:
CHAPTER 6 :
EXPRESSION OF BIOLOGICAL
INFORMATION
(2L + 8T)
6.0 Expression of Biological Information
6.1
6.2
6.3
6.4
DNA & Genetic Information
DNA Replication
Protein Synthesis: Transcription & Translation
Gene Regulation & Expression
Learning outcomes
6.1 DNA & Genetic Information
a)
State the concept of Central Dogma.
6.2 DNA Replication
a)
Explain semi-conservative replication of DNA.
b)
Explain the enzymes and proteins that involve
in DNA replication.
c)
Explain the mechanism of DNA replication.
Learning
Outcomes
:
Learning
Outcomes
:
6.1 (a) State the concept of Central Dogma
6.1 a)
Explain DNA as the carrier of genetic information
LECTURE
Central Dogma
⮚ Dogma: a belief that is accepted without doubt or question
⮚ Central Dogma of molecular biology explain the flow of genetic
information within biological system
⮚ Genetic information flows from:
i. DNA to DNA (during replication before a cell divides)
ii. DNA to protein (during its phenotypic expression in an organism)
transcription
DNA
RNA
reverse transcription
replication
translation
protein
Learning
Outcomes
:
Learning
Outcomes
:
6.1 (a) State the concept of Central Dogma
6.1 a)
Explain DNA as the carrier of genetic information
LECTURE
Central Dogma
⮚ Genetic information which flows from DNA to protein involves 2
steps:
i. Transcription, the transfer of information from DNA to RNA
ii. Translation, the transfer of information from RNA to protein
transcription
DNA
RNA
reverse transcription
replication
translation
protein
Learning
Outcomes
:
Learning
Outcomes
:
6.1 (a) State the concept of Central Dogma
6.1 a)
Explain DNA as the carrier of genetic information
LECTURE
Central Dogma
⮚ Genetic information also flows from RNA to DNA during the
conversion of genomes of viral RNA to their DNA proviral forms
⮚ The transfer of genetic information from DNA to RNA is
sometimes reversible
⮚ But the transfer of genetic information from RNA to protein is
irreversible
⮚ The transfer of protein to protein or from protein to nucleic acid
(DNA / RNA) also are not possible
transcription
DNA
RNA
reverse transcription
replication
translation
protein
Learning Outcomes :
6.2 (a) Explain semi-conservative replication of DNA
LECTURE
DNA Replication
•
•
•
DNA is copied precisely before cell division → replication
During S phase (interphase)
1 chromosome consists of 2 sister chromatids bound to a
centromere
Sister chromatids
DNA replication
Learning Outcomes :
6.2 (a) Explain semi-conservative replication of DNA
LECTURE
DNA Replication
•
•
3 models of replication
Meselson & Stahl (1957) suspected that DNA replication is
semi-conservative
DNA Replication Models
Conservative
Semi-conservative
Dispersive
Learning Outcomes :
6.2 (a) Explain semi-conservative replication of DNA
LECTURE
Model 1 : Conservative
Double stranded DNA
Conservative Model
*Note:
• Blue (parental strand)
• Red (new strand)
• 2 parental strands may remain
together after acting as templates
• 2 newly synthesized strand form
the second double helix
Learning Outcomes :
6.2 (a) Explain semi-conservative replication of DNA
LECTURE
Model 2 : Semi-conservative
Double stranded DNA
Semi-conservative Model
*Note:
• Blue (parental strand)
• Red (new strand)
• Both DNA molecules consist of 1
parental strand and 1 newly synthesized
strand
Learning Outcomes :
6.2 (a) Explain semi-conservative replication of DNA
LECTURE
Model 3 : Dispersive
Double stranded DNA
Dispersive Model
*Note:
• Blue (parental strand)
• Red (new strand)
• Each strand of double stranded
DNA contains a mixture of
parental and new strands
Learning Outcomes :
6.2 (a) Explain semi-conservative replication of DNA
LECTURE
DNA Replication
Conservative Model
• Parental strand complements
with parental strand
• Newly synthesised strand
complements with new strand
Semi-conservative Model
• Parental strand
complements with new
strand
Dispersive Model
• Each strand of double
stranded DNA has
combination of parental &
new strands
Learning Outcomes :
6.2 (a) Explain semi-conservative replication of DNA
ASSIGNMENT
Meselson & Stahl Experiment (1957)
• Use bacteria ~ Escherichia coli
• Use 2 nitrogen isotopes:
1) 14N (normal light isotope of nitrogen)
2) 15N (heavy isotope of nitrogen)
Learning Outcomes :
6.2 (a) Explain semi-conservative replication of DNA
ASSIGNMENT
Meselson & Stahl Experiment (1957)
•
•
•
Nitrogen isotopes are used to
form NH4Cl (added in the
medium culture)
Bacteria uses nitrogen to
synthesize nitrogenous bases
Which is used to synthesize
DNA during replication before
a cell divides
Learning Outcomes :
6.2 (a) Explain semi-conservative replication of DNA
ASSIGNMENT
Meselson & Stahl Experiment (1957)
•
2 isotopes are used to label & differentiate parental strand
DNA & newly synthesized DNA
Learning Outcomes :
6.2 (a) Explain semi-conservative replication of DNA
ASSIGNMENT
Meselson & Stahl Experiment (1957)
•
•
E. coli is grown in
a medium
containing 15NH4Cl
for several
generations
Some bacteria are
transferred to a
medium containing
14N
Learning Outcomes :
6.2 (a) Explain semi-conservative replication of DNA
ASSIGNMENT
Meselson & Stahl Experiment (1957)
Escherichia coli
Cultured in
medium 15N
F1 generation
Transfer
•
•
•
E. coli is grown in a medium containing 15NH4Cl for several
generations.
Some bacterial samples were taken (F0).
The rest are transferred to a medium containing 14N.
ASSIGNMENT
Learning Outcomes :
6.2 (a) Explain semi-conservative replication of DNA
Meselson & Stahl Experiment (1957)
Cultured in
medium 15N
Transfer
•
•
•
F1 generation
F1 generation
Extract DNA
Centrifuge
After one generation, some bacteria is isolated (F1
generation) while others continue to grow in 14N medium
DNA is extracted from the bacterial sample (F1)
Purified DNA is subjected to density gradient centrifugation
Learning Outcomes :
6.2 (a) Explain semi-conservative replication of DNA
ASSIGNMENT
Meselson & Stahl Experiment (1957)
Cultured in
medium 15N
F1 generation
Extract DNA
F2 generation
Transfer
F3 generation
•
•
Bacteria in medium 14N is allowed to replicate twice to
produce F2 generations & thrice (F3 generations)
Repeat DNA extraction & centrifugation steps for F2 and
F3 generations
ASSIGNMENT
Learning Outcomes :
6.2 (a) Explain semi-conservative replication of DNA
Meselson & Stahl Experiment (1957)
E. coli cultured in medium 15N
Transfer to medium 14N
DNA extraction
Generation F0
One cell division
DNA extraction
Generation F1
DNA extraction
Generation F2
2 cell divisions
3 cell divisions
DNA extraction
Generation F3
Learning Outcomes :
6.2 (a) Explain semi-conservative replication of DNA
ASSIGNMENT
ASSIGNMENT
Learning Outcomes :
6.2 (a) Explain semi-conservative replication of DNA
Meselson & Stahl Experiment (1957)
F0 generation
Control
14N14N
15N15N
• Molecules of different density can be separated by density gradient
centrifugation
• DNA of cells grown in medium containing 14N have a lower density than
DNA of cells grown in 15N
• Parental DNA (generation F0) accumulates in high-density region
Model 1 : Conservative
ASSIGNMENT
14N14N
14N14N
14N14N
14N14N
14N14N
14N14N
14N14N
14N14N
15N15N
14N14N
14N14N
15N15N
14N14N
15N15N
15N15N
Parental / F0
generation
F1 generation
F2 generation
F3 generation
14N14N
Hybrid DNA 14N/15N
15N15N
Model 2 : Semi-conservative
ASSIGNMENT
14N14N
14N14N
14N14N
14N15N
14N15N
14N15N
14N14N
15N15N
14N14N
14N15N
F1 generation
14N14N
14N14N
14N15N
Parental / F0
generation
14N14N
F2 generation
14N15N
F3 generation
14N14N
Hybrid DNA 14N/15N
15N15N
ASSIGNMENT
Model 3 : Dispersive
14N15N
14N15N
14N15N
14N15N
14N15N
14N15N
14N15N
15N15N
14N15N
14N15N
F1 generation
14N15N
14N15N
14N15N
Parental / F0
generation
14N15N
F2 generation
14N15N
F3 generation
14N14N
Hybrid DNA 14N/15N
15N15N
Learning Outcomes :
6.2 (a) Explain semi-conservative replication of DNA
ASSIGNMENT
Meselson & Stahl Experiment - Results
RESULT
14N14N
Hybrid DNA
14N/15N
15N15N
14N14N
Hybrid DNA
14N/15N
15N15N
14N14N
Hybrid DNA
15N15N
14N/15N
ASSIGNMENT
Learning Outcomes :
6.2 (a) Explain semi-conservative replication of DNA
Meselson & Stahl Experiment (1957)
14N14N
F0 generation
Control
14N14N
15N15N
15N15N
• Molecules of different density can be separated by density gradient
centrifugation
• DNA of cells grown in medium containing 14N have a lower density than
DNA of cells grown in 15N
• Parental DNA (generation F0) accumulates in high-density region
Learning Outcomes :
6.2 (a) Explain semi-conservative replication of DNA
ASSIGNMENT
Meselson & Stahl Experiment (1957)
Learning Outcomes :
6.2 (a) Explain semi-conservative replication of DNA
ASSIGNMENT
Meselson & Stahl Experiment (1957)
Learning Outcomes :
6.2 (a) Explain semi-conservative replication of DNA
ASSIGNMENT
Meselson & Stahl Experiment (1957)
FO
generation
F1
generation
F2
generation
F3
generation
ASSIGNMENT
Learning Outcomes :
6.2 (a) Explain semi-conservative replication of DNA
ASSIGNMENT
Meselson & Stahl Experiment - Results
Parental / F0
generation
F1 generation
F2 generation
F3 generation
14N14N
Hybrid DNA 14N/15N
15N15N
•
•
•
For F1 generation, 1 band is obtained (have intermediate
density known as hybrid density)
For F2 generations, 2 bands exist. 1 band is hybrid density &
the other band was light
After 3 generations, 2 bands exist: intensity of ¼ hybrid, ¾
light band
Learning Outcomes :
6.2 (a) Explain semi-conservative replication of DNA
ASSIGNMENT
Meselson & Stahl Experiment - Results
•
•
•
For F1 generation, only 1 intermediate density band is obtained
(between the density of light & heavy DNA a.k.a hybrid density)
For F2 generations, 2 bands exist. 1 band is hybrid density &
the other band was light
After 3 generations, DNA band was ¼ hybrid, ¾ light band
Learning Outcomes :
6.2 (a) Explain semi-conservative replication of DNA
Meselson & Stahl Experiment
Conclusion:
•
DNA replicates by semi-conservative
ASSIGNMENT
ASSIGNMENT
Learning Outcomes :
6.2 (a) Explain semi-conservative replication of DNA
DNA Replication
Hydrogen bonds formed
between complementary
bases
DNA is a double helix of
two polynucleotide strands
Complementary base pairs hold
the two DNA35strands together
Learning Outcomes
Outcomes ::
Learning
6.2(c)
(c) Explain
Explain the
of DNA replication
and the enzymes
4.1
the mechanism
Mendel’s experiments
on monohybrid
cross involved
DNA Replication
•
The process involves are :
Step 1: Separation of DNA double helix
Step 2: Building RNA primer
Step 3: DNA synthesis
Step 4: Replacing RNA primer
Step 5: Joining of Okazaki fragments (ligation)
Learning Outcomes :
6.2 (c) Explain the mechanism of DNA replication and the enzymes involved
• Initiator protein binds to origin of replication
• Helicase unwind the DNA double helix
Learning Outcomes :
6.2 (c) Explain the mechanism of DNA replication and the enzymes involved
• DNA replication is bidirectional from the origin of replication
(but will consider events at just one replication fork)
Learning Outcomes :
6.2 (c) Explain the mechanism of DNA replication and the enzymes involved
Step 1: Separation of DNA double helix
5’
G A T C C G T T
3’
A A C C G T A A T C T G C T A T C G G A C A T C T
T T G G C A T T A G AC G A T A G C C T G T A G A
C T A G G C A A
5’
Replication fork
3’
• Helicase unwind DNA double helix & break the hydrogen bond
between complementary nitrogenous bases
• 2 polynucleotide strands separate
• Each strand act as a template for the synthesis of a new strand
Learning Outcomes :
6.2 (c) Explain the mechanism of DNA replication and the enzymes involved
SSB Proteins
5’
G A T C C G T T
3’
A A C C G T A A T C T G C T A T C G G A C A T C T
T T G G C A T T A G AC G A T A G C C T G T A G A
C T A G G C A A
3’
5’
Replication fork
• Single-strand binding proteins (SSB proteins) bind to the
separated DNA strand to prevent them from rejoining
Learning Outcomes :
6.2 (c) Explain the mechanism of DNA replication and the enzymes involved
overwinding
Overwinding
supercoil
supercoil
• When double stranded DNA is unwound, it adds extra helical twist
beyond the unwound region
• DNA become more tightly twisted (overwinding/supercoil), creating
a strain to DNA strand
Learning Outcomes :
6.2 (c) Explain the mechanism of DNA replication and the enzymes involved
overwinding
Overwinding
supercoil
topoisomerase
nick
supercoil
• Topoisomerase makes a nick by breaking the phosphodiester bond
• Topoisomerase helps relieve the strain of unwound double
stranded DNA and rejoin the DNA strands by reforming
phosphodiester bond
Learning Outcomes :
6.2 (c) Explain the mechanism of DNA replication and the enzymes involved
Step 1: Separation of DNA double helix
Step 2: Building RNA primer
U
C
G
U
G
A
A
A
C
G
A
T
RNA nucleotide
G
C
G
G
A
T
C
Primase
5’
T
A
T
DNA nucleotide
DNA polymerase III
G A T C C G T T
3’
A A C C G T A A T C T G C T A T C G G A C A T C T
5’
T T G G C A T T A G AC G A T A G C C T G T A G A
G A U
5’
C T A G G C A A
Replication fork
3’
G
A
U
C
A
G
C
G
A
T
A
T
C
A
C
G
A
T
T
Learning Outcomes :
6.2 (c) Explain the mechanism of DNA replication and the enzymes involved
Step 2: Building RNA primer
• DNA polymerase III cannot begin the synthesis of a new
strand
• It can only extend an existing strand paired with a template
strand
• To begin synthesis, a short fragment of RNA, called RNA
primer, must be created and paired with the template DNA
strand
Learning Outcomes :
6.2 (c) Explain the mechanism of DNA replication and the enzymes involved
Step 2: Building RNA primer
• Each strand act as a template
• Primase catalyze the synthesis of a short RNA primer that are
complementary to the template DNA strand
• RNA primer is about 10-60 nucleotides long
• Elongation of a new strand is from 5’ to 3’
Learning Outcomes :
6.2 (c) Explain the mechanism of DNA replication and the enzymes involved
Step 2: Building RNA primer
• RNA primer is synthesized to provide free 3’-hydroxyls end
• Which allow DNA polymerase III to start the synthesis of DNA
strand
• by adding DNA nucleotide to the free 3’-hydroxyls end of
these RNA primers
Learning Outcomes :
6.2 (c) Explain the mechanism of DNA replication and the enzymes involved
Step 2: Building RNA primer
U
C
G
U
G
A
A
C
A
G
A
T
RNA nucleotide
G
C
T
G
A
T
C
Primase
G
A
T
DNA nucleotide
5’
DNA polymerase III
G A T C C G T T
3’
A A C C G T A A T C T G C T A T C G G A C A T C T
5’
T T G G C A T T A G AC G A T A G C C T G T A G A
G AUC C G T T
5’
C T A G G C A A
Replication fork
3’
G
A
U
C
A
G
C
G
A
T
A
T
C
A
C
G
A
T
T
Learning Outcomes :
6.2 (c) Explain the mechanism of DNA replication and the enzymes involved
Step 3: DNA synthesis
•
•
DNA polymerase III catalyze the synthesis of new DNA
strand by adding DNA nucleotides to the short RNA primer
The adding of DNA nucleotides is complementary to the
bases of the template
Learning Outcomes :
6.2 (c) Explain the mechanism of DNA replication and the enzymes involved
Step 3: DNA synthesis
Primase
5’
DNA polymerase III
G A T C C G T T
3’
C T A G G C A A
3’
5’
5’
T T G G C A T T A G AC G A T A G C C T G T A G A
G AUC C G T T
C T A G G C A A
3’
A A C C G T A A T C T G C T A T C G G A C A T C T
5’
Replication fork
Learning Outcomes :
6.2 (c) Explain the mechanism of DNA replication and the enzymes involved
Step 3: DNA synthesis
Primase
5’
DNA polymerase III
G A T C C G T T
3’
C T A G G C A A
3’
5’
5’
T T G G C A T T A G AC G A T A G C C T G T A G A
G AUC C G T T
C T A G G C A A
3’
A A C C G T A A T C T G C T A T C G G A C A T C T
5’
Replication fork
Learning Outcomes :
6.2 (c) Explain the mechanism of DNA replication and the enzymes involved
Step 3: DNA synthesis
5’
G A T C C G T T A G G A A A C C G T A A T C
3’
C T A G G C A A T C C T T T G G C AT UA G
3’
5’
5’
G AUC C G T T A G G A A AC C G T AA T C
C T A G G C A A T C C T T T G G C A T T A G
3’
Primase
T C G G A C A T C T
A G C C T G T A G A
5’
Replication fork
DNA polymerase III
Learning Outcomes :
6.2 (c) Explain the mechanism of DNA replication and the enzymes involved
Step 3: DNA synthesis
5’
3’
G A T C C G T T A G G A A A C C G T A A T C T G C T A T C G G A C A T C T
C T A G G C A A T C C T T T G G C A T U A G AC G A T A G C C T G T A G A
3’
5’
G A T C C G T T A G G A A A C C G T A A T C T G C T A T C G G A C A T C T
5’
3’
C T A G G C A A T C CT T T G G C A T T A G AC G A T A G C C T G T A G A
3’
5’
Primase
DNA polymerase III
Learning Outcomes :
6.2 (c) Explain the mechanism of DNA replication and the enzymes involved
Step 3: DNA synthesis
•
•
•
DNA is synthesized continuously / towards the replication
fork is called as the leading strand
On another strand, DNA is synthesized discontinuously /
away from the replication fork is called as the lagging strand
Producing short Okazaki fragments
•
DNA polymerase I replace RNA primer with DNA
nucleotides
Learning Outcomes :
6.2 (c) Explain the mechanism of DNA replication and the enzymes involved
Step 4: Replacing RNA primer
DNA polymerase I
5’
3’
G A T C C G T T A G G A A A C C G T A A T C T G C T A T C G G A C A T C T
C T A G G CC A
AA
A T C C T T T G G C AT U
TA
A GG A C G A T A G C C T G T A G A
3’
5’
G A T C C G T T A G G A A A C C G T A A T C T G C T A T C G G A C A T C T
5’
3’
C T A G G C A A T C CT T T G G C A T T A G AC G A T A G C C T G T A G A
3’
•
5’
DNA polymerase I replace RNA primer with DNA
nucleotides
Learning Outcomes :
6.2 (c) Explain the mechanism of DNA replication and the enzymes involved
Step 5: Joining of Okazaki fragments (ligation)
• DNA ligase joins Okazaki fragments
• By catalyzing the formation of phosphodiester bond
between adjacent Okazaki fragments
• 2 identical copies of DNA are produced
DNA ligase
5’
3’
G A T C C G T T A G G A A A C C G T A A T C T G C T A T C G G A C A T C T
C T A G G C A A T C C T T T G G C A T T A G AC G A T A G C C T G T A G A
3’
5’
G A T C C G T T AG G A AA C C G T A A T C T G C T A T C G G A C A T C T
5’
3’
C T A G G C A A T C CT T T G G C A T T A G AC G A T A G C C T G T A G A
3’
5’
Learning Outcomes
Outcomes ::
Learning
6.2(c)
(c) Explain
Explain the
of DNA replication
and the enzymes
4.1
the mechanism
Mendel’s experiments
on monohybrid
cross involved
DNA Replication
•
The process involves are :
Step 1: Separation of DNA double helix
Step 2: Building RNA primer
Step 3: DNA synthesis
Step 4: Replacing RNA primer
Step 5: Joining of Okazaki fragments (ligation)
Learning Outcomes :
6.2 (c) Explain the mechanism of DNA replication and the enzymes involved
• Initiator protein binds to origin of replication
• Helicase unwind the DNA double helix
Learning Outcomes :
6.2 (c) Explain the mechanism of DNA replication and the enzymes involved
Learning Outcomes :
6.2 (c) Explain the mechanism of DNA replication and the enzymes involved
Learning Outcomes :
6.2 (c) Explain the mechanism of DNA replication and the enzymes involved
5’
3’
Learning Outcomes :
6.2 (c) Explain the mechanism of DNA replication and the enzymes involved
Learning Outcomes :
6.2 (b) Explain the enzymes and protein involved in DNA replication
Function of Enzymes & Proteins in DNA Replication
Enzymes or Protein
Function
1. Helicase
Unwind the parental DNA double helix
and separate DNA strands by breaking
hydrogen bonds between
complementary bases
2. Single Strand Binding
Proteins (SSB protein)
Binds to the separated DNA strands to
prevent them from rejoining
3. Topoisomerase
Relieve the strain of unwound DNA
by breaking, swiveling and rejoining
DNA strands
4. Primase
Catalyze the synthesis of a short RNA
primer
Learning Outcomes :
6.2 (b) Explain the enzymes and protein involved in DNA replication
Function of Enzymes & Proteins in DNA Replication
Enzymes
5. DNA
polymerase III
Function
Catalyze the synthesize of new DNA
strand by adding complementary DNA
nucleotide to the 3’end of RNA primer
(which must be complementary to the
template)
6. DNA
polymerase I
7. DNA ligase
Replace RNA primers with DNA
nucleotides
Catalyze the formation of phosphodiester
bond between Okazaki fragments
Learning outcomes
6.3 Protein Synthesis: Transcription & Translation
a) Explain briefly transcription and translation.
b) Introduce codon & its relationship with sequence of
amino acid using genetic code table.
a) Explain transcription and the stages involved in the
formation of mRNA strand from 5’ to 3’.
a) Explain translation and the stages involved.
Learning Outcomes :
Learning
Outcomes :
6.3 a) Explain briefly transcription and translation
6.1 a) Explain DNA as the carrier of genetic information
LECTURE
Central Dogma
•
Genetic information flows from DNA to protein involves 2
steps:
i. Transcription, the transfer of information from DNA to RNA
ii. Translation, the transfer of information from RNA to protein
•
This flow of genetic information a.k.a as gene expression
transcription
DNA
RNA
reverse transcription
replication
translation
protein
Learning Outcomes :
6.3 a) Explain briefly transcription and translation
LECTURE
Protein Synthesis
1. Transcription
• synthesise mRNA
A process by which genetic information in DNA base
sequence is copied or transcribed into complementary
sequence of mRNA.
2. Translation
• synthesise protein
The sequence of nucleotides in mRNA act as a template
and is translated into sequence of amino acids.
Learning Outcomes :
6.3 c) Introduce codon and its relationship with sequence
of amino acid using genetic code table
LECTURE
Properties of Genetic Code
Functional mRNA
5’
3’
A U G C A U U U G C GA G A C U A A
1st Codon
•
•
2nd Codon 3rd Codon 4th Codon
5th Codon
6th Codon
3 base sequence of mRNA (triplet code) is called codon
1 codon on mRNA codes for 1 amino acid
Learning Outcomes :
6.3 c) Introduce codon and its relationship with sequence
of amino acid using genetic code table
LECTURE
1st base
(5’ end)
U
2nd base (5’ end)
U
UUU
UUC
UUA
UUG
C
Phe
Leu
UCU
UCC
UCA
UCG
C
CUU
CUC
CUA
CUG
CCU
CCC
Leu
CCA
CCG
A
AUU
AUC
AUA
AUG
ACU
ACC
ACA
ACG
G
GU U
GU C
GU A
GU G
Ile
Met/star
t
Val
GC U
GC C
GC A
GC G
A
Ser
Pro
Thr
Ala
3rd base
(5’ end)
G
UAU
Tyr
UAC
U A A STO
STO
UAG P
P
CAU
His
CAC
CAA
Gln
CAG
U GU
Cys
U GC
U G A STO
U G G P Trp
U
C
A
G
C GU
C GC
C GA
C GG
U
C
A
G
AAU
AAC
AAA
AAG
A GU
A GC
A GA
A GG
GA U
GA C
GA A
GA G
Asn
Lys
Asp
Glu
GGU
GGC
GGA
GGG
Arg
Ser
Arg
Gly
U
C
A
G
U
C
A
G
Learning Outcomes :
6.3 c) Introduce codon and its relationship with sequence
of amino acid using genetic code table
Properties of Genetic Code
1. Composed of nucleotide triplets
(triplet code)
• 3 nucleotides in mRNA is called a
codon
• If 2 nucleotides per codon
• will result in only 42 or 16 possible
codon (rejected)
LECTURE
Learning Outcomes :
6.3 c) Introduce codon and its relationship with sequence
of amino acid using genetic code table
Properties of Genetic Code
1. Composed of nucleotide triplets
(triplet code)
•
•
3 bases per codon
will result in 43 or 64 possible
codons
LECTURE
Learning Outcomes :
6.3 c) Introduce codon and its relationship with sequence
of amino acid using genetic code table
Properties of Genetic Code
2. Each codon represents
1 specific amino acid
• 20 different amino acids
commonly found in
proteins
• At least 20 different
codons
LECTURE
Learning Outcomes :
6.3 c) Introduce codon and its relationship with sequence
of amino acid using genetic code table
Properties of Genetic Code
3. The code is degenerate
• Most amino acids are
coded by more than one
codon
• Only 2 amino acids are
coded by 1 codon
• 61 codons represent 20
amino acids
LECTURE
Learning Outcomes :
6.3 c) Introduce codon and its relationship with sequence
of amino acid using genetic code table
Properties of Genetic Code
4. Contain start & stop codons
• AUG is the start codon.
• UAG, UAA, UGA are the stop codons.
• Do not code for any amino acid.
• Signals to terminate the coding sequence.
LECTURE
Learning Outcomes :
6.3 c) Introduce codon and its relationship with sequence
of amino acid using genetic code table
Properties of Genetic Code
5. Almost universal
• The same codon codes for the same
amino acids in all organisms
LECTURE
Learning Outcomes :
6.3 c) Introduce codon and its relationship with sequence
of amino acid using genetic code table
Properties of Genetic Code
6. Non-overlapping
•
•
•
•
Each codon consists of 3 nucleotides
Sequence of codon is read continuously
from a fixed starting point & proceed
until it reaches the termination point .
LECTURE
Learning Outcomes :
6.3 c) Introduce codon and its relationship with sequence
of amino acid using genetic code table
Properties of Genetic Code
7. Comma-free (commaless)
•
•
•
•
There are no commas
or other forms of punctuation
within the coding regions of mRNA
Codons are read consecutively
LECTURE
Learning Outcomes :
6.3 c) Explain transcription and the stages involved in the formation of mRNA strand
Transcription
•
•
•
Occurs in nucleus (for
eukaryote).
Involves RNA polymerase
enzyme.
During transcription, one
DNA strand acts as a
template to synthesise a
complementary strand of
mRNA.
ASSIGNMENT
Learning Outcomes :
6.3 c) Explain transcription and the stages involved in the formation of mRNA strand
Transcription
•
The processes involved are :
Step 1: Initiation
Step 2: Elongation
Step 3: Termination
ASSIGNMENT
Learning Outcomes :
6.3 c) Explain transcription and the stages involved in the formation of mRNA strand
ASSIGNMENT
Transcription
•
•
•
DNA consists of many genes.
Each gene has a promoter, coding region and a terminator.
Promoter:
✔ marks the beginning of a gene
✔ determines which strands of DNA is used as a template
• Coding region: base triplet that codes for amino acid
• Terminator: marks the end of a gene
Learning Outcomes :
6.3 c) Explain transcription and the stages involved in the formation of mRNA strand
ASSIGNMENT
Stage 1 : Initiation
•
•
•
RNA polymerase binds to promoter on DNA.
It moves along the DNA in the direction of transcription.
It catalyses the unwinding of the double helix DNA.
Learning Outcomes :
6.3 c) Explain transcription and the stages involved in the formation of mRNA strand
ASSIGNMENT
Stage 1 : Initiation
•
•
•
RNA polymerase catalyses the breaking of hydrogen
bonds between complementary nitrogenous bases.
Both DNA strands are separated.
One of DNA strand is used as a template for transcription.
Learning Outcomes :
6.3 c) Explain transcription and the stages involved in the formation of mRNA strand
ASSIGNMENT
Stage 2 : Elongation
•
•
RNA polymerase catalyses addition of nucleotides in the
direction of 5’ to 3’ end.
RNA polymerase catalyses addition of free ribonucleotides
which are complementary to the bases on the DNA template
strand.
Learning Outcomes :
6.3 c) Explain transcription and the stages involved in the formation of mRNA strand
ASSIGNMENT
Stage 2 : Elongation
•
•
Uracil pairs with Adenine (on DNA). Cytosine pairs with Guanine.
DNA-RNA hybrid is temporarily formed.
Stage 2 : Elongation
•
•
ASSIGNMENT
Newly synthesised mRNA (upstream) detaches from DNA template.
DNA double helix re-forms / rewinds, leaving only few DNA-RNA
hybrid regions.
ASSIGNMENT
Transcription: Stage 2 (Elongation)
Learning Outcomes :
6.3 c) Explain transcription and the stages involved in the formation of mRNA strand
ASSIGNMENT
Stage 3 : Termination
•
•
•
•
Elongation proceeds until RNA polymerase reaches termination
site.
Terminator sequence gives signal to RNA polymerase to stop
adding nucleotides to the mRNA strand.
RNA polymerase detaches.
mRNA molecule is released.
Learning Outcomes :
6.3 c) Explain transcription and the stages involved in the formation of mRNA strand
ASSIGNMENT
Stage 3 : Termination
•
•
•
•
•
RNA polymerase detaches.
mRNA molecule is released from DNA template.
Double stranded DNA rejoins to form double helix again.
In prokaryote, mRNA doesn’t need any modification.
In eukaryote, further modification is needed; thus after
transcription, its mRNA is called as pre-mRNA.
Learning Outcomes :
6.3 c) Explain transcription and the stages involved in the formation of mRNA strand
Transcription
A single gene can be
transcribed
simultaneously by several
molecules of RNA
polymerase,
• increases the number
of mRNA molecules
• allows a cell to
produce specific
protein in large
amount
ASSIGNMENT
Learning Outcomes :
6.3 c) Explain transcription and the stages involved in the formation of mRNA strand
ASSIGNMENT
Differences between DNA Replication & Transcription
DNA Replication
Transcription
Both DNA strands act as template Only one strand act as a template
Whole DNA genome is copied
Only a small part of DNA genome is
copied
Produce two molecules of (double Produce a single stranded mRNA
stranded) DNA
strand
Nitrogenous bases used is
guanine, adenine, cytosine and
thymine
Nitrogenous bases used is guanine,
adenine, cytosine and uracil
RNA primer is required
RNA primer is not required
Enzyme involved is DNA
polymerase
Enzyme involved is RNA
polymerase
Only occurs during S phase of
interphase
Occurs throughout the life of a cell
Learning Outcomes :
6.3 c) Explain transcription and the stages involved in the formation of mRNA strand
ASSIGNMENT
mRNA Modification (RNA Splicing) in Eukaryotes
•
•
•
In eukaryotes, both coding sequences (exons) and noncoding sequences (introns) are found along a DNA.
Coding sequences are not continuous.
Introns are found between coding regions (exons).
Learning Outcomes :
6.3 c) Explain transcription and the stages involved in the formation of mRNA strand
ASSIGNMENT
mRNA Modification (RNA Splicing) in Eukaryotes
•
•
Pre-mRNA contains exon & intron regions.
Process of removing the introns and recombining all the
coding exon regions is RNA splicing (in nucleus).
Learning Outcomes :
6.3 c) Explain transcription and the stages involved in the formation of mRNA strand
RNA Splicing in Eukaryotes
•
•
•
In RNA splicing, the introns
are cut out from pre-mRNA
by spliceosome
Spliceosome is made up of
proteins and small RNA
Spliceosome binds to
specific nucleotide
sequence along an intron
ASSIGNMENT
Learning Outcomes :
6.3 c) Explain transcription and the stages involved in the formation of mRNA strand
RNA Splicing in Eukaryotes
•
•
•
•
The intron is cut out and
released.
The exons that flanked the
intron together are joined.
Functional mRNA is
produced.
mRNA moves from
nucleus into cytoplasm
(ribosome) to be
translated.
ASSIGNMENT
LECTURE
Learning Outcomes :
6.3 b) Introduce codon and its relationship with sequence of amino acid using genetic code table
Properties of Genetic Code
Functional mRNA
5’
3’
A U G C A U U U G C GA G A C U A A
1st Codon
•
•
2nd Codon 3rd Codon 4th Codon
5th Codon
6th Codon
3-base sequence of mRNA (triplet code) is called codon.
1 codon on mRNA codes for 1 amino acid.
LECTURE
Learning Outcomes :
6.3 b) Introduce codon and its relationship with sequence of amino acid using genetic code table
1st base
(5’ end)
U
2nd base (5’ end)
U
UUU
UUC
UUA
UUG
C
Phe
Leu
UCU
UCC
UCA
UCG
C
CUU
CUC
CUA
CUG
CCU
CCC
Leu
CCA
CCG
A
AUU
AUC
AUA
AUG
ACU
ACC
ACA
ACG
G
GU U
GU C
GU A
GU G
Ile
Met/star
t
Val
GC U
GC C
GC A
GC G
A
Ser
Pro
Thr
Ala
3rd base
(5’ end)
G
UAU
Tyr
UAC
U A A STO
STO
UAG P
P
CAU
His
CAC
CAA
Gln
CAG
U GU
Cys
U GC
U G A STO
U G G P Trp
U
C
A
G
C GU
C GC
C GA
C GG
U
C
A
G
AAU
AAC
AAA
AAG
A GU
A GC
A GA
A GG
GA U
GA C
GA A
GA G
Asn
Lys
Asp
Glu
GGU
GGC
GGA
GGG
Arg
Ser
Arg
Gly
U
C
A
G
U
C
A
G
LECTURE
Learning Outcomes :
6.3 b) Introduce codon and its relationship with sequence of amino acid using genetic code table
Properties of Genetic Code
1. Composed of nucleotide
triplets (triplet code)
•
•
•
3 nucleotides in mRNA is
called a codon
If 2 nucleotides per codon
will result in only 42 or 16
possible codon (rejected)
3 bases per codon will
result in 43 or 64 possible
codons
LECTURE
Learning Outcomes :
6.3 b) Introduce codon and its relationship with sequence of amino acid using genetic code table
Properties of Genetic Code
2. Each codon represents 1
specific amino acid
• 20 different amino acids
commonly found in proteins.
• At least 20 different codons.
3. The code is degenerate
• Most amino acids are coded
by more than one codon.
• Only 2 amino acids are
coded by 1 codon.
• 61 codons represent 20
amino acids.
LECTURE
Learning Outcomes :
6.3 b) Introduce codon and its relationship with sequence of amino acid using genetic code table
Properties of Genetic Code
4. Contain start & stop codons
• AUG is the start codon.
• UAG, UAA, UGA are the
stop codons.
• Signals to terminate the
coding sequence (do not
code for any amino acid).
5. Almost universal
• With few exceptions, the
same codon codes for the
same amino acids in all
organisms.
LECTURE
Learning Outcomes :
6.3 b) Introduce codon and its relationship with sequence of amino acid using genetic code table
Properties of Genetic Code
6. Non-overlapping
• Each codon consists of 3 nucleotides and the following codons
are represented by the following triplets.
• Sequence of codon is read continuously from a fixed starting
point & proceed until it reaches the termination point at the
other end
LECTURE
Learning Outcomes :
6.3 b) Introduce codon and its relationship with sequence of amino acid using genetic code table
Properties of Genetic Code
7. Comma-free (commaless)
•
•
There are no commas, or other forms of punctuation within the
coding regions of mRNA.
Codons are read consecutively.
QUESTION
•
•
•
•
•
What is the base sequence of the Codon 1? [1 mark]
What is the anticodon sequence for codon 1? [1 mark]
Name the enzyme involved in the production of Y. [1 mark]
Name X and give its function in protein synthesis. [2 marks]
Give THREE differences between molecules X and Y. [3 marks]
Reference
• Campbell N.A et. al., Biology, 11th ed. (2018),
Pearson Education Limited. (page 392-397)
• Snustad D. P., Simmons M. J and Jenkins, J. B
Principles of Genetics, (1997), John Wiley & Sons, Inc,
(page 297-299, 303-305)
Learning Outcomes :
6.3 d) Explain translation and the stages involved
LECTURE
Protein Synthesis
1. Transcription
• synthesise mRNA
Process by which genetic information in DNA base
sequence is copied or transcribed into complementary
sequence of mRNA
2. Translation
• synthesise protein
The sequence of nucleotides in mRNA act as a template
and is translated into sequence of amino acids
Learning Outcomes :
6.3 d) Explain translation and the stages involved
LECTURE
Types of RNA
a) mRNA (Messenger RNA)
• single stranded RNA
• carry genetic information
transcribed from DNA in nucleus to
ribosomes for translation in
cytoplasm.
• The triplet bases on the mRNA are
called codon
Learning Outcomes :
6.3 d) Explain translation and the stages involved
Types of RNA
b) rRNA (Ribosomal RNA)
• Major component of ribosome
• Combines with protein to form
ribosomes
LECTURE
Learning Outcomes :
6.3 d) Explain translation and the stages involved
Ribosome
•
•
•
•
Site of protein
synthesis
Has 2 subunits
Small subunits –
contain mRNA
binding site
Large subunits –
contain 3 tRNA
binding site i.e:
⮚ E site
⮚ P site
⮚ A site
LECTURE
Learning Outcomes :
6.3 d) Explain translation and the stages involved
LECTURE
Ribosome
•
•
•
E site (exit site)
P site (peptidyl-tRNA binding site)
A site (aminoacyl-tRNA binding site)
E
Large subunit
Small subunit
P
site
A
site
P
A
Learning Outcomes :
6.3 d) Explain translation and the stages involved
Types of RNA
c) tRNA (Transfer RNA)
•
•
•
•
•
Single strand RNA (~70-80
nucleotides long).
It folds back on itself to form a
compact 3D shaped.
Forms a cloverleaf shape, held
by hydrogen bonds between
complementary bases.
The middle loop contains 3
bases called anticodon.
The protruding 3’ end (5’ CCA
3’ ending) act as attachment
site of a specific amino acid.
LECTURE
Learning Outcomes :
6.3 d) Explain translation and the stages involved
ASSIGNMENT
Activation of Amino Acids
•
•
•
mRNA codon carries the code for amino acids.
Anticodon (of tRNA) has a specific complementary binding
sequence for the codon (of mRNA).
tRNA that binds to the codon in mRNA must carry the amino
acid encoded by that codon to the ribosome to ensure
accurate translation.
Learning Outcomes :
6.3 d) Explain translation and the stages involved
ASSIGNMENT
Activation of Amino Acids
•
•
tRNA will bind with the amino acid that matches the code for
the correct mRNA codon.
Each kind of tRNA binds to a specific amino acid.
Learning Outcomes :
6.3 d) Explain translation and the stages involved
Activation of Amino Acids
Activation is a process in which a
specific amino acid is attached to a
tRNA molecule
• to form aminoacyl-tRNA,
• catalysed by aminoacyl-tRNA
synthetase.
• ATP is used as energy source.
• There are 20 different
aminoacyl-tRNA synthetases.
• tRNA is re-activated many
times.
ASSIGNMENT
Learning Outcomes :
6.3 d) Explain translation and the stages involved
Activation of Amino Acids
•
•
Amino acids are linked
to their respective tRNA
molecules by the
formation of ester bond,
at the 3’end of tRNA.
ASSIGNMENT
Learning Outcomes :
6.3 d) Explain translation and the stages involved
Activation of Amino Acids
ASSIGNMENT
Learning Outcomes :
6.3 d) Explain translation and the stages involved
Translation
•
The process involves are :
Step 1: Initiation
Step 2: Elongation
Step 3: Termination
ASSIGNMENT
Learning Outcomes :
6.3 d) Explain translation and the stages involved
ASSIGNMENT
Step 1 : Initiation
•
•
•
•
Small ribosomal subunit binds to mRNA.
It binds to the specific nucleotide sequence just upstream
of the start codon, AUG (in bacterial cell).
Initiator aminoacyl-tRNA that carries methionine (met)
binds to the start codon (held by hydrogen bonds).
The sequence of start codon (5’AUG3’) is complementary
with anticodon (3’UAC5’) found in initiator tRNA.
Learning Outcomes :
6.3 d) Explain translation and the stages involved
ASSIGNMENT
Step 1 : Initiation
•
•
•
•
Then, large ribosomal subunit attaches with the initiation complex
Completing the translation initiation complex.
Protein called initiation factors are needed to assemble all these
components together.
Energy is provided by the hydrolysis of GTP.
Learning Outcomes :
6.3 d) Explain translation and the stages involved
ASSIGNMENT
Step 1 : Initiation
•
•
•
Initiator tRNA is located in the P site of the ribosome.
A site is empty.
Available for the next aminoacyl-tRNA to enter (that has a
complementary sequence of anticodon to the second codon on
mRNA).
Learning Outcomes :
6.3 d) Explain translation and the stages involved
ASSIGNMENT
Step 2 : Elongation
•
In the elongation stage, amino acids are added one by
one to the previous amino acid at the C-terminus of the
growing chain..
Learning Outcomes :
6.3 d) Explain translation and the stages involved
Step 2 : Elongation
•
Each addition involves a
3-step cycle:1) Codon recognition
2) Peptide bond formation
3) Translocation
•
Energy is supplied by
hydrolysis of GTP in
steps 1 & 3
ASSIGNMENT
Learning Outcomes :
6.3 d) Explain translation and the stages involved
ASSIGNMENT
Step 2 : Elongation
•
During codon recognition, aminoacyl-tRNA with anticodon
which is complementary to the mRNA codon enters to the
empty A site.
Learning Outcomes :
6.3 d) Explain translation and the stages involved
ASSIGNMENT
Step 2 : Elongation
•
Peptide bond is formed between the carboxyl group of
amino acid in the P site & the amino group of the new
amino acid in A site.
Learning Outcomes :
6.3 d) Explain translation and the stages involved
ASSIGNMENT
Step 2 : Elongation
•
•
Catalysed by peptidyl transferase, a large subunit rRNA
that acts as enzyme, thus called ribozyme.
This step removes the growing polypeptide from the tRNA
in the P site and binds it to the new amino acid on the
tRNA in the A site by a peptide bond.
Learning Outcomes :
6.3 d) Explain translation and the stages involved
ASSIGNMENT
Step 2 : Elongation
•
•
Ribosome moves a codon ahead from 5’ to 3’ direction.
tRNA in the A site is now translocated to the P site.
•
This is called
translocation.
Initiator-tRNA is now
in E site, exits from
ribosome.
A site is now empty.
•
•
Learning Outcomes :
6.3 d) Explain translation and the stages involved
Step 2 : Elongation
•
•
•
•
The 3rd aminoacyl-tRNA
enters A site.
Peptide bond is formed
between 2nd & 3rd amino
acids.
The growing polypeptide
chain is released from tRNA
in the P site,
and binds to the amino acid
attached to the tRNA in the
A site by a peptide bond.
ASSIGNMENT
Learning Outcomes :
6.3 d) Explain translation and the stages involved
ASSIGNMENT
Learning Outcomes :
6.3 d) Explain translation and the stages involved
Step 2 : Elongation
ASSIGNMENT
Learning Outcomes :
6.3 d) Explain translation and the stages involved
ASSIGNMENT
Step 2 : Elongation
•
•
Amino acids are added one by one to the previous amino
acid.
The whole process is repeated until all the codons of
mRNA is translated.
Learning Outcomes :
6.3 d) Explain translation and the stages involved
ASSIGNMENT
Step 3 : Termination
•
•
•
•
Elongation continues until a stop codon in the mRNA reaches
the A site.
The stop codons (UAG, UAA, UGA) do not code for amino
acids; thus no tRNA will bind to a stop codon.
Stop codon acts as a signal to stop translation.
A release factor - a protein that recognises the stop codon binds to the stop codon in the A site.
Learning Outcomes :
6.3 d) Explain translation and the stages involved
ASSIGNMENT
Step 3 : Termination
•
•
•
•
The release factor causes the addition of water molecule to
the polypeptide chain.
This reaction hydrolyses the bond between the last amino acid
of the polypeptide chain and the tRNA in the P site.
Newly synthesised protein is released.
Ribosomal subunits dissociates.
Learning Outcomes :
6.3 d) Explain translation and the stages involved
Protein Synthesis
ASSIGNMENT
Learning Outcomes :
6.3 d) Explain translation and the stages involved
Polyribosomes Structure
ASSIGNMENT
Learning Outcomes :
6.3 d) Explain translation and the stages involved
ASSIGNMENT
Polyribosomes Structure
•
•
•
A single mRNA strand is
used to make many copies
of polypeptide.
Multiple ribosomes may
attach on the same mRNA
at a time.
It forms a structure called
polyribosomes / polysome.
QUESTION
• Name the stage of translation process shown in the
above figure. [1 mark]
• Which is the first codon used in protein synthesis from
this mRNA? [1 mark]
• What is the anti-codon sequence in tRNA 1? [1 mark]
QUESTION
• tRNA 1 has a 5’ – phosphate end and a 3’ – hydroxyl
end. What is the function of 3’OH end in tRNA 1? [1
mark]
• Name the enzyme that catalyzes the formation of the
peptide bond between amino acids carried by tRNA 1
and tRNA 2. [1 mark]
Learning outcomes
6.4 Gene Regulation & Expression: lac operon
a) Explain the concept of operon and gene regulation.
b) State the components of operon.
c) Explain the components of lac operon and their functions
in E. coli.
a) Explain the mechanism of the operon in the absence and
presence of lactose.
Learning Outcomes :
6.4 a) Explain the concept of operon & gene regulation
LECTURE
lac Operon
•
•
•
Operon model describes the mechanism for the control
of gene expression in bacteria.
Some of the bacterial genes will be expressed only
when it is necessary.
The operon model was proposed by Jacob and Monod
(in 1961) on E. coli
Learning Outcomes :
6.4 a) Explain the concept of operon & gene regulation
LECTURE
Concept of Operon
What is operon?
• Operon is a complete unit of gene expression in prokaryote.
• Operon consist of a group of two or more structural genes
having a related function.
• Plus DNA sequences responsible for controlling them.
Learning Outcomes :
6.4 a) Explain the concept of operon & gene regulation
LECTURE
Concept of Operon
•
•
•
Operon is regulated as a single transcriptional unit by a
single promoter.
These genes are transcribed into a single mRNA.
To allow bacteria to synthesize all enzymes needed to
metabolise a substrate at once if the substrate is
present.
Learning Outcomes :
6.4 a) Explain the concept of operon & gene regulation
LECTURE
Concept of Operon
•
1)
2)
3)
•
Operon consists of:
Promoter
Operator
Structural genes
Operon is controlled by a regulatory gene (upstream
from the operon).
Learning Outcomes :
6.4 a) Explain the concept of operon & gene regulation
ASSIGNMENT
Gene Regulation & Expression (lac Operon)
•
•
•
Francois Jacob & Jacques Monod (1961) conduct an
experiment by using E. coli
E. coli lives in the human’s colon & depends on glucose or
lactose consumed by human.
When both glucose & lactose are present, E. coli prefer to
metabolise glucose.
Learning Outcomes :
6.4 a) Explain the concept of operon & gene regulation
ASSIGNMENT
Gene Regulation & Expression (lac Operon)
•
•
•
Until glucose is depleted;
then only it uses lactose,
but it does not grow
immediately.
After a short delay, it starts
to use lactose as energy
source & grows rapidly on
the glucose medium.
E. coli responds to
changing environmental
conditions.
Learning Outcomes :
6.4 a) Explain the concept of operon & gene regulation
ASSIGNMENT
lac Operon
•
•
•
lac operon is a group of structural genes (3 genes)
together with its promoter and operator, that involves in
the metabolism of lactose which is controlled by regulatory
gene
In lac operon, the 3 genes (that codes for 3 enzymes) are
controlled as a single unit on E. coli
Allow E. coli to synthesize all 3 enzymes needed to
metabolise lactose rapidly.
lac Operon
Regulatory gene
lacI
Promoter Operator
P
O
Structural genes
lacZ
lacY
lacA
Learning Outcomes :
6.4 a) Explain the concept of operon & gene regulation
ASSIGNMENT
lac Operon
•
•
When lactose is absent → lac operon is inactivated.
When lactose is present (glucose absent) → lac operon
is activated → structural genes are transcribed.
lac Operon
Regulator
lacI
Promoter Operator
P
O
Structural genes
lacZ
lacY
lacA
Learning Outcomes :
6.4 b) State the components of lac operon
LECTURE
Components of lac Operon & Its Function
•
Operon consist of:
1) Promoter
2) Operator
3) Structural genes (lacZ, lacY & lacA)
•
Promoter & Operator are binding sites on DNA; they are
not transcribed
lac Operon
Regulator
lacI
Promoter Operator
P
O
Structural genes
lacZ
lacY
lacA
Learning Outcomes :
6.4 b) Explain the components of lac operon & their functions in E. coli
ASSIGNMENT
Components of lac Operon & Its Function
•
•
•
Structural genes are transcribed into a single mRNA.
mRNA contains coding information for 3 enzymes & is
translated to form 3 separate enzymes.
Regulator gene carries the genetic code to produce
repressor molecule that prevents the structural genes
from becoming active.
lac Operon
Regulatory gene
lacI
Promoter Operator
P
O
Structural genes
lacZ
lacY
lacA
Learning Outcomes :
6.4 b) Explain the components of lac operon & their functions in E. coli
LECTURE
Components of lac Operon & Its Function
•
Operon consists of:
Gene
Function
1. Regulatory gene lacI
Gene that encodes for repressor protein
2. Promoter (P)
Attachment site for RNA polymerase to
start the transcription of structural genes
3. Operator (O)
Attachment site of repressor protein
4. Structural genes lacZ Gene that encodes for β-galactosidase
lac
Y
Gene that encodes for lactose permease
lac
A
Gene that encodes for transacetylase
Learning Outcomes :
6.4 b) Explain the components of lac operon & their functions in E. coli
LECTURE
Function of Proteins Expressed by lac Operon
Protein
Function
1. Repressor protein
Controls whether the operon is activated or
not
2. β-galactosidase
Catalyses the hydrolysis of lactose into
glucose & galactose
3. lactose permease
Transports lactose into the cell efficiently
(increase the permeability of cell membrane
towards lactose)
4. transacetylase
Transfers acetyl group from acetyl co-A to
lactose
Learning Outcomes :
6.4 d) Explain the mechanism of lac operon in the absence and presence of lactose
ASSIGNMENT
Absence of Lactose (And Glucose Absent)
lacI
P
Transcription
5’
3’
Binding site for
RNA polymerase
O
lacZ
lacY
Binding site for repressor
Translation
Repressor
Protein
•
•
•
•
Regulatory gene (lacI) is expressed continuously
(constitutive).
A repressor protein is produced.
When lactose is absent, repressor protein is active.
Repressor protein binds to operator.
lacA
Learning Outcomes :
6.4 d) Explain the mechanism of lac operon in the absence and presence of lactose
ASSIGNMENT
Absence of Lactose (And Glucose Absent)
Binding site for repressor
lacI
RNA polymerase
•
•
•
•
P
O
lacZ
lacY
lacA
Repressor
Protein
Repressor protein blocks RNA polymerase from
transcribing the structural genes.
Causes lac operon to be inactivated (“switched off”).
Transcription of structural genes cannot occur.
Enzymes needed to digest lactose cannot be produced.
Learning Outcomes :
6.4 d) Explain the mechanism of lac operon in the absence and presence of lactose
ASSIGNMENT
Presence of Lactose (Glucose Absent)
Binding site for repressor
lacI
RNA polymerase
•
•
•
P
O
lacZ
lacY
lacA
Repressor
Protein
When lactose is present, a few lactose enter the cells &
converted to allolactose (isomer of lactose).
Allolactose binds to the repressor protein.
Thus, alter the shape of repressor protein (becomes
inactive).
Learning Outcomes :
6.4 d) Explain the mechanism of lac operon in the absence and presence of lactose
ASSIGNMENT
Presence of Lactose (Glucose Absent)
Binding site for repressor
lacI
P
O
lacZ
lacY
lacA
RNA polymerase
Repressor
Protein
•
•
•
Repressor protein can no longer bind to the operator; it
detaches from operator.
RNA polymerase can transcribe the structural genes
(transcription can occur).
Allolactose acts as inducer that induces the transcription
of lac operon.
Learning Outcomes :
6.4 d) Explain the mechanism of lac operon in the absence and presence of lactose
ASSIGNMENT
Presence of Lactose (Glucose Absent)
Binding site for repressor
lacI
P
O
lacZ
lacY
RNA polymerase
lacA
Transcription
3’
5’
•
•
•
•
Transcription of all 3 structural
genes occur,
to synthesize a single mRNA.
Translation occurs to produce
3 enzymes:
β-galactosidase, lactose
permease & transacetylase.
Translation
β-galactosidase
transacetylase
lactose permease
Learning Outcomes :
6.4 d) Explain the mechanism of lac operon in the absence and presence of lactose
ASSIGNMENT
Presence of Lactose (Glucose Absent)
Transcription occur
Lac I
P
O
lac Z
lac Y
lac A
3’
5’
Translation
β-galactosidase
transacetylase
lactose permease
Learning Outcomes :
6.4 d) Explain the mechanism of lac operon in the absence and presence of lactose
ASSIGNMENT
Function of Enzymes
•
•
•
β-galactosidase → to hydrolyse lactose molecule into
glucose & galactose
Lactose permease → facilitate lactose to diffuse into
the cell membrane
Transacetylase → transfer acetyl group from acetyl coA to lactose
ASSIGNMENT
Reference
• Neil A. Campbell, Lisa A. Urry, Michael L. Cain,
Steven A. Wasserman, Peter V. Minorsky & Rebecca
B. Orr (2021), Campbell Biology, 12th ed., Pearson
Education Limited.(page 385-406)
• Solomon E.P., Martin,C., Martin, D.W. & Berg, L.R.
(2019), Biology, 11th ed. Cengage Learning Asia Pte
Ltd.
• Snustad D. P., Simmons M. J and Jenkins, J. B
Principles of Genetics, (1997), John Wiley & Sons, Inc,
(page 297-299, 303-305)