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
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
3
Q
What is the bio-synthetic pathway leading to arginine synthesis?
A
pre >(E)> ornithine >(F)> citruline >(H)> arginine
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
5
Q
How do polyacrylamide and agarose gel electorphoresis differ?
A
Polyacrylamide for small DNA fragments (<1000bp)
Agarose for larger DNA fragments
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
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
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)
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.
10
Q
What is the purpose of a spread plate?
A
To distribute microcolonies, change distribution of cells.
11
Q
How do mean and variance relate to mutation selection and adaptation hypotheses?
A
Mutation selection- higher variance and mean
Adaptation
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)
13
Q
What is the nitrate assimilation pathway?
A
Nitrate > Nitrite > Ammonia
NO3 > NO2 > NH3
14
Q
What is the basic life cycle of Aspergillus?
A
15
Q
Which enzymes work in the nitrate assimilation pathway?
A
Nitrate Reductase (niaD) Nitrite Reductase (niiA)
16
Q
What is cross feeding?
A
Nitrite was supplied to niaD mutants because niiA mutants were nearby and reducing Nitrate to Nitrite
17
Q
In what situation will a positive NED test arise?
A
When nitrite is detected.
E.g. Nitrite reductase mutants (niiA)
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)
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 - )
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)
21
Q
Which pathways lead to brown and red pigment synthesis for drosophila eye colour?
A
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
23
Q
What kind of mutation was found in cross 2 in the drosophila recombination experiment?
A
Paracentric inversion
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
25
How can heritability be estimated?
Mean value of offspring (y) against mean value for parents (x)
Regression coefficient (b) measures slope and h2
26
How can you work out if a correlation is significant?
* Work out the correlation coeffecient (r).
* -1 meas perfect negative correlation
* +1 means perfect positive correlation
* 0 means no correlation
27
How and when can you work out regression?
If significant correlation between X and Y, y=a+bx
b= slope = regression coefficient
28
How can you calculate Pattern Intensity Index?
Number of tri radii on all digits
29
What are the steps in gene cloning?
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
What is the role of cutE?
* Produces protein for transport copper ions in E. coli
31
Why were white colonies chosen during the recom DNA experiment?
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
What were the transformation mixtures plated with and why?
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
Why did the blue:white ratios differ between the single and double digest?
Single Digest: More blue because re-ligation occurs more readily
34
What are satellite colonies?
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
How can bacteria be made competent?
Calcium Chloride treatment.
36
What differences were observed when cutting with EcoRI or Bam/HindIII?
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+
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37
What are the distinct bands seen in gel of uncut pBSK+ DNA?
* concatomers (large)
* naked open circular
* supercoiled/closed circular (smallest)
38
What can you see on the agarose gel electrophoresis of aspergillus DNA?
Smear, faitn discrete bands due to repetetive sequences (eukaryotic)
39
How can the average fragment length be calculated?
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
How could you confirm that the cutE gene had been successfully cloned?
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
What are the features of pBSK+?
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
How does the lysogenic cycle arise?
* 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
What were the role of E.coli W3110 and C600 in the lysogeny experiment?
W3110
* Allows the lamda to multiply
C600
* Allows lamda to act on it (lysed ones are killed)
44
How can titre (pfu/ml) be calculated?
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
What does the absence of plagues indicate?
Indicates the lytic cycle is occuring (cells killed)
46
How can the viable count (cfu/ml) be calculated?
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
How can the % of lysogenic cells be calculated?
(Number Induced (plagues, pfu/ml) - number plagues on no UV control plate)
\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_
(viable count (colonies, cfu/ml) )
48
How are some bacteria immune to super infection?
The CI repressor protein is present in the cell being affected so prevents the lytic cycle occuring
49
What makes up the lac operon?
* 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
What are the features of IPTG and ONPG?
IPTG: gratuitous inducer of lac operon, not substrate
ONPG: colourless substrate, not inducer, comverted to yellow (more yellow = more activity)
51
Which strain of bacteria was WT and which was mutant (what kind of mutant?) in the B-galactosidase prac?
WT
* CSH62
Mutant
* CSH36
* Constitutive mutant
* lac I - or
* lac O (c)
52
How can you work out co-transduction frequency?
number of co-transductants
\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_
total number of transductants
53
How can you work out RF through transduction?
Number co-transductants
\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_
Total Transductants X (co-transduction frequency/100)
* Via reciprocal crossing experiments, us appropriate cross data
54
How can you work out gene order using transduction?
* Reciprocal crossing experiments (most rare crossover has least cotransductants)
* Highest co-transduction frequency = closest togther
leu ----(65%)---- trp -(72%)- met
55
what were the genotypes of the donor and recipient in the conjugation prac?
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
How do ex-conjugants arise?
* 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
Why is streptomycin in selection plates in the conjugation prac?
To prevent Hfr donor strain (CSH62) growing
58
Why is vortexing important in conjugation experiments?
It breaks the conjugation tubes
59
What was the gene order shown in the conjugation prac?
leu - pro - trp
* Most colonies grew on leu then pro then trp
60
what is asynchronous conjugational mapping?
The mating is continuous and not interupted at regular intervals
61
How do allozymes and isozymes differ?
Allozymes: enzyme products of different alleles
Isozymes: enzymes of similar function produced by different GENES
62
What genotypes and phenotypes were demonstrated in the fly adh prac?
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
What is the charge of UF, F and S allozymes?
UF: -1
F: 0
S: +1
64
How is the gel stained to differentiate ADH enzymes?
* 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
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65
What zymogram patterns are seen for enzymes that are monomers, dimers etc?
Monomer
* Heterozygote has 2 bands
Dimer
* Heterozygote has 3 bands
Due to co-dominance
66
Is modification heritable?
No, it is imposed to phage by the host bacterial cell
67
What are r and m and how do they interact?
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
What did the table in the restriction modification prac show?
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)
