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Text Book of Molecular Biology
Department of molecular biology and biochemistry
Contents
Chapter One
Nucleotides and Nucleic Acids
Chapter Two
The Nucleotide Metabolism
Chapter Three
DNA Biosynthesis and DNA Damage Repair
Chapter Four
RNA Biosynthesis and Posttranscriptional Processing
Chapter Five
Protein Synthesis, Folding and Processing
Chapter Six
Regulation of Gene Expression in Prokaryotes
Chapter Seven
Chapter Eight
Regulation of Gene Expression in Eukaryotes
Recombinant DNA Technology
Chapter Nine
Signal Transducer and Cellular Signal Transduction
Chapter one
Nucleotides and nucleic acids
Section one
The generality of nucleic acids
Ⅰ.The building blocks of nucleic acids are nucleotides
Ⅱ. The nucleotides is composed of base, pentose and phosphate
Bases
There are four bases adenine (A),Guanine (G),cytosine(C) and thymine (T) in
DNA
There are also four bases A, G, C and U (uracil )in RNA.
A and G are pruines ,C,T,and U are pyrimidines.
P118 fig
Pentoses
In RNA the pentose is ribose whereas in DNA it is 2-deoxyribose (b-d-2
P118 fig
Nucleosides
A nucleoside consist of a base covalently bonded by the glycosidic bond to the
1-position of a ribose or 2-deoxyribose. There are ribonucleosides and
deoxynucleosides.
P119 fig
Nucleotides
A nucleotide is a nucleoside with one or more phosphate groups bound
covalently to the 3-，5- or (in ribonucleoside only，the 2-position,)by
phosphoester bond.
There are ribonucleotides and deoxynucleotides.
In the case of the 5-position, up to three phosphates may be attached ,to
form ,for example adenosine 5-triphosphate (ATP) or deoxyadenosine
5-monophoshate (dAMP) etc.
P120 fig 6-1
Cyclic nucleotides
cAMP and cGMP are formed from ATP and GTP respectively . they act as
intracellular second messengers.
P120 fig 6-1
Ⅲ. Nucleic acids
Deoxynucleotides or ribonucleotides respectively are joined into a polymer by
the covalent linkage of a phosphate group between the 5-hydroxyl of one
deoxyribose (or ribose) and the 3-hydroxyl of the next.
This kind of bond or linkage is called a phosphodiester bond, since the
phosphate is chemically in the form of a diester.
Polydeoxynucleotides and polyribonucleotides are nucleic acids . the former
are known as DNA, and the latter RNA.
A nucleic acid chain can be seen to have a direction. Any nucleic acid chain,
unless it is circular, has a free 5-end and a free 3-end.
The 5-end to 3-end arrangement of the nucleotide (or base) in a nucleic acid
chain is called the sequence of the nucleic acid chain.
The sequence of a nucleic acid chain is the primary structure of the nucleic
acid chain.
There is a convention to whrite the sequence with the 5-end at the left.
A stretch of DNA sequence might be written 5-GTCA-3 or even just GTCA
P121 fig 6-2
Section two
The three dimensional structure of DNA
Ⅰ.DNA double-helix structure is the secondary structure of DNA
DNA double-helix structure model was put forth in 1953 by Watson and Crick.
DNA double-helix model
1 Two separate and anti-parallel chains of DNA are wound around each other
in a right-handed helical path, with the sugar-phosphate backbones on the
outside and the bases, paired by hydrogen bonding and stacked on each other,
on the inside.
Adenine pairs with thymine by two hydrogen bonds; guanine pairs with
cytosine by three hydrogen bonds.
The two chains are complementary; one specifies the sequence of the other.
P122 fig 6-3
2. Between the backbone strands run the major and minor grooves ,which also
follow a helical path.
A single turn of the DNA double-helix contains ten base pairs. The distance
spanned by one turn of the helix (pitch) is 3.4nm .the diameter of the helix is
2nm.
3. The stability of a DNA double-helix is maintained by the hydrophobic base
stacking and the hydrogen bonding.
Multi-forms of double-helix DNA
The above-mentioned DNA double-helix is the B–form of DNA double-helix
(B-DNA). the other well-known forms of double-stranded DNA are A-DNA and
Z-DNA.
P124 fig 6-5
Triplex DNA, tetraplex DNA and DNA hairpin
1. Hoogsteen base pair
It is a nucleic acid base pair that differs from the Waston-Crick base pair.
In the Hoogsteen adenine-thymine base pair the 6-NH2 and N-7 of the adenine
are hydrogen bonded respectively to the 4-O and H-3 of the thymine. That
forms T*A=T.
The Hoogsteen guanine-cytosine pair requires that N-3 of the cytosine is
protonated .In this base pair the 6-O and N-7 of the guanine are hydrogen
bonded respectively to the 4-NH2 and the protonated N-3 of cytosine.(C*G≡C)
P123 fig 6-4
2. triplex DNA
It is a three-stranded structure formed by duplex DNA in association with an
oligonucleotide, when purine or pyrimidine bases occupy the major groove of
the DNA double-helix adopting Hoogsteen base pairing with the Waston-Crick
base-pairs.
P-125 fig 6-6
3.mirror repeat
5’------- TC TC TC TC
CT CT CT CT------------3’
3’------- AG AGAG A G GAGA GAGA----------5’
4. coditions of triplex DNA formation
(1) double stranded DNA ,one strand is polypurine, the other strand
polypyrimidine
(2)mirror repeat
(3)acidic pH
If the Hoogsteen pairing occurs , triplex DNA is formed.
For example, the above mirror repeat, one strand is polypyrimidine the other
strand polypuine. Under acidic pH condition it is denatured. The polypyrimidine
strand folds back and participates in the formation of triplex DNA.
P 125 fig 6-6
5. tetraplex DNA or quadruplex DNA
Tetraplex DNA is a four-stranded DNA structure adopted by sequences rich in
quanine bases.
There are two major classes of tetraplex DNA. The first involves the folding
back of a repetitive sequence of quanines resulting in anti-parallel strands. The
other is characterized by the association of four independent parallel strands
5-GGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGG-3
5
3
One strand folds back three times:
5-GGGGGGGG-3, 5-GGGGGGGG-3, 5-GGGGGGGG-3,
5-GGGGGGGG-3
Four strands associate
5---------------------3
5---------------------3
5---------------------3
5---------------------3
P123fig 6-4 and p125 fig 6-7
6.DNA hairpin
Palindrome structure or inverted repeat for example
----------------------------------
------
-----
5-TTAGCAC∣GTGCTAA-3
5-TGCGATACTCATCGCA-3
3-AATCGTG∣CACGATT-5
3-ACGCTATGAGTAGCGT-5
---------------------------------Perfect IR
-----
-----
partial IR
If the inverted repeat DNA is denatured the single stranded DNA may form
DNA hairpin after renaturation.
P126 fig 6-8
Ⅱ.The supercoil structure is the DNA tertiary structure
Closed-circular DNA and superciol
1 .The E coli chromosome is a closed-circular DNA of length 4639Kb,which
resides in a region of the cell called the nucleoid.
2.The eukaryotic mtDNA and ctDNA are also closed-circular DNAs.
3.Closed-circular DNA means that the two complementary single strands in
the DNA molecule are each joined 5’ to 3’ and form a double stranded circle.
4. Supercoiling is the coiling of the DNA axis upon itself. That results in
supercoil structure of DNA.
P126 fig 6-9
5. E coli DNA, mtDNA, ctDNA are closed –circular supercoil DNA
Topoisomerases
Topoisomerases regulate the type and level of supercoiling of DNA molecules.
A closed-circular DNA molecule may have different topoisomers.
If we link together the two ends of a piece of a double-stranded DNA molecule
with 2000 bp (~200 helical turns),that forms closed-circular DNA.The DNA
molecule is a relaxed closed-circular DNA molecule.
Its linking number (LK,the interwinding number of the two DNA strands in the
dsDNA) is 2000/10=200, If we twist the piece of double-stranded DNA first and
then link together the two ends,the twisting formed deformation is locked into
the system.
This deformation is shown as supercoiling, since it manifests itself as a coiling
of the DNA axis around itself in a higher-order coil, and corresponds to a
change in linking number from the relaxed circular situation.
If the twisting of the DNA is in the same direction as that of the double helix
(right-handed) that is the helix is twisted up before closure, then the
supercoiling formed is positive (the left-handed positive supercoil).
For example, when the piece of dsDNA with 200 helical turns is twisted up two
turns before closure, the formed closed-circular DNA molecule will produce
two left-handed positive supercoils.
If the helix is untwisted , then the supercoiling is negative.
For example, when the piece of dsDNA with 200 helical turns is untwisted two
turns before closure, the formed closed-circular DNA molecule will produce
two right-handed negative supercoils.
The above three forms of the closed-circular DNA molecule are topoisomers of
the molecule.
P 127 fig 6-10
In 1966 Vinograd and Lebowitz put forth an equation that describes the
relationship between linking number (Lk),twisting number (Tw) and writhing
number (Wr):
Lk=Tw+Wr
For B-DNA. Tw=the number of bp/10
Wr is the number of the coiling of the helix axis upon itself.
For a relaxed closed-circular DNA molecule
LK0=Tw + 0, in the former example: LK0=200+0=200. LK=Tw+Wr in the
twisting up example, Wr=+2, LK=200+2=202, in the untwisting example, Wr=-2
LK=200-2=198
Specific superhelix (superhelical density) is used to evaluate the supercoil
number and direction:
sigma=(Lk-Tw)/Tw or sigma =delta Lk/ LK0
The negative supercoil form is a preferred form of the chromosomal DNA,
because it is beneficial to separate the two strands in dsDNA in the events of
DNA replication, and transcription.
Topoisomerases catalyze topologic changes of DNA,(altering the linking
number of DNA). They can relax or insert supercoils. There are two types of at
least four topoisomerases in E coli.
Type I
Topo I (ω protein or nicking-closing enzyme)
Topo III
to remove negative supercoils
Type II
TopoII (gyrase)
TopoIV
to insert negative supercoils
Type I enzymes break one strand of the dsDNA , and change the Lk number of
+1 by passing the other strand through the break. Then they close the nick
requiring no hydrolysis of ATP. Type II enzymes catalyze negative supercoiling
of DNA with the concomitant hydrolysis of ATP. They act by cutting both
strands of dsDNA , passing the dsDNA through the gap and resealing the
break. One action changes the linking number by two.
Ⅲ. The highly ordered DNA-protein complex makes up the eukaryotic
chromosomes.
1. Nucleosome is the basic unit of the chromatin. The major protein
components of chromatin are histones. They are small basic proteins which
bind tightly to DNA.
There are five families of histones:H1 ,H2A,H2B,H3 and H4.
H2A,H2B,H3 and
H4 are known as core histones.
Two molecules of each families of core histones : (H2A)2 (H2B)2(H3)2(H4)2 ,
make up the octameric histone core.
P129 fig 11-6 A
The histone core with 146 bp of DNA wrapped 1.8 times in a left-handed
fashion around it is the nucleosome core (core particle).
P129 fig 11-6 B
A single molecule of H1 stabilizes the DNA at the point at which it enters and
leaves the nucleosome core.
There is the linker DNA of 60 bp or so between neighbouring nucleosome
cores.
P129 fig 11-6 C
Nucleosomes are arranged in a ‘beads on a string’ structure.
P129 fig 11-6 D
2. The ‘beads on a string’ structure is organized into chromatin/chromosome
3. The mitotic chromosome has one centromere and two telomeres.
P129 fig 6-12
The ends of the linear chromosomal DNA are protected from degradation and
gradual shortening by the telomeres. The telomere consists of up to hundreds
of copies of a short repeated sequence 5’ (TxGy)n and DNA binding proteins .
The short repeated DNA sequence is synthesized by the enzyme telomerase.
It is a ribonucleoprotein, contains not only protein subunits but also an RNA
molecule, which serves as the template for telomere DNA elongation.
Telomerase belongs to reverse transcriptase.
P130 fig 6-13
The centromere is the region where the two sister chromatins are joined in the
metaphase chromosome.
Ⅳ. DNA is the vehicle of inheritance.
1. DNA is the carrier of genetic information , which is demonstrated by Avery’s
experiment of pneumococcus transformation.
2.The whole genetic information of a virus or a cell is the genome of the virus
or the cell.
3.Chromatin in interphase can be seen under microscope in two types.The
less densely packed variety is termed euchromatin, which is genetically
expressed, and the more densely packed variety is termed heterochromatin,
which is transcriptionally inactive.
Section Three
RNA structure and function
Ⅰ. There are four major types of RNA in the animal cell .
P131 table 6-5
Ⅱ. Most eukaryotic mRNA contains five parts:the cap, the 5’UTR,the coding
region, the 3’UTR and the poly A tail.
P133 fig 6-15
Ⅲ. structural features of tRNA
1. The tRNA molecule is rich in rare bases.
P133 fig 6-16
2. The secondary structure of a tRNA molecule is like the cloveleaf.
P134 fig 6-17
3. The tertiary structure of a tRNA molecule is L-shaped.
P135 fig 6-18
Ⅳ.The structure of rRNAs is more complicate than that of mRNA or tRNA.
There are three types of prokaryotic rRNAs:5SrRNA,16SrRNA and 23SrRNA,
while there are four types of eukaryotic rRNAs:5SrRNA,5.8SrRNA,18SrRNA
and 28SrRNA.
P135 table 6-6
rRNAs are important components of the ribosome.The ribosome is the location
where the protein is synthesized.
Ⅴ.There are a lot of snmRNAs in the cell
RNomics studies the type, the structure, the function ,and the expression
profile of snmRNAs in the cell.
1.SnRNAs participate in pre-mRNA splicing
P136 table 6-7
2. Ribozymes or catalytic RNAs participate in splicing of particular RNAs.
3. snoRNAs participate in modification of rRNAs and tRNAs
4. siRNAs participate in post-translational regulation
Section four
Physico-chemical characters of nucleic acids
Ⅰ.Nucleic acids are susceptible to hydrolysis under acidic or basic
conditions,They are also hydrolyzed by nucleases
Ⅱ. Nucleic acids are amphipathic compounds
1. Bases are tautomeric
There are the keto form and the enolate form of G,.U and T, and the amino
form and the imino form of A and C.
p137 fig
P138 fig
2.Bases are basic under the physiologic condition
3.Nucleic acids are acidic compounds
Ⅲ. Nucleic acids solution esp. DNA solution has a high viscosity
Ⅳ. Nucleic acids absorb UV light due to the conjugated aromatic nature of the
bases.
The wavelength of maximum absorption of light by nucleic acids is
260nm,(lambda max=260) which is conveniently distinct from the lambda max
of protein(280nm). The absorption properties of nucleic acids can be used for
detection, quantitation and assessment of purity of nucleic acids.
Ⅴ. Denaturation,Renaturation and Molecular Hybridization of Nucleic Acids
1.When a solution of dsDNA is heated above a characteristic temperature,its
native structure collapses and its two complementary strands separate and
assume the random coil conformation.
The DNA denaturation process is accompanied by a qualitative change in the
DNA’s physical properties,such as decreasing viscosity and increasing UV260
absorption (hyperchromic effect) of the DNA solution and loss of the DNA
biological activity.
2. Tm (melting temperature)
DNA denaturation is a cooperative process.The collapse of one part of the
structure (at the ends of the dsDNA molecule or AT-rich internal regions)
destabilizes the remainder. The temperature corresponding to the mid-point at
the DNA thermal melting curve is known as Tm. It is a function of the G+C
content of the DNA sample.
3. Denatured DNA can be renatured. Slow cooling the denatured DNA allows
time for the wholly complementary DNA strands to find each other and the
sample can become fully double-stranded by base pairing (annealing).This
process is called DNA renaturation. The renaturation of regions of
complementary between different nucleic acid strands is known as
hybridization.
Southern blotting is used to examine DNA samples.
The Southern blotting procedure is as follows:
a.DNA molecules are separated by agarose gel electrophoresis.
b.They are denatured with alkali.
c.Transferred to a nylon or NC membrane.
d.Hybridized with labeled probe
e.Detection
Northern blotting is used to examine RNA sample.
Section five Nucleases
Ⅰ. grouping nucleases
1.according to the substrate specificity: DNases and RNases
2.acording to the hydrolytic style: exonucleases and endonucleases
3.according to the substrate secondary structure specificity ssDNA specific
nucleases and dsDNA specific nucleases.
Ⅱ. RNases are RNA endonucleases
Ⅲ.DNases hydrolyze DNA
Ⅳ. N-glycosidases remove the base from the nucleotide
Section Six
Technique for Isolation and Identification of Nucleic Acids
Ⅰ.There are several methods of isolation of nucleic acids
1. phenol extraction
2. ultracentrifugation
3. gel electrophoresis
4. chromatography
Ⅱ. Two methods of DNA sequencing
1. Sanger’s enzymic method (dideoxy termination method)
It uses dideoxynucleotides as chain terminators to produce a ladder of
molecules generated by polymerase extension of a primer
2. Maxam & Gilbert chemical method
The labeled DNA is subjected to base-specific cleavage reactions prior to gel
separation.
3. RNA sequencing is achieved by base-specific cleavage of 5’end-labeled
RNA using RNases that cleave 3’ to a particular nucleotide.
HOMEWORK
1. What are the differences between DNA and RAN?
2. Describe the structure and function of DNA.
3.Translate the summary on P143 into English.
Chapter Two
The Nucleotide Metabolism
Section One
Biological Functions of Nucleotides
Ⅰ. Nucleic acids from foods are digested in the alimentary canal to nucleotides
or smaller molecules.
Degradation products of nucleic acids or nucleotides in the cell can be further
degraded or participate in anabolism.
Ⅱ. Nucleotides have a lot of physiological and biochemical functions.
1.Nucleotides are building blocks of DNA or RNA
2.Nucleotide derivatives, such as UDPG,CDP-DG etc, are metabolic
intermediates , which participate in macromolecule biosynthesis.
3.ATP is the direct energy-supplier.
4.AMP and ADP are components of many co-enzymes
5.GTP participates in the synthesis of the cap of the eukaryotic mRNA and
tetrahydropteridine.
6.cAMP and cGMP are secondary messengers.
7.ppGpp, 2’-5’ oligo A , etc participate in physiological and biochemical
regulation.
Nucleotides participate in rapid regulation of enzymes.
Section Two
Anabolism and Catabolism of Purine Nucleotides
Ⅰ. There are two pathways for the formation of nucleotides , the de novo
synthesis and the salvage pathway.
Synthesis of nucleotides from relatively simple precursors such as amino
acids , ribose phosphate, one carbon unit , etc ,is known as de novo synthesis.
Using free bases or nucleosides for the synthesis of nucleotides is the salvage
pathway.
Ⅱ. de novo synthesis of purine nucleotides
1.the biosynthetic origins of purine ring atoms
P284 fig 13-3
2. Three features of synthesis of purine nucleotides
a. Liver, small intestine and thymus are the three organs in which purine
nucleotides are mainly de novo synthesized.
b.Synthesis of purine nucleotides begins with R-5-P and ATP ,and the initially
synthesized purine nucleotide is IMP
c.AMP and GMP are synthesized from IMP
3. IMP is synthesized in a pathway comprised of 11 reactions
R5P ─ PRPP ─ PRA ─ ─-------─IMP
P285 fig 13-4
4.from IMP to AMP
IMP+Aspartate ─ AS ─ AMP+fumarate
5. form IMP to GMP
IMP ─ XMP
XMP+glutamine ─ GMP+glutamate
P286 fig 13-5
Ⅲ. Regulaion of Purine Nucleotide Synthesis
1. The IMP pathway is regulated at its first two reactions:those catalyzing the
synthesis of PRPP and PRA.
a. Feedback inhibition:
Synthesis of PRPP is inhibited by both ADP and GDP .
Synthesis of PRA is inhibited by IMP,AMP and GMP.
b. Feedforward activation
R-5-P and ATP activate PRPP synthesis.
R-5-P and ATP activate PRA synthesis.
2.Regulation of the IMP to AMP or GMP pathway
a. Feedfack inhibition
AMP and GMP are competitive inhibitors of IMP and inhibit AMP and GMP
pathway respectively.
b.Reciprocal activation
GTP activates the synthesis of AMP,whereas ATP activates the synthesis of
GMP. This reciprocity serves to balance the production of AMP and GMP.
P286 fig 13-6
Ⅳ. Salvage Pathway of Purine Nucleotides
1.Using purine and PRPP as precursors to synthesize AMP, GMP and IMP.
APRT
PRPP+A ───────AMP+PPi
HGPRT
PRPP+G───────GMP+PPi
HGPRT
PRPP+I───────IMP+PPi
2. AK catalyzes the AR phosphorylation
AK
AR+ATP ─────── ADP+AMP
Ⅴ. Purine nucleotide is degraded to the end product uric acid , in the human
body.
P288 fig 13-7
Section Three
Anabolism and Catabolism of Pyrimidine Nucleotides
Ⅰ.The biosynthetic origins of pyrimidine ring atoms
P289 fig 13-9
1.denovo synthesis of pyrimidine nucleotides begins with CO2 and glutamine
CPSⅡ
CO2 + GLn ─────── CP
There are two different types of CPS: CPSⅠ and CPSⅡ
CPSⅠ is located in the mitochondria of the hepatocyte, which participates in
the urea synthesis.
CPSⅡ is a part of a multifunctional enzyme located in the cytosol. The product
CP is utilized to synthesize pyrimidine nucleotides.
2.The initially synthesized pyrimidine ring containing product is OA.
OA and PRPP form OMP. Decarboylation of OMP yields UMP
P290 fig 13-10
3.UTP is synthesized form UMP and CTP is formed by amination of UTP.
4.UDP is formed from UMP . dUDP is formed by reduction of UDP by
ribonucleotide reductase . dUDP is hydrolyzed to dUMP . Methylation of
dUMP yields dTMP (TMP).
p291 fig 13-11
Ⅱ. Regulation of synthesis of pyrimidine nucleotides
1. Feedforward activation
a. ATP activates PRPPK and CPSⅡ .
b. PRPP activates OPRT.
2. Feedback inhibition
a. UMP feedback inhibits CPSⅡ
b. UMP and CTP feedback inhibit ATCase
c. ADP and GDP feedback inhibit PRPPK
d. CTP feedback inhibits CTPS
P292 fig 13-12
Ⅲ. salvage pathway of pyrimidine nucleotides
1. Using U or OA as precursors to synthesize UMP or OMP .
PRPP+U(OA) ───────UMP(OMP) + PPi
2 Using UR,CR,or TdR to synthesize UMP,CMP or TMP
UR(CR) + ATP───────UMP(CMP) + ADP
TdR + ATP---------------TMP + ADP
Ⅳ. Catabolism of pyrimidine nucleotides
Products of U and C catabolism are NH3, CO2 and B-alanine ,while products
of T catabolism are NH3 , CO2 and β-aminoisobutyric acid.
P292 fig 13-13, fig 13-14
Section Four
The Conversion of Nucleotides in vivo
Ⅰ.Reduction of NDPs (N=A,G,U,C) forms dNDPs.
Reduction of the 2’-hydroxyl of NDPs catalyzed by ribonucleotide reductase,
forms the corresponding dNDPs.
Reduction requires thioredoxin,thiredoxin reductase NADPH and FAD besides
ribonucleotide reductase.
P294 fig 13-15
Ⅱ. NDPs and NTPs are interconvertible
1. from UMP to UDP, AMP to ADP
UMP+ATP─────UDP+ADP
AMP+ATP─────ADP+ADP
2.Interconversion of NDP and NTP
XDP+YTP ───────XTP+YDP
Section Five
Nucleotide Metabolism in Diseases
Ⅰ. Metabolic disorders of purine nucleotides may cause diseases.
1. A defect in HGPRT causes Lesch-Nyhan Syndrome.
2. Gout is a metabolic disorder of purine catabolism.
3. ADA deficiency is associated with an immunodeficiency disease.
4. Metabolic disorder of pyrimidine nucleotides is rarely associated with
clinically significant abnormalities.
Ⅱ.The Mechanism of antimetabolites is blocking the synthesis of nucleotides
1. Antimetabolites are analogues of metabolites or coenzymes of the
nucleotide synthesis pathways.
P296 fig 13-16
2.There are key reactions in the nucleotide synthesis pathways , which are
antimetabolites’ targets.
a. 6MP,the adenine analogue, is the competitive inhibitor of HGPRT .
b. 6MP nucleotide,the analogue of IMP, inhibits the conversion of IMP to AMP
or GMP.
c. Azaserine, the analogue of glutamine, inhibits de novo synthesis of purine
nucleotides.
d. MTX and aminopterin are the analogues of folate and inhibit DHFR.
P297 fig 13-17
e.Arabinosylcytosine inhibits the conversion of dCDP from CDP and the
biosynthesis of DNA.
f.5FU is an irreversible inhibitor of thymidylate synthase .
p297 fig 13-18
HOMEWORK
Summarize Chapter Two in English.
Chapter Three
DNA Biosynthesis and DNA Damage Repair
Section One
The General Features of Replication of Chromosomal DNA
Ⅰ.DNA replication is semiconservative.
In 1958 Meselson and Stahl demonstrated experimentally that DNA replication
is semiconservative.
P316 fig 15-1
Ⅱ. The point at which separation of the DNA strands and synthesis of new DNA
takes place is known as the replication fork.
P316 fig 15-2
Ⅲ. DNA replication is bidirectional.
1. replicon : Any piece of DNA which replicates as a single unit, is called a
replicon.
2. origin : The initiation within a replicon always occurs at a fixed point known
as the origin.
3. bidirectional replication : Usually two replication forks proceed
bidirectionally away from the origin and the strands are copied as they
separate until the terminus is reached.
Ⅳ. DNA replication is semi-discontinuous.
Because DNA can only be synthesized in a 5’3’ direction, at each replication
fork, one strand ( the leading strand ) is synthesized as one continuous piece,
while the other strand ( the lagging strand ) is made discontinuously as short
fragments in the reverse direction.
These short DNA fragments are called Okazaki fragments. They are joined by
DNA ligase and form the lagging strand.
Ⅴ. Origin contains short AT-rich repeat sequences. E coli’s origin is called oriC.
It is 245bp long and contains three 13bp-direct repeat and four 9bp-inverted
repeat sequences.
Ⅵ. DNA replication needs priming.
1. Most DNA replications are primed by RNA
A RNA primer is synthesized and DNA polymerase elongates with DNA at the
3’-end of the RNA primer.
2. There are also DNA priming or nucleotide priming.
Ⅶ. Multi-enzymes and proteins participate in DNA replication.
Ⅷ. DNA replication is of high fidelity.
Section Two
Features of DNA Polymerases
Ⅰ. The substrates of DNA polymerases are dNTPs and the primer-template
junction.
3’_____________________________________________________5’
---------⊙ here + dNTP
( dNMP )n + dNTP ─────── ( dNMP )n+1 + PPi
The template directs the DNA polymerase to synthesize a daughter strand
following the base-pairing rules.
Ⅱ. The active center of DNA polymerases catalyzes DNA synthesis.
Ⅲ. The semi-closed right-handed structure of the DNA polymerase is
composed of three domains.
Ⅳ. The protein of “ sliding clamp “ encircles the DNA template and accounts
for the processivity of the DNA polymerase.
Ⅴ. The 3’  5’exonuclease active center of the DNA polymerase is responsible
for the proofreading.
Section Three
DNA replication in E coli
Ⅰ. E coli DNA replication is initiated at oriC in a process mediated by a multiprotein complex including DnaA, DnaB, DnaC etc.
It involves wrapping of the DNA around Dna- ATP complex and separation of
the strands at AT-rich sequences with the help of DnaC. The helicase DnaB
then binds and extends the single stranded region for copying.
P321 fig 15-4
Ⅱ. Primase synthesizes RNA primer.
DnaG, the primase, a specific RNA polymerase, attaches to the DNA and
synthesized a short RNA primer.
Ⅲ. DNA polymerase Ⅲ synthesizes DNA and DNA polymerase Ⅰ removes
the primer.
The holoenzyme of DNA polymerase Ⅲ:
P323 fig 15-5 table 15-1
1. dsDNA unwinding:
As the parental DNA is unwound by DNA helicases and single-stranded
binding protein( SSB ), the resulting positive supercoiling has to be relieved
by the topoisomerase DNA gyrase .
2. elongation:
A mobile primosome ( DnaB-DnaG complex ) synthesizes multiple primers on
the lagging strand’s template. DNA polymerase Ⅲ holoenzyme elongates
both leading and lagging strands ( Okazaki fragments ).
P324 fig 15-6
3. formation of lagging strand:
The 5’  3’exonuclease activity of DNA polymerase Ⅰ removes the RNA
primer of the Okazaki fragment and the polymerase function simultaneously
fills the gap with DNA. DNA ligase joins Okazaki fragments to form lagging
strand.
Ⅳ. Tus recognizes and binds to the TER sites. That terminates replication.
Ⅴ. Only at the full methylated oriC can initiate replication.
Ⅵ. Two types of topoisomerases are required in DNA replication.
Section Four
DNA Replication in Eukaryotes
Ⅰ. There are five types of common eukaryotic DNA polymerases:
pol.alpha,pol.beta,pol.garmma, pol.delta, and pol. ipsilon.
Ⅱ. Eukaryotic and prokaryotic enzymes and protein factors that participate in
DNA replication at the replication fork are comparable.
P327 table 15-2
Ⅲ. After the pol.alpha/primase complex initiates replication pol.delta starts
elongation.
Ⅳ. There are two mechanisms of removing primers:
1. RNase HI and FEN1-dependent mechanism
2. helicase Dna2 and FEN1-dependent mechanism
P331 fig 15-11
Ⅴ. The eukaryotic chromosome replicates only once in a cell cycle.
1. pre-RC forms in the G1-phase and is activated in the S-phase.
2. Cdk controls the formation and activation of pre-RC.
P332 fig 15-12, fig 15-13
Ⅵ. Telomerase participates in the replication of telomere DNA.
1. The problem of replicating the ends of linear chromosomes :
The ends of linear chromosomes cannot be fully replicated by
semi-discontinuous replication as there is no DNA to elongate to replace
the RNA removed from the 5’end of the lagging strand. Thus genetic
information could be lost from the DNA.
P333 fig 15-14
2. Telomere and telomerase solve the problem :
To overcome this, the end of eukaryotic chromosomes ( telomeres ) consist of
hundreds of copies of a simple, non-informational repeat sequence ( TTAGGG
in humans ) with the 3’end overhanging the 5’end. The enzyme telomerase
contains a short RNA molecule, part of whose sequence is complimentary to
this repeat. This RNA acts as a template for the addition of these repeats to the
3’overhang by repeated cycles of elongation. The complimentary strand is
then synthesized by normal lagging strand synthesis leaving a 3’overhang.
P334 fig 15-15
Section Five
DNA Replication in Mitochondria and Phages
Ⅰ.MtDNA is replicated in D-loops.
Ⅱ.Phage’s circular DNA is replicated in rolling circle.
Section Six
The Repair of DNA Damage
Ⅰ. Physical or chemical agents may cause DNA damage. There are also
replication errors. DNA damages must be repaired. There are several DNA
repair systems in both prokaryotic and eukaryotic cells.
Ⅱ. Mismatch repair system repairs the replication errors.
Replication errors which escape proofreading have a mismatch in the
daughter strand. Hemi-methylation of the DNA after replication allows the
daughter strand to be distinguished from the parental strand. The mismatched
base is recognized and bound by MutS, MutL and MutH participate in cleavage
the daughter strand. A piece of the error containing daughter strand is
removed by exonucleases. The gap is filled by DNA polymerase Ⅲ and the
nick is ligated by DNA ligase.
P336 fig 15-16 p333 fig 15-17
Ⅲ. There are different kinds of DNA lesions caused by physical or chemical
agents.
Ⅳ. There are at least six DNA
damage repair systems in E coli.
P338 table 15-3
1. Photoreactivation
2. BER
3. NER
4. Recombinational repair
5. SOS repair
6. Mismatch repair
Section Seven
Reverse Transcription
Ⅰ. The discovery of reverse transcriptase develops the central dogma put forth
by Crick.
Ⅱ. Reverse transcriptase has three enzymatic activities :
1. RDDP activity : It transcribes the RNA template onto a complimentary
DNA strand to form an RNA-DNA hybrid.
2. RNase H activity : The enzyme then degrades the RNA template.
3. DDDP activity : It replicates the resulting single-stranded DNA to form the
duplex DNA.
Ⅲ. Retroviruses have diploid, positive sense RNA ( mRNA
) genomes, and
replicate via a dsDNA intermediate, the provirus. The provirus is inserted into
the host cell’s genome. The provirus genome is expressible. The gene
products are mRNA, the retrovirus genome, and various proteins, including
reverse transcriptase. The retrovirus particle is packaged in the host cell and
the mature virion buds from the host cell surface.
HOMEWORK
1.Summarize the contents on pages 328 and 329 in less than ten sentences.
2.Summarize NER on page 340 in five sentences.
3.Summarize recombinational repair in five sentences.
Chapter Four
RNA Biosynthesis and Posttranscriptional
Processing
Section One
DNA-dependent RNA Synthesis
Ⅰ. RNA polymerases catalyze RNA synthesis.
1.DDRPs initiates RNA synthesis directly.
RNA biosynthesis is also called transcription. It is the synthesis of a
single-stranded RNA from a double-stranded DNA. RNA synthesis occurs in
the 5’→ 3’direction and its sequence is complimentary to that of the DNA
template strand and corresponds to that of the DNA coding strand.
P350 fig 16-4
RNA polymerase moves along the DNA and sequentially synthesizes the
RNA chain. DNA is unwound ahead of the moving polymerase, and the helix
is reformed behind it.
P349 fig 16-2
The reaction is as follows :
( NMP)n＋NTP ────── ( NMP )n+1 + PPi
P349 fig 16-1
2.There is only one kind of prokaryotic RNA pol. while there are three kinds of
eukaryotic RNA pols: RNA pol I, RNA pol II and RNA pol III. RNA pol I is
located in the nucleoli. It is responsible for the synthesis of the precursors of
18SrRNA, 28SrRNA and 5.8SrRNA. RNA pol II is located in the nucleoplasm
and is responsible for the synthesis of mRNA precursors and some snRNAs.
RNA pol III is located in the nucleoplasm. It is responsible for the synthesis of
5SrRNA, tRNAs, snRNAs and scRNAs.
3.The holoenzyme of E coli RNA pol. is composed of core enzyme ( α2ββ’
ω )and σ subunit.
Each eukaryotic RNA pol. has 12 or more different subunits. The largest two
subunits are similar to each other and to the β and β’ subunits of E coli RNA
pol. Other subunits in each enzyme have homology to the α subunit of the E
coli enzyme .Five additional subunits are common to all three pols , and
others are polymerase specific.
The largest subunit of RNA pol II has a seven amino acid repeat at the C
terminus called carboxyl-terminal domain (CTD).This sequence
YSPTSPS is repeated 52 times in the mouse RNA
to phosphorylation.
pol II . CTD is subject
Ⅱ.RNA pol recognizes and binds to the promoter thus inhibites transcription.
1.
subunit is responsible for recognizing the promoter.
Promoters contain conserved sequences which are required for specific
binding of RNA pol and transcription initiation.
The
σ70 promoter of E coli consists of a sequence of between 40 and 60 bp.
With this sequence, there are short regions of extensive conservation which
are critical for promoter function. The -10 sequence is a 6 bp region present in
almost all promoters .The consensus -10 sequence is TATAAT. It is generally
10 bp upstream from the start site. The -35 sequence is also a 6bp region
recognizable in most promoters . The consensus -35 sequence is TTGACA.
The base at the start site of transcription is almost always a purine and is
assigned as position +1 .There is another consensus sequences, the AT rich
UP element, locates between -40 to -60 bp in some genes of high expression.
P 352 fig 16-6
Many prokaryotes, including E coli , have multiple sigma subunits .The most
common sigma
subunit in E coli is σ70 . Alternative classes of consensus
promoter sequences (e.g heat shock promoters ) are recognized only by an
RNA pol containing an alternative σ subunit .
The σ subunit enhances the specificity of the core enzyme α 2 ββ’ω, for
promoter binding . The polymerase finds the promoter -35 and -10 sequences
by sliding along the DNA and forming a closed complex with the promoter DNA.
Around 17bp of the DNA is unwound by the polymerase, forming an open
complex.
The first several bases are added without the enzyme movement along the
DNA. When initiation succeeds the enzyme releases the σ subunit and forms
a ternary complex of polymerase (the core enzyme), DNA and nascent RNA
and elongation begins.
P352 fig 16-7
2.
Eukaryotic RNA pol Ⅱ alone can not initiate transcription. It has a lot of
helpers called transcription factors such as TFⅡA, TFⅡB…etc.
P353 table 16-1
3.
Transcription initiation of eukaryotic RNA pol Ⅰ or pol Ⅲ is similar to that
of RNA pol Ⅱ.
Ⅲ. There are two types of prokaryotic transcriptional termination.
1. ρ(rho)–independent termination
Self-complementary sequence at the 3’-end of genes causes formation of
hairpin structure in the RNA which act as the terminator. The stem of the
hairpin often has a high content of G.C base pairs giving high stability,
causing the polymerase to pause. The hairpin is often followed by several
Us which result in weak RNA template DNA strand binding. This favor
dissociation of the RNA strand, causing transcription termination.
2. ρ(rho)–dependent terminator sequence which requires an additional
protein factor rho, for efficient transcription termination. rho binds to specific
sites in single-stranded RNA. It hydrolytes ATP and moves along the RNA
towards the transcription complex, where it enables the polymerase to
terminate transcription.
Section Two
RNA Processing
Ⅰ.Eukaryotic mRNA processing includes 5’capping, 3’ polyadenylation and
splicing.
1.5’-capping of pre mRNA (hnRNA)
This is the addition of a 7-methylguanosine nucleotide (m7G) to the 5’-end of
an RNA pol Ⅱ transcription when it is about 25-30 nt long .The m7G or cap, is
added in reverse polarity (5’to 5’) , thus acting a barrier to 5’-exonuclease
attack, but it also promotes splicing , transport and translation.
P357 fig 16-14,16-15
2. 3’ cleavage and polyadenylation
Most eukaryotic pre-mRNAs are cleaved at a polyadeylation site and poly(A)
polymerase (PAP) then adds a poly A tail around 250nt to generate the mature
3’-end.
P358 fig 16-16
3. Splicing
In eukaryotic pre-mRNA processing, intervening sequences (introns) that
interrupt the coding regions (exons) are removed (spiced out) and the two
flanking exons are joined .This spicing reaction occurs in the nucleus
and
requires the intron to have a 5’-GU and an AG-3’ and a branch point sequence.
In a two-step reaction, the intron is removed as a tailed circular molecule, or
lariat, and is degraded.
Splicing involves the building of sn RNPs to the conserved sequences to form
a spliceosome in which cleavage and ligation reactions take place.
P358 fig16-17 ,P.359 fig 16-18
P360 fig 16-49, fig 16-20
4. Alterative mRNA processing is the conversion of pre-mRNA species into
more then one type of mature mRNA . This can result from the use of
different poly(A) sites (alternative polyadenylation )or different patterns of
splicing (alternative splicing).
P.361 fig 16-21, fig 16-22
Ⅱ.Nucleases participation in tRNA and rRNA processing
1. An initial 30S transcript is made in E coli by RNA pol transcribing one of the
seven rRNA operons. Each contains one copy of the 5S,16S and 23S
rRNA coding regions , together with some tRNA sequences. This 6500 nt
transcription folds and complexes with some proteins, becomes methylated
and is then cleaved by specific nucleases to release the mature rRNAs.
2. In eukaryotes RNA pol transcribes the rRNA genes , which exist in tandem
repeats, to yield a 45S pre-RNA
which contains one copy each of the
18S ,5.8S and 28S sequences . Various spacer sequences are removed
from the long pre-rRNA molecule by a series of specific cleavages. Many
specific ribose methylation take place directed by snRNAs, and the
maturing rRNA molecules fold and complex with ribosome proteins.
P362 fig 16-23. 26-24
P363 fig 16-25
3. Prokaryotic and eukaryotic pre-tRNA processing are similar. It includes:
a.5’end and 3’end cleavages by specific RNases.
b.base modifications, such as methylation, reduction, deamination etc.
c.splicing: many eukaryotic pre-tRNA are synthesized with an intron. It is
removed by enzymatic cutting and joining mechanism.
d.3’ terminal CCA is added by the enzyme tRNA nucleotidyl transferase if
there is no CCA sequence at the 3’terminus.
P364 fig 16-26, P365 figs16-27,16-28
Ⅲ.Catalytic RNAs (ribozymes )can perform self-splicing reactions
Several biomedical reactions can be carried out by RNA enzymes or
ribozymes. They can cleave themselves or other RNA molecules, or perform
self-splicing reactions.
There is an intron in the large subunit rRNA of Tetrahymena that can remove
itself from the transcription in vitro in the absence of protein. The process is
called self-splicing and requires guanosine , or a phosphrylated derivative , as
cofactor.
P366 fig 16-29
Ⅳ. Some m RNA undergo editing
An unusual form of mRNA processing in which the sequence of the mRNA is
altered is called RNA editing. In man, editing cause a single base change from
C to U in the apolipoprotein B mRNA, creating a stop codon in the mRNA in
intestinal cells at position 6666 in the 14500nt molecule. The unedited RNA in
the liver makes apolipoprotein B100, a 512 KD protein , but in the intestine
editing causes the truncated apolipoprotein B45(241 KD) to be made.
P367 fig 16-31
Section Three
RNA Dependent RNA Synthesis
Most viruses have RNA as the genome . They are called RNA viruses .RNA
viruses require virus-encoded RNA dependent RNA polymerases (RDRPs) for
their replication. Retroviruses also belong to RNA viruses . They use an RDDP
(reverse transcriptase) to replicate via a DNA intermediate.
HOMEWORK
1. Summarize the contents of initiation of RNA pol Ⅱ, pol Ⅰ and pol Ⅲ on
pages 353, 354 and 356.The following questions may help you do the
homework.
a. What are the structural features of RNA pol Ⅱ(RNA pol Ⅰ and pol Ⅲ)
promoters?
b. What are the transcription factors of RNA pol Ⅱ(RNA pol Ⅰ and pol Ⅲ)?
c. What are the functions of the transcription factors?
2. Summarize the contents of chapter four.
Chapter Five
Protein Synthesis, Folding and Processing
Section One
Components Required for Protein Biosynthesis
Ⅰ.mRNA is the template of protein biosynthesis.
1. The genetic code contains 64 codons. 61 codons are sense codons,
whereas 3 codons UAG, UGA and UAA are stop codons
2. Five features of the genetic code:
a. Universality or almost universality
All organisms, from bacteria to human being use the universal genetic code.
However, since 1980, it has been discovered that mitochondria, which have
their own small genome, utilize a genetic code that differs slightly from the
standard universal code.
b. Direction: Codons in the coding sequence of mRNA are arranged from 5’ to
3’direction.
c. Commaless: The triplet codons arranged in the coding sequence of mRNA
are commaless, one after the other . Deletion or insertion one or two
nucleotides of the triplet will cause frame shift mutation.
d. Degeneracy: There are 61sense codons, while there are only 20amino
acids. 18 out of 20 amino acids have more than one codon to specify them.
e. Wobble: The codon in mRNA base pairing with the anticodon in tRNA may
not observe the standard base pairing rules (A pairs with T , U pairs with A
and C pairs with G).G-U, I-A ,I-C and I-U are wobble base pairs.
Ⅱ.tRNAs transfer amino acids.
When charged by attachment of a specific amino acid to their 3’-end , to
become aminoacyl-tRNAs, tRNA molecules act as adapter molecules in
protein synthesis.
Ⅲ. Ribosomes are the locations where protein synthesis takes place.
Ribosomes are complexes of rRNA molecules and specific ribosomal
proteins ,and these large RNPs are the machines the cell uses to carry out
translation.
The E coli 70S ribosome is formed from a large 50S and a small 30S subunit.
The large subunit contains one each of the 23S and 5SrRNA and more than 30
different proteins. The small subunit contains a 16SrRNA molecule and more
than 20 different proteins. The mammalian 80S ribosome is composed of one
large 60S subunit and one small 40S subunit. The small subunit contains one
18SrRNA molecule and the large subunit contains one 5SrRNA , one 5.8S
rRNA, one 28 SrRNA and more than forty proteins.
P372 fig 17-1
Section Two
Proteins synthesis Takes Place in Five stages
The five stages are formation of aminoacyl tRNAs, initiations, elongation,
termination and peptide folding and post-translational processing .
Ⅰ.Aminoacyl tRNAs are synthesized from amino acids and tRNAs by aminoacyl
tRNA synthetases. The reaction takes place in two steps:
P374 fig17-2
Proofreading occurs when a synthetase carries out step 1. of the
aminoacylation reaction with the wrong , but chemically similar amino acid. It
will not carry out step 2, but will hydrolyze the aminoacyl adenylate (E-AA-AMP)
instead.
Ⅱ. Initiation
A special tRNA(initiator tRNA), recognizing the AUG start codons, is used to
initiate protein synthesis in both prokaryotes and eukaryotes. In prokaryotes,
the initiator tRNA is first charged with Met by methionyl- tRNA synthetase. The
methionine residue is then converted to N-formyl-methionine by transformylase. In eukaryotes, the methionine on the initiator tRNA is not modified . There
are structural differences between E coli initiator tRNA and the tRNA that
inserts internal Met residues.
In prokaryotes, initiation requires the large and small ribosome subunits, the m
RNA, the initiator tRNA, three initiation factors(IFs) and GTP. IF1 and IF3 bind
to the 30S subunit and prevent the large subunit binding. IF2 + GTP can then
bind and will help the charged initiator tRNA to bind later. This small subunit
complex can now attach to a mRNA molecule via its ribosome-binding site.
The initiator tRNA can then base-pair with the AUG initiation codon which
releases IF3 thus creating the 30S initiation complex. The large subunit then
binds, displacing IF1 and IF2+GDP , giving the 70S initiation complex which is
fully assembled ribosome at the correct position on the mRNA (the initiator
tRNA with the initiation start codon AUG occupies the P site )
P376 fig17-3
Ⅲ.Elongation
Elongation involves the three elongation factors, (EF-Tu, EF-Ts and
EF-G),GTP, charged tRNA and the 70S initiation complex (or its equivalent ).It
takes place in three steps.
1.entrance (registration)
A charged tRNA is delivered to the A-site as a complex with EF.Tu and GTP.
The GTP is hydrolyzed and EF-Tu-GDP is released which can be re-used with
the help of EF-Ts and GTP(via the EF-Tu-EF-Ts exchange cycle)
P378 fig17-6
2. peptide bond formation
23SrRNA has the peptidyl transferase activity . It makes a peptide bond by
joining the two adjacent amino acids without the input of more energy.
P379 fig 17-7
3.translocation
Translocase (EF-G) with energy from GTP, moves the ribosome one codon
along the mRNA, ejecting the uncharged tRNA and transferring the growing
peptide chain from the A-site to the P-site. Now ,the A-site is vacant and ready
for another round of the three –step ribosomal cycle.
Ⅳ.Termination
When any one of the stop codons enters into the A-site, RF1 recognizes stop
codons UAA and GAG, RF2 recognizes UAA and UGA. They are helped by
RF3 which hydrolyzes GTP and make peptidyl transferase join the peptide
chain to a water molecule, thus releasing it.
P380 fig 17-9
Ⅴ. Folding and Posttranslational Processing See next section.
Section Three
Folding and Posttranslational Processing
Ⅰ. The nascent peptide undergoes posttranslational modification.
1. trim or modification of N or C terminal
2. restricted hydrolysis; polyprotein processing and protein splicing
3. various modification types of amino acid residues
Ⅱ .Peptide folding results in formation of higher orders of polypeptides’
structure.
1. Peptide folding is a complex process
2. Molecular chaperones participate in peptide folding
Molecular chaperones are proteins that assist in protein folding. They have
following functions:
a. binding to short hydrophobic segments in nascent polypeptides, thereby
preventing aggregation.
b. establishing
an
isolated
environment
for
protein
folding
without
interference.
c. promoting protein folding and disaggregation.
d. unfolding folded proteins in case of stress.
There are two classes of molecular chaperones: ribosome-binding molecular
chaperones and non-ribosome-binding ones.
Ⅲ .Assembly
of
protomers
(subunits)
forms
oligomers
or
polymers
(multi-subunit proteins).
Section Four
Clinical Relatives in Protein Synthesis
Ⅰ. Many viruses co-opt the host cell protein synthesis machinery.
Ⅱ. Many antibiotics work because they selectively inhibit protein synthesis in
bacteria.
Ⅲ. Certain toxins or cytokines inhibit the host cell or virus protein synthesis
1. Diphtheria toxin is an enzyme. It transfers the ADP-ribosyl group from
NAD+ to a certain histidine residue in eEF-2. That results in inactivation of
eEF-2 and inhibition of protein synthesis.
P387 fig 17-16
2.The interferons are a group of glycoproteins produces by a variety of cells in
response to several stimuli such as viral infections, antigens, mitogens etc.
Type Ⅰ interferons protect cells from viral infection by two mechanisms.
a. stimulating the phosphorylation and inactivation of eIF2 which is required to
initiate protein synthesis.
b. inducing synthesis of 2’-5’ oligo A which activates RNase L and causes
degradation of viral RNA.
P387 fig 17-17
HOMEWORK
1. explain the following terms:
RBS, ribosomal cycle, initiator tRNA, polysome , molecular chaperone.
2. summarize eukaryotic protein synthesis (initiation, elongation and
termination).
3. Tell the story of Hsp70 on pages 382, 383 and 384 in several sentences.
Chapter six
Regulation of Gene Expression in Prokaryotes
Section One
Principles of Gene Regulation
Gene expression is the process by which gene products are made.
Ⅰ.The common rules of Gene expression in both prokaryotes and eukaryotes.
1. The temporal and spatial specificity of gene expression
2. The cis-acting elements and trans-acting factors in gene regulation
3. The positive and negative regulation of gene expression
4. DNA-protein interaction is the molecular basis of gene regulation.
Ⅱ. Regulation of transcriptional initiation is the key step of gene regulation.
1. Structural gene is a gene that codes for the synthesis of a protein or a RNA
molecule with a non-regulatory function (e.g. rRNA, tRNA)
2. Regulation of gene is the result of a complex control hierarchy which
include chromatin remodeling, translational control, RNA processing
translational control, post translational processing, protein or enzyme
activation, protein or enzyme inactivation or degradation.
Ⅲ. Induction and repression are the basic rules of regulation of prokaryotic
transcriptional initiation.
1. An operon is a unit of prokaryotic gene expression which includes
co-ordinately regulated structural genes and control elements which are
recognized by regulatory gene products.
2. Induction: an inducer (usually a substrate for a catabolic pathway) turns on
gene expression.
3. Repression: a repressor (usually a product of a biosynthetic
pathway)
turns off gene expression.
4. Features of prokaryotic gene expression.
a. Regulation of transcription is the key step of gene regulation.
b. Operon regulation.
c. Negative controls are predominant.
d. Genes in an operon are all transcribed together on the mRNA, called a
polycistronic mRNA.
e. Many genes are constitutive genes (or house keeping genes ). They code
for gene products required for almost all cells and are routinely transcribed and
expressed.
Section Two
The Operon of Bactria
Ⅰ. The lactose operon
The lacZ, lacY and lacA gene are transcribed from a lacZYA transcription unit
under the control of a single promote. They encode enzymes required for the
use of lactose as a unique carbon source. The lac I product, the lac repressor
is expressed from a separate gene upstream from lac operon.
P393 fig 18-2
Ⅱ.The lac operon is turned off by the lac repressor
The lac repressor is made up of four identical protein subunits, it binds to
operator DNA sequence that overlaps lac promoter, and blocks transcription
from lac promoter.
The turned off lac operon can be turned on by lac inducer.
When lac repressor binds to the inducer (such as allolactose or IPTG), it
changes conformation and can not bind to the operator. The lac operon is
turned on.
P 394 fig 18-3
cAMP-CAP acts as the positive regulator of lac operon.
The lac promoter is a weak promoter.
RNA polymerase binding to the promoter and transcription initiation need
activation by cAMP-CAP complex.
CAP is a transcriptional activator which is activated by binding to cAMP.
cAMP levels rise when glucose is lacking.
cAMP-CAP complex binds to a site just upstream from lac PO region and
activates RNA polymerase binding to the promoter and transcription initiation.
P.395 fig 18-4
Ⅲ. Other types of operon regulation.
Section Three
Regulation of Gene Expression in Phage
The life cycle of phage lambda.
Lysogenic and lytic pathways.
P397 fig18-7
Bacteriophage lambda infects bacteria.
In lytic infection, phage particles are released from the cell by lysis. However in
lysogenic infection phage particles integrate their genome into that of the host
cell.
Such a phage is called the prophage and such a process, the lysogeny, and
such a host cell, lysogen.
The prophage may be stably inherited through several generation before
returning to lytic infection by inductions.
Section Four
Regulation of Translational Gene Expression
I. Two mechanisms of trp operon regulation.
1.Repression
The trp repressor binds L-tryptophan, the end product of trp synthesis
pathway, to form a complex that specifically binds to trp operator , so as to
negatively regulate operon transcription.
The trp repressor-operator system dose not fully account for the trp synthesis
in E coli.
2.Attenuation
There is a 161-nt-leader sequence (trp L) between the operator and the
structure gene trp E.
P400 fig 18-10
The trp leader RNA transcribed from trp L contains four regions of
complementary sequence which are capable of forming alternative hairpin
structures.
One of these is a kind of p-independent terminator, called attenuator.
P400 fig 18-11
The leader RNA contains an efficient ribosome-binding site and encodes a
14-amino acid peptide (lead peptide) which contains two trp residues.
When trp is low the ribosome will pause at the codon , while transcription of trp
operon by RNA polymerase goes on because of no attenuator formation.
When trp is high the leader peptide is synthesized, while transcription of trp
operon terminates because of attenuator formation.
Attenuation depends on the fact that transcription and translation are tightly
coupled in prokaryotes.
Translation can occur as a mRNA is being transcribed.
It depends on the conditions of leader peptide synthesis whether or not the
full-length mRNA is transcribed.
Ⅱ. SOS response
Agents that damage DNA induce a complex system of cellular changes in E
coli known as SOS response.
SOS response results from the expression of a network of operons.
P400 table 18-3
These operons are coordinately controlled and form a regulon.
Lex A protein represses all the operons for SOS response.
RecA gene is induced by heavy DNA damages and RecA protein is
synthesized.
RecA protein is a kind of protease.
It hydrolyzes the Lex A protein, thus SOS regulon is turned on.
P 401 fig 18-12
That results in SOS response. SOS repair is error-prone.
Ⅲ. The synthesis rate of ribosomal proteins is coordinated with that of r RNAs
ⅣRegulation of mRNA stability provides another control mechanism of gene
expression.
HOMEWORK
1. Summarize Lac operon regulation.
2. Tell the story of the lytic pathway of phage lambda.
Chapter
Seven
Regulation of Gene Expression is Eukaryotic
Section One
Control of Transcription of Initiation
I. Eukaryotic cis-acting elements are composed of promoter and regulatory
DNA sequences.
1.TATA box /initiator or CpG island is the eukaryotic core promoter.
Some RNA Pol Ⅱ promoters contain a sequence celled a TATA Box which is
situated 25-35 bp upstream from the start site. Other genes contain an initiator
element which overlaps the start site .These elements are required for basal
transcription complex formation and transcription initiation. Many RNA pol Ⅱ
promoters consist of both TATA box and initiator.
Some RNA pol Ⅱ promoters contains several transcription start sites
distributed in a 20 to 200 bp region.
These promoters have no TATA box and initiator .They have clusters of CG
sequences called CpG islands upstream from the start sites.
2.Elements reside 100-200 bp upstream from the promoter are generally
required for efficient transcription. Examples include CAAT and GC boxes.
3. Enhancers and silencers are sequence elements which can regulate
transcription from thousands of base pairs upstream or downstream. They
may contain a variety of sequence motif which are recognized and bound by
regulatory proteins. Enhancers activate transcription while silencers inhibit
transcription.
P 406 fig 19-1
Ⅱ.Transcription factors are composed of different functional domains. DNA
binding domains, and transcription activation domains are found both in
prokaryotes and eukaryotes.
DNA binding domains.
a. HTH: a recognition alpha-helix interacts with the DNA and is separated from
another alpha-helix by a characteristic right angle beta-turn.
b.Zinc finger domains include the C2H2 Zinc fingers which bind Zn2+ through
two Cys and two His residues and also the C4 fingers which bind Zn2+ through
four Cys residues.
P408 fig19-3
c. Basic domains are associated with leucine zipper at helix-loop-helix
dimerization domains. Dimerization is generally necessary for basic
domain binding to DNA.
3. Transcription activation domains
a. Acidic activation domains contain a high proportion of acidic amino aids
and are present in many transcription activators.
b. Glutamine-ride domains and proline-rich domains contain a high proportion
of glutamine residues and a continuous run of proline residues
respectively.
They can activate transcription.
Ⅲ.The “turn-on” mechanism of eukaryotic genes is more complex than that of
prokaryotic genes.
1. Chromatin
remodeling
includes
two
important
steps,
nucleosome
remodeling and modification of histones.
2. Assembly of transcription inhibition complex on the core promoter needs
not only the RNA pol, but also various general transcription factors.
P411 fig19-7
Section Two
Posttranscriptional Control
I. 5’ end capping and 3’end polyadenylation of eukaryotic mRNAs.
Ⅱ.Splicing and alterative splicing.
Ⅲ. mRNA editing.
Ⅳ. mRNA transporting from the nucleus to the cytoplasm.
Ⅴ. mRNA compartmentation in the cytoplasm.
Ⅵ. mRNA stability.
Ⅶ. mRNA polyadenylation in the cytoplasm.
Ⅷ. nonsense-mediated mRNA decay.
Ⅸ. RNA interference can make posttranscriptional gene silence.
Section Three
Translational Control
I.Translational initiation is controlled by phosphorylation /dephosphorylation of eIF2.
Ⅱ.5’ UTR and 3’ UTR binding proteins may negatively regulate translation.
Ⅲ. Some eukaryotic mRNA contains more than one start codons
HOMEWORK
1. Summary the contents of RNAi on page417.
2. Explain the following terms:
a.leucing zipper
b. enhancer
c. TATA box
d. CpG island
e.mRNA editing
Chapter eight
Recombinant DNA Technology
Section one
Overview of Recombinant DNA Technology
Ⅰ. Recombinant DNA technology is also called DNA cloning or molecular
cloning.
1. Cloning is a lab procedure that produce clones.
2. Clones are individuals formed by an asexual process so that they are
genetically identical to the original individual, for example, bacterial
colonies, multiple identical copies of a gene, etc.
Ⅱ.DNA cloning facilitates the isolation and manipulation of fragments of an
organism’s genome by replicating them independently as part of an
autonomous vector.
The procedure:
1. Select the cloning vector such as the plasmid and isolate the DNA fragment
of interest.
2. make the recombinant DNA molecule
3. transform the host cell such as E coli
4. recombinant DNA molecule replicates in the bacteria
5. select the transformed bacteria
6. isolate the recombinant DNA molecules and analyze them
P423 fig 20-1
Section Two
Molecular Tools in Recombinant DNA Technology
Ⅰ.The useful tool enzymes
1. Restriction endonuclease:
Restriction endonucleases are bacterial enzymes which can cleave DNA
symmetrically in both strands at specific recognition sequences to leave a
5’-phosphate and a 3’-OH.They leave blunt ends, or protruding 5’ or 3’ terminal
(cohesive or sticky ends)
REs form part of the restriction-modification defense mechanism against
foreign DNA.
They are the basic tool enzymes of molecular cloning.
P424 table 20-1
P425 fig 20-3
2. DNA ligase joins breaks in a dsDNA backbone with a 5’-phosphate and a
3’-OH in a ATP+ or NAD+ dependent reaction. Requires that the ends of the
DNA be compatible, i.e. blut with blut, or complementary sticky ends.
3. CIP and Klenow flagment of DNA pol. I are also required in molecular
cloning.
Ⅱ .Cloning vectors must be capable of being replicated and isolated
independently of the host’s genome and have enough capacity to clone foreign
DNA fragments.
1. Three essential elements of plasmid DNA
a. There is at least one origin of replication.
b. It must contain a selectable marker such as resistance to antibiotics.
c. It must contain cloning sites.
P427 fig 20-8
2. Phage DNA is used as the vector to construct libraries.
3. Cosmid, BAC, and YAC can be used to clone large or very large genomic
fragments from humans and other species.
4. In cultured animal cells, vectors have often been based on viruses which
naturally infect the required species ,either by maintaining their DNA extra
chromosomally (such as adenovirus ) or by integration into the host
genome (such as retrovirus).
5. Most of the vectors for use in eukaryotic cells are constructed
vectors.
as shuttle
They incorporate the sequences required for replication and selection in E coli
(ori, ampr) as well as in the desired host cells.
Ⅲ.Host cells provide the place for replication and expression.
Section Three
DNA cloning
Ⅰ .The core process of DNA cloning is construction of recombinant DNA
molecules.
1. Construction of recombinant DNA molecules.
a. Select the appropriate DNA cloning strategy.
b. Use restriction endonucleases to cut the vector at a unique site for the
cloning of foreign DNA.
c. Make the recombinant DNA molecule by joining the vector DNA and the
foreign DNA fragment.
2. Amplification, screening and isolation of recombinant
DNA molecules.
a. It is the process of take up of foreign DNA ,usually plasmids, by bacteria
( transformation ).
b. The introduction of foreign DNA into animal cells is called transfection.
c. The introduction of foreign DNA packaged in the virus particle into host
cells is called infection.
d. The process of identify particular clones containing the gene of interest
from among the very large numbers of others is called screening.
3. Marker screening , probe hybridization, and DNA sequencing are used to
isolate and identify recombinants.
a. Marker
screening
includes
antibiotic
resistance
screening,
alpha-
complementation and marker rescue.
b. Colony , plaque and molecular hybridization screening require the labeled
probe.
c. Expression screening is based on the fusion protein with the vector
encoded beta-galactosidase or the detection of gene product by specific
antibody.
Ⅱ. DNA cloning can be utilized to isolate and clone a single gene
1. Candidate genes may be derived from genomic or cDNA library.
A gene library is a collection of different DNA sequences from an organism
each of which has been cloned into a vector for ease of purification, storage
and analysis.
There are two types of gene library that can be made depending on the source
of the DNA used . If the DNA is genomic DNA, the library is called a genomic
library.
If the DNA is cDNA ,the library is called a cDNA library.
P432 fig 20-10
2. Screening to isolate one particular clone from a gene library routinely
involves using a nucleic acid probe for hybridization. The probe will bind to its
complementary sequence allowing the required clone to be identified.
3. Yeast one-hybrid system is used to isolate coding sequences of
transcription factors.
P433 fig20-11
4. Yeast two-hybrid system is used to isolate coding sequences of proteins
that interact with the “bait” protein.
P433 fig 20-12
Section four
Analysis of Cloned DNA
Ⅰ. Digesting recombinant DNA molecules with restriction enzymes, alone and
in combinations, allows the construction of a diagram (restriction map) of the
molecule indicating the cleavage positions and fragment sites.
Ⅱ. DNA sequencing tells the exact sequence of a DNA molecule . Sanger’s
enzymic method uses dideoxynucleotides as chain terminators to produce a
ladder of molecules generated by polymerase extension of a primer.
Ⅲ .Genomics
and
Proteomics
require
methods
of
Bioinformatics.
Bioinformatics describes the development and use of softwares to analyze
biological data.
1. detecting ORFs of a genome
2. determing intron-exon boundaries
3. screening of promoters
4. determing motifs and domains of proteins
5. sequence alignment
Ⅳ. Molecular hybridization and blotting technique can detect specific genes or
their expression.
1. Southern blotting detects specific genes on nylon membrane hybridize to a
particular probe.
P436 fig 20-13
2. Northern blotting detects RNA.
3. There are other methods to detect target genes or gene expression.
a. RNase protection experiment can show which parts of the RNA are
protected by bound probe.
P436 fig 20-14
b. RT-PCR is a method of rapid detection of mRNA.
P437 fig 20-15
c.Real time PCR detects gene expression quantitatively
d.In situ hybridization and FISH can locate the position of gene expression in
the cell and on the chromosome respectively.
e.DNA microarray can detect gene expression profile in a cell.
P437 fig 20-1
Section Five
Application of Recombinant DNA Technology in Biology and
Medicine
Ⅰ.Bioactive proteins may be expressed by cloned genes.
1. E coli gene expression system is easy to manipulate and very efficient and
hence economy.
2. Modified proteins may be expressed in eukaryotic expression systems.
3. There are cell free translation systems such as wheat germ extract or rabbit
reticulocyte lysate etc. They are commonly used to check the activity of an
mRNA.
Ⅱ. It is very useful to change just one or a few specific nucleotides in a
sequence (site-directed mutagenesis) to test the effect, such as the
importance of each residue in a transcription factor binding site, suspected
critical amino acids in a protein ( by altering codons in the cDNA molecule ).
Originally, site-directed mutagenesis used phage M13 mediated method. Now
much site-directed mutagenesis is carried out using PCR.
Ⅲ .RNAi can be used to study gene function.
RNA interference is a mechanism of gene regulation in organisms. When a
dsRNA molecule which is homologous to an endogenous mRNA coding
sequence, is introduced in to a cell, it will cause degradation of the mRNA and
result in posttranscriptional gene silence. The dsRNA used is the siRNA.
P440 fig 20-17
Ⅳ.Transgene and gene targeting are useful techniques of studying gene
function.
1. Transgenic mice are made by using microinjection.
2. Gene targeting includes gene knock-out and gene knock-in.
3. Transgene technique is used to establish animal models of human
diseases.
P441 fig 20-18
Ⅴ. Molecular cloning accelerates development of pharmacological and medical
recombinant proteins and vaccines.
Ⅵ. Applications of recombinant DNA technology in medical diagnosis and
gene therapy.
1. DNA cloning is a useful tool in molecular medical diagnosis such as
diagnosis of hereditary diseases, identification of microbe species,
diagnosis and analysis of tumors etc.
2. Attempts to correct a genetic disorder by delivering a gene to a patient is
described as gene therapy. Although gene therapy is still in its infancy, it
has great potential.
HOMEWORK
Translate the summary into English.
Chapter Nine
Signal Transducer and Cellular Signal Transduction
Section One
The General of Signal Transduction
Ⅰ. Extracellular chemical signal molecules are divided into two types: soluble
and membrane binding signal molecules.
1. Chemical signal communication undergoes evolution from simplicity to
complexity.
2. Soluble chemical signal molecules consist of three classes : hormones
( endocrine signals ), cytokines ( paracrine or autocrine signals ), and
neurotransmitters.
P522 table 25-2
3. Cell-surface molecules are also important extracellular signal molecules.
Ⅱ. Specific receptors receive extracellular signals.
1.Chemical signals are transduced into the cell through ligand-receiptor
interaction.
2. Receptors may be cell-surface molecules or intracellular molecules.
Ⅲ. Structural, quantitative and distribution changes of signal molecules are the
working basis of the signal transduction network.
1. Signal transduction pathways are composed of second messengers and
signal transducers. Signal transduction pathways form signal transduction
network through crosstalking.
2. Signal transduction is based on protein conformational changes.
P524 fig 25-1
3. Quantitative changes and translocation of intracellular transducers are the
important working style of regulation of signal transduction.
Section Two
Small Chemicals as Transducing Messengers
Ⅰ. The necessary features of the second messenger
Ⅱ. Cyclic nucleotides ( cAMP and cGMP ) are important second messengers.
P525 fig 25-2
Cyclic nucleotides are allosteric effectors and can regulate enzyme or protein
activity.
P526 fig 25-4 table 25-3
Ⅲ. Certain lipids can act as second messengers
1. PLC and PI3K can catalyze the synthesis of DG and PIP3 respectively. DG
and PIP3 are lipid second messengers.
P527 fig 25-5
2. Lipid messengers may be also allosteric effecters.
3. The target molecule of PIP3 is PKB.
Ⅳ. The second messenger Ca2+ has a lot of target molecules.
Ⅴ. Effects of nitric oxide are mediated by cGMP.
1. Nitric oxide synthases catalyze NO formation.
2. NO has many physical and patho-physical effects.
Section Three
Proteins Function as Signal Transducers
Ⅰ. Protein kinases and protein phosphatases regulate the activity of receptors
and protein transducers.
1. Phosphorylation and dephosphorylation may regulate protein activity.
2. Most protein kinases are serine/threonine kinases and tyrosine kinases.
3. MAPK cascade is the core of many signal transduction pathways :
a. Growth factor signaling : Ras, Raf and MAPK pathway that regulates cell
proliferation and differentiation.
b. JNK/SAPK pathway is activated by stresses caused by radiation, osmotic
pressure, temperature etc.
c. P38MAPK pathway responds to inflammation and apoptosis signals.
4. PTK transduces signals of cell proliferation and differentiation
a. Growth factor receptors belong to PTKs
P533 fig 25-9
b. There are also non-receptor PTKs
P533 fig 25-10
5. Phosphatidases ( phosphatases ) attenuate signals stimulated by protein
kinases.
Ⅱ. G protein and small G protein are activated by GTP and performs a function,
in the meanwhile GTP is hydrolyzed to GDP.
1. Hetero-trimeric G proteins mediate
signal transduction directly from
G-protein coupled receptors to downstream effecter molecules.
2. Members of Ras superfamily are important signal transducers.
Section Four
The Structural Basis of Signal Transduction Network
Ⅰ. Signal transduction complexes are characteristic.
1. Signal transduction complexes ensure that signal transduction is highly
efficient, precise and diverse.
2. Signal transduction complexes are located on membrane or cytoskeleton.
3. The composition of a signal transduction complex is changeable
dynamically.
Ⅱ. Protein-protein interaction domains are the basis of complex formation.
1. Protein interaction domains mediate protein-protein interaction.
P536 fig 25-12
2. SH2 domain is a Src homology 2 domain, that interacts with the specific
phosphotyrosine residue in target proteins.
3. SH3 domain is a Src homology 3 domain, that interacts with the specific
proline-rich motif in target proteins.
4. PH domain binds to membrane bound phospholipids.
Ⅲ. Adapter proteins and scaffolding proteins
1. Adapter proteins link transducers of signal transduction pathways.
P537 fig 25-13
2. Scaffolding proteins ensure that signal transduction is specific and highly
efficient.
Section Five
General Signal Pathway Mediated by Membrane Receptors
Ⅰ. Ion channel coupled membrane receptors receive signals that activate the
flow of ions across the membrane.
Ⅱ . GPCRs activate a G protein that performs a function via a second
messenger.
1. Activation of a G protein initiates signal transduction
P539 fig 25-15
2. Activated G protein regulates downstream enzyme activity.
3. Glucagon receptor mediates AC-cAMP-PKA pathway.
P540 fig 25-16
4. Angiotensin II receptor mediates PLC-IP3/DG-PKC pathway.
P541 fig 25-17
Ⅲ. Protein kinase coupled receptors perform functions via a protein kinase
cascade.
1. Protein tyrosine kinase coupled receptors have similar signal transduction
pathways.
2. Ras-MAPK pathway is the main pathway of EGFR.
P542 fig 25-18
3. There are other enzyme coupled receptors in the cell.
HOMEWORK
Explain the following terms in English :
1.second messenger
2. receptor
3. signal molecule
4. signal transduction
5.G protein
6. adapter protein
7. scaffolding protein
8. MAPK cascade
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