This working example is based on a
JC69 genetic distance matrix computed from the
5S ribosomal RNA sequence alignment of five bacteria:
Bacillus subtilis (a),
Bacillus stearothermophilus (b),
Lactobacillus viridescens (c),
Acholeplasma modicum (d), and
Micrococcus luteus (e).
First step •
First clustering Let us assume that we have five elements (a,b,c,d,e) and the following matrix D_1 of pairwise distances between them : In this example, D_1 (a,b)=17 is the smallest value of D_1, so we join elements a and b. •
First branch length estimation Let u denote the node to which a and b are now connected. Setting \delta(a,u)=\delta(b,u)=D_1(a,b)/2 ensures that elements a and b are equidistant from u. This corresponds to the expectation of the
ultrametricity hypothesis. The branches joining a and b to u then have lengths \delta(a,u)=\delta(b,u)=17/2=8.5 (
see the final dendrogram) •
First distance matrix update We then proceed to update the initial distance matrix D_1 into a new distance matrix D_2 (see below), reduced in size by one row and one column because of the clustering of a with b. Bold values in D_2 correspond to the new distances, calculated by
averaging distances between each element of the first cluster (a,b) and each of the remaining elements: D_2((a,b),c)=(D_1(a,c) \times 1 + D_1(b,c) \times 1)/(1+1)=(21+30)/2=25.5 D_2((a,b),d)=(D_1(a,d) + D_1(b,d))/2=(31+34)/2=32.5 D_2((a,b),e)=(D_1(a,e) + D_1(b,e))/2=(23+21)/2=22 Italicized values in D_2 are not affected by the matrix update as they correspond to distances between elements not involved in the first cluster.
Second step •
Second clustering We now reiterate the three previous steps, starting from the new distance matrix D_2 Here, D_2 ((a,b),e)=22 is the smallest value of D_2, so we join cluster (a,b) and element e. •
Second branch length estimation Let v denote the node to which (a,b) and e are now connected. Because of the ultrametricity constraint, the branches joining a or b to v, and e to v are equal and have the following length: \delta(a,v)=\delta(b,v)=\delta(e,v)=22/2=11 We deduce the missing branch length: \delta(u,v)=\delta(e,v)-\delta(a,u)=\delta(e,v)-\delta(b,u)=11-8.5=2.5 (
see the final dendrogram) •
Second distance matrix update We then proceed to update D_2 into a new distance matrix D_3 (see below), reduced in size by one row and one column because of the clustering of (a,b) with e. Bold values in D_3 correspond to the new distances, calculated by
proportional averaging: D_3(((a,b),e),c)=(D_2((a,b),c) \times 2 + D_2(e,c) \times 1)/(2+1)=(25.5 \times 2 + 39 \times 1)/3=30 Thanks to this proportional average, the calculation of this new distance accounts for the larger size of the (a,b) cluster (two elements) with respect to e (one element). Similarly: D_3(((a,b),e),d)=(D_2((a,b),d) \times 2 + D_2(e,d) \times 1)/(2+1)=(32.5 \times 2 + 43 \times 1)/3=36 Proportional averaging therefore gives equal weight to the initial distances of matrix D_1. This is the reason why the method is
unweighted, not with respect to the mathematical procedure but with respect to the initial distances.
Third step •
Third clustering We again reiterate the three previous steps, starting from the updated distance matrix D_3. Here, D_3 (c,d)=28 is the smallest value of D_3, so we join elements c and d. •
Third branch length estimation Let w denote the node to which c and d are now connected. The branches joining c and d to w then have lengths \delta(c,w)=\delta(d,w)=28/2=14 (
see the final dendrogram) •
Third distance matrix update There is a single entry to update, keeping in mind that the two elements c and d each have a contribution of 1 in the
average computation: D_4((c,d),((a,b),e))=(D_3(c,((a,b),e)) \times 1 + D_3(d,((a,b),e)) \times 1)/(1+1)=(30 \times 1 + 36 \times 1)/2=33
Final step The final D_4 matrix is: So we join clusters ((a,b),e) and (c,d). Let r denote the (root) node to which ((a,b),e) and (c,d) are now connected. The branches joining ((a,b),e) and (c,d) to r then have lengths: \delta(((a,b),e),r)=\delta((c,d),r)=33/2=16.5 We deduce the two remaining branch lengths: \delta(v,r)=\delta(((a,b),e),r)-\delta(e,v)=16.5-11=5.5 \delta(w,r)=\delta((c,d),r)-\delta(c,w)=16.5-14=2.5
The UPGMA dendrogram The dendrogram is now complete. It is ultrametric because all tips (a to e) are equidistant from r : \delta(a,r)=\delta(b,r)=\delta(e,r)=\delta(c,r)=\delta(d,r)=16.5 The dendrogram is therefore rooted by r, its deepest node.
Comparison with other linkages Alternative linkage schemes include
single linkage clustering,
complete linkage clustering, and
WPGMA average linkage clustering. Implementing a different linkage is simply a matter of using a different formula to calculate inter-cluster distances during the distance matrix update steps of the above algorithm. Complete linkage clustering avoids a drawback of the alternative single linkage clustering method - the so-called
chaining phenomenon, where clusters formed via single linkage clustering may be forced together due to single elements being close to each other, even though many of the elements in each cluster may be very distant to each other. Complete linkage tends to find compact clusters of approximately equal diameters. ==Uses==