Practice_CRISPR KEY - Biol14 CRISPR practice problems 1...

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Unformatted text preview: Biol14 CRISPR practice problems 1. What is the difference between gene editing and traditional methods for constructing transgenic organisms? The key difference is that gene editing involves targeted, specific cutting of the phosphodiester backbone of the DNA at a specific site in the target gene. Like older methods, it relies on the cell’s own DNA repair processes (non-­‐homologous end joining –NHEJ-­‐ or homologous recombination –HR) to remove or add DNA in the genome. If you just add a homology-­‐flanked transgene, you are depending on the cell to find the homology, cut the target DNA, and integrate the transgene into its genome at the target site. If you also use a gene editing nuclease such as Cas9, you are helping the cell toward this desired pathway, by also cutting the target sequence directly (with the nuclease). In fact, if your goal is only to disrupt the gene, you do not even have to add in a transgene. Simply cutting the target sequence will initiate NHEJ at the the target site, resulting in a small deletion within the target gene. 2. How is CRISPR/Cas better for gene editing than Zinc finger nucleases (ZFN)? To use a ZFN to target a particular gene, you first have to develop the specific ZFN protein that will bind to and cut your target DNA sequence. To use CRISPR/Cas, you simply have to design and synthesize an RNA molecule, about 100nt long, with 20 nt of homology to the target site adjacent to the PAM motif located in the target DNA sequence: the same Cas9 protein can cut any DNA sequence, as long as the correct guide RNA is present. 3. Your goal is to use CRISPR/cas to target the Catbts gene in mouse. The PAM, 5’-­‐ngg-­‐3’, is shown in red in the sequence. Given the target sequence below, which of the following could serve as the sequence-­‐specific part of the guide RNA? Choose the best one, and explain your reasoning. 5’-atctttgagcattatgagtccttaagtgaaggcaaattc-3’ 3’-tagaaactcgtaatactcaggaattcacttccgtttaag-5’ a. b. c. d. e. 3’-tcc-5’ 3’-atactcaggaattcacttcc-5’ 3’-gtaatactcaggaattcact-5’ 3’-gtaatactcaggaattcacttcc-5’ 3’-tagaaactcgtaatactcag-5’ Explain briefly: C is correct, because this sequence is 20nt long and anneals to the target DNA immediately adjacent to the PAM in the DNA, but does not include the PAM. The PAM is a DNA sequence motif that is recognized by the Cas9 protein, not the guide RNA. Biol14 CRISPR practice problems 4. a. b. c. Suppose you want to use CRISPR/cas to generate two different types of mice: one with a disruption (knock-­‐out) of the gene Catbts, the other with the EGFP gene inserted within the Catbts locus. For each molecule identified below, indicate whether it would be required for genome editing to create the knock-­‐out, the EGFP insertion, both, or neither. Cas9 protein knock-­‐out -­‐-­‐ EGFP insertion -­‐-­‐ both -­‐-­‐ neither An RNA including complementarity to Catbts knock-­‐out -­‐-­‐ EGFP insertion -­‐-­‐ both -­‐-­‐ neither A DNA including complementarity to EGFP knock-­‐out -­‐-­‐ EGFP insertion – both -­‐-­‐ neither 5. You believe you have successfully generated a knock-­‐in mouse. You will use a Southern blot to confirm the mouse’s genotype. Use the diagram of the gene, shown below, to predict what band(s) you will see in the control correct knock-­‐in sample, and choose the correct “blot results” from the options shown below the diagram. Blot results: WT is shown. Add the expected bands to lane 2 (knock-­‐in). Choose one. Answer: correct choice is the first on the left. The knockin cell is still heterozygous, hence the presence of the WT band. The knockin allele itself, after digest with HindIII, produces two bands (one about 10kb, one about 5kb), but the probe (shown on the left in the diagram) will only anneal to one of these probes; therefore, we will only see one of these two bands (the larger one), which is a bit smaller than the HindIII digest product recognized by the probe from the WT allele. Biol14 CRISPR practice problems ...
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