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Genetics > Experiments in Genetics Prac Reviews > Flashcards

Experiments in Genetics Prac Reviews Flashcards

1
Q

How do rich and minimal media differ? What kind is NA?

A

Rich/Complete: carbon, nitrogen, AAs, vitamins, purines/pyrimidines, essential salts

Minimal: carbon, nitrogen, essential salts

NA is rich media

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2
Q

What is the difference between an auxotroph and utilisation mutant?

A

Auxotroph: Needs a compund to grow

Utilisation: Won’t break down a compound and grow

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3
Q

What is the bio-synthetic pathway leading to arginine synthesis?

A

pre >(E)> ornithine >(F)> citruline >(H)> arginine

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4
Q

What is the purpose of cheslex resin?

A

Binds metal ions that would usually allow nucleases to degrade DNA when cells are lysed.

Used so we could isolate our DNA in polymorphism prac

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5
Q

How do polyacrylamide and agarose gel electorphoresis differ?

A

Polyacrylamide for small DNA fragments (<1000bp)

Agarose for larger DNA fragments

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6
Q

What is D1S80?

A
  • VNTR locus (microsatellite)
  • Highly variable in population
  • Chromosome 1, non coding
  • Repeat sequence is 16bp
  • 14 - 41 copies in people
  • 29 differnt alleles
  • Each person has 2 copies of the locus
  • Used for DNA fingerprinting
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7
Q

How can you work out the number of repeats present in fragments from the D1S80 locus?

A

n = (total bp - 115 bp - 145 bp - 32 bp) / 16

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8
Q

What is the Newcombe experiment?

A
  • Use E.coli (we use P. aeruginosa) and BOT1 resitance (we use streptomycin resistance)
  • Grow bacteria, some plates are spread to seperate micro-colonies
  • Non-spread = control
  • Introduce the selective agent
  • Adaptation hypothesis would suggest all bactieria would be phage sensitive at time of re spreading - no phage contact
  • Mutation Selection hypothesis would suggest both suseptible and resistant colonies would be present and spreading would increase number of resistant colonies (this was true)
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9
Q

How do mutation selection and adaptation hypotheses differ?

A

Mutation Selection: Mutation comes before selection and is spontaneous and random.

Adaptation: Mutation happens in response to selection and is not spontaneous or random.

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10
Q

What is the purpose of a spread plate?

A

To distribute microcolonies, change distribution of cells.

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11
Q

How do mean and variance relate to mutation selection and adaptation hypotheses?

A

Mutation selection- higher variance and mean

Adaptation

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12
Q

What is a heterokaryon and what can it be used for?

A
  • Gentically different nuclei in Aspergillus cells
  • Used for complementation
  • Parasexual cycle for haplodization (mapping genes TO chromosomes)
  • Sexual cycle for transient diploid and meiosis (mapping WITHIN chromosome)
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13
Q

What is the nitrate assimilation pathway?

A

Nitrate > Nitrite > Ammonia

NO3 > NO2 > NH3

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14
Q

What is the basic life cycle of Aspergillus?

A
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15
Q

Which enzymes work in the nitrate assimilation pathway?

A 
Nitrate Reductase (niaD)
Nitrite Reductase (niiA)
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16
Q

What is cross feeding?

A

Nitrite was supplied to niaD mutants because niiA mutants were nearby and reducing Nitrate to Nitrite

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17
Q

In what situation will a positive NED test arise?

A

When nitrite is detected.

E.g. Nitrite reductase mutants (niiA)

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18
Q

Which mutants affect the nitrate assimilation pathway and where?

A

NiaD - Nitrate reductase

NiiA - Nitrite reductase

NirA - Both parts of pathway (regulator gene, transcription factor)

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19
Q

Where are the genes for the assimilation pathway located?

A

niiA, niaD linked to same chromosome

nirA unlinked as a regulatory gene

nirA mutants episatic to all others
( phenotypes: nirA- = nirA- niiA- = nirA- niaD- = niiA- niaD - )

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20
Q

What is haplodization and how can it be used?

A
  • Random mitotic loss of chromosomes
  • No recombination
  • Map to known gene markers
  • No order, RF
  • Work out if genes on same chromosome (recover non parentals and parentals means differnt chromsomes)
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21
Q

Which pathways lead to brown and red pigment synthesis for drosophila eye colour?

A
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22
Q

What is the mode of inheritance for forked bristles, white eyes and lozenge eyes?

A

Recessive and on X chromosome.

w+ lz+ f+ = red

w- lz- f- = white

w -(24.6)- lz -(26.9)- f

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23
Q

What kind of mutation was found in cross 2 in the drosophila recombination experiment?

A

Paracentric inversion

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24
Q

What are polytene chromosomes and where are they found?

A
  • In certain somatic tissue
  • Chromosomes replicate without mitosis, giant chromosomes form (endomitosis)
  • Occurs in euchromatic regions (not tight)
  • Heterochromatic regions form chromocentre
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25
Q

How can heritability be estimated?

A

Mean value of offspring (y) against mean value for parents (x)

Regression coefficient (b) measures slope and h2

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26
Q

How can you work out if a correlation is significant?

A
  • Work out the correlation coeffecient (r).
  • -1 meas perfect negative correlation
  • +1 means perfect positive correlation
  • 0 means no correlation
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27
Q

How and when can you work out regression?

A

If significant correlation between X and Y, y=a+bx

b= slope = regression coefficient

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28
Q

How can you calculate Pattern Intensity Index?

A

Number of tri radii on all digits

29
Q

What are the steps in gene cloning?

A

Day One

  • Ligation
  • Ligate cutE+ (e.coli) with pBluescriptSK+ plasmid
  • pBluescript part plasmid, part phage (E.coli genome and bacteriophage plasmids from f1, M13, T3, T7)
  • Transformation
  • Transform the DNA into competent E.coli cells and place on indicator medium

Day Two

  • Plasmid DNA Isolation
  • Isolate plasmid DNA from transformed colony

Day Three

  • Restriction Enzyme Digests

Day Four

  • Gel Electrophoresis
  • Electrophorese digested and undigested DNA plasmid through agarose
  • See if gene successfully cloned
30
Q

What is the role of cutE?

A
  • Produces protein for transport copper ions in E. coli
31
Q

Why were white colonies chosen during the recom DNA experiment?

A

White colonies mean the gene has been inserted into the plasmid and disrupted the LacZ gene.

IPTG is inducer, X gal is cleaved to become blue when B-galactosidase is present.

32
Q

What were the transformation mixtures plated with and why?

A

A: Ligation Reaction

  • TCM buffer
  • Insert DNA
  • Vector DNA
  • Reaction buffer WITH ligase
  • Should allow ligation to occur

B: Zero DNA Ligase Control

  • TCM buffer
  • Insert DNA
  • Vector DNA
  • No ligase so no colonies should grow, work out efficiency of ligase enzyme

C: Uncut Vector DNA Control

  • TCM buffer
  • Uncut Vector (pBSK+) DNA
  • Should only see blue colonies

D: Competent Cells Only Control

  • TCM buffer

All:

  • Competent E.coli cells
  • Luria Broth (rich media)
  • Ampicillin (select for E.coli that took up pBSK+)
  • IPTG (incuder, stimulate lacZ in non-transformed)
  • X-gal (broken down/blue by non-transformed with functional lacZ)
33
Q

Why did the blue:white ratios differ between the single and double digest?

A

Single Digest: More blue because re-ligation occurs more readily

34
Q

What are satellite colonies?

A

Grow around larger transformant colonies. They grow becuase beta lactamase is secreted by other bacteria and so they’re not affected by the ampicillin.

35
Q

How can bacteria be made competent?

A

Calcium Chloride treatment.

36
Q

What differences were observed when cutting with EcoRI or Bam/HindIII?

A

EcoRI

  • pBSK+ 2.9kb
  • cutE insert 2.3kb
  • The pCutE plasmid has 2 EcoRI sites inside cutE (the gene gets cut down to 2.3kb)

Bam/Hind

  • pBSK+ 2.9kb
  • cutE insert 3.6kb
  • Bam and Hind each only appear once in the pCutE plasmid as the points where the gene is inserted into pBSK+
37
Q

What are the distinct bands seen in gel of uncut pBSK+ DNA?

A
  • concatomers (large)
  • naked open circular
  • supercoiled/closed circular (smallest)
38
Q

What can you see on the agarose gel electrophoresis of aspergillus DNA?

A

Smear, faitn discrete bands due to repetetive sequences (eukaryotic)

39
Q

How can the average fragment length be calculated?

A

Using %GC content.

e.g. 50% GC means (1/4) G, C, T, A

  • HindII: 5’ A|AGCTT 3’
  • Av fragment length = ( (1/4)^6) ^-1 = 4096 bp
  • Divide by size of genome to work out number of fragments
40
Q

How could you confirm that the cutE gene had been successfully cloned?

A

Transformation experiment : Functional Complementation

  • Transform E.coli cutE- strain with pBSK+/cutE+ plasmid DNA
  • Isolate/purify transformant
  • Insert, grow on culture
  • Perform survival assay with copper on agar plates and compare to controls
  • WT = better survival than cutE- as [copper] increases
41
Q

What are the features of pBSK+?

A

2.95 kb

ColE1 ORI

  • Autno0mous replication from bacterial chromosome
  • High copy number (relaxed), lots of replication

Ampicillin Resistance Gene (selection)

  • Bla gene (B-lactamase breaks down amp in media)
  • Select for transformants because they grow (blue and white)

LacZ gene (selection) (in MCS)

  • Can select transformants with insert DNA
  • Insertional Inactivation
  • DNA inserted to MCS, disrupt lacZ, no B-galactosidase made, colonies that are transformant are white
  • MCS has 23 unique restriction enzyme sites within lacZ fragment

LacI+ gene

  • Encodes repressor protein
42
Q

How does the lysogenic cycle arise?

A
  • Temperate bacteriophage intergrate into bacterial chromosome (site specific recombination)
  • int = integrase
  • attB, attP
  • CI repressor protein repressors lytic cycle
  • UV light induces lysogeny
    • Damages bacterial DNA
    • Bacterial SOS response, recA cleaves CI repressor proteins
    • xis gene produced by lamda, lytic cycle occurs
    • Infect C600 bacteria (lamda sensitive) and plagues formed on induction plates
43
Q

What were the role of E.coli W3110 and C600 in the lysogeny experiment?

A

W3110

  • Allows the lamda to multiply

C600

  • Allows lamda to act on it (lysed ones are killed)
44
Q

How can titre (pfu/ml) be calculated?

A

e.g.

100ul
14 plagues on dilution 10^-6
14 X 10^6
= 1.4 X 10^7 per 100ul (X 10^1 to get to ml)
= 1.4 X 10^8 per ml (1000ul)

45
Q

What does the absence of plagues indicate?

A

Indicates the lytic cycle is occuring (cells killed)

46
Q

How can the viable count (cfu/ml) be calculated?

A

e.g.

100ul
376 colonies on dilution 10^-6
376 X 10^6
= 3.76 X 10^8 per 100ul (X 10^1 to get to ml)
= 3.76 X 10^9 per ml (1000ul)

47
Q

How can the % of lysogenic cells be calculated?

A

(Number Induced (plagues, pfu/ml) - number plagues on no UV control plate)

_________________________________

(viable count (colonies, cfu/ml) )

48
Q

How are some bacteria immune to super infection?

A

The CI repressor protein is present in the cell being affected so prevents the lytic cycle occuring

49
Q

What makes up the lac operon?

A
  • Lac I: codes repressor protein (homo-tetrameric)
  • Plac : codes promoter sequence
  • Lac O: codes operator sequence
  • Lac Z: codes for the β-Galactosidase
  • Lac Y: codes permease (brings lactose into cell)
  • Lac A: codes transacetylase
  • Operon itself has Structural genes (Z, Y, A), Promoter (Plac) and Operator (LacO)
50
Q

What are the features of IPTG and ONPG?

A

IPTG: gratuitous inducer of lac operon, not substrate

ONPG: colourless substrate, not inducer, comverted to yellow (more yellow = more activity)

51
Q

Which strain of bacteria was WT and which was mutant (what kind of mutant?) in the B-galactosidase prac?

A

WT

  • CSH62

Mutant

  • CSH36
  • Constitutive mutant
  • lac I - or
  • lac O (c)
52
Q

How can you work out co-transduction frequency?

A

number of co-transductants

__________________________________

total number of transductants

53
Q

How can you work out RF through transduction?

A

Number co-transductants

_____________________________________

Total Transductants X (co-transduction frequency/100)

  • Via reciprocal crossing experiments, us appropriate cross data
54
Q

How can you work out gene order using transduction?

A
  • Reciprocal crossing experiments (most rare crossover has least cotransductants)
  • Highest co-transduction frequency = closest togther

leu —-(65%)—- trp -(72%)- met

55
Q

what were the genotypes of the donor and recipient in the conjugation prac?

A

Donor (E.coli Hfr CSH62)

  • Hfr, thi-, strs
  • Has integrated F factor, sensitive to streptomycin

Recepient (E.coli AB1515)

  • F-, this-, leu-, pro-, trp-, ade-, lac-, strr
  • No F factor, won’t utilise lactose, resitant to streptomycin
56
Q

How do ex-conjugants arise?

A
  • Rolling circle replication and DNA transfer (hfr to f)
  • Hfr = high frequency of recombination
  • F factor RANDOMLY integrates into bacterial genome (orientation, place varies)
  • Transfer from origin of transfer (F factor) – not boundary between F factor and chromosome
  • F transfers can carries bacterial chromosome with it
  • Order of transfer = order of genes
  • Use timing to work out order (Interrupted mapping, time of entry mapping)
  • Plate on medium which kills Hfr donor
  • Ex-conjugants selected for WT allele of each marker gene
  • E.g. leu+ only gene evident after 5mins, must come first after the F factor (F factor is engine, leu first carraige)
57
Q

Why is streptomycin in selection plates in the conjugation prac?

A

To prevent Hfr donor strain (CSH62) growing

58
Q

Why is vortexing important in conjugation experiments?

A

It breaks the conjugation tubes

59
Q

What was the gene order shown in the conjugation prac?

A

leu - pro - trp

  • Most colonies grew on leu then pro then trp
60
Q

what is asynchronous conjugational mapping?

A

The mating is continuous and not interupted at regular intervals

61
Q

How do allozymes and isozymes differ?

A

Allozymes: enzyme products of different alleles

Isozymes: enzymes of similar function produced by different GENES

62
Q

What genotypes and phenotypes were demonstrated in the fly adh prac?

A

AdhF: fast

  • 0 charge

AdhUF: ultra-fast

  • -1, moves to +ve

AdhS: slow

  • +1, moves to -ve

AdhNull: loss of function

AdhFT: thermo-stable

63
Q

What is the charge of UF, F and S allozymes?

A

UF: -1

F: 0

S: +1

64
Q

How is the gel stained to differentiate ADH enzymes?

A
  • Makes use of the excisting cycle which converts ethanol to acetaldehyde
  • PMS: light sensitve, darkens when exposed to light
  • NADH to NAD and NBT to Formazan means yellow to purple
65
Q

What zymogram patterns are seen for enzymes that are monomers, dimers etc?

A

Monomer

  • Heterozygote has 2 bands

Dimer

  • Heterozygote has 3 bands

Due to co-dominance

66
Q

Is modification heritable?

A

No, it is imposed to phage by the host bacterial cell

67
Q

What are r and m and how do they interact?

A

r = restriction endonuclease

m = modification enzyme (protects)

r cuts the DNA and m can prevent the DNA from being cut (e.g. methylation - dam methylase)

  • r+m+ = wildtype, protected from restriction
  • r+m- = not protected from restriction (0% efficiency and no plagues)
  • r-m- = no restriction and no modification
  • r-m+ = no restriction, but would be protected if there was
68
Q

What did the table in the restriction modification prac show?

A

C600 was WT (r+m+)

  • plagues formed when infected with lamda C6 (100% efficiency) meaning cells were surviving
  • no plagues formed when infected with lamda E8 (0%) suggesting cells were being restricted

ED8739 was mutant (r-m-)

  • plagues formed when infected with lamda C6 (25%) suggesting DNA can’t be modified (m-)
  • plagues formed when infected with lamda E8 (100%) suggesting cells were not resticted or protected (therefore r-m- because it would be 0% if r+ becuase the other late shows m is mutant and there is no protection)

Genetics (3 decks)

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