Molecular Techniques 3 Flashcards

1
Q

Gene libraries - SCREENING DNA LIBRARIES

PROCEDURE DEPENDS ON WHAT IS KNOWN ABOU THE GENE: STEP 1 AND 2

A
  1. PLATING CLONES OF THE LIBRARY
    PLASMID = cells diluted and plated so each bacterium grows into a distinct colony
    PHAGE = Allowed to infect lawn of bacteria. each plaque contains single cloned DNA fragment.
  2. PROBING PLATED COLOBNIES OR PLAQUES
    - similar gene from another organism
    synthetic probes
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2
Q

Genomic and cDNA can be screened with a probe to find the gene of interest..HOW = 7

A
  1. a disc of nitrocellulose or other membrane is laid on top of the bacterial colonies
  2. a few from each colony adhere to the nitrocellulose filter
  3. the cells are disrupted and their DNA is denatured and fixed to the filter.
  4. A labeled probe hybridises with any complementary DNA.
  5. Excess probe is washed off and the membrane is overlaid with X-ray film,..
  6. …Which detects the presence of the probe.
  7. comparison of the membrane with the master plate reveals which bacterial colonies have the DNA of interest
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3
Q

What happens after the gene has been Isolated?

A

After have isolated a clone of
interest, next step is to find
out more about the DNA.

Gel electrophoresis can be
used to separate DNA
molecules on basis of size
charge

E.g. determine REs in
genomic copy of a newly
discovered gene
Compare these sites among diverse human populations

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4
Q
A
  1. DNA samples containing fragments of different sizes are placed in wells in an agarose gel.
  2. An electrical current is passed through the gel.
  3. All DNA fragments move toward the positive pole; small fragments migrate faster than large fragments. After electrophoresis, fragments of different sizes have migrated different distances.
  4. A dye specific for nucleic acids is added to the gel.
  5. DNA fragments appear orange under UV light.
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5
Q

1st gene responsible for
human genetic disease
identified by positional
cloning was cystic
fibrosis’(CF).
EXPLAIN

A

Geneticists attempted
to isolate gene for CF.
Once linkage analysis
found a narrow enough
region, isolated clones
of sequences.
Identified clones from
human genomic library
covered the region of
interest

Examination of
sequences within
clones revealed four
genes = candidate
genes.

Gene for CF encodes
a membrane protein
that controls chloride
movement into & out
of cells.

Mutations cause
channel to stay closed
= chloride builds up
in cells, thick mucous
build up.

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6
Q

CF analysis and steps + conclusion

A
  1. Linage analysis was carried out on families with CF..
  2. ..and a asscoiation was found between the inheritance of molecular markers from Chr 7 and CF.
  3. linkage studies with additional marker indicated that the CF locus is close to markers D7S340 and D7S122.
  4. Clones from the region were isolated by chromosome walking and jumping.
  5. Analysis of DNA sequences within clones revealed 4 candidate genes.
  6. Additional studies eliminated three of the canditate genes.
  7. DNA sequencing revealed the presence of a 3-bp deletion in the gene of patient with CF
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7
Q

DNA sequences can be determined and analysed: how? 3

A

Molecular techniques also used to analyse DNA molecules through determined of their sequence:

  1. Restriction fragment length polymorphisms (RFLPs)
  2. DN Sequencing
  3. Next-Generation Sequencing technologies
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8
Q

Example of using RFLPs

A
  1. DNA sequence has 2 HaeIII restriction sites
  2. A mutation creates a polymorphism. Some copies have both restriction sites and other only one.
  3. When DNA from 2 persons is digested by HaeIII,…
  4. …2 different patterns appear on the autoradiograph of the gel.
  5. Bob’s DNA is cut into 3 bands because his chromosomes posses both resctriction sites.
  6. Joe’s DNA is cut into only 2 bands because his chromosomes posses only 1 of two sites.
  7. this example assumes that bob is homozygous for the A patten and joe is homozygous for the B patten. A person heterozygous for the RFLP would display bands seen in both the A and the B patterns.
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9
Q

What are RFLPs?

A

Markers essential to success
of positional cloning.

One such marker: restriction
fragment length
polymorphisms (RFLPs).

Variations in pattern of
fragment produced when
DNA molecules are cut with
the same RE.

Differences are inherited,
used in mapping, similar to
allelic differences used to
map conventional genes.

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10
Q

How can RFLPs be used to detect linkage? Example..

A

RFLPs can be used to detect linkage.

e.g.Close correspondence in inheritance of RFLP alleles the presence of Huntington disease

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11
Q

Understanding NEXT GENERATION SEQUENCING TECHNOLOGIES…
PROSEQUENICING

A

Made sequencing hundreds of times faster and cheaper.

Sequence in parallel, hundreds of 1000s of DNA fragments simultaneously.

EXMAPLE : PROSEQUENCING:
Nucleotides added one at a time, addition of particular nucleotide is detected by a flash of light.

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12
Q

Prosequencing steps

A
  1. DNA is brojen up into fragments,…

2… and adpaters containing primer sequences are attached to each fragment. The DNA is made single stranded.

  1. Each DNA fragment is attached to a bead and surrounded by a drop of solution containing reagents for PCR.
  2. The DNA fragment is amplified by PCR.
  3. DNA is made single stranded. Each bead is forced into a well and layered with sequencing reagents.
  4. DNA synthesis takes place within each well.
  5. A solution containing a specific deoxynucleoside triphosphate (dATP) flows over the plate.
  6. When a nucleotide is added to the growing chain, PPi is released and produces a light-emitting chemical reaction
  7. An instrument records the light emitted by each well.
  8. the amount of light is proportional to the number of nucleotides
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13
Q

Understanding Sequencing the genome: Large genomes
- STEPS 3

A

Large genomes requires robotics and automation.

Assembly of consensus sequence:
- ideally, one consensus sequence per chromosome
- standard is ten independent reads of each nucleotide
- assembled by computer

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14
Q

Why does the Sequencing Genome process require automation:

A

Consider human genome = 3x10^9 bp DNA

If could sequence each chromosome separately from one end to other C easy to generate sequence, but individual sequences generate ~600bp = 0.00002% entire genome.

One major challenge facing genome project is sequence
assembly, i.e. building up all individual sequences into
consensus - an authentic representation of sequence in genome.

Also error rate in sequencing: need 10 independent reads
for each base pair.

Given 600bp per sequence & human genome = 3 Gbp, 50
million reads required for 10 times coverage of all basepairs.

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15
Q

Gene technology now:

A

Preparation of clones, DNA, electrophoresis, sequence protocols
adapted to automation: production9line sequencing machines run
around clock. Produce 96 reads every 3 hours. A genome centre
has 200 machines, can produce 150,000 reads in*a single day.
3Gbp in 192 years.

Now faster with Next Gen technology 9
genome in 1 week, $1000 per genome.

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16
Q

When Assembling genomes Gaps usually occur: why?

A

Gaps usually occur, regardless of technique

  • short gaps filled by PCR
  • long gaps require additional cloning, sometimes in different host
17
Q

If genome is rich in repetitive elements, problem assembling consensus…EXPLAIN WHY 2

A

If genome is rich in repetitive elements, problem assembling consensus

  • Often repetitive sequence is in total longer than the length of
    the maximum sequence read (i.e. 600bp). In that case no way to bridge the gap between adjacent unique sequences.
  • Dispersed repetitive sequence can cause erroneous alignment of reads from different chromosomes.
18
Q

Repeated sequences in genome:

Make assembly of sequences difficult:

Categories of repeated sequences: tandem vs mobile

A

Categories of repeated sequences
* tandem repeat arrays
- multiple, adjacent copies
- large micro-satellites
- gene families
- telomeres

  • mobile elements: transposons
  • dispersed repeats, may be in high copy
19
Q

Human genome lessons

A

~45% is composed of repetitive transposon
transposon - derived sequences

  • Regions of large: scale repeat structure involving protein coding genes
  • <5% is protein encoding
  • exons typically ~150 bp
  • average transcript has 10 exons
  • ~60% genes have alternative splice sites:
  • indicates much larger proteome
  • many potential splice&sites need to be experimentally verified
20
Q

After sequences are
compiled, annotation of is
performed then
identification of candidate&
genes by scanning for ORFs.

A
  1. Bacterial chromosome
  2. Fragment by sonication
  3. Sonicated fragments
  4. prepare clone library
  5. Cloned library
  6. Sequence clone
  7. Computer with compiler software
  8. Assemble sequence
21
Q

Most of the human genome does not encode proteins

A

24 pairs of linear chromosomes.
genome size 3.4 billion bp
number of genes : aprox 32 000
g + c content 41 %