23-s rRNA induces the formation of peptide bond. This r–RNA acts as an enzyme so it is called‘‘Ribozyme’’.
* After formation of peptide bond t–RNA of P site released from ribosome via E-site and dipeptideattaches with A site.
* Now t–RNA of A site is transferred to P site and \ A site becomes empty.
* Now ribosome slides over m–RNA strand in 5' ® 3' direction. Due to sliding of ribosome on m–RNA,new codon of m–RNA continuously available at A site of ribosome and according to new codon ofm–RNA new amino acid attaches in polypeptide chain.
* Translocase enzyme is helpful in movement of ribosome (translocation). GTP provides energy forsliding of ribosome.
AA1
A U G5 3
P A
AA1
A U G5 3
P A
AA2
Fig. (1) Fig. (2)
E E
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Pre-Medical : BiologyALLEN* In elongation process some protein factors are also helpful, which are known as ‘Elongation factors’.
* In prokaryotes three ‘Elongation factors’ are present – EF–Tu, EF–Ts, EF–G.
(C) Chain – Termination :–
* Due to sliding of ribsome over m–RNA when any Nonsense codon (UAA, UAG, UGA) available
at A site of ribosome, then polypeptide chain terminates.
* The linkage between the last t–RNA and the polypeptide chain is broken by three release factorcalled RF1, RF2, RF3 with the help of GTP.
AA1
AA2 AA
3
P A
5AUG
3
E
Fig. (5)
AA1
AA2
AA3
AA4 AA
5
A
UAA
5 3
Fig. (6)
PE
transloca-
tionP
5 3
A
AA2AA
1
peptide bond
5AUG
3
AA1
AA2
P A
Fig. (3) Fig. (4)
E
AA3
E
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Pre-Medical : Biology ALLEN* An mRNA also have some additional sequences that are not translated and are referred as untranslated regions
(UTR). The UTRs are present at both 5'end (before start codon) and at 3'end (after stop codon).
* The UTR(untranslated regions) present on mRNA are required. for efficient translation process (by recognising
the smaller subunit of ribosome by mRNA)
Some Inhibitors of Bacterial Protein Synthesis :
Antibiotic Effect
Tetracycline Inhibits binding of amino-acyl tRNA to ribosome
Streptomycin Inhibits initiation of translation and causes misreading
Chloramphenicol Inhibits peptidyl transferase and so formation of peptide bonds
Erythromycin Inhibits translocation of ribosome along mRNA
Neomycin Inhibits interaction between tRNA and mRNA
BEGINNER'S BOX-2
1. Unidirectional flow of information called central dogma was given by
(1) F.H.C. Crick (2) Temin (3) Baltimore (4) Dulbecco
2. In eukaryotes, RNAPIII catalyses the synthesis of
(1) All rRNA and tRNA (2) mRNA, HnRNA and SnRNA
(3) 5S rRNA, tRNA and SnRNA (4) 28S, 18S and 5S rRNA
3. The core enzyme requires a factor for termination of RNA synthesis at some sites. This is known as
(1) Sigma factor (2) Rho factor
(3) Gamma factor (4) Alpha particle
4. If one strand of DNA has the base sequence ATCCACGACTAG and the second strand undergoes transcription
what would be the base sequence on mRNA ?
(1) TACGTGCTGATC (2) ATCCACGACTAG
(3) AUCCACGACUAG (4) AUGCACGACTAG
5. During protein synthesis, amino acid gets attached to tRNA with the help of
(1) mRNA (2) Aminoacyl synthetase
(3) Ribosome (4) rRNA
6. The first amino acid in any polypeptide chain of prokaryote is always
(1) Formylated methionine (2) Formylated arginine
(3) Lysine (4) Methionine
7. Which site of a tRNA molecule forms hydrogen bonds with mRNA molecule ?
(1) Codon (2) Anticodon
(3) 5' end of the t-RNA molecule (4) 3' end of the t-RNA molecule
8. To code the 50 aminoacids in a polypeptide chain, what will be the minimum number of nucleotides in its
cistron?
(1) 50 (2) 153 (3) 306 (4) 309
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Pre-Medical : BiologyALLEN9. A single anticodon can recognize more than one codon of m-RNA. This phenomenon is termed as
(1) Richmond and Lang effect (2) Gene flow hypothesis
(3) Wobble hypothesis (4) Transposability
10. The genetic code is called a degenerate code because
(1) One codon has many meanings
(2) More than one codon has the same meaning
(3) One codon has one meaning
(4) There are 64 codons present
REGULATION OF GENE EXPRESSION
– Constitutive genes (House-keeping genes) – These genes are expressed constantly, because their products
are constant needed for cellular activity. e.g. genes for glycolysis, gene of ATPase enzyme.
– Non-constitutive genes (Smart gene or Luxary gene) – These genes remain silent and are expressed only
when the gene product is needed. They are switched ‘on’ or ‘off’ according to the requirement of cellular
activities. Non-constitutive genes are of two types; inducible and repressible. The inducible genes are switched-
on in presence of a chemical substance called inducer, required for the functioning of gene activity. The
repressible genes continue to express themselves till a chemical, often an end product of the metabolism inhibits
or represses their activity. Such type of inhibition is called feed back inhibition or feed back repression.
– The mechanism which stimulates the expression of certain genes and inhibits that of others is called regulationof gene expression.
– It is possible only if the organism has a mechanism of regulating gene activity by allowing some to function and
others to restrain their activity through switching on and switching off system. This means, the genes are turned
‘on’ or ‘off’ as per requirement.
– A set of genes is ‘switched on’ when enzymes are required to metabolise a new substrate. The enzymes pro-
duced by these genes metabolise the substrate.
– The molecules of metabolite that come to switch on of the genes are termed as inducers and the phenomenon
is called induction.
– Similarly, certain genes which are in their ‘switch on’ state, continue to synthesise a metabolite till the later is
produced in amount more than required or else, it is supplied to the cell from outside. In other words, certain
genes continue to express themselves till the end product of inhibits or repress their expression. Inhibition by an
end product is known as ‘feed back repression’.
– Regulation of gene expression refers to a very broad term that may occur at various levels. Considering that
gene expression results in the formation of a polypeptide, it can be regulated at several levels. In eukaryotes,
the regulation could be exerted at
(i) transcriptional level (formation of primary transcript),
(ii) processing level (regulation of splicing),
(iii) transport of mRNA from nucleus to the cytoplasm,
(iv) translational level.
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Pre-Medical : Biology ALLENOPERON CONCEPT
– In 1961, two French microbiologist Francis Jocob and Jacques Monad at the Pasteur Institute in Paris,
proposed a mechanism called operon model for the regulation of gene action in E. coli.
– An operon is a part of genetic material or DNA, which acts as a single regulated unit having one or more
structural genes-an operator gene, a promoter gene, a regulator gene.
– Operons are of two types (i) inducible (ii) repressible.
1. Inducible System (Lac operon of E. coli)
p i p z y a In absence of inducer
Repressor binds to the operator region(o) and prevents RNA polymerase from transcribing the operon
Repressor mRNA
Repressor
p i p z y ao
Repressor mRNA
Inducer
(Inactive repressor)
TranscriptionIn presence of inducer
Translationlac mRNA
b-galactosidase permease transacetylase
– An inducible operon system normally remains in switched off condition and begins to work only when the
substance to be metabolised by it is present in the cell. Inducible operon system generally occurs in catabolic
pathways. e.g. Lac operon of E. coli.
Active repressor + inducer = inactive repressor
An inducible operon system consists of four types of genes
(i) Structural genes - These genes synthesise mRNAs, which in turn synthesise polypeptide or enzyme
over the ribosomes. An operon may have one or more structural genes. Each structural gene of an
operon is called cistron. The lac operon (lactose operon) of Escherichia coli contains three structural
genes (Z, Y and A). These genes occur adjacent to each other and thus are linked. They transcribe a
polycistronic mRNA molecule (a single stretch of mRNA covering all the three genes), that helps in the
synthesis of three enzymes-b galactosidase (breaks lactose into glucose and galactose), lactose permease
(helps in entry of lactose in cell from outside) and transacetylase (transfers an acetyl group from acetyl
Co A to b galactosidase).
(ii) Operator gene - It lies adjacent to the structural genes and directly controls the synthesis of mRNA over
the structural genes. It is switched off by the presence of a repressor. An inducer can take away the
repressor and switch on the gene that directs the structural genes to transcribe.
(iii) Promoter gene - This gene is the site for initial binding of RNA polymerase. When the operator gene is
turned on, the enzyme RNA polymerase moves over it and reaches the structural genes to perform
transcription.
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Pre-Medical : BiologyALLEN(iv) Regulator gene - It produces a repressor that binds to operator gene and stops the working of the
operator gene.
Repressor - It is a protein, produced by the regulator gene. It binds to the operator gene so that the
transcription of structural gene stops. Repressor has two binding site (1) operator gene (2) effective
molecule (inducer/corepressor)
Inducer - It is a chemical (substrate, hormone or some other metabolite) which after coming in contact
with the repressor, forms an inducer repressor complex. This complex cannot bind with the operator
gene, which is thus switched on. The free operator gene allows the structural gene to transcribe
mRNA to synthesise the enzymes.
The inducer for lac operon of Escherichia coli is lactose (in fact allolactose an isomer of lactose). When the sugar
lactose is added to the culture of E. coli, a few molecules of lactose gets into the bacterial cells by the action of
the enzyme permease, a small amount of this enzyme is present in the cell even when the operon is not working.
These few lactose molecules are then converted into an active form which acts as an inducer and binds to the
repressor protein. The inducer repressor complex fails to join with the operator, which is turned on. The three
genes are expressed as three enzymes to metabolise lactose. Allolactose is real inducer of lac operon.
2. Repressible System (Tryptophan operon of E. coli.)
– A repressible operon system is normally in it's switch on state and continue to synthesise a metabolise till the
latter is produced in amount more than required, or else it becomes available to the cell from outside. Repressible
operon system is commonly found in anabolic pathway. e.g. Tryptophan operon of E. coli.
[Inactive repressor + co-repressor = acitve repressor]
Tryptophan operon of Escherichia coli is an example of repressible system. It consists of the following:
(i) Structural genes. These genes are meant for transcription of mRNA, which in turn synthesise enzyme.
Tryptophan operon has five structural genes E, D, C, B and A. They lie in continuation and synthesise
enzymes for five steps of tryptophan synthesis.
(ii) Operater gene (trp O). It lies adjacent to the structural genes and controls the functioning of the
structural genes. Normally, it is kept switched on, because the apo-repressor produced by the regulator
gene does not bind to it. The operator gene is switched off when a co-repressor is available alongwith
apo-repressor.
(iii) Promoter gene (trp P). It marks the site at which the RNA polymerase enzyme binds. When the
operator gene is switched on, it moves from promotor gene to structural genes for transcription.
(iv) Regulator gene (trp R). It produces a regulatory protein called apo-repressor for (Inactive repressor)
possible blocking the activity of operator gene.
(v) Apo-repressor. It is a regulatory protein synthesised by regulator gene. When a co-repressor substrate
is available in the cell, the apo-repressor combines with the co-repressor to form a apo-repressor
co-repressor complex. This complex binds with the operator gene and switches it off. Presence of
apo-repressor alone, the operator gene is kept switched on because, by itself the apo-repressor is unable
to block the working of operator gene.
(vi) Co-repressor. It is an end product of reactions catalysed by enzymes produced by the structural genes.
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Pre-Medical : Biology ALLENIn the presence of tryptophan some molecules of tryptophan act as co-repressor, co-repressor-bind with
inactive repressor. co-repressor repressor complex bind with-operator region and prevent the binding of
RNA polymerase to the promoter, the trp-operon is off.
The repressor molecule has key role in regulation of lac-operon. Repressor molecule active or inactive. Active
repressor may be rendered inactive by addition of an inducer while the inactive repressor can be made active by
addition of a co-repressor.
Because the product of regulator gene the repressor act by shutling off the transcripition of structural gene the
operon model, as originally proposed by Jocob & Monad is referred as –negative control system.
MUTATION* Sudden heritable change in genetic material of an organism is called as Mutation.
* Mutation are source of discontinuous variation.
* Only those mutation are heritable which occur in germinal cell of an organism. While somatic mutations are
non heritable. Somatic mutations are also heritable in vegetative propagated plants.
* Mutation word was given by Hugo De Vries.
* De Vries studied mutations in the plant Oenothera lamarckiana (evening primrose).
* Mutation was first observed by Seth Wright. He observed some short legged sheep (Ancon) variety in a population
of long legged sheep.
* Beadle and Tatum induced mutations in Neurospora by the help of U. V. rays. or X-rays.
Wild Neurospora U. V. rays¾¾¾¾® Mutant Neurospora.
(PROTOTROPH) (AUXOTROPH)
P.G. O.G. E D C B A
¯ ¯ ¯ ¯ ¯
R.G.
¯No–TranscriptionNo–Translation
Aporepressor+
Co–repressor(Tryptophan)
6 7444 8444Structural gene
[OPERON – OFF]
P.G. O.G. E D C B A
¯ ¯ ¯ ¯ ¯
R.G.
Polycistronic m -RNA
¯ ¯ ¯ ¯ ¯
¯Aporepressor
¯(inactive
repressor)[OPERON – ON]
6 7444 8444Structural gene
Transcription
TranslationPolypeptides
E1 E2 E3 Enzymes
TryptophanChorismic acid
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Pre-Medical : BiologyALLENNormal-Neurospora can be grown in minimal medium because Neurospora can make all essential nutrients
required for it. This is known as Prototroph.
Mutant Neurospora doesn't has capability to grow in minimal medium because due to mutation it loses those
genes which codes for the enzyme that helps to prepare some special nutrients for it. They gave "one gene-one
enzyme" concept. This form is known as Auxotroph.
* M.S. Swaminathan induced mutations in wheat by the help of g-rays to obtain good varieties for eg. Sharbati
Sonora, Pusa Lerma. Swaminathan established g garden in IARI-New Delhi (Pusa Institute).
Types of mutation :i. CHROMOSOMAL MUTATION
ii. GENE MUTATIONS
(I) Chromosomal Mutations :* Change in number or structure of chromosome.
Types of chromosomal mutation
(1) Heteroploidy/Genomatic mutation ® change in chromosome number.
(2) Chromosomal aberration ® change in structure of chromosome.
1. Heteroploidy / Genomatic mutation* Change in number of one or few chromosomes in a set or number of entire set of chromosome.
It is of two types :
(i) Euploidy ® Change in number of chromosome sets.
(ii) Aneuploidy ® Change in number of chromosome in a set.
Euploidy :* Change in number of sets of chromosome i.e. either loss or addition of sets of chromosomes.
* Monoploidy (x) - Presence of one set of chromosomes.
* Diploidy (2x) - Presence of two sets of chromosomes.
* Polyploidy - Presence of more than two sets of chromosomes.
It may be :-
Triploidy (3x)
Tetraploidy (4x)
Pentaploidy (5x)
Hexaploidy (6x)
Heptaploidy (7x)
Octaploidy (8x)
* Polyploid plants with even number of sets are always fertile, reproduce sexually and form seeds.
* Polyploid plants with odd number of sets are always sterile don't reproduce by sexual reproduction, They don't
produce seeds but they may produce seedless fruits by parthenocarpy.
eg. Banana and seedless grapes.
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Pre-Medical : Biology ALLENExample
Triticale : Triticum aestivum X Rye (Secale Cereale)6x = 42 2x = 14
Triticale
4x = 28 Sterile
Colchicine
Triticale Fertile
8x = 56
Aneuploidy :Loss or addition of chromosomes in a set of chromosomes.
Types of Aneuploidy :
1. Hypoaneuploidy (loss)
* 2n – 1 = Monosomy :- (loss of one chromosome in one set).
* 2n – 1 – 1 = Double monosomy (loss of one chromosome from each set, but these are non homologus.)
* 2n – 2 = Nullisomy (loss of two homologus chromosome)
2. Hyperaneuploidy (addition)
* 2n + 1 = Trisomy : addition of one chromosome in one set.
* 2n + 1 + 1 = Double Trisomy : addition of one chromosome in each set.
* 2n + 2 = Tetrasomy : addition of two chromosome in one set.
* Cause of aneuploidy is chromosomal nondisjunction means chromosomes fail to separate during meiosis.
* Chances of aneuploidy are more in higher age female due to less activity of oocyte, so chances of syndrome
increase in children who are born from higher age female.
2. Chromosomal Aberrations :Change in structure of chromosome.
(i) Deletion :
Loss of a part or segment of chromosome which leads to loss of some gene is called as deletion.
It is of 2 types :-
(a) Terminal deletion - Loss of chromosomal segment from one or both ends.
eg. The cry -du-chat syndrome is an example of terminal deletion in short arm of5th chromosome.
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Pre-Medical : BiologyALLEN(b) Intercalary deletion - Loss of chromosomal part between the ends.
(ii) Inversion :Breakage of chromosomal segment but reunion on same chromosome in reverse orders. It leads to change in
distance between genes on chromosome or sequence of genes on chromosome so crossing over is affected.
It is of 2 types :-
(a) Paracentric - If inversion occur only in one arm and inverted segment does not include centromere.
1 2 3 4 5 6 7 8 ® 1 2 3 4 5 7 6 8
(b) Pericentric - In this type of inversion inverted segment include centromere.
1 2 3 4 5 6 7 8 ® 1 2 6 5 4 3 7 8
(iii) Duplication :Occurence of a chromosomal segment twice on a chromosome.
Example : In drosophila "Bar eye character" is observed due to duplication in X-chromosome. Bar eye is a
character where eyes are narrower as compared to normal eye shape.
A B C D E F A B C D¾¾®
A B C D E F A B C D E F E F
(iv) Translocation :In this, a part of the chromosome is broken and may be joined with non homologous chromosome. This is also
known as Illegitimate crossing over (illegeal crossing over)
Types of translocation -
(A) Simple Translocation® When a chromosomal segment breaks and attached to the terminal end of a
non- homologous chromosome.
1 2 3 4 5 6 1 2 3 4
¾®
A B C D E F A B C D E F 5 6
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Pre-Medical : Biology ALLEN(B) Reciprocal Translocation® Exchange of segments between two non-homologous chromosome.
1 2 3 4 5 6 1 2 3 4 E F
¾®
A B C D E F A B C D 5 6
eg. Chronic myloid leukemia [C M L] is a type of blood cancer. This disease is a result of reciprocaltransiocation between 22 and 9 chromosome.
Note : If exchange of segments takes place in between homologous chromosomes then it is called crossing over.
(II) Gene Mutation or point mutationTwo types :-
1. Substitution
2. Frame shift mutation.
A. Substitution :Replacement of one nitrogenous base by another nitrogenous base is called as substitution.
* It causes change in one codon in genetic code which leads to change in one amino acid in structure of protein.
eg. Sickle cell anaemia
* Change may not occur some time because for one animo acid more than one type of codons are present.
Substitution is of two types :-
1. Transition :
Replacement of one purine by another purine or replacement of pyrimidine by another pyrimidine.
2. Transversion :
Replacement of purine by pyrimidine or pyrimidine by purine is called transversion.
B. Frame shift mutation/Gibberish mutation :Loss or addition of one or rarely more than one nitrogenous bases in structure of DNA.
Frame shift mutation is of two types
1. Addition
Addition of one or rarely more than one nitrogenous bases in structure of DNA.
2. Deletion
Loss of one or rarely more than one nitrogenous bases in structure of DNA.
Due to frame shift mutation complete reading of genetic code is changed. It leads to change in all animo acids
in structure of protein so a new protein is formed which is completely different from previous protein.
* So frame shift mutations are more harmful as compared to substitution.
eg : Thalassemia (lethal genetic disorder)
MUTAGENS :* Mutagens are those substances which cause mutations.
Non ionising :- U. V. rays.
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Pre-Medical : BiologyALLEN
Mut
atio
n(
)
Sud
den
herit
able
cha
nge
in g
enet
ic m
ater
ial
(Cha
nge
in N
ucle
otid
e)
Fram
e sh
ift
mut
atio
n(L
oss
or a
dditi
on o
f on
e or
ra
rely
mor
e th
an o
ne n
itrog
enou
s ba
ses
in s
truc
ture
of D
NA
Ex
: T
hala
ssem
ia)
(Rep
lace
men
t of on
e ni
trog
enou
s ba
se b
y an
othe
r ni
trog
enou
s ba
se
Ex.
:
)Si
ckle
cel
l ana
emia
Het
erop
loid
y m
utat
ion
(Cha
nge
in n
umbe
r of
chr
omos
ome)
Chr
omos
omal
abe
rrat
ion
(Cha
nge
in s
truc
ture
of ch
rom
osom
e)
Del
etio
nD
uplic
atio
nIn
vers
ion
Tran
sloc
atio
n
Term
inal
Inte
rcal
ary
orin
ters
titi
alPar
acen
tric
Per
icen
tric
1 2 3 4 5 6 7 8
3 4 5 6 7 8
1 2 3 4 5 6 7 8
1 2 3 4 5 6 7 8
5 6 7 8
2 1 3 4 5 6 7 8
1 2 3 4 5 6 7 8
1 2 6 5 4 3 7 8+
+1 2
3 4
1 2 5 6 7 8
Sim
ple
Tran
sloc
atio
nR
ecip
roca
lTr
ansl
ocat
ion
1 2
3 4
5 6
7 8
AB
CD
EF
GH
1 2
3 4
5 6
AB
CD
EF
GH
78
1 2
3 4
5 6
7 8
AB
CD
EF
GH
1 2
3 4
5 6
G H
A B
C D
E F
7 8
Euplo
idy
(cha
nge
in n
umbe
r of
sets
of ch
rom
osom
es)
Ane
uplo
idy
(cha
nge
in n
umbe
r of
chro
mos
omes
in
sets
)
Mon
oplo
idy
[Los
s of
set
of c
hrom
ose]
(2
n–n)
Pol
yplo
idy
(Add
ition
of se
tsof
chro
mos
omes
)
2n
+ n
= 3
n tr
iplo
id2n
+ 2
n =
4n
tetr
aplo
id2n
+ 3
n =
5n
pen
tapl
oid
2n
+ 4
n =
6n
hexa
ploi
d2n
+ 5
n =
7n
hept
aplo
id2n
+ 6
n =
8n
octa
ploi
d
eg.
n nn
Hyp
oane
uplo
idy
(Los
s of
chr
om
osom
es in
set
s)H
yper
aneu
ploi
dy(A
dd.o
f ch
rom
osom
es in
set
s)
Mon
osom
y[2
n –
1]D
oubl
e m
onos
omy
[2n
– 1 –
1]
Nul
lisom
y[2
n –
2]
Trisom
y[2
n +
1]
Dou
ble
tris
omy
[2n
+ 1
+1]
Tetr
asom
y[2
n +
2]
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Pre-Medical : Biology ALLEN
GOLDEN KEY POINTS* Mostly mutations are harmful.
* Sometimes they are lethal which leads to death of organisms.
* But sometimes they are beneficial which are used to obtain good varieties of plants and animals. It is called as
Mutation Breeding.
* Mostly mutations are recessive and they never eliminate from a population.
Forward and Backward Mutation :
* Wild geneForward
Backward�������������� Mutant gene
Muton (unit of mutation) :
* Smallest part of DNA which undergoes mutation.
* It is one nucleotide.
Mis-sense mutation :-
* When a nucleotide change in genetic code cause the change of one amino acid of a polypeptide chain it is called
mis-sense mutation.
Non-sense mutation :-
* When a nucleotide change in one codon causes termination of polypepetide synthesis by producing non-sense codon.
Same sense codon :-
* A change in one nucleotide in a codon does not change amino acid in polypeptide chain, because both codons
code same amino acid.
BEGINNER'S BOX-31. (2n–1) condition of chromosomes is called :-
(1) tetrasomy (2) trisomy (3) monosomy (4) nullisomy
2. Given below is the representation of a kind of chromosomal mutation. It is :-A B C D E F G
A D C B E F G
(1) Deletion (2) Inversion (3) Duplication (4) Reciprocal translocation
3. Which of the following has normal sex chromosome complements.
(1) Down's syndrome (2) Klinefelter's syndrome (3) Super female (4) Turner's syndrome
4. Addition or deletion of a nitrogenous base causes
(1) Frameshift mutation (2) Inversion (3) Transformation (4) Translocation
5. Trisomy of which chromosome is involved in Down's syndrome.
(1) 8th (2) 13th (3) 21st (4) 22nd
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Pre-Medical : BiologyALLEN
DNA FINGER PRINTING / DNA TYPING / DNA PROFILING/ DNA TEST
– It is technique to identify a person on the basis of his/her DNA specificity.This technique was invented by sir Alec. Jeffery (1984).
– In India DNA Finger printing has been started by Dr. V.K. Kashyap & Dr. Lal Ji Singh.– DNA of human is almost the same for all individuals but very small amount that differs from person to person
that forensic scientists analyze to identify people.These differences are called Polymorphism (many forms) and are the key of DNA typing. Polymorphisms aremost useful to forensic scientist. It is consist of variation in the length of DNA at specific loci is called Restrictedfragment. It is most important segment for DNA test made up of short repetitive nucleotide sequences, theseare called VNTRs (variable number of tandem repeat).VNTR's also called minisatellites were discovered by Alec Jeffery. Restricted fragment consist of hypervariablerepeat region of DNA having a basic repeat sequence of 11-60 bp and flanked on both sites by restriction site.
– The number and position of minisatellites or VNTR in restriction fragmnt is different for each DNA and lengthof restricted fragment is depend on number of VNTR.
– Therefore, when the genome of two people are cut using the same restriction enzyme the length of fragmentsobtained is different for both the people.
– These variations in length of restricted fragment is called RFLP or Restriction fragment length polymorphism.– Restriction Fragment Length Polymorphism distributed throughout human genomes are useful for DNA Finger printing.– DNA Fingerprint can be prepared from extremely minute amount of blood, semen, hair bulb or any other cell
of the body.DNA content of 1 - Microgram is sufficient.Technique of DNA Finger printing involves the following major stpes.
1. Extraction – DNA extracted from the cell by cell lysis. If the content of DNA is limited then DNA can beamplified by Polymerase chain reaction (PCR). This process is amplification.
2. Restriction Enzyme Digestion : Restriction enzyme cuts DNA at specific 4 or 6 base pair sequences calledrestriction site.Hae III (Haemophilus aegyptius) is most commonly used enzyme. It cuts the DNA, every where the bases arearranged in the sequence GGCC. These restricated fragment transferred to Agarose Polymer gel.
3. Gel Electrophoresis :– Gel electrophoresis is a method that separates macromolecules-either nucleic acid or proteins-on the basis of
size, electric charge.– Gel electrophoresis refers to the technique in which molecules are forced across a span of gel, motivated by an
electrical current. Activated electrodes at either end of the gel provides the driving force. A molecule's proper-ties determine, how rapidly an electric field can move the molecule through a gelatinous medium.
– Nowadays the most commonly used matrix is agarose which is a natural polymer extracted from sea weeds. TheDNA fragments separate (resolve) according to their size through sieving effect provided by the agarose gel.
– Many important biological molecules such as amino acids, peptides, proteins, nucleotides, and nucleic acidsposses ionisable groups and, therefore, at any given pH, exist in solution as electrically charged species eitheras cation (+) or anions (–). Depending on the nature of the net charge, the charged particles will migrate eitherto the cathode or to the anode.
– By the gel electrophoresis these restricted fragments move towards the positive electrode (anode) because DNAhas –ve electric charge (PO4
–3).– Smaller Fragment more move towards the positive pole due to less molecular weight. So after the gel electro-
phoresis DNA fragment arranged according to molecular weight.– These separated fragments can be visualized by staining them with a dye that fluoresces ultraviolet radiation.4. Southern transfer / Southern blotting :
The gel is fragile. It is necessary to remove the DNA from the gel and permanently attaches it to a solid support.This is accomplished by the process of Southern blotting. The first step is to denature the DNA in the gel whichmeans that the double-stranded restriction fragments are chemically separated into the single stranded form.
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Pre-Medical : Biology ALLEN
RESTRICTIONENZYME
DIGESTION
GEL ELECTROPHORESIS(Agargose polymer gel)
DENATURATIONBY ALKALI SOLUTION
SOUTHERN BLOTTING
NYLON OR NITROCELLULOSE SHEET
HYBRIDIZATION
DNA PROBES
Steps of DNA Fingerprinting
The DNA then is transferred by the process of blotting to a sheet of nylon. The nylon acts like an ink blotter and"blots" up the separated DNA fragments, the restriction fragments, invisible at this stage are irreversibly at-tached to the nylon membrane the "blot".This process is called Southern blot by the name of Edward Southern (1970).
5. Hybridization : To detect VNTR locus on restricted fragment, we use single stranded Radioactive (P32) DNAprobe which have the base pair sequences complimentary to the DNA sequences at the VNTR locus. Com-monly we use a combination of at least 4 to 6 separate DNA probes.Labelled Probes are attached with the VNTR loci of restricted DNA Fragments, this process is called Hybridization.
6. Autoradiography : Nylon membrane containing radio active probe exposed to X-ray. Specific bands appearon X-ray film. These bands are the areas where the radioactive probe bind with the VNTR.
– This appears the specific restricted fragment length pattern. This length pattern is different in different indi-vidual. This is called Restricted Fragment length Polymorphism (RFLP).These allow analyzer to identify a particular person DNA, the occurance and frequency of a particular geneticpattern contained in this X-ray film. These x-ray film called DNA signature of a person which is specific for eachindividual.The probability of two unrelated individual having same pattern of location and repeat number of minisatellite(VNTR) is one in ten billion (world population 6.1 billion)In India the centre for DNA finger printing and diagnosis (CDFD - center for DNA finger printing & diagnosis)located at Hyderabad.
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Pre-Medical : BiologyALLENApplication of DNA Finger printing
1. Paternity tests. The major application of DNA finger printing is in determining family relationships. For
identifying the true (biological) father, DNA samples of Child, mother and possible fathers are taken and their
DNA finger prints are obtained. The prints of child DNA match to the prints of biological parents.
2. Identification of the criminal. DNA finger printing has now become useful technique in forensic (crime
detecting) science, specially when serious crimes such as murders and rapes are involved. For identifying a
criminal, the DNA fingerprints of the suspects from blood or hair or semen picked up from the scene of crime
are prepared and compared. The DNA fingerprint of the person matching the one obtained from sample
collected from scene of crime can give a clue to the actual criminal.
Chromosome 7 Chromosome 7
Chromosome 2 Chromosome 2
Chromosome 16 Chromosome 16
Paternal chromosome
Maternal chromosome
DNA from individual A DNA from individual B
Chromosome 7
Chromosome 2
Chromosome 16
0 11
DNA from crime scene (C)Amplified repeats, separatd by size on a gel,
geve a DNA fingerprint
C A B
12
11
10
9
8
7
6
5
4
3
2
1
Num
ber
of s
hort
tan
dem
rep
eats
Schematic representation of DNA fingerprinting : Few representative chromosomes have been shown to contain different copy number of VNTR. For the sake of understanding colour schemes have been used to trace the origin of each band in the gel. The two alleles (paternal and maternal) of chromosome also contain different copy numbers of VNTR. It is clear that the banding pattern of DNA from crime sceme matches with individual B, and not with A.
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HUMAN GENOME PROJECT
Genetic make-up of an organism or an individual lies in the DNA sequences. If two individuals differ, then their
DNA sequences should also be different, at least at some places. These assumptions led to the quest of finding out
the complete DNA sequence of human genome. With the establishment of genetic engineering techniques where
it was possible to isolate and clone any piece of DNA and availability of simple and fast techniques for determining
DNA sequences, a very ambitious project of sequencing human genome was launched in the year 1990.
Human Genome Project (HGP) was called a mega project. You can imagine the magnitude and the requirements
for the project if we simply define the aims of the project as follows :
Human genome is said to have approximately 3 × 109 bp, and if the cost of sequencing required is US $ 3 per
bp (the estimated cost in the beginning), the total estimated cost of the project would be aproximately 9 billion
US dollars. Further, if the obtained sequences were to be stored in typed form in books, and if each page of the
book contained 1000 letters and each book contained 1000 pages, then 3300 such books would be required
to store the information of DNA sequence from a single human cell. HGP was closely associated with the rapid
development of a new area in biology called as Bioinformatics.
Goals of HGP
Some of the important goals of HGP are as follows :
(i) Identify all the genes in human DNA.
(ii) Determine the sequences of the 3 billion chemical base pairs that make up human DNA.
(iii) Store this information in databases.
(iv) Improve tools for data analysis.
(v) Transfer related technologies to other sectors, such as industries.
(vi) Address the ethical, legal, and social issues (ELSI) that may arise from the project.
The project was completed in 2003. Knowledge about the effects of DNA variations among individuals can lead
to revolutionary new ways to diagnose, treat and someday prevent the thousands of disorders that affect human
beings. Besides providing clues to understanding human biology, learning about non-human organisms, DNA
sequences can lead to an understanding of their natural capabilities that can be applied toward solving challenges
in health care, agriculture, energy production, environmental remediation. Many non-human model organisms,
such as bacteria, yeast, Caenorhabditis elegans (a freeliving non-pathogenic nematode), Drosophila (the fruit
fly), plants (rice and Arabidopsis), etc., have also been sequenced.
Methodologies : The methods involved two major approaches. (1) Expressed Sequence Tags (ESTs) -
Identifying all the genes that expressed as RNA. (2) Sequence Annotation - The blind approach of simply
sequencing the whole set of genome that contained all the coding and non-coding sequence, and later assigning
different regions in the sequence with functions. For sequencing, the total DNA from a cell is isolated and
converted into random fragments of relatively smaller sizes (recall DNA is a very long polymer, and there are
technical limitations in sequencing very long pieces of DNA) and cloned in suitable host using specialised vectors.
The cloning resulted into amplification of each piece of DNA fragment so, that is subsequently could be sequenced
with ease. The commonly used hosts were bacteria and yeast, and the vectors were called as BAC (bacterial
artificial chromosomes), and YAC (yeast artificial chromosomes).
The fragments were sequenced using automated DNA sequencers that worked on the principle of a method
developed by Frederick Sanger. (Remember, Sanger is also credited for developing method for determination
of amino acid sequences in proteins). These sequences were then arranged based on some overlapping regionspresent in them. This required generation of overlapping fragments for sequencing. Alignment of these sequences
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Pre-Medical : BiologyALLENwas humanly not possible. Therefore, specialised computer based programmes were developed. These sequences
were subsequently annotated and were assigned to each chromosome. The sequence of chromosome I was
completed only in May 2006 (this was the last of the 24 human chromosomes -22 autosomes and X and Y- to
be sequenced). Another challenging task was assigning the genetic and physical maps on the genome. This was
generated using information on polymorphism of restriction endonuclease recognition sites, and some repetitive
DNA sequences known as microsatellites.
Salient Features of Human Genome –
Some of the salient observations drawn from human genome project are as follows :
(i) The human genome contains 3164.7 million nucleotide bases.
(ii) The average gene consists of 3000 bases, but sizes vary greatly, with the largest known human gene
being dystrophin at 2.4 million bases.
(iii) The total number of genes is estimated at 30,000-much lower than previous estimates of 80,000 to
1,40,000 genes. Almost all (99.9 per cent) nucleotide bases are exactly the same in all people.
(iv) The functions are unknown for over 50 per cent of discovered genes.
(v) Less than 2 per cent of the genome codes for proteins.
(vi) Repeated sequences make up very large portion of the human genome.
(vii) Repetitive sequences are stretches of DNA sequences that are repeated many times, sometimes hundred
to thousand times. They are thought to have no direct coding functions, but they shed light on chromosome
structure, dynamics and evolution.
(viii) Chromosome 1 has most genes (2968). and the Y has the fewest (231).
(ix) Scientists have identified about 1.4 million locations where single-base DNA differences (SNPs- singlenucleotide polymorphism, pronounced as 'snips') occur in humans, This information promises to
revolutionise the processes of finding chromosomal locations for disease-associated sequences and tracing
human history.
Organisms Base pair Gene No.
Bateriophage 10,000 ------/-----
Lily 106 Billion B.P. ------
E.coli 4.7 million B.P. 4,000
S. cerevisiae 12 Million B.P. 6,000
D. melangaster 180 Million B.P. 13,000
Caenorhabditis elegans 97 Million B.P. 18,000
Human 3 Billion B.P. 30,000
(a) First prokaryotes in which complete genome was sequenced is Haemophilus influenzae.
(b) First Eukaryote in which complete genome was sequenced is Saccharomyces cerviceae (Yeast).
(c) First plant in which complete genome was sequenced is Arabidopsis thaliana (Small mustard plant).
(d) First animal in which complete genome was sequenced is Caenorhabditis elegans (Nematode).
– b – globin and insulin gene are less than 10 kilo base pair T.D.F. gene is the smallest gene (14 base pair)
and Duchenne muscular Dystrophy gene is made up of 2400 kilo base pair.(Longest gene)
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Pre-Medical : Biology ALLEN
ANSWER KEY
BEGINNER'S BOX-4
1. DNA finger printing involves identifying differences in some specific
(1) Repetitive DNA (2) Non repetitive DNA
(3) Selfish DNA (4) All of the above
2. Which of the following is produced by E-Coli in the lactose operon.
(1) B galactosidase (2) Transacetylase
(3) Permease (4) All of the above
3. Maximum number of gene present on which chromosome number in human.
(1) 1st (2) X (3) Y (4) 10th
4. In lac operon RNA polymerase binds with
(1) Promoter gene (2) Operator gene
(3) Structural gane (4) Regulator gene
5. Fill the gap in following statement
Human genome have approximately _____________ and the cost of sequencing was ___________ per base
pair.
(1) 4 × 109 bp, 9 billion US dollars (2) 9 billion US dollars, 4 × 199 bp
(3) 3 × 109 bp, 3 US dollars (4) 4.7 million bp, 9 billion US dollars
Que. 1 2 3 4 5 6 7 8 9 10Ans. 4 2 1 2 4 2 2 4 4 2
Que. 1 2 3 4 5 6 7 8 9 10Ans. 1 3 2 3 2 1 2 3 3 2
Que. 1 2 3 4 5Ans. 3 2 1 1 3
Que. 1 2 3 4 5Ans. 1 4 1 1 3
BEGINNER'S BOX-4
BEGINNER'S BOX-1
BEGINNER'S BOX-3
BEGINNER'S BOX-2
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Pre-Medical : BiologyALLENEXERCISE-I (Conceptual Questions) Build Up Your Understanding
GENETIC MATERIAL, DNA
1. Circular and double stranded DNA occurs in -
(1) Golgibody (2) Mitochondria
(3) Nucleus (4) Cytoplasm
2. Double helix model of DNA which was proposed
by watson and crick was of-
(1) C-DNA (2) B-DNA
(3) D-DNA (4) Z-DNA
3. If there are 10,000 nitrogenous base pairs in a
DNA then how many nucleotides are there-
(1) 500 (2) 10,000
(3) 20,000 (4) 40,000
4. Double helix model of DNA is proposed by-
(1). Watson and Crick
(2) Schleiden schwann
(3) Singer and Nicholson
(4) Kornberg and Khurana
5. Back bone in structure of DNA molecule is made
up of -
(1) Pentose Sugar and phosphate
(2) Hexose sugar and phosphate
(3) Purine and pyrimidine
(4) Sugar and phosphate
6. Substance common in DNA and RNA -
(1) Hexose Sugar (2) Histamine
(3) Thymine (4) Phosphate groups
7. Nucleotide is -
(1) N2 - base, pentose sugar and phosphoric acid
(2) Nitrogen, Hexose sugar and phosphoric acid
(3) Nitrogen base, pentose sugar
(4) Nitrogen base, trioses and phosphoric acid
8. Unit of nucleic acids are-
(1) Phosphoric acid
(2) Nitrogenous bases
(3) Pentose Sugar
(4) Nucleotides
9. Which element is not found in nitrogenous base :-
(1) Nitrogen (2) Hydrogen
(3) Carbon (4) Phosphorus
10. DNA was first discovered by-
(1) Meischer (2) Robert Brown
(3) Flemming (4) Watson & Crick
11. Nucleic acid (DNA) is not found in-
(1) Nucleus & nucleolus
(2) Peroxysome & ribosome
(3) Mitochondria & plastid
(4) Chloroplast & nucleosome
12. DNA is not present in -
(1) Mitochondria (2) Chloroplast
(3) Bacteriophage (4) TMV
13. A nucleic acid contains thymine or methylated
uracil then it should be -
(1) DNA
(2) RNA
(3) Either DNA or RNA
(4) RNA of bacteria
14. Prokaryotic genetic system contains -
(1) DNA & histones
(2) RNA & histones
(3) Either DNA or histones
(4) DNA but no histones
15. A N2- base together with pentose sugar and
phosphte forms (or) building - block unit of nucleic
acid is :-
(1) Nucleoside (2) Polypeptide
(3) Nucleotide (4) Aminoacid
16. Which of the following is not a pyrimidine N2
base -
(1) Thymine (2) Cytosine
(3) Guanine (4) Uracil
17. The purine & pyrimidine pairs of complementry
strands of DNA are held together by –
(1) H - bonds (2) O - bonds
(3) C - bonds (4) N - bonds
18. Number of H - bonds between guanine and
cytosine are -
(1) One (2) Two
(3) Three (4) Four
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Pre-Medical : Biology ALLEN19. Which purine & pyrimidine bases are paired
together by H - bonds in DNA –
(1) AC & GT (2) GC & AT
(3) GA & TC (4) None of the above
20. A single stranded DNA is present in-
(1) TMV
(2) Salmonella
(3) f x174
(4) Bacteria
21. What is the nature of the 2 strands of a DNA
duplex :-
(1) Identical & Complimentary
(2) Antiparallel & complimentary
(3) Dissimilar & non complimentary
(4) Antiparallel & non complimentary
22. On an average, how many purine N2 bases are
present in single coil of DNA
(1) Four (2) Five
(3) Ten (4) Uncertain
23. Distance between two nucleotide pairs of DNA
is -
(1) 0.34 nm (2) 34 A0
(3) 3.4 m (4) 34 nm
24. In a double strand DNA amount of Guanine is
35% then what will be the amount of cytosine?
(1) 70% (2) 15%
(3) 30% (4) 35%
25. Chargaaf 's rule is given as -
(1) Purines ¹ Pyrimidines
(2) A + G = T+ C
(3) A+ U = G + C
(4) A + T / G + C = Const.
26. In RNA , Nucleotides are bonded by -
(1) H - bonds
(2) Phospo diester bonds
(3) Ionic bonds
(4) Salt linkage
27. A nucleoside differs from a nucleotide is nothaving -(1) Phosphate (2) Sugar(3) Phosphate & sugar (4) Nitrogen base
28. Wilkins X - ray diffraction showed the diameter ofthe DNA helix is -(1) 10 Å (2) 20 Å(3) 30 Å (4) 40 Å
29. In the DNA of an animal percentage of Adenineis 30 then percentage of Guanine will be -(1) 40 (2) 30 (3) 20 (4) 70
30. Similarity in DNA and RNA-(1) Both are polymer of nucleotides(2) Both have similar pyrimidine(3) Both have similar sugar(4) Both are genetic material
31. Length of one loop of B- DNA-(1) 3.4 nm. (2) 0.34 nm.(3) 20 nm. (4) 10 nm.
32. If base order in one chain of DNA is "ATCGA"then how many no. of H–bond found in DNAduplex :–(1) 20 (2) 12(3) 10 (4) 11
33. In DNA purine nitrogen bases are :-(1) Uracil and Guanine(2) Guanine and Adenine(3) Adenine and cytosine(4) None
34. Two free ribonucleotide units are interlinked with :(1) Peptide bond(2) Disulphide bond(3) Hydrogen bond(4) Phosphodiester bond
35. Short DNA segment has 80 thymine and 90guanine bases. The total number of nucleotides are(1) 160 (2) 40 (3) 80 (4) 340
36. Prokaryotic DNA is :-(1) double stranded circular(2) single stranded circular(3) double stranded linear
(4) double stranded RNA as nucleic acid
37. Nucleoside is :-(1) Polymer of nucleic acid(2) Phosphoric acid + base(3) Phosphoric acid + sugar + base(4) Sugar + base
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Pre-Medical : BiologyALLEN38. The back bone of RNA is consists of which of the
following sugar :-
(1) Deoxyribose (2) Ribose
(3) Sucrose (4) Maltose
39. Retrovirus have genetic material :-
(1) DNA only (2) RNA only
(3) DNA or RNA only (4) None
40. In a nucleotide H3PO4 binds to which carbon atom
of pentose sugar :-
(1) Only Ist carbon
(2) Only 3rd carbon
(3) Only 5th carbon
(4) Both 3rd and 5th carbon
41. DNA is acidic due to :-
(1) Sugar (2) Phosphoric acid
(3) Purine (4) Pyrimidine
42. T.M.V. contains :-
(1) D.N.A.
(2) R.N.A. + Protein
(3) D.N.A. + R.N.A.
(4) D.N.A. + Protein
43. R.N.A contains which of the following base, in place
of Thymine of D.N.A. :-
(1) Thymine (2) Uracil
(3) Adenine (4) None of these
44. Genetic information are transferred from nucleus
to cytoplasm of cell through :-
(1) DNA (2) RNA
(3) Lysosomes (4) ACTH
45. If one strand of double stranded DNA, consists of
the sequence 3'–ATTCGTAC–5', then the
complementary sequence must be –
(1) 5'–UAAGCAUG–3'
(2) 3'–TAAGCATG–5'
(3) 5'–TAAGCATG–3'
(4) 5'–TAAGCATG–3' in the reverse direction
46. Which of the following is a false statements ?
(1) DNA is chemically less reactive, as compared
to RNA
(2) RNA mutate at a faster rate, as compared
to DNA
(3) Guanyl transferase enzyme helps in capping
process during splicing of hn-RNA
(4) r RNA is less aboundant RNA in an animal cell
47. DNA molecule has uniform diameter due to ?
(1) Double stranded
(2) Presence of phosphate
(3) Specific base pairing between purine and
pyrimidine
(4) Specific base pairing between purine and
purine
48. Following structure is related to which compound?
(1) AdenineH–N
O
ON
H
(2) Guanine
(3) Uracil
(4) Thymine
49. If the sequence of bases in one strand of DNA is
known then the sequence in other strand can be
predicted on the basis of–
(1) Antiparallel (2) Complementary
(3) Polarity (4) Coiling
50. The unequivocal proof that DNA is the genetic
material came from the experiments of –
(1) Hershey and chaese (1952)
(2) Frederic Griffith (1928)
(3) Watson and Crick
(4) Meselson and Stal (1958)
DNA REPLICATION51. In process of replication deoxyribonucleoside
triphosphate
(1) acting as substrate
(2) providing energy for polymerisation reaction
(3) acting as an enzyme
(4) both (1) & (2)
52. DNA duplication occurs at
(1) Meiosis - II
(2) Mitotic interphase
(3) Mitosis only
(4) Meiosis and mitosis both
53. DNA Replication occurs at -
(1) G0 & G1 (2) G2 - stage
(3) S - Stage (4) Mitotic phase
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Pre-Medical : Biology ALLEN54. A DNA molecule in which both strands have
radioactive thymidine is allowed to duplicate in
an environment containing non- radioactive
thymidine. What will be the exact number of DNA
molecules that contains the radio active thymidine
after 3 duplications -
(1) One (2) Two
(3) Four (4) Eight
55. A bacterium with completely radioactive DNA
was allowed to replicate in a non- radioactive
medium for two generation what % of the bacteria
should contain radioactive DNA :-
(1) 100 % (2) 50 %
(3) 25 % (4) 12.5 %
56. In the base sequence of one strand of DNA is
GAT , TAG ,CAT , GAC what shall be the
sequence of its complementary strand-
(1) CAT, CTG, ATC, GTA
(2) GTA, ATC, CTG, GTA
(3) ATC, GTA, CTG, GTA
(4) CTA, ATC, GTA, CTG
57. Method of DNA replication in which two strands
of DNA separates and synthesize new strands:-
(1) Dispersive
(2) Conservative
(3) Semiconservative
(4) Non conservative
58. During replication of a bacterial chromosome DNA
synthesis starts from a replication origin site and
(1) RNA primers are not involved
(2) is facilitated by telomerase
(3) moves in one direction of the site
(4) moves in bi-directional way
59. Which one of the following hydrolyses internal
phosphodiester bonds in a polynucleotide
chain –
(1) Lipase (2) Protease
(3) Exonuclease (4) Endonuclease
60. The nature of DNA replication is :-(1) Conservation(2) Non conservative(3) Semi-consurvative(4) Cyanobacteria
61. The direction of D.N.A. replication is :
(1) From 5' end towards 3' end
(2) From 3' end towards 5' end
(3) Amino terminus to carboxy terminus
(4) Carboxy terminus to amino terminus
62. Semiconservation replication of DNA was given by
(1) Watson and Crick
(2) Bateson and Punnett
(3) Messelson and Stahl
(4) Avery, McCarty and Mactleod
63. Which of the following enzyme is used in DNA
multiplication :-
(1) RNA polymerase
(2) DNA endonuclease
(3) Exonuclease
(4) DNA Polymerase
64. Mode of DNA replication in E. coli is :-
(1) Conservative and unidirectional
(2) Semi conservative and unidirectional
(3) conservative and bidirectional
(4) Semi conservative and bidrectional
65. Which of the following enzyme is used to join DNA
fragments :-
(1) Terminase (2) Endonuclease
(3) Ligase (4) DNA polymerase
66. Okazaki fragments are synthesised on :-
(1) Leading strands of DNA only
(2) Lagging strands of DNA only
(3) Both leading and lagging strands of DNA
(4) Complementary DNA
67. DNA replication includes :-
(1) DNA ligase
(2) DNA polymerase and ligase
(3) RNA polymerase and ligase
(4) All of these
68. In DNA replication, the primer is :-
(1) A small deoxyribonucleotide polymer
(2) A small ribonucleotide polymer
(3) Helix destabilizing protein
(4) Enzyme taking part in joining nucleotides of
new strand
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Pre-Medical : BiologyALLEN69. The strand of DNA, which does not code for anything
is referred to as :-
(1) Template strand
(2) Antisense strand
(3) Coding strand
(4) Noncoding strand
70. During DNA replication discontinuosly synthesized
fragments are later joined by the enzyme –
(1) Ligase
(2) DNA polymerage
(3) RNA primer
(4) Primase
71. Replication fork is –
(1) Large opening of the DNA helix
(2) Small opening of the DNA helix
(3) Tightly coiled part of DNA helix
(4) Loosely coiled part of DNA helix
72. The DNA dependent DNA polymerase catalyse
polymerisation in–
(1) 3' ® 5' direction
(2) 5' ® 3' direction
(3) Depend on the nature of template strand
(4) both (1) & (2)
73. Main enzyme of DNA replication is –
(1) DNA dependent RNA polymerase
(2) DNA dependent DNA polymerase
(3) RNA dependent RNA polymerase
(4) RNA dependent DNA polymerase
RNA, TRANSCRIPTION
74. The Process of copying genetic information from
one strand of DNA into Y is termed as Z .
Y Z
(1) Transcription RNA
(2) RNA Transcription
(3) DNA Translation
(4) Replication RNA
75. Code in RNA corresponding to AGCT in DNA-
(1) TACA (2) UCGA
(3) TCGA (4) AGUC
76. Which of the following is called adaptor molecule-
(1) DNA (2) m-RNA
(3) t-RNA (4) RNA
77. Which may be attached with Adenine base in
RNA -
(1) Guanine (2) Cytosine
(3) Uracil (4) Thymine
78. In the base sequence of one starand of DNA is
CAT, TAG , CAT , CAT, GAC what would be the
base sequence of its complementary m-RNA-
(1) GUA, GUA, CUG, AUC, CUG
(2) AUG, CUG, CUC, GUA, CUG
(3) GUA, AUC, GUA, GUA, CUG
(4) GUC, CUG, CUG, CUA, CUU
79. The process by which DNA of the nucleus passes
genetic information to m-RNA is called-
(1) Transcription (2) Translocation
(3) Translation (4) Transportation
80. A sequence of three consecutive bases in a t- RNAmolecule which specifically binds to a
complementary codon sequence in m RNA is
known as -
(1) Triplet
(2) Non - sense codon
(3) Anti codon
(4) Termination codon
81. t - RNA attach to larger subunit of ribosomes with
the help of which loop -
(1) DHU - loop (2) T y C loop
(3) Anticodon loop (4) Minor loop
82. In bacteria the codon AUG stands for -
(1) Glycine (2) Methionine
(3) N- formyl methionine (4) Alanine
83. In three dimensional view the molecule of t-RNA is
(1) L-shaped (2) S-shaped
(3) Y- shaped (4) E-shaped
84. During transcription, the DNA site at which RNA
polymerase binds is called :-
(1) Promoter (2) Regulator
(3) Receptor (4) Enhancer
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Pre-Medical : Biology ALLEN85. During transcription, if the nucleotide sequence of
the DNA strand that is being coded is ATACG, then
the nucleotide sequence in the mRNA would be
(1) TATGC (2) TCTGG
(3) UAUGC (4) UATGC
86. Which form of RNA has a structure resembling
clover leaf ?
(1) rRNA (2) hnRNA
(3) mRNA (4) tRNA
87. Which one of the following makes use of RNA as
a template to synthesize DNA –
(1) DNA dependant RNA polymerase
(2) DNA polymerase
(3) Reverse transcriptase
(4) RNA polymerase
88. cDNA probes are copied from the messenger
RNA molecules with the help of :
(1) Restriction enzymes
(2) Reverse transcriptase
(3) DNA polymerase
(4) Adenosine deaminase
89. The enzyme responsible for transcription is :-
(1) D.N.A polymerase-I (2) R.N.A. polymerase
(3) Reverse transcriptase (4) D.NA. polymerase-III
90. If the base sequence in DNA is 5' AAAA 3' then
the bases sequence in m-RNA is :-
(1) 5' UUUU 3' (2) 3' UUUU 5'
(3) 5' AAAA 3' (4) 3' TTTT 5'
91. Correct order of molecular weight is :-
(1) DNA < r-RNA < t-RNA
(2) DNA < m-RNA < r-RNA
(3) t-RNA < m-RNA < DNA
(4) t-RNA < DNA < m-RNA
92. The genes are responsible for growth and
differentiation in an organism through regulation
of :-
(1) Translocation
(2) Transformation
(3) Transduction and translation
(4) Translation and transcription
93. Method by which information reaches from DNA
to RNA is :-
(1) Transcription (2) Translation
(3) Transformation (4) Transduction
94. DNA acts as a template for synthesis of :-
(1) RNA (2) DNA
(3) Both '1' and '2' (4) Protein
95. Which is soluble RNA :-
(1) hnRNA (2) rRNA
(3) mRNA (4) tRNA
96. Portion of gene which is transcribed but not
translated is :-
(1) exon (2) intron
(3) cistron (4) codon
97. The smallest RNA is :-
(1) r-RNA (2) m-RNA
(3) t-RNA (4) nuclear RNA
98. The most abundant RNA of cell is :-
(1) r-RNA (2) t-RNA
(3) m-RNA (4) None of these
99. One strand of DNA (non template) has base
sequence CAG, TCG, GAT. What will be the
sequence of bases in m-RNA :-
(1) AGC, CTA, CTA
(2) GTC, AGC, CTC
(3) CAG. UCG. GAU
(4) GAC. TAG. CTA
100. Inverse transcription was discovered by :-
(1) Watson and Crick
(2) Khorana
(3) Temin an Baltimore
(4) Meischer
101. Mature eucaryotic m–RNA is recognised by
(1) Shine dalgarno sequence at 5' end
(2) 7–methyl guanosine at 5' end and polyadenine
bases at 3' end
(3) Anti shine dalgarno sequence at 5' end
(4) Presence of coding and noncoding sequence
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Pre-Medical : BiologyALLEN102. Transcription unit in DNA is –
(1) Promoter (2) Structural gene
(3) Terminator (4) All
103. In DNA promoter is the site for the initiation of
(1) Replication (2) Translation
(3) Transcription (4) Both (2) & (3)
104. Main enzyme of transcription–
(1) DNA dependent DNA polymerase
(2) DNA dependent RNA polymerase
(3) RNA dependent RNA polymerase
(4) RNA dependent DNA polymerase
105. Removal of introns and joining of exons is called
(1) Capping (2) Tailing
(3) Splicing (4) All
GENETIC CODE, TRANSLATION
106. A codon in m-RNA has :-
(1) 3-bases
(2) 2-bases
(3) 1-base
(4) Number of bases vary
107. A DNA strand is directly involved in the synthesis
of all the following except-
(1) Another DNA (2) t-RNA & m-RNA
(3) r-RNA (4) Protein
108. Genetic code was discovered by-
(1) Nirenberg & Mathei
(2) Kornberg & Crick
(3) Khorana & Kornberg
(4) Gamow
109. Genetic code was deciphered by chemically
synthesizing the trinucleotides by-
(1) Watson & Crick
(2) Beadle & Tatum
(3) Briggs & King
(4) M.W. Nirenberg
110. Nirenberg synthesized an m-RNA containing 34
poly-Adenine (A-A-A-A-A-A-----) and found a
polypeptide formed of 11 poly-lysine this proved
that genetic code for lysine was
(1) one-adenine (2) A-A doublet
(3) A-A-A triplet (4) Many adenines
111. 64 Codons constitute genetic code because-
(1) There was 64 types of amino acid
(2) 64 types of t-RNA
(3) Genetic code is triplet
(4) There are 64 enzymes
112. Which codon gives signal for the start of
polypeptide (protein) chain synthesis-
(1) AUG (2) UGA
(3) GUA (4) UAG
113. The function of non-sense codons is-
(1) To release polypeptide chain from t-RNA
(2) To form an unspecified amino acid
(3) To terminate the message of a gene controlled
protein synthesis-
(4) To convert a sense DNA into non sense DNA
114. Termination of chain growth in protein synthesis
is brought about by-
(1) UUG, UGC, UCA
(2) UCG, GCG, ACC
(3) UAA, UAG, UGA
(4) UUG, UAG, UCG
115. Genetic code determines-
(1) Structural pattern of an organism
(2) Sequence of amino acid in protein chain
(3) Variation in offsprings
(4) constancy of morphological trait
116. m - RNA is attached with -
(1) E.R. (2) Ribosome
(3) Nucleus (4) Lysosome
117. Sometimes the starting codon is GUG in place of
AUG, GUG normally stands for:-
(1) Valine (2) Glycine
(3) Methionine (4) Tyrosine
118. Which one of the following triplet codes, is
correctly matched with its specificity for an amino
acid in protein synthesis or as 'start' or 'stop'
codon :-
(1) UCG – Start (2) UUU – Stop
(3) UGU – Leusine (4) UAC – Tyrosine
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Pre-Medical : Biology ALLEN119. During translation initiation in prokaryotes, a GTP
molecule is needed in :-
(1) Formation of formyl-met-tRNA
(2) Binding of 30S subunit of ribosome with mRNA
(3) Association of 30 S-mRNA with formyl-met
tRNA
(4) Association of 50 S subunit of ribosome with
initiation complex
120. Degeneration of a genetic code is attributed to
the :-
(1) First member of a codon
(2) Second member of a codon
(3) Entire codon
(4) Third member of a codon
121. What would happen if in a gene encoding a
polypeptide of 50 amino acids, 25th codon (UAU)
is mutated to UAA :-
(1) A polypeptide of 24 amino acids will be formed
(2) Two polypeptides of 24 and 25 amino acids
will be formed
(3) A polypeptide of 49 amino acids will be formed
(4) A polypeptide of 25 amino acids will be formed
122. A sequence of how many nucleotides in
messenger RNA makes a codon for an amino
acid ?
(1) Three (2) Four
(3) One (4) Two
123. A strand of DNA has following base sequence 3'–
AAAAGTGACTAGTGA–5'. On transcription, it
produces an m–RNA which of the following
anticodon of t–RNA recognizes the third codon of
this mRNA :–
(1) AAA (2) CUG
(3) GAC (4) CTG
124. Protein synthesis in an animal cell occurs –
(1) On ribosomes present in cytoplasm as well asin mitochondria
(2) On ribosomes present in the nucleolus as wellas in cytoplasm
(3) Only on ribosomes attached to the nuclearenvelope and endoplasmic reticulum
(4) Only on the ribosomes present in cytosol
125. Which one of the following statement is true forprotein synthesis (translation) :
(1) Amino acids are directly recognized by m-RNA
(2) The third base of the codon is less specific
(3) Only one codon codes for an amino acid
(4) Every t-RNA molecule has more than one aminoacid attachment site
126. The drug streptomycin inhibits the process of :-
(1) Prokaryotic translation(2) Eukaryotic translation
(3) Prokaryotic transcripion
(4) Eukaryotic transcription
127. Translation is the process in which :-
(1) D.N.A. is formed on D.N.A template
(2) R.N.A. is formed on D.N.A. template
(3) D.N.A. is formed on R.N.A. template
(4) Protein is formed from R.N.A. message
128. In a polypeptide chain of 125 amino acids, if the25th amino acid is mutated to UAA, then :-(1) A polypeptide of 124 amino acid is formed(2) A polypeptide of 25 amino acid is formed(3) A polypeptide of 24 amino acid is formed(4) Any of the above can be possible
129. The first codon discovered by Nirenberg and Mathiiwas :-(1) CCC (2) GGG(3) UUU (4) AAA
130. Which of the following is not correct abouttranslation :-(1) It starts with AUG(2) Stopped at termination codon(3) Based on operon model(4) Occurs in nucleus
131. t-RNA attaches, amino acid at its :-(1) 3' end (2) 5' end(3) Anticodon (4) Loop
132. Out of 64 codons only 61 codes for the 20 differentamino acids.. This character of genetic code is called(1) Degeneracy(2) Non ambiguous nature(3) Redundancy(4) Overlapping
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Pre-Medical : BiologyALLEN133. Anticodons are found in :-
(1) m RNA (2) t RNA(3) r RNA (4) In all
134. One-gene-one enzyme hypothesis was proposed by:-(1) Beadle and Tatum (2) Jacob and Monod
(3) Lederberg (4) Watson and Crick
135. How many ATP and GTP molecules are requiredrespectively for incorporation of 25 amino acids inpeptide chain ?(1) 20 ATP, 20 GTP (2) 25 ATP, 25 GTP(3) 50 ATP, 50 GTP (4) 25 ATP 50 GTP
136. Which of the following RNA play structural andcatalytic role during translation.(1) m-RNA (2) t-RNA
(3) r-RNA (4) All
137. Transfer of genetic information from a polymer ofnucleotides to a polymer of amino acid is –(1) Replication
(2) Transcription
(3) Translation(4) Reverse transcription
138. Translation refers to the process of –
(1) Polymerisation of nitrogen bases
(2) Polymerisation of nucleotides
(3) Polymerisation of nucleosides
(4) Polymerisation of amino acids
139. Khorana & his collegeous synthesized an RNA
molecule with repeating sequences of U G N2–
bases. The RNA with "UGU GUG UGU GUG"
produced a tetra peptide with alternating
sequences of cystein and valine. This prove that
codon for cystein & valine is
(1) UGG, GUU (2) UUG, GGU
(3) UGU & GUG (4) GUG & UGU
GENE REGULATION140. Gene and cistron words are sometimes used
synonymously because–
(1) One cistron contains many genes
(2) One gene contains many cistrons
(3) One gene contains one cistron
(4) One gene contains no cistron
141. A gene containing multiple exons and at least one
intron is termed as :-
(1) split gene
(2) operator gene
(3) synthetic gene
(4) epistatic gene
142. Gene which is responsible for the synthesis of a
polypeptide chain is called :-
(1) Promotor gene (2) Structural gene
(3) Regulator gene (4) Operator gene
143. Which is true for tryptophan operon :-
(1) It is the example of inducible operon
(2) It is example of repressible operon
(3) on co repressor-¾ ®¾¾¾ off
(4) (2) and (3) both are correct
144. Which is true for repressible operon :–
(1) Off Inducer¾¾¾¾® on
(2) Inactive repressor + Co-repressor = active
repressor
(3) Active repressor + Inducer = inactive repressor
(4) On Inducer¾ ®¾¾ off
145. What does "lac" refer to, in what we call the lac
operon :-
(1) Lactose
(2) Lactase
(3) Lac insect
(4) The number 1,00,000
146. Which of the following is not produced by E.Coli
in the lactose operon –
(1) b galactosidase
(2) Thiogalactoside transacetylase
(3) Lactose dehydrogenase
(4) Lactose permease
147. A functional complex comprising a cluster of genes
including structural gene, a promoter gene, an
operator gene and a regulator gene was discovered
by :-
(1) Beadle and Tatum (1958)
(2) Watson and crick (1953)
(3) Jacob and Monad (1961)
(4) Britten and Davidson (1961)
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Pre-Medical : Biology ALLEN148. Who explained the operon model for the first time
(1) Francois Jacob
(2) Jacques Monod
(3) Francois Jacob and Jacques Monod
(4) Beadle & Tatum
149. The accessibility of promotor regions of prokaryotic
DNA by RNA polymerase is in many cases regulated
by the interaction of some protein with sequences
termed as –
(1) Promoter (2) Operator
(3) Regulator (4) Cistron
150. Regulation of lac operon by repressor is referred
to as–
(1) Positive regulation
(2) Nagative regulation
(3) Both (1) and (2)
(4) None
151. Which is incorrect
(1) i-gene codes for the repressor of lac operon
(2) z-gene codes for the beta-galactosidase
(3) y-gene codes for transacetylase
(4) three gene products are required for metabolism
of lactose
152. Which is the primary step for regulation of gene
expression.
(1) Transport of m-RNA from nucleus to the
cytoplasm
(2) Translational level
(3) Processing level
(4) Transcriptional level
153. Find out the correct sequence of structural gene in
lac operon
(1) y, a, z (2) a, z, y
(3) z, y, a (4) z, a y
MUTATION154. The concept of sudden genetic change which breeds
true in an organism is visualized as :-
(1) Natural selection
(2) Inheritance of acquired characters
(3) Mutation
(4) Independent assortment
155. Mutation is :-
(1) An abrupt or discontinuous change which is
inherited
(2) A factor for plant growth
(3) A change which affects parents only and is never
inherited
(4) A change which affects the offspring of F2
generation
156. The change of chromosomal parts between non
homologous pairs of chromosome :-
(1) Crossing over/Transduction
(2) Translocation
(3) Inversion
(4) Transition
157. Which of the following can be called a mutation :-
(1) The halfing of the chromosome number at meiosis
(2) The doubling of the chromosome after syngamy
(3) The possession of an additional chromosome
(4) All the above
158. Mutations are generally :-
(1) Dominant
(2) Recessive
(3) Codominant
(4) Incompeletely dominant
159. Genetic mutations occur in :-
(1) DNA (2) RNA
(3) Protein (4) RNA & protein both
160. Which of the following undergoes change in
mutation :-
(1) Chromosome (2) Structure of gene
(3) Sequence of gene (4) Any of the above
161. The locus of mutation is :-
(1) Gene (2) Chromosome
(3) Centromere (4) Nucleus
162. In the octaploid wheat, the haploid (n) and basic
numbers (x) of chromosomes are :-
(1) n=21, x=7 (2) n=28, x=7
(3) n=7 x=28 (4) n=7, x=21
163. Non-ionizing radiations commonly used for inducing
mutations in organisms are :-
(1) UV-rays (2) Beta-rays
(3) X-rays (4) Gamma-rays
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Pre-Medical : BiologyALLEN164. The smallest unit of genetic material which upon
mutation produce a phenotypic effect is :-
(1) Mutons (2) Recon
(3) Gene (4) Cistron
165. Ultimate source of genetic variation is (OR) the process
which provides raw material for evolution is:-
(1) Sexual reproduction
(2) Meiosis
(3) Mutation
(4) Independent assortment
166. Haploids are preferred over diploids for mutation
studies because :-
(1) Recessive mutation is expressed in F1
(2) Recessive mutation is expressed in F2
(3) Dominant phenotype is expressed
(4) Dominant phenotype is suppressed
167. Type of gene mutation which involves replacement
of purine with pyrimidine or vice versa (OR) The
substitution of one type of base with another type
of base is :-
(1) Transduction (2) Transversion
(3) Translocation (4) Transcription
168. The minimum requirement for mutation is :-
(1) Change of triplet codon
(2) Change in single nucleotide
(3) Change in whole DNA
(4) Change in single strand of DNA
169. Mutations are :-
(1) Always useful (2) Mostly useful
(3) Never useful (4) Rarely useful
170. Sickle cell anaemia is an example of :-
(1) Frame shift mutation
(2) Point mutation
(3) Segmental mutation
(4) Gibberish mutation
171. The most striking example of frame shift mutation
was found in a disease called :-
(1) Sickle cell anaemia
(2) Colour blindness
(3) Laesh-Nyhn Syndrome
(4) Thallesemia
172. A nutritionally wild type organism, which does not
require any additional growth supplement is known
as :-
(1) Holotype (2) Auxotroph
(3) Prototroph (4) Phenotype
173. Given below is the representation of a kind of
chromosomal mutation :
What is the kind of mutation represented
(1) deletion (2) duplication
(3) inversion (4) reciprocal translocation
174. The "cri-du-chat" syndrome is caused by change in
chromosome structure involving:-
(1) Deletion (2) Duplication
(3) Inversion (4) Translocation
175. A class of mutation induced by addition or deletion
of a nucleotide is called :-
(1) Missense
(2) Non-sense
(3) Substitution
(4) frame shift
176. Chromosomes with genes abcdefg becoming
abedcfg is :(1) duplication (2) deletion
(3) translocation (4) inversion
177. Gene mutation is :
(1) mutation in the genes of DNA
(2) mutation in the phosphodiester linkage
(3) mutation in the chromosomes
(4) change in the sequence of nitrogenous bases
178. Chromosome number 2n-1 is an example of
(1) trisomy (2) euploidy
(3) polyploidy (4) monosomy
179. After a mutation at a genetic locus the character
of an organism changes due to the change in
(1) protein structure
(2) DNA replication
(3) protein synthesis pattern
(4) RNA transcription pattern.
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Pre-Medical : Biology ALLENDNA FINGER PRINTING, HUMAN GENOME PROJECT
180. DNA finger printing was invented by :–
(1) Kary Mullis (2) Alec Jeffery
(3) Dr. Paul Berg (4) Francis Collins
181. Which one of the following pairs of terms/names
mean one and the same thing
(1) Gene pool - genome
(2) Codon - gene
(3) Cistron - triplet
(4) DNA Fingerprinting - DNA profiling
182. Which step does not involve in DNA finger printing
(1) Southern blotting
(2) Gel electrophoresis
(3) Restriction enzyme digestion
(4) Northern blotting
183. The technique of transferring DNA fragment
separated on agarose gel to a synthetic membrane
such as nitrocellulose is known as
(1) Northern blotting (2) Southern blotting
(3) Western blotting (4) Dot blotting
184. Western blotting is used for the identification of:-
(1) DNA (2) RNA
(3) Protein (4) All the above
185. Which of the following techniques are used inanalyzing restriction fragment length polymorphism(RFLP) :-
(a) Electrophoresis
(b) Electroporation
(c) Methylation
(d) Restriction digestion
(1) 'a' and 'c' (2) 'c' and 'd'
(3) 'a' and 'd' (4) 'b' and 'd'
186. The approximate number of genes contained in thegenome of Kalpana Chawla was
(1) 40,000 (2) 30,000
(3) 80,000 (4) 1,00,000
187. The transfer of protein from electrophoretic gelto nitrocellulose membrane is known as :-
(1) transferase
(2) northern blotting
(3) western blotting
(4) southern blotting
188. Which of the following is not associated with HGP–
(1) Bioinformatics
(2) Cloning vectors BAC & YAC
(3) Automated DNA sequencers
(4) VNTR
189. In density gradient centrifugation , the bulk DNAforms_____ while satellite DNA forms________.
(1) Major peak; Minor peak
(2) Minor peak; Major peak
(3) Major peak; Major peak
(4) Minor peak; Minor peak
190. Which step is not correct in DNA finger printing–
(1) Isolation of DNA
(2) Digestion of DNA by DNA ligase enzyme
(3) Separation of DNA by electophoresis
(4) Hybridisation using labelled VNTR probe
191. DNA fingerprinting method is very useful for -
(1) DNA tests for identity & relation ships
(2) Forensic studies
(3) Polymorphism
(4) All of the above
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Pre-Medical : BiologyALLENEXERCISE-I (Conceptual Questions) ANSWER KEY
Que. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
Ans. 2 2 3 1 1 4 1 4 4 1 2 4 1 4 3Que. 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30
Ans. 3 1 3 2 3 2 3 1 4 2 2 1 2 3 1Que. 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45
Ans. 1 2 2 4 4 1 4 2 2 3 2 2 2 2 3Que. 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60
Ans. 4 3 3 2 1 4 4 3 2 2 4 3 4 4 3Que. 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75
Ans. 1 1 4 4 3 2 4 2 3 1 2 2 2 2 2Que. 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90
Ans. 3 3 3 1 3 2 3 1 1 3 4 3 2 2 3Que. 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105
Ans. 3 4 1 3 4 2 3 1 3 3 2 4 3 2 3Que. 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120
Ans. 1 4 1 4 3 3 1 3 3 2 2 1 4 3 4Que. 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135
Ans. 1 1 3 1 2 1 4 3 3 4 1 1 2 1 4Que. 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150
Ans. 3 3 4 3 3 1 2 4 2 1 3 3 3 2 2Que. 151 152 153 154 155 156 157 158 159 160 161 162 163 164 165
Ans. 3 4 3 3 1 2 3 2 1 4 1 2 1 1 3Que. 166 167 168 169 170 171 172 173 174 175 176 177 178 179 180
Ans. 1 2 2 4 2 4 3 3 1 4 4 4 4 1 2Que. 181 182 183 184 185 186 187 188 189 190 191
Ans. 4 4 2 3 3 2 3 4 1 2 4
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Pre-Medical : Biology ALLEN
AIPMT 20061. Which antibiotic inhibits interaction between tRNA
and mRNA during bacterial protein synthesis ?
(1) Erythromycin (2) Neomycin(3) Streptomycin (4) Tetracycline
2. Amino acid sequence, in protein synthesis is de-cided by the sequence of(1) tRNA (2) mRNA(3) cDNA (4) rRNA
3. One gene-one enzyme hypothesis was postulatedby(1) R. Franklin (2) Hershey and Chase(3) A.Garrod (4) Beadle and Tatum
4. One turn of the helix in a B-form DNA is ap-proximately(1) 20 nm (2) 0.34 nm(3) 3.4 nm (4) 2 nm
5. Antiparallel strands of a DNA molecule meansthat(1) one strand turns anti-clockwise(2 the phosphate groups of two DNA strands, at
their ends, share the same position(3) the phosphate groups at the start of two DNA
strands are in opposite position (pole)(4) one strand turns clockwise
6. Cri-du-chat syndrome in humans is caused by the
(1) Fertilization of an XX egg by a normal Y-bearingsperm
(2) Loss of half of the short arm of chromosome 5(3) Loss of half of the long arm of chromosome 5(4) Trisomy of 21stchromosome
7. Triticale, the first man-made cereal crop, has beenobtained by crossing wheat with –(1) Rye (2) Pearl millet(3) Sugarcane (4) Barley
AIIMS 2006
8. During protein synthesis in an organism, at one
point the process comes to a halt. Select the group
of the three codons from the following from which
any one of the three could bring about this halt –
(1) UUU, UCC, UAU (2) UUC, IIA, UAC
(3) UAG, UGA, UAA (4) UUG, UCA, UCG
9. Thymine is –
(1) 5–Methyl uracil (2) 4–Methyl uracil
(3) 3–Methyl uracil (4) 1–Methyl uracil
10. In which one of the following combinations (1–4) of
the number of the chromosomes is the present day
hexaploid whaet correctly represented
Combination (1) (2) (3) (4)Monosomic 21 7 21 41
Haploid 28 28 7 21
Nullisomic 42 40 42 40
Trisomic 43 42 43 43
AIPMT 2007
11. Differentiation of organs and tissues in a developing
organism, is association with :-
(1) Developmental mutations
(2) Differential expression of genes
(3) Lethal mutations
(4) Deletion of genes
12. Molecular basis of organ differentiation depends
on the modulation in transcription by :
(1) RNA polymerase (2) Ribosome
(3) Transcription factor (4) Anticodon
13. The Okazaki fragments in DNA chain growth :
(1) Result in transcription
(2) Polymerize in the 3'-to-5' direction and forms
replication fork
(3) Prove semi-conservative nature of DNA repli-
cation
(4) Polymerize in the 5'-to-3' direction and explain
3'-to-5' DNA replication
14. The two polynucleotide chains in DNA are :
(1) Parallel
(2) Discontinuous
(3) Antiparallel
(4) Semiconservative
15. In the hexaploid wheat, the haploid (n) and basic
(x) numbers of chromosomes are :-
(1) n=7 and x=21 (2) n=21 and x=21
(3) n=21 and x=14 (4) n=21 and x=7
EXERCISE-II (Previous Year Questions) AIPMT/NEET & AIIMS (2006-2018)
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Pre-Medical : BiologyALLENAIPMT 2008
16. Which of the following nitrogen base is not found
in DNA-
(1) Thymine (2) Cytosine
(3) Guanine (4) Uracil
17. Polysome is formed by :-
(1) A ribosome with several subunits
(2) Ribosomes attached to each other in a linear
arrangement
(3) Several ribosomes attached to a single mRNA
(4) Many ribosomes attached to a strand of
endoplasmic reticulum
18. Which one of the following pairs of nitrogenous
bases of nucleic acids, is wrongly matched with
the category mentioned against it ?
(1) Guanine, Adenine – Purines
(2) Adenine, Thymine – Purines
(3) Thymine, Uracil – Pyrimidines
(4) Uracil, Cytosine – Pyrimidines
19. In the DNA molecule:-
(1) the proportion of Adenine in relation to thymine
varies with the organism
(2) there are two strands which run antiparallel one
in 5' ®3' direction and other in 3' ® 5'
(3) the total amount of purine nucleotides and
pyrimidine nucleotides is not always equal
(4) there are two strands which run parallel in the
5' ®3' direction
20. Which one of the following pairs of codons is
correctly matched with their function or the signal
for the particular amino acid ?
(1) AUG,ACG – Start/Methionine
(2) UUA, UCA –Leucine
(3) GUU, GCU –Alanine
(4) UAG, UGA – Stop
21. Which of the following bond is not related to
nucleic acid :
(1) H-bond
(2) Ester bond
(3) Glycosidic bond
(4) Peptide bond
22. Haploids are more suitable for mutation studies than
the diploids. This is because :-
(1) haploids are more abundant in nature than diploids
(2) All mutations, whether dominant or recessive are
expressed in haploids
(3) Haploids are reproductively more stable than
diploids
(4) Mutagens penetrate in haploids more effectively
than in diploids
AIPMT 2009
23. What is not true for genetic code :-
(1) It is unambiguous
(2) A codon in mRNA is read in a non-contiguous
fashion
(3) It is nearly universal
(4) It is degenerate
24. Removal of introns and joining the exons in a defined
order in a transcription unit is called :-
(1) Capping
(2) Splicing
(3) Tailing
(4) Transformation
25. Semiconservative replication of DNA was first
demonstrated in :-
(1) Salmonella typhimurium
(2) Drosophila melanogaster
(3) Escherichia coli
(4) Streptococcus pneumoniae
26. Whose experiments cracked the DNA and
discovered unequivocally that a genetic code is a
"triplet" :-
(1) Beadle and tatum
(2) Nirenberg and Mathaei
(3) Hershey and Chase
(4) Morgan and Sturtevant
27. Point mutation involves :-
(1) Deletion
(2) Insertion
(3) Change in single base pair
(4) Duplication
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Pre-Medical : Biology ALLENAIPMT 2010
28. Select the two correct statements out of the four (a-
d) given below about lac operon :
(a) Glucose or galactose may bind with the repressor
and inactivated
(b) In the absence of lactose the repressor binds
witht the operator region
(c) The z-gene codes for permease
(d) This was elucidated by Francois Jacob and Jacque
Monod
The correct statements are :
(1) (b) and (d) (2) (a) and (b)
(3) (b) and (c) (4) (a) and (c)
29. Satellite DNA is useful tool in :
(1) Forensic science (2) Genetic engineering
(3) Organ transplantation (4) Sex detemination
30. The one aspect which is not a salient feature of
genetic code, is its being :
(1) Universal (2) Specific
(3) Degenerate (4) Ambiguous
31. Which one of the following does not follow the
central dogma of molecular bilogy?
(1) Chlamydomonas (2) HIV
(3) Pea (4) Mucor
AIPMT (Pre.) 2012
32. PCR and Restriction Fragment Length
Polymorphism are the methods for :-
(1) DNA sequencing
(2) Genetic fingerprinting
(3) Study of enzymes
(4) Genetic transformation
AIPMT (Mains) 2012
33. What is it that forms the basis of DNA Fingerprinting?
(1) The relative amount of DNA in the ridges and
grooves of the fingerprints.
(2) Satellite DNA occurring as highly repeated short
DNA segments
(3) The relative proportions of purines and
pyrimidines in DNA
(4) The relative difference in the DNA occurence in
blood, skin and saliva
34. Read the following four statements (A-D):
(A) In transcription, adenosine pairs with uracil.
(B) Regulation of lac operon by repressor is referred
to as positive regulation.
(C) The human genome has approximately 50,000
genes.
(D) Haemophilia is a sex-linked recessive disease.
How many of the above statements are right?
(1) Four (2) One
(3) Two (4) Three
35. Which one of the following is a wrong statement
regarding mutations?
(1) UV and Gamma rays are mutagens
(2) Change in a single base pair of DNA does not
cause mutation
(3) Deletion and insertion of base pairs cause frame-
shift mutations.
(4) Cancer cells commonly show chromosomal
aberrations.
NEET-UG 201336. Which enzyme/s will be produced in a cell in which
there is a nonsense mutation in the lac Y gene ?
(1) Lactose permease and transacetylase
(2) b-galactosidase
(3) Lactose permease
(4) Transacetylase
37. DNA fragments generated by the restriction
endonucleases in a chemical reaction can be
separated by :
(1) Restriction mapping
(2) Centrifugation
(3) Polymerase chain reaction
(4) Electrophoresis
AIPMT 2014
38. Commonly used vectors for human genome
sequencing are :-
(1) T-DNA
(2) BAC and YAC
(3) Expression Vectors
(4) T/A Cloning Vectors
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Pre-Medical : BiologyALLENAIIMS 2014
39. What is the correct sequence of DNA finger
printing?
a-seperation of desired DNA by gel electrophoresis
b-Digestion by restriction endonuclease
c- Isolation of DNA
d- Hybridisation using labelled VNTR probe
e- Southern blotting
(1) a ® b ® c ® d ® e
(2) b ® d ® e ® a ® c
(3) c ® b ® a ® d ® e
(4) c ® b ® a ® e ® d
AIPMT 201540. In sea urchin DNA, which is double stranded, 17%
of the bases were shown to be cytosine. The
percentages of the other three bases expected to
be present in this DNA are :-
(1) G 17%, A 16.5%, T 32.5%
(2) G 17%, A 33%, T 33%
(3) G 8.5%, A 50%, T 24.5%
(4) G 34%, A 24.5%, T 24.5%
41. Gene regulation governing lactose operon of E.colithat involves the lac I gene product is :
(1) Negative and inducible because repressor protein
prevents transcription
(2) Negative and repressible because repressor
protein prevents transcription
(3) Feedback inhibition because excess of
b-galactosidase can switch off trascription
(4) Positive and inducible because it can be induced
by lactose
42. The movement of a gene from one linkage group
to another is called :-
(1) Duplication (2) Translocation
(3) Crossing over (4) Inversion
Re-AIPMT 201543. Which of the following biomolecules does have a
phosphodiester bond ?
(1) Nucleic acids in a nucleotide
(2) Fatty acids in a diglyceride
(3) Monosaccharides in a polysaccharide
(4) Amino acids in a polypeptide
44. Which one of the following is not applicable to RNA?
(1) Chargaff's rule
(2) Complementary base pairing
(3) 5' phosphoryl and 3' hydroxyl ends
(4) Heterocyclic nitrogenous bases
45. Satellite DNA is important because it :
(1) Codes for enzymes needed for DNA replication
(2) Codes for proteins needed in cell cycle
(3) Shows high degree of polymorphism in
population and also the same degree of
polymorphism in an individual, which is heritable
from parents to children
(4) Does not code for proteins and is same in all
members of the population
46. Identify the correct order of organisation of genetic
material from largest to smallest :
(1) Chromosome, genome, nucleotide, gene
(2) Chromosome, gene, genome, nucleotide
(3) Genome, chromosome, nucleotide, gene
(4) Genome, chromosome, gene, nucleotide
AIIMS 2015
47. rRNA is synthesised in :-
(1) Nucleus (2) Glogi body
(3) Cytoplasm (4) Centriole
48. Which of the following is involved in translation:-
(1) DNA
(2) mRNA, tRNA, DNA
(3) mRNA, tRNA
(4) Only mRNA
49. Which set of RNA are involved in protein synthesis.(1) tRNA, mRNA, rRNA(2) tRNA, mRNA, hnRNA(3) hnRNA, mRNA, rRNA(4) hnRNA, tRNA, rRNA
NEET-I 2016
50. Which of the following is required as inducer(s) for
the expression of Lac operon ?
(1) Glucose
(2) Galactose
(3) Lactose
(4) Lactose and galactose
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Pre-Medical : Biology ALLEN51. A complex of ribosomes attached to a single strand
of RNA is known as :-
(1) Polysome (2) Polymer
(3) Polypeptide (4) Okazaki fragment
52. Which of the following is not required for any of thetechniques of DNA fingerprinting available atpresent ?(1) Polymerase chain reaction(2) Zinc finger analysis(3) Restriction enzymes(4) DNA–DNA hybridization
53. Which one of the following is the starter codon ?
(1) AUG (2) UGA (3) UAA (4) UAGNEET-II 2016
54. Taylor conducted the experiment to provesemiconservative mode of chromosome replicationon :(1) Drosophila melanogaster(2) E. coli(3) Vinca rosea(4) Vicia faba
55. The equivalent of a structural gene is :(1) Operon (2) Recon(3) Muton (4) Cistron
56. Which of the following rRNAs acts as structural RNAas well as ribozyme in bacteria ?(1) 23 S rRNA (2) 5.8 S rRNA(3) 5 S rRNA (4) 18 S rRNA
57. A non-proteinaceous enzyme is :-(1) Ligase (2) Deoxyribonuclease
(3) Lysozyme (4) Ribozyme
58. A molecule that can act as a genetic material mustfulfill the traits given below, except :-(1) It should be unstable structurally and chemically
(2) It should provide the scope for slow changes thatare required for evolution
(3) It should be able to express itself in the form of'Mendelian characters'
(4) It should be able to generate its replica
59. DNA-dependent RNA polymerase catalyzestranscription on one strand of the DNA which iscalled the :-(1) Alpha strand (2) Antistrand
(3) Template strand (4) Coding strand
60. The mechanism that causes a gene to move fromone linkage group to another is called :(1) Translocation (2) Crossing-over(3) Inversion (4) Duplication
AIIMS 2016
61. Reverse transcriptase using RNA, forms which ofthe following ?
(1) Double stranded DNA
(2) Double stranded RNA
(3) DNA & RNA
(4) Single stranded RNA
62. An immature stop codon leads to :-
(1) Mutation
(2) Non-sense mutation
(3) Variation
(4) Intron
63.
A B C D K L(Parents) (Children)
On the basis of given DNA fingerprint, match thechildrens (K & L) with their respective parent (A,B,Cand D) :-
(1) K-A, L-A (2) K-A, L-D
(3) K-A, L-C (4) K-B, L-A
64. What occurs in point mutaion ?
(1) Change in single base pair in DNA
(2) Change in single base pair in RNA
(3) Change in double base pair in DNA
(4) Change in double base pair in RNA
65. In most of the plant viruses genetic material is :-
(1) ssDNA (2) ssRNA
(3) dsRNA (4) ssRNA + ssDNA
66. Biological name of wheat is :-
(1) Triticum aestivum (2) Triticum triticale(3) Triticum sativum (4) Triticum tuberosum
NEET(UG) 201767. The final proof for DNA as the genetic material
came from the experiments of :
(1) Hershey and Chase
(2) Avery, Mcleod and McCarty
(3) Hargobind Khorana
(4) Griffith
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Pre-Medical : BiologyALLEN68. DNA fragments are:
(1) Negatively charged
(2) Neutral
(3) Either positively or negatively charged
depending on their size
(4) Positively charged
69. If there are 999 bases in an RNA that codes for a
protein with 333 amino acids, and the base at
position 901 is deleted such that the length of the
RNA becomes 998 bases, how many codons will be
altered ?
(1) 11 (2) 33
(3) 333 (4) 1
70. During DNA replication, Okazaki fragments are
used to elongate:
(1) The lagging strand towards replication fork.
(2) The leading strand away from replication fork.
(3) The lagging strand away from the replication
fork.
(4) The leading strand towards replication fork.
71. Which of the following RNAs should be most
abundant in animal cell ?
(1) t-RNA (2) m-RNA
(3) mi-RNA (4) r-RNA
72. What is the criterion for DN A fragments movement
on agarose gel during gel electrophoresis ?
(1) The smaller the fragment size, the farther it
moves
(2) Positively charged fragments move to farther end
(3) Negatively charged fragments do not move
(4) The larger the fragment size, the farther it moves
73. DNA replication in bacteria occurs:
(1) Within nucleolus
(2) Prior to fission
(3) Just before transcription
(4) During S phase
74. Spliceosomes are not found in cells of;
(1) Fungi (2) Animals
(3) Bacteria (4) Plants
75. The association of histone H1 with a nucleosome
indicates:
(1) DNA replication is occurring.
(2) The DNA is condensed into a Chromatin Fibre.
(3) The DNA double helix is exposed.
(4) Transcription is occurring.
AIIMS 2017
76. The process of formation of RNA from DNA is
known as
(1) Transcription
(2) Translation
(3) Replication
(4) Transformation
77. If base sequence in m-RNA is 5' UAC GUA CGU
ACG UAC GUA CGU ACG 3' then what will be
sequence of template strand?
(1) 5' CGT ACG TAC GTA CGT ACG TAC GTA 3'
(2) 5' TAC GTA CGT ACG TAC GTA CGT ACG 3'
(3) 5' ATG CAT GCA TGC ATG CAT GCA TGC 3'
(4) 5' GTA TAC ACG TGC GTA GTA CAG GCA 3'
78.a b
c
d
Choose correct one :-
(1) a-DNA, b-H1-histone, c-histone octamer,d-core of histone
(2) a-core of histone, b-DNA, c-H1 histone,d-histone octamer
(3) a-Histone octamer, b-core of histone,c-DNA, d-H1 histone
(4) a-H1 histone, b-histone octamer, c-core of histone,d-DNA
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Pre-Medical : Biology ALLEN79. Below digram represent the DNA finger printing,
Which is true about diagram ?
A b c d
(1) All are graphically similar
(2) All are graphically dissimilar
(3) Only A and C are similar
(4) Only B and D are similar
80. Match the column-I with column-II :-
Codon Amino acid
a AUG i Methionine
b UUU ii Arginine
c AGA iii Phenylalanine
d GUG iv Serine
v Valine
Options :-
(1) a-i, b-iii, c-ii, d-v (2) a-ii, b-iv, c-iii, d-i
(3) a-iv, b-ii, c-iii, d-i (4) a-i, b-ii, c-iii, d-iv
81. The following samples are collected from sugarcanefield. Which of the following is true?
A B C D
(1) All are dissimilar
(2) All are similar
(3) A and D are similar
(4) A and B are similar
82. As the codon, which is made by four N2 basessimilary protein is made up by how many aminoacids?
(1) 20 (2) 10 (3) 11 (4) 25
83. Match column-I to column-II :-
Codon Amino acid
A AAA i Glycine
B CCC ii Lysine
C UUU iii Phenylalanine
D GGG iv Proline
v Tyrosine
Options :-(1) A-ii, B-v, C-iii, D-iv
(2) A-v, B-iii, C-i, D-iv
(3) A-ii, B-iv, C-iii, D-i
(4) A-iii, B-ii, C-v, D-i
84. Formation of protein on RNA-strand is called:-
(1) Transcription (2) Translation
(3) Replication (4) TransformationNEET(UG) 2018
85. The experimental proof for semiconservativereplication of DNA was first shown in a
(1) Fungus (2) Bacterium
(3) Plant (4) Virus
86. Select the correct match :(1) Alec Jeffreys – Streptococcus
pneumoniae(2) Alfred Hershey and – TMV
Martha Chase(3) Matthew Meselson – Pisum sativum
and F. Stahl(4) Francois Jacob and – Lac operon
Jacques Monod
87. Select the correct Match :
(1) Ribozyme - Nucleic acid
(2) F2 × Recessive parent - Dihybrid cross
(3) T.H. Morgan - Transduction
(4) G. Mendel - Transformation
88. AGGTATCGCAT is a sequence from the coding
strand of a gene. What will be the corresponding
sequence of the transcribed mRNA ?
(1) AGGUAUCGCAU
(2) UGGTUTCGCAT
(3) ACCUAUGCGAU
(4) UCCAUAGCGUA
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Pre-Medical : BiologyALLEN89. All of the following are part of an operon
except :-(1) an operator (2) structural genes(3) an enhancer (4) a promoter
AIIMS 201890. 'Nuclein' term was coined by :-
(1) Meischer and Crick (2) Altman
(3) Griffith (4) Meischer
91. Codons of proline amino acid are :-
(1) CCU CCC CCA
(2) CAU CAC CAA
(3) GGU GGC GGA
(4) AAU AAA AAC
92. Which of the following is incorrect ?
(1) RNA polymerase II ® Heterogeneous nuclear RNA
(2) RNA polymerase III ® t-RNA, Sn-RNA,5 S-rRNA
(3) RNA polymerase I ® r-RNA
(4) RNA polymerase I ® t-RNA, Sn-RNA,5 S-rRNA
93. p i p o z y a
Choose the correct option regarding abovediagram:-
(1) i-inhibitor, z-b-galactosidase, y-permease,a-transacetylase
(2) i-inducer, z-transacetylase, y-b-galactosidase, a-permease
(3) i-repressor, z-permease, y-transacetylase,a-b-galactosidase
(4) i-inhibitor, z-transacetylase, y-permease,a-b-galactosidase
94. t-RNA, m-RNA and r-RNA is used in synthesis of :-
(1) Carbohydrate (2) Protein
(3) Lipid (4) Steroid
95. Which one is incorrect ?
(1) DNA ® Ribose sugar
(2) Splicing ® Removal of introns
(3) Capping ® Adding of 7 mG at 5'end
(4) RNA ® Uracil
96. Match the following columns :-
Column-I Column-II
a f × 174bacteriophase i 48502
b Human(haploid) ii 5386
c E.Coli iii 3.3 × 109
d l-phase iv 4.6 × 106
(1) a-iv, b-iii, c-ii, d-i
(2) a-iii, b-iv, c-i, d-ii
(3) a-ii, b-iii, c-iv, d-i
(4) a-i, b-ii, c-iii, d-iv
97. Codon of Arginine amino acid are :-
(1) CGU, CGC, CGA
(2) UGU, UGC, UGA
(3) AGU, AGC, AGA
(4) AUA, AUC, AUG
98. Select the correct statement regarding prokaryotes:-
(1) DNA dependent RNA polymerase form all types
of RNA.
(2) Different RNA polymerase form different types
of RNA.
(3) Single DNA dependent DNA polymerase form
one type of RNA.
(4) Different DNA dependent DNA polymerase
form one type of RNA.
99. Which of the following is incorrect for satellite DNA?
(1) High degree of polymorphism.
(2) It is not inherited from parent to children.
(3) It is used in DNA fingerprinting for genetic
diversity.
(4) Does not code for any protein.
100. f-174 has which type of genome ?
(1) ss-RNA (2) ss-DNA
(3) ds-DNA (4) ds-RNA
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Pre-Medical : Biology ALLEN
EXERCISE-II (Previous Year Questions) ANSWER KEY
101. All types of RNA are formed by transcription of :-
(1) DNA (2) m-RNA
(3) r-RNA (4) t-RNA
102. Codon for alanine are :-
(1) GCU GCC GCA (2) ACU ACC ACG
(3) GAG GAC GAU (4) GUG GUC GUA
Que. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
Ans. 2 2 4 3 3 2 1 3 1 4 2 3 4 3 4Que. 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30
Ans. 4 3 2 2 4 4 2 2 2 3 2 3 1 1 4Que. 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45
Ans. 2 2 2 3 2 2 4 2 4 2 1 2 1 1 3Que. 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60
Ans. 4 1 3 1 3 1 2 1 4 4 1 4 1 3 1Que. 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75
Ans. 1 2 2 1 2 1 1 1 2 3 4 1 2 3 2Que. 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90
Ans. 1 1 1 2 1 4 1 3 4 2 4 1 4 3 2Que. 91 92 93 94 95 96 97 98 99 100 101 102 103Ans. 1 4 1 2 1 3 1 1 2 2 1 1 2
103. t-RNA, m-RNA and r-RNA is used in synthesis of :-
(1) Carbohydrate (2) Protein
(3) Lipid (4) Steroid
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Pre-Medical : BiologyALLENEXERCISE-III (Analytical Questions) Check Your Understanding
1. A point mutation which involves change of A®G,
C®T, C®G and T®A in DNA are :-
(1) Transit ion, Transit ion, Transversion,
Transversion
(2) Transit ion, Transversion, Transition,
Transversion
(3) Transversion, Transition, Transversion,
Transition
(4) None of the above
2. A segment of DNA has a base sequence : AAG,
GAG, GAC, CAA, CCA–, Which one of the following
sequence represents a frame shift mutation :-
(1) AAG, GAG, GAC, CAA, CCA–
(2) AAG, AGG, ACC, AAC, CA–
(3) ACG, GAG, GAC, CAG, CC–
(4) AAG, GCG, GAC, CAG, CC–
3. If the DNA-codons are ATGATGATG and a cytosine
base is inserted at the begining which of the following
would be the result :-
(1) A non sense mutation
(2) CA, TGA, TGA, TG
(3) CAT, GAT, GAT, G
(4) C, ATG, ATG, ATG
4. A completely radioactive double stranded DNA
molecule undergoes two round of raplication in a
non radioactive medium. What will be the radioactive
status of the four daughter molecules :-
(1) All four still contain radioactivity
(2) Radioactivity is lost from all four
(3) Out of four, three contain radioacitivity
(4) Half of the number contain no radioacitivity
5. Consider the following sequence on m–RNA
AUGGCAGUGCCA. Assuming that genetic code is
overlap then how many number of codon may be
present on this genetic code
(1) 9 (2) 10
(3) 8 (4) 11
6. A normal DNA molecule is continuously replicate
in N15 medium than what is the % of lighter DNA in
4th generation.
(1) 12·5% (2) 25%
(3) 0% (4) 6·25%
7. Find out the sequence of binding of the following
amino acyl – t–RNA complexes during translation
to an m–RNA transcribed by a DNA segment having
the base sequence. 3' ATACCCATGGGG 5'. Choose
the answer showing the correct order of alphabets:–
(a)
Amino acid
C C CAnticodon
(b)
Amino acid
A U GAnticodon
(c)
Amino acid
G G GAnticodon
(d)
Amino acid
A U AAnticodon
(1) a, b, c, d (2) d, a, b, c
(3) a, b, d, c (4) b, a, c, d
8. KHORANA synthesized two RNAs (a) with repeat
sequence of AB and (b) with repeat sequence of
ABC the polypeptides coded by (a) & (b) are
respectively :-
(1) Homopolypeptides in both (a) and (b)
(2) Heteropolypeptides in both
(3) Homopolypeptide in (a) & peptide heteropoly
in (b)
(4) Heteropolypeptide in (a) & peptide homopoly
in (b)
9. Which of the following m–RNA is translated
completely :–
(A) 5' AUG UGA UUA AAG AAA 3'
(B) 5' AUG AUA UUG CCC UGA 3'
(C) 5' AGU UCC AGA CUC UAA 3'
(D) 5' AUG UAC AGU AAC UAG 3'
(1) (A) and (B) (2) (B) and (D)
(3) (C) and (D) (4) (A) and (D)
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Pre-Medical : Biology ALLEN10. In a m–RNA sequence of N2–base is 5' AUG GUG
CUC AAA' 3'. What is the correct sequence of
anticodons which recognizes codons of m–RNA :-
UUU GAG (a) (b)
UAC CAC (c) (d)
(1) a, b, c, d (2) d, a, b, c
(3) c, d, b, a (4) d, c, b, a
11. Suppose evolution on earth has occurred in such a
way that there are 96 amino acids instead of 20.
DNA has 12 different types of bases and DNA syn-
thesis occur in the same way as today. The mini-
mum number of bases per DNA condon would be
(1) 12 (2) 8
(3) 2 (4) 3
12. Assume that their are 6 types of nitrogen bases
available and 40 types of amino acid are available
for protein synthesis, then in genetic code each
codon made up by minimum how many nitrogen
bases ?
(1) 3 (2) 4
(3) 5 (4) 2
13. In a segment of DNA 3.2 kelobases are present.
If DNA segment has 820 adenine molecules, then
what will be number of cytosine ?
(1) 1560 (2) 1480
(3) 780 (4) 740
14. Which statement is correct ?
(a) Degeneracy of code is related to, third member
of codon
(b) Single codon, codes for more than one amino
acid
(c) In codon first two bases are more specific
(d) In codons, third base is wobble
(e) Code is universal
(1) a, b, c, d, e
(2) a, b, d
(3) a, c, d
(4) a, c, d, e
15. Both the strand of DNA are not copied during
transcription because :-
(1) If both strands act as a template, they would
code for RNA with different sequence
(2) The two RNA molecules, i f produced
simultaneously would be complementary to
each other, hence would form a double
stranded RNA
(3) They would code, for RNA molecules with
same sequences
(4) Both (1) and (2) are correct
16. The salient feature of DNA are –
(i) It is made of two polynucleotide chain
(ii) Back bone is constituted by sugar and nitrogen
base
(iii) Two chain have parallel polarity
(iv) Bases in two strands are paired through H-bonds
(v) The two chain are coiled in a left handed fashion
(1) i, iv, v (2) i, iv
(3) i, ii, v (4) i, ii, iii, iv, v
17. Which is incorrect for genetic code–
(i) The codon is triplet
(ii) 64 codons code for amino acids
(iii) Genetic code is unambiguous
(iv) Genetic code is nearly universal
(v) AUG has dual functions
(1) only ii
(2) ii & iii
(3) iii, iv + v
(4) All are correct
18. Which is correct –
(i) t-RNA has an anticodon loop that has bases
complementary to the code
(ii) t-RNA has an amino acid acceptor end
(iii) t-RNA are specific for each amino acid
(iv) For initiation, there is specific t-RNA that is
reffered to as initiator t-RNA
(v) For termination there is specific t-RNA that is
reffered to as terminator t-RNA
(1) i, ii
(2) i, ii, iii
(3) i, ii, iii, iv
(4) i, ii, iii, iv, v
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Pre-Medical : BiologyALLEN19. E. coli cells with a mutated z gene of the lac operon
cannot grow in medium containing only lactose as
the source of energy because –
(1)They cannot synthesize functional
betagalactosidase
(2) They cannot transport lactose from the medium
into the cell
(3) The lac operon is constitutively active in these
cells
(4) In the presence of glucose, E. coli cells do not
utilize lactose
20. Select the incorrect statement.
(1) DNA from single cell is enough to perform DNA
fingerprinting analysis
(2) DNA fingerprinting has much wider applications
in determining population & genetic diversities.
(3) The VNTR belongs to a class of satellite DNA
referred as microsatellite.
(4) DNA fingerprint differs from individual to
individual in a population except in th case of
monozygotic twins.
21. DNA fragments separated by gel electrophoresis
are shown. Mark the correct statement :-
(1) Band 3 contains more positively charged DNA
molecule than 1
(2) Band '3' indicates more charge density than 1
and 2
(3) Band 1 has longer DNA fragment than 2 and 3
(4) All the bands have equal length and charges
but differ in base composition
EXERCISE-III (Analytical Questions) ANSWER KEYQue. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
Ans. 1 2 3 4 2 3 2 4 2 3 3 1 3 4 4Que. 16 17 18 19 20 21
Ans. 2 1 3 1 3 3