Download Class VII Using cloned DNA fragments to study gene expression 1

Class VII
Using cloned DNA fragments to study gene expression
1. Polymerase chain reaction – PCR
2. Cloning of DNA fragments
3. Sequencing of cloned DNA
4. Using cloned DNA fragments to study
gene expression
By cloning and sequencing of our genomic DNA as well as
all our mRNA we characterized the sequence of our genome,
as well as where all the genes in the genome are.
Why did we do this?
Answers:
Differences among genes determine whether we have a
disease, whether we will inherit a trait (can be a disease),
and even whether we can develop a disease in the future.
Differences among genes also explain why we look different
from each other, why we look similar to some animals, and
how evolution might have happened.
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These differences among genes can be primarily of two
types:
A. Differences in the sequence of the same gene present
in two different individuals or organisms. (These can be
alleles or mutations)
B. The “activity” of the same gene might be different in
two different individuals (or tissues or organisms). This
would cause differences in “gene expression”, and this in
turn would cause differences in mRNA levels, which
causes differences in protein quantities. Since proteins
determine function, we can expect to see different
functions in a tissue if its protein levels are different from
that of another.
If the expression of a gene can be “up-regulated” or “down-regulated” the
gene is a “differentially expressed” gene: it is expressed at different levels in
different cells
Gene A
I
mRNA
Gene B
Gene A: Less function (less
mRNA, less protein) in cell I
compared to cell II
Gene B: Same amount of
function in both cell I & II
Gene A
II
mRNA
Gene A: More function in
cell II compared to cell I
Gene B
The regulation of bacterial “lac operon” is a good example to see
how a cell regulates the mRNA levels of a gene
Gene A
mRNA
Gene B
Gene A
Gene A: lac gene
(makes the lactase
enzyme (lac Z+Y+A))
mRNA
Gene B: lac repressor
Gene B
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The lac operon
I (inhibitor) gene
lac gene (Z+Y+A)
I mRNA
RNA
polymerase
lac repressor
(I protein)
-“I” gene is always is transcribed and translated, producing an “inhibitor”
protein called the “lac repressor”
- The expression of the “lac” gene results in the generation of a protein called
“β-galactosidase” (lactase). β-galactosidase is an enzyme which the bacteria
use to digest “lactose” (a sugar composed of glucose and galactose)
The lac operon
I (inhibitor) gene
lac gene (Z+Y+A)
I mRNA
RNA
polymerase
lac repressor
- lactose is not a sugar that is normally found where this bacteria lives (it
generally uses glucose and can use galactose).
- “lac” gene is a “big” gene, and its protein is “big” too. i.e. the bacteria does
not need to waste nucleotides and aminoacids to make an enzyme it does
not always need
- But there are situations when only lactose is available and no other sugar.
So sometimes the bacteria needs lactase
The lac operon
I (inhibitor) gene
lac gene (Z+Y+A)
I mRNA
RNA
polymerase
lac repressor
- So in the absence of lactose, the bacteria makes just a few numbers (about
10) of a “small” protein (lac repressor) from a “small” gene (I). When there is
no lactose in the environment, the lac repressor binds immediately 3’ to
those sequences where the RNA polymerase binds, preventing transcription
of the gene.
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The lac operon
I (inhibitor) gene
lac gene (Z+Y+A)
I mRNA
lactose (glucose+galactose)
- However, if lactose is present, “lac” gene needs to be actively transcribed so lactose binds to the “lac repressor” and changes its shape. When the
shape of “lac repressor” is changed, it no longer can bind to DNA and RNA
polymerase is free to transcribe “lac” mRNA
The lac operon
I (inhibitor) gene
lac gene (Z+Y+A)
I mRNA
lactose (glucose+galactose)
- lac mRNA is translated into 3 proteins including β-galactosidase which
hydrolizes lactose into glucose and galactose. When all lactose is digested,
the inhibitor regains its original shape, binds its DNA sequence and stops lac
gene expression.
Differential gene expression is also responsible for the generation
of different tissues from the zygote. Cells of the adult tissues
transcribe different genes
Gene A
Gene B
Gene C
Gene A
Gene B
Gene C
Genes
expressed:
mRNAs
Muscle
A&B
Genes
expressed:
mRNAs
Skin
B&C
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Abnormal induction of a gene’s expression can result in abnormal
cells like cancer too.
Gene A
Gene B
Gene C
Gene A
Gene B
Gene C
Genes
expressed:
mRNAs
Muscle
A&B
Genes
expressed:
mRNAs
Muscle
cancer
A&
B
How can we visualize and study differential gene expression
A. We can do our own experiments in the laboratory
B. We can use bioinformatic analyses (we can use the NCBI
cancer genome anatomy project (CGAP) data)
How can we visualize and study differential gene expression
A. We can do our own experiments in the laboratory
B. We can use bioinformatic analyses (we can use the NCBI
cancer genome anatomy project (CGAP) data)
We will use Northern blotting
(as well as other methods like PCR: next class)
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How can we visualize and study differential gene expression
A. We can do our own experiments in the laboratory
B. We can use bioinformatic analyses (we can use the NCBI
cancer genome anatomy project (CGAP) data)
Northern blotting is based on “Southern blotting”. So
we’ll learn the latter first.
Southern Blotting
Southern blotting is a method by which we can visualize a
specific DNA molecule among many others.
The method relies on the fact that DNA molecules generated
by restriction endonuclease digestion will be of various
lengths and that they will be separated by gel
electrophoresis.
Once separated, the DNA fragments will be transferred to
paper (blotted onto paper), and visualized using a probe.
The “probe” is a piece of DNA or RNA whose sequence we
know and which we have labeled radioactively.
Southern Blotting
L 5-26
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Southern Blotting
Separate DNA cut by R.E.s by gel electrophoresis
(+)
(-)
Transfer (blot) separated DNA onto a special paper
(nitrocellulose or special plastic) by capillary
diffusion
Southern Blotting
Dry absorbent (filter) paper
Nitrocellulose filter paper
Wet absorbent (filter) paper
solid block (e.g. glass)
Buffer suitable for transferring DNA
Southern Blotting
DNA blotted on
nitrocellulose paper
Radioactive probe
hybridized to DNA
Radioactively
labeled Probe
hybridizes only
to its
complementary
DNA
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Southern Blotting
5’-ATGCTACATGATCGATCGATATAGCCGCGAG-3’
3’-ATGTACTAGCTAGCTATATCGGCG-5’
5’-CGCGCTTTACTGTACGTG-3’
The “probe” we use for Southern blotting is a radioactively labeled DNA that
was previously cloned into a plasmid and sequenced. We, therefore, know
from which gene it is derived. At the “hybridization” step of Southern blotting,
our probe only hybridizes to its complementary DNA, which can be thus
visualized.
Southern Blotting
Expose nitrocellulose
paper to film
Develop film
(Autoradiogram)
Southern Blotting
How long should our probe be to ensure specificity?
Length of Probe
(chances of finding the same
sequence in the human genome)
46 = (1/4,096)
48 = (1/65,536)
410 =(1/1,048,576)
412 =(1/16,777,216)
414 =(1/268,435,456)
415 =(1/1,073,741,824)
number of sites the
probe will bind to in the
genome
732421
45776
2861
178
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<3
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Northern Blotting
5’-AUGCUACAUGAUCGAUCGAUAUAGCCGCGAGAAA-3
3’-ATGTACTAGCTAGCTATATCGGCG-5’
5’-CGCGCUUUACUGUACGUGAAAAAAA-3’
Northern blotting borrow its full methodology from Southern
blotting, the difference is that mRNA is separated by gel
electrophoresis instead of DNA
Northern Blotting
A
B
C
D
A
B
C
D
Northern blotting can help us visualize how much of a given mRNA there is in
a cell or tissue. In the above figure, mRNA was prepared from four different
tissues (A-D). The red quantities in each tissue is shown on the left and the
image obtained by Northern analysis is on the right. As shown here, the image
directly reflects how much “red” mRNA there is in each tissue
Northern Blotting
In the below shown experiment, mRNAs obtained from the
indicated tissues were separated by gel electrophoresis, blotted,
and visualized using radioactive DNA fragments that specifically
hybridized with either mRNA-A or -B
mRNA-A
mRNA-B
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Northern Blotting
The pattern we observe for mRNA-A is typical for a gene that is differentially
regulated (mRNA is present in some tissues but not others), while the pattern
we see for mRNA-B is typical for that of a “house-keeping” gene. Housekeeping genes are transcribed in almost every cell because they perform
functions that are essential for all cells, like nucleic acid synthesis.
mRNA-A
mRNA-B
How can we visualize and study differential gene expression
A. We can do our own experiments in the laboratory
B. We can use bioinformatic analyses (we can use the NCBI
cancer genome anatomy project (CGAP) data)
The CGAP Project:
1. make cDNA libraries from all normal tissues as well as
different types of cancer tissues in separate experiments
2. sequence all clones and determine expressed genes
3. count the numbers of cDNAs obtained for each gene in each
library to determine how actively that gene is transcribed
compared to other genes from the same tissue
4. Repeat this experiment for cDNA libraries obtained from all
human tissues, and determine which genes are differentially
expressed among each tissue
5. Share this data with the scientific community
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http://cgap.nci.nih.gov/Tissues/xProfiler
http://cgap.nci.nih.gov/Tissues/xProfiler
http://cgap.nci.nih.gov/Tissues/xProfiler
adipose
brain
0+29 = 29 gene expressed
only in adipose tissue
4756+742 = 5498 genes
expressed in adipose but not
brain tissue (but also
expressed in some nonebrain tissues)
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http://cgap.nci.nih.gov/Tissues/xProfiler
http://cgap.nci.nih.gov/Tissues/xProfiler
?, very important !
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http://cgap.nci.nih.gov/Tissues/xProfiler
http://cgap.nci.nih.gov/Tissues/xProfiler
1
2
3
http://cgap.nci.nih.gov/Tissues/xProfiler
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expressed sequence tag
P value
number of cDNAs
corresponding to this
gene in brain libraries
/ all cDNAs in all
libraries made from
brain tissue
“P” (probability) value:
Is a value that is generated by various statistical tests that measure if an
observation is “significant”.
What the value tells us is the probability of the observed results being the
result of pure coincidence:
e.g.
p = 0.01 : the results of the data analyzed can occur by pure chance
coincidence 1% of the time.
p= 0.8 : the results of the data analyzed can occur by pure chance
coincidence 80% of the time. (if this experiment was repeated 100 times, you
could get this result in 80 experiments by random coincidence)
p = 0.00001 : the results of the data analyzed can occur by pure chance
coincidence 0.001% of the time.
In biology, we generally consider p ≤ 0.05 “significant” ( i.e. not due to chance)
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Required reading:
Lodish: Chapter: 5.3
Lodish: “Lac operon” section in Chapter 7.1
Brown: page 142-144
Suggested reading:
Brown chapter 8.4 (hybridization)
http://www.dartmouth.edu/~cbbc/courses/movies/LacOperon.html
1. An alternative to Northern analysis is “dot blot” analysis where you spot mRNAs
obtained from tissues on to nitrocellulose directly without separating them by gel
electrophoresis, and you proceed as you would with Northern blotting. What is the
advantage of electrophoresis used for Northern analysis. Can you give an
example to where your Northern analysis results would be different from your dot
blot experiments?
2. What is an EST (expressed sequence tag)?
3. The virtual northern analysis provided by NCBI’s CGAP project might sometimes
not be the best way to analyze gene expression because it is based on
information collected from cDNA. Why can this be a disadvantage compared to
Northern analysis? Can you an example were you would rather perform Northern
blotting than virtual Northern analysis.
4. We used the “5-azacytidine induced mRNA” as an example in this class. Assume
that you would like to study this gene further by Northern blotting in the laboratory.
For this you want to clone the cDNA of this gene. Which link in those web pages
we covered in class contains the cDNA sequence of this gene?
5. Once you obtain the cDNA sequence, you would like to generate a DNA fragment
from this sequence to be used as a probe for Northern analysis in the laboratory.
You decide to use PCR to produce this DNA and you then want to clone it into a
plasmid. Please describe the experiments you would be doing.
6. If a statistical test tells you that the results you get can be obtained by chance in
every 10 of 1000 experiments, what will its “p value” be?
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