MST 2 Revision- Lectures 10-21 Flashcards by Bronwen Delbridge

What is not involved in preparing a karyotype?

a. Centrifugation if using blood in order to separate RBC from WBC
b. Colchicine to bind tubulin and prevent spindle formation
c. Staining with heparin if blood
d. Tissue culture with phytohaemogluttin to promote mitosis

c. Staining with heparin if blood

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What is false about banding techniques?

a. NOR banding is the reverse to G banding
b. Q banding requires a fluorescent dye
c. T banding is specialised R banding for telomeres
d. C banding can pick up centromeres and heterochromatin

a. NOR banding is the reverse to G banding

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What does 1q21 refer to?

a. The band is in region 1 on the q arm of chromosome 2
b. The band of interest is on chromosome 21
c. There is only one copy of chromosome 2
d. The band is in region 2 on the q arm of chromosome 1

d. The band is in region 2 on the q arm of chromosome 1

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What is not an example of a common polymorphism observed in normal karyotypes?

a. Heterochromatin levels can vary greatly around chromosome 9
b. Satellite size in chromosomes 13, 14, 15, 21 and 22
c. Spontaneous fragile sites that cannot be passed on to offspring
d. The length of the Y chromosome q arm

c. Spontaneous fragile sites that cannot be passed on to offspring

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What is false about fragile X syndrome?

a. The site is at Xq27
b. There is a triplet repeat (CGG) in the 5’ UTR of FMR1 which results in promoter methylation
c. Females have stronger phenotypes (mental retardation) than males
d. >200 repeats of CGG are required for an individual to be affected with fragile X

c. Females have stronger phenotypes (mental retardation) than males

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Which describes a female with a terminal deletion in region 2, band 5 on the q arm chromosome 3?

a. 46, XX, del (3) q25
b. Del, 46, XX, (3) q25
c. XX, del, (3), qq, 2, 5
d. 46, XX, del, 2q5, (3)

a. 46, XX, del (3) q25

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What does 46, XY, del (6)(q13q22) refer to?

a. A male has a terminal deletion at q13 and q22
b. There is a interstitial deletion on the q arm of chromosome 6
c. There is an interstitial deletion that has resulted in losing all parts of the chromosome except for the section between region 1, band 3 and region 2, band 2 on the q arm
d. The section between region 1, band 3 and region 2, band 2 on the q arm of chromosome 6 has been replaced

b. There is a interstitial deletion on the q arm of chromosome 6

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What does 46, XY, dup(6) (q21 → q22) refer to?

a. Chromosome 6 has been duplicated and lost region 2 on the q arm
b. This male has twice as many copies of chromosome 6 compared to a healthy individual
c. There has been a duplication of the section between q arm, region 2, band 1 and q arm region 2, band 2 on chromosome 6
d. Region 2, band 2 has been replaced with region 2, band 1 on the q arm of chromosome 6

c. There has been a duplication of the section between q arm, region 2, band 1 and q arm region 2, band 2 on chromosome 6

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What are the products of a pericentric inversion event?

a. 2 non-recombinant chromosomes, a dicentric chromosome and an acentric chromosome
b. 3 deletion products and 1 viable chromosome
c. 2 non-recombinant chromosomes and 2 imbalanced chromosomes
d. An inversion product, normal product and 2 deletion products

c. 2 non-recombinant chromosomes and 2 imbalanced chromosomes

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What does 46, XY, inv ins (3;4) (q13;q26q13) mean?

a. Region 2, band 6 is closer to the chromosome 3 centromere than region 1, band 3 on the q arm because of inversion and insertion
b. Region 2, band 6 is closer to the chromosome 4 centromere than region 1, band 3 on the q arm
c. Chromosome 3 has donated the section of the q arm between region 2, band 6 and region 1, band 3 to chromosome 4
d. This individual has Cru du Chat syndrome

a. Region 2, band 6 is closer to the chromosome 3 centromere than region 1, band 3 on the q arm because of inversion and insertion

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What causes complete androgen insensitivity?

a. Deletion of the Xq11.2-12 region where the androgen receptor gene is housed
b. Duplication of the Xq11.2-12 region on both X chromosomes
c. An inversion with a break point in the Xq11.2-12 region
d. 46, XY, inv ins (3;4) (q13;q26q13)

c. An inversion with a break point in the Xq11.2-12 region

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• To study chromosomes, you must use dividing cells.

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• FISH can label whole chromosomes but not single segments.

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• Karyograms show a set of banded chromosomes and Karyotypes show the chromosome complement of an individual.

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• Most karyograms relate to Q banding.

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• The long arm of the chromosome is called the P arm.

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• Chromosomes are divided into regions and numbered out from the centromere.

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• A satellite site is a non-staining gap in a chromosome that is inherited according to Mendelian inheritance.

F (fragile)

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• Paracentric inversions include the centromere.

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• Inversions are caused by recombination events within the inversion loop.

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Which is the most significant cause of cancer?

a. Infection
b. Inheritance
c. Diet
d. Pollution

c. Diet

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What kind of cancer is likely to have an early onset and be due to a single gene mutation?

a. Hereditary
b. Familial
c. Sporadic
d. Polygenic

a. Hereditary

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What is not considered to increase the risk of developing breast or ovarian cancer?

a. Having multiple affected relatives
b. Ashkenazi Jewish ancestry
c. Having relatives diagnosed at a late age
d. Having relatives with both breast and ovarian cancer

c. Having relatives diagnosed at a late age

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When would mutation detection be used?

a. Once a mutation has been found within a family
b. To find a mutation in an affected individual
c. To find out if a mutation is testable
d. To locate a mutation in unaffected individuals

b. To find a mutation in an affected individual

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What is not used to detect mutations? a. Sequencing b. NGS c. MLPA d. Southern Blot

d. Southern Blot

What is a benefit of genetic testing? a. There is no burden of disclosing the information to other family members b. It can aid future decision making c. It can impact future employment and insurance choices d. It can provide accurate information about when you will develop cancer

b. It can aid future decision making

What can be found through genetic testing? a. A polymorphism that reveals the disease causing mutation b. No mutation which excludes the possibility that the disease has a genetic cause c. A pathogenic mutation which results in the disease d. A variant of unknown significance that is likely normal genetic variation

c. A pathogenic mutation which results in the disease

What would not be used by a BRCA1 carrier to manage their cancer risk? a. Prophylactic mastectomy and/or bilateral salpingo-oophorectomy b. Tamoxifen taken as a risk reducing drug c. Annual MRIs and mammograms d. Yearly MLPA mutation detection

d. Yearly MLPA mutation detection

• Most cancers are inherited.

• Familial Adenomatous Polyposis is due to a mutation in one gene and often results in prophylactic bowel removal.

• 10-15% of referrals are knocked back for genetic counselling appointments.

• A BOADICEA score is the only factor that needs to be analysed when assessing risk for genetic testing.

• A blood sample can take 2 months to undergo genetic testing.

• Whole genome sequencing is the current preferred technology used for genetic testing.

F (Gene panels)

What does ‘dominant male’ system refer to? a. The absence of SRY leads to the development of the testis b. The male phenotype is the default pathway c. If SRY is present, a male should develop d. Someone needs two copies of SRY to develop a male phenotype

c. If SRY is present, a male should develop

What is a feature of the SRY location? a. It is within the PAR on the q arm of the Y chromosome b. The location means that crossing over between X and Y chromosomes will never result in the exchange of SRY c. It is close to the PAR on the Y chromosome p arm d. It is in region 1A2 on the q arm of the Y chromosome

c. It is close to the PAR on the Y chromosome p arm

What was involved in cloning and identifying the SRY location? a. Looking for the smallest deleted region on Y chromosomes in male carriers b. Looking for the smallest fragments of Y chromosomes added to X chromosomes in the XY females c. Finding the region that was present in the XY females and absent in the XX males d. Determining that the SRY was in the region where the smallest insert made XX male and shortest deletion made XY female overlap

d. Determining that the SRY was in the region where the smallest insert made XX male and shortest deletion made XY female overlap

What is involved in the development of testes or ovaries? a. After 5 weeks, a gonadal ridge forms from the mesonephros and germ cells migrate to the ridge b. Sex is majorly determined by growth rate, which is more rapid in females c. Ovary differentiation begins at 6 weeks, weeks before testes tissue differentiation d. Females secret androgens and antimullerian hormone (AMH) to prevent to development of the mullerian duct

a. After 5 weeks, a gonadal ridge forms from the mesonephros and germ cells migrate to the ridge

What makes up the SRY gene? a. It has several TATA boxes that are conserved between species b. It has a conserved 80AA HMG box that shares high homology to the SRY of other mammals c. It has several exons but little homology between species d. It has a conserved SOX-box which shares high homology with other human genes

b. It has a conserved 80AA HMG box that shares high homology to the SRY of other mammals

What is false about SOX genes? a. They have a Sox box which is an HMG box 70% homologous to that of SRY b. SOX stands for SRY related box c. SOX 3 is thought to be an ancestor of SRY and functions in brain development d. 50% of XY females have mutation in SOX9 or SOX8

d. 50% of XY females have mutation in SOX9 or SOX8

What is false about the action of SRY as a transcription factor? a. It binds to the minor groove of DNA and bends it at a 60 degree angle b. The TAACAATAG binding site is also recognised by SOX genes c. In mice, it contains a glutamine repeat region essential for testis development d. The Sip-1 sequence in SRY is an important nuclear localisation signal

d. The Sip-1 sequence in SRY is an important nuclear localisation signal

• On day 11 of development, mice have 1500 sexually dimorphic genes expressed even though the gonad is morphologically indistinct.

• Female mice express one gene that alters the expression of 1500 genes in the ovary.

F (200)

• The events for correct gonadal development of secondary characteristics for a male or female phenotype are classified as part of sex determination.

F (differentiation)

• 50-70% of genes are expressed in a sex biased manner, although the mean difference is only 10%.

• Bovines have 1274 transcribed Y chromosome genes which are mostly expressed during testis development.

• The male phenotype is the default pathway in humans.

• XY females and XX males can result from a crossover that transfers the SRY locus to the X chromosome.

• Homology between mammals is greatest at the level of DNA.

• For the first 5 weeks, there is no detectable morphological difference between male and female destined embryos.

• Male reproductive tracts develop from the Mullerian duct and female reproductive tracts develop from the Wolfian duct.

• Secondary sex characteristics develop at puberty for both males and females.

• The SRY protein product is much smaller in mice (204AA) than in humans (395AA).

• Removing the glutamine repeat region in the mouse prevents the transcription of Sry, yet testis still develop.

What is an important developmental event that occurs by the sixth week in male development? a. Mullerian ducts begin to develop b. The testis develop and secret sertoli cells c. Sertoli cells of the testis develop d. Development of sertoli cells inhibits wolifian duct formation

c. Sertoli cells of the testis develop

What is the significance of meiosis during sex determination? a. An ovary cannot develop once a germ cell has entered meiosis b. Sertoli cells block the onset of meiosis so testis can form c. Ovary to testis formation can only occur during a brief window in early development d. The XX embryo grows rapidly due to high levels of meiosis

b. Sertoli cells block the onset of meiosis so testis can form

What is false about parentally imprinted genes? a. The genes are always on the Y chromosome b. X genes with maternal imprint function under the restraint of that imprint in male cells c. Female cells have half their genes imprinted by the female parent and half imprinted by the male parent d. Parental imprinting of X chromosome genes can influence adult cognitive and social behaviours

a. The genes are always on the Y chromosome

What is not a feature of WT1? a. It has 4 zinc fingers b. It functions downstream of SRY c. It induces the outgrowth of the gonadal ridge in males and females d. Mutating its zinc fingers can result in XY individuals developing as female

b. It functions downstream of SRY

What is true about the alternative forms of WT1? a. -KTS is involved in RNA processing and stabilising the SRY transcript b. Knocking out +KTS leads to an XY ovary and no AMH production c. +KTS is a classic transcription factor that activates SRY d. Overexpression of –KTS results in an undifferentiated gonad

b. Knocking out +KTS leads to an XY ovary and no AMH production

What induces the AMH gene? a. Sry b. Wt1 c. SOX9 d. TES

c. SOX9

What describes a result of a mutation of SOX9? a. Duplication can lead to campomelic dysplasia b. Deletion can lead to XX males c. Too much SOX9 can cause male development even in the absence of SRY d. A lack of SOX9 results of gonadal dysgenesis and skeletal malformation

c. Too much SOX9 can cause male development even in the absence of SRY

What is the role of Dax1? a. It prevents NR5A1 binding to TES b. It plays a crucial role in ovary development c. It regulates steroid biosynthesis by binding SOX9 d. It acts with SRY to upregulate SOX9

a. It prevents NR5A1 binding to TES

How do DAX1, WNT, RSPO1 and FOXL2 influence sexual development? a. DAX1, WNT and RSPO1 suppress SOX9 in the XY gonad b. Knocking out DAX1, WNT and RSPO1 represses FOXL2 c. When FOXL2 in knocked out, SOX9 is upregulated and a testis can develop d. Knocking out WNT, RSPO1, FOXL2, SOX9 in the XX gonad results in undifferentiated gonads

c. When FOXL2 in knocked out, SOX9 is upregulated and a testis can develop

How might an XY ovary develop? a. Due to a mutation in DMRT1 which usually suppresses FOXL2 b. Due to the absence of SRY c. Following upregulation of SOX9 and SOX3 d. Due to over production of sertoli cells

a. Due to a mutation in DMRT1 which usually suppresses FOXL2

• Male embryos grow more rapidly than female embryos.

• Unlike mammals, the sexual phenotype of the gonads and body of drosophila are controlled by the same genes.

• Y chromosome genes have male and female determining effects.

• X chromosome genes impart their effects after the onset of X inactivation when there is a 1:1 X chromosome ratio in females.

• XIST inactivates genes that cause female somatic cells to be different to male cells and the barr body is produced.

• Midbrain cells between sexes exhibit different levels of tyrosine hydroxylase expression only after gonads begin making hormones.

• Sex determining genes do not have other roles in development.

• Sertoli cells can only be XY and the absence of Sry leads to them developing as pregranulosa (ovarian) cells instead.

• NR5A1 inhibits binding of SRY to target DNA such as WDR5.

• SRY and NR5A1 bind to the TES of SOX9 and cofactors WDR5 and CITED2 to upregulate SOX9.

• SOX9 is involved in feedback regulation to block SRY and auto-regulate itself.

• FOXl2 null mice are sterile and show defects of early ovarian development.

F (no development defects)

• The OD model suggests that a gene is required for ovary development rather than the pathway just being default.

Which enzyme is not involved in epigenetics? a. Writers which add histone modifications b. Extenders that duplicate histone modifications c. Erasers that remove histone modifications d. Readers that bind histone modifications

b. Extenders that duplicate histone modifications

What is not an epigenetic mechanism? a. RNA splicing b. Histone modification c. LncRNA d. DNA methylation

a. RNA splicing

What is a feature of DNA methylation? a. Guanine residues are methylated b. DNA is activated c. Changes are heritable d. Chromatin is not affected

c. Changes are heritable

What is a feature of long non coding RNA? a. They bind chromatin modifying proteins b. They interact with the genome in a non-specific manner c. They cannot influence mRNA splicing or stability d. PRC2 is encoded for by a lncRNA

a. They bind chromatin modifying proteins

What is a feature of an epigenetic phenomenon? a. X inactivation is found in marsupials only b. Genomic imprinting is localised gene silencing on autosomes c. Genomic imprinting is conserved in all mammals d. X inactivation allows gene expression to be higher in males than females

b. Genomic imprinting is localised gene silencing on autosomes

What is a not a feature of XIST? a. It encodes a 17kb non coding RNA b. It is the only gene expressed on the X chromosome to be inactivated c. It coats the X chromosome that will remain active d. It is cis acting and recruits epigenetic machinery

c. It coats the X chromosome that will remain active

What happens if the A loop of XIST is deleted? a. PRC2 and PRC1 are overexpressed b. PRC2 cannot bind and carry out methylation c. H3K27 will be trimethylated d. The gene state will become inactive

b. PRC2 cannot bind and carry out methylation

What is a feature of TSIX? a. It causes the promoter of XIST to be methylated b. It is downregulated on the active X chromosome c. It is expressed on both X chromosomes d. It is upregulated on the inactive X chromosome

a. It causes the promoter of XIST to be methylated

What is a feature of genomic imprinting? a. About 100 genes are imprinted and therefore dizygotic b. It can only result from histone modification c. Imprints are reproduced during cell divisions and erased in the germline d. Inherited maternal and paternal genomes are functionally equivalent

c. Imprints are reproduced during cell divisions and erased in the germline

In terms of genomic imprinting, what is the result of a two paternal genome? a. A large foetus and large placenta b. A small foetus and large placenta c. A large foetus and small placenta d. A small foetus and small placenta

b. A small foetus and large placenta

• DNA methylation depends on a cytosine residue preceding a guanine residue.

• Many epigenetic diseases are due to mutations in genes affecting methyltransferases and Demethylases

• A foetus can still be viable with two active X chromosomes.

• The barr body is distinct as a dense heterochromatin structure during interphase in XX females.

• XIST is one of 3 genes expressed on the inactivated X chromosome.

• TSIX is expressed from both X chromosomes.

F (only the active)

• XACT is an LncRNA that coats the active X chromosome in human pluripotent cells.

• Marsupials express XIST and XACT.

F (neither)

What is not a feature of the H19 imprinting cluster? a. H19 is a lncRNA b. H19 is only expressed on the maternal chromosome c. IGF2 is expressed on the maternal and paternal chromosome d. IGF2 and H19 have separate enhancers upstream of H19

c. IGF2 is expressed on the maternal and paternal chromosome

What is the role of CTCF and the DMR? a. The DMR binds regions of methylated DNA to block transcription b. CTCF binds to the DMR on the maternal chromosome to block IGF2 expression c. The paternal DMR is methylated to encourage CTCF binding d. CTCF binds to un-methylated chromatin on the maternal and paternal chromosome

b. CTCF binds to the DMR on the maternal chromosome to block IGF2 expression

How can IGF2 imprinting be linked to disease? a. When the maternal and paternal chromosomes are methylated at the ICR, Wilms tumours can form b. When neither chromosome is methylated at the ICR, there is excess cell proliferation c. When the maternal chromosome is methylated and the paternal chromosome is un-methylated, MET is demonstrated d. When the maternal and paternal chromosomes are methylated at the ICR there is a loss of cellular adhesion

a. When the maternal and paternal chromosomes are methylated at the ICR, Wilms tumours can form

Which statement is correct in regards to SNRPN imprinting? a. The most common origin of angelman syndrome is through imprinting mutation b. A deletion in the paternal chromosome results in a silenced PWS region and Prader Willi syndrome c. Deletion mutations are rarely the cause of Angelman and Prader Willi syndrome d. A deletion in the maternal chromosome results in an upregulated AS region and Prader Willi syndrome

b. A deletion in the paternal chromosome results in a silenced PWS region and Prader Willi syndrome

What is not a function of a lncRNA? a. Protein synthesis shown by rRNA b. X inactivation shown by XIST RNA c. Epigenetic modification and chromatin remodelling d. Splicesome formation shown by snRNA

d. Splicesome formation shown by snRNA (small non coding RNA)

What is a feature of hypospadias? a. It can be seen in sons of women exposed to DES b. It is now rarely seen in Australian births c. The increase is occurring slower than genetic changes d. In utero exposure to estrogenic chemicals and decrease the chance of the disease

a. It can be seen in sons of women exposed to DES

What is involved in trans-generational inheritance? a. It can be proven by observing defects in the F2 and parents only b. Reprogramming of the germ line occurs in the gonads c. Only the parent generation receives direct exposure d. The use of DES in the 1930s has had no trans generational consequences

b. Reprogramming of the germ line occurs in the gonads

• Imprint genes are often found in clusters.

• Snrpn is active only on the maternal chromosome.

• H19 recruits MBD1 which modifies chromatin on the paternal chromosome to silence IGF2.

• A H3K4me3 modification is associated with active promoters and a H3K27me3 modification is associated with inactive chromatin.

• 30% on lncRNAs in ESCs are for chromatin modification and 23% interact with PRC2.

• In agouti mice, higher levels of methylation can result in a more yellow coat.

F (more meth = brown)

• Multiple generations can be affected by unfavourable in utero conditions, not just the F2.

What is involved in cell division? a. Cells maintain their size during cleavage due to the G1 and G2 phases b. An original cell divides into many cells and each daughter cell receives a full copy of the genome c. Cleavage does not occur until after differentiation and determination d. The embryo remains the same size during later cell divisions due to the absence of the G1 and G2 phase

b. An original cell divides into many cells and each daughter cell receives a full copy of the genome

What does not occur during differentiation? a. All genes are expressed to the same degree in all cells b. Internal structure and outward appearance of cells is established c. Cells can become polarised d. Differential gene expression occurs

a. All genes are expressed to the same degree in all cells

How do drosophila neural stem cells divide? a. GMCs divide into daughter neuroblasts b. Symmetric division due to secretion of wingless c. Asymmetric division due to the localisation of prospero d. By inhibiting the BMP signalling pathway

c. Asymmetric division due to the localisation of prospero

Which is not a common signalling pathway? a. TGFB/BMP b. Hedgehog c. FGF (MapK) d. Hippo/Yorkie

d. Hippo/Yorkie

What is involved in morphogenesis? a. Apoptosis is not a normal part of morphogenesis b. Cells develop through cell divisions due to cleavage and growth c. Once morphogenesis begins, cells cannot migrate d. Morphogenesis only involves MET

b. Cells develop through cell divisions due to cleavage and growth

What is not an example of collective cell behaviour? a. Epithelial folding b. Convergent extension c. Cleavage and Growth d. Epithelial branching

c. Cleavage and Growth

What does twist do? a. It is a determinant in the neuroblasts of drosophila during asymmetric division b. It regulates genes for epithelial folding, EMT and migration c. It causes MET of cells in the drosophila embryo d. It is expressed in the dorsal side of human embryos to regulate many genes at once

b. It regulates genes for epithelial folding, EMT and migration

What is not a benefit of drosophila as a model organism? a. It has a large genome like humans b. Cultivation is short and easy c. The embryo is easy to visualise d. Ready access to genetic resources

a. It has a large genome like humans

What accurately describes a benefit of a specific model organism? a. Flies can be frozen and kept alive to be studied at a later date b. Worms can be genetically manipulated via mRNA injections to the zygote c. The cell linage of chicks has been determined d. Zebrafish have a clear embryo and are genetically tractable vertebrates

d. Zebrafish have a clear embryo and are genetically tractable vertebrates mRNA injections = frog not worms

• To form the blastocyst, the morula must escape the zona pellucida in order to contact the uterus epithelium.

• Cell division of B and T lymphocytes and sperm cells results in daughter cells receiving a full copy of the genome.

• Cells in the presumptive eye region in drosophila are determined at the gastrula stage.

F (neurala)

• In propspero mutants, GMCs are transformed into self-renewing neural stem cells.

• There is no G1 and G2 stage during cleavage.

• EMT involves the formation of a polarised epithelium and MET involves the formation of migratory cells.

• Most cases of achondraplasia result from a glycine to arginine mutation in the FGFR3 gene.

• Cancer can develop when regulatory genes are mutated and cells escape the strict controls of development.

What correctly describes the events from ovulation to implantation? a. The morula forms in the ovary b. First cleavage occurs in the uterus c. Fertilisation occurs in the fallopian tube d. The blastocyst develops in the ovary

c. Fertilisation occurs in the fallopian tube

What occurs during fertilisation? a. When a spermatozoa contacts the zona pellucida, the egg release proteolytic enzymes to break down the zona pellucida. b. A spermatozoa contacts the oocyte after the second meiotic division and creation of the 2nd polar body and female pronucleus c. Once the sperm pronucleus is delivered into the oocyte, the egg releases enzymes that modify the surface and prevent more sperm from entering d. The male and female pronuclei fuse to form the first haploid cell known as the zygote

c. Once the sperm pronucleus is delivered into the oocyte, the egg releases enzymes that modify the surface and prevent more sperm from entering a. Sperm releases enzymes, not egg

What is a feature of the morula? a. It describes the embryo at the 8 cell stage b. It develops following cleavage and rapid G1 and G2 cycles c. It describes the embryo at the 16 cell stage d. It develops following cleavage of the blastocyst

a. It describes the embryo at the 8 cell stage

Epithelial cells: a. Are polarised along the dorsal-ventral axis b. Are detached from neighbour cells c. Have tight junctions to prevent molecules and water passing between them d. Have gap junctions for adhesion

c. Have tight junctions to prevent molecules and water passing between them

What happens during compaction? a. E-cadherin is down-regulated in cells b. The first epithelium is formed c. Cells lose polarity in order to maximise contact d. Microvilli can be restricted to the basal surface

b. The first epithelium is formed

How does the inner cell mass form? a. Through symmetric divisions within the morula b. During cleavage and compaction of the blastocyst c. From the unpolarised inner cell that results from asymmetric division in the morula d. When E-cadherin expression is turned off in the morula

c. From the unpolarised inner cell that results from asymmetric division in the morula

How do the first major cell linages split? a. Outer cells in the morula express Oct4 b. Oct4 and Cdx2 mutually activate one another c. Inner cells in the moruala express Cdx2 d. Mutual repression of Cdx2 and Oct4 stabilises cell differentiation

d. Mutual repression of Cdx2 and Oct4 stabilises cell differentiation

How do the epiblast and hypoblast develop? a. Epiblast cells express nanog which represses GATA6 b. Hypoblast cells lack GATA6 expression c. FGF4 directly activates nanog expression in epiblast cells d. Hypoblast cells express nanog in order to up-regulate GATA6

a. Epiblast cells express nanog which represses GATA6

What is not a feature of gastrulation? a. It is the first event of morphogenesis and involves body axes being defined b. The Primitive streak forms in a region where epiblast cells undergo EMT and enter the space between epiblast and hypoblast layers c. Cells exit the primitive streak and intercalate in outer hypoblast layer to form endoderm d. The endoderm only contains epiblast cells

d. The endoderm only contains epiblast cells

What is a feature of the germ layers? a. Epiblast cells coming through the primitive streak take on a mesodermal fate b. The mesoderm can give rise to the gut and lungs c. Endoderm forms from epiblast cells that don’t go through the primitive streak d. Ectoderm gives rise to muscle and connective tissue

a. Epiblast cells coming through the primitive streak take on a mesodermal fate c. That's how ectoderm forms

What occurs during mesodermal cell migration? a. Ecad is up regulated to encourage a MET transition b. FGF4, FGF8 are expressed in the primitive streak c. FGFR1 interactions lead to Ecad expression via Snail d. FGFR1 mutants demonstrate repressed Ecad expression

b. FGF4, FGF8 are expressed in the primitive streak

What occurs through the FGF pathway? a. It leads to Brachyury expression in mesodermal cells b. It signals for sperm to release proteolytic enzymes upon contact with the zona pellucida c. Snail is repressed so that Ecad can be expressed d. It represses Brachyury which is a regulator of endodermal cell fate

a. It leads to Brachyury expression in mesodermal cells

• Ecadherin (epithelial cadherin) is the major cell-cell adhesion molecules in adherens junctions.

• Cancer metastasis can result when epithelial cell polarity factors are lost.

• During blastocyst formation, the blastocyst appears to pulse as water is pumped out of the morula.

• The zona pellucida must be intact for the blastocyst to make contact with the uterine wall epithelium.

• Embryonic stem cells are derived from the hypoblast of the inner cell mass.

• The amniotic cavity forms within epiblast cells and the primitive yolk sac forms within hypoblast cells.

• During gastrulation, the ectoderm is the first germ layer to form.

F (endoderm)

What defines the body axes and determines the body plan? a. The neural tube b. Epidermis/Mesoderm junctions c. The primitive streak d. Presomitic mesoderm

c. The primitive streak

What occurs in neurulation? a. The neural tube forms the neural plate b. Epithelial folding of the neural plate leads to the formation of the neural tube c. A region of mesoderm thickens to form the neural plate d. The neural tube results from convergent extension

b. Epithelial folding of the neural plate leads to the formation of the neural tube

What is a feature of neural tube defects? a. An open anterior neuropore results in spina bifida b. They affect 1 in 100 pregancies phaly d. They can involve mutations in genes in the PCP pathway

d. They can involve mutations in genes in the PCP pathway | b. wrong, 1/1000

What is not a pathway associated with neural tube defects? a. Wnt/B-catenin pathway b. PCP pathway c. Hedgehog pathway d. TGFB/BMP pathway

a. Wnt/B-catenin pathway

What is not a feature of neural crest cells? a. They undergo SLUG dependent EMT b. They arise from the region between the neural plate and epidermis c. They form epithelial cells which insulate the spinal cord d. They are called the “fourth germ layer” due to their ability to differentiate to many cell types

c. They form epithelial cells which insulate the spinal cord

What is false about Waardenburg Syndrome? a. The phenotype is hypo-pigmentation and deafness b. Mutations to the MITF gene are dominant negative c. The mutation can be mapped to the long arm of chromosome 3 or 13 d. The white forelock arises due to a failure in melanocyte migration

c. The mutation can be mapped to the long arm of chromosome 3 or 13 (just 3)

Which mutation could not cause a Waardenburg Syndrome phenotype in mice? a. BMP- b. KIT- c. SLUG- d. MITF-

a. BMP-

What is a feature of neural stem cells? a. They stretch from the basal side of the neural tube to the apical side b. They exist in a multi layered epithelium c. The nuclei move up and down in the ventricular zone during the cell cycle d. Symmetric divisions help to generate diversity

c. The nuclei move up and down in the ventricular zone during the cell cycle

How is the neural tube patterned? a. BMP4 localises at the notochord and floor plate b. The local level of BMP4 and Shh determines neuron differentiation c. Shh is expressed in the epidermis and roof plate d. Neuron differentiation solely depends on BMP4 levels

b. The local level of BMP4 and Shh determines neuron differentiation

What is a feature of HPE? a. It is a structural anomaly of the developing hind brain b. The phenotype is always severe disfigurement c. 1:250 live births are impacted d. It results from incomplete cleavage of the forebrain

d. It results from incomplete cleavage of the forebrain

What is a feature of Spondylocostal Dysostosis (SCDO1)? a. It results in incomplete cleavage of the forebrain b. It can be caused by a mutation in DLL3 which is involved in the notch pathway c. The disease is usually mild and leads to malformed muscle tissue d. The disease cannot occur in vertebrates

b. It can be caused by a mutation in DLL3 which is involved in the notch pathway

What happens during somitogenesis? a. Notch signalling is turned on and off up the spine in a wave like pattern b. Mesodermal cells undergo an EMT transition c. Somites form in the region where notch activation begins d. Notch degrades Lunatic Fringe proteins

a. Notch signalling is turned on and off up the spine in a wave like pattern

How can notch signalling travel like a wave up the spin during somite formation? a. Once notch is activated, Lfng is activated and turns off notch b. The notch protein degrades quickly following translation c. Lunatic Fringe is repressed by notch activation d. Notch and Lfng interact in a linear pathway

a. Once notch is activated, Lfng is activated and turns off notch

What is not a feature of DCC? a. A lack of DCC means that neurons cannot cross the midline and connect with motor neurons on the same side b. It is a receptor for the diffusible axonal chemoattractant Netrin1 c. Neurons expressing DCC extend to the floor plate where netrin is expressed d. When growing towards the source of Netrin, neurons do not cross the midline

d. When growing towards the source of Netrin, neurons do not cross the midline

• Grafting the primitive streak from one chick embryo to another can lead to the formation of a second body axis.

• Neural epithelial cells divide at the basal surface of the neural tube.

• Shh helps to pattern the neural tube by localising at the epidermis and roof plate.

• The ventral midline is specified by the sonic hedgehog pathway.

• Bilateral structures, such as the eye field and nostril, fail to separate when the Shh activator Jervine is introduced to mouse embryos.

• Congenital mirror movements disorder can be caused from a mutation in the DCC gene where the splice donor sequence is mutated, exon 6 is skipped and a frame shift mutation results in an early stop codon.

Which type of cancer is correctly matched with its tissue of origin? a. Melanoma forms in blood cells b. Sarcoma forms in connective tissue c. Carcinoma forms in the nervous system d. Leukaemia forms in the lymphatic system

b. Sarcoma forms in connective tissue

What is not a hallmark of cancer? a. Sustaining proliferative signalling b. Invade and metastasise c. Replicative immortality d. Promote cell death

d. Promote cell death

What is senescence? a. A cell is alive but not actively proliferating and in an irreversible state of arrest b. A cell is actively proliferating to increase the cell population c. A cell is exhibiting increased cell division and decreased apoptosis d. A cell is expressing at least 3 of the hallmarks of cancer

a. A cell is alive but not actively proliferating and in an irreversible state of arrest

What is a feature of an oncogene? a. It’s normal action is to prevent the formation of a cancer cell b. They usually promote differentiation and cell death c. They usually promote cell proliferation d. Examples are p53, Rb and ECadh

c. They usually promote cell proliferation

What is a feature of a tumour suppressor gene? a. Ras and Myc are well known examples b. They can limit cell proliferation and promote cell death c. They are often involved in promoting motility and cell growth d. Only one allele needs to be mutated for a cell to become cancerous

b. They can limit cell proliferation and promote cell death

What happens in Burkitt’s Lymphoma? a. There is a T(8;14)(q24;q32) double deletion b. Translocation of the myc gene results in it being regulated by the Ig heavy chain enhancer c. There is a fusion between BCR and the c-abl oncogene d. Myc expression is down-regulated due to a premature stop codon

b. Translocation of the myc gene results in it being regulated by the Ig heavy chain enhancer

What happens in chronic myeloid leukaemia? a. C-ABL and BCR fuse and oligermisation leads to constitutive activation of the ABL tyrosine kinase domain b. A T(9;22)(q34;q11) translocation leads to ABL being constitutively activated by the BCR enhancer c. A chimeric c-ABL-bcr protein is produced with down-regulated tyrosine kinase activity d. The constitutively active BCR-ABL onco-protein drives apoptosis of hematopoietic cells

a. C-ABL and BCR fuse and oligermisation leads to constitutive activation of the ABL tyrosine kinase domain

What is most likely to cause thyroid cancer? a. Gene amplification leading to the overproduction of a normal protein b. Fusion of actively transcribed gene rpdoicts to produce a hyperactive fusion protein c. A deletion or point mutation in coding sequence that leads to a hyperactive protein made in normal amounts d. Chromosome rearrangement that causes a gene to be regulated by a constitutive enhancer

c. A deletion or point mutation in coding sequence that leads to a hyperactive protein made in normal amounts

What is involved in the Knudson two hit model? a. Loss of heterozygosity occurs following the first hit/mutation of a tumour suppressor gene b. Inactivation of a tumour suppressor gene can only occur as a result of genetic mutations c. Mutations to tumour suppressor genes are gain of function d. Both alleles of a tumour suppresser gene must be mutated for cell transformation to occur

d. Both alleles of a tumour suppresser gene must be mutated for cell transformation to occur

What are Rb and p53? a. Rb is a oncogene whilse p53 is a tumour suppressor gene b. Both are tumour suppressor genes which regulate the cell cycle c. Rb promotes apoptosis and DNA repair whilst p53 regulates the cell cycle d. Both are proto-oncogenes that encourage cell proliferation

b. Both are tumour suppressor genes which regulate the cell cycle

What does Rb do? a. It phosphorylates and actives E2F to promote S phase b. When a faulty copy is inherited, children usually go on to form a tumour in one eye c. It is phosphorylated by G1-CdK which leads to the activation of E2F d. An absence of Rb would result in cell cycle arrest

c. It is phosphorylated by G1-CdK which leads to the activation of E2F

What is a feature of p53? a. P53 mutations are only associated with rare cancers b. It encodes a DNA binding protein that controls genes for apoptosis and DNA repair c. It can prevent senescence and apoptosis d. It leads to the expression of genes that lead to telomere shortening

b. It encodes a DNA binding protein that controls genes for apoptosis and DNA repair

How can HPV cause cancer? a. The viral E6 protein binds to p53 and encourages its degradation b. Viral E7 proteins sequester and inhibit Rb in the cytoplasm c. The viral E6 protein ubiquitinates Rb and encourages its degradation d. Viral E7 is a proteolytic enzyme that cleaves p53

a. The viral E6 protein binds to p53 and encourages its degradation

Which statement about mutation patterns is correct? a. Most oncogene mutations are truncating b. Oncogenes and tumour suppressor genes follow identical patterns of mutation c. Truncating mutations are rarely found in tumour suppressor genes d. In oncogenes, repeated mutations are often found at a key amino acid residue

d. In oncogenes, repeated mutations are often found at a key amino acid residue

What is a feature of Li-Fraumeni syndrome (LFS)? a. It results from loss of function germline mutations inherited in both p53 alleles b. Females and males with LFS have a 100% chance of developing some form of cancer c. Cancer predisposition results due a germline mutation in one copy of p53 d. Males with LFS are more likey to develop cancer as p53 is an X linked gene and they only have one copy

c. Cancer predisposition results due a germline mutation in one copy of p53

• Cancer is a genetic and cellular disease.

• About 80% of human cancers are sarcomas.

• Angiogenesis is when the tumour mass is invaded with blood vessels and leads to the cancer receiving nutrients.

• The promotion of DNA repair systems results from the action of tumour suppressor genes such as p53.

• For a mutation in a proto-oncogene to result in cancer both alleles must be mutated and this is referred to as the Knudson two hit theory.

F (Tumour suppressor gene)

• Myc promotes cell proliferation and angiogenesis.

• The majority of p53 mutations occur in the DNA binding domain.

What is involved in metastasis of the liver? a. Cells grow as a benign tumour in capillaries b. Micro-metastasis occurs in the blood stream c. After traveling in the blood, cells can adhere to the blood vessel wall in the liver d. Cells escape blood vessels to colonise the epithelium

c. After traveling in the blood, cells can adhere to the blood vessel wall in the liver

How can hypoxia link to angiogenesis? a. Cells on the boundaries of tumours become starved of oxygen and are sensed by HIF1a b. In hypoxic conditions, VEGF is activated and induces vesicular sprouting c. When oxygen level are high, HIF1a builds up and promotes the transcription of VEGF d. A lack of oxygen leads to the degradation of HIF1a

b. In hypoxic conditions, VEGF is activated and induces vesicular sprouting

What supports the ability of circulating tumour micoembolis to metastasise? a. In mice, multi coloured CTM clusters were only present in 2% of the blood yet caused over 50% of metastatic events b. CTM clusters directly activate VEGF to induce sprouting of vessels c. In mice multi coloured CTM clusters were present in 50% of the blood yet caused only 2% of metastatic events d. CTCs are much more common in secondary tumours and CTMs are more common in primary tumours

a. In mice, multi coloured CTM clusters were only present in 2% of the blood yet caused over 50% of metastatic events

What is a feature of the CellSearch system? a. It is used to detect cancer stem cells and transit amplifying cells b. It recognises cytoplasmic molecules that specifically mark CTCs c. It can only be used to detect CTCs from carcinomas as it detect epithelial cell surface markers d. Patients with high CTCs had higher survival rates than those with low CTC levels

c. It can only be used to detect CTCs from carcinomas as it detect epithelial cell surface markers

What is a feature of chromosome instability? a. It refers to having an abnormal number of chromosomes b. It can occur due to mutations in genes involved in the DNA repair response c. It often results in polyploidy d. It is a specific term used to describe gaining chromosomes via translocation

b. It can occur due to mutations in genes involved in the DNA repair response

What kind of tissue is most susceptible to cancer? a. Critical organs that do not undergo self-renewal b. Epidermal tissue which doesn’t contain stem cells c. Tissue that generates blood cells involved in immunity d. Self-renewing tissues that contain stem cells

d. Self-renewing tissues that contain stem cells

Which statement about cancer stem cells is true? a. They can divide indefinitely but do not have the capacity to self-renew b. They divide quickly and are therefore resistant to chemotherapy c. They provide an explanation for relapse as they survive cancer treatment d. Drugs that target a solid tumour will also remove CSCs

c. They provide an explanation for relapse as they survive cancer treatment

What transcription factor promotes metastasis and invasiveness? a. DCC b. Ecadherin c. pVHL d. Twist

d. Twist

What does Twist do? a. It drives mesenchymal to epithelial transition (MET) b. It causes EMT of cells on the ventral side of the embryo c. It is only induced after embryonic development d. It is down-regulated in many cancers and used as a prognostic

b. It causes EMT of cells on the ventral side of the embryo

Which genes are often mutated in colorectal cancer? a. P53 mutations cause an overactive Wnt pathway b. Kras mutants cause a lack of DNA repair c. APC mutants destroy B-catenin and down-regulate the Wnt pathway d. P53 mutants lack efficient DNA repair

d. P53 mutants lack efficient DNA repair

What is a feature DCC? a. It is a netrin receptor that represses the growth of neurons b. It triggers apoptosis when netrin is absent c. There is a loss of heterozygosity of DCC in 20% of colorectal cancers d. When netrin binds DCC, a cell undergoes apoptosis

b. It triggers apoptosis when netrin is absent

How do netrin and DCC interact in the colon? a. Both are expressed at the vilus core where there are high levels of apoptosis b. Netrin is expressed in the crypt where new cells are produced c. DCC is degraded when epithelial cells reach the villus tip to signal apoptosis d. Netrin is absent in the crypt which leads to the apoptosis of epithelial cells expressing DCC

b. Netrin is expressed in the crypt where new cells are produced

How do BRAF mutations link to melanoma? a. Mutant BRAF can lead to constitutive activation of the MapK pathway b. Successful BRAF inhibitors often result in full remission c. Mutations in BRAF result in down regulation of cell DNA repair pathways d. Inhibiting constitutive BRAF selects for new repressing mutations in upstream MapK pathway components

a. Mutant BRAF can lead to constitutive activation of the MapK pathway

• Following EMT, tumour cells can undergo intravasation and become circulating tumour cells.

• Secondary tumour formation requires extravasation and MET.

• CTCs are more likely to cause metastasis than CTMs.

• Cancers are most likely to arise in childhood.

• Many cancers are due to mutations in genes involved in the DNA damage response.

• Netrin inhibits metastasis and DCC promotes metastasis.

• Cells that express their own netrin are capable of metastasis even in the presence of DCC.

What cannot be used to measure levels of gene expression? a. Microarray b. Southern Blotting c. RNAseq d. qPCR

b. Southern Blotting

What is a feature of gene expression in inner cell mass cells? a. Sox2 shows a bimodal expression pattern during the whole developmental period b. By day 4, all cells express gata6 c. Most genes are expressed stochastically in the early stage d. Gene expression in individual cells follows a strict pattern from the early stages of development

c. Most genes are expressed stochastically in the early stage

What can RNAseq show about glioblastoma? a. There are low levels of variability at the chromosome level b. Chromosome 7 is often gained c. Each tumour cell will have an identical RNAseq profile d. Chromosome 10 is often gained

b. Chromosome 7 is often gained

What is not used to construct gene ontology? a. Biological processes b. Amino Acid composition c. Cellular components d. Molecular functions

b. Amino Acid composition

Which best describes a technique used to analyse a list of genes? a. STRING shows the relationship between genes b. chEA detects orthologues in other mammals c. Enrichr detects common transcription factors that interact with all of the listed genes d. STRING and Enrichr detect pathways likely to interact with the genes

a. STRING shows the relationship between genes

• Nanog is a crucial element of a stem cell and is central in regulating the network of genes that mark pluripotency.

F (not essential)

• In vitro, intestinal stem cells that express LGR5 develop into organoids which form crypt like structures and differentiated cell types like paneth cells.