Session 4.2 - Group Work Flashcards
Thursday 26th October 2017 09:00-10:30
- The ends of a linear chromosome are not readily replicated by cellular DNA polymerases. The image below shows the end of a chromosome undergoing replication.
See image
a) Why might there be a problem with producing the final Okazaki fragment on the lagging strand?
DNA polymerases work from a 5’ to 3’ direction. When making the leading strand, this is no problem because the template DNA opens 3’ to 5’, so the complementary DNA can be made 5’ to 3’.
However, on the lagging strand, the DNA template is being opened 5’ to 3’, so the complementary DNA will be from the 3’ to 5’ direction. However, DNA polymerase cannot work in this manner, and so needs to have a primer to allow it to jumpstart ‘backwards’ into its normal 5’ to 3’ direction and allow normal function. However, once the DNA helicase reaches the end of the DNA needing to be replicated, the lagging strand loses the last Okazaki fragment, because there is no more DNA for the DNA polymerase to attach to (the 5’ primer required would be further along the chromosome than exists). Thus, the final Okazaki fragment is not created and a small portion of the DNA at the end is lost and not conserved.
The end that is lost are the chromosome’s telomere’s (repeated sequences), hence, telomeres shorten with age (each subsequent division).
- The ends of a linear chromosome are not readily replicated by cellular DNA polymerases. The image below shows the end of a chromosome undergoing replication.
See image
a - The final Okazaki fragment on the lagging strand is not produced.
b) What would be the consequence of this in:
i) the daughter DNA produced
On the leading strand, the DNA would be copied no problem and the daughter DNA will be the exact same as the template.
On the lagging strand, however, due to the single-stranded overhang from lack of final Okazaki fragment being produced, the DNA will then be slightly shorter as the last end bit of DNA is lost. So the daughter DNA will be slightly shorter than the parent DNA.
Hence, some telomeres will be lost.
- The ends of a linear chromosome are not readily replicated by cellular DNA polymerases. The image below shows the end of a chromosome undergoing replication.
See image
a - The final Okazaki fragment on the lagging strand is not produced.
b) What would be the consequence of this in:
ii) the granddaughter DNA (i.e. the DNA copied form the daughter DNA)
The daughter DNA would produce one DNA of the same length (leading strand) and one DNA that is slightly shorter (lagging strand)
The granddaughter DNA of the leading strand therefore would produce one DNA that is same length as the original template, and also a slightly shorter one (such as that of the daughter lagging DNA)
The granddaughter DNA of the lagging strand would produce one copy that is the same as the daughter template (leading strand - but slightly shorter than the original template as this has already been lost), and the other copy even shorter still (lagging strand).
The consequence of this is that a little bit of DNA at the end (telomere) is lost with each round of replication.
- The ends of a linear chromosome are not readily replicated by cellular DNA polymerases. The image below shows the end of a chromosome undergoing replication.
See image
a + b: the ends of the DNA is lost in replication.
c) What molecular mechanism do certain eukaryotic cells have to allow them to overcome this problem? Explain how this works.
Eukaryotes have repeated sequences at the ends of their chromosomes, known as telomeres (TTAGGG), so that when the chromosome is truncated during replication, coding DNA (genes) are not lost but part of the telomere length. These telomere ends become shorter with each subsequent division.
The reverse transcriptase, telomerase, is available at the ends of normal stem cells to lengthen the telomere. It is normally absent, or operates at very low levels in most somatic cells.
- The ends of a linear chromosome are not readily replicated by cellular DNA polymerases. The image below shows the end of a chromosome undergoing replication.
See image
a, b + c: the ends of DNA (telomere) is lost in replication. These can be lengthened by the telomerase enzyme.
d) How might interfering with this system allow treatment of cancer?
As the telomerase enzyme adds telomeres to the ends of cells, and therefore makes them ‘immortal’, we may be able to interfere with the telomerase enzyme in cancer cells. This would mean the cancerous cell is no longer functioning, and therefore unable to continue to add telomeres to the end, so the malignant cell would undergo ageing and eventually cell death, just like a normal human somatic cell.
- a) What is xeroderma pigmentosa (XP)?
Xeroderma pigmentosum (XP) is an autosomal recessive (genetic) disorder which leads to an increased sensitivity to sunlight, due to a mutation causing decreased ability to repair DNA damage.
2) b) What genes are associated with the development of this disease [Xeroderma pigmentosum] and what do they code for?
There are many different genes associated with this disease, such as XPA, XPC, DDB2, POLH, ERCC2, ERCC3, ERCC4, ERCC5 etc.
These genes all code for enzymes involved in nucleotide excision repair. This means any damage to nucleotides via these mechanisms (such as by sunlight/UV light) are unable to be fixed by the normal enzymes, hence the increased susceptibility to UV light/sunlight.
2) c) Why are individuals with this disease [Xeroderma pigmentosum] highly sensitive to sunlight?
Nucleotides in DNA can become damaged from exposure to UV light, which is found in sunlight, as well as other causes. These damages are normally repaired by an internal physiological mechanism known as nucleotide excision repair, in which the damaged nucleotides are ‘excised’ and then ‘repaired’.
In patients with XP, these enzyme(s) are mutated, and therefore this mechanism doesn’t function properly. This means that any damage to the DNA via this pathway is unable to be fixed, so patients with XP that are exposed to sunlight and therefore UV light are exposed to slight DNA damage. Although this is normally fixed in a healthy person with no significant consequences, in a patient with XP, this cannot be fixed, hence leading to an increased sensitivity to sunlight.
2) d) Cultures of fibroblast from healthy individuals and from patients with XPG (a type of XP associated with mutations in one of the genes) are irradiated with UV light. The DNA is isolated and denatured and the resulting single-stranded DNA is analysed by ultra-centrifugation.
i) Samples from the normal fibroblasts show a significant reduction in the average molecular weight of the single-stranded DNA after irradiation but samples from the XPG cells show no such reduction. Can you explain why?
DNA is damaged by UV light, and in normal cells, DNA damage of this type is removed by a process known as nucleotide excision repair. This process has many enzymes involved, including an enzyme encoded for by the ERCC5 gene. The ERCC5 gene is responsible for incising the damaged nucleotides for removal.
When DNA from healthy patients is exposed to UV light, the resulting damaged DNA is excised from the sequence. This results in a molecular weight loss (due to removal of nucleotides).
Patients with XPG have a mutation in ERCC5. This means that they are unable to excise the damaged DNA, hence having no removed nucleotides, thus the molecular weight of these cells stay the same.
2) d) Cultures of fibroblast from healthy individuals and from patients with XPG (a type of XP associated with mutations in one of the genes) are irradiated with UV light. The DNA is isolated and denatured and the resulting single-stranded DNA is analysed by ultra-centrifugation.
ii) Which step of the encoded DNA repair system is most likely to be defective in these cells?
The lack of weight loss in cells from the XPG system shows that damaged DNA is not being lost.
This suggests that it is the incision step of the pathway that is defective, and it is likely that the gene for XPG (ERCC5) has endonuclease activity, which are enzymes that cleave phosphodiester bonds within polynucleotide chains (incision).
- Can inhibitors of DNA synthesis be used as antibiotics? If so, can you find any examples and what are they used for?
Yes - Quinolones are a group of antibiotics that inhibit DNA synthesis (topoisomerase II, a DNA gyrase [relaxes supercoiled DNA molecules]). In the UK, only fluoroquinolones are available, with the most common example being ciprofloxacin, and these are very active against Gram negative bacteria or atypical pathogens (e.g. Legionella for Legionnaire’s disease and Rickettsia for tick bite fever). Infections that can be treated include UTIs, typhoid, dysentery, TB and gonorrhoea.
Other DNA inhibitors include metronidazole, which is used for anaerobic infections, such as from Helicobacter pylori or Clostridium difficile infections; and nitrofurantoin, which can be used in treatment for UTIs (cystitis).
- Look at Cancer Research UK’s website on testicular cancer - http://www.cancerresearchuk.org/aboutcancer/testicular-cancer.
Find the different types of chemotherapeutic agents that are used to treat this disease.
Compare and contrast the different modes of actions of these molecules. Are there any that are commonly used targets for these compounds?
Actinomycin - inhibits transcription
Bleomycin - induction of DNA strand breaks
Cisplatin - cross-links DNA
Cyclophosphamide - cross-links DNA in cells with low levels of ALDH [nitrogen mustard]
Etoposide - blocks topoisomerase II (prevents re-ligation)
Ifosfamide - blocks DNA replication
Methotrexate - inhibits DHFR
Paclitaxel - targets tubulin
Platinum e.g. carboplatin - cross-linking of DNA
Vinblastine - suppress microtubule interactions
Vincristine - binds to tubulin
These all inhibit DNA synthesis in some way or another, for example:
- inhibits transcription so cell can’t replicate (actinomycin, ifosfamide, methotrexate)
- actively breaks DNA (bleomycin)
- cross-links DNA thereby blocking replication (cisplatin,, platinum)
- DNA alkylation (cyclophosphamide)
- prevents DNA re-ligation (etoposide)
- interferes with microtubule assembly so chromosomes can’t separate in mitosis (paclitaxel, vinblastine, vincristine)
- It is observed that in some cases tumours seem to respond well to chemotherapy; growth slows down and the tumour reduces in size significantly. However, after some time the tumour growth resumes, and the cancer is back as aggressively as before (and sometimes even more destructive).
Describe two ways how this observation can be explained.
CANCER THERAPY RESISTANCE - cancers are not homogeneous, but in fact, heterogeneous. This means that if you look at cells from the SAME tumour mass, the cells within these masses are not all the same - so although the tumour is the same the individual cells may not be. This is due to mutations driving cancer evolution, and these mutations then following different pathways to further promote cancer evolution. The phenomenon of these different cell types is known as intra-tumour heterogeneity.
This causes problems when it comes to treating tumours. Although at first a patient may seem to respond well to chemotherapy, what may actually occur is that the chemotherapy is simply only targeting subclones that are particularly sensitive to chemotherapy. These cells are killed, but what’s left are the subclones of cells within the tumour that are resistant to this therapy, and these cells are able to divide and grow further, especially in the absence of competing with other cells. So in fact, all you’ve done is select for the resistant clone (much like in antibiotic resistance), meaning the patient will no longer respond to the current chemotherapy treatment.
CANCER STEM CELLS - another reason this might occur is due to the cancer coming into recurrence. So although the initial tumour may appear to be killed, there is a new discovery of cancer stem cells, which means that cancers are able to create new cells. This means that although the initial ones were killed, these new ones are able to replicate and become new cells, although the cancer is bigger and worse than before.
- Non-homologous end joining is one of the double stranded break repair mechanisms.
a) Describe in simple terms how it works.
1) Double-stranded DNA is broken (DSB = double-strand break)
2) Factors (see below) are recruited to the ends of the DSB to protect, process and recruit other factors for repair
3) DNA bases are removed that are broken/damaged
4) Once repaired, further factors are recruited for ligation of the two ends back together
(break, repairman, fix)
- KU70/80 is recruited to the ends of the DSB to protect them. This will recruit other factors for processing/repair
- DNA-PKcs (DNA-PK catalytic subunit) is recruited to form DNA-PK complex. This processes the ends if necessary to remove the damage DNA components.
- Recruitment of XRCC4, DNA ligase IV and XLF into a complex occurs to ligase the 2 broken ends back together.
- Non-homologous end joining is one of the double stranded break repair mechanisms.
b) What is a potential problem with this type of DNA repair?
Although the most simple pathway of the two in the genotoxic double-strand break repairs, the process can be error prone leading to problems with DNA repair. These include:
- chromosome fusing together in the wrong way
- removal of bases may falsely cause nucleotide deletions
- polymerase activity may accidentally cause nucleotide additions
- in extreme cases, NHEJ can fuse two non-homologous chromosomes together, causing different genes. This is known as a translocation.
