Lecture 5 - Tumour supressor genes Flashcards Preview

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Flashcards in Lecture 5 - Tumour supressor genes Deck (15)
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1
Q

How were cell fusion studies used to show that most human cancer phenotypes are recessive?

A

Monkey kidney cells and NIH3T3 fibroblasts growing adjacent to each other were presented with inactivated Sendai virus. Two cells will fuse and form a heterokaryon with multiple nuclei. These cells were then inserted into a mouse model. If a cancer cell and a non-cancer cell were fused together we would expect there to be two possible outcomes, one where the hybrid cell is tumorigenic and another when it is not. Most of the time no tumour formed on the mouse model suggesting most cancers to be recessive.

2
Q

What was the initial argument for the existence of tumour suppressor genes

A

A loss of function is more likely to occur than a gain of function mutation. I.e it’s more likely to lose tumour suppression genes then gain a functional mutant copy of Ras. Although the loss of both alleles of a tumour suppressor gene is also highly unlikely.

3
Q

Which condition gave insight into the existence of tumour suppressor genes?

A

Retinoblastoma - Diagnosed by observing red eyes in photographs of children.

4
Q

How does retinoblastoma arise?

A

Sporadically and familiarly Sporadic only affects one eye usually, with familial effecting both eyes. (A recessive phenotype)

5
Q

How does the log curve of cases not diagnosed at a certain age differ between unilateral and bilateral patients.

A

Unilateral curve conforms to a 2-hit theoretical curve, i.e two things have to have happened for the disease to manifest, a second order reaction. The bilateral patients appear on a one hit curve, a first order linear reaction.

6
Q

What is required of a patient with heterozygous retinoblastoma for them to present with the disease.

A

One of the alleles is already mutated, all it then takes is for a mutation in the second allele to produce the disease.

7
Q

How is the frequency of loss of the second WT allele explained

A

Loss of heterozygosity - by which the process of mitotic recombination and segregation give rise to two mutated alleles.

8
Q

What did zymography prove about retinoblastoma

A

The occurance of LOH was at Ch13q14. D esterase is on the same chromosome arm as retinoblastoma. There are two forms of D esterase and it’s easy to see which a patient has by zymography. Retinoblastoma tissue compared to normal always has a LOH of D esterase. LOH is therefore responsible for the formation of 2 mutated alleles when a patient already has one mutated copy.

9
Q

What is chromosome walking and how did it help discover the genetics of retinoblastoma

A

Focused on the area of chromosome 13. You “walk” along the DNA sequence and compare the normal and tumour DNA. The tumour cell line will have a missing gene at some point. They found a human DNA segment which predisposes an individual to retinoblastoma and osteosarcoma. This was concluded because they found a piece of DNA that was missing in the tumou that encoded a gene.

10
Q

How are RFLPs used to show LOH has occured?

A

Using the pic>.

The maternal DNA can be cleaved by a restrictive enzyme, while the homologous paternal which has a single base-pair sub is unable to be cleaved as a result. The presence or absence of this base substitution therefore represents an RFLP.

A radioactive probe recognising the right end of the paternal DNA can then be used in southern blot analysis to determine if the cleavage has occured or not.

Below - Two cancer patients have normal tissue which has preserved its heterozygosity. However their tumour DNAs have a LOH with one losing the paternal DNA and another losing the maternal.

11
Q

What did RFLP tracking show about colorectal cancer

A

When the percentage of colorectal cancer tumours with allele deletions was compared with respect to the long and short arms of each chromosome, it showed there to be a LOH events that can occur in a variety of chromosomes - but mainly the short arm Chr.17 and long arm Chr.18

12
Q

What is the technique that surpassed RFLP analysis

A

Single nucleotide polymorphism analysis

13
Q

How is SNP analysis performed

A

A PCR based approach in which two primers are used to amplify the amount of the target allele.

If there is a change in the DNA compared to the second allele, the first primer is unable to anneal so amplification will fail to occur.

This shows there is a SNP in the second allele and indicates a loss of heterozygosity.

14
Q

Which tumour supression gene mutant causes familial bowel cancer.

A

APC on 5p21 - its normal function is to degrade beta catenin.

When beta catenin activity is not supressed then the target genes induced by Wnt signalling are expressed. This leads to proliferation cells and lack of differentiation in the intestinal crypt. This generates an adenomatous polyp.

Beta catenin induced proliferation is dependent on its ability to associate with TcF4 protein to form the active transcription factor

15
Q
A