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CMMB 461 > Reverse genetics 1 > Flashcards

Reverse genetics 1 Flashcards

1
Q

What are the two main approaches to determining the function of genes?

A
  1. Loss-of-function experiments: Delete/mutate a gene and analyze the consequence (phenotype)
  2. Gain-of-function experiments: Mutate/amplify a gene and analyze the consequence (phenotype)

Might expect the phenotypes of LOF and GOF to be opposite of each other.

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2
Q

Genes that function in the same pathway/biological process produce…

A

Similar phenotypes

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3
Q

Genes that function in distinct biological processes may show what?

A

Multiple phenotypes (pleiotrophy)
-e.g. eyeless gene also regulates blood sugar levels. So eyeless (LOF) mutants exhibit phenotypes consistent with loss of insulin signalling such as smaller body size and abnormal carbohydrate levels.

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4
Q

Amorphs/null alleles

A

No gene function, usually entire gene is deleted
- Recessive mutation

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5
Q

Hypomorphs

A

Reduced gene function relative to wild type (doesn’t obliterate the gene function)
- Recessive mutation in gene that partially affects gene function

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6
Q

Hypermorphs

A

Enhanced gene function relative to the wildtype
- Dominant mutation

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7
Q

Antimorphs

A

No/reduced gene function relative to wild type
- Dominant loss-of-function mutation where the mutated protein interferes with function of the wildtype protein

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8
Q

How to determine if a mutation is dominant or recessive?

A

Make the heterozygote of the mutation, with one wildtype allele and one mutated allele. If the mutation is recessive, then the wildtype phenotype would appear. If the mutation is dominant, any phenotype but the wildtype would appear.

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9
Q

Forward genetics

A

Determine the identity of gene causing a mutant phenotype (“phenotype-driven-screen”)
- Phenotype is chosen, find the gene that is mutated

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10
Q

Reverse genetics

A

Determine a phenotype from a systematic perturbation of genes (“gene-driven”)
- Mutate gene first, then determine observe the phenotype

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11
Q

What are the 5 steps involved in forward genetics?

A
  1. Identify an interesting phenotype that addresses the biological problem
  2. Mutate genomic DNA (chemical mutagens, DNA insertions)
  3. Identify individuals exhibiting desired phenotype (represent multiple genes involved in the same biological process/molecular pathway)
  4. Identify the genes that were mutated (very laborious)
  5. Confirm that these genes are contribution to the phenotype (e.g. GOF experiments)
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12
Q

Plasmid complementation

A

Isolation of a DNA fragment from a wild type genomic library that can rescue/suppress the mutant phenotype

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13
Q

Describe how to identify a mutated gene in a chemical mutagenesis forward screen (4 steps)

A
  1. Transform wildtype genomic library (bunch of wildtype gene fragments cloned into vectors) into temperature-sensitive mutant strain at the lower temperature. This will restore the bacteria’s ability to survive at high temperatures in the mutant bacteria through plasmid complementation.
  2. Select for cells containing a plasmid from library (leucine prototroph/transformants, can make leucine itself)
  3. Test transformants for survival at the higher temperature
  4. Isolate plasmid from high-temp survivors and sequence the genomic fragment that contains the wild-type copy of the gene. By using plasmid complementation, you narrow down the candidate to a single gene (the one in the plasmid that rescued the phenotype), making it much easier to find the mutation.
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14
Q

What are the potential problems with a chemical mutagenesis forward screen? (3)

A
  1. Gene may not be in the genomic library
  2. More than one gene can suppress the mutant phenotype
  3. Overexpression of the gene can be toxic to the cell
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15
Q

Give an example of how more than one gene can suppress the mutant phenotype

A

In a null cell, where MAPKKK is knocked out, MAPKKK is restored when the plasmid is transformed into the cell. But it is not restored when MAPKK is added into the cell because the MAPKK still can’t be activated by MAPKKK.
However, a hypomorph with MAPKKK knocked out will be restored with MAPKK plasmid because some of the MAPKKK in the cell can still activate the MAPKK.

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16
Q

Transposable elements aka transposons

A

Transposable elements are allowed to excise and insert into various locations in the genome (usually in viruses)

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17
Q

How do transposable elements cause mutagenesis?

A

Mutagenesis occurs when transposons insert into a gene or cis-regulatory regions (e.g. promoters) that disrupt gene function

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18
Q

Transposable elements engineered for mutagenesis will include what?

A

A marker to select for insertion (antibiotic resistance, GFP, lacZ)

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19
Q

DNA insertion within the genome is _______, but easier to ______

A

DNA insertion within the genome is pseudorandom (inserts preferentially into certain sequences), but easier to identify mutated gene (all you need to do is determine where the transposon has been inserted into the genome.

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20
Q

What is DNA insertional mutagenesis (transposon tagging) used for?

A

Used to identify the genomic sequences flanking the transposon

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21
Q

What are the steps for DNA insertional mutagenesis (transposon tagging)?

A
  1. Restriction digest of genomic DNA (makes many fragments). Primers around transposon are facing opposite directions, away from the transposon.
  2. Self ligation of the fragments (due to sticky ends). Only one circular fragment will contain the transposon.
  3. Primers are now facing each other after circularization, allowing for inverse PCR. PCR primers are located at the ends of the transposon and so only the flanking genomic sequences will be amplified.
  4. Sequence PCR product and determine genomic location of insertion by BLAST.
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22
Q

In terms of reverse genetics, what do genome sequences allow for?

A

Targeted gene perturbation (inactivation/overexpression) and creation of large mutant collections

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23
Q

What does reverse genetics bypass?

A

The need to identify the mutated gene by forward genetic approaches since you know which gene is mutated in every mutant in the collection

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24
Q

True or false: it’s difficult to do LOF and GOF reverse genetics experiments on budding yeast S. cerevisiae

A

False
- it’s easy to do LOF and GOF experiments

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24
Q

True or false: many of the fundamental cellular processes are conserved in complex multicellular eukaryotes

A

True

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25
Q

Describe targeted gene deletion

A

Antibiotic-resistance selectable marker (KanR) is transformed through a double cross-over to budding yeast ORF, which replaces the ORF. This now allows for selection of the kanamycin-resistant transformants.

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26
Q

Describe the yeast deletion mutant collection (4)

A
  • Attempted systematic single deletion of 5,916 genes (96.5% of all genes)
  • 18.7% (1,105) genes were essential for growth in rich glucose medium (essential genes)
  • 81.3% (4811) genes were viable for growth in rich glucose media (nonessential genes)
  • High proportion of nonessential genes my indicate functional redundancy of genes or that its function is required in a speific perturbation or stress not yet identified.
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27
Q

Describe the type of yeast deletion mutant collections (3)

A
  1. ~6000 heterozygous deletion mutants (mutant essential and nonessential genes)
  2. ~4800 homozygous diploid mutants (mutant nonessential genes because the cell dies if the essential genes were knocked out)
  3. ~4800 haploid mutants (mutant nonessential genes)

Assemble mutant collection into high density yeast arrays where each colony is a unique deletion mutant

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28
Q

How can you test the sensitivity of phenotypes of deletion yeast mutants?

A
  • For low density arrays, can do manual pinning onto a stress condition to test which deletion yeast mutants are sensitive to stress.
  • For high density arrays, can do robotic pinning.
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29
Q

Describe the inference of gene function by phenotypic characterization of mutant collection (galactose example)

A
  • The utilization of galactose as a carbon source is one of the best studied pathways in yeast.
  • Identified 10 previously unknown genes to be required for optimal growth on this carbon source.
  • Therefore, fitness profiling (looking at how fast a cell proliferates) can discover genes involved in pathways; even those that are well studied
30
Q

Mutants with resembling fitness profiles…

A

Likely represent genes functioning in a similar biological process/pathway

31
Q

During the drug development pipeline, each drug trial is several million dollars and you want to die if a drug turns out to not even work after a bunch of trials. How can you elucidate the mode-of-action of the drug compound (target, side effects) to avoid wasting a bunch of money? (2)

A
  1. Microarray expression profiling of drug-treated cells
    - Compare with/without the drug, then see which genes are upregulated or downregulated -> can tell you potentially what the drug is targetting, and if it’s working
  2. Chemical-genetic profiling (drug-sensitive mutant strains)
    - Exposing yeast mutant collections to the drug to see if the mutants are sensitive or not -> gives us info on what the drug is doing, and potentially what the drug target it
32
Q

Usually drugs (activate/inhibit) target molecules

A

Inhibit

33
Q

What can finding out what the target gene is and the side effects of a drug help with?

A

Helps determine the best candidates to develop the drug and carry it to human trials

34
Q

Drug-induced haploinsufficiency profiling (HIP) in general

A

Identifies small molecules that target gene products essential for growth by measuring the drug sensitivity of heterozygous deletion mutants (possible drug target)

35
Q

Homozygous profiling (HOP) in general

A

Screening of drug sensitivity of homozygous diploids or haploid deletions to identify pathways that buffer drug sensitivity (possible drug side effects)

36
Q

Describe drug-target identification by induced haploinsufficiency

A

Heterozygotes that are sensitive to a particular drug may contain the ORF that encodes the drug target
- Reduced fitness seen in drug-induced haploinsufficiency because the drug inhibits some of the protein being made by WT allele
- When you see sensitivity of a heterozygote to a drug, the essential gene is likely the protein target of the drug
- Have to use heterozygotes because a wildtype organism with two functional copies of the gene often still leads to enough protein being produced, even if a drug partially inhibits it (WT organism would be less sensitive to the drug, making it harder to detect which gene the drug is targeting)

37
Q

Describe HOP assays

A

Looks at the non-essential genes that regulate buffering pathways, which try to neutralize the cytotoxic effects of the drug
- Wildtype cell has normal growth when exposed to drug
- Mutant cell (signal transduction gene deleted, nonessential gene so cell still alive) results in no buffering pathway to adapt to drug cytotoxic effects, leading to cell death or hypersensitivity
- Helps identify pathways that compensate for drug toxicity and can provide insight into drug action and resistance mechanisms

38
Q

Describe chemical-genetic profiling using molecular barcodes

A
  • Each yeast deletion strain is marked by two unique 20 bp molecular barcodes flanked by universal primers
  • Barcodes from a culture of pooled mutants can be PCR-amplified with the degree of barcode amplification of a particular mutant to be proportional to its relative density in the culture (b/c if the cell is sensitive to the drug, the cell dies so there’s less barcode amplification in the culture)
  • Therefore, this approach allows the simultaneous determinatino of relative proliferate rate/fitness of each mutant strain among all strains in an identical condition.
39
Q

How can you identify relative amounts of 6000 unique barcodes/mutant hypersensitivities simultaneously?

A

Using a microarray

40
Q

Describe the barcode microarray procedure (6 steps)

A
  1. Pool tagged deletion strains
  2. Grow deletion pool in condition of choice (some of the mutants will grow, and some will be sensitive)
  3. Purify genomic DNA
  4. PCR- amplify uptags and downtags
  5. Hybridize PCR products to microarray chip
  6. Based on fluorescence, you can figure out which strains are sensitive
41
Q

On a barcode microarray, a ____ spot would appear if there is no change

A

yellow

42
Q

Describe the drug-induced haploinsufficiency profiling (HIP) assay done using methotrexate (anticancer compound)

A
  • Methotrexate known target is dihydrofolate reductase encoded by the DFR1 gene
  • FOL1/FOL2 are essential genes and involved in folic acid synthesis (not a target of methotrexate but may be rate limiting in the drug target pathway)
  • YBT1 and YOR072W are nonessential genes likely involved in the availability of the drug in the cell (small molecular transporters) and pump the drug out of the cell
  • Upregulation of YBT1 (GOF) in human cancer cells causes methotrexate resistance (better drug pumping out of the cell, cancer cells are often resistant to anticancer compounds)
  • Fitness defect score was highest for DFR1 which confirms that cells that lack DFR1 are the most sensitive to the drug
43
Q

What are high density tiling microarrays and what are two examples of them?

A

Several million probes/microarrays can allow more thorough investigation of the genome than just measuring expression of known genes.
1. ORF microarray
2. Tiling microarray

44
Q

Tiling microarray and limitation

A

Cover the entire genome, including non-coding regions, at regular intervals (high resolution)
Limitation: Resolution isn’t perfect compared to RNA-seq. If probes are positioned 300 nucleotides apart, some nucleotides will remain unsequenced.

45
Q

ORF microarray and limitation

A

Probing for sequences of known and predicted genes
Limitation: non-coding genes are harder to find because there’s no start/stop codons, splicing sites or exons. There’s no way that these are all the genes in the genome, so we’re limited to only the genes that we know about.

46
Q

What do tiling microarrays allow for in terms of UTRs?

A

Allows for defining the length of 5’-UTRs and 3’-UTRs for every mRNA

47
Q

What can comparing the signal intensities on a tiling array to annotated ORFs allow?

A

Allows for researchers to determine whether transcription is happening where they expect (inside ORFs) or if there’s transcription in regions not previously annotated, indicating the presence of novel transcripts or non-coding RNAs

48
Q

Tiling microarrays led to the discovery of what type of RNAs?

A

Long noncoding RNAs (lncRNA)

49
Q

LncRNA

A

Noncoding RNA molecules longer than 200 bp (different from miRNA, siRNA, snoRNA, etc.)
- Involved in different levels of gene regulation and numerous diseases

50
Q

Hox transcript antisense RNA (HOTAIR, example of LncRNA) function

A

Silences transcription across 40 kb of the HOXD locus by recruiting PRC2 and inducing a repressive chromatin state

51
Q

HOTAIR in breast cancer cells

A

LncRNA HOTAIR reprograms chromatin state to promote cancer metastasis
- HOTAIR is upregulataed in breast cancer cells, good prognostic marker for metastasis and survival and promotes invasion of breast carcinoma cells in mice
- Knocking out HOTAIR in mice decreases metastasis
- Can use HOTAIR as a potential gene target

52
Q

Why would a HIP assay not be informative for cisplatin?

A

Cisplatin, and anticancer drug that acts to inhibit cell proliferation by covalently binding to the N-7 of purines, has no protein target (binds DNA instead), so a HIP assay would not be informative.
- HOP assay can be performed with cisplatin to uncover the mechanism of action.

53
Q

Mutants compromised in which process are sensitive to the cisplatin drug?

A

Mutants compromised in DNA repair.

54
Q

Clustering chemical-genomic profiles (cell response to drug) for compounds reveals…

A

Similar modes-of-action.

55
Q

Give an example (in general) of chemical-genomic profile clustering with the drugs amiodarone and tamoxifen

A
  • Amiodarone: Anti-arrhythmic drug that perturbs calcium homeostasis
  • Tamoxifen: A competitive inhibitor of estradiol binding to the estrogen receptor and common breast cancer drug
  • Resemblance of chemical-genetic profiles indicate that tamoxifen may produce an increase in cytosolic calcium in yeast (disruption of calcium homeostasis in the cell)
56
Q

Crz1p

A

A transcription factor that functions in the induced expression of genes needed to respond to various external conditions including high Ca2+

57
Q

Amiodarone and tamoxifen have (upregulated/downregulated) Crz1-regulated genes

A

Upregulated

58
Q

Amiodarone and tamoxifen have (higher/lower) lacZ (gene regulated by Crz1)

A

Higher

59
Q

When Crz1 is labelled with GFP, fluorescence is seen in…

A

Cells after both amiodarone and tamoxifen administration

60
Q

_______, as well as mode-of-action may be predicted by clustering a compendium of chemical genetic profiles

A

Drug structure

61
Q

Describe the chemical-genetic profiles of papuamide B and alamethicin

A
  • Papuamide B: isolated from marine sponge with antifungal and anti-HIV, but didn’t know how it worked.
  • Chemical profile resembles alamethicin, an antimicrobial pore-forming peptide (perhaps papuamide B disrupts the cell membrane of pathogens by making peptide)
62
Q

Elucidate papuamide B’s mechanism of action

A
  • Papuamide B reacts with the lipid phosphatidylserine disrupting the yeast cell membrane, thereby lysing the cells
  • Interestingly, phosphatidylserine is also found on the HIV membrane, possibly explaining the anti HIV activity of papuamide B
63
Q

More difficult to study (essential/nonessential) genes because most mutations are (LOF/GOF) so cell is more difficult to isolate

A

essential, LOF

64
Q

What are the two types of essential mutants?

A
  1. Conditional: mutant strain grows at the permissive condition and displays lethal phenotype when exposed to the restrictive condition (reversible)
  2. Constitutive hypomorph: mutation with lower gene activity but sufficient for cell viability and visible phenotype
65
Q

Describe an example of a constitutive hypomorph (3)

A
  • “DAmPing” involves disrupting the 3’-untranslated region immediately downstream of a gene’s stop codon by inserting an antibiotic resistance cassette.
  • Destabilization of the gene’s mRNA and lowering abundance of mRNA 2-10 fold of wildtype depending on the gene
  • Sufficient gene activity in DAmP allele to keep cells viable, but may exhibit mutant phenotype due to lowered gene activity
65
Q

Give an example of a conditional mutant

A

Temperature-sensitive mutants
- Mutant cells viable at 25 degrees C (permissive condition), but lethal phenotype exhibited at 35 degrees C. (restrictive condition).
- Point mutation alleles that compromise the stability, structure and function of the protein at the restrictive temperature.

66
Q

In DAmP alleles, why is there variation (2-10 fold) in how much wild type expression is deecreased?

A

Variation is due to initial stability/half-life duration of mRNA to begin with:
- more stable mRNA might have 2-fold decrease, but mRNA with short half life might have 10-fold decrease

67
Q

In Breslow et al. experiment, 842 DaMP alleles of 1,100 essential genes were produced, where 43% of DAmP alleles have an observable phenotype under unstressed conditions while 57% of DAmP alleles do not. What are 2 reasons why this 57% of alleles did not show expression?

A
  1. Might not have been knocked down enough for life (might have been knocked down too much)
  2. Some of the genes might have been more essential under other conditions.
68
Q

(Essential/nonessential) genes are involved in fundamental cellular processes (processes required for life and found in all cells)

A

Essential

69
Q

(Essential/nonessential) genes are more likely to have homologs in other organisms (are more conserved)

A

Essential
-more conserved because these essential genes are required for life.

70
Q

(Essential/nonessential) genes are more likely to be associated with species-specific functions
- give an example of this

A

Nonessential
- e.g. pheromone genes getting knocked out just prevent the organism from mating, doesn’t cause death.

71
Q

(Essential/nonessential) genes are less likely to be duplicated genes with little/no functional redundancy

A

Essential
- So nothing to rescue cell if essential gene is knocked out

72
Q

(Essential/nonessential) genes tend to interact more with other genes (hubs)

A

Essential
- So if you knock out an essential gene with higher connectivity, there will be a bigger impact.

CMMB 461 (6 decks)

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