Lecture 5 Tumour Suppressor Genes

Question 1

Viral oncogenes exert a dominant effect T or F

Answer 1

T

Question 2

What is significant about the Sendai virus with regards to cancer

Answer 2

Sendai virus can cause mammalian cells to fuse and operates as a fusagenic agent.

Question 3

What happens to the chromosomal content of cells fused with the Sendai virus

Answer 3

After the second cell cycle of these fused cells the daughters had doubled their chromosomal contents

Question 4

What experiments showed that tumour genes don’t always exert a dominant effect

Answer 4

Rao and Johnson fused transformed monkey kidney cells with non-cancerous mouse fibroblasts before injecting these cells into nude mice. Unlike predicted this didn’t lead to the production of tumours in the mice suggesting that the cancer was recessive

Question 5

If a cancer was dominant what would be the effect of injecting nude mice with cancer cells that had been fused with mouse fibroblasts

Answer 5

Tumours would develop in the mice

Question 6

What assumption was made by the result that injection of fused tumour cells into nude mice didn’t result in tumorigenesis

Answer 6

The idea that something was acting to inhibit the cancer cell phenotype a tumour suppressor gene

Question 7

What rationale supports the idea for a tumour suppressor gene over a proto-oncogene

Answer 7

Loss of a growth suppressor gene is much more more likely than gain of function of another gene that causes cancer. Loss of function mutations are far more common than gain of function mutations

Question 8

What rationale goes against the idea that loss of a tumour suppressor gene is a mechanism by which cancers form

Answer 8

Loss of both alleles of a putative growth suppressor gene is unlikely

Question 9

Describe the two kinds of retinoblastoma and how they differ

Answer 9

Sporadic or unilateral retinoblastoma usually only affects one eye whereas familial or bilateral retinoblastoma almost always affects both eyes. Unilateral retinoblastoma can be either non-hereditary or hereditary but all bilateral retinoblastomas are hereditary

Question 10

How does hereditary retinoblastoma occur

Answer 10

Hereditary retinoblastoma is caused by recessive mutation in Rb gene. This gene is a tumour suppressor gene that normally prevents cells from becoming cancerous. In these patients’ retinoblastoma will therefore only occur due to the inactivation of both alleles of the tumour suppressing gene

Question 11

How may heterozygotes for the retinoblastoma gene develop tumours in both eyes

Answer 11

Heterozygous individuals for a recessive mutation in the Rb gene will only develop tumours in both eyes if independent secondary mutations inactivate the remaining wild-type copy in cells of both eyes

Question 12

Outline the results that lead to the development of the 1-hit/2-hit hypothesis

Answer 12

Knudson plotted a log of the percentage of both unilateral and bilateral retinoblastoma cases that were not yet diagnosed against age. The curve generated for the bilateral retinoblastoma cases was a straight line indicative of a first order reaction dependent on one factor. In comparison the curve for unilateral retinoblastoma resembled a second order reaction dependent on two factors. This lead to the idea that somehow unilateral (sporadic) retinoblastoma patients need to pick up two separate mutations in order to get the cancer whereas bilateral (familial) retinoblastoma patients need to only pick up one mutation in order to get the cancer

Question 13

How was the 1-hit/2-hit hypothesis explained in terms of why there was a difference in the number of mutations required to develop the cancer

Answer 13

The bilateral retinoblastoma patients must inherit a non-functional copy of the gene which is what renders them only needing one mutation to develop the tumours

Question 14

What process can account for loss of heterozygosity

Answer 14

Homologous recombination during G2 phase of the cell cycle. During DNA replication cells undergo homologous recombination during G2 phase. Following homologous recombination there are several different ways in which sister chromatids can separate

Question 15

What is meant by loss of heterozygosity

Answer 15

Where daughter cells inherit two functional copies or two non-functional copies of a gene which the parent cell was heterozygous for (with one functional and one non-functional allele)

Question 16

Explain how loss of heterozygosity occurs in retinoblastoma

Answer 16

Homologous recombination in a cell heterozygous for the retinoblastoma loss of function allele results in this mutant gene being present in a chromatid in both the maternal and paternal chromosomes. Due to the fact that there are several different ways in which sister chromatids can separate there are subsequently three different combinations of daughter cells that can be formed. More commonly daughter cells retain their heterozygosity each inheriting 1 wild type chromatid and 1 chromatid with the mutant retinoblastoma gene. However occasionally daughters can end up with two functional copies of Rb (LoH) whereby wild type levels are restored. Alternatively in terms of tumorigenesis segregation of the chromosomes can also generate daughters completely null for functional Rb. This loss of heterozygosity will result in no functional Rb genes and hence will cause tumours.

Question 17

How was the retinoblastoma gene idenitified

Answer 17

Cytogenetics revealed that patients with retinoblastoma had a missing band in the q-arm of chromosome 13 (13q14)

Question 18

Describe the technique of zymography

Answer 18

Zymography involves using protein electrophoresis whereby the protein is an enzyme and the substrate for the protein is placed in the gel in which it is run. The protein migrates to its position based on its size and then digests the region around where it migrates to. This creates a region of digested gel that represents where the protein is in the gel which can be seen and used to study loss of heterozygosity

Question 19

What is the significance of esterase D in retinoblastoma

Answer 19

Esterase D (an enzyme that hydrolyses esters) is on the same chromosome arm as retinoblastoma and lies very near its locus. There are two forms of esterase D type 1 and type 2. As it is an enzyme you can see which form of D esterase an individual has by using Zymography. This also allows you to determine if a loss of heterozygosity has occurred in that gene which can then in turn be used to indicate a loss of heterozygosity in retinoblastoma

Question 20

How can esterase D zymography be used to determine a retinoblastoma gene loss of heterozygosity

Answer 20

D Esterase type 1 slightly smaller than D Esterase type 2 and hence it forms a band lower down the gel in zymography. Patients heterozygous for the esterase D gene will have two bands in the zymography gel indicating the two forms of the gene. However in cells in which loss of heterozygosity has occurred there will now be only one band. This can indicate as loss of heterozygosity in the Rb gene in these patients if they have developed tumours in the retina

Question 21

What is significant about the genotype in patients used in zymography

Answer 21

They need to be heterozygous for both the tumour suppressor gene and the reference gene

Question 22

What mechanism is the main way in which tumour suppressor genes are lost in cancer

Answer 22

Loss of heterozygosity

Question 23

How can loss of heterozygosity be determined using restriction fragment length polymorphisms

Answer 23

Loss of heterozygosity can be inferred by comparing the digestion of maternal and paternal chromosomes by restriction enzymes. Heterozygotes for a gene containing a restriction enzyme recognition sequence where the base changes(s) occur within that recognition sequence will have two different bands on a southern blot when interrogated for a probe for that gene (probe still binds to the different alleles). This is because base changes that occur in alternative alleles may render these restriction sites unrecognisable by a restriction enzyme and hence this allele won’t be cleaved. If loss of heterozygosity has occurred in these patients southern blots carried out on tumour samples will show only a single band compared to non-cancerous tissue.

Question 24

Explain how single nucleotide polymorphism analysis can be used to determine whether loss of heterozygosity has occurred in a tumour

Answer 24

Single nucleotide polymorphism analysis uses primers that recognise particular regions of a chromosome and generate a PCR product from amplification of each region. These primers recognise a sequence known to vary in the population. Hence you can use these primers to determine if alleles differ in single nucleotides by determining if you generate a PCR product or not. If a PCR product is not generated it means that there must be one or more SNPs in that allele. Thus this technique can be used to determine if loss of heterozygosity has occurred based on determining which alleles are present in a tumour and the number of PCR products for a known gene primer

Question 25

Other than retinoblastoma list some examples of other cancers where loss of heterozygosity of a tumour suppressor gene has occurred

Answer 25

DPC4 in juvenile polyposis NF1 in neurofibromatosis and APC in familial adenomatous polyposis (FAP)

Question 26

Neurofibromatosis is a cancer associated with the loss of heterozygosity of the tumour suppressor gene NF1 that affects 1 in 3500 people. What is significant about this gene

Answer 26

NF1 is a neuronal specific RasGAP

Question 27

Outline how loss of heterozygosity in the APC gene causes FAP

Answer 27

At the bottom of the crypt in the gut wnt signalling is on causing proliferation and the upregulation of TCF genes. As the cells move up the crypt wnt signalling stops and transcription of TCF target genes is downregulated. Patients with FAP have a non-functional APC gene which means that GSK-3β cannot induce the degradation of β-catenin. This means that wnt signalling remains on and hence causes an over-proliferation of cells in the crypt which presents a polyps which can then become malignant

Question 28

Explain the role of the Von Hippel Landau gene in angiogenesis

Answer 28

VHL is a negative regulator of the transcription factor hypoxia-inducible-factor-1 (HIF1). Activated HIF1 translocate to the nucleus and regulates the transcription of growth factor genes such as VEGF.

Question 29

Describe how VEGF levels are regulated by oxygen levels

Answer 29

VEGF is upregulated in hypoxia and downregulated in normoxia. In the presence of normoxia HIF1 is modified by proline hydroxylase. This hydroxylation of proline residues means that HIF1 is recognised by VHL (and other proteins) and is degraded. However under hypoxic conditions proline hydroxylase cannot hydroxylate proline residues on HIF1 meaning it isn’t recognised by VHL. Hence HIF1 isn’t degraded and can translocate to the nucleus and upregulate transcription

Question 30

How does a loss of heterozygosity in VHL lead to tumorigenesis

Answer 30

Loss of both alleles of VHL mean that HIF1 is never degraded. This means that it can continue to upregulate genes involved in angiogenesis which is one of the hallmarks of cancer