Lineweaver Burk plot data analysis

Рет қаралды 159,574

13 жыл бұрын

using a Lineweaver Burk plot to analyse enzyme data
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Пікірлер: 63

@SuperZwiebertje 12 жыл бұрын

Finally a professor that takes time in teaching you the whole thing instead of expecting that you already know 80% of it.

@jameshigdon100 8 жыл бұрын

This went from "we can find these numbers" to a solved problem.

@CLEANDrumCovers 8 жыл бұрын

Thank you, David Attenborough.

@amandar3299 3 жыл бұрын

I was SO confused about this - you are a life saver! :)

@makhzania 11 жыл бұрын

Thank you very much for the explanation, I have found it extremely helpful in my biochemistry lab work. You explain things clearly and in simple terms which many lecturers can't see to manage. Either that or as others have noted here, they seem to assume you know most of it already!

@issie339 11 жыл бұрын

Thank you so much! I wish you were my professor! So very helpful and easy to understand. Please keep making these videos, I learn more from you, than from my lectures and books.

@xanaanak55 11 жыл бұрын

That was so unbelievably helpful. Thankyou so much. Please keep making more videos!

@robynbryant2800 9 жыл бұрын

Love it, thank you so much for being so helpful!!

@chocolatepoodle6028 6 жыл бұрын

Thank you! This really helped me in writing my biochem report. :-)

@davidrenteria9585 Жыл бұрын

This is gold. Thank GOD 🙏 And thank you teacher.

@Boomerangpie 12 жыл бұрын

THANK YOU!! this was SO helpful - a topic that has confused me for weeks :-)

@sofieStormen 11 жыл бұрын

you saved my life :) thanks for so many details and the good explanetions :)

@EdisonvsTesla Ай бұрын

Thanks so much, may God protect 🙏 long live

@GHOST3940136 6 жыл бұрын

nice videos Peter, thx, that cleared some things up.

@magictrees101 2 жыл бұрын

Use a computer software. Alot easier these days to get the values 😁 I love how you explained this fully! 💯

@rarasrizkialviani8379 4 жыл бұрын

Hello, I'm from Indonesia, i'm biology collage student, i want to say thank you so much because this video helpfully.

@srikantadash4603 3 жыл бұрын

U r just a life saver for me.... Thank you ❤😘

@kathxlin 9 жыл бұрын

Thanks so much!!! This really helps heaps, Dr Klappa. Cheers! =)

@apvert 12 жыл бұрын

Extremely helpful, thanks!

@sweetestseed 11 жыл бұрын

Thank youuuuuu you're a life saver! God bless you

@Nadasistschoen 11 жыл бұрын

Thank you very much, this was incredibly useful!!!!

@SUNNI3DAY 11 жыл бұрын

Very helpful! Saved my lab write up.

@ssoblasted 12 жыл бұрын

thank you, this really helped my lap report.

@tsmanchester 6 жыл бұрын

How did the gradient all of a sudden come up how does one work out where the gradient goes

@frannnnn 12 жыл бұрын

GRANDE VIEJOO SOS UN GROSO ...

@1705ika 10 жыл бұрын

thaaankk youuu so much Peter Kappa, you really help me, you safe my life :')

@premiumedjr 13 жыл бұрын

Very helpful tutorial! What say about Dixon's plot in determination of inhibition constants? It will be better appropriate than lineweaver-burk, once it take in to consideration just the inverse of velocity, while lineweaver utilizes both the inverse of velocity as the inverse of substrate concentration? Thank you!

@jmse7en77 10 жыл бұрын

Thank you, it was very helpful!

@ABDULGTV Жыл бұрын

Thank you.

@Wandelaarke 12 жыл бұрын

Finally a professor that manages to explain this perfectly!

@jonahtristian5729 2 жыл бұрын

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@owenduncan760 2 жыл бұрын

@Jonah Tristian instablaster =)

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@jonahtristian5729 2 жыл бұрын

@Owen Duncan It worked and I now got access to my account again. I'm so happy! Thank you so much you really help me out :D

@owenduncan760 2 жыл бұрын

@Jonah Tristian happy to help =)

@dbwun 10 жыл бұрын

Great video, thanks!

@alvaro811994 13 жыл бұрын

thanks a lot...it really helped me..keep it up!!!

@maevecahalan1938 11 жыл бұрын

Excellent!!!!!!!

@Merhawit10 9 жыл бұрын

I'm still a bit confused on how Km and Vmax were calculated?

@lukechess7477 9 жыл бұрын

Reading off the Graph 1/Vmax= 0.016. This is the reciprocal, to reverse it you 1/ 1/Vmax (1/ 0.016 = 62.5) So Vmax= 62.5. You do the same for 1/ Km (1/ 0.07 = 14.3). Once you have Vmax and Km you can work out the Gradient (14.3/ 62.5 = 0.23). Again you Reverse this giving (1/ 0.23 = 4.3)

@dunyalaftamohammedkufa6908 Жыл бұрын

What about ki of non competitive inhibitors

@MrFilu13 12 жыл бұрын

thank u , it is very helpful

@user-qx6nh6sj2s 4 жыл бұрын

Thank you so much

@Brisbane09 10 жыл бұрын

there's one thing I don't get. Our teacher told us that Km has a unit mmol/l. But the reciproke of mmol/l would be 1/mmol/l = l*mmol^-1 wouldn't it?

@tajikamman 11 жыл бұрын

very helpful thanks so much! :-)

@KimHjalmarsson 11 жыл бұрын

This was helpful - I don't understand some of the basics still though.. Wasn't Km supposed to be half of Vmax simply? How can it be a totally different number than Vmax in the Lineweaver-Burk-plot?

@88abuk 12 жыл бұрын

thnx very much really helpfull

@misschocolate383 8 жыл бұрын

thank you sir

@getfiasco 12 жыл бұрын

i get how you work vmax but how do you work out km :s

@gamzeyazgeldi6925 5 жыл бұрын

Thanks!!

@EddyKackboon 7 жыл бұрын

very cool =)

@asumangulsumkaya2887 6 жыл бұрын

still I couldn't understand where Vmax come from? He said that it is 0,016 from graph but later he says it is 62,5??? Btw thank you for this enlightening video

@SuperYersinia 5 жыл бұрын

1/0,016 = 62,5

@rajkishore5334 10 жыл бұрын

nice talk..

@getfiasco 12 жыл бұрын

oh i get it now. very helpful :)

@BioScienceClass 4 жыл бұрын

For more videos on Enzyme Kinetics:kzfaq.info/get/bejne/raeHjKx4ubi8gXk.html

@poppyflower333 10 жыл бұрын

Thanks,

@adrianculasing1279 6 жыл бұрын

Bruh thank you.

@039stephen 5 жыл бұрын

40:41 jack, its called sleet

@Rob-ug4nb 5 жыл бұрын

why wouldn't he use the largest value of V for Vmax instead of just making a number up for Vmax????

@pk11kent 5 жыл бұрын

Because the largest number for v does not represent Vmax. If you take the Michaelis Menten equation you see that v only reaches Vmax when the substrate concentration reaches infinity (which in practice it never will).