What does Cas9 do in CRISPR-Cas9 editing? O It makes double stranded breaks at specific locations in the genome. O It is complementary to the spacer DNA and serves to determine where the endonuclease will make breaks. O It is complementary to the guide DNA and serves to determine where the endonuclease will make breaks. O It inserts at random places in the genome. O It joins freshly cut strands of DNA.
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- Why are antibiotic resistance markers such as ampR important components of bacterial plasmid cloning vectors? a. The plasmid must have resistance to accept DNA inserts. b. They allow the detection of plasmids that contain an inserted DNA fragment. c. They ensure the presence of the ori site. d. They ensure that the plasmid can be cut by a restriction enzyme. e. They allow identification of bacteria that have taken up a plasmid.6. Below are several DNA sequences that are mutated compared with the wild-type sequence: 3'-TACT GACTG ACGAT C-5'. Envision that each is a section of a DNA molecule that has separated in preparation for transcription, so you are only seeing the template strand. Construct the complementary DNA sequences (indicating 5' and 3' ends) for each mutated DNA sequence, then transcribe (indicating 5' and 3' ends) the template strands, and translate the mRNA molecules using the genetic code, recording the resulting amino acid sequence (indicating the N and C termini). What type of mutation is each?6. Below are several DNA sequences that are mutated compared with the wild-type sequence: 3'-TACT GACTG ACGAT C-5'. Envision that each is a section of a DNA molecule that has separated in preparation for transcription, so you are only seeing the template strand. Construct the complementary DNA sequences (indicating 5' and 3' ends) for each mutated DNA sequence, then transcribe (indicating 5' and 3' ends) the template strands, and translate the mRNA molecules using the genetic code, recording the resulting amino acid sequence (indicating the N and C termini). What type of mutation is each? 6.a. Mutated DNA Template Strand #1: 3'-TACTGTCT GACGATC-5' Complementary DNA sequence: mRNA sequence transcribed from template: Amino acid sequence of peptide: Type of mutation: 6.b. Mutated DNA Template Strand #2: 3'-TACG GACT GAC GATC-5' Complementary DNA sequence: mRNA sequence transcribed from template: Amino acid sequence of peptide: Type of mutation:
- 5. What are some PROS and Cons of CRISPR? List at least 3 of ceach Pros of CRISPR Cons of CRISPR 6. What do you think the future of CRISPR is? How do you think humans will use it in a positive way? Negative?12. You are working with a picce of DNA of the sequence: 5'-TATTGAGCTCCCCGGAT-3 3'-ATAACTCGAGGGGCCTA-5 You cut the above piece of DNA with a restriction enzyme that recognizes the sequence 5'GAGCTC and cuts on the 3' side of the A within this sequence. Please, draw all products that you get after digestion. Label all 5' and 3' ends.5. Design a 10-bp primer that could be used to amplify the following sequence of DNA: 5'-AGTCGATCCCTGATCGTACGCTACGGTAACGT-3'
- 9. The CRISPR associated protein Cas9 is now being used as a molecular tool for genome editing C. In a cell, what happens to a gene that has been cut by a genome editing system? It is repaired using either non-homologous end joining (NHEJ) or homology dependent repair (HDR) D. The Cas9 protein requires an RNA molecule to carry out genome editing. What sequence requirements does this RNA have? E. Name two limitations of CRISPR/Cas9 genome editing.6. Below are several DNA sequences that are mutated compared with the wild-type sequence: 3’-T A C T G A C T GA C G A T C-5’. Envision that each is a section of a DNA molecule that has separated in preparation for transcription, so you are only seeing the template strand. Construct the complementary DNA sequences (indicating 5’ and 3’ ends) for each mutated DNA sequence, then transcribe (indicating 5’ and 3’ ends) the template strands, and translate the mRNA molecules using the genetic code, recording the resulting amino acid sequence (indicating the N and C termini). What type of mutation is each? 6.d. Mutated DNA Template Strand #4: 3’-T A C G A C T G A C T A T C-5’Complementary DNA sequence:mRNA sequence transcribed from template:Amino acid sequence of peptide:Type of mutation: 6.a. Mutated DNA Template Strand #1: 3’-T A C T G T C T G A C G A T C-5’Complementary DNA sequence:mRNA sequence transcribed from template:Amino acid sequence of peptide:Type of mutation: 6.b. Mutated DNA…6. Use the image to determine what type of mutation is found in each of the DNA strands below and what type of effect that would have on the protein made. OPTIONS: substitution, deletion, translocation, and insertion. DNA: CCC GGG mRNA: CCC Protein: Pro DNA: GAA CTT mRNA: GAA Protein: Glu DNA: TTA AAT mRNA: UUA Protein: Leu CCA GGT CCA Pro GTA CAT GUA Val TAA ATT UAA Stop Type of Mutation Substitution Substitution Substitution What happens when this type of mutation occurs?
- 3. In future lectures we will describe a technique known as Northern blotting that can be used to detect RNA transcribed from a particular gene. Briefly, in this method a specific RNA is detected using a short segment of cloned DNA as a probe. The DNA probe, which is radioactive, is complementary to the RNA that the researcher wishes to detect. After the radioactive probe DNA basepairs to the RNA, the RNA is visualized as a dark (radioactive) band on an X-ray film. As shown here, the method of Northern blotting can be used to determine the amount of a particular RNA transcribed in a given cell type. If one type of cell produces twice as much of a particular mRNA as occurs in another cell, the band will appear twice as intense. Also, the method can distinguish if alternative RNA splicing has occurred to produce an RNA that has a different molecular mass. 875 nucleotides 675 nucleotides 1 Northern blot 2 - 3 Lane 1 is a sample of RNA isolated from nerve cells. Lane 2 is a sample of RNA…10. Why does a cDNA copy of a eukaryotic mRNA gene need to be made in order to determine the expressed protein sequence? Group of answer choices Genomic DNA is methylated and cannot be used as a template Genomic DNA is less stable than cDNA and is therefore harder to work with The genomic DNA sequence in eukaryotes contains introns which need to be removed first Genomic DNA is too long so the shorter mRNA copy is used The 5’ cap gets in the way of protein sequencing16. Why is re-circularized insert generally not a problem in cloning procedures, while re- circularized vector can be a problem? Is it because re-circularized insert: A: is detrimental to bacteria B: is a kind of product that DNA ligase cannot form C: makes E. coli sensitive to antibiotics D: lacks a replication origin E: is beneficial to bacteria om