Molecular Genetics 3 - Lecture Quiz's
1 Chromosomes are numbered according to size, numbering [1] first, and arms are labelled either “p” or “q” with p being the [2] arm. Chromosomes containing arms of equal size are labelled [3], while chromosomes with two arms that are slightly unequal in length are known as [4]. A chromosome with one very short arm is termed [5], where the [6] is located at one extreme end.
[1] smallest, [2] largest, [3] acrocentric, [4] submetacentric, [5] metacentric, [6] centromere
[1] largest, [2] smallest, [3] metacentric, [4] submetacentric, [5] acrocentric, [6] telomere
[1] largets, [2] smallest, [3] metacentric, [4] submetacentric, [5] acrocentric, [6] centromere
[1] smallest, [2] largest, [3]metacentric, [4] submetacentric, [5] acrocentric, [6] centromere
The image below shows fluorescence in situ hybridisation (FISH) to detect the Philadelphia chromosom (Ph), which causes chronic myelogenous leukemia (CML) and results from a reciprocal translocation between 9q and 22q. In this FISH experiment a red fluorescently labelled DNA probe that can hybridize to 9q and a green labelled DNA probe than can hybdridise to 22q were used. Fluorescence signal from the two DNA probes that are close together can fuse to form one dot. Using the image below as a reference alongside your understanding of chromosome nomenclature and FISH, select ALL the correct statements from the list below. Note: partial credit will be awarded for each correct answer, while points will be detracted from incorrect choices.
In a normal somatic cell during metaphase, I would expect to see a maximum of 1 green and 1 red dot.
In a normal somatic cell during metaphase, I would expect to see a maximum of 2 green and 2 red dots.
In a normal somatic cell during metaphase, I would expect to see a maximum of 4 green and 4 red dots.
The 2 green dots that are very close to each other represent DNA from two homologs.
Without translocation, chromosome 9 should be larger than chromosome 22.
Without translocation, chromosome 22 should be larger than chromosome 9.
FISH from the figure above suggests that the red DNA probe can only detect the exchanged are of 9q.
Both the exchanged and non-exchanged areas of 22q are detected by the green DNA probe used for FISH in the figure above.
The figure above suggests that the red DNA probe can only detect non-exchanged are of 9q using FISH.
The 2 red dots that are very close to each other represent DNA from two sister chromatids.
The red and green dots can only co-localise in the philadelphia chromosome
Starting with a visual representation of the human karyotype (2n= 46), it is possible to reconstruct a similar representation of the cat karyotype (2n= 38) by making 7 autosomal chromosome breaks (each producing 1 extra block) and 11 autosomal chromosome fusions (combining two pre-existing blocks). How many major conserved syntenic blocks are shared between the female human and cat karyotypes? (Note that we are also ignoring any smaller changes and inversions).
Rare intrachromosomal rearrangement takes place as a result of homologous recombination between two copies of the Alu repeat, flanking sequences 1-2-3-4. In allele (I) the SINE sequences are oriented in the same direction whereas in allele (II) they have opposite orientation to each other (see figure below). Dotted lines signify the upstream (X) and the downstream (Y) regions.
After recombination between the Alu sequences within allele I the sequence 1-2-3-4 will be [1].
After recombination between the Alu sequences within allele II, the sequence 1-2-3-4 will be [2].
[1] duplicated, [2] inverted
[1] deleted, [2] translocated
[1] inverted, [2] duplicated
[1] deleted, [2] inverted
Shark populations are being severely depleted due to the worrying practice of 'finning' (cutting the fins off live sharks and leaving them to die) to make shark fin soup. Local conservation work aims to analyse DNA from the removed fins to identify the species under attack. Which method of genome analysis is likely to be most informative.
Sequencing of SINE elements.
Sequencing of CpG islands.
Counting microsatellite repeat number.
Analaysing minisatellite length.
Sequencing the most highly conserved genes.
Minisatellite analysis of one individual was carried out using a multilocus DNA probe that can detect three minisatellites: (a), (b) and (c). The DNA was digested with an enzyme that conveniently (surprisingly) cut each minisatellite locus exactly 50 bp away from both ends (50 bp upstream and 50 bp downstream). The digested DNA fragments were separated by agarose gel electrophoresis and probed with the multilocus probe, showing that the individual had all 3 minisatellites shown here. However, the individual was heterozygous for minisatellite loci (a) and (c) and homozygous for locus (b). For loci (a) and (c), this person's second allele had 2 extra minisatellite copies, resulting in a total of 20 and 16 repeat units, respectively.
Select ALL the band sizes you would expect to detect from this individual's DNA fingerprinting using this multilocus probe.
238
272
288
322
338
372
464
516
564
594
644
660
694
710
760
1128
Select all of the chromatin types or modifications that are broadly considered to be activating or associated with open chromatin.
Heterochromatin
H3K9 acetylation
H3K9 tri-methylation
H3K4 tri-methyltation
Histone acetyltransferases (HAts) enrichment
Histone deacetylases (HDACs) enrichment
CpG methylation
Xist RNA decoration
Cfp1 enrichment (Cfp1 recognises unmethylated CpG)
H3K27 acetylation
An antibody that can specifically recognize trimethylated H3K4 residues (H3K4me3) and Immunoglobulin G (IgG) antibody, which does not exhibit specific interaction with H3K4me3, were used to precipitate fragmented human chromatin in a ChIP experiment. Precipitated DNA was analysed by qPCR at two distinct loci, one within centromeric heterochromatin and the other within the promoter of an active gene. Match the results shown for samples numbered 1-4 with their likely explanation.
Sample 2
Gene promoter, H3K4me3 enrichment
Sample 3
Centromeric sequence, negative control
Sample 4
Centromeric sequence, H3K4me3 enrichment
Sample 1
Gene promoter, negative control
Human chromosomal DNA was digested either with restriction enzyme HpaII or MspI. Both enzymes recognise and cut the sequence CCGG but DNA methylation inhibits HpaII activity. A cosmid DNA probe was used to specifically detect a 10 kb DNA fragment generated by NotI digestion, whose map is shown below (no internal NotI sites within this fragment). There are 2 genes within this region both of which have a CpG island at their promoters. CCGG sites across the region are shown as blue vertical lines. DNA fragments can be classed as tiny (100 bp or less), medium (~1-2 Kbp) or larger fragments of 8 Kbp or 10 Kbp. Assuming that all CpGs are methylated in non CpG island, finish the following two statements by typing in one of possible answers below.
MspI digestion is expected to generate [1] detected by the cosmid probe.
HpaII is expected to generate [2] detected by the cosmid probe.
[1] tiny and medium fragments, [2] tiny and 8kb fragments
[1] only the 10kb fragment, [2] only medium fragments
[1]tiny and medium fragments, [2] tiny and 10kb fragments
[1]only 8kb fragments, [2] all fragments
Please identify the factor that carries out the following roles/functions:
Recognises the BRE element in DNA at the core promoter
TFIIB
Enables interactions between transcriptional activators and RNA polymerase II
TFIIH
Acts as a helicase to unwind DNA at the core promoter
Mediator
Recognises TATA sequence in DNA at the core promoter
TFIID
Please answer the following questions about the schematic shown below (the scale is shown below the schematic).
Region A-C are found (i) upstream or ii) downstream) of the transcriptional start site. Region A could contain a recognition element for (i) activator, ii) repressor or iii) either activator or repressor). Region C is within the core promoter, approx. 30 nt from the transcription start site (TSS). What element (motif) would be expected to be found here? (i) exon, ii) intron, iii) enhancer, iv) TATA box).
I,iii,ii
Ii,i,iV
Ii,iii,iV
I,iii,iV
In order to use a footprinting assay to determine the binding site of a protein on a DNA, you must have an antibody against the protein of interest.
True
False
Electrophoretic mobility assays use a native (non-denaturing) gel in order to detect protein binding to the DNA.
True
False
Check all the factors that interact with the CTD tail of RNA polymerase II:
CPSF
U1 RNA
MicroRNAs
Capping enzyme
The below wt DNA sequence encodes a messenger RNA with 3 exons that can be alternatively spliced to result in 2 different mature messenger RNAs. However, some people carry mutations in the DNA that abolish the 3’splice site of the first intron (shown with “X”).
A - Which two mRNAs would be generating from splicing of the wt DNA?
B - Which mRNA would be generated from splicing the DNA with mutation 'X'?
A - C, B - C
A - B, B - C+A
A - A+C, B - C
A - A+B, B - A
MicroRNAs can be encoded within the introns of messenger RNA genes and they get processed by the microprocessor in the nucleus
True
False
MicroRNAs regulate messenger RNAs in the cytoplasm by binding directly to sequences in the messenger RNA that are complementary to the microRNA
True
False
Which factors promote messenger RNA stability and which promote decay?
5'Cap
Stability
Exosome
Decay
MicroRNAs
Stability
HuR
Decay
Reporter assays can be used to measure the ½ lives of a messenger RNA in a cell
True
False
Regulatory motifs that effect a messenger RNAs half-lives are commonly found in introns
True
False
A messenger RNA encoding gene X is found to increase 100-fold when a cell is infected with a virus. Which scenario could explain how this happens:
A transcriptional repressor of gene X is induced upon virus infection
A transcriptional activator of gene X is induced upon virus infection
A microRNA is induced upon virus infection that regulates gene X through interactions with the 3'UTR
TTP is induced upon virus infection that binds to motifs in the the 3’UTR of gene X
Choose ALL correct statements
Cohesin forms a ring-like structure embracing a pair of homologous chromosomes.
Cohesin is destroyed at onset of anaphase in mitosis by a pulling force generated from microtubule depolymerisation.
The synaptonemal complex is responsible to hold homologous chromosomes together during meiotic metaphase I.
Cohesin molecules between the centromere and the chiasma are responsible for holding homologous chromosomes together during meiotic metaphase I.
Some cohesin molecules are still maintained on chromosomes in meiotic anaphase I.
Cohesin molecules in the centromere regions are important to hold sisterchromtids together in meiotic metaphase II.
You examine oocytes in the progeny (F1) of a cross between two homozygous strains (A b and a B). Which diagram shows a possible arrangement of chromosomes in meiotic prophase I in the oocytes? (see figure in folder if it is not showing within the question). A and a (or B and b) are alleles on the same locus. Each line represents a sister-chromatid. Circles represent centromeres which connect a pair of sister-chromatids.
A
B
C
D
E
F
G
H
I
If 12% of the prophase I oocytes have one crossover between the two loci as shown above, what percentage of unfertlised eggs will have the genotype A b? Type in the nearest whole number without percentage.
To explain maternal age effect, which observations or assumptions is the cohesin fatigue hypothesis based on? Choose ALL correct answers.
All oocytes are generated before the birth of a mother.
An arrest in metaphase II in oocytes lasts longer when the mother becomes older.
Cohesin rings that hold sisterchromatids are established only during S phase.
The number of crossovers between homologous chromosomes increases with the mother's age.
The number of chiasma in oocytes decreases with the mother's age.
The synaptonemal complex, which contains cohesin, is slowly decayed with time.
The chromosome constitution of a Down syndrome child and both of his parents was determined and is shown in the diagrams below. Nine molecular markers are shown in order from the centromere at the top of the diagram to the telomere at the foot. The numbers identify alternative alleles at various microsatellite and minisatellite loci along the chromosome. + and – are alternative alleles at restriction sites on the chromosome.
When did a chromosome missegregation error occurred to generate trisomy?
the first meiotic division in the mother
the second meiotic division in the mother
the first meiotic division in the father
The second meiotic division in the father
a mitotic division in the fertilised egg
Account for the chromosomes found in the Down syndrome child (Figure in question 2) by re-constructing the meiotic events that gave rise to the disomic gamete. At minimum, how many crossovers took place during recombination?
0
1
2
3
4
5
You have isolated DNA from an unknown type of bacteriophage and digested this with the restriction enzyme Sau3A. (Sau3A recognises the sequence 5’/GATC3’, where / denotes the strand cleavage position). You then ligate the digested DNA, which will probably contain many different Sau3A restriction fragments, into the BamHI site of a plasmid vector. (BamHI recognises the sequence 5’G/GATCC3’, where / denotes the strand cleavage position). After transfection of the ligation into E. Coli you screen for colonies containing a plasmid vector with an insert by checking the size of uncut plasmid DNAs compared to the uncut starting plasmid vector DNA to identity those that are larger in size. You now decide to recover the restriction fragment insert from these larger plasmids by digestion with BamHI. Assuming that the DNA you isolated has a random sequence with equal composition of each of the four nucleotides, what fraction of the plasmids recovered with an insert do you predict will be cut to result in an excised BamHI fragment containing bacteriophage DNA free from the vector DNA?
100%
50%
25%
12.5%
6.25%
Long range PCR can enable amplifications of up to and over approximately 25 Kb to be routinely performed. This PCR typically uses a blended mixture of two thermostable DNA polymerases: Taq DNA polymerase, which is capable of very fast elongation rates, but has no proof reading activity (a 3’ to 5’ exonuclease activity); and a thermostable polymerase with an efficient proof reading activity but a slower elongation rate. Neither DNA polymerase alone is capable of such long distance PCR amplification. Select the statement below that correctly explains why the blended mixture works:
Taq DNA polymerase is fast but not stable for long enough to remain active over the long periods required for DNA polymerisation of 25 kb distances. The other DNA polymerase is more stable and assists the elongation process as the Taq DNA polymerase activity declines during later stages of the PCR.
Over long distances there is more chance of Taq DNA polymerase stalling in regions of secondary structure of the single strand DNA template, whereas the other DNA polymerase is less susceptible to this problem of secondary structure and can take over the elongation process in these regions.
Taq DNA polymerase can extend the 3’ end of an incorrectly annealed primer even if it is unstably annealed to the template DNA, and over many cycles this mispriming expends enzyme activity on non-productive polynucleotide chain elongations. The other DNA polymerase removes the incorrectly annealed primers using its 3’ to 5’ exonuclease activity and these primers then no longer bind to the template to cause mispriming.
During DNA polymerisation, if nucleotides are misincorporated into the elongating polynucleotide chain the Taq DNA polymerase cannot use the resulting improperly base paired 3’ end as a substrate for continuing DNA synthesis, and this therefore pauses or prohibits further chain elongation. The other DNA polymerase can removes the misincorporated nucleotide at the 3’ end with its proof reading activity, and Taq DNA polymerase can then recommence polynucleotide chain elongation.
You are making a large complex plasmid vector containing an ampicillin resistance gene plus other genes of interest with a total size of 8 Kb, and you need to add to this a further genetic component in the form of a 2 Kb BamHI restriction fragment. The plasmid already contains several BamHI sites, so inserting the BamHI fragment into one of these sites will be technically difficult. You identify a single NotI site in the plasmid at a suitable location, and therefore you decide to insert the BamHI fragment into this position. To do this, you decide to add adapters to the ends generated by NotI digestion so that these will be complementary to the cohesive ends on the BamHI fragment. (NotI recognises and cuts at the sequence 5’GC/GGCCGC3’ where / indicates the strand cleavage position; BamHI recognises and cuts the sequence 5’G/GATCC3’ where / indicates the strand cleavage position). You cut the plasmid with NotI and then efficiently modify the ends with T4 DNA polymerase plus dNTPs to make them blunt by filling in the recessed 3’ end. You design two oligonucleotides with complementary sequence that when annealed will make an adapter with a single strand 5’ protrusion complementary to the cohesive end generated by BamHI digestion: Oligonucleotide 1 – 5’OH-GATCCCCGGC-3’OH (Note the absence of a P at the 5’ end) Oligonucleotide 2 - 5’P-GCCGGG-3’OH (P = phosphate group; OH = hydroxyl group) You mix the oligonucleotides 1 and 2, anneal them and ligate to the NotI cut plasmid DNA (now with blunt ends). This reaction is very efficient as it is driven by a high concentration of the annealed oligonucleotides so that 100% of ends get ligated to the adapter. You then ligate the plasmid DNA with added adapters (the vector) to the BamHI fragment (the insert); this is called the vector plus insert ligation. You also set up a ligation of the vector alone without the insert; this is called the vector-only ligation. You transfect competent E. Coli cells with these ligations and spread cells onto ampicillin plates. Choose which one of the following predictions below will be correct:
There will be colonies from both the vector plus insert ligation and from the vector-only ligation. On analysis of the plasmid DNAs of colonies from the vector plus insert ligation some of these plasmids will be cut with NotI to give just a 8 Kb restriction fragment consistent with the size of the original plasmid, and some plasmids will be cut with NotI to result in a 2 Kb restriction fragment in addition to a restriction fragment of 8 Kb. Plasmids of colonies from the vector-only ligation will all be cut with NotI to produce only a 8 Kb restriction fragment and no 2 Kb restriction fragment. Plasmids of colonies from the vector-only ligation will contain the same number of BamHI sites as in the starting plasmid. Those plasmids of colonies from the vector plus insert ligation that give a 2 Kb NotI fragment will have a new size of BamHI fragment that is 2 Kb larger than one of the original BamHI fragments.
There will be colonies from the vector plus insert ligation, but no colonies from the vector-only ligation. On analysis of the plasmid DNAs of colonies from the vector plus insert ligation all the plasmids will be cut with NotI to result in a 2 Kb restriction fragment and a 8 Kb restriction fragment; but none of the plasmids digested with BamHI will result in a 2 Kb restriction fragment, and instead one of the BamHI restriction fragments of the original plasmid will increase by 2 Kb.
There will be colonies from the vector plus insert ligation but no colonies from the vector-only ligation. On analysis of the plasmid DNAs of colonies from the vector plus insert ligation all the plasmids will be cut with NotI to result in 2 Kb and 8 Kb restriction fragments; all the plasmids digested with BamHI will also result in a new 2 Kb restriction fragment, but there will be a BamHI fragment in the original plasmid missing and replaced by two new BamHI fragment sizes.
There will be colonies from the vector plus insert ligation but no colonies from the vector-only ligation. On analysis of the plasmid DNAs of colonies from the vector plus insert ligation none of the plasmids will be cut with NotI, but uncut DNAs will appear larger than the original plasmid DNA; all the plasmids digested with BamHI will result in a new 2 Kb restriction fragment, but there will be a BamHI restriction fragment in the original plasmid missing and replaced by two new BamHI restriction fragment sizes.
You have isolated a new gene, possibly belonging to the SET-domain containing methyl transferases (HTMases) family. To understand if your new gene encodes an active enzyme, you want to mutate the conserved Histidine (H), which normally collaborates with the catalytic Tyrosine (T) within the active site, into an Alanine (A); see nucleotide and amino acid sequences below:
L10 Q1
his new SET gene is inserted into an appropriate plasmid expression vector, which is amplified in E.coli, and the plasmid DNA is then used as a template for PCR-based mutagenesis. Choose the answer below summarising the correct combination of steps necessary in order to obtain pure plasmid DNA encoding for the Histidine -to-Alanine mutated SET protein.
A PCR is set up using the following primer pair, together with the template plasmid encoding for the wild type SET gene: 5’-GCG GAT CTT ATT AAC GCG AGC GCG GGA GTG ACC ACC-3’ together with 5’-CGC CTA GAA TAA TTG CGC TCG CGC CCT CAC TGG TGG-3’ The PCR product is subsequently digested with the restriction enzyme DpnI, and the whole reaction mix is then transfected into E. Coli to amplify the mutated plasmid.
A PCR is set up using the following primer pair, together with the template plasmid encoding for the wild type SET gene: 5’-GCG GAT CTT ATT AAC GCG AGC GCG GGA GTG ACC ACC-3’ together with 5’-GGT GGT CAC TCC CGC GCT CGC GTT AAT AAG ATC CGC-3’ The PCR product is subsequently digested with the restriction enzyme DpnI, and the whole reaction mix is then transfected into E. Coli to amplify the mutated plasmid.
A PCR is set up using the following primer pair, together with the template plasmid encoding for the wild type SET gene: 5’-GCG GAT CTT ATT AAC GCG AGC GCG GGA GTG ACC ACC-3’ together with 5’-CGC CTA GAA TAA TTG CGC TCG CGC CCT CAC TGG TGG-3’ The PCR product is subsequently digested with the restriction enzyme MboI, and the whole reaction mix is then transfected into E. Coli to amplify the mutated plasmid.
PCR is set up using the following primer pair, together with the template plasmid encoding for the wild type SET gene: 5’-GCG GAT CTT ATT AAC GCG AGC GCG GGA GTG ACC ACC-3’ together with 5’-GGT GGT CAC TCC CGC GCT CGC GTT AAT AAG ATC CGC-3’ The PCR product is subsequently digested with the restriction enzyme MboI, and the whole reaction mix is then transfected into E. Coli to amplify the mutated plasmid.
You have two mouse cell lines, one (A) that is sensitive to 6-thioguanine (6-TG), the other (B) that is resistant. Sensitive cells (A) convert 6-TG into a nucleotide form that is incorporated into the DNA, killing the cells. To identify the gene responsible for the metabolism of 6-TG into the nucleotide form, which strategy below would be the best choice to implement?
Perform a functional screen, by creating a genomic library from cell line A in a plasmid shuttle expression vector and transfect it into cell line B. The clones that will express the responsible gene will become sensitive to the presence of 6-TG in the growth medium.
Perform a functional screen, by creating a cDNA library from cell line A in a plasmid shuttle expression vector and transfect it into cell line B. The clones that will express the responsible gene will become sensitive to the presence of 6-TG in the growth medium.
Perform a functional screen, by creating a genomic library from cell line A in a cosmid vector and transfect it into cell line B. The clones that will express the responsible gene will become sensitive to the presence of 6-TG in the growth medium.
Perform a functional screen, by creating a cDNA library from cell line A in a cosmid vector and transfect it into cell line B. The clones that will express the responsible gene will become sensitive to the presence of 6-TG in the growth medium.
You want to construct a full length cDNA library in a λgt10 expression vector that contains a single EcoRI site for insertion of the cDNA embedded within the λcI repressor gene. After making double strand cDNA by reverse transcription of mRNA and second strand synthesis, what would you do? Select the correct option:
The cDNA will be treated with EcoRI methyltransferase and S-adenosylmethionine. EcoRI linkers will then be ligated onto the ends of the cDNA. The product will be cut with EcoRI and ligated into the EcoRI-cut vector. Ligated DNA will be packaged in vitro into λ phage heads and these used to infect E. Coli cells. Clear lytic plaques on a confluent lawn of E. Coli cells will indicate clones carrying a cDNA insert.
EcoRI linkers will be ligated onto the ends of the cDNA and the product of the ligation then treated with EcoRI methyltransferase and S-adenosylmethionine. The product will be cut with EcoRI and ligated into the EcoRI-cut vector. Ligated DNA will be packaged in vitro into λ phage heads and these used to infect E. Coli cells. Clear lytic plaques on a confluent lawn of E. Coli cells will indicate clones carrying a cDNA insert.
The cDNA will be treated with EcoRI methyltransferase and S-adenosylmethionine. EcoRI linkers will then be ligated to the ends of the cDNA. The product will be cut with EcoRI and ligated into the EcoRI-cut vector. Ligated DNA will be packaged in vitro into λ phage heads and these used to infect E. Coli cells. Growth of bacterial colonies will indicate clones carrying a cDNA insert.
EcoRI linkers will be ligated onto the ends of the cDNA and the product of the ligation then treated with EcoRI methyltransferase and S-adenosylmethionine. The product will be cut with EcoRI and ligated into the EcoRI-cut vector. Ligated DNA will be packaged in vitro into λ phage heads and these used to infect E. Coli cells. Growth of bacterial colonies will indicate clones carrying a cDNA insert.
Bone morphogenetic proteins (BMPs) are secreted during development to induce chondrocytes differentiation and promote the expression of cartilage-specific genes. One of these genes is Fgfr3. In order to study the dynamics of nucleosome remodelling on the Fgfr3 promoter upon BMP-2 stimulation, DNase I sensitivity analysis was performed. Cells are stimulated or not with BMP-2, nuclei are isolated and digested with increasing amounts of DNAse I. Genomic DNA is extracted, digested with HindIII, run on an agarose gel and Southern blotting is performed with a probe recognising Exon 2 of Fgfr3 to visualise the Fgfr3 promoter region.
What does the additional band appearing upon BMP-2 stimulation represents?
The removal of a nucleosome during chromatin remodelling to activate the promoter.
A transcription factor associating with the Fgfr3 promoter during the stimulation, changing the mobility of the DNA.
An additional DNA fragment generated by decreased DNA accessibility, due to the association of a transcription factor with the Fgfr3 promoter during the stimulation.
An additional DNA fragment, generated by increased DNA accessibility, due to the exposure of single strand DNA when transcription starts.
In Drosophila melanogaster the Polycomb complex PRC2 is responsible for tri-methylation of histone H3K27. How PRC2 is recruited to specific sites in the genome is under investigation. In this work, the authors analysed the role of the Zn-finger transcription factors Pho and Spps.
Results are shown above of analysis of DNA sequencing of chromatin immunoprecipitations (CHIP-seq) done for H3K27me3 and Pho for the region encompassing the exex locus in wild type (WT), Spps and Pho mutant larvae. Which conclusion(s) (A to D below) based on these CHIP-seq results is/are correct?
That Pho is required for the recruitment of Spps to the exex locus.
That Spps is required for the recruitment of Pho to the exex locus.
That Spps is required for tri-methylation of H3K27 at the exex locus.
That H3K27me3 is required for Spps and Pho binding to the exex locus.
These are three examples 1, 2 and 3 above of polonies (clusters) in a flow cell of an Illumina DNA sequencer slide. What could be the explanation for the significantly different densities of polonies in the three examples?
The relative efficiencies of the DNA sequence amplification step during polony generation in the three examples.
The relative efficiencies of the hybridization of the denatured library with the chemically bonded primers to initiate polony generation in the three examples.
The relative concentrations of the DNA fragment library initially loaded onto the slide in the three examples.
The relative efficiencies of the incorporation of dNTPs during the sequencing step in the three examples.
The DNA molecule X has been formed by resolution of two Holliday junctions by RuvABC. Where did RuvABC cleave the DNA intermediate as indicated by the triangles (select all possible answers)?
A
B
C
D
What molecule(s) may be created by site specific recombination between molecules X and Y. The arrows represent recombination sites and the star, circle and triangle are genetic markers (select all possible answers)?
A
B
C
D
E
Scientists isolated bacteriophages A that lyse E. coli strain K-1 but not K-9. They carried out a PCR of 6 kb and mapped the amplified fragment using restriction enzymes NotI, BamHI and HindIII. They observed the restriction pattern shown below on an electrophoresis gel.
Part 1: Type how many cutting sites for each enzyme are in the PCR fragment, based on the restriction pattern shown in the gel in Figure 1. Type your answer as a number: 1, 2, 3...
NotI
BamHI
HindIII
NotI - 1, BamH - 3, HindII - 1
NotI - 1, BamH - 1, HindII - 3
NotI - 1, BamH - 1, HindII - 1
Scientists carried out a heat-shock on bacteriophage A and noticed that some of the heat-shocked bacteriophages acquired the ability to also lyse E. coli strain K-9. They called this population bacteriophages H and carried out on bacteriophages H the same PCR and restriction mapping as on bacteriophages A. They observed the restriction pattern below on an electrophoresis gel.
What type of DNA sequence(s) should be present in this region to explain the result of this recombination event (select all possible answers)?
An origin of replication
A head-to-head repeat
A tus sequence
A G/C rich region
A head-to-tail repeat
A A/T rich region
Heteroduplex DNA
A Chi sequence
If a double digestion with NotI and BamHI of the PCR fragment from bacteriophages A results in the restriction pattern shown on the electrophoresis gel below, what would be the expected band sizes of the same digestion of the PCR fragment from bacteriophages H (select all possible answers)?
A band of double intensity at 1 kb and a band at 4 kb
A band at 0.5 kb, one at 1.5 kb and one at 4 kb
A band at 1 kb, one at 1.5 kb and one at 3.5 kb
A band at 0.5 kb, one at 1 kb and one at 4.5 kb
A band of triple intensity at 2 kb
A band of double intensity at 1.5 kb and a band at 3 kb
A band at 1 kb, one at 2 kb and one at 3 kb
What is/are the biochemical activitie(s) of RecBCD on a DNA molecule with double-strand end that does not contain a Chi site? Select all that apply
Double-strand DNA endonuclease (ATP-dependent)
5’ strand exonuclease
5’ ATP-dependent helicase
Double-strand DNA-dependent ATPase
3’ ATP-dependent helicase
Select all TRUE statements about the molecular process described in the figure below
The schematic represents the reversal of a replication fork
The DNA break is usually repaired by site specific recombination
RecBCD loads on the DNA double-strand end to start the repair process
RecA forms a nucleo-protein filament on the double-stranded DNA end
The repair of the DNA double-strand break will lead to replication re-start
Which one of the following statements is TRUE?
When replication needs to re-start away from OriC, DnaA is loaded spontaneously as the replication fork is already formed
When replication needs to re-start away from OriC, DnaC is loaded in a DnaA-dependent manner on a replication structure
When replication needs to re-start away from OriC, DnaB is loaded in a PriA-dependent manner on a replication or recombination structure
When replication needs to re-start away from OriC, DnaA is loaded in a PriA-dependent manner on a replication or recombination structure
When replication needs to restart from OriC, DnaC is loaded in a DnaA-dependent manner on a replication or recombination structure
Match the answers to the correct result:
RecBCD-dependent processing of a DNA double-strand end
Formation of a “chicken foot” and therefore a free DNA double strand end
PriA loading on a replication structure
Replication re-start
Replication fork reversal
Stimulation of loading of RecA on single strand DNA
Gap in DNA molecule
Replication fork collapse
What are the consequences of a ∆recA mutation (i. e. the RecA protein is not produced) on replication fork collapse and repair in E. coli? Select all true statements
The ∆ recA mutation will increase the frequency of replication fork collapse
The ∆ recA mutation will prevent RecBCD-dependent processing of DNA double strand ends formed after collapse
The ∆ recA mutation will prevent PriA-dependent re-start at collapsed forks
The ∆ recA mutation will stimulate repair by non homologous end-joining
What is/are the consequence(s) of a ∆ruvABC mutation (i. e. the RuvABC complex is not produced) on replication fork reversal and repair in E. coli? Select all true statements.
The frequency of replication fork reversal will be affected as the major pathway of fork reversal will not be active
The reversed forks will be repaired by RecA-dependent DNA degradation
The reversed forks will be cleaved and converted into DNA double-strand breaks
The reversed fork will be processed by RecBCD exonuclease activity to be returned to a normal configuration
Why is it generally thought that transposable elements have been maintained long-term within genomes?
TEs occasionally provide beneficial mutations
TEs provide a net benefit to the rest of the genome
TEs replicate more than the rest of the genome
TEs are not a net cost to the rest of the genome
TEs may be useful for the species in the future
As the number of TEs in a genome doubles, it is thought that the overall cost of TEs increases by more than double. Which source of TE-induced damage is most consistent with this hypothesis?
Mutations caused directly by insertion
Double strand breaks caused by excision
Insertion of new promoter sequences
Induction of local ectopic chromatin formation
Ectopic recombination between TEs
You have sequenced the genomes to two adult Drosophila, a male and a female, and the genomes of six of their virgin offspring. A sequence-based search identifies a new transposable element (TE) in many of these flies, but the putative TE is present only once in each parental genome. Other genetic variation is used to assign offspring chromosomes to their parent of origin, and you can assume that there is no variation in TE status among the somatic cells of each individual. The figure below shows the same two homologous autosome arms for each individual parent and offspring, with the position of the TE marked as a coloured box. The maternal copies of the chromosome arm are shown in blue (one light blue, one dark blue) and the paternal copies of the chromosome arm are shown in orange (one light orange, one dark orange). Quiz_15.3.png Based on the figure and information above, select the individual(s) inside which one or more detectable TE insertion events took place.
Mother
Father
Offspring A
Offspring B
Offspring C
Offspring D
Offspring E
Offspring F
DNA Polinton Maverick-like
DNA Helitron Hellraiser-like
DNA TIR Mariner-like
RNA LTR Gypsy-like
RNA LINE-like
RNA Penelope-like
On sequencing the genome of a previously unstudied eukaryotic species, you discover a moderately repetitive sequence distributed across the genome. This sequence possesses an open reading frame, but you notice that there are multiple truncated copies in which the 5' end of the coding sequence, but not the 3' end, is missing. Some copies of the sequence are associated with a target site duplication, but not all of them, and there are no other repeats flanking the sequence. Which of the following repetitive sequences is this most likely to be?
Non-LTR retrotransposon
LTR retrotransposon
Retrovirus insertion
TIR transposon
Helitron
Crypton
Polinton
Microsatellite
VNTR (Variable Number Tandem Repeat)
Which of these is an important difference between LTR retrotransposons and retroviruses?
Encoding of a GAG protein
Possession of a Long Terminal Repeat
Possession of an integrase gene
Frequent horizontal transmission among individuals
Replication within virus-like particles
The lexA3 mutant encodes a non-cleavable LexA protein. A lexA3 E. Coli strain subjected to a high level of DNA damage would become highly mutated thanks to the action of SOS induced error-prone polymerases.
True
False
It would depend whether the strain is also recA mutant
Associate the cellular consequence with the mechanism that caused it (one answer for each)?
Positive feedback
Allows more time for repair
Delay cell division
Permits faithful DNA repair
Homologous recombination
Permits unfaithful DNA repair
Error-prone translesion synthesis
Permits to stop the system to avoid formation of new stress
Negative feedback
Increases the strength of the reaction
Using a specific endonuclease, double-strand breaks are formed at a precise sequence on every copies of the chromosome of an E. Coli strain. Part 1 – Which mechanism will the cell try to use to repair this DNA damage?
Homologous recombination
Transposition
Translesion synthesis
Site-specific recombination
Nucleotide excision repair
DNA repair
With reference to the last question - Will the SOS response be activated?
Yes, but only the early response.
Yes, up to the late response.
No
In the Meselson and Weigle experiment (1961), if the scientists had looked at the density of the (c,+) single recombinants rather than the (+,mi), what population(s) of phages would they have observed?
Three populations: one with heavy DNA, one with DNA that is half heavy/half light, one with light DNA
Two populations: one with heavy DNA, one with DNA that is half heavy/half light.
Two populations: one with heavy DNA, one with light DNA
Two populations: one with DNA that is half heavy/half light, one with light DNA
One population, with heavy DNA
One population, with DNA that is half heavy/half light
One population, with light DNA
A three-factor cross (between factors d, e and f, placed in alphabetical order on the chromosome) was carried out. The progeny was as shown below (please note that we know that the markers do not affect the frequency of recombination and there is reciprocality in the population). What are the type and level of interference in this cross?
|
Genotype |
DeF |
def |
DEf |
DEF |
Def |
dEf |
deF |
dEF |
|
Number in the progeny |
46 |
200 |
250 |
190 |
4 |
44 |
260 |
6 |
Positive interference of 3
Negative interference of -3
No interference
Positive interference of 0.25
Negative interference of 0.25
Positive interference of 0.75
Negative interference of -0.75
Two dsDNA molecules differ in three markers (A, B and C). An event creates a single strand break in both molecules between markers A and B. A cross-over recombination event happens in front of marker A. The cross-over recombination event is resolved by cleavage. What can you say about this process (select all possible answers)?
This process leads to positive interference
This process leads to negative interference
This process leads to the production of heteroduplex DNA
This process is non-reciprocal
This process is reciprocal
After resolution and replication, this process leads to the formation of double recombinants
After resolution and replication, this process leads to the formation of single recombinants
Mice is one of preferable models for analysis of mechanisms of development. After genetic modification of one genetic locus and breeding a 2-week old female was identified with an interesting phenotype and a wild type male was put immediately into the same cage to produce progeny. How many new-born animals are expected to be generated by this female during the next 2 months if the male remains sitting in the same cage? Consider average litter size = 6 littermates.
6
9
12
15
18
24
Identify correct statements in the topic:
Early mouse embryos but not sperm can be cryopreserved.
Mouse sperm but not embryos can be cryopreserved.
Derivation of inbred mouse strains is problematic due to increase of genetic disorders.
Inbred mouse strains are usually heterozygous for approximately 20% of genes.
Each gene in the inbred mouse strain is represented by two identical copies.
ENU mutagenesis generated a mouse with an abnormally small kidney. Backcrossing confirmed that this mutation produces a dominant phenotype. Serial backcrossing identified a genetic interval in the mouse genome that contained several candidate genes potentially responsible for this phenotype. Westen blot showed reduced amount of protein encoded by one of these genes. Analysis of RNA isoforms encoded by this gene showed the same range of isoforms as in wild type mice. After discussion researchers came to a conclusion that more than one type of mutations could be assigned to this phenotype. Which type(s) of mutation could be reliably excluded as non-relevant?
Missense
Nonsense
Hypomorph
Antimorph
Neomorph
Expression pattern of gene of interest can be studied by generation of transgenic mice using pronuclear injection. For this purpose the transgene was designed to contain the gene promoter driving a LacZ reporter. Which of the statements below are correct?
The LacZ reporter always shows correct pattern of expression as long as the integrity of the transgene is maintained.
LacZ expression in the founder transgenic mouse can occur only if the promoter sequence in the transgene remained intact.
In some founders, the transgene can activate an endogenous gene expression.
Mice in which both the transgene promoter and LacZ sequences are mutated cannot affect the phenotype of the mouse.
Copy number of the transgene in the genome can be calculated using FISH.
Copy number of the transgene can be calculated by Northern blot using LacZ probe.
Copy number of the transgene can be estimated by Southern blot using LacZ probe.
A transgenic mouse line has been generated by pronuclear injection and researchers decided to investigate integration sites of the transgene using Southern blotting. They digested mouse genomic DNA with Pvu1 (unique restriction site in the transgene, which cuts mouse genome with low frequency = 0.036/ 10 kb) and DraI, which does not cut the transgene but cuts mouse genome with high frequency 1/ 1 kb. The transgene is shown in Fig. 1a. Consider four transgene integration patterns (A, B, C, D) in Fig. 1b and match them with 4 variants from Southern blots represented in Fig. 1c.
Transgene integration pattern B
Southern Blot 6
Transgene integration pattern C
Southern Blot 5
Transgene integration pattern A
Southern Blot 4
Transgene integration pattern D
Southern Blot 1
Seven DNA constructs were used to target a genetic locus. Each construct contained left and right homology arms (LHA and RHA, respectively) and selection markers, each driven by a ubiquitous promoter: Puror (Puromycin resistance cassette, similar to Neor) and HSV-tk in different positions. Some constructs also contained a GFP reporter.
After electroporation with individual constructs in separate experiments, the cells were cultured in selection media containing first puromycin and subsequently ganciclovir. At the end of each culture, researchers obtained one of the following results (only cells with intact transgenes were considered):
- No live cells
- Correctly targeted clones only
- Correctly targeted clones and clones with random transgene integration
- Cells with random integration only
For each of the 7 constructs, indicate which result (a,b,c, or d) would be obtained after targeting?
Construct 4
No live cells
Construct 7
No live cells
Construct 1
No live cells
Construct 2
Cells with random integration only
Construct 6
Correctly targeted clones only
Construct 5
Correctly targeted clones and clones with random transgene integration
Construct 3
No live cells
Mouse ES cells are planned to be used for the generation of mutant mice using CRISPR/Cas9 system. An essential sequence chosen for mutation of gene of interest is shown in Fig. 1. Eight variants of potential protospacer sequence within gRNA have been considered corresponding to coloured horizontal fragments, boundaries of which are delineated by thin vertical lines. Which of these potential protospacers can be considered for inclusion in gRNA to cut genomic DNA?
1
2
3
4
5
6
7
8
A gene targeting experiment was performed to introduce an insert (red) into a genetic locus (blue) using targeting construct with left and right homology arms (LHA and RHA) as shown in Fig. 2. There are two unique restriction sites (EcoRI; HindIII) in the locus near the site of insert, which are also present in LHA and RHA, respectively. Which of the following restriction digests will identify correctly targeted clones by Southern blot using the probe indicated in Fig. 2? Choose variants in Table below.
HINT: you should be able to discriminate correctly targeted clones from wild type clones and from clones with random transgene integration.
|
HindIII; EcoRI; BamHI |
|
EcoRI; BamHI |
|
HindIII; BamHI |
|
HindIII; EcoRI |
HindIII; EcoRI; BamHI
EcoRI; BamHI
HindIII; BamHI
HindIII; EcoRI
To study expression of gene A, researchers built a DNA construct in which LacZ reporter is driven by locus A regulatory sequences. Transgenic mice were generated by pronuclear injection and transgene integration was validated by PCR. Crossing 7 different founder males (A - J) with wild type (WT) females gave embryos, which were studied. The following results were obtained: Founders A, B, C, H, I, J: LacZ expression observed both in the lung and brain; Founder D: LacZ expression was observed in the eye; Founder F: some embryos expressed LacZ in the lung and brain and other embryos - in the brain only. Founder E: complete absence of LacZ expression. Part 1: Which of the following reasons may cause the aforementioned LacZ expression patterns in the progeny of individual founders for result 1? Note: I. More than 1 reason may be applicable to individual results
Transgene expression is faithfully driven by locus A regulatory elements
Transgene expression is influenced by surrounding chromatin
Transgene is nibbled or rearranged
Transgene is present as multicopy array
Transgene is integrated in more than 1 site of the genome
To study expression of gene A, researchers built a DNA construct in which LacZ reporter is driven by locus A regulatory sequences. Transgenic mice were generated by pronuclear injection and transgene integration was validated by PCR. Crossing 7 different founder males (A - J) with wild type (WT) females gave embryos, which were studied. The following results were obtained: Founders A, B, C, H, I, J: LacZ expression observed both in the lung and brain; Founder D: LacZ expression was observed in the eye; Founder F: some embryos expressed LacZ in the lung and brain and other embryos - in the brain only. Founder E: complete absence of LacZ expression. Part 2: Which of the following reasons may cause the aforementioned LacZ expression patterns in the progeny of individual founders for result 2? Note: I. More than 1 reason may be applicable to individual results;
Transgene expression is faithfully driven by locus A regulatory elements
Transgene expression is influenced by surrounding chromatin
Transgene is nibbled or rearranged
Transgene is present as multicopy array
Transgene is integrated in more than 1 site of the genome
To study expression of gene A, researchers built a DNA construct in which LacZ reporter is driven by locus A regulatory sequences. Transgenic mice were generated by pronuclear injection and transgene integration was validated by PCR. Crossing 7 different founder males (A - J) with wild type (WT) females gave embryos, which were studied. The following results were obtained: Founders A, B, C, H, I, J: LacZ expression observed both in the lung and brain; Founder D: LacZ expression was observed in the eye; Founder F: some embryos expressed LacZ in the lung and brain and other embryos - in the brain only. Founder E: complete absence of LacZ expression. Part 3: Which of the following reasons may cause the aforementioned LacZ expression patterns in the progeny of individual founders for result 3? Note: I. Please choose only one essential reason.
Transgene expression is faithfully driven by locus A regulatory elements
Transgene expression is influenced by surrounding chromatin
Transgene is nibbled or rearranged
Transgene is present as multicopy array
Transgene is integrated in more than 1 site of the genome
To study expression of gene A, researchers built a DNA construct in which LacZ reporter is driven by locus A regulatory sequences. Transgenic mice were generated by pronuclear injection and transgene integration was validated by PCR. Crossing 7 different founder males (A - J) with wild type (WT) females gave embryos, which were studied. The following results were obtained: Founders A, B, C, H, I, J: LacZ expression observed both in the lung and brain; Founder D: LacZ expression was observed in the eye; Founder F: some embryos expressed LacZ in the lung and brain and other embryos - in the brain only. Founder E: complete absence of LacZ expression. Part 4: Which of the following reasons may cause the aforementioned LacZ expression patterns in the progeny of individual founders for result 4? Note: I. More than 1 reason may be applicable to individual results
Transgene expression is faithfully driven by locus A regulatory elements
Transgene expression is influenced by surrounding chromatin
Transgene is nibbled or rearranged
Transgene is present as multicopy array
Transgene is integrated in more than 1 site of the genome
{"name":"Molecular Genetics 3 - Lecture Quiz's", "url":"https://www.quiz-maker.com/QPREVIEW","txt":"1 Chromosomes are numbered according to size, numbering [1] first, and arms are labelled either “p” or “q” with p being the [2] arm. Chromosomes containing arms of equal size are labelled [3], while chromosomes with two arms that are slightly unequal in length are known as [4]. A chromosome with one very short arm is termed [5], where the [6] is located at one extreme end., The image below shows fluorescence in situ hybridisation (FISH) to detect the Philadelphia chromosom (Ph), which causes chronic myelogenous leukemia (CML) and results from a reciprocal translocation between 9q and 22q. in this FISH experiment a red fluorescently labelled DNA probe that can hybridize to 9q and a green labelled DNA probe than can hybdridise to 22q were used. Fluorescence signal from the two DNA probes that are close together can fuse to form one dot. Using the image below as a reference alongside your understanding of chromosome nomenclature and FISH, select ALL the correct statements from the list below. Note: partial credit will be awarded for each correct answer, while points will be detracted from incorrect choices., Starting with a visual representation of the human karyotype (2n= 46), it is possible to reconstruct a similar representation of the cat karyotype (2n= 38) by making 7 autosomal chromosome breaks (each producing 1 extra block) and 11 autosomal chromosome fusions (combining two pre-existing blocks). How many major conserved syntenic blocks are shared between the female human and cat karyotypes? (Note that we are also ignoring any smaller changes and inversions).","img":"https://www.quiz-maker.com/3012/images/ogquiz.png"}