Bacteriophage Genetics

Question 1

Bacteriophage

Answer 1

Bacteriophage are viruses that infect bacteria and the ones that infect Ecoli can be either virulent or temperate:

• Virulent phage do not integrate their DNA into the cell and usually just kill it. Virulent phage include the “T” series phage T1, T2, T3 etc.
• Temperate phage integrate their DNA into the host DNA causing a permanently infected state. Examples of temperate phage are phage λ and φ80.

Different phage have different forms of genetic material such as dsDNA (double stranded), ssDNA, and RNA. It can be either circular or linear.

Question 2

Examining Bacteriopahge

Answer 2

To demonstrate the effect of phage on bacteria we use agar plates covered in a bacterial lawn. When phage infect the bacteria they cause holes in the lawn called plaques, these holes are where the phage have successfully propagated and can be turbid or clear depending on which cycle is used.

Bacteriophage anatomy = different phage have different morphologies;

the T4 phage has a head that contains the DNA, a tail with a base plate, and also tail fibers.

The λ phage is very similar except for it lacks tail fibers

Question 3

Lytic cycle

Answer 3

the phage attaches to the cell wall of the bacteria and injects DNA into the cell.

The DNA forms into a circle and recruits the host machinery to produce more and more phage components.

Eventually the cell wall ruptures and all the new phage come out after being assembled and having DNA packaged into them.

Once this has occurred the cell is considered to have been lysed.

Produces clear plaques

most infections lead to lytic cycle, ~20% lysogenic

Question 4

Lysogenic cycle

Answer 4

begins similarly to the lytic cycle by attaching to the cell wall of the bacteria, injecting nucleic acid, and circularising.

It then starts to differ because the C1 gene produces a repressor protein that binds to its transcription factor and blocks the lytic pathway.

Then the int gene produces integrase which allows recombination with the host DNA and eventually integration into it.

The attachment points are called attP (for prophage) and attB (for bacteria), and they are complementary.

The integrated phage is known as a prophage and the bacteria housing it is called the lysogen.

Produces turbid plaques

lambdah DNA slows down growth of E.coli DNA

Question 5

Superinfection

Answer 5

• Lysogens are immune to superinfection from phage of the same type. This is because the excess C1 protein floating around will bind to any more phage DNA that enters the cell and cause it to be lost.
• This means it is almost beneficial to be lysogenic if you are the bacteria, like an immune system

Question 6

Induced lytic cycle

Answer 6

• UV light damages DNA and protein, need to be repaired
• UV light induce lambdah phage to undergo lytic cycle, especially in cells that are more damaged
• same for other DNA damaging agent
• survival mechanism
• protease triggered to clean up damaged protein
• C1 sensitive to protease, gets broken down, no C1 >> lytic cycle not blocked
• expression of gene xis, antagonist to int, reverses recombination event that int generated, xis separates two circles >> prophage moves out of E.coli genome, all lytic gene expressed, makes more phage protein, kills E.coli, progeny comes out

Question 7

The choice between lytic and lysogenic cycles

Answer 7

The choice of the phage to enter the lytic or lysogenic cycle comes down to the physiological condition of the bacteria.

If the bacterium is healthy enough to sustain the phage it will enter the lysogenic cycle and use the bacteria’s machinery.

However, if the bacterium is lacking in nutrients and unhealthy the phage will go down the lytic pathway since there is no point sticking around.

Question 8

Conjugation

Answer 8

can also lead to prophage induction.

This is because phage DNA is replicated and transferred through a conjugal bridge.

If there is no C1 in the partner cell, then the balance will be tipped and the Ecoli will enter the lytic cycle.

• lambdah excises itself (xis expressed) from E.coli DNA

Prophage induction that is induced through conjugation is called zygotic induction.

Conjugation between two lysogens will not result in zygotic induction because C1 is present in both.

Question 9

Host specificity

Answer 9

Phage has proteins that specifically interact with proteins on its host. Interactionbetween phage tail protein and receptor on host cell.

These allow it to determine whether it is a good host or not.

The recognition mechanism is a protein- protein interaction.

Machinery of host must be compatible with phage, co-evolved

Question 10

Circularisation

Answer 10

λ phage
- linear DNA is inserted into the cell and once inside it circularises. It does this with what are called “sticky ends”.

• Sticky ends are small single stranded sequences of DNA sticking out from either end of the linear molecule. These ends are also sometimes referred to as cos ends (cohesive) and are complementary to each other.
• The two cos ends find each other and base pair to form the molecule into a circle. An enzyme called ligase is needed to form phosphodiester bonds between the two ends.

Question 11

Bidirectional Replication (theta replication)

Answer 11

A type of replication that involves two replication forks proceeding in opposite direction from the specific origin of replication.

The two original strands serve as templates for the new strands and the result is two identical daughter molecules.

This is what Lambda phage do initially upon entering the bacterial cell.

Eventually it has to switch to another strategy so it can make linear DNA that can be packaged.

Question 12

Rolling circle (sigma replication)

Answer 12

• creates linear DNA

A nick is made in one strand only by endonucleus (at a different origin) and complementary DNA is added to this.

Okazaki fragment formed

The now linear strand is pulled out as DNA is added causing the inner circle to turn (hence the name).

DNA is also added to the inner circle in a leading strand fashion which allows the replication to continue indefinitely into a long concatamer.

Concatamers consist of multiple genomes joined by cohesive sites (cos sites).

Question 13

Packaging

Answer 13

The packaging of concatamers centers around the recognition of cos sites.

The DNA is fed into the head of the phage, the cos sites are recognised by head proteins, the concatamer is cut, and the phage tail is added.

Genes are clustered by function in Lambda bacteriophages (next to each other, efficient). This is because it means things can be regulated sequentially i.e. proteins need to be synthesised sequentially.

Question 14

T4

Answer 14

• can undergo lytic cycle

Quite similar to Lambda however there are a few differences:

• DNA replication proceeds linearly most of the time. To combat the loss of DNA at either ends caused by okazaki fragments T4 have adapted.

linear DNA hard to replicate, loss of DNA >> telomere

inject host with DNA >> replicate >> multiple linear chromosomes >> concatemer >> crossing over >> resolve concatemer

• Replication is bidirectional however there is recombination between the phage genomes which creates concatamers. These are around 166kb long and are fed into the phage head (headful method) which can contain 171kb. This results in a phenomena called terminal redundancy, reels in more than 1 gene, 5 extra bp, double copies of A&B, nexy phage starts at C (circularly permuted).

Terminal redundancy = a little more than an entire genome can be inserted into the head which results in repetition of some genes. The parts that are repeated result in a partial diploid DNA molecule, DNA does not get shorter. Terminal redundancy can be shown in experiments with heterozygote T4 phenotypes showing gene duplication.

Question 15

T4 vs lambdah

Answer 15

• T4 genome is 166kb whilst the λ genome is 48kb.
• The T4 head contains 171kb whilst the λ head contains 48kb.
• The T4 genome is terminally redundant and circularly permuted (order of gene in each progeny is different) and the λ genome is neither.

Question 16

Phage and virus

Answer 16

Phage are viruses that infect prokaryotes and thus they have similarities to other viruses such as:

• use similar strategies for infection and replication in hosts
• both show host specificity
• both use host machinery for their propagation
• can be temperate (latent) or virulent
• can have DNA/RNA as genetic material
• single stranded mammalian RNA virus known as retrovirus
• viruses can be as diverse as phage

Question 17

Retroviruses

Answer 17

Single stranded mammalian RNA viruses are called retroviruses.

They only code for 3 genes (differential expression = more than 3 gene products) and have LTR (long terminal repeats) at either end and they are different to ssRNA phages in that they change the ssRNA to dsDNA whilst phages leave theirs as RNA.

Retroviruses do this using reverse transcriptase (RNA dependent DNA polymerase) and DNA polymerase (DNA dependent DNA polymerase).

The dsDNA form then inserts into the host genome in a random location and once inserted is called a provirus.

Viruses that integrate like this are called slow viruses because they lie dormant in the host genome until they reappear if they excise and undergo the infectious cycle.

This is a similar situation to lambda lysogens. It is widely believed that retrotransposons (sequences that move around the genome) could be defective retroviruses (partially degraded retrovirus).

Question 18

Phage can exist in number of different forms and their ode of replication reflects these differences

Answer 18

• dsDNA phage = can be either virulent or temperate
• ssDNA phage = requires dsDNA replicative intermediate
• ssRNA phage = have dsRNA replicative intermediate and infect only male bacteria

same is true of mammalian viruses

• both dsDNA and ssRNA viruses exist but replication of ssRNA viruses very different from ssRNA phage as it requires a dsDNA intermediate

Question 19

Cloning vectors and gene therapy

Answer 19

There are ways to take advantage of the viruses’ habit of assimilating into the host genome.

If we cut out the dangerous functions and insert a gene of interest many good things can be done.

Lambda cloning vectors = lambda packages DNA by recognising cos sites and we can cut out some genes and put our gene of interest in between cos sites then clone it.

This can be done in vitro (outside the living), does not ahve to proceed in bacterial cell. You can do the same type of thing with other viruses. HOWEVER, recom. of transposon and provirus from vector may trigger infection - pitfall

Phage derived vectors = phage, cosmid, artificial chromosome

Viral vectors = gene therapy

Question 20

Phage genetics

Answer 20

Phage are haploid, they have no nuclei, and they don’t undergo mitosis or meiosis.

Normal mendelian genetics does not apply however recombination and complementation are features of phage and transient diploid states do occur so genetics can be done with this, need to have genetic variation in order to genetic crosses

Question 21

Genetic variation in phage

Answer 21

Unconditional and condition mutants:

• Unconditional mutants (plaque morphology mutants) are those that don’t change their phenotype even for changing conditions e.g. clear plaque mutants - lost ability to form lysogens, lysogen negative mutants
• Conditional mutants are those that do change phenotype in different conditions
  e. g. temperature-sensitive mutants that can be heat sensitive or cold sensitive

host range specific mutants that only grow on certain strains of bacteria.

Conditional mutants grown under one set of conditions called the permissive conditions. They will not grow in non-permissive or restrictive conditions.

• Common mutations in lambda phage result from changes to genes such as C1, int, and attP and these usually result in a clear plaque mutant which is unconditional.

Question 22

Rapid lysis (r-) mutants of phage T4

Answer 22

• create plaques on E.coli B that have very sharp edges as compared with WT fuzzy edged plaque
• r- unable to grow on E.coli K but r+ can
• r- has 2 genotypes, dependent on assay

Question 23

Recombination in phage

Answer 23

By doing recombination analysis in phage we can map their genes.

The first step is to infect an Ecoli with a lot of phage i.e. a high multiplicity of infection (MOI = ratio of infectants to hosts, high = MOI>1, low = MOI<1).

Coinfection = when there are two or more mutant phage in an Ecoli creating a diploid state. This allows for recombination between phage DNA which will create two genotypes distinct from the parents in the progeny - wild type or double mutant.

To conduct genetic analysis with phages you can simply coinfect bacteria with two mutants, collect the progeny phage, and plate them onto bacteria to score the number of recombinants.

RF = (2 x WT recombinants)/ (Total progeny)

Conducting analyses =

mutant 1 and 2 are crossed using coinfection and the progeny are plated on both Ecoli B (permissive host) and Ecoli K (restrictive host) to determine the number of recombinants. High numbers of phage can be tested which allows geneticists to account for fine details on the map.

Question 24

To find out genotype

Answer 24

• infect E.coli at low multiplicity, on average each E.coli only infected by 1 progeny phage
• can be WT, double recom., double mutant, parent
• should only have 1 plaque on E.coli K, WT
• on E.coli B everything should make a plaque, count amount of progeny, E.coli B is the permissive host
• E.coli K the restrictive host, counted to give WT recombinant progeny

Question 25

3 factor cross

Answer 25

• determines the order of genes
• determines which gene is in the middle
• select for 2 markers; is the 3rd marker included/excluded form phage

Question 26

Genetic fine structure

Answer 26

Genetic fine structure allows you to map mutations within a gene.

This is possible with phage because we can score 10^8 progeny and set up crosses so only recombinants survive. This technique is what allow Benzer to discover that the nucleotide is the smallest functional unit of a gene.

Question 27

The RII system and Seymour Benzer

Answer 27

Benzer used complementation analysis to identify that the RII system of phage consisted of 2 separate genes.

Some of the mutations failed to complement each other and Benzer theorised that this was because of deletions spanning across the genes.

The method:

• Benzer isolated thousands of point mutations and mapped a few of them instead of all to save time. He reasoned that if there were deletion strains a region then there’d be no crossover and no WT. If WT appeared this meant the mutation was outside the deletion region. Benzer used this to map all of the deletion mutation endpoints.
• He then crossed all of the mutants to each of the deletion mutations and if WT appeared it meant it was outside the deletion region, if there were only mutants then it was within the deletion region.

Conclusions: Benzer was able to completely map the RII system genes and show that there are “hot” and “cold” spots for mutations which is due to mutagens favouring certain sections. He also showed that some mutations are silent.

Question 28

Complementation

Answer 28

Designed to test whether two mutations are allelic (same gene) or not.

If two mutations are in two different genes, then when you combine them they will complement to form a wild type (complementation event).

To do this with phage you coinfect Ecoli and select against plaques, if you see plaques then it means that the mutations have complemented.

You have to set it up such that recombination is a minor component.

To do this we genotype the diploid cell, not the progeny of that cell. You might see 1 or 2 plaques from rare recombination events so for complementation you’re looking for lots of plaques because every cell will complement.

Coinfection: when you coinfect bacteria with a phage this creates a diploid state. If mutations are in different genes, then their gene products will complement and give a wild type phenotype. If a mutation appears to be in two genes at once this generally means that it is a deletion (best guess). This also indicates that the two genes must be relatively close.

Question 29

Genetic variation in bacteria

Answer 29

Types of mutants:

• Morphology mutants e.g. rough and smooth
• Auxotrophic mutants = unable to grow in the absence of a nutrient
• Utilisation mutants = unable to grow from some types of nutrient sources e.g. arabinose as a carbon source.
• Resistance mutants = able to grow in the presence of antibiotics etc.
• Temperature sensitive mutants = cold sensitive - can’t grow in cold temperature, heat sensitive - can’t grow in hot temperature.

Screening = look at a large number of potential mutants to find one with the characteristics you need

Selection = only allows the survival of the mutants you’re interested in.

Replica plating = plating the bacteria once onto a plate then transferring these individuals to another plate that has been treated differently.

Question 30

Auxotrophic mutants

Answer 30

Prototrophs are bacteria that can grow on minimal medium whilst auxotrophs need a nutrient to be added to this before they can grow, usually due to a gap in the biosynthetic pathway.

Identified through a process called screening in which you plate colonies on rich medium, then minimal medium and compare the two.

1. mutagenise bacterium
2. place on rich medium dish
3. place fabric on dish, transfer bacteria to fabric = duplicate
4. place duplicate on minimal media dish
5. add additional compounds to minimal media plate until colony grows to see what auxotroph is missing e.g. Tyr, Ser i.e. if it grows on Try = Tyr auxotroph, cannot produce own Tyr

Question 31

Utilisation mutant

Answer 31

Mutants can be isolated that are unable to grow on a particular compound.

They’re also identified by screening. e.g. lactose non-utilising mutants can’t reduce lactose to other sugars so they need glucose + galactose as their carbon source

Can grow on minimal media but cannot grow on base media other than glucose

Question 32

Resistance Mutants

Answer 32

Mutants that are resistant to the toxic effect of a particular compound. These are easier to identify because you simply need to select for them. Throw them on a plate with the toxic compound e.g. antibiotics and see if they grow because the wild type will not grow

straight into restrictive conditions

Question 33

Nomenclature

Answer 33

Genes are given names based on the phenotype they display when they are mutated:

• Galactose utilisation mutants are denoted as gal mutants. Alleles are notated by a letter e.g. galA+ for wild type and the mutant allele can need a number e.g. galA1.

Question 34

Bacterial genetics

Answer 34

The three types of bacterial genetics involve:
Conjugation - bacterial sex, the transfer of one bacterial genome into another

Transduction - phage mediated gene transfer

Transformation - when the bacteria uptake DNA from their environment.

Question 35

Bacterial genetics - conjugation

Answer 35

In Ecoli there are two types of cells: donor cells (F+ cells) and recipient cells (F- cells).

The donor cells have pili which allow it to adhere to other bacteria and transfer its genome.

The pili are encoded by the F factor gene in the donor cell, in fact everything driving the conjugation comes down to the F factor genes. These genes are on episomal DNA (sometimes called a plasmid, replicates independently of chromosomal DNA).

Process:
- Pili of F+ attach to F- cell. Then a conjugal tube (a sort of cytoplasmic bridge) forms and the F factor replicates itself due to the action of its own genes.

• An entire copy of the F factor gets transferred. This results in the F- cell becoming an F+ cell.
• All the cells aren’t F factor because it costs energy, it is more easily infected (F+ cell sensitive to MS2, infect cells via pilli), and random partitioning during binary fission.

This phenomenon allows you to do genetics by mixing two cell types, and observing

recombinants e.g. F+ met- leu+ X F- met+ leu-

At certain frequencies the F factor can integrate into the Ecoli chromosome. Homologous recombination can then occur because there are some homologous transposon regions on both the F factor and the bacterial genome. Strains of bacteria with the integrated F factor are called hfr (high frequency recombination) strains. In these strains the copying of the F factor can lead to the transfer of the entire Ecoli chromosome.

You can also stop the process at a certain point by breaking the conjugal tube and this allows you to conduct genetic analysis because the time a gene takes to be transferred is a representation of its location on the chromosome. The length of time is a proxy for distance (time is a unit of genetic distance).

- different amounts of DNA transferred dependent on how long the cells are joined

Question 36

Transduction

Answer 36

A phage mediated transfer of DNA from the donor host to the recipient host.

Generalised transduction - used to map genes anywhere on the bacterial chromosome

Specialised transduction - limited to certain regions of the bacterial chromosome when mapping

Transduction uses phage that act as vectors to carry DNA. The size of the phage means that it can only carry a small portion of the chromosome. This means you’re mapping genes that are relatively close on the chromosome.

Question 37

Generalised transduction

Answer 37

Most commonly done with P1 phage and Ecoli.

P1 is about 100kb (2% of the entire bacterial genome). P1 is temperate (it can undergo lytic OR lysogenic cycle) like Lambda, however the DNA in the lysogenic cycle is autonomous which means it maintains itself as an episome.

Variants of P1 have lost the ability to recognise their own DNA so they package the host DNA instead at low multiplicities of infection.

This phage then infects a new Ecoli cell, the piece of DNA that comes out of it cannot circularise so it has to integrate with the host genome by undergoing recombination. The recombination events are double because you can’t undergo single recombination with a circular and linear molecule.

Phage P1 particles can pick up any piece of DNA and recombine it into the host genome. This means genes that are closely linked are more likely to be cotransduced together and this allows us to map close genes with relatively fine detail.

Cotransduction frequency is thus another unit for measuring genetic distance between genes. The frequency of cotransduction also measures the likeliness that a cross over will occur between two closely linked genes.

decreased distance, increased transduction

co-transduced - must be closer than 100kb apart to co-transduce

increase transfer = increase retention = 2 markers are close

Transduction allows the mapping of genes within a 5-minute segment of the Ecoli chromosome that cannot be distinguished with conjugation.

Question 38

Further transduction

Answer 38

as well as mapping two mutations in different genes you can also map mutations within a gene relative to an outside marker.

To do this we set up a pair of reciprocal crosses e.g. cys+, trpA- to cys-, trpB- as well as the reverse.

The crossover events needed will be different depending on the order i.e. double crossover for a certain gene order in cross 1 whilst quadruple crossover for the opposing gene order for cross 1. This means the relative frequencies in the crosses will be different for each order.

Origin of replication:

Transduction can also be used to map the origin of replication of a bacterial chromosome.

If we sample the DNA at various times after the initiation of replication and count the number of copies of each gene, then you can identify where the replication must have started.

This is because if a gene is before the fork then there will be two copies and if it is after the fork then there will be one copy. The genes with 2 copies will be transduced twice as often.

You need to be able to synchronise the time of replication among all the cells used for the experiment or else the data will be meaningless. We do this by using replication inhibitors or mutations such as heat sensitive.

Question 39

Transformation

Answer 39

Direct transfer of DNA from a donor into the recipient, not by a conjugal tube but because the DNA has been released into the environment. Some bacteria do this naturally. Transformation is genetically similar to transduction but physically quite different.

Whilst some bacteria do it naturally, others need a bit of prodding so to speak. This is done by using divalent cations (charge of 2+) or through electroporation. This makes the Ecoli competent to take up DNA by changing the cell wall

Genetic exchange:

Transformation leads to genetic exchange which requires the DNA to recombine with the host cell. Whilst transformation is genetically similar to transduction it becomes very useful for bacteria with poorly defined conjugation or transduction systems.

Question 40

Specialised transductoin

Answer 40

Limited to certain regions of the genome because you use temperate phage that integrate into the host chromosome (such as lambda and fi80 phage).

This proceeds similar to the F factor because the phage integrate and when they leave the host chromosome they may take some of it with them on their journey.

• firstly, a recombination event between the two att sites results in the phage genome being integrated into the host chromosome.
• there are two clusters of Ecoli genes that flank this site: gal and bio.
• most of the time the lambda genome leaves the chromosome as it came however occasionally the adjacent gal and bio genes are transduced with it.
• the phage usually has to leave some of its own genes so it becomes defective. The only way it can infect again is with a helper phage that makes up for the functions the defective phage has lost.
• the new phage with Ecoli genes can insert into a new host chromosome with a helper phage which creates a partial diploid and allows us to do genetics. You can do both complementation and recombination tests.

Question 41

Specialised Eukaryotic Genetics

Answer 41

In many instances haploid eukaryotes are small and have experimental advantages that are similar to bacteria in terms of answering genetic questions.

Generic life cycle of haploid eukaryotes:

• The vegetative state is haploid
• Occasionally two different haploid cells fuse and cytoplasmic and nuclear fusion results in a transient diploid.
• The diploid then undergoes meiosis to create 4 haploid spores which germinate and undergo mitosis to get to the vegetative stage.

These spores are equivalent to the gametes of diploid organisms. Some haploid organisms create unordered meiotic products whilst some create ordered products (i.e. spores are produced in the order of the meiotic divisions that can be observed).

Tetrads are the 4 base meiotic products whilst octads are 8 meiotic products (additional mitotic event after meiosis) created when tetrad products undergo mitosis once (producing 4 copies).

Ordered tetrads: meiosis in organisms such as N.crassa is followed immediately by a single mitotic division which produces an octad of spores. The order of the spores reflects the order in which the individual chromosomes segregated at meiosis 1.

If recombination occurs between an allele and the centromere, then the order of the tetrads will change. This allows the placement of the centromere on the genetic map because crossovers between centromere and locus will produce different phenotypes to no crossover between them.

The RF that results must be multiplied by 0.5 because only half of the spores are actually recombinant to give a different order, the other two didn’t crossover.

1:1 parental ditype to non-parental ditype ratio = unlinekd

Question 42

Genetics of haploid eukaryotes

Answer 42

The types of mutations used to analyse haploid eukaryotes include auxotrophic, utilisation, morphological, and developmental phenotypes. There are no requirements for a test cross because the phenotype is the genotype.

Ditype: Asci (the cells that bear the spores) that contain two genotypes are called ditype e.g. AB, AB, ab, ab OR Ab, Ab, aB, aB. Parental ditypes are produced by no crossovers, or a double crossover between two strands. Non-parental ditypes are produced by a double crossover involving the 4 strands.

Tetratypes: Asci that contain four genotypes are called tetratypes e.g. AB, aB, Ab, ab. Tetratypes are produced by single crossovers between two strands, or three stranded double crossovers (hold one crossover constant and vary the position of the second).

Question 43

Mapping using tetrads

Answer 43

Could be calculated using simple RF from parental and recombinant genotypes. However, for large distances this is not accurate because double crossovers between the genes are more likely and parental ditypes can also be double crossovers between two strands.

The key is the non-parental ditype which can only be produced through a double crossover involving the 4 strands.

This leads to the formula:

Map distance = 50 (Tetratypes + 6 x non-parental ditypes)

This is much more accurate than simply looking at the total number of recombinant progeny.

Mapping function:

The Poisson distribution describes events that are potentially frequent but mostly infrequent.

First you find a recombination frequency, then to adjust it so it is more accurate you insert it into this equation:

• *RF = 1/2 (1 - e^-m)**
• *where m is the mean number of crossovers**

m=-ln(1-2RF)

You then multiply m by 50 (the maximum map distance) to get the corrected map distance

Question 44

Somatic cell genetics

Answer 44

• no meiosis, only mitosis
• used to map genes to chromosomes
• identify protein products of genes, biochemical variation

Electrophoretic variation: involves the use of electrophoresis to separate proteins based on their charge and size. The gel is then stained on one side for total protein, and on the other side for a specific enzyme/protein.

The reason we can do genetics on this is because if you fuse cells such as a mouse cell and a human cell the hybrid resulting cell will be unstable and randomly lose human chromosomes. Each resulting cell has a different subset of human chromosomes which allows us to conduct analyses using a probe for a certain gene product. This allows you to map genes to a particular human chromosome by seeing if the product appears in each of the cell lines.

You can also conduct finer mapping by taking a cell line containing the chromosome you’re interested in and exposing it to radiation so that smaller subsets of the chromosome can be studied (radiation causes DS DNA to break>>lose bits of chromosome = sub-collection of chromosome).

Question 45

Mitotic recombination

Answer 45

Homologous chromosomes do not pair in mitosis however sometimes they can synapse when they are close and undergo recombination.

This was first discovered a long time ago in drosophila. See example from lecture slides.

If you are studying the order of two genes relative to the centromere you can use crossover frequencies. Only one crossover is required for recombination to produce the outer gene mutant. However, two crossovers are required to produce the mutant closer to the centromere. This means the mutant closer to the centromere will appear in lower numbers.

Mitotic analysis in fungi - Aspergillus nidulans:

Haploidisation: sometimes cell lines in the fungi lose a chromosome and undergo haploidisation where they shed one of all the other homologous chromosomes (which one chosen is random) until they are haploid.

Parasexual cycle:

• can lose chromosome during meiosis (at random), can use to map genes = 2n-1 until it reaches stable state = haploid, but must retain one of the each chromosome