Quiz Submissions Quiz 4
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1 Quiz Submissions Quiz 4 Attempt 1 Written: Nov 1, :35 Nov 1, :19 Submission View Released: Nov 4, :24 Question 1 0 / 1 point Three RNA polymerases synthesize most of the RNA present in the cell. Which of the following is true for RNA Polymerase II: it is present in the nucleolus It only transcribes protein coding genes (synthesizes mrna exclusively) Its large subunit has a carboxy terminal domain that becomes phosporylated A and B A and C Question 2 An antibody against the phosphorylated form of RNA Polymerase II was used in a ChIP experiment to determine its genome wide distribution. In a parallel GROseq experiment only half of the genes that were positive for RNA Polymerase II occupancy were actively transcribing. Based on this discrepancy which information can we conclude from the ChIP data? Due to a limiting concentration of nucleotides RNA Polymerase II is often blocked in a preinitiation complex prior to transcriptional initiation The RNA Polymerase II antibody can recognize the enzyme even when it is not bound to DNA
2 RNA Polymerase II can bind to DNA non specifically and almost everywhere in the genome suggesting widespread transcription in non protein coding genes The bound but non transcribing enzymes may be paused elongation complexes. RNA Polymerase II seems to take pauses during transcription Question 3 0 / 1 point You would like to add an HA tag (the peptide sequence is YPYDVPDYA) onto your protein of interest so that you could use affinity purification to enrich a sample for further characterisation. Which of the following options would be the best downstream oligonucleotide primer sequence for PCR amplification in combination with a perfectly complementary upstream primer in order to eliminate the natural stop (a UAA (ochre) stop) of your protein while adding on the extra amino acids that correspond to the HA sequence along with a new stop codon. 5' TAT TAC CCA TAC GAT GTT CCA GAT TAC GCT TAG 3' Where the 5' TAT disrupts the natural stop and provides read through while the 3' TAG will generate a new UAG amber stop codon at the end of the HA tag 5' CTA AGC GTA ATC TGG AAC ATC GTA TGG GTA ATA ATA 3' Where the 3' ATA provides readthrough from the natural stop and the 5' CTA will generate a new UAG (amber) stop codon at the end of the HA tag 5' TAT TAT CCA TAT GAT GTT CCA GAT TAT GCT TAG 3' Where the 5' most TAT disrupts the natural stop and provides readthrough while the 3' TAG will generate a new UAG amber stop codon at the end of the HA tag A and C only A, B and C Question 4
3 TBP is a relatively small protein (~35kDa) that can bind DNA sequences (the TATA box) on its own with relatively high affinity. However if nuclear extracts are subjected to Gel Filtration Chromatography or Zonal Centrifugation and each eluted/collected fraction is tested in a western blot it appears that TBP is consistently found at a position that corresponds to approximately 500kDa. This fraction is still competent to drive transcription in vitro using a core promoter. Why might TBP behave like such a large protein? TBP exists in a large transcriptionally competent multi subunit complex it can dimerize so its molecular weight would appear greater than predicted from its sequence There may be other TBP like sequences that are recognised by the antibody A and B Question 5 cdna libraries are important to maintain an accurate permanent record of the mrnas present in a given sample at a given time/context when the mrna was isolated. Which of the following primers would be essential for priming first strand synthesis (reverse transcriptase) of all mrnas >cdna in your RNA sample. dgggggg (poly dg) dcccccc (polydc) dttttttt (poly dt) Reverse Transcriptase does not require a primer daaaaaa (poly da) Question 6
4 Transcription initiation by RNA Polymerase II requires unwinding of the DNA helix at the promoter. This process involves a specific general transcription factor and ATP. Which of the following activities best describes the "enzymatic" properties of this factor required for formation of the open complex? DNA damage recognition Nucleotide polymerization DNA helicase DNA phosphorylation Question 7 Which of the following molecular approaches can provide information regarding the levels of mrna present in a given sample for each gene in the genome in one single experiment? GROseq RNAseq Northern Analysis A and B only A, B and C Question 8 The maximum rate of cdna being formed from an RT reaction using 1 unit of enzyme is around 1000pmol/min, providing that dntps and template are
5 not limiting. What is the rate that corresponds to the K m of the RT enzyme in this reaction? 1000 pmol/min 500 pmol/min 100 pmol/min 1 pmol/min Question 9 Titin is a giant protein that has a molecular weight of over 3 x 10 6 Daltons and is considered the largest human protein produced. Based on this information what type of purification method would be the most efficient (cheap and rapid) to obtain a fraction enriched for the protein Titin. 2D gel electrophoresis ion exchange chromatography affinity chromatography gel filtration chromatography differential centrifugation Question 10 A student made a radioactive probe from a DNA fragment using PCR. The probe was very radioactive and was to be used to analyse RFLPs between family members that potentially carried a mutant copy of a gene involved in familial neurodegeneration. When the student's Southern blot was hybridized with the radiolabelled probe nothing appeared on the x ray film (autoradiogram). However, when a single stranded oligonucleotide was
6 radiolabelled using polynucleotide kinase (PNK) it detected a number of bands that corresponded to the patients' DNA. Which of these explanations provides a possible reason for why the probe made using PCR did not work, while the oligonucleotide probe did work? the patients' DNA was not properly denatured in the gel (forgot to add alkali) an exonuclease activity in the solutions removed the label probes made using PCR are uniquely for RNA detection the mutant sequence was modified so radically that it could not bind the probe The PCR generated probe was not denatured prior to use so it couldn't base pair with its single stranded (denatured) DNA target Attempt Score: Overall Grade (first attempt): Done
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