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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ND STONY BROOK UNIVERSITY LOOSELEAF GENETICS: FROM GENES TO GENOMES
- Shown below is the genomic structure of the human B-globin gene. The numbers within the boxes indicate the length in nucleotides of each region. Question 6: How many amino acids are present in the wild-type human B-globin protein? = exons = introns Transcription termination site (also poly A site) Promoter Start of transcription 3' 5' ATG 50 TẠC TAA 126 132 |ATT 90 130 222 850 3 5' Start codon Stop codon А. 438 В. 146 C. 620 D. 206 © 2013 John Wiley & Sons, Inc. All rights reserved.arrow_forwardThe yeast gene SER3, whose product has a role in serine biosynthesis, is repressed during growth in nutrient-rich medium, so little transcription takes place, and little SER3 enzyme is produced, under these conditions. In an investigation of the repression of the SER3 gene, a region of DNA upstream of SER3 was found to be heavily transcribed when SER3 is repressed ). Within this upstream region is a promoter that stimulates the transcription of an RNA molecule called SRG1 RNA (for SER3 regulatory gene 1). This RNA molecule has none of the sequences necessary for translation. Mutations in the promoter for SRG1 result in the disappearance of SRG1 RNA, and these mutations remove the repression of SER3. When RNA polymerase binds to the SRG1 promoter, the polymerase travels downstream, transcribing the SGR1 RNA, and passes through and transcribes the promoter for SER3. This activity leads to the repression of SER3. Propose a possible explanation for how the transcription of SGR1 might…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_forward
- The locations of the TATA box in two species of yeast, Saccharomyces pombe and Saccharomyces cerevisiae, differ dramatically. The TATA box of S. pombe is about 30 nucleotides upstream of the transcription start site, similar to the location in most other eukaryotic cells. However, the TATA box of S. cerevisiae is 40 to 120 nucleotides upstream of the start site. To better understand what sets the start site in these organisms, researchers at Stanford University conducted a series of experiments to determine which components of the transcription apparatus of these two species could be interchanged (Y. Li et al. 1994. Science 263:805–807). In these experiments, different general transcription factors and RNA polymerases were switched in S. pombe and S. cerevisiae, and the effects of each switch on the level of RNA synthesis and on the starting point of transcription were observed. The results from one set of experiments are shown in the table below. Components cTFIIB, cTFIIE, cTFIIF,…arrow_forwardThe locations of the TATA box in two species of yeast, Saccharomyces pombe and Saccharomyces cerevisiae, differ dramatically. The TATA box of S. pombe is about 30 nucleotides upstream of the transcription start site, similar to the location in most other eukaryotic cells. However, the TATA box of S. cerevisiae is 40 to 120 nucleotides upstream of the start site. To better understand what sets the start site in these organisms, researchers at Stanford University conducted a series of experiments to determine which components of the transcription apparatus of these two species could be interchanged (Y. Li et al. 1994. Science 263:805–807). In these experiments, different general transcription factors and RNA polymerases were switched in S. pombe and S. cerevisiae, and the effects of each switch on the level of RNA synthesis and on the starting point of transcription were observed. The results from one set of experiments are shown in the table below. Components cTFIIB, cTFIIE, cTFIIF,…arrow_forwardThe locations of the TATA box in two species of yeast, Saccharomyces pombe and Saccharomyces cerevisiae, differ dramatically. The TATA box of S. pombe is about 30 nucleotides upstream of the transcription start site, similar to the location in most other eukaryotic cells. However, the TATA box of S. cerevisiae is 40 to 120 nucleotides upstream of the start site. To better understand what sets the start site in these organisms, researchers at Stanford University conducted a series of experiments to determine which components of the transcription apparatus of these two species could be interchanged (Y. Li et al. 1994. Science 263:805–807). In these experiments, different general transcription factors and RNA polymerases were switched in S. pombe and S. cerevisiae, and the effects of each switch on the level of RNA synthesis and on the starting point of transcription were observed. The results from one set of experiments are shown in the table below. Components cTFIIB, cTFIIE, cTFIIF,…arrow_forward
- The IMD2 promoter contains three upstream transcription start sites (TSS) that are utilized under high GTP conditions and a single downstream TSS (-106) that is normally only utilized under low GTP conditions. In a wild type cell, expression of IMD2 mRNA only occurs if transcription initiates from the -106 TSS. In 300 words or less, describe: 1.) The normal function of Ssl2, and 2.) why a mutation in Ssl2, that increases its catalytic rate, would allow expression of the IMD2 ORF under high GTP conditions. (Conditions under which the IMD2 ORF is NOT expressed in the wild type.)arrow_forwardThe following diagram represents a transcription unit on a DNA molecule. a. Assume that this DNA molecule is from a bacterial cell. Draw the approximate locations of the promoter and terminator for this transcription unit. b. Assume that this DNA molecule is from a eukaryotic cell. Draw the approximate location of an RNA polymerase II promoter.arrow_forwardAn 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_forward
- To 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_forwardBelow is the double stranded DNA sequence of part of a hypothetical yeast genome encoding a very small gene. Transcription starts at nucleotide immediately following the promoter. The termination sequence is TATCTC. How many amino acids will this protein have? 5' TCATGAGATA GCCATGCACTA AGGCATCTGA GTTTATATCT CA 3' 3' AGTACTCTAT CGGTACGTGAT TCCGTAGACT CAAATATAGA GT 5'arrow_forwardYou would like to add a nuclear localization sequence (NLS) of Lys-Lys-Lys-Arg-Lys to a protein that is usually found in the cytoplasm of a yeast cell. To accomplish this, you introduce the nucleotide sequence encoding the NLS into the gene that encodes the cytoplasmic protein of interest. a. What is the size of the nucleotide insert that will encode the NLS? Briefly explain. 5' 3' b. Below is a diagram of the gene encoding the cytoplasmic protein of interest in the yeast genome. If your goal is to put the NLS at the carboxyl (C) terminus of the protein, at which location (A-E) should the NLS be inserted? Briefly explain. A TATAA ATATT promoter +1 B ATG TAC D TAA ATT stop codon E 3' 5'arrow_forward
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