MST 2 Revision- Lectures 10-21 Flashcards
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
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
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
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
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
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
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
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
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
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
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
• To study chromosomes, you must use dividing cells.
T
• FISH can label whole chromosomes but not single segments.
F
• Karyograms show a set of banded chromosomes and Karyotypes show the chromosome complement of an individual.
T
• Most karyograms relate to Q banding.
F
• The long arm of the chromosome is called the P arm.
F
• Chromosomes are divided into regions and numbered out from the centromere.
T
• A satellite site is a non-staining gap in a chromosome that is inherited according to Mendelian inheritance.
F (fragile)
• Paracentric inversions include the centromere.
F
• Inversions are caused by recombination events within the inversion loop.
T
Which is the most significant cause of cancer?
a. Infection
b. Inheritance
c. Diet
d. Pollution
c. Diet
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
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
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
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.
F
• Familial Adenomatous Polyposis is due to a mutation in one gene and often results in prophylactic bowel removal.
T
• 10-15% of referrals are knocked back for genetic counselling appointments.
T
• A BOADICEA score is the only factor that needs to be analysed when assessing risk for genetic testing.
F
• A blood sample can take 2 months to undergo genetic testing.
T
• 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.
T
• 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%.
T
• Bovines have 1274 transcribed Y chromosome genes which are mostly expressed during testis development.
T
• The male phenotype is the default pathway in humans.
F
• XY females and XX males can result from a crossover that transfers the SRY locus to the X chromosome.
T
• Homology between mammals is greatest at the level of DNA.
F
• For the first 5 weeks, there is no detectable morphological difference between male and female destined embryos.
T
• Male reproductive tracts develop from the Mullerian duct and female reproductive tracts develop from the Wolfian duct.
F
• Secondary sex characteristics develop at puberty for both males and females.
T
• The SRY protein product is much smaller in mice (204AA) than in humans (395AA).
F
• Removing the glutamine repeat region in the mouse prevents the transcription of Sry, yet testis still develop.
F
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.
T
• Unlike mammals, the sexual phenotype of the gonads and body of drosophila are controlled by the same genes.
T
• Y chromosome genes have male and female determining effects.
F
• X chromosome genes impart their effects after the onset of X inactivation when there is a 1:1 X chromosome ratio in females.
F
• XIST inactivates genes that cause female somatic cells to be different to male cells and the barr body is produced.
T
• Midbrain cells between sexes exhibit different levels of tyrosine hydroxylase expression only after gonads begin making hormones.
F
• Sex determining genes do not have other roles in development.
F
• Sertoli cells can only be XY and the absence of Sry leads to them developing as pregranulosa (ovarian) cells instead.
T
• NR5A1 inhibits binding of SRY to target DNA such as WDR5.
F
• SRY and NR5A1 bind to the TES of SOX9 and cofactors WDR5 and CITED2 to upregulate SOX9.
T
• SOX9 is involved in feedback regulation to block SRY and auto-regulate itself.
T
• 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.
T
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