Concept explainers
F. Port and S. Bullock at the University of Cambridge (UK) designed the elegant plasmid vector pCFD3 for the expression of sgRNAs in Drosophila. The following figure shows a part of this vector. The orange sequences are part of a strong promoter (transcription from this promoter starts at the G in bold—which must be present—and goes from left to right). The purple sequences are a portion of the tracrRNA component of the sgRNA. After cutting the pCFD3 plasmid with the restriction enzyme BbsI (whose recognition site is also shown in the following figure), you will replace the blue sequences in the figure with sequences that will allow the expression of an sgRNA that targets a Drosophila gene called NiPp1.
The last part of the jigsaw puzzle you will need is the following sequence, which shows part of the NiPp1 gene including the triplet corresponding to the start codon. Capital letters are in the gene’s first exon with the coding region in blue; lowercase letters are in the first intron. The NiPp1 protein is 383 amino acids long. Your assignment is to generate a knockout allele of this gene by inducing Cas9 to produce a double-strand break into the gene that will be repaired imprecisely by nonhomologous end-joining (NHEJ).
a. | Identify the two PAM sites in this sequence. Which of these PAM sites would you want to use in order to produce a null allele of the NiPp1 gene? Why would you prefer this site? |
b. | If you targeted Cas9 to the proper location in the NiPp1 gene, and the resultant double-strand break was repaired imprecisely by NHEJ (so that a few—usually ≤6 bp are deleted or added at that location), about what percentage of the imprecisely repaired genes could you say with confidence would be null alleles? Explain. |
c. | Diagram the pCFD3 vector after it has been cut with the BbsI enzyme. Don’t worry about the small blue fragment that will be removed; the emphasis here is to show the 5′-overhangs that will be made. |
d. | Design two 24-nt DNA oligonucleotides that you could anneal together and clone into BbsI-cut pCFD3 vector so that the recombinant plasmid could express an sgRNA useful for making null mutations in the NiPp1 gene. |
e. | Show exactly where Cas9 would cut the NiPp1 gene. |
f. | Briefly outline what you would do with your recombinant plasmid to make a null mutation in the fly NiPp1 gene. |
g. | Briefly outline how you would modify this technique to generate a knockin allele in which the first amino acid in the NiPp1 protein after the initiating Met (that is, Thr) would be changed to Ala. |
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Genetics: From Genes to Genomes
- The tac promoter, an artificial promoter made from portions of the trp and lacUV5 promoters, has been introduced into a plasmid. It is a hybrid of the lac and trp (tryptophan) promoters, containing the −35 region of one and the −10 region of the other. This promoter directs transcription initiation more efficiently than either the trp or lac promoters. Why?arrow_forwardThe genes that encode the enzymes for arginine biosynthesis are located at several positions around the genome of E. coli, and they are regulated coordinately by a transcription regulator encoded by the ArgR gene. The activity of the Argr protein is modulated by arginine. upon binding arginine, Argr alters its conformation, dramatically changing its affinity for the DNA sequences in the promoters of the genes for the arginine biosynthetic enzymes. given that ArgR is a repressor protein, would you expect that ArgR would bind more tightly or less tightly to the DNA sequences when arginine is abundant? if ArgR functioned instead as an activator protein, would you expect the binding of arginine to increase or to decrease its affinity for its regulatory DNA sequences? explain your answers.arrow_forwardTo identify the following types of genetic occurrences, would acomputer program use sequence recognition, pattern recognition,or both?A. Whether a segment of Drosophila DNA contains a P element(which is a specific type of transposable element)B. Whether a segment of DNA contains a stop codonC. In a comparison of two DNA segments, whether there is aninversion in one segment compared with the other segmentD. Whether a long segment of bacterial DNA contains one ormore genesarrow_forward
- An electrophoretic mobility shift assay can be used to study the binding of proteins to a segment of DNA. In the results shown here, an EMSA was used to examine the requirements for the binding of RNA polymerase |l (from eukaryotic cells) to the promoter of a protein-encoding gene. The assembly of general transcription factors and RNA polymerase Il at the core promoter is described in Week 4. In this experiment, the segment of DNA containing a promoter sequence was 1100 bp in length. The fragment was mixed with various combinations of proteins and then subjected to an EMSA. Lane 1: No proteins added Lane 2: TFIID Lane 3: TFIIB Lane 4: RNA polymerase IIl Lane 5: TFIID + TFIIB Lane 6: TFIID + RNA 1 2 3 4 5 6. 7 polymerase II Lane 7: TFIID + TFIIB + RNA polymerase Il 1100 bp Explain the results.arrow_forwardThe DNA sequence of the promoter region of E. coli xyzA gene is shown below. Transcription start site is the A (in bold) at position 43. 5 10 15 20 25 30 35 40 45 50 GAGCT GTTGA CAATT AATCA TCGAA CTAGT TAACT AGTAC GCAAG TTCAC Mutations were introduced in the sequence to identify residues important for gene expression. Indicate the effect of the following mutations on xyzA expression (increase, decrease, no effect, cannot be predicted). Provide reasoning for each answer. A. G3A (G at position 3 was changed to A) G9A Deletion of TCA at position 18-20 C22A T31A, A32T double mutant T35G G45C C48A B. What are the promoter sequences of the gene?arrow_forwarda) what is a promoter and give the element and their functions of E.coli promoter b) what are eukaryotic transcription factor and list the class 2 general transcription factors and state their functionsarrow_forward
- Negative supercoiling of DNA favors the transcription of genes because it facilitates unwinding. However, not all promoter sites are stimulated by negative supercoiling. The promoter site for topoisomerase II itself is a noteworthy exception. Negative supercoiling decreases the rate of transcription of this gene. Propose a possible mechanism for this effect and suggest a reason why it may occur.arrow_forwardThe sequence of the lac promoter and two mutant promoters (Mutn 1 and Mutn 2) are shown below. The activity of these promoters is measured by fusing the them to the gene for Green Fluorescent Protein (GFP) and measuring the production of green fluorescence by GFP protein. What would you expect the GFP signal to be higher, lower or same for mutant promoters. Mutn 1 (enter higher, lower or same) Mutn 2 (enter higher, lower or same) - 42 ? ? +1 Lac CCAGGCTTTACACTTTATGCTTCCGGCTCGTATGTTGTGTGGA CCAGGCTTAAAACTTTATGCTTCCGGCTCGTATGTTGTGTGGA Lac Mutl Lac Mut 2 CCAGGCTTTACACTTTATGCTTCCGGCTCGTATAATGTGTGGAarrow_forwardThe following diagram show what is required for an active promoter of a gene of interest, where:A1 = Activator 1A2 = Activator 2Med = MediatorRep = Repressor Based on the following data, predict: Chromatin conformation Methylation state of the proximal promoter If protein A1 is present or absent If protein A2 is present or absent If the mediator is present or absent If the repressor is absence or present If this gene is likely to be transcribed or notPlease note that this is an "all or nothing" bonus question, and no partial credit will be awarded. Selecting all answers will result in zero points awarded. Selecting at least one incorrect answer will result in zero points being awarded. Question 28 options: Euchromatin Heterochromatin Methylated promoter Unmethylated promoter Activator A1 present Activator A1 absent Activator A2 present…arrow_forward
- You made four mutants for a promoter sequence in DNA and studied them for transcription. The results of the amount of gene expression or transcription (based on beta-Gal activity shown on Y-axis) for these DNAs (X-axis) are shown. The sequence of the wild-type and mutant DNAs, and consensus sequence from many promoters are shown here for your convenience. From this experiment you can conclude that: Nucleotide substitution can identify important bases of the binding sites or promoter in DNA (e.g., -10 and -35 promoter sequences of lac operon). True or false: Spacer (a) -10 region -35 region TTGACA Consensus sequence TATAAT Wild-type Lac promoter GGCTTTACACTTTATGCTTCCGGCTCGTATGTTGTGTGGAATT Mutant 1 GGCTTTACACTTTATG-TTCCGGCTCGTATGTTGTGTGGAATT Mutant 2 GGCTTTACACTTTATGCTTCCGGCTCGTATAATGTGTGGAATT Mutant 3 GGCTTTACACTTTATG-TTCCGGCTCGTATAATGTGTGGAATT Mutant 4 GGCTTGACACTTTATG-TTCCGGCTCGTATAATGTGTGGAATT (b) 700 600- 500- 400- 300- 200- 100. 0 ● True O False B-Galactosidase activity Wild-type…arrow_forwardTo characterize the promoter of the gadA operon you made a series of deletion mutants removing pieces of the promoter to see what would happen. The results are found below: gad promoter gada gadX gadz 450 +1 lacz activity transcription start site pH 2.0 pH 7.0 A gad promoter beta-galactosidase (lacZ) +++ 450 gad promoter beta galactosidase (lacZ) +++ +++ 300 +1 gad promoter beta galactosidase (lacZ) 150 D gad beta galactosidase (lacz) -450 150 E gad promoter beta-galactosidase (lacZ) -450 -300 Based on these results, what can you conclude about the gad promoter? O a. The promoter is only regulated by repression Ob. The promoter is regulated by a mix of activation and repression O c. The promoter is only regulated by activation O d. The promoter has multiple operators and multiple enhancersarrow_forwardSuppose you had isolated a new transcription factor and wanted to know which genes this protein might regulate. Is there any way that you could use a cDNA microarray of the type shown in the picture to approach this question?arrow_forward
- Human Heredity: Principles and Issues (MindTap Co...BiologyISBN:9781305251052Author:Michael CummingsPublisher:Cengage Learning