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

Cl (

N is the heavy

isotope of nitrogen) as the only nitrogen source for many

generations. The result was that

N was incorporated into newly

synthesised DNA (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

N is not a radioactive isotope, and it can be

separated from

N only based on densities).

(ii) Then they transferred the cells into a medium with normal

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

N to

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:

)

They grew

E. coli

in a medium containin

(

N is the heavy

isotope of nitrogen) a

the

only nitrogen source for many

generations. The result was that

N was incorporated into newly

nthesi

ed DNA

(

as well as other nitro

n containi

com

unds

] hy

intermediate densi

. DNA extracted from the cult

ure after

another generation [that is after 40 minutes, II generation] was

202

2-2

110055

Figure 6.7

Meselson and Stahl’s Experiment

( 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

N is not a radioactive isotope, and it can be

separated from

N only based on densities)

(ii)

Then the

transferred the cells into a medium with normal

Cl and took samples at various definite time intervals as

the cells mult

lied, and extracted the DNA that remained as

double-stranded helices. The various sa

les were s

arated

independentl

on CsCl

adients to measure the densities of

DNA (Figure 6.7).

Can you recall what centri

gal

rce is, and think why

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

N to

N medium [that is

after 20 minutes;

E. coli

divid

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

bp (compare

it with human whose diploid content is 6.6 × 10

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’

If E. coli was allowed to grow for 80 minutes then what would be the

proportions of light and hybrid densities DNA molecule?

ery similar experiments involving use of radioactive thymidine to

detect distribution of newly synthesised DNA in the chromosomes was

There is a definite region in

DNA where the replication

originates.

Such regions are termed as

origin of replication

. It is

202

2-2

110066

performed on

Vicia fa

(faba beans) by Taylor and colleagues in 1958.

The experiments proved that the DNA in chromosomes also re

icate

semiconservatively.

6.4.2 The Machinery and the Enzymes

In livi

cells, such as

E. coli

, the process of replication requires a set of

cata

sts (enz

es). The main enz

e is referred to as DNA-de

ndent

DNA polymerase

, since it uses a DNA te

late to cata

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.

that has only 4.

bp (compare

it with human whose diploid content is 6.6 ×

bp), completes the

process of replication within 18 minutes; that means the average rate of

merisation has to be approximate

2000 bp per second. Not only do

these po

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

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

ication fork

. The DNA-de

ndent DNA

polymerases catalyse polymerisation only in one direction, that is 5

This creates some additional complications at the replicating fork.

Consequent

, on one strand (the tem

ate with polarity 3

), the

replication is

continuous

, while on the other (the template with

polarity

), 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

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

uirement of

the

origin of

replication that a piece of DNA if needed to b

propagated during recombinant DNA procedures,

equir

es a vector

The vectors pr

ovide

origin of

replication.

Further

, not every detail of r

lication is

The Structural gen

(iii)

A Terminator

202

2-2

110077

understood well. In eukaryotes, the replication of

DNA takes place at S-phase of the cell-cycle. T

ication of DNA and cell division cycle should

highly coordinated. A failure in cell division after

DNA replication results into polyploidy(a

omosomal anomaly). Y

ou will lear

n the detailed

nature of origin and the processes occurring at th

site, in higher classes.

6.5 T

The process of copying genetic information from on

strand of the DNA into RNA is termed a

transcription

. Here also, the principle of

complementarity governs the process of transcription, except the adenosine

complements now for

ms base

ir with uracil instead of th

ine. 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 answe

. 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

code for

oteins, the se

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

event

RNA from bei

translated into

rotein and the exercise of transcr

tion

uld be

6.5.1 Tran

A transcri

ion unit in DNA is defined

rimari

the three re

ons in

DNDN

(i)

Figur

plpl

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:

-ATGCATGCATGCATGCATGCATGC-5

Template Strand

-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

merisation in on

one direction, that is,

→

, the strand that has

the polarity

→

acts as a template, and is also referred to as

template

strand

. The other strand which has the polarity (5

→

) and the se

ence

same as RNA (except thymine at the place of uracil), is displaced during

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

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:

-A

TGCATGCATGCATGCATGCATGC-

Tem

ate Strand

-T

G-

Coding Stra

Can you now write the sequence of RNA transcribed from the above DNA?

Figure 6.9

tic struct

ofofof

t t

nscr

tion unit

The

promoter

and

terminator

flank the

structural gene

in a

transcription unit. The promoter is said to be located towards 5

-e

(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 ter

minato

, 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

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

109

MOLECULAR BASIS OF INHERITANCE

define a gene in terms of DNA sequence. The DNA sequence coding for

tRNA or rRNA molecule also define a gene. However by defining a cistron

as a segment of DNA coding for a polypeptide, the structural gene in a

transcription unit could be said as monocistronic (mostly in eukaryotes)

or polycistronic (mostly in bacteria or prokaryotes). In eukaryotes, the

monocistronic structural genes have interrupted coding sequences – the

genes in eukaryotes are split. The coding sequences or expressed

sequences are defined as exons. Exons are said to be those sequence

that appear in mature or processed RNA. The exons are interrupted by

introns. Introns or intervening sequences do not appear in mature or

processed RNA. The split-gene arrangement further complicates the

definition of a gene in terms of a DNA segment.

Inheritance of a character is also affected by promoter and regulatory

sequences of a structural gene. Hence, sometime the regulatory sequences

are loosely defined as regulatory genes, even though these sequences do

not code for any RNA or protein.

6.5.3 Types of RNA and the process of Transcription

In bacteria, there are three major types of RNAs: mRNA (messenger RNA),

tRNA (transfer RNA), and rRNA (ribosomal RNA). All three RNAs are

needed to synthesise a protein in a cell. The mRNA provides the template,

tRNA brings aminoacids and reads the genetic code, and rRNAs play

structural and catalytic role during translation. There is single

DNA-dependent RNA polymerase that catalyses transcription of all types

of RNA in bacteria. RNA polymerase binds to promoter and initiates

transcription (Initiation). It uses nucleoside triphosphates as substrate

Figure 6.10 Process of Transcription in Bacteria

2022-23

MOLECULAR BASIS OF INHERITANCE

define a

ene in terms of DNA se

ence. The DNA se

ence codin

for

tRNA or rRNA molecule also define a gene. However by defining a

as a segment of DNA coding for a polypeptide, the structural gene in a

transcri

ion unit could be said as

monocistronic

(most

in euka

otes)

cistronic

(mostly in bacteria or prokaryotes). In eukaryotes, the

monocistronic structural genes have interrupted coding sequences – the

Figure 6.10

Process of Transcription in Bacteria

202

2-2

110099

genes in eukaryotes are split. The coding sequences or expressed

sequences are defined as

exons

. Exons are said to be those sequence

that a

ear in mature or

rocessed RNA. The exons are interru

ed b

. Introns or intervening sequences do not appear in mature or

processed RNA. The split-gene arrangement further complicates the

definition of a gene in terms of a DNA segment.

Inheritance of a character is also affected by promoter and regulatory

sequences of a structural gene. Hence, sometime the regulatory sequences

are loosely defined as regulatory genes, even though these sequences do

not code for any RNA or

protein.

6.5.3 Types of RNA and the process of Tr

riri

ion

In bacteria, there are three major types of RNAs: mRNA (messenger RNA),

tRNA (transfer RNA), and rRNA (ribosomal RNA). All three RNAs are

needed to synthesise a protein in a cell. The mRNA provides the template,

tRNA brings aminoacids and reads the genetic code, and rRNAs play

structural and catalytic role during translation. There is single

DNA-dependent RNA polymerase that catalyses transcription of all types

of RNA in bacteria. RNA polymerase binds to promoter and initiates

transcription (

Initiation

). It uses nucleoside triphosphates as substrate

110

BIOLOGY

and polymerises in a template depended fashion following the rule of

complementarity. It somehow also facilitates opening of the helix and

continues elongation. Only a short stretch of RNA remains bound to the

enzyme. Once the polymerases reaches the terminator region, the nascent

RNA falls off, so also the RNA polymerase. This results in termination of

transcription.

An intriguing question is that how is the RNA polymerases able

to catalyse all the three steps, which are initiation, elongation and

termination. The RNA polymerase is only capable of catalysing the

process of elongation. It associates transiently with initiation-factor (σ)

and termination-factor (ρ) to initiate and terminate the transcription,

respectively. Association with these factors alter the specificity of the

RNA polymerase to either initiate or terminate (Figure 6.10).

In bacteria, since the mRNA does not require any processing to become

active, and also since transcription and translation take place in the same

compartment (there is no separation of cytosol and nucleus in bacteria),

many times the translation can begin much before the mRNA is fully

transcribed. Consequently, the transcription and translation can be coupled

in bacteria.

In eukaryotes, there are two additional complexities –

(i) There are at least three RNA polymerases in the nucleus (in addition

to the RNA polymerase found in the organelles). There is a clear

cut division of labour. The RNA polymerase I transcribes rRNAs

Figure 6.11 Process of Transcription in Eukaryotes

2022-23

BIOLOGY

and polymerises in a template depended fashion following the rule of

complementarity. It somehow also facilitates opening of the helix and

continues elongation. Only a short stretch of RNA remains bound to the

enzyme. Once the polymerases reaches the terminator region, the nascent

RNA falls off, so also the RNA polymerase. This results in

termination

transcri

ion.

able

Figure 6.11

Process of Transcription in Eukaryotes

202

2-2

111100

An intriguing question is that how

s the RNA polymerase

able

to cataly

e all the three steps, which are initiation, elongation and

termination. The RNA po

merase is on

capable of cata

ing

the

process of elongation. It associates transiently with

initiation-factor

(

)

and

termination-factor

(

)

to initiate and terminate the transcription,

respectively.

Association with these factors alter

the specificity of the

RNA polymerase to either initiate or terminate (Figure 6.10).

since the mRNA does not require any processing to become

active, and also since transcription and translation take place in the same

compartment (there is no separation of cytosol and nucleus in bacteria),

many times the translation can begin much before the

mRNA is fully

Consequently, the transcription and translation can be coupled

in bacteria.

In eukaryotes, there are two additional complexiti

–

(i)

There are at least three RNA polymerases in the nucleus (in addition

to the RNA polymerase found in the organelles).

There is a clear

f labo

. The RNA polymerase

111

MOLECULAR BASIS OF INHERITANCE

(28S, 18S, and 5.8S), whereas the RNA polymerase III is responsible

for transcription of tRNA, 5srRNA, and snRNAs (small nuclear

RNAs). The RNA polymerase II transcribes precursor of mRNA, the

heterogeneous nuclear RNA (hnRNA).

(ii) The second complexity is that the primary transcripts contain both

the exons and the introns and are non-functional. Hence, it is

subjected to a process called splicing where the introns are removed

and exons are joined in a defined order. hnRNA undergoes

additional processing called as capping and tailing. In capping an

unusual nucleotide (methyl guanosine triphosphate) is added to

the 5

-end of hnRNA. In tailing, adenylate residues (200-300) are

added at 3

-end in a template independent manner. It is the fully

processed hnRNA,

now called mRNA, that is transported out of the

nucleus for translation (Figure 6.11).

The significance of such complexities is now beginning to be

understood. The split-gene arrangements represent probably an ancient

feature of the genome. The presence of introns is reminiscent of antiquity,

and the process of splicing represents the dominance of RNA-world. In

recent times, the understanding of RNA and RNA-dependent processes

in the living system have assumed more importance.

6.6 GENETIC CODE

During replication and transcription a nucleic acid was copied to form

another nucleic acid. Hence, these processes are easy to conceptualise

on the basis of complementarity. The process of translation requires

transfer of genetic information from a polymer of nucleotides to synthesise

a polymer of amino acids. Neither does any complementarity exist between

nucleotides and amino acids, nor could any be drawn theoretically. There

existed ample evidences, though, to support the notion that change in

nucleic acids (genetic material) were responsible for change in amino acids

in proteins. This led to the proposition of a genetic code that could direct

the sequence of amino acids during synthesis of proteins.

If determining the biochemical nature of genetic material and the

structure of DNA was very exciting, the proposition and deciphering of

genetic code were most challenging. In a very true sense, it required

involvement of scientists from several disciplines – physicists, organic

chemists, biochemists and geneticists. It was George Gamow, a physicist,

who argued that since there are only 4 bases and if they have to code for

20 amino acids, the code should constitute a combination of bases. He

suggested that in order to code for all the 20 amino acids, the code should

be made up of three nucleotides. This was a very bold proposition, because

a permutation combination of 4

(4 × 4 × 4) would generate 64 codons;

generating many more codons than required.

Providing proof that the codon was a triplet, was a more daunting

task. The chemical method developed by Har Gobind Khorana was

2022-23

MOLECULAR BASIS OF INHERITANCE

(28

, 18

, and 5.8

), whereas the RNA polymerase III is responsible

for transcription

tRNA

5srRNA

, an

snRNAs

(

small nuclear

The RNA polymerase II transcribes precursor of mRNA, the

heterogeneous nuclear RN

(

hnRNA

(

) The second complexity is that the primary transcripts contain both

the exons and the introns

and are no

functional.

Hence, it is

nerati

man

more codons than r

uired.

Providing proof that the codon was a triplet, was a more daunting

task. The chemical method developed by Har Gobind Khorana was

202

2-2

111111

subjected to a process called

splici

where the introns are removed

e joined in a defined

additional processing called as capping and tailing. In

capping

unusual nucleotide (methyl guanosine triphosphate) is added to

the 5

end of hnRNA. In

tailing

, adenylate residues (200-300) are

added at 3

end in a template independent mann

It is the

fully

processed hnRN

alled mR

that is transported out of the

nucleus for translation (Figure 6.11).

The significance of such complexities is now beginning to be

understood. The split-gene arrangements represent probably an ancient

feature of the genome. The presence of introns is reminiscent of antiquity,

and the process of splicing represents the dominance of

A-

rld

recent times, the understanding of RNA and RNA-dependent processes

in the living system have assumed more importance.

6.6 G

ENET

ODE

During replication and transcription a nucleic acid was copied to form

another nucleic acid. Hence, these processes are easy to conceptualise

on the basis of complementarity. The process of translation requires

transfer of

netic information from a

er of nucleotides to s

thesise

mer of amino acids. Neither does a

com

ementari

exist between

nucleotides and amino acids, nor could any be drawn theoretically. There

existed ample evidences, though, to support the notion that change in

nucleic acids (genetic material) were responsible for change in amino acids

roteins. This led to the

osition of a

enetic code that could direct

the sequence of amino acids during synthesis of proteins.

If determining the biochemical nature of genetic material and the

structure of DNA was very exciting, the proposition and deciphering of

genetic code were most challenging. In a very true sense, it required

involvement of scientists from several disci

ines –

sicists, o

anic

chemists, biochemists and geneticists. It was George Gamow, a physicist,

who argued that since there are only 4 bases and if they have to code for

20 amino acids

the code should constitute a combination of bases. He

suggested that in order to code for all the 20 amino acids, the code should

be made

of three nucleotides. This was a ve

bold

osition, because

a permutation combination of 4

(4 × 4 × 4) would generate 64 codons;

generating many more codons than required.

112

BIOLOGY

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). Finally a checker-board for genetic

code was prepared which is given in Table 6.1.

Table 6.1: The Codons for the Various Amino Acids

The salient features of genetic code are as follows:

(i) The codon is triplet. 61 codons code for amino acids and 3 codons do

not code for any amino acids, hence they function as stop codons.

(ii) Some amino acids are coded by more than one codon, hence

the code is degenerate.

(iii) The codon is read in mRNA in a contiguous fashion. There are

no punctuations.

(iv) The code is nearly universal: for example, from bacteria to human

UUU would code for Phenylalanine (phe). Some exceptions to this

rule have been found in mitochondrial codons, and in some

protozoans.

(v) AUG has dual functions. It codes for Methionine (met) , and it

also act as initiator

codon.

(vi) UAA, UAG, UGA are stop terminator codons.

If following is the sequence of nucleotides in mRNA, predict the

sequence of amino acid coded by it (take help of the checkerboard):

-AUG UUU UUC UUC UUU UUU UUC-

2022-23

BIOLOGY

instrumental in

nthesisi

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 (en

matic

nthesis of RNA). Final

a checker

-boar

d for

netic

-AUG UUU UUC UUC UUU UUU UUC-

202

2-2

111122

code was prepared which is given in Table 6.1.

able 6.1: The Codons for the V

arious Amino Acids

The salient fea

tures of genetic code are as follows

(i)

The codon is tri

et.

61 codons code for amino acids and 3 codons do

not code for a

amino acids, hence th

function

as sto

codons.

(ii)

Some amino acids are coded b

more than one codon, hence

the code is

degenerate

(iii)

The codon is read in mRNA in a conti

ous fashion. There are

no punctuations.

(iv)

The code is near

universal

or example, from bacteria to human

UUU would code for

Phenylalanine (phe).

Some exceptions to this

rule have been found in mitochondrial codons, and in some

protozoans.

)

AUG has dual functions. It codes for Methionine

)

, and it

also act as

initiator

codon

UAA, UAG, UGA are stop terminator codons.

ollowing is the sequence o

nucleotides in mRNA, predict the

sequence of amino acid coded by it (take help of the checkerboar