Thank you sir.... I didn't understood this tehnique before.. And now my exam is coming.. I don't have any more doubts about this topic... Thank you for your hard work... May Allah bless you...
I also had the doubt on how to get 4 for distance between AB and CD... He is using the second matrix he obtained in Round 2 (where CD appears as a single taxa) to get that value. So, d(AB,CD) = (dA,CD + dB,CD) - dA,B / 2 = (17 + 8 ) - 17 / 2 = 4
I am very thankful to you sir... I didn't understood this technique before and now my exam is coming... I don't have any more doubts thank you so much sir for your hardwork... May Allah bless you
I also had the doubt on how to get 4 for distance between AB and CD... He is using the second matrix he obtained in Round 2 (where CD appears as a single taxa) to get that value. So, d(AB,CD) = (dA,CD + dB,CD) - dA,B / 2 = (17 + 8 ) - 17 / 2 = 4
Dear, in round 3, you will generate new distance matrix based on values in the previous distance matrix. By employing the formula d'(ij,k)=(di,k + dj,k - di,j)/2 you will get the value 4.
Maybe you could help me, I'm trying to understand what something means, I have a neighbor joining tree and there are certain values on the branches (i.e. 100, 75, 52) and I don't understand what those mean and there's a scale (it goes from 0.250 to 0) and that I also don't understand, maybe you could help me? What does that mean?
the values with the branches (100, 75, 52) are bootstrap values. Please google that. The scale is a distance scale, in principle the number of inferred mutations normalized by sequence length,
@@marzanbinteabid7538 To find r'i value, it is not necessary to divide by 2; rather, it's supposed to be divided by n-2. Here, the n=3 (A, B & CD). so n-2=1. Therefore those're actually divided by 1.
thanks sir. better understanding video never saw for complicated sums
Thank you sir.... I didn't understood this tehnique before.. And now my exam is coming.. I don't have any more doubts about this topic... Thank you for your hard work... May Allah bless you...
YOU'RE A LITERAL SAVIOURRRR SIR. Thank you for breaking this all down in simpler ways.
PS: really appreciate you starting the video with bismillah :)
I have never understand it before. Thank you so much
Very grateful for your videos watching from India..💥
Sir, there is discrepancy in round 2 in calc ris. Coz in round 1 you divided by n-2 but not in r2.
I also had the doubt on how to get 4 for distance between AB and CD... He is using the second matrix he obtained in Round 2 (where CD appears as a single taxa) to get that value. So, d(AB,CD) = (dA,CD + dB,CD) - dA,B / 2 = (17 + 8 ) - 17 / 2 = 4
Thank you so much!
I was looking for this explanation. Thanks a bunch, @Manu4Rusher!
thank you so much
I never understood this technique before. Thank u for hard work.
Glad to know !!!
@@FarhanHaqj I will show your all videos
@@rsonlinebiologyclasses7697 :)
Spread the word with others as well :)
I am very thankful to you sir... I didn't understood this technique before and now my exam is coming... I don't have any more doubts thank you so much sir for your hardwork... May Allah bless you
Very well explained! thank you for the video!
Thank you very much for this excellent video!
For round 2, why the r'i value for A is not (17+17)/2 and for other two also?
cause for ri values we have (17+17)/n-2; and for this case n=3 so its divide by 1.
Thank u so much sir.ur way of teaching is very helpful to me .
So nice of you. Please subscribe and share with others. Good day!
Sure sir 🙏
According for what that you can get distance matrix A-B is 17 in this matrix table. 2:10 ? i still can not understand.
Love and respect from Bangladesh 🇧🇩🇧🇩
Thank you Thank you !! Love from Pakistan
Thank you so much for this amazing video.....to get understood very easily bt i have one doubt-
How to calculate the distance between AB and CD ??
I have the same questions.... How can we get 4 from AB to CD at round 3.
I also had the doubt on how to get 4 for distance between AB and CD... He is using the second matrix he obtained in Round 2 (where CD appears as a single taxa) to get that value. So, d(AB,CD) = (dA,CD + dB,CD) - dA,B / 2 = (17 + 8 ) - 17 / 2 = 4
@@Manu4Rusher Ohhh oks thnk you.
@@Manu4Rusher Please can you tell how the ri' values got calculated in 18:02 where CD are clustered??
@@umarhasnain7369same doubt i have
Sir, in round two shouldn't ri' be 34 and 16?
Thank you so much sir the vedio was very helpful ... I never understood this before
You are most welcome
can you explain the round 3 how we got 4?
Can you explain how to caculate in round 3, i do not know why we have 4 in round 3
Dear, in round 3, you will generate new distance matrix based on values in the previous distance matrix. By employing the formula d'(ij,k)=(di,k + dj,k - di,j)/2 you will get the value 4.
Hassaan Awan thank you so much, this video is very useful
d(ij,k)=d(i,K)+d(j,k)-d(i,j/2) here ij is ab and k is cd. now use distances obtained in round 2 i.e, 17+8-17/ 2= 4
Maybe you could help me, I'm trying to understand what something means, I have a neighbor joining tree and there are certain values on the branches (i.e. 100, 75, 52) and I don't understand what those mean and there's a scale (it goes from 0.250 to 0) and that I also don't understand, maybe you could help me? What does that mean?
the values with the branches (100, 75, 52) are bootstrap values. Please google that. The scale is a distance scale, in principle the number of inferred mutations normalized by sequence length,
Thank you so much, this was more than helpful! :)
Thank you so much sirrr......❤️❤️❤️
Do u use fb or insta I have a question picture how to send you ?
Very nice explanation sr
Hii I have question for you
jizakAllah , very nicely explained...
sir can we share your slide?
Yes
Nicely explained sir. Thank you. Jai shree ram
Very big problem
From India ❤
I think rd value is wrong
May u correct or suggest.
18:05 when you displayed r'i, maybe you forgot to divide it by 2.
@@marzanbinteabid7538 To find r'i value, it is not necessary to divide by 2; rather, it's supposed to be divided by n-2. Here, the n=3 (A, B & CD). so n-2=1. Therefore those're actually divided by 1.
@@arafathossen7084 Ok. Thanks.
Привет
too lengthy and complicated
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