DNA and Replication

Question 1

What are chromosomes and why are they useful?

Answer 1

Chromosomes are genomic information storage units
It is a highly coiled fibre of chromatin. Packing of DNA into chromatin gives a flexible substrate which allows for the reliable expression of the genome and faithful replication and transmission of the genome to daughter cells.

Question 2

How and when were chromosomes discovered?

Answer 2

In 1902, Boveri and Sutton, chromosomes were first seen as coloured bodies inside the cells and were suggested to carry genes. This was elaborated on by Morgan in 1915

Question 3

What are chromosomes comprised of and what specialised functions do they have?

Answer 3

Comprised of linear DNA and proteins (histones)
They are able to package and unfold DNA within the nucleus, control replication, repair and recombination, maintain chromosome integrity, undergo proper segregation during cell division and regulate gene expression. Mitochondria and chloroplasts also contain small circular chromosomes.

Question 4

When can individual chromosomes be most easily distinguished?

Answer 4

At the metaphase stage when they are condensed.

Question 5

What is a karyotype?

Answer 5

An organized representation of all the chromosomes in a eukaryotic cell at metaphase. Each type of chromosome is labelled with a different fluorescence. This allows abnormalities to be seen easily.
Individual chromosomes occupy distinct sub-nuclear territories even in interphase (uncondensed)

Question 6

When happens to DNA when it becomes transcriptionally active?

Answer 6

The nuclear periphery in interphase cells is composed of transcriptionally inactive DNA. When a gene becomes active, it moves to the centre of the nucleus and becomes less condensed

Question 7

What does interphase chromatin resemble? What is a 30nm fibre?

Answer 7

Beads on a string (uncondensed DNA on nucleosome). A 30nm fibre is the supercoiled version.

Question 8

What is a nucleosome, what is their structure?

Answer 8

The ‘beads’ that the DNA is wrapped around. The protein subunits of the nucleosome are core histones. The N-terminus tails of the 8 core histone subunits project out from the nucleosome core and are free to interact with other proteins. This facilitates the regulation of chromatin structure and function. The core histones tails include 2x H2A, 2x H2B, 2xH3 and 2xH4. Linker histones like H1 strap DNA onto the octamer and limit movement of DNA relative to the histone octamer. This stabilises the formation of the 30nm fibre.

Question 9

What is the telomere on a chromosome and what is its function?

Answer 9

Telomeres are specialised DNA sequences at the ends of linear chromosomes which maintain chromosomal integrity. They are replicated by DNA telomerase. They are repeats of single-stranded 3 prime overhangs (TTAGGG) repeat arrays. They can be several hundred nucleotides long.

Question 10

What are replication origins?

Answer 10

DNA sequence where the DNA replication is initiated.

Question 11

What is a centromere and what is its structure/function?

Answer 11

The centromere is where the kinetochore forms and mediates segregation. Centromeres contain specialised proteins and DNA sequences that facilitate chromosome segregation during cell division. They contain alpha-satellite DNA repeats which form condensed heterochromatin with histone octamers which contain unusual subunits - Has chromatin containing normal H3 and then another specific version of H3 (which as a methylated lysine 4) (CENA-P). This funny H3 is where the inner plate of the kinetochore binds to.

Question 12

What is the kinetochore?

Answer 12

A structure made of an inner plate and outer plate that binds to the centromere and mitotic spindles (microtubules) for chromosome segregation etc

Question 13

What is the kinetochore like in yeast?

Answer 13

Its a basket that linkes a signel nucleosome of centromeric chromatin to a single microtubule.

Question 14

What are transposons?

Answer 14

DNA sequences that can amplify themselves and then insert into other areas of the genome. They move by a cut and paste mechanism without self-duplication, requiring the transposon-encoded enzyme transposase. They are important in gene function and evolution. Also in research, transposons from other organisms like flies, maize or e.coli are important mutagens.
There are 3 types:
- DNA transposons
- retrotransposons (They are made into RNA via transcription and then self-encoded reverse transcriptase turns them back into cDNA, they are then reincorporated into the genome in a different place.)
- Non-retroviral polyA retrotransposons
Transposons make up about 50% of DNA (only 1.5% of our DNA actually codes for cellular proteins)
Some L1 insertions (a polyA insertion) are known to disrupt genes and cause human disease like haemophillia.

Question 15

How does our DNA differ from simple organisms and why?

Answer 15

More complex organisms have more protein-coding genes and more non-protein coding genes. Some of the non-protein coding DNA encodes transcriptional regulatory information which determines expression. Increasing biological complexity thus depends on an increasing no. of protein genes and an increasing amount of non-protein coding cis-regulatory DNA. Non-retroviral retrotransposons have expanded hugely in numbers during the evolution of higher mammals, originally evolving from a single copy of the 7SL RNA gene.

Question 16

What is meant by semi-conservative DNA replication?

Answer 16

When 1 double strand can be used to make 2 new ones. The new ones are made up of 1 strand of the old double. (old strands used as a template)

Question 17

What direction does DNA replication occur in and why?

Answer 17

5’ to 3’ due to the formation of phosphodiester bonds.. The 5’ end has a phosphate group and the 3’ end has the sugar. When a nucleotide adds, its phosphate binds to the OH group on the 5’. The template strand is anti-parallel

Question 18

What reaction occurs when a nucleotide is added onto the polymer, when does this make DNA replication irreversible?

Answer 18

To add a nucleotide, you take PPi off the triphosphate, leaving it free to bind to the OH group. The PPi (pyrophosphate) then breaks into 2Pi, via pyrophosphatase, which a very exothermic reaction (therefore energetically favourable). This is a coupled reaction (when 2 reactions happen at the same time)

Question 19

What is the first step of DNA replication?

Answer 19

The creation of a replication form where the DNA strands are separated - this is done by DNA helicase. It uses ATP to separate the parental strands at the replication fork and more the fork forward.

Question 20

What are Okazaki fragments and why do they occur?

Answer 20

On the leading strand, continuous synthesis can occur because it is in a 5’-3’ direction. The antiparallel orientation of parental strands and unidirectional orientation of new DNA synthesis means that both new strands cannot be synthesised continuously. Therefore, you get Okazaki fragments on the lagging strand. These are small strands of un-continuous new DNA that have been synthesised and then stopped

Question 21

How are RNA primers made?

Answer 21

All DNA synthesis is initiated by extension of a short primer of RNA. The short RNA primer is synthesised by DNA Primase and only requires a DNA template and NTPs (nucleoside triphosphate)

Question 22

Explain how DNA synthesis occurs on the lagging strand.

Answer 22

It occurs more slowly than on the leading strand.

1. DNA primase makes RNA primer
2. DNA polymerase adds to the new RNA primer to start the new Okazaki fragment. (requires a primer-template junction)
3. Ribonuclease H removes the RNA primer so it can be replaced with DNA (by DNA polymerase)
4. DNA Ligase seals all the nicks - it uses the energy of ATP hydrolysis in a 2 step reaction (ATP hydrolysed, then ADP hydrolysed to leave AMP) (ATP+5’ -> P-P + 5’ P-AMP) The ligation process is rendered energetically highly favourable by the conversion of PPi to 2Pi by pyrophosphatase

Question 23

What diseases are caused by mutations in DNA Helicases?

Answer 23

Werner’s syndrome (premature ageing). Autosomal recessive. Mutations in RECQ gene WRN
Bloom syndrome - a rare cancer syndrome caused by a loss of function mutation in RECQ as well (its role is to maintain genome integrity)

Question 24

What are sliding clamps and how do they work?

Answer 24

Sliding clamps are molecules that help improve the processivity of DNA polymerase. The sliding clamp is positioned close to the primer-template junction by a clamp loader. The energy of ATP hydrolysis is used to position it. They circle the DNA like a nut of a bolt and help to push the polymerase forward. The human sliding clamp (PCNA) is nearly identical to the e.coli one. This shows it is well conserved and therefore must be important.

Question 25

What is the function of SSBPs?

Answer 25

Single-stranded binding proteins expose single-stranded DNA in the replication fork, making it available for templating syntheses of new DNA strand and easing the replication fork process. (they basically stop the single strand that hasn’t been replicated joining wrongly to other single strands.

Question 26

What are topoisomerases and howd= do they work?

Answer 26

DNA topoisomerases prevent DNA from becoming tangled during replication. The unwinding of parental DNA strands at the replication form introduces superhelical tension into the DNA helix. This tension is relaxed by the topoisomerases which nick and reseal the backbone of the parental helix.
Type 1s nick and reseal one of the 2 strands (no ATP required) and type 2 nick and reseal both strands (ATP is required)

Question 27

How is DNA replication first initiated?

Answer 27

Initiation of DNA replication happens at the replication origin
Specific DNA sequences (replicators and origins) direct the initiation of DNA replication by recruiting replication initiator proteins (ARS in yeast) In humans, DNA sequences near LMNB2, Myc, Hbb can act as inhibitors. But initiators are also defined by chromatin-free structures e.g. nucleosome-free regions, rather than a specific sequences.
Initiation of DNA replication is biphasic
1. replicator selection - formation of a pre-replicative complex (occurs in G1 phase)
2. Origin activation - the unwinding of DNA and recruitment of polymerase - occurs in S phase
Temporal separation of these events means that each origin is used and each chromosome is only replicated exactly once per cycle.

Question 28

What is the molecular pathway for the initiation of DNA replication?

Answer 28

Eukaryotic replicator selection occurs in G1 and leads to the formation of a pre-replicative complex (pre-RC)
1. Origin recognition sequence binds to the replicator sequence
2. Helicase loading proteins Cdc6 and Cdt1 bind to ORC
3. The helicase Mcm2-7 binds to complete the formation of the pre-RC
High levels of Cdk activity in S-phase activates exisitng pre-RC but prevents formation of new ones. Cdk activity is low in G1 and high in S phase.
Close relationships between pre-RC function, Cdk levels and the cell cycle ensure that chromosomes are replicated exactly once per cell cycle.

Question 29

How is DNA replication finished? Why are telomeres needed?

Answer 29

The need for an RNA primer for initiation of DNA synthesis creates an end replication problem for linear chromosomes. Ribonuclease H removes the primers and the gaps are closed by ligase etc. All but 1 gap is closed, so the end sequence is lost. so without telomeres, the DNA would get shorter and shorter with each replication and crucial coding would be lost. Telomeres prevent this because it is them that get shorter.

Question 30

What is the telomerase shuffle? How are telomeres maintained?

Answer 30

Telomerase contains an RNA component that specifies telomere sequence. Telomerase is a ribonucleoprotein with an intrinsic RNA component that acts as a template on which telomere repeat sequences are synthesised in a step-like process (the telomerase shuffle) (telomere RNA has Us not Ts) The telomerase RNA allows the addition of multiple TTAGGG repeats to the 3’ OH at each telomere. 3 new nucleotides are added, then 6, then 6.

Question 31

In what ways can DNA be damaged?

Answer 31

By oxidation (occurs more readily at guanine because of its high oxidative potential - OH gets added onto it CH group), hydrolysis or uncontrolled random methylation of any of the 4 nucleotides (opposed to normal methylation used for controlling transcription)
Depurination and deamination

Question 32

What are the most common forms of spontaneous forms of DNA damaged?

Answer 32

Hydrolytic depurination (adenine/guanine are cleaved off) or deamination (removal of an amine group by an enzyme) or bases - this causes a different base to bind when replicated e.g. a deaminated C has a very similar structure to a U so binds to an A instead of a G

Question 33

How are deaminated and depurinated bases repaired?

Answer 33

Base Excision Repair.
1. The damaged base is removed by DNA glycosylase
2. Sugar is removed by Apurinic/Apyrimidic endonuclease
3 Phosphate removed by phosphodiesterase
4. DNA polymerase extends the primer-template junction and DNA ligase seals the nick

Question 34

What damage does Ultraviolet radiation cause and how can it be repaired?

Answer 34

UV causes the formation of pyrimidine dimers (basically 2 pyrimidines next to each other sort of become linked)
This can arrest DNA replication or cause misreading of the sequence by the polymerase. This and many other types of DNA damage can be repaired by nucleotide excision repair:
1. Excision nuclease cleaves the single-stranded DNA sequence containing the defect
2. DNA helicase removes damaged segment
3. DNA polymerase extends primer-template junction
4. DNA ligase seals the nick

Question 35

What disease can occur when NER mechanisms go wrong?

Answer 35

Defective nucleotide excision repair mechanisms can cause xeoderma pigmentosum which makes patients extremely sensitive to sunlight-induced skin cancer. (because they can’t repair damage from UV
There are 7 different genes have been identified including XPA, XPC, XPD, XPF and XPG.
They all encode proteins that participate in the NER pathway for correction of UV induced DNA lesions. They’re homologs of E.Coli Uvr proteins (conserved and therefore important). XP and Uvr NER mechanisms are transcription coupled (removed damage from transcriptionally active genes)

Question 36

Why are double-stranded breaks in the DNA bad and how are they repaired?

Answer 36

Because large fragments of chromosomes can be lost. This type of damage is typically caused by ionising radiation ( X rays etc)
They can be repaired by Non-homologous end joining or Homologous recombination
HR is often thought of as the last line of defence to DNA damage because single-stranded breaks produced by BER and NER may weaken DNA structure, making it prone to double-stranded breaks.

Question 37

Explain how non-homologous end joining works

Answer 37

Non-homologous end joining is ‘quick and dirty’ - the ends at the break are degraded to make them clean and then the pieces are stuck back together. But this causes a loss of DNA, which is fine if the region is non-coding and the fixing stops a larger amount of DNA being lost. But if the break is in a coding area, it can cause problems, so HR is the prefered method but takes longer.

Question 38

Explain how homologous recombination works.

Answer 38

It is more accurate and then preferred to non-homologous end joining. It makes use of the DNA sequence information in the undamaged homologous chromosome.
The broken DNA is resected at its 5’ ends to create single strands that can be used to prime DNA synthesis when annealed to a template. The DNA binding protein RecA promotes ‘strand invasion’ of the undamaged template molecule by one strand from the damaged DNA molecule that acts as a primer. The newly synthesised DNA dissociates from its template, reanneals to its original partner strand, allowing second strand synthesis and formation of a pair of staggered single-stranded nicks which are repaired by DNA ligase. So the information is taken from the undamaged DNA molecule and used to repair the damaged DNA molecules.

Question 39

What diseases are caused by mutations in genes encoding HR DNA repair mechanisms?

Answer 39

They usually cause types of cancer.
E.g. BRCA2 - breast, ovarian and prostate cancer
ATM - ataxia telangiectasia (Loius-Bar syndrome)
Fanconi anaemia- Complicated with leukaemia - 13 different FANC genes
Mutations in this pathway mean DNA repair becomes dependant on other repair pathways like BER. If multiple DNA repair pathways become non-functional, then cells that suffer DNA damage are likely to die (Synthetic lethality)
BRCA2 mutated cancers are highly sensitive to drugs that inhibit BER because they rely on this pathway so much for survival e.g. Oliparib/lynparaza (inhibitors of poy ADO ribose polymerase)

Question 40

What else other than DNA repair is homologous recombination involved in?

Answer 40

HR in meiotic cells means breakage of each DNA strand is required for the crossing over of paired chromosomes. This mediated genetic recombination during meiosis.
Spo11 endonucleases make the initial cleavage and Mre11 exonucleases resect the 5’ ends of one chromosome. Exonucleases then expose a single stranded 3’ end. RecA promoted strand invasion. DNA synthesis occurs on both strands to create a double Holliday junction (when the chromosomes are intertwined with new DNA). The junction can be theoretically resolved in 1 of 2 ways to recombine the DNA.

Question 41

What are the 2 ways a double Holliday junction can be resolved?

Answer 41

1. Only internal strands are broken and they rejoin internally. I.e. the same strands are broken and rejoined at each junction. Recombination between A/a and B/b loci is NOT achieved. (chromosomes don’t cross over)
2. The external and internal strands are broken and rejoined partly internally and partly externally. I.e. different DNA strands are broken and rejoined at each junction. recombination between A/a and B/b is achieved and the chromosomes cross over.

Question 42

What happens in G1 phase of the cell cycle

Answer 42

The 1st stage of the cell cycle. Can be very long of permanent (quiescent cells like neurones and retinal cells) In this phase cells grow and make the proteins the will need for the rest of the cell cycle (unless they are in G0 phase)

Question 43

What happens in M phase of the cell cycle?

Answer 43

Mitosis after DNA replication. It includes nuclear division which can be broken up into prophase, metaphase, anaphase and telophase/cytokinesis
Prophase = condensation of sister chromatids (the identical copies)
Metaphase - attachment to the mitotic spindle
Anaphase - Separation of sister chromatids
telophase - chromatin decondensed after separation and cytokinesis separated the cells

Question 44

How can yeast be used as a genetic model for the cell cycle?

Answer 44

Cell cycle mechanisms are highly conserved due to them being crucial, so its very similar in small single-celled organisms like yeast. The cell cycle varies slightly with different types of yeast. Budding yeat have no G2 and produce differently sized daughter cells, and fission yeast spends a long time in G2 and stay joined together after cytokinesis. They’re useful genetic models because they have a rapid division rate (their cycle is less than an hour), they are highly conserved and yeast can also be grown as haploid or diploid. Genetic tricks allow identification of potentially lethal mutations because they’re recessive. Diploid yeast can be used as a store for lethal mutations which can then be studied as haploid. Temperature sensitive mutations allow growth at permissive temperatures - the yeats don’t regulate their own temp, so you can get a colony to stoop and start their cycle. Means you can get all cells in a colony to the same stage and then watch them grow together at the same stage (can see how certain mutations/drugs effect timing etc)

Question 45

How can xenopus be used for cell cycle research?

Answer 45

Their oocytes are large and grow quickly. Therefore, they are good for biochemical analysis rather than genetic (what proteins are being expressed and where etc).

Question 46

Explain cell-free mitosis and how it can be used for research.

Answer 46

When you take the cytoplasm (purified via centrifugation) from an egg and the nuclei from a sperm, the chromosome will begin replicating without the need to be enclosed with a cell. You can deplete the cytoplasm of different proteins using antibodies so you can identify their function. You can also remove the cytoplasm at different stages to study changes e.g. in protein phosphorylation over time.

Question 47

What checkpoints are in the cell cycle and what do they check?

Answer 47

There is a checkpoint at the end of G1 which checks if the environment is favourable and if there is enough energy available to proceed. There is a checkpoint at the start of mitosis to check if the DNA has replicated accurately and there is also one before cytokinesis occurs to check if the chromosomes have been separated properly

Question 48

What are cyclins?

Answer 48

Cyclins are proteins that are expressed at different levels during the cell cycle. When present, cyclins bind to specific CDKs to activate them. CDKs phosphorylate many proteins that are specific to certain stages of the cell cycle.

Question 49

What other proteins have the ability to modify Cdk activity apart from cyclins?

Answer 49

Cdk activity can be modified through phosphorylation and binding
E.g. Wee1 can add an extra phosphate to inactivate the CDK/cyclin complex, and Cdc25 phosphatase can take off the extra phosphate to activate it again. p27 can inhibit the whole complex when it binds to it.
Anaphase-promoting-complex (APC) is a ubiquitin ligase. When APC has activated it recruits to the CDK/cyclin complex causing degradation of M-cyclin in the proteosome which promotes metaphase to anaphase.

Question 50

What happens in meiosis and why is it needed?

Answer 50

Diploid organisms have 2 homologs of a chromosome (one maternal and one paternal). However, only one homolog of each chromosome is packed into a gamete. Meiosis resembles mitosis except that there are extra steps that separate the chromosomes. The pairing of homologs before segregation allows for crossing over (homologous recombination)

Question 51

What is the process of meiosis 1?

Answer 51

In meiosis 1, the cell with duplicated chromosomes splits into 2 daugther cells -This is when crossing over and segregation occurs. In mitosis, the daughter cells are identical, with 1 maternal and one paternal chromosome in each daughter. This is because they line up vertically. In meiosis, the chromosomes line up next to each other and recombine, then when the cell splits, the daughter cells are not identical. one contains the mostly maternal copies and one contains the mostly paternal copies. But they have swapped over little bits to make new combinations of alleles. HOWEVER, the sex chromosomes behave like homologs during sperm cormation due to small regions of homology

Question 52

What happens in meiosis 2?

Answer 52

The daughter cells produced in meiosis 1 are still diploid, so each cell splits again without replicating so the daughter cells have 1 copy of the chromosome each but have different alleles.

Question 53

What does homologous recombination allow for?

Answer 53

Genetic recombination between parental and maternal DNA on the same chromosome. This is important for genetic variance.

Question 54

Why is genetic variance important?

Answer 54

It is crucial for evolution because it allows for new advantages and selection pressures between species. However, it can also cause genetic conditions. mistakes during meiosis 1 result in gametes with an extra chromosome or lacking a homolog. this is called non-disjunction and the cells that arise from these gametes will be aneuploid. 4% of mammalian sperm and 20% of mammalian eggs are aneuploid.

Question 55

What is the difference between permanent and non permanent genetic alterations?

Answer 55

Genetic alterations to DNA sequence e.g. base mutations can permanently affect gene expression, whereas epigenetic changes to chromatin structure can modulate gene expression but they do not alter DNA sequence and are reversible. Epigenetic modifications facilitate stable changes to gene expression, which may persist for the life of a cell or organisms, but they can be erased in the germ line. Permanent changes to DNA may be carried through the germ line to offspring.

Question 56

What part of the chromosome is modified during epigenetic changes?

Answer 56

The nucleosomes, which can be covalently modified. These structural changed to chromatin affect gene transcription. The N Terminus of the lysine rich tails of core histones project radially from the nucleosomal core and are covalently modified. The N termini differ between ‘condensed’ inactive chromatin and ‘open’ active chromatin

Question 57

What are the different types of histone modifications and where do they happen?

Answer 57

Acetylation (addition of a acetyl group) and methylation (addition of a methyl group, can be mono-, di-, or tri-) Covalent modifications are added to the termini of lysine (and arginine) and serine side chains. Methylation and acetylation both occur on lysine, they’re competing chemical modifications. Methylation occurs on lysine and arginine. Acetylation loosens the chromatin structure (because it basically removes a positive charge from lysine and is therefore less attracted to the negative DNA)

Question 58

What are the enzymes that control histone modification?

Answer 58

Histone Acetyl Transferases (HATs) can modify many different lysine residues in core histones (non specific) whereas Histone Methyl Transferases (HMTs) exhibit site specificity (Histone code writers) Therefore, methylation of distinct residues in core histones is mediated by different enzymes- that may be subject to different regulation and the perception of different methylation by cell machinery can also be selective. This ,means that different methylation marks have different biological meanings. Methylation and acetylation of some lysines are mutually exclusive.
A variety of histone modifying enzymes affect gene transcription in distinct ways - code writers and code erasers. Histone modifications signify distinct states of transcriptional activity or competence for transcriptional activity. These histone modifications are reversible and can be undone by histone demethylases and histone deacetylases.

Question 59

How does the modification of different histones effect the gene transcription?

Answer 59

Methylation (on Lysine (K) or arginine (R)) effects genes in the following way:
- H3-K4 - makes genes active
- H3-K9 - makes genes inactive
- H3-K27- Makes genes inactive
-H3-R17- makes genes active
(most methylation occurs on H3)
Aceylation:
Activates many lysines on H2A, H2B, H3 and H4 ( so all lol) Acetylation of histones creates binding sites for transcriptional ACTIVATION factors that contain a bromodomain
In general, acetylation and methylation marks create binding sites for TFs

Question 60

What types of proteins do histone modifications create binding sites for?

Answer 60

Acetylation of histones creates binding sites for transcriptional ACTIVATION factors that contain a bromodomain
In general, acetylation and methylation marks create binding sites for TFs
Methylation of core histones can create binding sites for :- transcriptional repressors which contain a chromodomain (to H3K27, H3K9) OR
Transcriptional activators that contain a PHD Zinc finger domain (depending on the particular lysine amino acid residue modified)

Question 61

How do transcription activator proteins work in chromatin

Answer 61

Selective histone remodelling (change the structure of the chromatin to make genes more accessible)
Selective histone removal (via histone chaperones) - remove histones that the gene is wrapped arpund
Selective histone replacement (makes the gene less accessible for transcription)
Activator proteins typically induce combinations of these effects - all of which promote RNA polymerase II recruitment

Question 62

WHat actions do transcriptional repressors have?

Answer 62

Recruitment of chromatin remodelling complexes, recruitment of histone deacetylases, recruitment of HMTs. The Polycomb group of proteins (Polycomb repressive complexes-PRC) includes proteins that can generate or recognise repressive chromatin modifications n- a histone code writing and reading system. E.g. H3K27 (inactivates genes) methylation is mediated by the Enhancer of Zeste (Ezh2) component of PRC (code writer). The recruitment of PRC1 via Polycomb (Pc) chromodomain (code reader) leads to the formation of silence, repressed chromatin.

Question 63

What is the relationship between histone and DNA modification?

Answer 63

A close relationship exists between transcriptionally repressive histone methylation and DNA methylation. Transcriptionally inactive promoters are frequently rich in methylated CpG dinulcleotides. Addition of methyl groups to cytosine residues (on the DNA not histones) is medited by DNA methyltransferases (DNMTs). The histone methyltransferase Ezh2 physically interacts with DNMTs and together these enzymes mutually reinforce each others effects.
Proteins which bind to methylated CpG dinucleotides like methyl-CpG-binding protein 2 and MeCP2, interact with histone deacetylases and HMTs and help to transform acetylated nucleosomes into methylated nucleosomes.

Question 64

What is meant by the term X-inactivation?

Answer 64

Mammalian X-chromosome inactivation equalises the levels of X-chromosome derived gene products in males and females. Males have XY so have 1 dose of X genes, females have XX, so have 2. SO in females, one X chromosome is inactivated to even this out. One chromosome is silenced in each somatic cell during early development of the female embryo (Xp - the paternal one, or Xm - the maternal one) The silencing decision is propagated clonally i.e. all progeny of each cell in which the silencing decision was taken inherit the same silenced X chromosome - epigenetic changes.

Question 65

Explain X-inactivation in the case of calico cats.

Answer 65

Orange allele = XA (dominant), Black allele= Xa (recessive). in males, XAY = orange, XaY = black. In females XAXA= Orange, XaXa = orange, XAXa = calico. Therefore, calico cats are exclusively female and re heterzygous for 2 alleles of an X-linked coat pigment gene. Patches of orange and black are because of random X-inactivation during embryogenesis (instead of all progeny inheriting the same active chromosome, some inherit one and some inherit the other.). The white is because there is no pigment cells in this area.

Question 66

How are chromosomes inactivated?

Answer 66

The mechanism of X-inactivation involves synthesis of a non-coding RNA (Xist) from the X-inactivation centre (XIC) on the chromosme destined for inactivation. Xist RNA binds to the chromosome in cis and promotes chromatin condensation via a process that preads away from the XIC in both directions. Xist promotes the formationof silent chromatin by recruiting histones, modifying enzymes and other Polycomb group components, leading to the H3K27 and H3K9 methylation of core histones in the X chromosome chromatin. The Polycomb protiens detect Xist transcripts on the X chromosome that is to be inactivated and cause its transcriptional silencing.

Question 67

What are Barr bodies?

Answer 67

The highly condensed inactive X chromosome at the periphery of the nucleus of the mutant female ( a big black dot)

Question 68

How can diet effect epigenetics?

Answer 68

Dietary components can affect DNA methylation patterns which regulate phenotypes. E.g. in the Agouti mouse. Agouti is a dominant mutant allele of the agouti gene which causes obesity and yellow fur due constiuitively high expression of the agouti protein. The Agouti allele contains an insertion of an IAP element which strongly activates the agouti gene. The IAP element is sensitive to DNA methylation: high levels of methyl donors (folate,choline, betaine) in the maternal diet represses Agouti transcription in developing progeny.

Question 69

What are restriction enzymes are what are they used for?

Answer 69

Used for cutting chromosomes into manageable chuncks for research (large DNA is very fragile and will disintegrate so is not usable). Most restriction enzymes used are from other organisms like E.Coli. They acts as dimers and recognise short palindromic DNA sequences. They have precise recognition sequences. Some leave overhangs and others cut DNA flush (in the middle) and are called blunt restriction enzymes. The DNA cut this way are know as restriction fragments and can be separated by size using gel electrophoresis. Use dyes like ethidum bromide or DAPI to visualise this.

Question 70

How can you use restriction enzymes to clone DNA?

Answer 70

Cloning DNA involves the ligation of 2 DNA fragments. DNA Ligase is an enzyme that joins fragments to make recombinant DNA. Cohesive termini (sticky ends) ligate because they can hybridise because they have complementary sequences.
Cloning DNA usually involves ligation o DNA fragments into a plasmid vector. Plasmids are small cicular, extrachromosomal DNA that occur naturally in bacteria. They have their own origin of replication that usually results in about 50 copies of the plasmid being made i each bacterium. They also ussually carry antibiotic resistant genes which can be used to select which plasmids have been taken up properly. Plasmid vectors are made from plasmids, ussually by adding a bunch of restriction enzyme sites in one part of the plasmid (called multiple cloning sites). Plasmid vectors only hold less than 30kbps of DNA.
Bacterial artifical chromosomes can hold 300kbps and yeast artificial chromosomes hold about 3 mega bps. To put the DNA into the bacterium, yyou need to make temporary holes in the cell membrane using electroporation or chemicals. Competent bacteria are bacteria that are ready to take up new DNA. This method is not very efficient (only about 1 in 1mill take up the plasmid right) Single colonies from the plate that have taken it up right are used to start liquid cultures which can be easily purified for research or libraries.
You normally originally source the DNA you clone using mRNA or cDNA libraries.

Question 71

What are expressed sequence tags?

Answer 71

They are the sequenced ends of clones that genomic labs often make to make it easier to identify new clones (means you don’t have to sequence the whole thing)

Question 72

How do you sequence DNA?

Answer 72

First, you have to denature the template so that its single-stranded. Then you allow it to cool and anneal with a primer. Then you start a DNA synthesis reaction with DNA polymerase and dNTPs (an in vitro reaction)
Primers are short single-stranded DNAs that can be easily synthesised. They are often designed to anneal onto the edges of the vector. If you add tagged dideoxy terminators in a smaller amount than normal dNTPs the strands will end at different positions because DNA polymerase cannot extend once a ddNTP has been added. You can then separate out the differently sized strands and look at the tags to see what each base is. Running all four ddNTP reactions on the same gel results in a nucleotide ladder. Usually, the gel is then transferred onto a photographic film to detect the radioactive primer. Automated sequencing (up to 1000 nucleotides read in one reaction) is used nowadays. ddNTPs are labelled with a fluorescence instead. The camera doesn’t take a pic of the whole gel it just measures one position over time - the results are presented as a graph showing intensity over time. If the DNA us larger than 1000bps progressive sequencing is used.

Question 73

What is progressive sequencing?

Answer 73

When the ends of clones (found in a genomic library) are sequenced using promoters from the vector. Primers are designed based on the new sequenced and another round of sequencing is performed, and so on until it meets in the middle.

Question 74

Explain shotgun sequencing

Answer 74

When you make a genomic plasmid library and just sequence the ends from each cone using primers from the vector. These short random sequences are assembled by a computer programme into a contig using the overlapping sequence. Requires no thought or planning and is automated. However, you need to sequence more than 6x the size of the genome to make sure you have enough to overlap and make the contig. Therefore it is inefficient and there is n guarantee you will complete it all. Both shotgun and progressive are usually used together.

Question 75

How are genes found within a nucleotide sequence?

Answer 75

1. • Gene prediction software - often involves scanning the sequence for promoters, start and stop codons and intron splice sites. However, this is obviously just a prediction.
2. Use a computer to translate DNA in all 6 reading frames then search for similarity to known proteins (Known as BLAST)

Question 76

Explain how BLAST protein alignment works

Answer 76

INput the amino acid sequence of the proposed protein. The BLAST programme searches a huge database for other proteins which have similar sequences. This shows an alignment of an uncharacterised protein (query) to protein (called zen)
Similarity found between protein sequence suggests that the proteins evolved from the same common ancestor and that the porteins have similar molecular functions.

Question 77

Explain how microarrays work and what they are used for

Answer 77

Microarrays allow us o compare the transcriptosomes of different tissues to each other e.g. healthy to diseased.
It’s a high-throughput test which is done on a small scale and is fast and automated. A very presice robot makes the array chip. Each position on its grid contains one cDNA (antisense strand). You purify the mRNA from the tissues and tag each tissue with a different colour.YOu put the mRNA onto the array and wash off the excess. IF the gene is expressed in one tissue there will be an mRNA for it, which will bind to its cDNA and show a colour. You use a reader with a very sensitive camera to detect which genes are on. Genes that are lost in the tumour tissue might be tumour suppressors and genes that are extra in the tumour tissue might be oncogenes.

Question 78

What are the different ways to identify genes?

Answer 78

1. Making a library of cDNA clones from mRNA ( a transcriptosome)
2. Making a library of genome clones then making predictions based upon genomic sequence
3. identify sets of interesting genes using microarrays.
   then to work out their function you can make KOs / KIs

Question 79

How do you make a knockout?

Answer 79

TO knockout a single gene in mice you first have to acquire a genomic clone of the gene and insert it directly into an exon which destroys the activity of the gene.
E.g. NEO is inserted into the gene and TK is placed off to one side. The sequences from the target gene s are called homologous arms because they are the only sequence with the homology to the mouse genome. Then you introduce the construct into mouse ES cells using cell culture techniques. Next, the cell’s DNA repair machinery recombines the construct into the genome - however, this is not very efficient. Homologous recombination occurs sometimes, which creates the knockout. When this occurs, the selecting gene (TK) is also lost, this is how you tell whether it has been knocked out properly. (cells with NEO will grow in neomycin and cells that have TK will die in GANC media) The selected line is then introduced into a mouse. Te first gen in mosaic. These are bred togteher to make non-mosaic carriers of the transgene. The carriers are then bred to make homo knockouts.

Question 80

Explain what forward genetics is

Answer 80

When you randomly mutated the genome and look for unusual phenotypes - then find the gene responsible. Because random mutagenesis effects the whole genome, you have to analyse many mutagenised animals to find interesting phenotypes. Therefore organisms usually used for this are things like C. elegans, Drosophila and zebrafish because they have short gen times. You mutagenise the male and breed it with a normal female. You then interbreed the offsrping to make homozygous embryos. To find if similar phenotypes are caused by the same gene you use complementation testing. If you breed 2 similar phenotypes, if the gene is the same you get 1/4 of the offspring expressing the phenotype. If they are different no offspring will show the phenotype.

Question 81

What are the different types of mutations?

Answer 81

Amorphic/non-functional - completely inactivates the protein
Hypomorphic/ weak - makes proteins function less
Hypermorphic- Makes the protein overactive
Antimorphic - a protein made stops wt one from working