Molecular Biology Tools

Question 1

Expression of Recombinant human insulin in E. Coli

Answer 1

1) insulin mRNA is copied by RT into cDNA
2) end of insulin cDNA is digested
3) insulin cDNA is ligated into bacterial vector that has also been digested to make the recombinant DNA.
4) bacterial cells get transformed
5) Bacterial cells produce insulin

Question 2

Reverse Transcriptase

Answer 2

RT binds

2) mRNA is annealed to synthetic primer
3) synthesis of cDNA in 5’ to 3’ direction
4) makes dDNA-mRNA hybrid
5) mRNA is digested with alkali
6) primer is ligated to 3’ end of cDNA
7) DNA pol extends from primer to yield ds DNA

Question 3

Restriction Enzymes

Answer 3

recognize and cleave recognition sequences in DNA
found in bacteria
makes both sticky ends and blunt ends

Question 4

Recognition sequences

Answer 4

usually 4-6 nts and are palindromic

Question 5

DNA ligase

Answer 5

joins 5’ phosphate with 3’ OH to form phosphodiester bond

can join fragments from two different DNA

Question 6

Plasmid

Answer 6

extra chromosomal DNA in bacteria that replicates independently, does not integrate into the chromosome
used in vectors
confers antibiotic resistance
contains: OR, restriction sites, selectable marker
simple to clone, but low transformation efficience

Question 7

Antibiotic resistance

Answer 7

plasmid vector contains antibiotic resistance gene making the transformed cell resistant. plated on agar medium containing antibiotic for selection

Question 8

Bacteriophage

Answer 8

infect E coli with high efficiency
introduction of recombinant vector
kills host
25kb insert

Question 9

Cosmids

Answer 9

hybrid of plasmid and bacteriophage
does not kill host
uses plasmid OR
carry up to 45kb DNA

Question 10

CRISPR

Answer 10

confers resistance to foreign plasmids and phages
cut exogenous genetic elements similar to RNAi
regulates Cas9 delivery

Question 11

Cas9

Answer 11

is an RNA guided DNA endonuclease
unwinds foreign DNA or plasmid and checks to see whether it is complimentary to 20bp spacer region of guide RNA.
if complimentary, Cas9 cuts it to cleave it.

Question 12

Restriction Digestion

Answer 12

1) bacteria containing recombinant plasmids are grown as clones
2) DNA is isolated from multiple clones and digested with enzymes
Pattern of restriction enzymes shows if DNA segment of interest has been inserted

Question 13

cathose is ______

Answer 13

negative

Question 14

anode is

Answer 14

positive

Question 15

Intercalating dye

Answer 15

insertion of dye between molecules of the planar bases of DNA
Ethidium bromide (exposed to UV)

Question 16

Hybidization

Answer 16

cones are blot transfered to membrane and DNA is denatured and fixed onto surface.
add radioactive probe and allow to hybridize (temp just below Tm)
exposed to X ray and see where the hybridization occurd.

Question 17

Tm =

Answer 17

2°(A+T) + 4°(C+G)

Question 18

Southern Blotting

Answer 18

DNA fragment separation

labeled with radioactive DNA probe

Question 19

Northern Blotting

Answer 19

RNA fragment separation

Question 20

SDS-Page

Answer 20

separates proteins based on size and forms a negatively charged complex with proteins to allow them to migrate through a polyacrylamide gel.

Question 21

Probes used in Blots

Answer 21

cDNA for Genes for Southern and Northern
Antibody for Western

1) specific length 2) specific sequence 3) radioactivity for detection 4) must be added in sufficient quantity to compete with other binding

Question 22

Microarrays

Answer 22

developed to simultaneously analyze expression of thousands of genes at once.

1) isolate total mRNA in each condition
2) reverse transcribe cNDA labeled with fluorescent dye
3) mix
4) hybridize to DNA in microarry
5) wash
6) measure fluoresce one each spot.

Question 23

traditional DNA seqeuncing

Answer 23

2’3” dideoxyribose which does not have 3’ OH, so it cann’t be extended.
1) combine unkown DNA with pol, primase, dNTS, in four containers each with a specific A, T, C, G, of ddNTP
2)DNA syntehsis occurs
3) separate fragment by electrophoresis
sequence is the compliment of the unknown strand

Question 24

PCR

Answer 24

1) DNA denatured at 95° and cooled to 55° to allow primers to hybridize
2) Temp is raised to 72° and polymerase makes copy to 3’ end
3) repeat cycle
gives 2^N copies

Question 25

taq

Answer 25

heat safe DNA polymerase used in PCR

Question 26

Restriction Fragment Length Polymorphisms

Answer 26

1) disease alleles cause either a gain or loss of restriction endonuclease cleavage
2) for screening of HbS mutation in sickle cell anmeia, destoys restriction site for MstII.
3) digest patient DNA with diagnostic RE followed by southern or PCR analysis to detect altered size.

Question 27

mutated MstII restriction site

Answer 27

in Sickle Cell anemia
Changed from A to G
produces two 1.3 kb segments
normal has 1.1 and .2 segment

Question 28

Allele Specific PCR

Answer 28

used to detect small mutations (point mutations, insertion or deletions)
primers are designed to hybridize to mutant but not normal allele
can test for multiple mutations at the same time by designing so that each PCR product is a different size.

Question 29

Cystic Fibrosis

Answer 29

CFTR gene the fourth Amino acid is normally TTT or Phe.
in delta F508, common mutation in cystic fibrosis, mutations gives rise to GGT and Glycine. must change the mutant primer.

Question 30

DNA genotyping

Answer 30

use microarray to determine the status of a single NT polymorphism at specific base location
determines mutant status (heterozygosity) as well as halotype status
1) add anneal specific probe to genomic DNA
2) fill gap and ligate and do PCR to create fluorescent bar coded probe for each allele
3) measure fluorescence

Question 31

Paternity Testing

Answer 31

DNA fingerprinting by variable number of tandem repeats.

PCR or restriction digestion followed by south blotting of region with repeats

Question 32

Short Read sequences

Answer 32

DNA sequencers that produce lots of short 100 bps reads in single run.
low error rate
used to provide high coverage over specific set of DNA template from complex mixture.
Illumina

Question 33

Long Read Sequences

Answer 33

can read 10,000Bp in length
high error rate
useful in human genetics for linking contiguous polymorphisms, measuring diveristy of mRNA isoforms,
watches single DNA pol incorporate fluorescent NTs.

Question 34

Read depth

Answer 34

Coverage

number of times a base in the genome was independently sequenced. high read depth improves error rate

Question 35

Error Rate

Answer 35

measurement of quality of converting signal into accurate prediction of NT.

Question 36

Continguity

Answer 36

see the recombination events between male and female gametes. Recombination cross over tend to be long distances apart.
short read are too short to link variants in the same haplotype block
long reads can sometimes link variants together

Question 37

Fragmentation of DNA

Answer 37

is pseudo-random, meaning ends vary

Question 38

Single nucleotide polymophisms

Answer 38

are detected through reading multiple fragmented DNA and seeing how often the mutation occurs.

Question 39

what if 50% of fragmented DNA has mutation

Answer 39

either high error rate or heterozygous mutation

Question 40

Exome Sequencing

Answer 40

used to isolate DNA fragments that contain protein coding sequences for sequencing
useful in sequencing mendelian where variants are rare and not likely to exist in database
1) use oligonucleotide probes that are complimentary to regions of interests that hybridize to DNA
2) fish out DNA fragments to be sequenced

Question 41

Variants of unknown significance

Answer 41

exome sequencing has produced large catalog of variants within protein coding regions, but it is difficult to predict with high confidence the functional impact.

Question 42

how do we interpret variants of unknown sigificance?

Answer 42

1) genetic analysis and looking at each variant (valid, but low throughput)
2) computation prediction - do a mutation and see if it does something to structure - not valid, but high throughput
3) massive parallel functional analysis - high validity and high throughput

Question 43

Parallel Functional Analysis

Answer 43

make a tube with all possible variants that have single mutation
measure how many variatns are in tube before and after. if this changes, it indicates gain or loss of function of variant.

Question 44

Genomic DNA footprintingq

Answer 44

treat cell with low concentration of DNAse I that cuts based on accessibility, DNA fragments are collected and sequenced to identify regulatory sites.