2 - inheritance + transfer of genetic material

Question 1

describe how plasmids replicate

Answer 1

plasmids are circular with a constant origin of replication

this means they can replicate autonomously (self-replicating)

Question 2

define F, R and Col plasmids

Answer 2

F = fertility plasmid
contains all genetic material required for conjugation

R = resistance plasmid
carry genes for antibody resistance

Col =
carry genes which express products to kill other bacteria —-> knock out competition when competing for resources

Question 3

describe the structure of bacterial genomes

Answer 3

core genome contains ~15000 genes
very large and incredibly diverse, showing that bacteria exhibit genomic plasticity

as genome size increases, gene content increases proportionally

bacterial gene is ~1,000 bp (~330 aa) long

bacterial chromosomes exist in super-coiled looped circles —> very compact

bacterial genomes described by their %GC pairing

contain many operon (cluster of genes with related function)
operon is expressed by single promoter —> operon either on or off

Question 4

describe the structure of viral genomes

Answer 4

outer capsid protein (head) enclosed genetic materials

variety of genetic materials: DNA, ssDNA, dsRNA, ssRNA

contain both linear and circular genomes

extremely high coding density (genome compressed to fit inside head)

Question 5

list the advantages of using bacteria and viruses to study genomics

Answer 5

1. small size —> want large population in small place
2. rapid reproduction —> bacterial replication takes ~20 mins
3. selective media (e.g., antibiotics)
4. simple structures and physiology
5. genetic variability
6. complete genome sequences

Question 6

virulent vs avirulent bacteria

Answer 6

virulent:

• capsule, therefore pathogenic
• smooth colonies (Type S)

avirulent:

• no capsule
• small, rough colonies (Type R)

type of virulent strain depends on specific genes —> which polysaccharide is expressed (type I, II, III)

Type R Type S happens at very low rate
Type I does not convert to II or III

Question 7

how was transformation discovered using S. pneumoniae?

Answer 7

S. pneumonia is an infectious bacterium which makes a capsule to protect itself from the host immune system

Griffith injected various types of S. pneumonia into a living mouse

• living type 3s —> kills mouse
• dead type 3s (heat killed) —> mouse lives
• living type 2r (no capsule) —> mouse lives
• dead type 3s + living type 2r —> dead mouse = living type 3s?

non-virulent (harmless) bacteria have been converted into virulent, killer bacteria

Dawson and Sia further investigated this in vitro

1) ADD PROTEASE
- all protein degraded
- colonies grown
- protein cannot be transforming factor
2) ADD RNase
- all RNA degraded
- colonies grown
- RNA cannot be transforming factor
3) ADD DNase
- colonies do not grow
- DNA must be transforming factor

DNA allows expression of capsule

DNA transferred between cells = TRANSFORMATION

Question 8

transformation uptake machinery

Answer 8

1) exogenous DNA bound to receptor on recipient cell via competence proteins (comEA + comG)
2) DNA pulled through comEC protein channel by comFA DNA translocase
3) as it passes through channel, one strand of DNA is degraded by deoxyribosenuclease
4) surviving single strand is stabilised by ssDNA binding protein
5) recA binds ssDNA to allow recombination
6) ssDNA fragment integrated into host chromosome —> produces heteroduplex with different alleles in the double strands (a+/a-
* note: this will produce two daughter cells, one a+ one a-. environment will favour one, often other strain does not survive

Question 9

frequency of co-transformation

Answer 9

cotransformation is very unlikely as very small pieces of DNA are transformed

two genes must be extreemly close together to be transformed

the frequency with which two genes are transformed can determine how far apart they are

Question 10

mechanism of conjugation

Answer 10

1) donor synthesises an F-pilus which contacts the recipient cell
2) pilus is retracted to bring the membranes of the two cells close together
3) TraG (membrane-anchored protein) forms a pore between the two membranes

4) TraI nicks F-factor at OriT and rolling circle replication used to transfer the plasmid into donor cell
- —> 5’ end is displaced into recipient cell, covalent extension begins at the 3’ end
- —> circular template strand continues to roll with 3’ covalent extension (creating ds)
- —> displaced end is cleaved and re-circularised to produce ss plasmid in recipient cell
- —> ds plasmid is formed via discontinuous synthesis (okazaki fragments)

recipient now becomes donor

Question 11

why can only some bacteria initiate transfer in conjugation?

Answer 11

cell is fertile only if it has the genetic fertility trait

F+ = fertile
F- = sterile

this F-factor can be transferred by itself as a plasmid or in conjunction with other chromosomal sequences

Question 12

what is the size of the F-factor

Answer 12

~100kb = significant length

Question 13

what are the advantages of having a free F-plasmid?

Answer 13

conjugative plasmids carry several genes required for conjugation to occur
e.g. pilin protein is necessary for the production of a sex pilus

Question 14

why can’t F+ transfer to F+ ?

Answer 14

only one F-plasmid is needed per cell

TraS is responsible for stopping further transfer of F-plasmid

If cell contains F-plasmid, it makes TraS

TraS binds to and blocks TraG —> prevents formation of membrane bridge between two cells —> inhibiting uptake of further DNA —> only one F-plasmid per cell

Question 15

how are Hfr cells formed?

Answer 15

Hfr = F-plasmid integrated into chromosome

integration of episome depends on insertion sequence (IS:

• present in both episome and bacterial chromosome
• IS elements are transposable

episome inserts into the chromosome at the IS site

the IS can become duplicated during insertion = target site duplication

Question 16

in E. coli, 1 minute corresponds to what length of DNA?

Answer 16

~40kb

Question 17

what in an F’ factor?

what is it called when an F’ factor is transferred to a new cell?

Answer 17

F-factor can be excised from Hfr cells via non-homologous recombination

this results in a portion of the host chromosome being included in the excised F-factor = F’ factor

when F’ factors a transferred, a partial diploid is created

this transfer is called sexduction

Question 18

life-cycle of bacteriophage T4

Answer 18

T4 = virulent = lytic

1) bacteriophage attaches to cell and injects it’s DNA
2) synthesis of phage-specific mRNAs. phage-encoded nucleases degrade host DNA (phage has alteration in cystine*, therefore is protected from degration)
3) phage replication via host mechanisms
4) phage particles reassemble
5) phage-encoded lysozyme ruptures bacterial cell and phages escape
* 5-hydroxymethyl-cytosine

Question 19

how does T4 package it’s chromosome?

Answer 19

T4 genome is circularly permutated and terminally redundant (duplicated, these areas undergo recombination)

packaged via headful mechanism

1) DNA is packaged into head
2) once head is full, it will contain the whole genome plus some of the terminally redundant area (some extra DNA)
3) nuclease cleaves DNA
4) next head will begin filling with continuing sequence

Question 20

life cycle of bacteriophage lambda

Answer 20

1) injection of phage DNA
2) circularisation of prophage

then enters either:
lytic pathway —> replication, viral assembly, cell lysis
lysogenic pathway —> progphage integrates into chromosome, silent

pathway choice depends on nutrient status of host cell

prophage can switch between two pathways

Question 21

how does the lambda phage integrate into the host chromosome?

Answer 21

integration occurs at homologous regions (~15bp) which are able to recognise eachother

attP site on lambda phage attaches to attB site on bacteria

• —> integrase binds to attP site
• —> resulting DNA-protein complex can now bind to attB

recombination event mediated by lambda integrase
—-> integrase catalyses cutting and resealing reactions

attachment sites also sites for excision —> sharing DNA

Question 22

generalised vs specialed transduction

Answer 22

generalised:

• occasionally, bacterial DNA is packed into a phage
• phage infects new cell, bacterial DNA in phage recombines into new host genome

specialised (lysogenic only):

• section of prophage DNA integrated into bacterial chromosome
• occasionally, prophage exits incorrectly and takes a fragment bacterial DNA
• phage infects new cell, bacterial DNA + phage DNA in phage recombines into new host genome

Question 23

how does the lambda phage excise from the host chromosome?

Answer 23

NORMAL EXCISION

1) prophage loops out, attBP pairs with attPB
2) site-specific recombination occurs between pairs, excising the phage chromosome

ANAMOLOUS EXCISION

1) prophage loops out abnormally, attBP and attPB are not paired
2) excision occurs, leaving the phage with some bacterial DNA and vice versa
3) in new cell, can either undergo site-specific recombination or crossover between bacterial genes

Question 24

list and describe three types of transposable elements

include examples

Answer 24

1) CUT AND PASTE TRANSPOSONS:
- element cut (excised) from one site and pasted (integrated) into a new site
- facilitated by transposon-encoded enzyme = transposase
- e.g. IS elements, composite transposons, Ac/Ds elements

2) REPLICATIVE TRANSPOSONS:
- an element is replicated and one copy is inserted into a new site
- e.g. Tn3 elements

3) RETROTRANSPOSITION:
- an element’s mRNA transcripts are used as a template to synthesise DNA molecules by reverse transcription
- these are then inserted into new genomic sites
- e.g. telomere-specific tetrosposons

bacteria only have cut and paste
replicative transposons + retrotransposons are only in eukaryotes

Question 25

how do insertion sequences insert? include structural information

Answer 25

1) endonuclease cleaves two strands of the target DNA at different sites –> usually at terminal inverted repeats (segments which only encode genes needed for movement)
2) IS element is inserted into the gap created by the staggered cleavage (sticky ends)
3) DNA polymerase repairs sticky ends —> results in duplicated DNA regions either side of transposon = target site duplications

the IS elements can now control expression of neighbouring genes

Question 26

what is composite transposition?

Answer 26

when two IS elements insert very close together, they can take the small DNA region between them and move as a whole

can carry antibiotic resistance genes
—> alter population very quickly

Question 27

what are Tn3 elements and how are they transposed?

Answer 27

Tn3 elements are replicative transposons found in bacteria

they are larger than IS elements

they have simple inverted repeats at each end

they produce target site duplication where they insert

like composite transposons, they contain genes that are not required for transposition

transposition of Tn3:
1) Tn3 encoded transposase catalyses the formation of a cointergrate between the donor and recipient plasmids

2) Tn3 replicated during step 1 such that there is a copy at each junction in the cointegrate
3) resolvase produced by tnpR gene resolves the cointegrate by mediating recombination between the two Tn3 plasmids
4) donor and recipient plasmids separate, each with a copy of Tn3

Question 28

what is staphylococcus aureus?

how has it become resistant to antibiotics?

Answer 28

s. aureus = “golden staph”

~30% of population is temporarily or permanently colonised

methicillin resistant staphylococcus aureus (MRSA) is treated via vancomycin

some MRSA is now becoming VRSA/VISA —> this arose ~10 years after the rise of vancomycin resistant enterococci (VRE)

how does MRSA become VRSA?

1) VRE carries Tn1546 (mediates vancomycin resistance)
2) Tn1546 can jump into a conjugative plasmid for multiple different species
3) s. aureus (MRSA or MSSA) produces a mating pheromone that is very similar to the mating pheromone of e. faecalis
4) s. aureus and e. faecalis mate and the conjugative plasmid + the Tn1546 is transferred
5) VRSA is created

Question 29

what is the Ac/Ds system?

Answer 29

Ac/Ds system is a cut-and-paste transposon found in Maize

colour on maize kernels is affected by the C’ allele, which encodes a dominant inhibitor of aleurone coloration (inhibits all colours but yellow)
—–> the genotype in yellow kernels is C’CC

mosaics with pigmented patches are caused by loss of the C’ allele
——> the genotype in pigmented kernels is –CC, where the – means that one of the copies of the gene has been lost

• the dissociation factor (Ds) is located at a site on chromosome 9 in mosaic kernels where chromosome breakage occurs
• the Ds transposable factor was discovered because of its ability to break chromosomes, but Ds cannot induce chromosome breakage by itself
• Ds activity is activated by another transposable element, Ac (Activator), which encodes a transposase (i.e. Ds not functional on its own)
• the Ac element encodes a transposase that is responsible for excision, transposition, mutation, and chromosome breakage
• the Ac transposase catalyses the movement of the Ac/Ds elements

Question 30

how does intrachromosomal recombination between transposons produce deletions / inversions

Answer 30

DELETIONS = TRANSPOSONS IN SAME ORIENTATION

1) chromosome loops out so transposons can pair with eachother
2) recombination between the paired transposons deletes the region

INVERSIONS = TRANSPOSONS IN OPPOSITE ORIENTATION

1) chromosome bends back on itself so transposons can pair with eachother
2) recombination between transposons inverts the region

Question 31

what happens when there is unequal crossing-over between transposons on sister chromatids?

Answer 31

unequal crossing-over between transposons on sister chromatids causes gene duplication

1) chromosome (w/ two transposons in same orientation) replicates to form two sister chromatids
2) transposons pair unequally, transposon recombination results in one chromosome with deletion and the other with duplication