1) Why is the design of single
1) Why is the design of single guide RNA limited to sequencesthat are next to PAMs?
a) PAM sites are necessary for endonucleases to create singlestrand breaks.
b) Cas9 recognizes PAM sites to create double strand breaks.
c) PAM sites are necessary for complementary base pairing.
d) Single guide RNAs bind to PAMs.
2) What repair mechanism allows for the generation of precisemutations?
a) base excision repair
b) nonhomologous end joining
c) double strand DNA break
d) homologous recombination
e) mismatch repair
Answer:
1)single guide RNA is known as sgRNA. sgRNA is a single RNA moleculethat contains both the custom-designed short crRNA sequence fusedto the scaffold tracrRNA sequence. sgRNA can be syntheticallygenerated or made in vitro or in vivo from a DNA template.
These gRNAs are non coding short RNA sequences which bind to thecomplementary target DNA sequences. Guide RNA first binds to theCas9 enzyme and the gRNA sequence guides the complex via pairing toa specific location on the DNA, where Cas9 performs itsendonuclease activity by cutting the target DNA strand.
PAM refers to protospacer adjacent motif. PAM is a 2–6-base pairDNA sequence immediately following the DNA sequence targeted by theCas9 nuclease in the CRISPR bacterial adaptive immune system. Cas9will not successfully bind to or cleave the target DNA sequence ifit is not followed by the PAM sequence.
Cas9–guide RNA complexes recognize 20-base-pair sequences in DNAand generate a site-specific double-strand break. The PAM isrequired for a Cas nuclease to cut and is generally found 3-4nucleotides downstream from the cut site.PAM is frequently used to mark proper target sites.
Hence, option “c” is the correct answer.
2)The CRISPR/Cas9 system enables targeted genome modifications acrossa range of eukaryotes. By applying this system, precise pointmutations and genomic deletions can be generated. They can serve asan alternative to the conventional homologous recombination-basedgene disruption. It results in DNA double-strand breaks (DSBs).
Hence, option “c” is the correct answer.
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