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BIOL 2320 - Genetics Lab > Yeast Cloning + Sequencing > Flashcards

Yeast Cloning + Sequencing Flashcards

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1
Q

Which organism did we clone DNA from?

A

yeast, Saccharomyces cerevisiae

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2
Q

Describe yeast

A

a unicellular, eukaryotic organism that can reproduce asexually or sexually

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3
Q

What is the number of homologous pairs of chromosomes in haploid yeast cells?

A

n = 16

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4
Q

What is the number of homologous pairs of chromosomes in diploid yeast cells?

A

n = 32

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5
Q

Why is yeast a model organism?

A

small organism

short life cycle

can be grown in large quantities in lab

cheap to grow

grows in liquid and solid medium

contains organelles

does mitosis for asexual reproduction

does meiosis for sexual reproduction

70% of genome is coding region

23% of genome is similar to humans

full genome has been sequenced

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6
Q

What is the overall purpose of this lab?

A

to clone a fragment of the S. cerevisiae genome into the pUC19 plasmid by creating recombinant DNA and to be sequenced and analyzed to determine which fragment was cloned and its function

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7
Q

What is a cloning vector?

A

a piece of DNA that can have a small fragment of another foreign DNA fragment inserted into it

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8
Q

What is the cloning vector in this experiment?

A

pUC19

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9
Q

What is pUC19?

A

a bacterial plasmid (plasmid University of California #19)

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10
Q

Describe bacterial plasmid

A

small, circular pieces of DNA that replicate independently from the bacterial genome

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11
Q

Why is pUC19 a useful cloning vector?

A

because it is a bacterial plasmid, and plasmids allow the isolation, analysis, and manipulation of DNA fragments because they replicate separately from the bacterial genome

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12
Q

What are the 4 major characteristics of pUC19 plasmid that make it useful for this experiment?

A

presence of:

AmpR gene
ORI
LacZ gene
Multiple cloning site (MCS)

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13
Q

Why is the AmpR gene important in this experiment? which organism has this?

A

AmpR gene codes for resistance to the antibiotic ampicillin which is used in the media for this experiment

pUC19 has this

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14
Q

Why is the ORI an important quality of pUC19?

A

the origin of replication allows for independent replication in bacterial host that does not require the entire bacterial genome to replicate - faster replication can occur

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15
Q

Why is the MCS an important quality of pUC19 plasmid?

A

the multiple cloning site is a region of the plasmid DNA that contains various restriction sites, one of which will be used to cut the circular DNA with a restriction enzyme so that the yeast fragment can be inserted

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16
Q

Why is the lacZ gene an important quality of pUC19 plasmid?

A

a gene located in the MCS that codes for the beta-galactosidase enzyme which is important for blue-white screening

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17
Q

Which restriction enzyme is used? why?

A

EcoRI because it recognizes restriction sites in both the pUC19 plasmid DNA and the yeast DNA

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18
Q

What is EcoRI isolated from?

A

bacteria

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19
Q

What is the purpose of EcoRI in bacteria?

A

restriction enzymes like EcoRI provide defence for the bacteria against foreign viruses as they can chop them up

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20
Q

How do bacteria protect themselves from being digested by their own restriction enzymes?

A

bacteria methylate their DNA so the restriction enzyme cannot recognize the restriction site sequences and cut up the genome

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21
Q

What sequence does EcoRI recognize in pUC19?

A

a 6 bp sequence: 5’ - GAATTC - 3’ located in the MCS

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22
Q

How does EcoRI cut the restriction sequence? What results?

A

it makes a staggered cut between the G and the A to create sticky, complimentary ends and a linear plasmid

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23
Q

What are the first steps of the project?

A

to isolate yeast genomic DNA

to digest pUC19 with EcoRI and to dephosphorylate the digested pUC19

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24
Q

What is the purpose of SDS solution (when it is used at any point in this procedure)?

A

to break down cell walls in yeast

to dissolve cell membranes in bacteria

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25
Q

What is the name of the kit used to isolate the yeast DNA?

A

GeneJet Genomic DNA Purification Kit

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26
Q

What reagents are included in the GeneJet Genomic DNA Purification Kit?

A

digestion solution

proteinase K

lysis solution

RNase A

purification column containing silica (not a reagent)

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27
Q

What is the purpose of the digestion solution in the isolation of yeast?

A

it prepares conditions for digestion

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28
Q

What is the purpose of proteinase K in the isolation of yeast? what kit is it apart of?

A

proteinase K digests the proteins in the yeast

GeneJet Genomic DNA Purification Kit

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29
Q

What is the purpose of RNase A in the isolation of yeast? What kit is it in?

A

it removes RNA from the yeast

GeneJet Genomic DNA Purification Kit

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30
Q

What is the purpose of lysis solution in the isolation of yeast? What kit is it in?

A

ensures the cell is entirely lysed + maintains the pH of required for DNA to bind to the silica membrane of the purification column

GeneJet Genomic DNA Purification Kit

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31
Q

What is ethanol used for while isolating yeast?

A

ethanol promotes the binding of the yeast DNA to the silica column

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32
Q

What is the purpose of the wash buffers in the GeneJet Genomic DNA purification kit used to isolate yeast?

A

as the DNA is bound to the silica column, they will wash away impurities

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33
Q

What is the purpose of the elution buffer in the isolation of the yeast DNA?

A

it allows the yeast DNA to be removed from the silica membrane of the purification column

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34
Q

What restriction enzyme is used to digest pUC19?

A

EcoRI

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35
Q

Why does pUC19 need to be dephosphorylated after it is digested?

A

to prevent the rebinding of the cut sticky ends during the ligation reaction

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36
Q

What is used to dephosphorylate pUC19?

A

alkaline phosphatase

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37
Q

How does alkaline phosphatase dephosphorylate pUC19?

A

it catalyzes the release of the 5’ phosphate group from the plasmid DNA to prevent cut ends from rebinding during ligation

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38
Q

What procedures were done after the yeast DNA was isolated and the pUC19 DNA was digested and DeP?

A

yeast DNA was digested with restriction enzyme

competent JM101 E. coli cells prepared

ligation reaction to produce recombinant DNA

transformation and plating

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39
Q

What restriction enzyme was used to digest the yeast DNA?

A

EcoRI

40
Q

Why is it important that both the yeast DNA and the pUC19 DNA be cut with the same restriction enzyme?

A

in order to create sticky ends that are complimentary and that will bind together during ligation to produce recombinant DNA molecules

41
Q

What organism was used to produce competent cells for transformation?

A

Escherichia coli

42
Q

What strand of E. coli was used?

A

JM101

43
Q

Why were JM101 E. coli cells used for transformation?

A

they have AmpS gene (cannot grow in ampicillin)

lack functional lacZ gene

44
Q

Describe transformation

A

the process of cells taking up DNA from their environment and incorporating it into their own genome and expressing that DNA

genetic alteration

45
Q

What does it mean for a cell to be competent? does this occur often in nature?

A

competent = cell is capable of taking up DNA from its environment (capable of transformation)

uncommon in nature

46
Q

How were the JM101 E. coli cells made competent?

A

By exposing them to a chemical solution = T-solution

47
Q

What are 2 other methods of making cells competent?

A
  1. exposing bacteria to ice cold salt and then heat shocking
  2. electroporation (exposure to electric field to change membrane permeability)
48
Q

What are the hypotheses for transformation?

A

DNA molecules pass through large membrane channels in bacterial membrane because:

  1. exposure to cold salt changes/weakens structure of cell surface
  2. salt solution contains cations which can reduce/eliminate the repulsion between the negative DNA and membrane phospholipids
  3. heat shocking produces temperature gradient
49
Q

What were the two controls used for transforming the competent E. coli cells?

A

transformation with uncut pUC19

transformation with no pUC19

50
Q

What is the purpose of transformation?

A

to make competent E. coli cells uptake recombinant DNA that can be selected and isolated with blue-white screening

51
Q

What reaction is required to make recombinant DNA?

A

ligation of the EcoRI digested yeast DNA and pUC19 DNA

52
Q

What kit was used for the ligation reaction? What major reagent was included in this?

A

Rapid DNA Ligation Kit

included T4 DNA ligase

53
Q

What is T4 DNA ligase?

A

an enzyme isolated from bacteriophage T4 that forms a phosphodiester linkage between the junctions of the hybridized molecules of yeast and pUC19 sticky ends to create recombinant DNA

54
Q

Why is DNA ligase from T4 bacteriophage used for this experiment?

A

because it is active at room temperature

55
Q

What kit was used for transformation? What major reagent was included?

A

TransformAid Bacterial Transformation Kit

includes T-solution

56
Q

What is T-solution?

A

a proprietary solution that induces competency in bacteria cells for transformation

57
Q

What are the 2 identifiable phenotypes of pUC19?

A

AmpR gene = can grow in ampicillin

lacZ = produces beta galactosidase which digests XGal in medium in the presence of IPTG to produce a blue pigment in the colonies

58
Q

What will happen to all the genomes of the JM101 E. coli cells that transform pUC19 plasmid?

A

any cell that transforms the pUC19 plasmid will convert their AmpS gene to AmpR and be able to grow in ampicillin

59
Q

What will happen to all the genomes of the JM101 E. coli cells that transform pUC19 ONLY?

A

AmpS –> AmpR (can grow in ampicillin)

lacZ- –> lacZ+ = will produce B galatosidase and produce blue pigmented colonies

60
Q

What will happen to all the genomes of the JM101 E. coli cells that transform pUC19 + yeast fragment (recombinant DNA)?

A

convert AmpS –> AmpR

translation of lacZ+ will not occur because the yeast fragment interrupts the plasmid’s MCS at the location of the lacZ gene = colony grows white

61
Q

What technique was used to plate the transformed cells?

A

spread-plate technique under alcohol burner

62
Q

What were the experimental ligations and what were they plated on?

A

ligated DeP/EcoRI-digested pUC19 + EcoRI-digested yeast

plated on

LB/Amp/XGAL/IPTG agar plates

63
Q

What is LB?

A

luria broth = a medium that contains all essential nutrients for E. coli

64
Q

What is XGAL?

A

an artificial substrate of beta-galactosidase that is hydrolyzed if a functional lacZ gene is able to produce B-galactosidase

65
Q

What is IPTG?

A

a synthetic analog of lactose which inactivates the lac repressor protein and allows for the synthesis of B-galactosidase

66
Q

How are IPTG, XGAL, and LacZ gene connected? Why is this important for this lab?

A

IPTG is a synthetic analog of lactose

a functional lacZ gene produces B-galactosidase only in the presence of lactose of an analog (IPTG)

B-galactosidase is an enzyme that hydrolyzes X-Gal in the medium and precipitates a blue colour and turns the colonies blue

this is important for blue-white screening

67
Q

What is ampicillin? why is it used in this experiment?

A

ampicillin is an antibiotic which E. coli cells do not naturally have a resistance to

it is used in this experiment to make sure that any bacterial cells that did not transform the pUC19 plasmid do not grow (no conversion of AmpS –> AmpR)

68
Q

What were the 3 controls that were plated (including what they were plated on)?

A

uncut pUC19 on LB/AMP/XGAL/IPTG

no pUC19 on LB only

no pUC19 on LB/AMP/XGAL/IPTG

69
Q

What was the purpose of the control with uncut pUC19 plated on LB/AMP/XGAL/IPTG? What is the expected result of this control?

A

E. coli cells that transformed uncut pUC19 were plated as a control to make sure the X-Gal and IPTG in the medium have not degraded and that when produced, B-galactosidase can hydrolyze XGal and turn the colonies blue

expected: only blue colonies

70
Q

What was the purpose of the control with no pUC19 plated on LB/AMP/XGAL/IPTG? What is the expected result of this control?

A

E. coli cells that transformed no pUC19 will not be able to grow in the presence of ampicillin (AmpS gene) so this makes sure the ampicillin is working

expected = no growth

71
Q

What was the purpose of the control with no pUC19 plated on LB only? What is the expected result of this control?

A

E. coli cells that transformed no pUC19 but were grown on LB only plates makes sure the LB has all the essential nutrients required for the growth of the bacteria

expected = lawn of white colonies

72
Q

Why would we expect to see a lawn of white colonies in the control with no pUC19 plated on LB only?

A

Because LB is the complete medium for E. coli so all they need to grow is provided

white because without pUC19, the bacterial cells do not have the lacZ+ gene to produce B-galactosidase, but also XGal and IPTG are not present in the medium so there would be no blue anyways

73
Q

Why would we expect to see no growth in the control with no pUC19 plated on LB/AMP/XGAL/IPTG?

A

E. coli cells that did not transform pUC19 will not have been able to convert their AmpS gene to AmpR and will therefore not be able to grow in the presence of ampicillin

74
Q

Why would we expect to see only blue colonies in the control with uncut pUC19 plated on LB/AMP/XGAL/IPTG?

A

E. coli cells that transformed uncut pUC19 DNA will have been able to convert their lacZ- gene to functional lacZ+ and in the presence of IPTG, be able to express beta-galactosidase that will hydrolyze the XGal in the medium and precipitate the blue colour, giving the colonies a blue pigment

75
Q

Why do we expect to see both blue and white colonies in the experimental ligation platings?

A

because some E. coli cells will transform:

only pUC19 DNA = blue
both pUC19 + yeast DNA = white

those that transform only yeast DNA or no DNA will not grow

76
Q

Why would E. coli that transformed recombinant DNA appear as white?t

A

the yeast DNA fragment interrupts the translation of the lacZ+ gene in the plasmid DNA because it is inserted in the MCS of the plasmid DNA (also where the lacZ gene is) so beta-galactosidase cannot be produced and XGal will not be broken down and the colony will not show the blue colour

77
Q

What is the purpose of blue-white screening?

A

to determine which colonies of E. coli transformed the recombinant DNA and can be selected for running the gel and sequencing

78
Q

What is transformation efficiency?

A

the number of transformant bacterial cells / ug of plasmid DNA

79
Q

Why are high transformation efficiencies needed?

A

to maximize the recovery of recombinant DNA molecules

80
Q

What is the range that transformation efficiencies should ideally be in for successful cloning?

A

at least 10^6 transformants/ ug of plasmid DNA

81
Q

What 3 factors effect transformation efficiency?

A

the amount of DNA used (Saturation can occur)

the form of the DNA (supercoiled is better than linear or single-stranded)

the purity of the DNA

82
Q

Which plate is used to calculate the transformation efficiency? why?

A

the uncut pUC19 plate because this DNA is circular and circular transforms better

83
Q

What is the calculation for transformation efficiency?

A

= # colonies on uncut pUC19 plate / ng of DNA plated x 1000 ng/ug

84
Q

How do you calculate the ng of DNA plated?

A

= pUC19 (ng) / volume of competent cells (uL) x volume plated (uL)

85
Q

Why were the chosen colonies amplified in LB/Amp liquid media?

A

Ampicillin used to kill off any bacteria that did not transform the pUC19 DNA

86
Q

How was the recombinant plasmid DNA isolated from the E.coli cells in LB/Amp liquid?

A

by Alkaline Lysis Method

87
Q

Why did we use the alkaline lysis method in this lab?

A

because it is the preferred technique for plasmid DNA extraction from bacteria

88
Q

Why is the alkaline lysis method popular?

A

it is fast and effective

it lyses bacterial cells and destroys genomic DNA while keeping the plasmid DNA intact

89
Q

Which alkaline lysis kit did we use?

A

AKA Miniprep

90
Q

How does the alkaline lysis method isolate the plasmid DNA?

A

it uses a high pH (~12) to denature large pieces of DNA (the bacterial genomic DNA) which cannot renature in a lower pH and will be precipitated out

plasmid DNA is smaller and can renature when pH is lowered

91
Q

How are cells lysed in the alkaline lysis method?

A

SDS dissolves membranes

NaOH provides the basic conditions to break up cell walls and denature the bacterial DNA

92
Q

Once the cells are lysed, how are the genomic and plasmid DNA separated in the alkaline lysis method?

A

NaOH provides the basic conditions to denature the DNA (both) but the large genomic DNA cannot renature when the solution is neutralized

when solution is neutralized with KOAc, the genomic DNA and proteins precipitate

93
Q

What happens during the neutralization reaction in the alkaline lysis method?

A

KOAc (potassium acetate) is added and the denatured genomic DNA and proteins precipitate out

the plasmid DNA renatures

94
Q

How is the renatured plasmid DNA precipitated out in the alkaline lysis method?

A

95% ice-cold ethanol + KOAc precipitate the renatured plasmid DNA

95
Q

How is the renatured and precipitated plasmid DNA purified completely in the alkaline lysis method?

A

70% ethanol is used to remove any residual salts

BIOL 2320 - Genetics Lab (5 decks)

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