Using CRISPR Gene Editing to Modify the LacZ gene in E.coli
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- čas přidán 29. 08. 2024
- In this investigation, CRISPR-Cas9 we be used to cut bacterial chromosomal DNA at a specific location within the lacZ gene. You will then take advantage of the cells’ ability to perform HDR to cause a desired change in the lacZ sequence. You will do this by providing the cells with large quantities of a donor template DNA, which includes an insert with a stop codon that will disrupt the gene function. In short, the chromosomal lacZ gene will be edited in E. coli by transforming them with plasmid-based expression system for donor template DNA and guide RNA.
Thank you sir 🙏
This was so well explained. Very thankful for this. Makes wish all my professors explained topics so efficiently. I truly understand this well now!
Thanks man. I'm giving a presentation on this exact experiment to my entire bio dept (students and professors) on what I found doing this experiment and I was having a real hard time explaining HDR and homologous recombination. How you explained it gave me a "aha moment". Thanks!!!!
Such an important and awesome session. Thanks a lot for the presentation.
Thank you so much sir for providing such a detailed presentation, you have saved my life! :)
very interesting and informative
Thanks Professor , Its been so useful for me !
This is a great video. I would like to do an experiment for lab genetics lab but i am unsure of what kind of experiment to run. I am not sure what could be changed or added to or what questions can be asked. Do you have any ideas or examples of experiments that can be done with this technology?
I would try this one first and then you might have an idea. www.bio-rad.com/en-us/category/crispr-gene-editing-kits?ID=Q0JG5VTU86LJ
@@raycinti5031 Thank you for responding! This is the exact kit my professor ordered but she wants us to make a few changes like the testing variable. I am thinking about changing the type of media by using different plates with different sugars and observing any changes in the Ecoli. I also thought about how UV light exposure may affect the Ecoli. What do you think about this? Do you have any ideas of what variables i could possibly do the experiment on?
I would think that A should become blue since there is no cut (because of the lack of sgRNA) and B should stay colorless or even die because the LacZ gene will be cut but not repaired since the lack of Arabinose. And since the sgRNA is missing for the cut C should remain colorless. If I got it right only D should successfully perform the insert of the stop codon which should end up as a colorless medium.
Is this correct?
Blue colony color indicates the hydrolysis of X-gal by functional beta galactosidase, which will be expressed only if the lac Z gene is functional and induced by IPTG. A lack of blue color in the presence of both X-gal and IPTG indicates no enzyme activity, which suggests that the lac Z gene is not functional. Both plates B and D show evidence of the lacZ gene having been cut by cas-9. Plate D shows evidence of DNA having been cut and repaired because the bacteria have survived the cut and are white. The bacteria on plate B have no functional repair machinery and have not survived the double stranded break. We also know that plates A and C cannot have been cut and then repaired because the donor template DNA introduces a stop codon, which prevent the expression of functional b-galactosidase. Plate B had no colonies on it. These bacteria express Cas-9 and were provided with sgRNA to guide Cas-9 to cut the lacZ gene. However, they were grown on starter plates without arabinose and so the repair system was not activated. Therefore, the cut could not be repaired and the bacteria died without forming colonies.
Hi .. can we find what kind of modification has happened in CRISPR sequence ...that is how to compare the target with the edited sequence
Blue colony color indicates the hydrolysis of X-gal by functional beta galactosidase,which will be expressed only if the lac Z gene is functional and induced by IPTG. Alack of blue color in the presence of both X-gal and IPTG indicates no enzyme activity,which suggests that the lac Z gene is not functional.Both plates B and D show evidence of the lacZ gene having been cut by cas-9. Plate D shows evidence of DNA having been cut and repaired because the bacteria have survived the cut and are white. The bacteria on plate B have no functional repair machinery and have not survived the double stranded break. We also know that plates A and C cannot have been cut and then repaired because the donor template DNA introduces a stop codon, which prevent the expression of functional b-galactosidase. Plate B had no colonies on it. These bacteria express Cas-9 and were provided with sgRNA to guide Cas-9 to cut the lacZ gene. However, they were grown on starter plates without arabinose and so the repair system was not activated. Therefore, the cut could not be repaired and the bacteria died without forming colonies.
@@raycinti5031 Can we identify or compare using online tools. that is comparison of the pre edited and edited sequences.
Hello Professor ,
Is this right ?
A: Blue
B:No colony
C:Blue
D:White
👍
Thank you Professor !