Molecular Biology

Molecular Biology
Aaser Abdelazim
Professor of Medical Biochemistry &Molecular Biology
FAIMER fellow 2021 (Medical Education)
Clinical Chemistry Consultant
aaserabdelazim@yahoo.com
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AASER ABDELAZIM BIOCHEMISTRY& MOLECULAR
BIOLOGY

Molecular Biology
1. Nucleic acid structures and functions (DNA &RNAs)
2. DNA replication
3. Transcription
4. Translation
5. DNA mutations
6. DNA repair
7. Gene expression regulation
8. Applications of Molecular biology
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Molecular organization of the cell
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What is Molecular Biology?
The study of gene function and structure at molecular level.
Gene
Allele
Dominant allele (A)
Recessive allele (a)
Heterozygous
Homozygous Homozygous
Locus

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Gene
Replication Produces
RNAs and proteins
Accumulates mutations
and allows evolution
DNA always be replicated
Transcription and translation
GAG
GTG
(glutamate)
(valine)
Sickle cell anemia
Note:
the genetic material could be DNA (human) or RNA (retroviruese)
dsRNA or ssRNA

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Cell cycle
S
G2
M
G0
G1
Interphase
DNA synthesis and replication
Post replicational stage
Mitosis
Some cells stop at G0
and did not complete
(nondividing cells)
Growth and metabolic activities
(prereplication stage)
Prophase
Metaphase
Anaphase
Telophase
Almost, cells remain
in interphase

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Mitosis
2n
2n
2n
Meiosis
2n
2n
2n
Meiosis I
n
n n
n
Meiosis II
Cell division

NUCLEIC ACIDS
qUnlike carbohydrates, proteins and lipids, nucleic acids are used for storage and expression of
genetic information.
qThere are two types of nucleic acids DNA &RNA.
qEukaryotes: organisms that whose cells contain a limiting membrane around the nucleus e.g
human cells
qProkaryotes: organisms whose cells contain no mitochondria and their DNA not enclosed with a
membrane and does not undergo mitosis.
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NUCLEIC ACIDS
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DNA RNA
Nucleotide is the structure unit of RNA and Deoxynucleotide is the structure
unit of DNA
Nucleotide Base
= + Sugar
+ P
Nucleoside Base
= + Sugar

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Nucleotide Structure:
Nucleotide Base
= + Sugar
+ P
Purines Pyrimidines

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Ribose
Deoxyribose
RNA
Sugar
DNA

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Purines Pyrimidines
Base
DNA and RNA purines
Adenine Guanine
DNA pyrimidines
RNA pyrimidines

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(A) Adenosine –monophosphate (adenylic acid)
(B) Adenosine –diphosphates
(C) Adenosine triphosphates
Nucleotide may be mono, di or tri phosphates
Nucleotide Structure:
(A)
(B)
(C)

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Nomenclature of nucleosides and nucleotides
Base Nucleosides Nucleotides
Adenine (A)
Adenosine Adenosine monophosphate (AMP) /adenylic acid(AA).
Deoxyadenosine
Deoxyadenosine monophosphate (deoxyadenylic acid)
d.AMP
Guanine (G)
Guanosine Guanosine monophosphate (GMP)/guanylic acid (GA)
Deoxyguanosine Deoxyguanosine monophsphate
Cytosine (C) Cytidine Cytidine monophosphate (CMP)/ cytidylic acid
Uracil (U) Uridine Uridine monophosphate (UMP)/uridylic acid (UA)
Thymine (T) Thymidine Thymidine monophosphate (TMP)/ thymidylic acid (TA)
Xanthine (X) Xanthosine Xanthosine monophospahte
Hypoxanythine (I) Inosine Inosine monophosphate

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Functions of nucleotides
Base Nucleotide Functions
Adenine
ATP
High energy phosphate compound give about 7000 calories. So it is the stored form of the energy inside the
cell. It is called the energy currency of the cell and used for
•Muscle contraction
•Nerve conduction
•Absorption of nutrients and their excretion.
•Activation of some compounds in the body like glucose.
cAMP •Synthesized mainly from ATP and used as a second messenger for some hormones (see hormones part).
SAM
It is called active methionine and perform the major role in the transmethylation reactions inside the cell
(see protein part).
Adenosine 3-
phosphate 5-
phosphosulfate (APPS)
It is called the active sulfate and used as sulfate donor inside the cell (see protein part).
Guanine
GTP In most cases acts like ATP as a source of energy.
cGMP Acts as a second messenger for hormones as ANF (see hormones part).
Cytosine CTP It is required for synthesis of phospholipids (recall the lipids part).
Uracil UDP
Attached to glucose and galactose during synthesis of glycogen and glycolipids.
UDP-glucouronic acid is used for synthesis of glycoproteins and in conjugation reactions.
Vitamins
NAD, NADP, FAD, FMN,
and Coenzyme A.
Recall vitamins part.

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USUAL , UNUSUAL BASES AND SYNTHETIC BASE ANALOGS
[1] Usual bases:
(a) Bases consisted in DNA &RNAs: Are bases that enter in the RNAs and
DNA structure include normal purines and pyrimidines.

(b) Bases present free in cells: it result from oxidation of purines
6
Hypoxanthine
(6-oxopurine)
Xanthine
(2,6-oxopurine)
6
2
Uric acid
(2,6,8-oxopurine)
6
2
8
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(c) Plant bases:
Caffeine (coffee)
(1,3,7 trimethyl xanthine)
Thiophylline (tea)
(1,3, dimethyl xanthine)
Thiobromine (cocoa)
(3,7 dimethyl xanthine)
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[2] Unusual bases: it is a modified bases present in RNAs. The modifications
include methylation, Hydroxymethylation, Glycoslation and alterations of
some atoms.
4-thiouracil 1-Methyl guanine
Pseudouracil(Ψ) 5- methylcytocine
Dihydrouracil (D)
N-6 isopentyl adenine
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[3] Synthetic base analogs: synthetic bases used mainly as treatment agents
Antiviral, antiuric acid and antitumor.
5- Fluorouracil 6- Thioguanine 6- Mercaptopurine
Allopurinol
Azathioprine
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DNA
DNA Eukaryotes Prokaryotes
Site 1. Nucleus
2. Mitochondria
1. Single chromosomes
2. Non chromosomal
DNA (plasmid)
Functions
(DNA four
roles)
1. Replication (cell division)
2. Expression of genetic information
3. Encoding to proteins
4. Mutation and recombination (evolution)
Mitochondrial DNA is used for
- Synthesis of proteins and enzymes inside
mitochondria
- Synthesis of proteins and enzymes of
respiratory chain.
- Detection of persons relation ships.
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[Four roles of DNA]

The following should be noted:
1. 5’ end and 3’ ends
2. DNA backbone structure
3. Bonds (phosphodiester bonds,
glycosidic linkage and hydrogen
bonds).
4. The two strands are antiparallel.
One is called template strand
(3à5) direction and the other is
called coding strand (5à3)
direction.
5. Arrangement of sugar and bases.
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DNA structure
Glycosidic linkage

Double helix
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ü Base from purines paired
with a base from
pyrimidine.
ü Adenine (A) paired with
thymine (T) with two
hydrogen bonds.
ü Guanine (G) paired with
cytosine (C) with three
hydrogen bonds.
ü Exposure of the two
strands to heat with
denture the two strands.
(DNA melting or DNA
denaturation).
ü Under proper conditions
they can come back
together again (DNA
reannealing or DNA
renaturation )
Base pairing role

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DNA renaturation
Factors Description
(1) Temperature The optimum temperature for renaturation of DNA is 25C.
The temperature is low enough to does not promote denaturation but
high enough to allow diffusion the two strands.
(2) DNA concentration The higher the concentration the higher the reannealing.
(3) Renaturation time The longer the time allowed for reannealing the more will occur.
Term (C0t) is used to describes the relationship between the
concentration of DNA (moles of nucleotides/L) in a given time per
seconds

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Problem 1
In the given figure the C0t curve of different DNAs. The blue curve
represents the renaturation of phage (lambda ⑁ DNA) and the red curve
represents the renaturation of DNA form nuclear polyhedrodsis virus
(NPV).
a)What are the C0t1/2 of the two DNAs?
b)Given the size of lambda ⑁ DNA is 50 kb, what is the size of NPV DNA?
(assuming no repetitive sequences in either DNAs and an equivalent GC content in both).
Solution 1
a)To calculate C0t1/2 draw a line from the midpoint of axis (y) till reach the
point on of lambda ⑁ DNA and NPV curve. Then read the point on axis (X)
which will be the C0t1/2 respectively it will be (0.08 for ⑁ DNA and 0.22 for
NPV DNA).
b)The size of NPV DNA = 0.22x50/0.08 = 138 kb
NOTE: C0t: is the product of initial DNA concentration
(C0) in moles of nucleotides/ Litre at time (t) in seconds.
0.001 0.01 0.1 1.0
80
60
40
20
0
⑁ DNA
NPV DNA
y
X
Percent
of
reassociation
C0t

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DNA forms
A form B form Z form

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Differences among DNA forms
Items A form B form Z form
Helix type Right –handled Right-handled Left-handled
Helix diameter
(nm)
2.55 2.37 1.84
Distance for each
complete turn
(nm)
3.2 3.4 4.5
Number of base-
pairs /turn
11 10.5 12
Rise of base pair 0.29 0.34 0.37
Sugar pucker
conformation
C3’ end C2’ end
C2’ end for
pyrimidines and C3’
end
For purines
Glycosyl bond
conformation
Anti Anti
Anti for pyrimidines
and syn for purines
Major groove Narrow deep Wide and deep Flat
Minor groove Wide and shallow Narrow and shallow Narrow and deep

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Problem 2
Phage P1 has a double stranded DNA with 91500 bp (91.5 kb)
a)How many full double-helical turns does this DNA contain?
b)How long is the DNA in microns (1 micron = 104 ºA)?
c)What is the molecular mass of this DNA?
d)How many phosphorus atoms does the DNA contain?
Solution 1
a) # of turns = total number of bases/10.5 (number of bases in each turn). = 91500/10.5 = 8714 turns.
b) The space between each base in DNA = 3.4 Aº or 3.4 x 10-4 µm so the long of DNA will equals Total number of bp x
3.4 x 10-4 = 91.500 X 3.4 x 10-4 = 31 µm.
c) Each base-pair has a molecular mass = 660 Daltons (D) so the molecular mass of this DNA = the total number of
bp x 660 = 91500x660= 60390000 Daltons.
d) Each base-pair has two P atoms (one in each strand) so the # of P atoms – total bp x 2 = 91500x2= 183000 P
atoms.

DNA organization
Chromosome
They are nucleoproteins formed from DNA and basic proteins, they are 46 in number and
act as the functional units of heredity.
Nucleosome Histones with DNA resemble beads on string
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Nucleosome structure

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Chromatin / Chromatid / Chromosome
Chromatin
DNA wrapped around
proteins
At the beginning of cell division,
DNA producing two identical
copies of DNA connected to
each others at Centromeres- this
X-like structure called a sister
Chromatid pair. So the
Chromatid is just one of these
strands
During mitosis the sister
chromatid pair condenses
further to give the fat X
chromosome therefore
chromosome can be found in
three forms 1- thread like
chromatin 2- thread like sister
chromatid 3- condensed visible
form.
Chromatid Chromosome
DNA
Centromere
Condensed DNA

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Items Euchromatin Heterochromatin
Staining Light ( less intense)
colored bands
Dark (more intense)
colored bands
Presence Eukaryotes and
prokaryotes
Only in eukaryotes
Packing Light packed Tight packed
Gene
concentrati
on
Rich in gene conc. And
under active transcription
Poor in gene
% 92% 8%
Location In the middle of the
nucleus
In the periphery of the
nucleus
Types constitutive type as
house keeping gene
1. Facultative
2. Constitutive
Functions 1. Form the
Nucleosome beads
2. Role in RNA
transcription
1. Form
Centromeres and
telomeres.
2. Role in gene
regulation
3. Protects the
chromosome
from
endonuclease.
Chromatin
• Chromatin is the complex between DNA and histones (contains as much twice proteins
as DNA).
Heterochromatin
Centromere
Euchromatin
Telomere
Heterochromatin
Euchromatin

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Junk DNA
ü Junk DNA was a collective label attributed to the portions of the DNA sequence of a chromosome or genome for
which no function has been identified.
ü About 80-90% or more of the human genome had been designated as “junk”, including most sequences within
introns and intergenic DNA.
ü Roles of junk DNA
oCapable to repair broken strands of DNA.
oReservoir for sequences from which useful genes could be emerge (important genetic basis of evolution).
oSource for antifreeze gene in some species of fish.
oPlay some role in the regulation of gene expression and promotion of gene diversity.
oSocial behavior in rodents and possibly in humans was affected by portions of the genetic code once thought to
be junk.
oImportant to the evolutionary survival of an organism.
oPlay specialized roles in cell behavior.
oRegulate protein production and could generate micro RNAs.
oProtective buffer against genetic damage and harmful mutations.

RNAs
RNA Eukaryotes Prokaryotes
Site 1. RNAs that synthesized in nucleus perform their
functions in cytoplasm.
2. RNAs that synthesized in mitochondria
performs their functions in mitochondria.
1. Cytoplasm
Functions 1. Protein biosynthesis (expression of genetic
information).
1. Genetic materials of
some viruses are
single or double
strands RNA.
2. Protein biosynthesis.
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TYPES OF RNA
RNAs for protein synthesis
Type Function Location
mRNA Carry the codes for protein synthesis All organisms
tRNA Transfer amino acids during translation All organisms
rRNA Attached to ribosomes during translation All organisms
RNAs in Posttranstional modifications
SnRNA Splicing of RNA Eukaryotes
Y RNA RNA processing and DNA replication Animals
Telomerase RNA Telomeres synthesis Most eukaryotes
miRNA
microRNA for regulation of gene expression by
cleavage of post transcriptional mRNA
All cells
siRNA
Small interfering RNA regulation of gene expression
by cleavage of post transcriptional mRNA
All tissues
Regulatory RNA
Antisense RNA (a
RNA)
1. Transcriptional attenuation
2. mRNA degradation
3. mRNA stabilization
4. Translation block
All organisms

RNA STRUCTURE
1. Coding region
2. Non coding region
3. 5’ Cap>>>>> 7 methyl-guanzine (Eukaryotes)
4. 3’ Tail >>>>>> polyadenine tail (Eukaryotes)
5. Constitutes only about 5% of total RNAs
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(1) mRNA
Structure

1. 4 loops (D loop, TΨC loop, variable loop and
anticodon loop)
2. Base pairing
3. Amino acid acceptor arm
4. Clover leave shape.
1. Its molecular weight equals 4 S (smallest)
2. Its length about 74-95 Nucleotides.
3. Each amino acid have its special tRNA(s).
4. 15% of total RNAs.
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TΨC loop
Anticodon loop
Amino acid acceptor arm
Variable loop
(extra arm)
D loop
D: dihydrouracil
Ψ (Epsi): pseudouracil
T: thymine
Structure
(2) tRNA
Characters

The characters
1. Synthesized as one strand then cleaved.
2. Associated with proteins to form ribosomes.
3. Factory for protein synthesis.
4. 80% of total RNAs.
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(3) rRNA
Eukaryotes rRNA Prokaryotes rRNA

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miRNA
MicroRNA: small single strand RNA non-coding RNA molecule.
Functions
- miRNAs play important roles in gene regulation.
- All known miRNAs cause inhibition of gene expression by decreasing specific protein production via distinct
mechanisms.
Structure
miRNAs are single stranded RNA molecules of about 21 to 23 nucleotides in length. They were first described in
1993 by Lee and colleagues. But the term micro RNA was introduced only in 2001.
Formation and Processing
• The genes encoding miRNAs are much longer than the processed mature miRNA molecules.
• miRNAs are first transcribed as primary transcripts or Pre-miRNAs with a cap and a poly-A tail and
processed to short 70 nucleotides stem loop structure known as Pre-miRNA in the cell nucleus. (This processing
is performed in animals by a protein complex known as the microprocessor complex, consisting of the Nuclease Drosha and the double
stranded RNA binding protein Pasha.
• These Pre-miRNAs are then processed to mature miRNAs in the cytoplasm by interaction with the
“Endonuclease dicer”, which also initiates the formation of the RNA induced silencing complex (RISC). [This
complex is responsible for the gene silencing observed due to miRNA expression and RNA interference.
• The small processed mature miRNAs typically hybridize via the formation of imperfect RNA-RNA duplexes
within the 3’-untranslated regions (3’UTR) of specific target m-RNAs leading via unknown mechanisms to
translation arrest.

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miRNA gene
Transcription
Pre-miRNA
Cleavage
• Nuclease Drosha
• Protein Pasha
Pre-miRNA
NUCLEUS
Nuclear export
Cleavage
(DICER)
CYTOPLASM
Duplex miRNA
RISC
Mature miRNA
RISC complex
FORMATION AND PROCESSING OF MIRNA

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siRNA
ü Si-RNAs also play an important role in gene regulation.
ü Si-RNAs are derived by the specific nucleolytic cleavage of larger double stranded RNAs to produce small
(21 to 23 nucleotides in length) products from the precursors.
ü These short Si-RNAs usually form perfect RNA-RNA hybrids with their distinct targets potentially
anywhere within the length of the m-RNA where the complementary sequence exists.
ü Formation of such duplexes Si-RNA and m-RNA results in reduced specific protein production because the
Si-RNA-m-RNA complexes are degraded by specific nucleolytic machinery; some or all of these m-RNA
degradation occurs in specific cytoplasmic organelles termed “p-bodies”.

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Mechanism of action of both miRNA and siRNA
Pairing to target mRNA
siRNA
miRNA
Target mRNA
Target mRNA

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Items miRNA siRNA
Synonyms microRNA Small interfering RNA
Origin Endogenous (from organism genome) Exogenous
(viral RNA, transposons, heterochromatin)
Existence Higher animals (mammals) and plants Lower animals and plants
Structure ssRNA (19-25 nucleotides) dsRNA (21-23) nucleotides
Complementary to
target mRNA
Partially complementary to target mRNA
(imperfect pairing)
Fully complementary to target mRNA
(perfect pairing)
Target mRNA Each miRNA has many mRNA
miRNA----à100 mRNAs
Every siRNA is specific for one mRNA
Mechanism of action Endonucleolytic cleavage through
formation of RISC complex
Endonucleolytic cleavage through formation
of RISC complex
Clinical application • Therapeutic drug development
• Has a role to decrease the development
of some types of cancer.
• Animals deficient in miRNA has a risk to
develop some heart diseases.
• Serve as a biomarker (diagnosis)
• Decrease or (knock down) specific protein.
• Killing cancer cells
Differentiation of miRNA and siRNA

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Comparison between DNA and RNA
Items DNA RNA
(1) Structure
•Sugar Deoxyribose Ribose
•Bases A, G, C & T A, G, C& U
•Unusual bases Not present Present
•Strands
Double strands
(except in some viruses
with single strand DNA
ssDNA )
Single strand
(except in some
viruses with double
strands RNA dsRNA )
(2) Site of synthesis Nucleus Nucleus
(3) Site of action
Nucleus and
mitochondria
Cytoplasm
(4) Types only one type
Three main types
(tRNA, mRNA& rRNA )
(5) Functions
1. Gene expression
2. Storage of genetic
information.
Synthesis of proteins

DNA REPLICATION
Introduction:
a) Two copies of genetic materials should be done before each cell division.
b) DNA is copied by semi conservative method this means each of daughter DNA molecules will
contain one parent strand and one new strand.
c) During replication double stranded DNA should be separated into separate two strand; one of them
used as template for synthesis of a new complementary strand.
d) The process is catalyzed by enzyme called DNA polymerase.
e) DNA synthesis in higher organisms is less well understood, but involves the same types of
mechanisms.
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Central dogma of molecular biology
We are here…..

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Steps of DNA replication/synthesis:
A. Separation of the two complementary DNA strands
qThe two strands of DNA should be separated before replication.
qIn prokaryotes: the separation occur at a single site called the origin of replication which
is rich in AT bases.
qIn eukaryotes: the separation occur at multiple sites [ due to huge size of eukaryotes
DNA –this enable rapid replication of human genome].
B. Formation of the replication fork
qThe unwinding of the dsDNA lead to formation what is called replication fork.
qAt which the replication proceed in both directions.

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Origin of replication in prokaryotes (A) and eukaryotes (B)

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1. Proteins required for DNA strand separation
1. Helix- destabilizing proteins
(HD) (Dna A protein)
qBinds to specific sequence at the origin of replication.
qMaintain the two strands of DNA separated in the place of replication.
2. Helix unwinding proteins
(DNA helicase).
qAct as scissors for separation of the two strands of DNA.
qThis function requires ATP (2 ATP are consumed to separate each base
pair)
3. Single strand binding
proteins (SSB)
qBinds to the single strand that generated by helicase.
qMaintain the two strands of DNA separated in the place of replication.
qProtects the ssDNA from the action of nucleases.
3. Topoisomerases qTopoisomerase I {swivelase}:- prevents the super twisting of the two
strands.
qTopoisomerase II {gyrase}:- prevents the rotation of the entire
chromosomes which consume energy
Single strand
binding proteins
(SSB)
Helix unwinding
protein (Helicase)
DnaA proteins

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(A) Topoisomerase I (B) Topoisomerase II
2. Solving the problem of supercoils:
Clinical note
Anticancer agents, such as etoposide, target human topoisomerase II.
Bacterial DNA gyrase is a unique target of a group of antimicrobial agents called quinolones, for example,
ciprofloxacin.

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C. Direction of DNA replication
Leading strand
Lagging strand
DNA polymerase
qDNA polymerases read the template DNA strand in the 3′→5′ direction. But they synthesize
the new strand in the 5′→3′ (antiparallel) direction.
q So the two strand will be synthesized in an opposite direction.
qLeading strand: Copied in the direction of the replication fork and is synthesized
continuously.
qLagging strand: Copied in the direction away from the replication fork, is synthesized
discontinuously, with small fragments of DNA termed Okazaki fragments.
RNA primer
Okazaki fragment

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D. RNA primer
qDNA polymerases unable to start the new strands synthesis with out presence of RNA
primer (approximately 10 nucleotides long) added on the 5 end of the new strand.
qRNA primers are synthesized by primase (specific RNA polymerase) at Primosome.
1. It is common in lagging strand
2. It is anti parallel to template strand.
3. Later will be removed
DNA replication is catalyzed by DNA polymerases
Prokaryotes Eukaryotes Catalytic activity
DNA polymerase I DNA polymerase α 1. Remove RNAs primers and fill gaps
2. DNA repair
DNA polymerase II DNA polymerase ᵋ Proofreading and repair
DNA polymerase ᵦ DNA repair
DNA polymerase ᵧ Mitochondrial DNA synthesis
DNA polymerase III DNA polymerase ᵟ 1. Synthesis of leading strand
2. Synthesis of Okazaki fragments
E. Chain elongation

(1) leading strand
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E. Chain elongation and excision of RNA primer

(2) Lagging strand
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RNA SYNTHESIS
A) Introduction
1. Copies a RNA from DNA is called RNA synthesis or transcription .
2. Transcription process produces the three types of RNA (tRNA, rRNA
and mRNA).
3. The process is catalyzed by RNA polymerase which work due to
signals on the DNA guide it to when starts the transcription, where
and when it stops it.
4. After synthesis of RNA molecules they will undergo many
modifications includes (terminal additions, base modifications,
trimming, and internal segment removal).
5. To start the RNA synthesis 3 major things should be present [1- RNA
polymerase, 2- transcription unit, 3- nucleotides phosphates]
Transcription
DNA
tRNA
rRNA
mRNA

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(1) RNA polymerases (RNAPs)
q In bacteria (prokaryotes) RNA polymerase synthesizes all types of RNA {except RNA
primers for DNA replication}.
qRNA polymerase is multisubunits enzyme
(1) Core enzyme: Formed from Two (α) alpha subunits, one Beta (ᵦ) and one beta(ᵦ').
(2) Holoenzyme : Formed from the core enzyme with sigma factor (σ) and ohm subunit
(Ω)
Subunits Size Function
(α) subunit 36.5 KD Chain elongation and interaction with
regulatory proteins (5′→3′ RNA
polymerase activity)
(ᵦ) subunit 151 KD Chain initiation and elongation (carry the
nucleotides to add to chain elongation)
(ᵦ') subunit 155KD Binds to the DNA template strand.
(σ) subunit 70KD Recognizes the promoter regions on DNA.
(Ω) subunit ---------- Its function is unclear.

(1) Promoter region
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(2) Transcription unit
(A) Prokaryotes
(B) Eukaryotes

1. Specific segment on DNA which control and
increase the transcription.
2. May be up-stream or down stream of the
promoter.
3. Many thousands of base pairs can separate the
enhancer from the gene which it regulate.
4. When specific protein attached it the
transcription will increase e.g., steroid hormones
(3)Transcribed region: that will be transcribed to
RNA.
(4) Termination: located at the end of the transcribed
region.
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(2) Enhancer region

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(3) Steps in RNA synthesis
(1) Initiation Starts by binding of the RNAP holoenzyme to promoter region each subunit
perform their action.
(2) Elongation qStarts after melting the DNA template strand
qRNAP starts to add NTPs with release of Ppi in the 5-3 direction
qSupercoiling also resolved by the action of topoisomerases I &II
(3) Termination Elongation is continue until the termination signals are reached. It may be
1.Spontaneous (intrinsic) - rho (ρ) –independent: look figure – depend on the
formation of hairpin (palindromes).
2.ρ-Dependent termination: needs rho = ρ factor (hexameric adenosine
triphosphatase (ATPase) with helicase activity. It binds a C-rich “rho recognition
site” near the 3′-end of the nascent RNA and, using its ATPase activity to pause
RNA polymerase at the termination site. The ATP-dependent RNA-DNA helicase
activity of rho separates the RNA-DNA hybrid helix, causing the release of the RNA.
Summary of the steps

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Termination of RNA synthesis
1. rho (ρ) –independent 2. rho (ρ) –dependent
ρ protein with ATPase
activity binds to C rich
site on nascent RNA
qρ protein ATPase activity
pause the RNAP action.
qIts helicase activity releases
RNA from DNA-RNA hybrid
RNA polymerase
DNA template strand
Nascent RNA

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Items DNA polymerase RNA polymerase
Role DNA synthesis (replication) RNA synthesis (transcription)
Nucleotides Add dNTPs and releases Ppi. Add NTPs and release Ppi.
Direction of its action 5ʹ→3ʹ direction 5ʹ→3ʹ direction
Proofreading Has a proofreading activity Have not proof reading activity
Primers Need RNA primer No need for primers
Its Termination of
action
Depends on the completion of
all genome replication
Depends on it self or presence
of (ρ) factor
Differences between DNA polymerases and RNA polymerases

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Action of antibiotics on RNA synthesis in prokaryotes
Antibiotic Action
(1) Rifampin q inhibits the initiation of transcription by binding to the β
subunit of prokaryotic RNA polymerase, thus interfering with
the formation of the first phosphodiester bond .
q Rifampin is useful in the treatment of tuberculosis.
(2) Dactinomycin
(actinomycin D)
q The first antibiotic to find therapeutic application in tumor
chemotherapy.
q It binds to the DNA template and interferes with the
movement of RNA polymerase along the DNA.

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TRANSCRIPTION OF EUKARYOTIC GENES
A) Introduction
1. The transcription of eukaryotic genes is more complicated process and different than transcription in
prokaryotes.
2. A number of proteins called transcription factors are involved that bind to distinct sites on the DNA—
either within the promoter region or some distance from it.
3. For transcription factors to recognize and bind to their specific DNA sequences, the chromatin structure
in that region must be altered to allow access to the DNA.
B) Chromatin structure and gene expression
(1) Euchromatin Contains the most active transcribed genes
(2) Heterochromatin Contains the inactive genes
Euchromatin
Heterochromatin
RNA to be transcribed – Heterochromatin should be converted to Euchromatin
(Chromatin remodeling)
Histone acetyltransferase (HAT)
Histone deacetylase(HDAC)

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C) RNA polymerases of eukaryotic cells
qThere are three distinct classes of RNA polymerase in the nucleus of eukaryotic cells.
qAll are large enzymes with multiple subunits.
q Each class of RNA polymerase recognizes particular types of genes.
Enzyme Action
RNA polymerase I Synthesizes the precursor of the 28S, 18S, and 5.8S rRNA(large
ribosomal RNA) in the nucleolus.
RNA polymerase II 1. Synthesizes the precursors of mRNA which translated to
produce proteins.
2. Synthesizes certain small nuclear RNAs, snRNA.
3. Recognizes the promoter regions through the action of many
transcription factors (TFII).
4. Inhibitors of RNA polymerase II: This enzyme is inhibited by α-
amanitin—a potent toxin produced by the poisonous
mushroom Amanita phalloides (sometimes called “death cap”
or “destroying angel”). α-Amanitin forms a tight complex with
the polymerase, thereby inhibiting mRNA synthesis and,
ultimately, protein synthesis.
RNA polymerase III The enzyme produces the small RNA, including tRNA, 5S ribosomal
RNA, and some snRNA.
Mitochondrial RNA
polymerase
Mitochondria contain a single RNA polymerase that resembles
bacterial RNA polymerase .
Eukaryotes RNA polymerases (RNAPs)

(1) Eukaryotes mRNA
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Posttranscriptional modifications of eukaryotes mRNA (pre-mature mRNA is called heteronuclear RNA (hnRNA)
(1) 5’ capping Guanyl transferase adds
7-methylguanozine 5 to 5 triphosphate linkage
Function
1.Protection
2.Facilitate initiation of translation
(2) Poly A tail -Poly A polymerase E adds poly adenine tail at the 3’ end. Function
1.Protection
2.Stabilization
3.Facilitate its exit from nucleus
(3) Splicing Exons – introns
Occur in spliceosomes
Spliceosomes formed from
1.hnRNA
2.5 nuclear RNAs(U1,2,5,U4/U6
3.More than 50 proteins
Function
1.Splicing
2.Transportation of mature RNA to cytoplasm

Transcripted as large one molecule 45S in nucleus.
Its modifications include
1. Methylation
2. Cleavage by specific Exonuclease
(2) rRNA
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Eukaryotes Ribosomes
[4 RNAs + 80 proteins]
5,23 +
32 proteins
16 S RNA
+21 proteins
Large
50s
Small 30s
Prokaryotes Ribosomes
[3 RNAs + 53 proteins]
5,5.8,28 +
50 proteins
18 s RNA
+30 proteins
Large
60s
Small 40s
Association of rRNA with ribosomes
Eukaryotes rRNA
[5,5.8, 18, 28]
Prokaryotes rRNA
[5, 16, 23]

Modifications are:
1. Reduction in size by ribonuleases
2. Attachment of CCA at 3’ end
3. Methylation of some bases
4. Removal of single intron (10-40 bases) present near anticodon loop.
(3) tRNA
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Reverse transcription
ssRNA
RT
cDNA
DNAP
dsDNA
Rnase H
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q it is the conversion of single strand RNA to double strands DNA.
qThe process is catalyzed by enzyme called reverse transcriptase.
qReverse transcription is involved in:
1. The replication of retroviruses, such as human immunodeficiency virus (HIV).
2.Transposons, DNA elements that can move about the genome In eukaryotes, such
elements are transcribed to RNA, the RNA is used as a template for DNA synthesis by a
reverse transcriptase encoded by the transposon, and the DNA is randomly inserted into
the genome.

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Points Transcription in prokaryotes Transcription in eukaryotes
Catalysis (RNAPs) Catalyzed by one RNAP
only. (see transcription in
prokaryotes).
Catalyzed by 4 types of
RNAPs (see transcription in
eukaryotes).
Promoter regions Formed from TTGACA box
and pribnow box (TATAAT)
Formed from CAAT box and
Hogness box (TATA)
Transcription factors Not involved Involved
Enhancers Uncommon Common (steroid hormones)
Complications Less complicated More complicated
Synthesized RNAs Active Inactive
Posttranscriptional
modifications
Not occur Common occurring
Effect of antibiotics 1. Rifampin binds the B-
subunit of RNAP.
2. Dactinomycin interferes
with the movement of
RNAP on the DNA
template.
Dactinomycin interferes with
the movement of RNAP on
the DNA template.
Reverse transcription Involved in the replication of
RNA viruses like retroviruses
(HIV).
Involved in the DNA elements
that move along the genome
(transposons).
Differentiations between transcription in prokaryotes and eukaryotes

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Problem 3
The given sequence is for template DNA fragment.
5-ATGCCGTGACTAATTCG-3
a) Write the sequence of this DNA fragment in conventional double –strands form?
b) Assuming the transcription of this DNA begins with the first nucleotide and ends with the last, write
the sequence of the transcript of this DNA in conventional form (5à3).
Solution 1
a) The template strand given here is written (5à3) direction. Template should written in (3à5) so it will be
3-GCTTAATCAGTGCCGTA-5
Now we can add the coding strand (5à3) direction.
(template) 3-GCTTAATCAGTGCCGTA-5
(Non-template or coding) 5-CGAATTAGTCACGGCAT-3
b) The transcript should originates from coding strand so the its sequence will be the same as coding
and complementary with template with replacement of (T) with (U) so its sequence will be
(transcript) 5-CGAAUUAGUCACGGCAU-3

Where we are now?!!
qDNA is not directly used as a template for protein
biosynthesis.
qA temporary RNA copy is synthesized for direct
synthesis of proteins this is called central dogma.
qTo understand protein synthesis we should
study GENETIC CODE
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TRANSLATION
(Protein biosynthesis)
AUGAGUAACGCG
ATGAGTAACGCG
TACT CATT GCGC
MetSerAsnAla

A collection of sequences (codons) on DNA which translated to proteins(# of codon ).
3 letters sequence on mRNA
Code
Codon
C. Character of genetic code
B. Is there is a difference between code and codon ?!
A. Introduction
The genetic information of the organism is contained in 3 letters sequence of bases called CODONS the collection of
codons make up what is called the GENETIC CODE.
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Characters Description
(1) Specificity
Each codon always codes for only one specific amino acid e.g. UUA codes only for Leucine and not for any other
amino acid.
(2) Degeneracy
(synonyms)
One amino acid can be coded by more than one codon e.g. Valine is coded by four codons, Leucine is coded by six
codons.
(3) Non- overlapping
the codons are not overlapped each others and comma less e.g. 5’-AUAAGAAAAUCA-3’ to read this start always
from 5’ ends to 3’ ends with out any punctuation. AUA AGA AAA UCA.
(4) Universality
Almost the genetic code is universal for all organisms (plants, animals) while mitochondria shows some
differences as AUA codes for Methionine instead of Isoleucine and UGA codes for Try instead of acting as stop
codon.
Nullomers: they are codons of natural amino acids but not present in some organisms genome. E.g AGA and CGA
are codons of arginine in all organisms, in some bacteria AGA not codes for arginine and if we artficially substitute
AGA for CGA it will be lethal to the organism.

The Genetic Codons
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1. Codon-anti codon recognition
Requirements of translation:
1. Pairing role
2. Wobble hypothesis
qTranslation is started by the linking of tRNA with mRNA.
qFor achieving this link the codons on mRNA should be
recognized by anti codons on tRNA
qThere are two theories explain this recognition:
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Codon anti codon recognition theories
(A) Pairing role
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Traditional base-pairing in first and
second codons

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(B) Wobble hypothesis
Non traditional base pairing between 3’ codon on mRNA and 5’ of anticodon on
tRNA. The importance for this is to accelerate the process of protein synthesis.

2. Charging of tRNA
qAmino acyl- transferase catalyzes
two step reaction results in covalent
attachment of an amino acid to the 3’
end of its corresponding tRNA.
qThis reaction requires ATP.
qBy this the tRNA is named as
charged or activated.
qThe 1st amino acid attached is
usually Methionine.
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NOTE: What we mean by second genetic code?
Second genetic code explain two events (1) how
amino acid residues in a protein determines the
secondary and tertiary structure. (2) the mechanism
of tRNA specificity to its correct amino acyl-tRNA
synthetase for the activation of amino acid during
initiation of translation process.

3. Ribosomes
Small units and large units must be dissociated (separated)
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PROCESS OF TRANSLATION
Steps Description
(I) Amino acid activation Achieved by amino acyl-transferase
(II) Initiation 1. Ribosomal dissociation: 80 S unit is dissociated to 40S and 60 S by action of eIF-3& eIF-1A.
2. Formation of 40S pre initiation complex= ternary complex (eIF-4+mRNA+met-tRNA+40S).
3. 40 S initiation complex: eIF-4F+mRNA+40S pre initiation complex + ATP lead to formation of 40S
initiation complex.
4. Formation of 80S initiation complex: 40S initiation complex+60S +eIF-5+eIF-2+eIF-3+GDP+eIF-1A
lead to formation of 80S with met-tRNA placed in P site of large (80S) ribosomal unit.
(III) Elongation 1. Binding of another tRNA with its amino acid in A site of 80S ribosomal unit.
2. Peptide bond formation: the COOH group of amino acid in P site is transferred to bind to NH2 group
of amino acid in A site. This reaction is catalyzed by peptidyle transferase in 60S unit.
3. Translocation:
q after peptide bond formation; ribosome movers three letters towards 3’ ends of mRNA this is called
translocation and need translocase +eEF-2+GTP.
q As a result of movement of ribosome the following occur (release of uncharged tRNA, transfer of
newly formed peptidyle-tRNA to P site, A site become free to accept another amino acyl-tRNA). Then
process repeated.
q 4 high energy compounds (2ATP+2GTP) are required for each peptide bond formation.
(IV) Termination qTermination occurs when ribosomes meet one of stop codons on mRNA (UAA/UAG/UGA).
qAt this moment releasing factors (eRF-1,2,3) identify stop codons and then release of both peptide and
tRNA and dissociation of 80S into 40S and 60S.
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40S
STEPS OF TRANSLATION
40S
eIF-3
eIF-1A
1
2
eIF-4
Formation of 40S pre-initiation complex
40S
3’
5’
mRNA
M
tRNA
60S
P A
60S
P A
Dissociation of 80S ribosome
[A] INITIATION

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3
eIF-4F
Formation of 40S initiation complex
3’
5’
mRNA
M
tRNA
ATP
40S
4 Formation of 80S initiation complex
eIF-4F
M
3’
5’
mRNA
ATP
40S
P A
eIF-1
eIF-2
eIF-3
eIF-5

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eIF-4F
M
3’
5’
mRNA
ATP
40S
P A
P
5 Binding of another aminoacyl-tRNA to A site
6 Peptide bond formation
eIF-4F
3’
5’
mRNA
ATP
40S
P A M
P
Peptide bond
Peptidyle transferase
C=O[OH]
M [H]NH P
P
O
II
C
M N
I
H
P
[HOH]
[B] ELONGATION

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3’
5’
mRNA
40S
P A
M
P
7 Translocation
GTP
eEF-2
1. Peptidyl-tRNA become in P site
2. A site become free and ready to
accept another amino acyl-tRNA
3. Uncharged tRNA become free.
4. For every peptide bond formed
(2ATP+2GTP) are needed.
3’
5’
mRNA
40S
P A
M
P
A

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3’
5’
mRNA
40S
P A
M
P A
3’
5’
mRNA
40S
P A
M
P
A
The steps will be repeated

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3’
5’
mRNA
40S
P A
3’
5’
mRNA
40S
P A
M
P
A
G
M
P
A

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3’
5’
mRNA
40S
P A
G
M
P
A
3’
5’
mRNA
40S
P A
G
M
P
A

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8 Termination [C] TERMINATION
3’
5’
mRNA
40S
P A
G
M
P
A
STOP
40S
P A
G
M
P
A
3’
5’
mRNA
STOP

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3’
5’
mRNA
40S
P A
STOP
G
M
P
A eRF-1
eRF-2
eRF-3
3’
5’
mRNA
40S
P A
60S
M
P
A
G
H
Polypeptide chain
UAA/UAG/UGA.
NOTE: What we mean by transcriptional
decoding ? It is a mechanism by which
stop codons are used for coding 21st
(selenocysteine) and 22nd (pyrolysine)
amino acids during protein synthesis.

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Problem 4
Bacteriophage T2 DNA contains 2x105 bp
How many genes of average size (encoding proteins of about 40,000 molecular weight) can this phage
contains?
Solution 4
a) The average molecular mass of each amino acid produced form translation = 100 D.
b) The given molecular weight of the protein = 40,000 Daltons
c) The # of amino acids in the given protein = 40,000/100 = 400 amino acids.
d) Each amino acid is coded by 3 bases (codon) so the need number of codons for this protein will be 3x400=1200
bp.
e) The # genes needed to produce such proteins = total bp/1200= 2x105 /1200 = 167 genes

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Inhibitor Mode of action
Rifamycin Inhibits DNA-dependent RNA polymerase by binding its beta-subunit.
Linezolid Prevents the formation of the initiation complex, although the mechanism is
not fully understood
Tetracyclines Block the A site on the ribosome, preventing the binding of aminoacyl -tRNAs.
Aminoglycosides Interfere with the proofreading process, causing increased rate of error in
synthesis with premature termination.
Chloramphenicol blocks the peptidyl transfer step of elongation on the 50S ribosomal subunit in
both bacteria and mitochondria.
Macrolides
&clindamycin
qBind to the 50s ribosomal subunits, inhibiting peptidyl transfer.
qEvidence of inhibition of ribosomal translocation.
qCause premature dissociation of the peptidyl-tRNA from the ribosome.
Quinupristin/
dalfopristin
qAct synergistically, with dalfopristin, enhancing the binding of quinupristin,
as well as inhibiting peptidyl transfer.
qQuinupristin binds to a nearby site on the 50S ribosomal subunit and
prevents elongation of the polypeptide as well as causing incomplete chains
to be released
Fusidic acid Prevents the turnover of elongation factor G (EF-G) from the ribosome.
Puromycin qHas a structure similar to that of the tyrosinyl aminoacyl-tRNA. Thus, it binds
to the ribosomal A site and participates in peptide bond formation,
producing peptidyl-puromycin.
qHowever, it does not engage in translocation and quickly dissociates from
the ribosome, causing a premature termination of polypeptide synthesis.
Streptogramins Causes premature release of the peptide chain
Inhibitors of translation

Post-translational modifications
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Modifications Description
(A) TRIMMING Removal of peptide sequences from the inactive proteins to be active e.g. inactive Pepsinogen to be
active pepsin it should loss some sequences from its amino acid contents.
(B) COVALENT MODIFICATIONS
1. Phosphorylation Addition of Phosphate group to the free OH group of tyrosine/or serine
2. Glycosylation Addition of sugar to free OH of tyrosine and serine
3. Biotinylated enzymes Biotin can be firmly attached to some enzymes as a prosthetic group .e.g lysyl residues of
carboxylases.
4. Hydroxylation OH group is added to proline to form hydroxy proline in some proteins like collagen.
5. Carboxylation Addition of COOH group to the amino acid glutamate in proteins e.g. clotting factors (VII, IX, X)
6. Farnesylated proteins Addition of farnesyl group to cysteine residue

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Covalent modifications of proteins
Covalent
modifications
of proteins
Phosphorylation
Glycosylation
Farnesylated
proteins
Biotin addition
Hydroxylation
Carboxylation

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Covalent modifications of proteins

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Points Translation in prokaryotes Translation in eukaryotes
Link to transcription Simultaneous with
transcription (occur
together)
Both processes are separated
mRNA life span Few seconds to two minutes Few hours to days
Ribosomes 70S in cytoplasm 80S attached to ER
Occurrence At any cell cycle phase At G1&G2 of cell cycle
Initiation factors IF1&IF2 and IF3 Involved 9 initiation factors
(1,2,3,4A,4B,4C,4D, 5 and 6)
Initiation amino acid Formylated methionine Methionine
Initiation site Shine delgarno sequence Kozak sequence
Rate of translation Faster process Slower
Releasing factor Only one RF1 Two (RF1&RF2)
Differentiations between translation in prokaryotes and eukaryotes

Defintion: Change the Deoxynucleotides sequence
Causes:
1.Fault replication
2.Anti-malignant drugs
3.Nitrous compounds
Types:
Point Frame shift
Transition Transversion Addition Deletion
A
T C
G A
T C
G
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DNA MUTATIONS

Effects of point mutation
1. Accepted
2. Partially accepted
3. Non accepted
Pre mature termination
of peptide chain
synthesis.
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C
U A
U C U
Codon for Serine
U A
A
Nonsense
mutation
Termination codon
Codon for Serine
C A
C
Codon for Proline
Silent
mutation
Missense
mutation
POINT MUTATION

Effects Frame shift mutation
1. Garbled translation.
2. Premature termination of peptide chain
synthesis.
3. Addition or deletion of amino acid
to/from the growing peptide chain.
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5- U C A U C C U A U G G C U-3
Ser Ser Tyr gly
5- U C A C C U A U G G C U -3
Ser Pro Met Ala
U Addition of base
Deletion of base
C
5- U C A C U A U G G C U -3
Ser Leu Trp
FRAME SHIFT MUTATION

q DNA variation is an essential factor to evolution (1000-10^6 lesions per day)
q DNA stability is important for the individual (less than 1/1000 mutations are permanent)
q A relatively large amount of genes are devoted to coding DNA repair functions.
q Replication occurs at a rate of approximately 1000 nucleotides per second
Error rate is 1/1,000,000 bases à approximately 1000 mutations every time a cell divides.
1. Heat
2. Metabolic accidents (free radicals)
3. Radiation (UV, X-Ray)
4. Exposure to substances (especially aromatic compounds)
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DNA REPAIR
A. INTRODUCTION
B. SOURCES OF DNA DAMAGE

C. TYPES OF DNA DAMAGE:
Damage Causes
(1) Single base damage 1. deamination of nucleotides
2. depurination of nucleotides
3. oxidation of bases
(2) Two bases alterations Thyamine dimer
(3) Chain breaks Breaks in DNA strands
q Single strand breaks: information is still backed up in the
other strand
q Double strand breaks: no backup and can cause the
chromosome to break up.
(4) Cross linkage -Between bases
-Between DNA and histones
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A. Single strand repair
D. MECHANISMS OF DNA REPAIR
1. Mismatch Repair
• Mismatch = when the wrong nucleotides
are paired
– ex. T with C
• Repair
1. Methylation of the correct strand
2. Special enzymes cut out the DNA and
replace it with the appropriate base pairs.
• GATC endonuclease (cuts out
DNA)
• DNA Polymerase (replaces DNA)
• Ligase (seals it with rest of DNA)
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(1) Base excision repair (repair of cytosine
deamination)
A base-specific DNA glycosylase detects an altered base and
removes it.
Endonuclease cuts phosphodiester bond on 5’ side and
removes sugar phosphate
DNA Polymerase I fills and DNA ligase seals the nick
U
Uracil-N-
glycosylase
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2. Excision repair
1
2
3
4
5

(2) Nucleotide excision repair (repair of thymine
dimmer)
1. A large multi enzyme compound scans the DNA
strand for abnormalities.
2. upon detection a UV-specific endonuclease cuts the
strand on both sides of the damage.
3. Exonuclease removes the oligonucleotide.
4. DNA poly I fill the gaps.
5. DNA ligase legate the two ends.
Effect of thymine dimmer
qPrevents the DNAP from replicating the DNA strand beyond the dimmer
qXeroderma pigmentosum
Autosomal recessive disease resulted from defects in the thymine dimmer
repair
Cause
Absence of UV specific endonuclease.
Symptoms
Patient sensitive to UV with malignant transformation
thymine dimmer can be repaired by PHOTOACTIVATION REPAIR
Use the visible light for the activation of photo activating enzyme
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Pyrimidine dimmer UV- specific
endonuclease
Nick
Nick
Removal of
damaged
oligonucleotide
DNA
polymerase I
Nick
DNA Ligase

A. Double strands repair
Repair Description
(1) Non homologous end-
joining
qonly in emergency situations
qtwo broken ends of DNA are joined together
qa couple of nucleotides are cut from both of the
strands
qligase joins the strands together
(2) Homologous end-joining damaged site is copied from the other chromosome
by special recombination proteins
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If the repair not occur,
what do you expect to be happen ?!!!
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(Premature aging, retarded growth)
(1) Werner syndrome

(2) Xeroderma Pigmentosum
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(3) Cockayne’s Syndrome
Occurrence:
1 per million population
Sensitivity:
ultraviolet radiation (sunlight)
Disorder:
•arrested development, mental retardation,
dwarfism, deafness, optic atrophy, intracranial
calcifications
Biochemical:
defect in nucleotide excision repair (NER)
Genetic:
autosomal recessive.
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(4) Trichothiodystrophy
Occurrence:
1-2 per million population
Sensitivity:
ultraviolet radiation (sunlight) in
subset of patients
Disorder:
sulfur deficient brittle hair, mental
and growth retardation, peculiar
face with receding chin.
Biochemical:
defect in NER
Genetic:
autosomal recessive.
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Introduction
Although all cells derived from one origin , not all produce insulin
Why?
Organism can carry genes codes for >3000 proteins but they not
expressed at the same time
Why?
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REGULATION OF GENE
EXPRESSION

Regulation of gene expression in Prokaryotes
At the level of transcription
Operon theory
Structure of Operon
Structural genes Control genes Regulatory gene
Codes for proteins Site of regulator
protein action
Codes for regulatory
proteins
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Lac operon theory Trp operon theory

Lac operon theory
Glucose Lactose
1. B-galactosidase
2. Lactose
permease
3. Galactose
acetylase
Synthesize
E coli
No need
Proper utilization and
metabolism
It metabolize glucose
efficiently
B-galactosidase Lactose
permease
Galactose
acetylase
Hydrolyze lactose Transports
lactose
inside the
cell
Un
known
function
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z y a
O
P
CAP
i
P
Structural genes
Control genes
Regulatory gene
Repressor
Inducers
Structure of Lac operon
B- galactosidase
Lactose permease
Galactose acetylase
RNAP
Catabolite activating protein/cAMP
Lactose /allolactose
cAMP
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Structure of trp operon
trp A
trp B
trp C
trp D
trp E
trp L
O
P
trp R
Attenuator
element
Inactive
repressor
Repressor mRNA 1. High level of
tryptophan
Tryptophan (trp)
+
Proteins of tryptophan synthesis pathway
+
NOTE
1.High tryptophan level will repress the synthesis of tryptophan through repression of
trp A, B, C, D and E.
2.Low tryptophan level will enhance the synthesis of tryptophan synthetase which
synthesizes tryptophan.
2. Low level of
tryptophan
Attenuated mRNA
+
trp mRNA
Regulatory gene Structural genes
Active
repressor
Blocked operator

Regulation of gene expression in eukaryotes
At level of gene itself
5
At level of transcription
At level of Translation
During RNA processing
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1. Gene loss (Partial loss) during differentiation of cells (RBCs)
2. Gene amplification : methotrexate induce hundreds of FH2 reductase copies
3. Gene movement (gene rearrangement)
4. Modification of bases
(1) At level of gene itself
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e.g.of base modifications
DNA methylation
Cytosine methylation e.g., globin genes more methylated in
nonthyroid cells than in thyroid cells
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1. Histones : nonspecific repressors
2. Inducers : steroid hormones
3. Some gene have more than one promoter
Hormone response element (HRE)
mRNA expression
(2) At level of transcription
Hormone
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1. Eukaryotic mRNA produced in an
immature State called hnRNA then
maturation occurs after addition of 5’
cap and poly A tail.
2. Regulatory mechanisms occurring
during these processes can control
the quantity of protein produced.
(3) During RNA processing
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RNA interference (SiRNA)
1. Double stranded RNA is cleaved by
endonuclease (Dicer) into fragments.
2. siRNA fragments hybridize with target
mRNA forming RNA induced slicing complex
(RISC).
3. RISC cleavages the target mRNA
SiRNA is Considered defense against
retroviruses
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1. Heme: stimulate globin
synthesis by prevents
phosphorylation of eIF-2
2. Interferon: stimulate
phosphorylation of eIF-2
inducing stop of protein
synthesis.
(4) At level of Translation
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Clinical application of Molecular Biology
1. Gene therapy: tumors, hypercholesterolemia, cystic fibrosis etc
2. Synthesis of recombinant proteins and vaccines
3. Diagnosis of neoplastic and infectious diseases
4. Prenatal diagnosis
5. Application in the forensic medicine
6. Preparation of antibodies: used in the pregnancy tests and hormonal assays
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Application Description
(1) In the field of molecular
medicine
1. Gene therapy
2. Easily detection of the diseases
3. Pharmacogenetics: studying how individual genes affected
his response to the drugs , age , diet and environment
Benefits of Pharmacogenetics:
•Creation of more specific to an individual
•Determine the more safe time and dose of a drug
(dosage based on person)
(2) In the field of microbial
genetics
1. Perfect diagnosis of microbial diseases
2. Complete sequencing of the M.Os genome
(3) Risk assessment Detection of risk to environmental pollution, toxins and
radiation
(4) In the field of forensic
medicine
DNA Finger printing:
qDNA analysis in the crimes
qDetermination of paternity
qMatching of organ donors and recipients
Applications of genome research
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124
References:
1)Harpers Biochemistry, 25th edition. RK. Murray, DK. Granner, PA. Mayes and VE. Rodwell, publisher:
Appleton and Lange ISBN: 0-8385-3684-0.
2)Tietz fundementals of clinical chemistry, 5th edition. Carl A. Burits, Edward R. Ashwood Publisher,
W.B saunders company last digit in print number: 987654321?
3)Clinical chemistry, principles, procedures, correlations- 4th edition. ML Bishop, JL. Duben-Engelkrik
and EP. Fody publisher: Lippincott Williams and Wikins ISBN: 0-7817-1776-0.
4)Text book of medical biochemistry (2012), 8th edition, MN Chatterjea and Rana Shinde, Jaypee
brothers medical publishers (p) ltd, New Delhi • Panama City • London.

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