105
MOLECULAR BASIS OF INHERITANCE
Figure 6.7 Meselson and Stahl’s Experiment
and human cells. Matthew Meselson and Franklin Stahl performed the
following experiment in 1958:
(i) They grew E. coli in a medium containing
15
NH
4
Cl (
15
N is the heavy
isotope of nitrogen) as the only nitrogen source for many
generations. The result was that
15
N 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
15
N is not a radioactive isotope, and it can be
separated from
14
N only based on densities).
(ii) Then they transferred the cells into a medium with normal
14
NH
4
Cl 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 (Figure 6.7).
Can you recall what centrifugal force is, and think why a
molecule with higher mass/density would sediment faster?
The results are shown in Figure 6.7.
(iii) Thus, the DNA that was extracted from the culture one
generation after the transfer from
15
N to
14
N 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
2022-23
MOLECULAR BASIS OF INHERITANCE
and human cells. Matthew Meselson and Franklin Stahl performed the
following experiment in 1958:
(i
)
They grew
E. coli
in a medium containin
g
i
15
NH
4
Cl
(
15
N is the heavy
isotope of nitrogen) a
s
the
only nitrogen source for many
generations. The result was that
15
N was incorporated into newly
sy
nthesi
s
ed DNA
(
as well as other nitro
ge
n containi
ng
com
po
unds
).
] hy
intermediate densi
ty
. DNA extracted from the cult
ure after
another generation [that is after 40 minutes, II generation] was
202
2-2
3
110055
Figure 6.7
Meselson and Stahl’s Experiment
sy
( ge ng po ).
This heavy DNA molecule could be distinguished from the normal
DNA by centrifugation in a cesium chloride (CsCl) density gradient
(Please note that
15
N is not a radioactive isotope, and it can be
separated from
14
N only based on densities)
.
(ii)
Then the
y
transferred the cells into a medium with normal
14
NH
4
Cl and took samples at various definite time intervals as
the cells mult
ip
lied, and extracted the DNA that remained as
double-stranded helices. The various sa
mp
les were s
ep
arated
independentl
y
on CsCl
gr
adients to measure the densities of
DNA (Figure 6.7).
Can you recall what centri
fu
gal
fo
rce is, and think why
a
molecule with higher mass/density would sediment faster?
The results are shown in Figure 6.7.
(iii
)
Thus, the DNA that was extracted from the culture one
generation after the transfer from
15
N to
14
N medium [that is
after 20 minutes;
E. coli
divid
i
es in 20 minutes] had a hybrid or
106
BIOLOGY
composed of equal amounts of this hybrid DNA and of ‘light’
DNA.
If E. coli was allowed to grow for 80 minutes then what would be the
proportions of light and hybrid densities DNA molecule?
Very similar experiments involving use of radioactive thymidine to
detect distribution of newly synthesised DNA in the chromosomes was
performed on Vicia faba (faba beans) by Taylor and colleagues in 1958.
The experiments proved that the DNA in chromosomes also replicate
semiconservatively.
6.4.2 The Machinery and the Enzymes
In living cells, such as E. coli, the process of replication requires a set of
catalysts (enzymes). The main enzyme is referred to as DNA-dependent
DNA polymerase, since it uses a DNA template to catalyse the
polymerisation of deoxynucleotides. These enzymes are highly efficient
enzymes as they have to catalyse polymerisation of a large number of
nucleotides in a very short time. E. coli that has only 4.6 ×10
6
bp (compare
it with human whose diploid content is 6.6 × 10
9
bp), completes the
process of replication within 18 minutes; that means the average rate of
polymerisation has to be approximately 2000 bp per second. Not only do
these polymerases have to be fast, but they also have to catalyse the reaction
with high degree of accuracy. Any mistake during replication would result
into mutations. Furthermore, energetically replication is a very expensive
process. Deoxyribonucleoside triphosphates serve dual purposes. In
addition to acting as substrates, they provide energy for polymerisation
reaction (the two terminal phosphates in a deoxynucleoside triphosphates
are high-energy phosphates, same as in case of ATP).
In addition to DNA-dependent DNA polymerases, many additional
enzymes are required to complete the process of replication with high
degree of accuracy. For long DNA molecules, since the two strands of
DNA cannot be separated in its entire length (due to very high energy
requirement), the replication occur within a small opening of the DNA
helix, referred to as replication fork. The DNA-dependent DNA
polymerases catalyse polymerisation only in one direction, that is 5
'
à3
'
.
This creates some additional complications at the replicating fork.
Consequently, on one strand (the template with polarity 3
'
à5
'
), the
replication is continuous, while on the other (the template with
polarity 5
'
à3
'
), it is discontinuous. The discontinuously synthesised
fragments are later joined by the enzyme DNA ligase (Figure 6.8).
The DNA polymerases on their own cannot initiate the process of
replication. Also the replication does not initiate randomly at any place
in DNA. There is a definite region in E. coli DNA where the replication
originates. Such regions are termed as origin of replication. It is
2022-23
BIOLOGY
composed of equal amounts of this hybrid DNA and of ‘light’
DN
A.
If E. coli was allowed to grow for 80 minutes then what would be the
proportions of light and hybrid densities DNA molecule?
V
ery similar experiments involving use of radioactive thymidine to
V
V
detect distribution of newly synthesised DNA in the chromosomes was
e
of
t
he
nt
of
re
e
of
do
on
lt
e
In
on
es
l
h
f
y
A
A
he
h
d
f
ce
in
D
NA
.
There is a definite region in
E.
c
ol
i
DNA where the replication
i
originates.
Such regions are termed as
origin of replication
. It is
202
2-2
3
110066
performed on
Vicia fa
ba
(faba beans) by Taylor and colleagues in 1958.
ba
The experiments proved that the DNA in chromosomes also re
pl
icate
semiconservatively.
6.4.2 The Machinery and the Enzymes
In livi
ng
cells, such as
E. coli
, the process of replication requires a set of
cata
ly
sts (enz
ym
es). The main enz
ym
e is referred to as DNA-de
pe
ndent
DNA polymerase
, since it uses a DNA te
mp
late to cata
ly
se the
polymerisation of deoxynucleotides. These enzymes are highly efficient
enzymes as they have to catalyse polymerisation of a large number of
nucleotides in a very short time.
E.
c
ol
i
that has only 4.
i
6
×
10
6
bp (compare
it with human whose diploid content is 6.6 ×
10
9
bp), completes the
process of replication within 18 minutes; that means the average rate of
po
ly
merisation has to be approximate
ly
2000 bp per second. Not only do
these po
ly
merases have to be fast, but they also have to catalyse the reaction
with high degree of accuracy. Any mistake during replication would result
into mutations. Furthermore, energetically replication is a very expensive
process. Deoxyribonucleoside triphosphates serve dual purposes. In
addition to acting as substrates, they provide energy for polymerisation
reaction (the two terminal phosphates in a deoxynucleoside tri
ph
os
ph
ates
are high-energy phosphates, same as in case of ATP).
In addition to DNA-dependent DNA polymerases, many additional
enzymes are required to complete the process of replication with high
degree of accuracy. For long DNA molecules, since the two strands of
DNA cannot be separated in its entire length (due to very high energy
requirement), the replication occur within a small opening of the DNA
helix, referred to as
re
pl
ication fork
. The DNA-de
pe
ndent DNA
polymerases catalyse polymerisation only in one direction, that is 5
'
à
3
'
.
This creates some additional complications at the replicating fork.
Consequent
ly
, on one strand (the tem
pl
ate with polarity 3
'
à
5
'
), the
replication is
continuous
, while on the other (the template with
polarity
5
'
à
3
'
), it is
discontinuous
. The discontinuously synthesised
fragments are later joined by the enzyme
DNA ligase
(Figure 6.8
).
The DNA polymerases on their own cannot initiate the process of
replication. Also the replication does not initiate randomly at any place
in DNA
Th is definit gi i
E. oli
DNA he the li ti
i
107
MOLECULAR BASIS OF INHERITANCE
because of the requirement of the origin of
replication that a piece of DNA if needed to be
propagated during recombinant DNA procedures,
requires a vector. The vectors provide the origin of
replication.
Further, not every detail of replication is
understood well. In eukaryotes, the replication of
DNA takes place at S-phase of the cell-cycle. The
replication of DNA and cell division cycle should be
highly coordinated. A failure in cell division after
DNA replication results into polyploidy(a
chromosomal anomaly). You will learn the detailed
nature of origin and the processes occurring at this
site, in higher classes.
6.5 TRANSCRIPTION
The process of copying genetic information from one
strand of the DNA into RNA is termed as
transcription. Here also, the principle of
complementarity governs the process of transcription, except the adenosine
complements now forms base pair with uracil instead of thymine. However,
unlike in the process of replication, which once set in, the total DNA of an
organism gets duplicated, in transcription only a segment of DNA and
only one of the strands is copied into RNA. This necessitates defining the
boundaries that would demarcate the region and the strand of DNA that
would be transcribed.
Why both the strands are not copied during transcription has the
simple answer. First, if both strands act as a template, they would code
for RNA molecule with different sequences (Remember complementarity
does not mean identical), and in turn, if they code for proteins, the sequence
of amino acids in the proteins would be different. Hence, one segment of
the DNA would be coding for two different proteins, and this would
complicate the genetic information transfer machinery. Second, the two
RNA molecules if produced simultaneously would be complementary to
each other, hence would form a double stranded RNA. This would prevent
RNA from being translated into protein and the exercise of transcription
would become a futile one.
6.5.1 Transcription Unit
A transcription unit in DNA is defined primarily by the three regions in
the DNA :
(i) A Promoter
(ii) The Structural gene
(iii) A Terminator
Figure 6.8 Replicating Fork
2022-23
MOLECULAR BASIS OF INHERITANCE
because of the r
eq
uirement of
the
origin of
replication that a piece of DNA if needed to b
e
propagated during recombinant DNA procedures,
r
equir
es a vector
.
The vectors pr
ovide
th
e
origin of
replication.
Further
, not every detail of r
ep
lication is
un
de
rs
to
od
w
el
l.
In
e
uk
ar
yo
te
s,
t
he
r
ep
li
ca
ti
on
o
f
(i
i)
The Structural gen
e
(iii)
A Terminator
202
2-2
3
110077
understood well. In eukaryotes, the replication of
DNA takes place at S-phase of the cell-cycle. T
he
re
pl
ication of DNA and cell division cycle should
be
highly coordinated. A failure in cell division after
DNA replication results into polyploidy(a
ch
r
omosomal anomaly). Y
ou will lear
Y
Y
n the detailed
nature of origin and the processes occurring at th
is
site, in higher classes.
6.5 T
RA
NS
CR
IP
TI
ON
The process of copying genetic information from on
e
strand of the DNA into RNA is termed a
s
transcription
. Here also, the principle of
complementarity governs the process of transcription, except the adenosine
complements now for
ms base
pa
ir with uracil instead of th
ym
ine. However
,
r
r
unlike in the process of replication, which once set in, the total DNA of an
organism gets duplicated, in transcription only a segment of DNA and
only one of the strands is copied into RNA. This necessitates defining the
boundaries that would demarcate the region and the strand of DNA that
would be transcribed.
Why both the strands are not copied during transcription has the
simple answe
r
. First, if both strands act as a template, they would code
for RNA molecule with different sequences (Remember complementarity
does not mean identical), and in turn, if th
ey
code for
pr
oteins, the se
qu
ence
of amino acids in the proteins would be different. Hence, one segment of
the DNA would be coding for two different proteins, and this would
complicate the genetic information transfer machinery. Second, the two
RNA molecules if produced simultaneously would be complementary to
each other
, hence would for
m a double stranded RNA. This would
pr
event
RNA from bei
ng
translated into
p
rotein and the exercise of transcr
ip
tion
wo
uld be
co
me
a
f
ut
il
e
on
e.
6.5.1 Tran
sc
sc
ri
ri
pt
pt
io
io
n
n
Un
it
A transcri
pt
ion unit in DNA is defined
p
rimari
ly
b
y
the three re
gi
ons in
th
e
DN
A
DNDN
:
(i)
A
Pr
om
ot
er
Figur
e
6.
6.
6.
8
8
Re
Re
Re
plpl
pl
ic
ic
ic
at
at
at
in
in
in
g
g
g
Fork
108
BIOLOGY
There is a convention in defining the two strands of the DNA in the
structural gene of a transcription unit. Since the two strands have opposite
polarity and the DNA-dependent RNA polymerase also catalyse the
polymerisation in only one direction, that is, 5
'
→3
'
, the strand that has
the polarity 3
'
→5
'
acts as a template, and is also referred to as template
strand. The other strand which has the polarity (5
'
→3
'
) and the sequence
same as RNA (except thymine at the place of uracil), is displaced during
transcription. Strangely, this strand (which does not code for anything)
is referred to as coding strand. All the reference point while defining a
transcription unit is made with coding strand. To explain the point, a
hypothetical sequence from a transcription unit is represented below:
3
'
-ATGCATGCATGCATGCATGCATGC-5
'
Template Strand
5
'
-TACGTACGTACGTACGTACGTACG-3
'
Coding Strand
Can you now write the sequence of RNA transcribed from the above DNA?
Figure 6.9 Schematic structure of a transcription unit
The promoter and terminator flank the structural gene in a
transcription unit. The promoter is said to be located towards 5
'
-end
(upstream) of the structural gene (the reference is made with respect to
the polarity of coding strand). It is a DNA sequence that provides binding
site for RNA polymerase, and it is the presence of a promoter in a
transcription unit that also defines the template and coding strands. By
switching its position with terminator, the definition of coding and template
strands could be reversed. The terminator is located towards 3
'
-end
(downstream) of the coding strand and it usually defines the end of the
process of transcription (Figure 6.9). There are additional regulatory
sequences that may be present further upstream or downstream to the
promoter. Some of the properties of these sequences shall be discussed
while dealing with regulation of gene expression.
6.5.2 Transcription Unit and the Gene
A gene is defined as the functional unit of inheritance. Though there is no
ambiguity that the genes are located on the DNA, it is difficult to literally
2022-23
BIOLOGY
There is a convention in defining the two strands of the DNA in the
structural gene of a transcription unit. Since the two strands have opposite
polarity and the
DNA-dependent RNA polymerase
also catalyse the
po
ly
merisation in on
ly
one direction, that is,
5
'
→
3
'
, the strand that has
the polarity
3
'
→
5
'
acts as a template, and is also referred to as
template
strand
. The other strand which has the polarity (5
'
→
3
'
) and the se
qu
ence
same as RNA (except thymine at the place of uracil), is displaced during
a
a
a
nd
to
g
a
y
e
nd
e
ry
e
ed
A gene is defined as the functional unit of inheritance. Though there is no
ambiguity that the genes are located on the DNA, it is difficult to literally
202
2-2
3
110088
same as RNA (except thymine at the place of uracil), is displaced during
transcription. Strangely, this strand (which does not code for anything)
is referred to as
coding strand
. All the reference point while defining a
transcription unit is made with coding strand. To explain the point, a
hypothetical sequence from a transcription unit is represented below:
3
'
-A
TGCATGCATGCATGCATGCATGC-
5
'
Tem
pl
ate Strand
5
'
-T
AC
GT
AC
GT
AC
GT
AC
GT
AC
GT
AC
G-
3
'
Coding Stra
nd
Can you now write the sequence of RNA transcribed from the above DNA?
Figure 6.9
Sc
he
he
he
ma
ma
ma
tic struct
ur
ur
ur
e
e
e
ofofof
a
a
a
t t
t
ra
ra
ra
nscr
ip
tion unit
The
promoter
and
terminator
flank the
structural gene
in a
transcription unit. The promoter is said to be located towards 5
'
-e
nd
(upstream) of the structural gene (the reference is made with respect to
the polarity of coding strand). It is a DNA sequence that provides binding
site for
RNA polymerase, and it is the presence of a promoter in a
or
transcription unit that also defines the template and coding strands. By
switching its position with ter
minato
r
, the definition of coding and template
strands could be reversed. The terminator is located towards 3
'
-end
(downstream) of the coding strand and it usually defines the end of the
process of transcription (Figure 6.9). There are additional regulatory
sequences that may be present further upstream or downstream to the
pr
omoter
. Some of the pr
operties of these sequences shall be discussed
while dealing with regulation of gene expression.
6.5.2 Transcription Unit and the Gene
A defi d he f l f he Tho h th