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Announcements
-A reading assignment on DNA hybridization has been
posted on the course website.
-The second midterm exam (Monday, May 19th) will cover
material through today’s lecture and next weeks quiz
section.
-My office hours: Fridays, by appointment between 1:00 &
2:30 PM.
-Emailing questions to course TA’s…
If you don’t have much background
with molecular biology techniques:
-The textbooks that are on reserve in the undergraduate
library cover these techniques very well:
-Chapter 20 (specifically, pages 715-735) in Griffiths, et al.
-Chapter 9 (specifically, pages 301-335) in Hartwell, et al.
-Chapter 19 in Klug, et al.
-I have provided links on the course website that direct you
to sites that provide information on gel electrophoresis,
Southern blotting and northern blotting.
-I have placed a file on the course website that describes
gel electrophoresis and Southern blotting .
-The TA’s can also help-but please, not by email.
Today…
-DNA cloning
-DNA hybridization
-Sequencing
-Polymerase chain reaction
The Yeast Adenine Biosynthetic Pathway
ADE4 ADE5*
A
B
ADE8 ADE6 ADE7 ADE2 ADE1
X
D
E
F
Y
G
C
ADE13
ADE3*
H
red
ADE17
pigment
ADE16
I
ADE17
ADE16
J
ADE12
K
ADE13
AMP
Given that many different genes are involved in adenine
biosynthesis, what do all of these enzymes “look” like?
--how are they different?
--what is the sequence of amino acids?
--what is their 3-D structure?
--how do the enzymes work?
--do humans have the same enzymes as yeast?
Bigger question:
How can we figure out what any gene actually encodes?
How can we get our hands on the piece of DNA so that we
can sequence it and determine the amino acid sequence of
the protein?
DNA cloning
Dolly
Clone: cells or organisms
that are genetically identical
because they are related by
non-sexual reproduction—i.e.,
by mitosis not meiosis.
Copy Cat
DNA clone: identical copies
yeast
of a DNA fragment, usually
generated by propagation in
bacteria
Each colony is >106 identical cells.
DNA cloning—the idea
A tiny fraction of a genome:
1. Break the genome into gene-sized pieces
2. Introduce DNA into E. coli
3. Culture each bacterium
separately to generate clones
4. Study each gene in
isolation
Problems to solve:
» How can the new DNA survive and replicate in E. coli?
» How can we “select” for bacteria with the new DNA?
» How do we find the one bacterium with the gene we
want to study--the one with the ADE2 gene?
Plasmids as vectors for transformation
Non-essential DNA
molecules that can
replicate independently of
the chromosome
Selectable marker
AmpR or TetR
ori
because they have an
origin of replication
A gene that confers resistance
to a drug such as ampicillin or
tetracycline.
For cloning: “recombine” the foreign DNA into a plasmid.
In principal, vectors solve two of the problems—maintenance
& selection.
But how does cloning work in practice?
Restriction endonucleases
Recognize and cut specific sequences in DNA
Restriction enzyme EcoRI:
QuickTime™ and a
TIFF (Uncompressed) decompressor
are needed to see this picture.
EcoRI site in DNA:
--GACTGAATTCTGATC---CTGACTTAAGACTAG-Reversed by DNA ligase
--GACTG
--CTGACTTAA
AATTCTGATC-GACTAG--
sticky ends
Using EcoRI to make recombinant DNA
(Watson & Crick strands shown)
cut with EcoRI
partial digest with EcoRI
--GACTG
--CTGACTTAA
AATTCTGATC-GACTAG--
sticky ends
Making a DNA library
Tetracycline-sensitive E. coli
DNA ligase to repair
phosphodiester
backbone
transformation
(CaCl2, heat shock)
This is a
clone
tetracycline plate
A DNA library
Each clone may have a different piece
of the“foreign” genome in it in the
form of a recombinant DNA molecule
“Genomic DNA library”:
• start with genomic DNA
• expect all the chromosomal DNA to be represented
• expect about equal representation of all sequences
Practice question
The yeast genome is ~12,000kb and the ADE2 gene is ~2kb.
1. If the average insert size in the yeast genomic library
is 12 kb, what fraction of the E. coli colonies are expected
1/1000
to contain the ADE2 gene? ___________
2. If you looked at 10,000 colonies how many are
10
expected to contain the ADE2 gene? ___________
3. At 200 colonies per plate, how many plates would be
needed to ensure that you have 10 different clones with the
50
ADE2 gene? ___________
This principle is called “coverage”--the number of times, on average,
that the genome (or particular gene) is represented in the library
Quiz Section this week and next:
Genetic Analysis in Caenorhabditis elegans
Goals: to illustrate the logic
of genetic analysis
Mutant analysis (AKA Genetic Analysis)
The use of mutants to understand how a
biological process normally works
• Start with “unknown” system (e.g., metabolic
pathway, embryonic development, behavior,
etc.)
Locomotion
• Generate mutations that affect the “unknown” unc-93,
unc-29,
system
(i.e., athat
“break” the “unknown”
Is the
mutation
loss-of-function
or gain-of-function?
The etc.
system)
answer
influences the interpretation of the phenotype
Paralyzed?
• Study the mutant phenotypes to reveal the
hyperactive?
functions of the genes
etc.
• Map the genes
• Identify the genes (more on this later) K+-channel, Ach
Receptor, etc
Loss-of-function mutations are usually recessive
Amount of
protein
activity
threshold
WT phenotype
mutant phenotype
AA
Aa
aa
Loss-of-function mutation simply decreases the activity of
the protein; threshold determines whether dominant or
not.
Gain-of-function mutations are usually dominant
threshold
Amount of
protein
activity
mutant phenotype
wt phenotype
aa
Aa
AA
One type of gain of gain-of-function mutation simply
increases the activity of the protein; threshold determines
whether dominant or not.
Example from QS - acetylcholine receptor (unc-29)
acetylcholine
+
outside
Wild-type
UNC-29
closed
open
inside
LOF - reduced gene function, GOF - increased gene function.
Phenotypes are often opposite.
What would be the phenotype
of a LOF mutation? - receptor
missing or defective
(equivalent to being always
closed)
What would be the phenotype of
a GOF mutation? - receptor
activated in absence of Ach
(equivalent to being always open)
A real ACh receptor (top view)
pentamer with four
types of subunit:
a2, b, d, g
All four types of subunits (a, b, d, g) are required for function.
What happens if any of them is knocked-out by mutation (LOF)?
What (probably) happens if one has GOF and another has LOF?
Example from QS: potassium channel gene unc-93
mostly
closed
unc-93 is expressed
in muscle (body and
egg-laying)
outside
Wild-type
UNC-93
inside
gate
K+
outside
unc-93
(cb1500)
mutant
How could you make new
LOF mutations in unc-93?
Excess K+ flow
prevents muscle
excitation
inside
K+
unc-93 is featured
again in QS 7.
Cloning by complementation!
Finding the yeast ADE2 gene in an E. coli library by
complementation of an E. coli adenine mutant phenotype.
Hypothesis: if the intermediates in these pathways are the
same, then the enzymes should be interchangeable.
E. coli purE+ gene
HC
N
CH
C
H2 N
N
R
(AIR)
*
Yeast ADE2 gene
-OOC
C
N
CH
C
H2 N
N
R
(CAIR)
*Both species need the same basic
enzyme for this step of adenine synthesis.
Cloning a gene based on its function
“cloning by complementation”
» Start with (recessive) E. coli purE- mutant strain
purE-
»Transform with DNA library made from (dominant) ADE2
wild-type yeast strain
»Ask: which plasmid clone rescues the mutant phenotype?
portion of
yeast genome
Library
ADE2
purE-
E. coli
no growth
purE-
purE-
no growth
-adenine plate
portion of
yeast genome
ADE2 must
be on this
plasmid!
purE-
growth!
-adenine plate
Practice Questions
Does that colony really have an ADE2
plasmid? Could it not be a purE+
revertant or a contaminant?
1. What phenotype would the E. coli have if the plasmid
was lost? (How to lose a plasmid? How exactly would
you do the experiment?)
back to being ade- (unable to grow on -adenine plates)!
2. What would they expect if the the original yeast
genomic DNA library had been made from ade2
mutant yeast? no colonies on the -adenine plate
3. Why is this type of functional cloning considered an
example of “complementation”?
4. Can this experiment be done using the human “ADE2”
gene?
The cell division cycle
Cell cycle regulation— MUCH of
what we know comes from genetic
analysis using temperaturesensitive yeast mutants
restrictive
temperature
QuickTime™ and a
TIFF (Uncompressed) decompressor
are needed to see this picture.
Can complement
some yeast
mitotic defects
with human DNA!
Lee Hartwell
2001 Nobel Prize
Working with DNA
What else might we
want to know about the
ADE2 gene?
QuickTime™ and a
TIFF (LZW) decompressor
are needed to see this picture.
What other questions
could we ask?
Wesequence
need to know
the
--what is the
of amino
sequence of the ADE2
acids?
gene3-D structure?
--what is their
--how do the enzymes work?
--how do the mutants differ from
But first, a little review. . . .
WT?
P
S
S
P
P
S
S
P
P
S
S
S
P
P
S
P
S
S
S
P
P
S
S
P
P
S
S
S
P
P
Review: DNA Structure
S
Nucleotide: repeating unit
phosphate
group
S
nitrogenous
base
P
O
P
S
OH
P
N
O
S
O
P
S
S
P
P
S
S
P
P
S
S
S
P
P
S
P
S
S
S
P
P
S
S
S
S
P
P
S
S
P
P
S
S
P
P
S
P
S
denaturation
breaks these
H-bonds
restriction
enzymes break these
covalent bonds
H C
2
4’
5’
N
O
C H
H
H C
C
3’
OH
H
C
1’
H
2’
deoxyribose
sugar
Review: DNA synthesis
DNA polymerase
requires . . .
template strand
3'
5'
A A T G C G A C A G T A G A
T T A C G C T G T C A T C T
OH
OH
3’ OH
primer
activated
nucleotides
5'
5'
G
OH
5'
5'
C
OH
T
5'
OH
A
to make
the
nascent
strand
OH
DNA denaturation (and renaturation)
The strands of DNA can come apart . . . and go back together!
..TAGAGTTCTGTACGTTCGTGATGCCATGCCTGGAAC..
..ATCTCAAGACATGCAAGCACTACGGTACGGACCTTG..
Heat to 95˚
(or NaOH)
Cool to ~65˚C (+ some salt)
5’..TAGAGTTCTGTACGTTCGTGATGCCATGCCTGGAAC..
+
..ATCTCAAGACATGCAAGCACTACGGTACGGACCTTG..5’
It’s reversible!
DNA denaturation (and renaturation)
In a complex mixture of different DNA molecules,
complementary ssDNA strands can find each other and
reform dsDNA.
denaturation
NaOH or
>95˚C
65˚C +
salt
renaturation
base pairing provides the specificity to reform dsDNA
DNA (or RNA) hybridization
Annealing single strands from different sources
..TAGAGTTCTGTACGTTCGTGATGCCATGCCTGGAAC..
..ATCTCAAGACATGCAAGCACTACGGTACGGACCTTG..
heat
5’..TAGAGTTCTGTACGTTCGTGATGCCATGCCTGGAAC..
+
..ATCTCAAGACATGCAAGCACTACGGTACGGACCTTG..5’
cool
5’GTACGTTCGTGAT
GTACGTTCGTGAT
..ATCTCAAGACATGCAAGCACTACGGTACGGACCTTG..5’
a “hybrid” DNA molecule.
Add excess
oligonucleotides
with a sequence
complementary
to one of the
strands
Some notes on hybridization
-Probe and target need not be the same length
both okay
-Small mismatches are tolerated if overall match length is
substantial
okay
-Small probes (25-30 bases) can work if conditions (salt,
temperature) are adjusted. Mismatches much more
significant for small probes.
DNA sequencing
Uses DNA polymerase
The sequence obtained is for the strand
being synthesized
Can determine ~600-800 bases in one “read”
With 3,000,000,000 bp in the human
genome, that’s a lot of sequencing
reactions!
Fred Sanger
Nobel Prize for
protein
sequencing
(1958) and for
DNA sequencing
(1980)!
Addition of a nucleotide to a 3’OH
new
nucleotide
at 3’ end
3’ end
activated
nucleotide
H2O
O
2
O
P
O-
O-
+
Reagents for Sanger DNA sequencing
• DNA fragment to be sequenced
heat
• 1 primer: short ssDNA complementary to ONE region
of the template DNA
• dNTPs
All 4: dATP, dCTP, dGTP, dTTP
• DNA polymerase from E. coli
Question: How does synthesizing a strand of DNA
give us information on its sequence of
bases??????
These are
• small amount of each dideoxyNTP
the key
• Split the mixture
among four tubes
C
A
G
T
• Add a small amount of 1 dideoxy nucleotide to each
reaction
What’s a dideoxy?
dye molecule
H H
A = green
T = red
C = blue
G = black
No 3’ OH. “chain terminator”
• Incubate until all synthesis is complete
• Analyze newly synthesized strands on a gel
QuickTime™ and a
Cinepak Codec by Radius decompressor
are needed to see this picture.
Products of the four synthesis reactions
“C”
reaction
Primer +
T C T T T T A A G C A T T C
A G A A >>>>>>>>>>
A GA A A A TTC
5
Primer
“A”
reaction
T C T T T T A A G C A T T C
A G A A >>>>>>>>>>
AGAAA
AGAAAA
A GA A A A TTCGTA
A GA A A A TTCGTA A
1
2
8
9
“G”
reaction
“T”
reaction
Primer +
T C T T T T A A G C A T T C
A G A A >>>>>>>>>>
A GA A A A TTCG
6
A GA A A A TTCGTA AG
10
T C T T T T A A G C A T T C
A G A A >>>>>>>>>>
AGAAAAT
3
A GA A A A TT
4
A GA A A A TTCGT
7
Analysis of the new strands on a gel
ddA ddC ddG ddT
+etc
+11
+10
+9
+8
+7
+6
+5
+4
+3
+2
+1
primer
-
mix
all 4
reactions
3’
A
T
G
G
A
C
T
T
A
5’-A
+
Read bottom up
Laser scan of a sequencing gel
ssDNA fragments run past a laser where the color is detected.
Sequencing our clone
The vector sequence is known; insert is to be sequenced
 use vector sequence for initial sequencing primers…
Ampr
ori end
primer
ori
vector
Amp end
primer
insert
+ insert
ADE2 gene
then use the new sequence info to extend the sequencing
What if . . .
. . . you wanted to know what was
different about the sequence of
the ADE2 gene from an ade2
mutant strain?
ori
ampR
vector
insert
Do we need to start with a library
from an ade2 mutant?
ADE2 gene
Not necessarily; if we already know the sequence of
ADE2 we can use PCR to obtain DNA for the ade2
mutant allele.
The polymerase chain reaction
Uses DNA polymerase
Makes unlimited quantities of the DNA of interest
Only requires a single template molecule—very sensitive
Uses two DNA primers
Reagents for PCR
1. target DNA
genomic DNA
2. Two specific primers—
AAGT
CACG
“ left” and “right”
short (17-20 nt) oligonucleotides that are complementary
to sequences on either side of the region of interest
3. All 4 dNTPs
dATP, dCTP, dGTP, dTTP
4. DNA polymerase from the heat-loving bacterium,
Thermus aquaticus doesn’t die at 95°C!
DNA products at successive cycles of PCR
starting dsDNA
denature
anneal
synthesize
cycle 1
cycle 3
repeat
cycle 2
cycle 4
After 4 cycles, half of
the products are DNA
fragments of a specific
size—the size of the
DNA that lies between
the two primers. By 30
cycles there would be
228 of the DNA
molecules from the
initial DNA molecule
Primers for PCR
…must “point inward” to the region to be amplified
5’  3’
“forward
primer”
5’
3’
region to be amplified
3’
5’
“reverse
primer”
Primers for PCR
…must “point inward” to the region to be amplified
5’  3’
region to be amplified
5’
3’
3’
5’
AACGTATTTTAACGATTTCGATATCCAGTTTAATTCCGAGTGTA
TTGCATAAAATTGCTAAAGCTATAGGTCAAATTAAGGCTCACAT
Primers for PCR
…must “point inward” to the region to be amplified
5’  3’
region to be amplified
5’
3’
3’
5’
AACGTATTTTAACGATTTCGATATCCAGTTTAATTCCGAGTGTA
TTGCATAAAATTGCTAAAGCTATAGGTCAAATTAAGGCTCACAT
Primers for PCR
…must “point inward” to the region to be amplified
5’  3’
region to be amplified
5’
3’
3’
5’
AACGTATTTTAACGATTTCGATATCCAGTTTAATTCCGAGTGTA
TTGCATAAAATTGCTAAAGCTATAGGTCAAATTAAGGCTCACAT5’
5’AACGTATTTTAACGATTTCGATATCCAGTTTAATTCCGAGTGTA
TTGCATAAAATTGCTAAAGCTATAGGTCAAATTAAGGCTCACAT
Practice question
DNA sequence from exon 1 of the huntingtin gene
5’
GTCCCTCAAGTCCTTCCAGCAGCAGCAGCTTGAGCCGCCAC
CGC
A. What
6 nucleotides from the gray boxes could be used
CAGGGAGTTCAGGAAGGTCGTCGTCGTCGTTGTCGGCGGTG
as primers
GCG to amplify the region of the genome shown
above?
“left (forward) primer”? 5’ ________3’
CCCTCA
GTGGCG
“right (reverse) primer”? 5’ ________3’
B. What size will the DNA fragment be after PCR
amplification using the two primers above? 33 base pairs
C. Huntington’s disease is a dominant autosomal trait that is
fatal. It usually strikes in middle age, by which time affected
individuals may have already passed on the disease allele.
Individuals have different numbers of the triplet CAGs in the
first exon in the “huntingtin” gene.
5’
GTCCCTCAAGTCCTTCCAGCAGCAGCAGCTTGAGCCGCCAC
CGC
CAGGGAGTTCAGGAAGGTCGTCGTCGTCGAACTCGGCGGT
GGCG
How many CAG repeats are present in this allele? 4
If the CAG were in the coding portion of exon 1, how
would this sequence appear in the huntingtin protein?
…S L K S F Q Q Q Q L E P P P…
D. Below is a gel in which the PCR fragments of different
individuals are shown. The region amplified used the
same primers as in part A. Why do people have 2 bands?
Because each individual represented is heterozygous for a
huntingtin gene with a different number of CAG repeats
Triplet
repeat
number
65
50
35
20
5
no disease
disease
E. What is the
correlation
between the
number of repeats
and the presence
of disease?
individuals with a
single Huntingtin
allele of more than
~ 35 CAG repeats
display the disease
phenotype.
F. Examine the pedigree below and determine the probability of disease
among the 3 females in generation III. Notice that the middle branch of
the family refused DNA testing. Numbers (i.e. 7,21) refer to the number
of CAG repeats.
45 years old
40
7,21
7,46 11,15
46,11
36
5
5
14,46 12,12 26
12,46 7,26
5
12,26
G. Discussion questions: Would you want test results if you were II-5 or II-6?
Results on the previous page indicate that individuals with a single Huntingtin
allele of more than ~ 35 CAG repeats will display the disease phenotype.
Individual II-5 will have this disease, whereas individual II-6 will not.
Would you want your child, III-3, tested if you weren’t willing to be tested?
What are the different concerns of each individual? These are personal
choices, and the answers given for the other questions should provide
sufficient information for you to consider these issues.