Science Molecular Biology Practice Questions

Molecular BiologyBIMM100-FA2022
Tue-Thu, 12:30-1:50 PM
Lecture #5:
DNA repair (part II)
Genome editing
Jose Pruneda-Paz
Announcements
Certify commencement of academic activity
by submitting #FinAid survey in Canvas
(Assignments tab)
By tomorrow Oct 7, 2022
DNA repair
DNA
replication
DNA
transcription
RNA
translation
Protein
1. DNA repair (during DNA
replication)
2. DNA repair (independent of
DNA replication)
3. DNA repair (CAS9 induced)
DNA damage can lead to a permanent change
on the DNA sequence (mutation)
Accumulation of
mutations can result in
improper cell function
C
C
Parental dsDNA
From Fig 5-34
What about DNA damage that arises
during DNA replication?
DNA polymerase may add an incorrect (non-complementary)
nucleotide during replication
daughter
template
3’
5’
GCGATG A
3’ CGCTACGTAA 5’
Replication errors => 1 in ≈10,000 (104) nucleotides (*)
Yet the overall mutation rate is only 1
in ≈1,000,000,000 (109) (*)
WHY?
(*) in bacteria
DNA Repair Mechanisms
mismatches that arise
DURING DNA
replication
Damage
INDEPENDENT of
DNA replication
1. Proofreading
2. DNA repair – one strand damaged
a. Mismatch excision repair
b. Base excision repair
c. Nucleotide excision repair
3. DNA repair – dsDNA break
a. Homologous recombination
b. End-joining
Endonucleases and Exonucleases hydrolyze
phosphodiester bonds
Endonuclease enzymes cleave bonds
within a nucleic acid molecule (they
do not remove nucleotides at the end
of the molecule)
Exonuclease enzymes cleave
nucleotides at the ends of a nucleic acid
molecule (they do require a nucleotide
with a free 3’-OH or 5’-PO4)
5’ exonucleases (5’>3’)
3’ exonucleases (3’>5’)
Endonuclease
Exonuclease
DNA polymerases introduce copying errors
A 3’-to-5’ exonucleolytic activity of DNA
polymerase can remove the last added
nucleotide (if a mismatch is detected)
DNA polymerase repairs errors
A 3’-to-5’ exonucleolytic
activity of DNA
polymerase can remove
the last added nucleotide
(if a mismatch is detected)
Replicating DNA polymerases have 3’-to-5’
exonucleolytic proofreading activity
Replication
Proofreading
Fig. 4-34
3’-to-5’ exonuclease site
3’
5’
dAMP5’
GCGATG A
3’ CGCTACGTAA 5’
3’
GCGATG
3’ CGCTACGTAA 5’
Proofreading (3’-to-5’ exo)
5’
3’
GCGATGCA
3’ CGCTACGTAA 5’
Continued replication
Fig 5-33
DNA Repair
DNA replication
What
if the Mechanisms
mutation is during
not proofread
by the
DNA polymerase?
mismatches that arise
DURING DNA
replication
Damage
INDEPENDENT of
DNA replication
1. Proofreading
2. DNA repair – one strand damaged
a. Mismatch excision repair
b. Base excision repair
c. Nucleotide excision repair
3. DNA repair – dsDNA break
a. Homologous recombination
b. End-joining
General strategy of mechanisms that repair a
single strand DNA damage/mutation
*
DNA damage/mutation
1) Damage/mutation recognized
2) Damaged/mutated DNA strand or
nucleotide excised (Nuclease)
3) Gap filled (DNA polymerase + DNA ligase)
Mismatch excision repair
(corrects replication errors right after replication)
1. Incorrect base in
daughter strand is
recognized
2. The mutated DNA
strand is removed
3. Gap is repaired by
DNA Pol + DNA ligase
HOW DOES IT KNOW WHICH IS
THE DAUGHTER STRAND?
Fig 5-36
Mismatch excision repair
(corrects replication errors right after replication)
1. Incorrect base in
daughter strand is
recognized
2. The mutated DNA
strand is removed
3. Gap is repaired by
DNA Pol + DNA ligase
For a short period after replication
only the template strand will be
methylated
Fig 5-36
=CH3
DNA repair mechanisms
during DNA replication
•
Proofreading mechanism
– repair a mismatch immediately after a wrong
nucleotide is added
– requires DNA polymerase 3’>5’ exonuclease
(proofreading) activity
•
Mismatch excision repair
– repair a mismatch as long as the newly synthesized
DNA strand is unmethylated
– requires several enzymatic activities: endonuclease,
helicase, exonuclease, DNA polymerase, DNA ligase
DNA repair
DNA
replication
DNA
transcription
RNA
translation
Protein
1. DNA repair (during DNA
replication)
2. DNA repair (independent of
DNA replication)
3. DNA repair (CAS9 induced)
What can damage DNA integrity?
1. DNA replication errors and some DNA repair mechanisms
2. Spontaneous cleavage of chemical bonds
3. Environmental agents (UV, ionizing radiation)
4. Reaction with geno-toxic chemicals
* By-products of normal cellular metabolism
* Environmental
What about DNA damage that arises
independent of DNA replication?
1. Nucleotide substitution
2. Thymine dimers
3. Abasic sites
4. Single-strand breaks
5. Double-strand breaks
Estimated 10,000 to 1,000,000 mutational events per cell per day!
DNA Repair Mechanisms independent of DNA replication
(single or double strand DNA damages)
mismatches that arise
DURING DNA
replication
Damage
INDEPENDENT of
DNA replication
1. Proofreading
2. DNA repair – one strand damaged
a. Mismatch excision repair
b. Base excision repair
c. Nucleotide excision repair
3. DNA repair – dsDNA break
a. Homologous recombination
b. End-joining
General strategy of mechanisms that repair a
single strand DNA damage/mutation
*
DNA damage/mutation
1) Damage/mutation recognized
2) Damaged/mutated DNA strand or
nucleotide excised (Nuclease)
3) Gap filled (DNA polymerase + DNA ligase)
Base Excision Repair
(replaces a chemically changed base)
DNA is chemically reactive (base deamination)
H2O
NH4
2-Deoxyribose
2-Deoxyribose
H2O
(5-mC, a common modified base in DNA)
NH4
Fig 5-34
Base Excision Repair
(replaces a chemically changed base)
Met
C
G
Met-C deamination
1. Remove incorrect base from
sugar phosphate backbone
HOW DOES DOES THE BASE-EXCISION
REPAIR MACHINERY KNOW WHICH
BASE TO REMOVE?
Human cells contain a battery of
glycosylases that recognize
different chemically modified
DNA bases
Fig 5-35
Base Excision Repair
(replaces a chemically changed base)
1. Remove incorrect base from
sugar phosphate backbone
2. & 3. Endonuclease/lyase
remove deoxyribose
phosphate
4. Gap is filled with correct
base by DNA Pol + ligase
3’
Fig 5-35
Nucleotide excision repair
(fixes distorted DNA)
For example: thymine-thymine dimers
Nucleotide excision repair
1. Distortion is recognized
2. Helix opened (helicase activity)
3. Endonucleases remove
damaged DNA strand
4. Gap is repaired by DNA
Pol + DNA ligase
Fig 5-38
Poll Question #1
Which of the following schematics
correctly illustrates how DNA
polymerase fills in the gap during
excision repair?
A.
– iClicker Question #1 –
B.
C.
Any additional enzymatic
activity required?
What if there is a double-stranded break?
1) Nonhomologous End-joining (NHEJ)- “error prone”
– Two separated ends rejoined
OR
– Different broken ends joined
=> translocations
2) Homologous recombination (HR) – “error free”
– Undamaged DNA (i.e. sister chromatid)
serves as template.
dsDNA Repair by Nonhomologous End Joining
(NHEJ)
Repair by end-joining
1. Recognition of broken
ends
2. Exonucleases remove
overhanging nucleotides
3. Ligation
END-JOINING CREATES ERRORS!
Fig 5-39
Broken ends from different chromosomes
are sometimes end-joined – translocations
LEUKEMIA!
Fig 8-38
Repair by Homologous recombination
1 & 2 dsDNA break is
recognized and
exonucleases remove
nucleotides, leaving
3’end ssDNA
overhangs
Based on Fig 5-41
Repair by Homologous recombination (cont.)
3. ssDNA searches for
homologous sequence
in sister chromatid (strand
invasion)
4. replication using 3’ end
of broken DNA as a
primer and sister DNA
as template
5. repaired strand base
pairs with other broken
strand
6. Gap filled by DNA Pol+ligase
Based on Fig 5-41
Poll Question #2
Why are dsDNA breaks not always repaired
by (error-free) homologous recombination?
A: Homologous recombination is less
efficient than end-joining.
– iClicker Question #2 –
B: Homologous recombination is more
complicated than end-joining.
C: Homologous recombination is restricted
to certain stages of the cell cycle.
D: It creates an evolutionary advantage to
have DNA errors in somatic (tissue) cells.
DNA repair mechanisms independent of DNA
replication
•
Base excision repair and Nucleotide excision repair
– repair single stranded DNA damages
– requires several enzymatic activities: endonuclease,
helicase, lyase, DNA polymerase, and/or DNA ligase
•
End-joining and Homologous recombination
– repair a double stranded DNA breaks
– NHEJ à error prone
– HR à error free, but requires an undamaged
homologous template DNA (i.e. sister chromatid)
What happens when DNA is damaged?
DNA Repair
1. Proofreading
2. DNA repair – one strand damaged
a. Mismatch excision repair
b. Base excision repair
c. Nucleotide excision repair
3. DNA repair – dsDNA break
a. Homologous recombination
b. End-joining (NHEJ)
mismatches that arise
DURING DNA
replication
Damage
INDEPENDENT of
DNA replication
Molecular Biology
BIMM100-FA2022
Tue-Thu, 12:30-1:50 PM
Lecture #4
DNA replication (part III):
Telomerase
DNA repair
Jose Pruneda-Paz
Announcements
1. Certify commencement of academic
activity by submitting #FinAid survey in
Canvas (Assignments tab)
By Oct 7, 2022
2. iClicker mobile enabled
DNA Replication
DNA
replication
DNA
transcription
RNA
translation
Protein
1. Telomerase
2. DNA repair (during DNA
replication)
Some organisms have circular chromosomes…
… but in many cases genomes (i.e. human) are
organized into linear chromosomes
Fig. 8-35
A replicated metaphase
chromosome (just before
chromatid separation)
The end replication problem
DNA would
progressively
shorten!
Fig. 8-43
The end replication problem is solved by extending
the lagging strand template
3’ end of template strand
is extended
3’
5’
Gap filled !!
HOW?
From Fig. 8-43
How do cells replicate the ends (telomeres) of linear
chromosomes?
?
?
Fig. 8-35
Telomeres – ends of linear chromosomes
Muller & McClintock (~1938)
“There is something different about
the ends of linear chromosomes
and double strand DNA breaks”
Barbara McClintock
1) Sequencing and in situ hybridization assays revealed that ends
of chromosomes consist of a short nucleotide sequence
(TTGGGG in Tetrahymena) that is repeated for several kilobases
Probe
(TTGGGG)4
Figs. from: Shakoori, A.R. (2017). Fluorescence In Situ Hybridization
(FISH) and Its Applications. Chromosome Structure and Aberrations.
Fluorophore
5’
3’
TTGGGG TTGGGG TTGGGG…..3’
AACCCC AACCCC AACCCC……5’
2) Identification of an enzyme activity that adds dT and dG to
DNA ends
HOW?
(Greider & Blackburn, Cell, 1985)
Reverse transcriptases catalyze DNA
synthesis using RNA as template
DNA
Reverse transcription
(Reverse transcriptase)
1. Primer strand (DNA)
2. dNTPs
3. RNA template strand
RNA
Replication
(DNA polymerase)
Transcription
(RNA polymerase)
Telomerase is a Reverse Transcriptase and contains
its own template RNA
Telomerase RNA
Telomerase Protein
Telomerase is a
protein-RNA complex
(from Lingner et al., 1997, Science)
Telomerase adds multiple copies of the telomeric
sequence by reverse transcribing telomerase RNA
Telomerase RNA
Lagging strand
3’
Telomerase Protein
(Reverse Transcriptase)
Annealing of Telomerase RNA to the single
stranded 3’ end of the template DNA strand
The source of the telomerase (not the source of the
chromosome being replicated) determines the telomere
sequence
Similar to Fig. 8-44
Telomerase adds multiple copies of the telomeric
sequence by reverse transcribing telomerase RNA
Telomerase RNA
Lagging strand
GGG 3’
Telomerase Protein
(Reverse Transcriptase)
G G G T T G 3’
The source of the telomerase (not the source of the
chromosome being replicated) determines the telomere
sequence
Similar to Fig. 8-44
Telomerase adds multiple copies of the telomeric
sequence by reverse transcribing telomerase RNA
Telomerase RNA
Lagging strand
GGG 3’
Telomerase Protein
(Reverse Transcriptase)
G G G T T G 3’
G G G T T G 3’
Telomerase RNA reanneals to the 3’ end of the newly
synthesized DNA
Similar to Fig. 8-44
Telomerase adds multiple copies of the telomeric
sequence by reverse transcribing telomerase RNA
Telomerase RNA
Lagging strand
GGG 3’
Telomerase Protein
(Reverse Transcriptase)
G G G T T G 3’
G G GT T GGGGTT G 3’
Repeated Elongation, Translocation, etc, etc
Similar to Fig. 8-44
Double-stranded Telomeric sequence subsequently
generated by the Replication machinery
new Okazaki
fragments
Similar to Fig. 8-44
Importance of Telomeres
In humans, telomeres consist of 3-20kb
of DNA of the repeated sequence 5’TTAGGG3’.
– Protection from nucleases
– Length of telomeres regulates cell replicative
capacity:
-Telomeres shorten with every cell division –
Aging!
– Telomerase is often upregulated in cancers
– Why?
DNA looping and multiple proteins protect telomeres
from degradation and recruit telomerase
Telomeric Proteins
Telomeric DNA looping
The mammalian telomeric complex
(from Verdun & Karlseder 2007)
Loss of telomeres normally limits the number of
rounds of cell division
Poll Question #1
While in Tetrahymena the telomeric sequence is
TTGGGG, human telomeres consists of 3-20k bp
of the repeated sequence TTAGGG.
Why Tetrahymena and human telomere
sequences are different?
A: structure of Tetrahymena and human chromosomes.
– iClicker Question
B: enzymatic activity of Tetrahymena
and human #1 telomerase proteins
C: sequence of Tetrahymena and human telomerase
RNAs.
D: sequence of Tetrahymena and human telomerase
protein.
E: ‘B’ and ‘C’.
(from Lingner et al., 1997, Science)
Structure and function
of Telomerase
• The telomerase enzyme is a protein-RNA complex
1) Protein has Reverse Transcriptase (RT) enzymatic activity
2) Associated telomerase RNA is the RT template
• Telomerase adds a repetitive sequence (telomere) to the ends of
linear chromosomes (multiple copies of the sequence repeats in
the telomerase RNA)
• Solves end replication problem by extending the DNA template
(3’ end) used for the lagging strand DNA synthesis
DNA Replication
DNA
replication
DNA
transcription
RNA
translation
Protein
1. Telomerase
2. DNA repair (during DNA
replication)
DNA damage can lead to a permanent change
on the DNA sequence (mutation)
Accumulation of
mutations can result in
improper cell function
C
C
Parental dsDNA
From Fig 5-34
What about DNA damage that arises
during DNA replication?
DNA polymerase may add an incorrect (non-complementary)
nucleotide during replication
daughter
template
3’
5’
GCGATG A
3’ CGCTACGTAA 5’
Replication errors => 1 in ≈10,000 (104) nucleotides (*)
Yet the overall mutation rate is only 1
in ≈1,000,000,000 (109) (*)
WHY?
(*) in bacteria
DNA Repair Mechanisms
mismatches that arise
DURING DNA
replication
Damage
INDEPENDENT of
DNA replication
1. Proofreading
2. DNA repair – one strand damaged
a. Mismatch excision repair
b. Base excision repair
c. Nucleotide excision repair
3. DNA repair – dsDNA break
a. Homologous recombination
b. End-joining
Endonucleases and Exonucleases hydrolyze
phosphodiester bonds
Endonuclease enzymes cleave bonds
within a nucleic acid molecule (they
do not remove nucleotides at the end
of the molecule)
Exonuclease enzymes cleave
nucleotides at the ends of a nucleic acid
molecule (they do require a nucleotide
with a free 3’-OH or 5’-PO4)
5’ exonucleases (5’>3’)
3’ exonucleases (3’>5’)
Endonuclease
Exonuclease
Replicating DNA polymerases have 3’-to-5’
exonucleolytic proofreading activity
Replication
Proofreading
Fig. 4-34
3’-to-5’ exonuclease site
3’
5’
dAMP5’
GCGATG A
3’ CGCTACGTAA 5’
3’
GCGATG
3’ CGCTACGTAA 5’
Proofreading (3’-to-5’ exo)
5’
3’
GCGATGCA
3’ CGCTACGTAA 5’
Continued replication
Fig 5-33
DNA polymerases introduce copying errors
DNA polymerase repairs errors
DNA Repair
DNA replication
What
if the Mechanisms
mutation is during
not proofread
by the
DNA polymerase?
mismatches that arise
DURING DNA
replication
Damage
INDEPENDENT of
DNA replication
1. Proofreading
2. DNA repair – one strand damaged
a. Mismatch excision repair
b. Base excision repair
c. Nucleotide excision repair
3. DNA repair – dsDNA break
a. Homologous recombination
b. End-joining
General strategy of mechanisms that repair a
single strand DNA damage/mutation
*
DNA damage/mutation
1) Damage/mutation recognized
2) Damaged/mutated DNA strand or
nucleotide excised (Nuclease)
3) Gap filled (DNA polymerase + DNA ligase)
Mismatch excision repair
(corrects replication errors right after replication)
1. Incorrect base in
daughter strand is
recognized
2. The mutated DNA
strand is removed
3. Gap is repaired by
DNA Pol + DNA ligase
HOW DOES IT KNOW WHICH IS
THE DAUGHTER STRAND?
Fig 5-36
Mismatch excision repair
(corrects replication errors right after replication)
1. Incorrect base in
daughter strand is
recognized
2. The mutated DNA
strand is removed
3. Gap is repaired by
DNA Pol + DNA ligase
For a short period after replication
only the template strand will be
methylated
Fig 5-36
=CH3
DNA repair mechanisms
during DNA replication
•
Proofreading mechanism
– repair a mismatch immediately after a wrong
nucleotide is added
– requires DNA polymerase 3’>5’ exonuclease
(proofreading) activity
•
Mismatch excision repair
– repair a mismatch as long as the newly synthesized
DNA strand is unmethylated
– requires several enzymatic activities: endonuclease,
helicase, exonuclease, DNA polymerase, DNA ligase
Molecular Biology
BIMM100-FA2022
Tue-Thu, 12:30-1:50 PM
Lecture #2:
DNA structure
DNA replication (part I)
Jose Pruneda-Paz
Structure of Nucleic Acids / DNA replication
DNA
replication
DNA
transcription
RNA
translation
Protein
1. DNA/RNA structure
2. DNA Replication (part I)
DNA/RNA molecules are nucleotide polymers
A nucleotide is composed
of a phosphate, a sugar
and a base
The sugars
RNA
DNA
Fig. 2-16
Two types of bases: Purines and Pyrimidines
(RNA + DNA)
Fig. 2-17
(RNA only)
(RNA + DNA)
(DNA only)
(RNA + DNA)
The chemical nature of DNA and RNA
DNA
RNA
Nitrogenous bases:
Nitrogenous bases:
Adenine (A)
Cytosine (C)
Guanine (G)
Thymine (T)
Adenine (A)
Cytosine (C)
Guanine (G)
Uracil (U)
Phosphate
Phosphate
Sugar: 2’ deoxyribose
Sugar: ribose
Nucleotides have chemically reactive groups in
C3’ and C5’
5’
1’
4’
3’
2’
Nucleotides are linked by phosphodiester
bonds
Fig. 4-2
Nucleotide polymers have an end-to-end
directionality
By convention, a polynucleotide sequence is always
written in the 5’->3’ direction
Fig. 4-2
Poll question #1
Is the polynucleotide chain
shown in the left a DNA or RNA
molecule?
A: DNA
B: RNA
– iClickerC:Question
#1 –or RNA
could be DNA
The 3-D Structure of DNA
•Chargaff’s rules (1947):
In any genomic DNA the #Purines = #Pyrimidines
[A] = [T] and [G] = [C]
However, [A + T] ≠ [G + C]
The 3-D Structure of DNA
•Franklin’s X-ray diffraction data (1952):
DNA is a helix, at least 2 strands
The 3-D Structure of DNA
•Watson and Crick (1953): model of double helix
with purines base-pairing with pyrimidines
Nobel Prize for Medicine 1962 – Watson, Crick, Wilkins
Adenine pairs with Thymine
Hydrogen
Bonding
(non-covalent)
A Hydrogen covalently bound to an electronegative atom (such as N or O) bears a large
partial positive charge and becomes electrostatically attracted to an electronegative
atom (such as N or O)
Guanine pairs with Cytosine
Hydrogen
Bonding
DNA forms a double-helix
Fig. 5-3
Complementary DNA strands are anti-parallel
3’
5’
Strands are
anti-parallel!
5’
3’
Fig. 5-3
Spaces between the intertwined strands form
two helical groves of different widths
Fig. 5-3
Atoms on the edges of each
base within these grooves are
accessible from outside the helix
https://upload.wikimedia.org/wikipedia/commons/0/0c/DNA_animation.gif
DNA molecules can be denatured and
renatured
What bonds
are disrupted?
What would happen
if you heat an RNA
molecule that has a
secondary
structure?
Temperature of melting
(Tm) is the temperature
at which half of the
bases in a double
stranded DNA molecule
have been denatured
How would the DNA
sequence affect the Tm?
Fig.5-7
Tm is a function of the GC content
Fig.5-7
Poll question #2
Which of the following single stranded DNA (ssDNA)
molecules is/are complementary to the ssDNA molecule
indicated in red ?
5’ ACTGAAGT
3’
– iClicker Question #2 –
A: 5’ ACTGAAGT 3’
B: 5’ ACTTCAGT 3’
C: 5’ TGAAGTCA 3’
D: Sequences A and C
E: Sequences B and C
DNA structure
Fig. 5-3
•
Nucleotide structure:
phosphate, sugar (deoxyribose)
and base (A,T,C,G)
•
Nucleotide polymer by
phosphodiester bonds
•
DNA has a double helical
structure:
> strands are directional and
antiparallel
> A-T and G-C paired
•
DNA can be denatured and
renatured
What about RNA 3D structure?
Unlike DNA (which has a conserved 3D structure
regardless of the nucleotide sequence), RNAs
may adopt different 3D conformations depending
on the nucleotide sequence.
Primary structure
5’ TTTTTTTTTTTTTTTTTTTTT – – – – – – – – TTTTTTTTTTTTTTTTTTTT 3’
AGCAACCUUGACUACCUAACG
CGAAGGGUAGCAAGGUCGUA
Complementary regions could form
double helices when the molecule folds
back onto itself
Cellular RNAs are
single-stranded
polynucleotides
What about RNA 3D structure?
Unlike DNA (which has a conserved 3D structure
regardless of the nucleotide sequence), RNAs
may adopt different 3D conformations depending
on the nucleotide sequence.
Primary structure
5’ TTTTTTTTTTTTTTTTTTTTT – – – – – – – – TTTTTTTTTTTTTTTTTTTT 3’
AGCAACCUUGACUACCUAACG
Secondary structure
5’
Cellular RNAs are
single-stranded
polynucleotides
3’
Similar to Fig 5-9
CGAAGGGUAGCAAGGUCGUA
RNAs may adopt different 3D conformations
depending on the nucleotide sequence.
RNA structure
•
Nucleotide structure: phosphate, sugar
(ribose) and base (A,U,C,G)
•
Nucleotide polymer by phosphodiester
bonds
•
Single stranded: directional (5’ & 3’
ends)
•
May adopt different conformations
depending on nucleotide sequence
Structure of Nucleic Acids / DNA replication
DNA
replication
DNA
transcription
RNA
translation
Protein
1. DNA/RNA structure
2. DNA Replication (part I)
DNA Replication has to occur prior to cell division
Eukaryotic cell cycle
WHY?
Mitosis
(cell division)
2
1
Fig. 1-8
Replication
Fig. 1-21
Most genome sequences comprise several
million nucleotides
DNA replication must occur with extraordinary fidelity,
speed and accurate regulation
How is DNA replicated?
Hypotheses:
DNA strand
DNA strand
Replication
Red: old
DNA strand
Blue: new
DNA strand
Is DNA replication conservative or semiconservative?
Late 1950s
Meselson-Stahl experiment (1958)
1 generation
Bacteria grown for
long time with 15N
“heavy” nitrogen
source.
2 generations
Switched to 14N
“light” nitrogen
source.
All DNA 15N
All newly synthesized DNA strands
14N
All old DNA strands 15N
Predictions from Meselson-Stahl’s experiment
IF
IF
Generations
0
1
2
How distinguish? 15N DNA heavy; 14N DNA light
Fig. 5-28a
Results of Meselson-Stahl’s experiment
H
Ultra-centrifugation
M
M:L
(2:2)
M:L
(2:6)
LH
Fig. 5-28b
Interpretation of the Meselson-Stahl experiment
Generations
DNA replication is
semiconservative
H
0
M
1
M:L
(1:1)
2
DNA replication – part I
•
DNA replication is needed early on
during cell division
•
DNA replication results in the
synthesis of identical dsDNA
copies (high fidelity)
•
The mechanism of DNA replication
is semiconservative, thus the
parental dsDNA must unwind prior
to the synthesis of daughter
strands
Molecular Biology
BIMM100-FA2022
Tue-Thu, 12:30-1:50 PM
Lecture #3
DNA replication (part II):
Molecular mechanism
Jose Pruneda-Paz
Announcements
• Assignment #1 will be posted today
• Discussion sections start next week
• Prof office Hours today @ 3PM
DNA Replication
DNA
replication
DNA
transcription
RNA
translation
Protein
1. Molecular mechanism
DNA replication – part I (lecture 2)
The mechanism of DNA replication is
semiconservative,
thus the parental dsDNA must unwind
prior to the synthesis of daughter
strands
HOW?
DNA replication is semiconservative
Initiation of DNA replication requires unwinding of the
template DNA, which is catalyzed by DNA helicases
Origin
DNA unwinding by
helicases (ATP)
Why ?
Replication bubble
Fig. 5-32
Why does DNA unwinding require ATP?
Denaturation: the process that separates DNA strands.
(Requires energy).
What bonds
are disrupted?
dsDNA
Energy
ssDNA
Annealing/Hybridization: uniting single stranded DNA with DNA (or RNA) that is
complementary in sequence
(Energetically favorable).
Is replication uni- or bi-directional?
Origin
DNA unwinding by
helicases (ATP)
Replication bubble
Fig. 5-32
Is replication uni- or bi-directional?
Experiment setup and predicted outcomes
Uni-directional
DNA synthesis
starting point
+ mix of labeled (1%) and nonlabeled nucleotides (99%)
(Light signal)
+ only labeled (100%)
nucleotides
(strong signal)
Unlabeled
DNA template
Labeled newly
synthesized DNA
Direction of
DNA synthesis
Is replication uni- or bi-directional?
Experiment setup and predicted outcomes
Bi-directional
Uni-directional
DNA synthesis
starting point
+ mix of labeled (1%) and nonlabeled nucleotides (99%)
(Light signal)
+ only labeled (100%)
nucleotides
(strong signal)
Unlabeled
DNA template
Labeled newly
synthesized DNA
Direction of
DNA synthesis
Is replication uni- or bi-directional?
Expected results
DNA synthesis
starting point
Uni-directional
Bi-directional
only the newly synthesized DNA is labeled!!
Uni-directional
Unlabeled
DNA template
Bi-directional
Labeled newly
synthesized DNA
Direction of
DNA synthesis
Is replication uni- or bi-directional?
Observed result:
(black signal corresponds to radioactive DNA)
DNA replication is bidirectional !!
Poll Question #1
The following picture provides a snapshot of an
active replication fork.
Considering the experiment results described in the
previous slides, where in this replication fork is DNA
being synthesized at the time the image was taken?
B
C
– iClicker Question #1 –
A
D: A and C
E: A, B and C
DNA replication (part I summary)
Origin
DNA unwinding by
helicases (ATP)
Fig. 5-32
Fork
Fork
Replication bubble (DNA synthesis)
•
DNA replication is semiconservative
•
DNA replication is bidirectional (two
replication forks per replication bubble)
DNA replication is bidirectional
DNA synthesis occurs at both replication forks in
opposite directions
Origin
DNA unwinding by
helicases (ATP)
Fork
Fork
Fig. 5-32
Unwinding the template DNA creates torsional
stress downstream of each replication fork
Origin
DNA unwinding by
helicases (ATP)
Fork
Fork
Fig. 5-32
Torsional stress is relieved by Topoisomerases
Topoisomerase I
Torsional stress due to DNA unwinding is
relieved by Topoisomerases
Origin
DNA unwinding by
helicases (ATP)
Fork
Torsional stress relieved by
topoisomerases (ATP)
Fork
How does DNA synthesis happen in the replication bubble?
Fig. 5-32
DNA is synthesized by an enzyme called
DNA polymerase
DNA replication is bidirectional so synthesis occurs at
both replication forks in opposite directions
Origin
DNA unwinding by
helicases (ATP)
Fork
Fork
How does DNA polymerase catalyze DNA synthesis?
Fig. 5-32
DNA Polymerase is a template-directed enzyme
Adds complementary bases using a single stranded DNA
molecule as template
DNA Polymerases catalyze the formation of
phosphodiester bonds
DNA polymerase catalyzes the nucleophilic attack of a polynucleotide
chain 3’OH terminus on the alfa-phosphate group of an incoming dNTP
Incoming
dNTP
5’
H
H
T
T
H
H
3’
3’
Base pairing
5’
DNA template strand
DNA Polymerases catalyze the formation of
phosphodiester bonds
DNA polymerase catalyzes the nucleophilic attack of a polynucleotide
chain 3’OH terminus on the alfa-phosphate group of an incoming dNTP
DNA Pol
Incoming
dNTP
5’
H
H
T
T
H
H
3’
5’-to-3’ DNA
strand growth
Base pairing
3’
5’
DNA template strand
NUCLEIC ACID SYNTHESIS: ALWAYS 5’ to 3’!!!
DNA Polymerases catalyze the formation of
phosphodiester bonds
Initiation of DNA replication requires an RNA primer
RNA primase: RNA polymerase
that catalyzes the synthesis of
the RNA primer during DNA
replication
DNA Pol
Incoming
dNTP
RNA Primer
5’
T
H
3’
5’-to-3’ DNA
strand growth
Base pairing
3’
5’
DNA template strand
NUCLEIC ACID SYNTHESIS: ALWAYS 5’ to 3’!!!
General rules of DNA synthesis by DNA polymerases
DNA Pol needs:
1.
2.
3.
ssDNA template strand
Primer strand (RNA)
dNTPs
DNA Pol
Incoming
dNTP
RNA Primer
5’
T
H
3’
5’-to-3’ DNA
strand growth
Base pairing
3’
5’
DNA template strand
NUCLEIC ACID SYNTHESIS: ALWAYS 5’ to 3’!!!
DNA Polymerases catalyze the
formation of phosphodiester bonds
DNA Pol needs:
1.
2.
3.
ssDNA template strand
Primer strand (RNA)
dNTPs
Replication initiation requires DNA unwinding
and synthesis of RNA primers
Origin
DNA unwinding by helicases (ATP)
Torsional stress relieved by
topoisomerases (ATP)
RNA primer made by Primase
DNA synthesis by DNA polymerase
Fork
Fork
Fig. 5-32
Poll Question #2
Where is DNA polymerase?
B
A
C
– iClicker Question #2 –
D: A and C
E: A, B and C
If polynucleic acid synthesis is always 5’-to-3’, how
can one strand be synthesized in the “3’-to-5’”
direction?
3’
3’-to-5’
3′
synthesis???
5’
3’
Fig. 5-29
5’
Replication occurs in Leading and Lagging strands
One leading and one lagging strand are synthesized in each replication fork
Replication
fork
Synthesis away from fork
Repeated priming
Synthesis towards fork
Single primer
Fig. 5-29
Multiple enzymes in replication (leading strand)
Relaxes supercoiling (ATP)
TOPOISOMERASE
Unwinds dsDNA (ATP)
HELICASE
DNA POLYMERASE
Synthesizes DNA
PRIMASE
Makes RNA primer
Lagging strand synthesis
NOTE:
Direction of fork
O
O
P
3’ T T
5’
O-
O-
OH
T 5’
3’
TTTTTTTT
DNA ligase + ATP
O
3’OH
O
O-
P
OH
O-
Phosphodiester bond
O
O
P
O
O-
3’ T T T T T T T T 5’
5’
3’
TTTTTTTT
(by DNA polymerase)
3’ T T T T T T T T 5’
5’
3’
TTTTTTTT
(by RNase)
Enzymes involved in lagging strand synthesis
1) Degrades RNA primers
1) RNase
2) Fills in gaps from degraded primers
2) DNA POLYMERASE
3) Ligates DNA fragments
3) DNA LIGASE
Replication
fork
TOPOISOMERASE
HELICASE
DNA POLYMERASE
PRIMASE
Fig. 5-29
Poll Question #3
Which of the following enzymatic activities (1
through 5) and in which order are required for DNA
replication on the lagging strand in eukaryotes?
(Each activity may be used more than once or not
at all)
1. DNA polymerase
2. Primase
A) 1, 2,Question
4, 5
– iClicker
#3 –
3. DNase
B) 2, 1, 3, 5
4. RNAse
C) 2, 1, 4, 5
5. DNA ligase
D) 2, 1, 4, 1, 5
E) 2, 1, 4, 5, 1
DNA Replication mechanism overview
Origin
Fork
Fork
Animation: http://sciprim.com/html/repFork.html
Fig. 5-32
DNA Replication
Fig. 5-32
•
DNA replication is semiconservative
•
DNA replication is bidirectional (two
replication forks per replication bubble)
•
DNA polymerase catalyzes DNA synthesis
•
Other enzymatic activities needed: helicase,
topoisomerase, Primase, DNA ligase, RNAse
•
DNA polymerase adds new nucleotides at the
3′ end of an existing strand using a single
stranded DNA molecule as template (added
nucleotides complementary to template)
•
One leading and one lagging strand are
synthesized in each replication fork
How are large chromosomes rapidly replicated?
DNA replication can start at multiple origins
Some antibiotics target
DNA replication
Molecular Biology
BIMM100-FA2022
Tue-Thu, 12:30-1:50 PM
Lecture #9:
Recombinant DNA
Jose Pruneda-Paz
BIMM100 – Midterm preparation
1) Midterm: Tuesday 10/25 (lectures 1-9)
Online Exam in Canvas
2) Regular IA & professor office hours
In person or Zoom as indicated in Syllabus
3) Midterm review session: Saturday 10/22 (10A-noon)
Zoom meeting link in class website
4) Additional professor office hours: Friday 10/21 (2-3:30 PM)
In person (Muir Bio bldg. #1102)
5) Optional Assignment (lectures 8-9) and iClicker questions
Will be posted in Canvas after class today
No discussion section next week
Major classes of nuclear eukaryotic DNA
1. Genes
a. RNA-coding only
b. Protein-coding
2. Repetitious DNA
a. Simple-sequence DNA
b. Interspersed DNA repeats
(mobile DNA elements)
3. Unclassified intergenic DNA
Roles of repetitious DNA in evolution ?
Repetitious DNA sequences are found multiple times across
genomes
Transposase
Cut&Paste
Reverse
transcriptase
DNA transposons
Copy&Paste
Retrotransposons
Transposase
Integrase
Fig 8-8
Repetitious DNA influenced genome evolution by
providing homologous sites for unequal crossing over
1. Generation of gene families – duplications
2. Creation of new genes – exon shuffling
Fig 8-18
3. Formation of complex regulatory regions that control gene expression
Major classes of nuclear eukaryotic DNA
1. Genes
2. Repetitious DNA
3. Unclassified DNA
How can we study the function of a specific
part of a genome?
To study the function of a specific part of a genome it is
typically necessary to isolate it and propagate it
Steps to clone a gene or DNA segment:
1) Select and prepare a small molecule
of DNA capable of self-replication
(cloning vector or plasmid)
2) Obtain DNA fragment to be cloned
(i.e. using restriction endonucleases)
3) Generate recombinant DNA molecule
by ligating the two DNA molecules
covalently
4) Transfer the recombinant DNA from
the test tube to a host organism
5) Identify cells containing the
recombinant DNA
Basic components of a bacterial replicating plasmid
1. Replication origin
2. Selection marker
3. Exogenous DNA
cloning region
(polylinker)
Contains single
restriction
endonuclease
sites
Fig. 6-13
Restriction enzymes or restriction endonucleases are
sequence-specific DNA endonucleases
Many bacteria contain restriction endonucleases to “restrict”
invasion by a foreign DNA
Invading DNA (e.g. virus)
Bacterial genome DNA
Modifying enzyme (methylase)
5’
3’
GAATTC
CTTAAG
3’
5’
EcoRI
5’
3’
G 3’
5’ AATTC
CTTAA 5’
3’ G
3’
5’
5’
3’
CH3
I
GAATTC
CTTAAG
I
CH3
3’
5’
EcoRI
Based on Fig. 6-11
Restriction enzymes cleave DNA at specific
(usually palindromic) sequences
Fig. 5-11
5’
3’
GAATTC
CTTAAG
3’
5’
EcoRI
5’
3’
G 3’
5’ AATTC
CTTAA 5’
3’ G
3’
5’
Complementary (“sticky”) ends
Based on Fig. 6-11
Digested fragments (with compatible ends) can be
ligated using DNA ligase
Fig. 5-11
5’
3’
GAATTC
CTTAAG
3’
5’
EcoRI
5’
3’
3’
5’
G 3’
5’ AATTC
CTTAA 5’
3’ G
Complementary (“sticky”) ends
+ DNA ligase
5’
3’
GAATTC
CTTAAG
3’
5’
EcoRI site restored
Based on Fig. 6-11/6-12
Generate a recombinant DNA molecule by ligating a
DNA fragment to a cloning vector
Vector
DNA fragment
DNA
ligase
Fig. 6-12
iClicker Question #1
Which fragment could be ligated
into plasmid X?
NotI
A:
NotI BamHI
EcoRI NotI
B: – iClicker Question #1 –
Plasmid X
C:
HindIII NotI
NotI NotI
D:
NotI SalI
E:
iClicker Question #2
Which fragment could be ligated into
plasmid X digested with SmaI and NotI?
A:
SmaI
NotI
SmaI SmaI
SmaI NotI
– iClicker
Question #2 –
B:
Plasmid X
C:
D:
E:
NotI NotI
NotI SmaI
B and D
Transforming ligated DNA to a host organism and
obtaining clones with your recombinant plasmid
All cells in a colony contain the
same recombinant DNA plasmid
Bacterial colonies
containing your plasmid
(only plasmid-containing
bacteria will survive on
ampicillin)
Fig. 6-14
Plate on selective medium (e.g. containing ampicillin)
Plasmid restriction digestion analysis could be used to
identify bacterial clones containing your plasmid
500 bp
4000 bp
Which clone (A-E) has the correct plasmid?
4,000 bp
3,000 bp
2,000 bp
1,000 bp
500 bp
A
B
Some clones may contain
“insert-free” religated plasmid
C
D
E
Pick colonies (A-E);
isolate plasmids;
restriction digest
isolated plasmids with
BamHI;
Separate in agarose gel
In general we want to propagate a specific DNA
segment (i.e. a protein coding sequence)
Gene of interest
Why?
Vector
– Express protein (for example in bacteria)
For protein function/structure studies.
For medical or nutritional production of a protein.
– Change the expression of a gene in an organism
(gain- or loss-of-function)
For gene function studies.
How do you obtain a specific DNA fragment
(for example a protein coding sequence)?
DNA fragments of interest a rarely flanked by the restriction
endonuclease sites available at the polylinker region of a cloning vector
From a DNA source
(e.g. genomic DNA)
From an RNA source
(e.g. total RNA)
How can we generate multiple dsDNA copies of a specific
genome region or a specific transcript?
dsDNA synthesis using the polymerase chain reaction (PCR)
The polymerase chain reaction (PCR) is used to
synthesize millions of copies of a target DNA
fragment from a complex source (i.e. genomic DNA)
1. Template dsDNA
2. DNA polymerase
3. dNTPs
4. Primers (two primers driving the synthesis of
complementary DNA strands)
The polymerase chain reaction (PCR) is used to
synthesize millions of copies of a target DNA
fragment from a complex source (i.e. genomic DNA)
Basic PCR reagents and steps
DNA
template
Primers
5’
3’
DNA
polymerase
5’ Primer 3’
3’Primer 5’
3’
95˚C
50-72˚C
5’
72˚C
Fig 5-23 (7e)
Fig 6-18 (8e)
Basics of PCR (polymerase chain reaction)
Target
DNA
molecules
Fig 5-23
Exponential amplification of
the target DNA segment
2
1
3
Animation at:
http://www.dnalc.org/view/15475-The-cyclesof-the-polymerase-chain-reaction-PCR-3D-
Denaturation step inactivates E.
coli DNA polymerase ..
2
A thermophilic bacterium
(Thermus aquaticus) isolated
from hot springs at Yellowstone
1
3
.. but Taq DNA polymerase is
stable at high temperatures
Animation at:
http://www.dnalc.org/view/15475-The-cyclesof-the-polymerase-chain-reaction-PCR-3D-
How do you obtain a specific DNA fragment (i.e. a gene)?
From a DNA source
(e.g. genomic DNA)
PCR
From an RNA source
(e.g. total RNA)
Genomic DNA
3’
5’
Primer 1
Primer 2
dNTPs
3’
5’
PCR
5’
3’
3’
5’
Ligation to
cloning vector
Primer 1
Primer 2
5’
PCR
– Transform to bacteria,
– Select
5’
3’
3’
5’
Recombinant DNA allows us to identify the function of DNA
sequences found in a genome
•
Used to understand the function of
DNA sequences (i.e research) or to
obtain gene products from
heterologous sources (i.e.
pharmaceutical or industry products)
•
Plasmids: used to propagate
recombinant DNA
•
Restriction enzymes: sequencespecific endonucleases (may create
5’ or 3’ overhang ends)
•
PCR: synthesize a specific segment
of DNA (millions of copies)
Molecular Biology
BIMM100-FA2022
Tue-Thu, 12:30-1:50 PM
Lecture #8:
Major classes of nuclear eukaryotic DNA
Jose Pruneda-Paz
BIMM100 – Midterm preparation
1) Midterm: Tuesday 10/25 (lectures 1-9)
Online Exam in Canvas
2) Regular IA & professor office hours
In person or Zoom as indicated in Syllabus
3) Midterm review session: Saturday 10/22 (10A-noon)
Zoom meeting link in class website
4) Additional professor office hours: Friday 10/21 (2-3:30 PM)
In person (Muir Bio bldg. #1102)
5) Optional Assignment (lectures 8-9) and iClicker questions
Will be posted on Canvas after class on Thursday
No discussion section next week
What is in a genome?
Major classes of nuclear eukaryotic DNA
1. Genes
a. RNA-coding only
b. Protein-coding
2. Repetitious DNA
a. Simple-sequence DNA
b. Interspersed DNA repeats
(mobile DNA elements)
3. Unclassified intergenic DNA
Some genes encode proteins…
Protein Coding Gene
+1
DNA:
5’UTR
PROMOTER
3’UTR
CODING
ATG
stop
(UTR: Untranslated Region)
Transcription
mRNA:
5’
3’
Translation
protein:
… but other genes produce RNAs that do not
encode proteins (non-coding RNAs)
Protein coding genes may encode a single protein
(eukaryotes) or multiple proteins (prokaryotes)
Most eukaryotic mRNAs
encode a single type of protein
(MONOCYSTRONIC)
Many bacterial genes are
organized in operons
(POLYCISTRONIC)
gene
Z
Transcription
mRNA
Start site for
protein synthesis
Translation
Z
Protein/s
Based on Fig. 5-13
In eukaryotes, most RNA transcripts are further
processed after transcription
RNA Coding Region
DNA:
PROMOTER
Transcription
Primary transcript
(RNA)
5’
3’
RNA processing
Processed “mature” transcript
(RNA)
5’
3’
(messenger RNA, transfer RNA, ribosomal RNA, small RNA posttranscriptional
processing will be discussed throughout the course)
Some RNAs require further processing after
transcription (for example most protein-coding mRNAs
in eukaryotes)…
DNA
gene
pre-mRNA
Pre-mRNA
Processing
5’ Capping,
3’
Polyadenylation,
Splicing
introns
RNA
exons
In Bacteria/archaea
Protein coding genes
don’t undergo
this kind of
processing!
mRNA
cap
poly(A) tail
3’UTR
5’UTR
Protein coding region
Protein
Based on Fig. 5-15
… but some genes encode RNAs that do not require
further processing after transcription (for example
protein-coding mRNAs in prokaryotes do not contain
introns)
gene
mRNA
Protein coding region
Some genes encode a single RNA product
(Simple genes: one gene, one product)…
Protein-coding gene
Fig. 8-3a
Nonprotein-coding RNA gene (e.g. 5S rRNA)
5S rRNA
… but other genes encode multiple RNA products
(Complex genes: one gene, multiple products)
Complex genes require further processing after transcription
Promoter
Poly(A)
Gene
mRNA1
5’
mRNA2
5’
mRNA3
5’
3’
Alternative splicing
3’
3’
translation
Different protein isoforms
Based on Fig. 8-3b
Alternative splicing
Example: a-tropomyosin gene is alternatively
spliced in different cell types/tissues
Some genes are found only once in the genome…
but others in multiple copies
SOLITARY GENES: Found only once in the genome
DUPLICATED GENES: Multiple genes of close but
usually not identical sequences (often within 5-50kb of
one another)
1) Heavily used genes.
2) Gene families (encode homologous proteins
that constitute a protein family of related functions)
Some duplicated genes encode heavily used gene
products
Number of rRNA gene repeats:
Yeasts: 100-200
Human: ≈300
Heavily used gene products are encoded by multiple
gene copies
Many new protein functions evolved through gene
duplications
Example: The human b-globin gene family:
expressed
in adult
expressed in
fetal stage
(higher O2
affinity)
expressed
in adult
1. Beneficial mutations
2. Pseudogenes – dead genes, no longer active
More than 95% of the sequence of human protein
coding genes consist of introns and flanking noncoding regions
RNA Coding Region
DNA:
PROMOTER
Transcription
Primary transcript
(RNA)
5’
3’
RNA processing
Processed transcript
(RNA)
5’
3’
More than 20,000 genes in our genome
(introns ~38% but exons only ~2% of our genome)
A gene is the entire DNA sequence needed to make a
functional gene product, i.e. RNA.
•
Not all genes encode proteins.
•
Some protein-coding genes
encode a single polypeptide but
others (i.e. in bacteria) encode
multiple ones.
•
Many gene products require
further processing but others
(i.e. in bacteria) do not.
•
Some genes encode a single
RNA product but others (i.e. via
alternative processing) encode
multiple RNA products.
•
Some genes are found only once
in a genome but others are
found in multiple copies.
Gene
+1
DNA
PROMOTER
PROMOTER
region
RNA-coding region
Transcription
RNA
RNA:
5’
3’
iClicker Question
What segments of DNA constitute the globin gene?
5’UTR
intron
exon
intron
exon
3’UTR
exon
DNA
mRNA
A: The coding region (all- red
regions).Question #1 –
iClicker
B: All exons (grey and red regions).
C: Coding region and introns (red and blue regions).
D: All exons and introns (grey, red and blue regions)
and the poly(A) site.
E: Exons, introns, poly(A) site and a region upstream
(left) of the 5’UTR.
iClicker Question
A, B, C, D, and E indicate positions of point
mutations found in alleles of the complex gene
below.
Which of these mutations is likely to change the
amino acid sequence of the protein encoded
from mRNA1 only?
– Promoter
iClicker Question #2 –
Promoter
A
Alternative
mRNAs
mRNA
3
mRNA
B
C
5
’
5
’
D
E Poly(A
)
3
’
3
’
Major classes of nuclear eukaryotic DNA
1. Genes
a. RNA-coding only
b. Protein-coding
2. Repetitious DNA
a. Simple-sequence DNA
b. Interspersed DNA repeats
(mobile DNA elements)
3. Unclassified intergenic DNA
Simple-sequence – repetitious DNA
(≈6% of human genome)
Simple-sequence – repetitious DNA
(≈6% of human genome)
Satellite DNA: 14-500 bp repeats in tandem, 20-100kb
Microsatellites: 1-13bp repeats in tandem, up to 2 kb
How do these repeats arise?
1. replication errors
2. unequal crossover
Much of this repetitious DNA at centromeres
The number of repeats can change through unequal
crossing over during meiosis
Meiosis I:
n-1
Meiosis II:
Result: # repeats can be highly
variable between individuals
Biology (Campbell, 9 ed) Fig 13-12
Major classes of nuclear eukaryotic DNA
1. Genes
a. RNA-coding only
b. Protein-coding
2. Repetitious DNA
a. Simple-sequence DNA
b. Interspersed DNA repeats
(mobile DNA elements)
3. Unclassified intergenic DNA
Mobile DNA elements = Transposable elements
Developed the transposon theory
by studying corn kernel coloration.
Barbara McClintock
MacArthur “Genius” Award, 1981; Lasker
Award, 1981; Nobel Prize, 1983
Wild-Type
Mutant
Revertants
DNA can move within and between chromosomes
~1940s – very controversial idea!
More than 3 million mobile elements in our genome!
(~45%)
Two types of transposons
1) DNA transposons (transpose directly as DNA)
2) Retrotransposons (transpose via an RNA
intermediate molecule)
-LTR
-non-LTR (LINE, SINE)
Two general types of transposition
Transposase
Cut&Paste
DNA transposons
Transposase
Fig 8-8
Structure of bacterial DNA transposons
Transposon sequence encodes the transposase enzyme
Fig. 10-9
Encodes
Transposase
Catalyzes DNA
transposition
Fig 8-9
DNA transposons can increase in copy number
during DNA replication
Fig 8-11
Two general types of transposition
Transposase
Cut&Paste
Reverse
transcriptase
DNA transposons
Copy&Paste
Retrotransposons
Transposase
Integrase
Fig 8-8
Structure of eukaryotic LTR retrotransposons
Transposon sequence encodes the Reverse transcriptase
and Integrase enzymes
Encodes
Reverse Transcriptase
and
Catalyze Retrotransposition
Integrase
LTR: Long terminal repeat
Fig 8-12
Transposons are kept (mostly) silent
In humans:
• only 1 new (germline) transposition in every 8 people
• rarely occurs in somatic cells
• accounts for only 0.1-0.2% spontaneous mutations
Strong silencing maintained at both the transcriptional and post
transcriptional levels: chromatin structure and RNAi (more on
this in future weeks)
Repetitious DNA influenced genome evolution by
providing homologous sites for unequal crossing over
1. Generation of gene families – duplications
2. Creation of new genes – exon shuffling
Fig 8-18
3. Formation of complex regulatory regions that control gene expression
Repetitious DNA and genome evolution
Example: The human b-globin gene family:
expressed
in adult
Sequence drift:
expressed in
fetal stage
(higher O2
affinity)
expressed
in adult
1. Beneficial mutations
2. Pseudogenes – dead genes, no longer active
Repetitious DNA is extensively found in eukaryotic genomes
•
Some repetitious DNA are mobile
elements (transposons) and some
are not (simple sequence repeats)
•
Repetitious DNA contributed to
the evolution of genomes
•
Simple sequence repeats could be
introduced by unequal crossover
(meiosis)
•
The # of simple sequence repeats
in specific genomic regions is
variable across individuals (DNA
fingerprint)
•
DNA and LTR transposons are
mobilized through different
mechanisms
•
Transposon sequences encode
enzyme/s needed for mobilization
Major classes of nuclear eukaryotic DNA
1. Genes
a. RNA-coding only
b. Protein-coding
2. Repetitious DNA
a. Simple-sequence DNA
b. Interspersed DNA repeats
(mobile DNA elements)
3. Unclassified intergenic DNA
25% of our genome is “unclassified”!
Potential functions of unclassified intergenic DNA
Role in gene expression?
• complex transcriptional control regions affecting promoter activity
Role in chromatin organization?
• may impact structure of chromosomes and how DNA is organized
in the nucleus
Unknown genes – new proteins or non-coding RNAs?
• Deep sequencing of RNA transcripts = lots of transcription
occurring in spacer DNA!
Molecular Biology
BIMM100-FA2022
Tue-Thu, 12:30-1:50 PM
Lecture #7:
DNA sequencing
Major classes of nuclear eukaryotic DNA
(genes)
Jose Pruneda-Paz
How can we determine the sequence of
nucleotides in DNA?
? ? ?
DNA sequencing by DNA synthesis
What is required to synthesize DNA “in vitro”?
1. Template DNA
2. Primer (DNA)
3. dNTPs
4. DNA polymerase
What controls where the DNA polymerase starts?
How can the primer get access?
DNA sequencing using chain terminators (Sanger)
What happens to DNA synthesis in
the presence of ddNTPs?
Chain termination by ddNTPs
5’->3’
synthesis
Incoming
ddNTP
3’
H
H
T
T
H
H
H
5’
No 3’-OH to
attack the next
incoming dNTP!
3’
Base pairing
5’
DNA template strand
Based on Fig. 5-10
iClicker Question
You anneal the primer and DNA template
shown below for a sequencing reaction.
Primer 5′-CTAAT-3’
DNA: 3’-GACGATTACCATTC-5’
You add DNA polymerase, and a mix of
dGTP, dTTP, dATP, and dCTP.
– iClicker Question #1 –
What is the product you expect?
A: 5’-CTGCTAAT-3’
B: 5’-CTAATGGTAAG-3’
iClicker Question
You anneal the primer and DNA template
shown below for a sequencing reaction.
Primer 5′-CTAAT-3’
DNA: 3’-GACGATTACCATTC-5’
You add DNA polymerase, a mix of dGTP,
dTTP, dCTP, and ddATP instead of dATP.
What of the following DNA
synthesisQuestion
product/s
iClicker
#2 –
you expect?
A: 5’-CTAATGGTAAG-3’
B: 5’-CTAATGGTA-3’
C: 5’-CTAATGGT-3‘
D: 5’-CTAATGGTA-3’ + 5’-CTAATGGTAA-3’
E: 5’-CTAATGGTA-3’ + 5’-CTAATGGTAA-3’ + 5’-CTAATGGTAAG-3’
iClicker Question
You anneal the primer and DNA template shown
below for a sequencing reaction.
Primer 5′-CTAAT-3’
DNA: 3’-GACGATTACCATTC-5’
You add DNA polymerase, a mix of dGTP, dTTP,
dCTP, and dATP, and also LOW amounts of ddATP.
What of the following DNA
synthesis product/s
you #3 –
– iClicker
Question
expect?
A: 5’-CTAATGGTAAG-3’
B: 5’-CTAATGGTA-3’
C: 5’-CTAATGGTA-3’ + 5’-CTAATGGTAA-3’ + 5’-CTAATGGTAAG-3’
D: 5’-CTAATGGT-3‘
E: 5’-CTAATGGTA-3’ + 5’-CTAATGGTAA-3’
iClicker Question
You anneal the primer and DNA template shown
below for a sequencing reaction.
Primer 5′-CTAAT-3’
DNA: 3’-GACGATTACCATTC-5’
You add DNA polymerase, a mix of dGTP, dTTP,
dCTP, and dATP, and also LOW amounts of ddATP,
ddTTP, ddGTP and ddCTP.
– iClicker
Question
What of the following DNA
synthesis product/s
you #4 –
expect?
A: 5’-CTAATGGTA-3’ + 5’-CTAATGGTAA-3’
B: 5’-CTAATG-3’ + 5’-CTAATGG-3‘
C: 5’-CTAATGGT-3‘
D: 5’-CTAATGGTAAG-3’ + 5’-CTAATGGTAA-3’
E: All the molecules mentioned above (in A, B, C, D)
DNA sequencing by DNA synthesis
(Sanger)
1. Template DNA
2. primer/s (controls where DNApol starts)
3. dNTPs
4. DNA polymerase
5. ddNTPs (in limiting concentration to randomly
terminate DNA synthesis)
Sanger sequencing using fluorescently labeled ddNTPs
template
primer
3’CATCGTTAGTACTGAACTC 5’
+ DNApol
5’GTAG
+ dNTPs
+ ddNTPs
(ddATP, ddCTP, ddTTP, ddGTP)
Sanger sequencing using fluorescently labeled ddNTPs
template
primer
3’CATCGTTAGTACTGAACTC 5’
+ DNApol
5’GTAG
+ dNTPs
+ ddNTPs
(ddATP, ddCTP, ddTTP, ddGTP)
5’GTAGC
5’GTAGC
5’GTAGC
5’GTAGC
5’GTAGC
5’GTAGC
5’GTAGC
5’GTAGC
Sanger sequencing using fluorescently labeled ddNTPs
template
primer
3’CATCGTTAGTACTGAACTC 5’
+ DNApol
5’GTAG
+ dNTPs
+ ddNTPs
(ddATP, ddCTP, ddTTP, ddGTP)
5’GTAGCA
5’GTAGCA
5’GTAGCA
5’GTAGCA
5’GTAGCA
5’GTAGCA
5’GTAGCA
5’GTAGCA
Sanger sequencing using fluorescently labeled ddNTPs
template
primer
3’CATCGTTAGTACTGAACTC 5’
+ DNApol
5’GTAG
+ dNTPs
+ ddNTPs
(ddATP, ddCTP, ddTTP, ddGTP)
5’GTAGCA
5’GTAGCAA
5’GTAGCAA
5’GTAGCAA
5’GTAGCAA
5’GTAGCAA
5’GTAGCAA
5’GTAGCAA
Sanger sequencing using fluorescently labeled ddNTPs
template
primer
3’CATCGTTAGTACTGAACTC 5’
+ DNApol
5’GTAG
+ dNTPs
+ ddNTPs
(ddATP, ddCTP, ddTTP, ddGTP)
5’GTAGCA
5’GTAGCAAT
5’GTAGCAAT
5’GTAGCAAT
5’GTAGCAAT
5’GTAGCAA
5’GTAGCAAT
5’GTAGCAAT
Sanger sequencing using fluorescently labeled ddNTPs
template
primer
3’CATCGTTAGTACTGAACTC 5’
+ DNApol
5’GTAG
+ dNTPs
+ ddNTPs
(ddATP, ddCTP, ddTTP, ddGTP)
5’GTAGCA
5’GTAGCAATC
5’GTAGCAAT
5’GTAGCAATC
5’GTAGCAATC
5’GTAGCAA
5’GTAGCAATC
5’GTAGCAATC
Sanger sequencing using fluorescently labeled ddNTPs
template
primer
3’CATCGTTAGTACTGAACTC 5’
+ DNApol
5’GTAG
+ dNTPs
+ ddNTPs
(ddATP, ddCTP, ddTTP, ddGTP)
5’GTAGCA
5’GTAGCAATCA
5’GTAGCAAT
5’GTAGCAATCA
5’GTAGCAATC
5’GTAGCAA
5’GTAGCAATCA
5’GTAGCAATCA
Sanger sequencing using fluorescently labeled ddNTPs
template
primer
3’CATCGTTAGTACTGAACTC 5’
+ DNApol
5’GTAG
+ dNTPs
+ ddNTPs
(ddATP, ddCTP, ddTTP, ddGTP)
5’GTAGCA
5’GTAGCAATCAT
5’GTAGCAAT
5’GTAGCAATCATG
5’GTAGCAATC
5’GTAGCAA
5’GTAGCAATCA
5’GTAGCAATCATGA
Electrophoresis
of ssDNA
fragments
Gel electrophoresis to separate DNA molecules
, monitor fluorescence
Fig 5-19
Sanger sequencing using fluorescently labeled ddNTPs
template
primer
3’CATCGTTAGTACTGAACTC 5’
+ DNApol
5’GTAG
+ dNTPs
+ ddNTPs
(ddATP, ddCTP, ddTTP, ddGTP)
5’GTAGCAATCATGA
5’GTAGCAATCATG
5’GTAGCAATCAT
5’GTAGCAATCA
5’GTAGCAATC
Electrophoresis
5’GTAGCAAT
of ssDNA
5’GTAGCAA
5’GTAGCA
Automation
+
Automated Sanger sequencing using fluorescently
labeled ddNTPs
http://bcs.whfreeman.com/lodish7e/#800911__812048__
Genome sequences can be determined by
assembling overlapping sequencing reads
Similar to Fig. 6-23
Sequencing of genomes
Human Genome Project: 1st draft (1990-2001)
New generations of DNA sequencing methods
significantly improved genome sequencing throughput
Next-generation reversible terminator sequencing
Key development #1: reversible fluorescently labeled-3’OH blocked dNTPs
fluor
fluor
New generations of DNA sequencing methods
significantly improved genome sequencing
Next-generation reversible terminator sequencing
Steps to determine each nucleotide in the template molecule
Add fluorescently
labeled-3’OH
blocked
dNTPs
Wash out
unincorporated
nucleotides
Read
fluorescence
Remove label &
3’ blocking
group
Similar to Fig. 6-22
New generations of DNA sequencing methods
significantly improved genome sequencing
Next-generation reversible terminator sequencing
Key development #2:
miniaturization and
improved detection
systems
Millions of genome
fragments are individually
immobilized in a solid
surface and sequenced in
parallel
All fragments are simultaneously sequenced
A complete human genome can now
be sequenced in 1-2 days!
Similar to Fig. 6-23
DNA sequencing
•
DNA synthesis is used for DNA
sequencing (similar principles as
DNA replication but “in vitro”)
•
The molecule to be sequenced
provides the template for
synthesis
•
Use a synthetic DNA primer
•
Chemically modified nucleotides
allow end-point (sanger) or
continuous (Next gen seq)
monitoring of DNA synthesis
•
Fluorescently labeled
nucleotides are used to
determine the template
sequence
Major classes of DNA sequences in eukaryotic
genomes
DNA
replication
DNA
transcription
1. Genes
2. Repetitious DNA
RNA
3. Unclassified DNA
translation
Protein
The DNA content varies between organisms
Haploid
The DNA content varies between organisms
Haploid
What is in a genome?
1. Genes
2. Repetitious DNA
3. Unclassified DNA
? ? ?
DNA sequencing
Major classes of nuclear eukaryotic DNA
1. Genes
a. RNA-coding only
b. Protein-coding
2. Repetitious DNA
a. Simple-sequence DNA
b. Interspersed DNA repeats
(mobile DNA elements)
3. Unclassified intergenic DNA
What is a gene?
A gene is the entire DNA sequence needed to
make a functional gene product, i.e. RNA.
Gene
+1
DNA
PROMOTER
PROMOTER
region
RNA-coding region
Transcription
RNA
RNA:
5’
3’
How do we know which parts of the
genome are transcribed into RNA?
Isolate RNA from cells
1. Analyze by Hybridization techniques
i.e.:Northern blotting or Microarrays
2. Analyze by Sequencing
i.e.: Sanger or Next generation sequencing
Northern blot
This technique is useful to detect a specific
RNA fragment in a complex mixture
RNA
Nitrocellulose
X-ray film
Agarose gel
Transfer buffer
Capillary action transfers
RNA from gel to nitrocellulose
Fig. 6-24
Labeled probe (ssDNA) hybridizes to
complementary sequence (ssDNA) in
membrane-bound fragments
ssDNA=single stranded DNA
New generations of DNA sequencing methods
These techniques can be use for genome wide identification
and quantification of RNA molecules in a sample
RNA (sample)
RT
cDNA
Millions of cDNA
fragments are
individually
immobilized in a
solid surface
All fragments are simultaneously sequenced
Similar to Fig. 6-23
High throughput DNA sequencing identify genomewide transcribed regions
RNA (sample)
DNA sequencing
RT
cDNA
Several million short sequencing reads
Alignment to
reference genome
Reference
genome
Gene A
Gene B
Gene C
Gene D
Reads only align to part of the gene sequence. Why?
More reads for some genes (none for gene A). Why?
Gene E
Some genes encode proteins…
Protein Coding Gene
+1
DNA:
5’UTR
PROMOTER
3’UTR
CODING
ATG
stop
(UTR: Untranslated Region)
Transcription
Messenger RNA
(mRNA)
5’
3’
Translation
protein:
A gene is the entire DNA sequence needed to make a
functional gene product, i.e. RNA.
•
Gene
+1
DNA
PROMOTER
PROMOTER
region
RNA-coding region
Transcription
RNA
RNA:
5’
3’
Not all genes encode proteins
Molecular Biology
BIMM100-FA2022
Tue-Thu, 12:30-1:50 PM
Lecture #6:
Chromatin
DNA sequencing
Jose Pruneda-Paz
Chromatin/DNA sequencing
DNA
replication
DNA
1. Chromatin structure
transcription
RNA
translation
Protein
2. DNA sequencing
Structural organization of eukaryotic chromosomes
Eukaryotic DNA is packed into nucleosomes
Fig. 8-23 (a)
Eukaryotic DNA is packed into nucleosomes
One nucleosome: ≈150 bp of DNA wrapped around
8 histones (2 each of Histones H2A, H2B, H3, H4).
Linker DNA
Fig
Fig.6-29
8-24 (a)
Linker DNA is more
susceptible to nuclease
digestion
The complex of histones and DNA is called CHROMATIN
Eukaryotic DNA is packed into nucleosomes
Linker DNA
Fig. 8-23 (a)
Linker DNA is more susceptible to nuclease digestion
Assay to monitor nucleosome occupancy
chromatin
Nuclease
(DNAse)
proteinase
DNA electrophoresis
Agarose Gel Electrophoresis for the Separation of
DNA Fragments
DNA
Add DNA
fluorescent
dye
http://www.jove.com/science-education/5057/dna-gel-electrophoresis
http://www.jove.com/video/3923/agarose-gel-electrophoresis-for-the-separation-of-dna-fragments
Assay to monitor nucleosome occupancy
chromatin
Nuclease
(DNAse)
well
proteinase
DNA electrophoresis
DNA ladder
Assay to monitor nucleosome occupancy
chromatin
no DNAse added
Nuclease
(DNAse)
Intact genomic DNA
proteinase
DNA electrophoresis
DNA ladder
Assay to monitor nucleosome occupancy
chromatin
DNAse added
Nuclease
(DNAse)
proteinase
~150 basepair DNA
fragments
DNA electrophoresis
Why?
DNA ladder
Assay to monitor nucleosome occupancy
chromatin
Increasing DNAse activity
Nuclease
(DNAse)
proteinase
DNA electrophoresis
DNA ladder
What would be the result if we do the proteinase treatment first?
lad
de
r
iClicker Question #1
1
2
3
DN
The following scheme
represents the results
obtained for 3 DNA
samples subject to agarose
gel electrophoresis.
A
sample
1000 bp
800 bp
600 bp
bp
– iClicker Question400#1
–
200 bp
Base on this scheme we can conclude that :
A: Sample 1 contains DNA fragments of ~500 bp
B: Sample 2 contains DNA fragments of ~500 and 800 bp
C: Sample 3 contains no DNA
D: All answers are correct
E: Only A and B are correct
Nucleosomes are arranged into a 30 nm fiber
Fig. 8-23
Nucleosomes are arranged into a 30 nm fiber
Fig. 8-23 (b)
Chromatin condensation status may vary across
different nuclear regions
Heterochromatin
– Generally inactive
Euchromatin
– Generally active
Fig. 8-27 (a)
Fig. 8-28 (a)
Chromatin in eukaryotic chromosomes is further
organized in higher-order condensed forms
Structural organization of eukaryotic chromosomes
• Nuclear DNA is found as a
complex with proteins: called
CHROMATIN.
• NUCLEOSOMES are the most
basic DNA-protein complex
found in chromatin.
• Chromatin can be organized
higher order CONDENSED
forms during cell division.
What about prokaryotes?
No chromatin
• In non-dividing cells some
chromatin sectors can be
CONDENSED while others
DECONDENSED.
Chromatin condensation status varies across
different nuclear regions
Heterochromatin
– Generally inactive
Euchromatin
– Generally active
Fig. 8-28 (a)
Fig. 8-27 (a)
Fig. 8-27 (a)
How can we determine if a specific chromatin
region is in a condensed or decondensed
chromatin region?
Assay to monitor chromatin condensation status
Decondensed chromatin
DNA
sequencing
Assay to monitor chromatin condensation status
Procedure
DNA
sequencing
Results and interpretation
(sequencing reads are mapped to the
genome sequence)