Question

I got to the part in bold and I have no idea where to go from here.

Step 1:Find the pair with the smallest distance.

Step 2:Join them in one group and draw those branches.

Step 3:Compute average distance between this group and all other taxa one-by-one and create new table with those values.

Step 4:Repeat Steps 1 to 3 using the new distance matrix.

Characters	A	B	C	D	E	F	G
A	-
B	12	-
C	8	7	-
D	37	26	2	-
E	17	41	35	30	-
F	28	21	5	11	19	-
G	42	16	15	32	46	10	-

Based on above data, find the first two characters with the shortest distance and draw those branches with the respective distances on the branches below according to the example given in class or online. This should be your first two branch tips.

C and D is the pair with the shortest distance as it is 2 units. A sub tree will be drawn with the branch point half way between the two given units. Each branch is 1 unit as this is 2 units divided equally and each point is 1 unit away from the center node.

Recalculate distances based on the combination of the last characters. This first table has been partially filled out for you. Determine and enter the missing values. (You need not fill out the boxes above the dashes as they are mirror values to the below-dash values.)

	A	B	CD	E	F	G
A	-
B	12.00	-
CD	22.50	(7+26)/2 = 16.5	-
E	17.00	41.00	32.50	-
F	28.00	21.00	(5+11)/2 = 8	19.00	-
G	42.00	16.00	(15+32)/2 = 23.5	46	10.00	-

Draw the next branch pair based on the shortest distance of your newly determined values and label each branch with its length and each tip with its character.

Recalculate new distances using the below table.

Draw new branches.

Recalculate distances.

Characters
	-
		-
			-
				-

Draw new branches.

Recalculate Distances.

Characters
	-
		-
			-

Draw new branches.

Answer 1

UPGMA method:

The pair with the lowest distance is clustered. Here C and D separated by the distance of 2, the branch point is separated at the distance of 1, the subtree is as follows.

dist(C,D),A = (distCA + distDA) / 2 = 8+37 = 45/2=22.5

dist(C,D),B = (distCB + distDB) / 2 = 7+26 = 33/2 = 16.5

dist(C,D),E = (distCE + distDE) / 2 = 35+30 = 65/2 = 32.5

dist(C,D),F = (distCF + distDF) / 2 = 5+11 = 16/2 = 8

dist(C,D),G = (distCG + distDG) / 2 = 15+32 = 47/2 = 23.5

second cycle:

The next lowest distance is between F and CD - 8.

dist(CD,F),A = (distCDA + distFA) / 2 = 22.5+28 = 50.5/2=25.25

dist(CD,F),B = (distCDB + distFB) / 2 = 16.5+21 = 37.5/2=18.75

dist(CD,F),E = (distCDE + distFE) / 2 = 32.5+19 = 51.5/2=25.75

dist(CD,F),G = (distCDG + distFG) / 2 = 23.5+10 = 33.5/2=16.75

Third cycle:

The next lowest distance is between A and B is 12

dist(A,B),CDF = (distACDF + distBCDF) / 2 = 25.25+18.75 = 44/2= 22

dist(A,B),E = (distAE + distBE) / 2 = 17+41 = 58/2=29

dist(A,B),G = (distAG + distBG) / 2 = 42+16 = 58/2=29

forth cycle:

The next minimum distance is with G and CDF - 16.75. The sub tree is

dist(G,CDF),AB = (distGAB + distCDFAB) / 2 = 29+22 = 51/2=25.5

dist(G,CDF),E = (distGE + distCDFE) / 2 = 46+25.75 = 71.75/2=35.8

Fifth cycle:

The next minimum distance is between AB and GCDF - 25.5

Final cycle:

dist(AB,GCDF),E = (distABE + distGCDFE) / 2 = 29+35.8 = 64.8/2=32.4

The final tree is: