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. 2021 Jun 26;14(1):evab151. doi: 10.1093/gbe/evab151

Table 2.

Sequences and SIFT Scores for the “Toy Example” Three-Codon Sequence

Codon 1
Codon 2
Codon 3
Nucleotides C C A G G T C A G
Degeneracy 0 0 4 0 0 4 0 0 4
SIFT for A TOL DEL TOL TOL DEL TOL TOL DEL TOL
SIFT for C TOL TOL TOL TOL DEL DEL TOL TOL TOL
SIFT for G DEL DEL TOL TOL TOL TOL TOL DEL TOL
SIFT for T DEL TOL TOL TOL DEL TOL DEL DEL TOL
Total
0 TOL → TOL 1/3 1/3 0 1 0 0 2/3 0 0 2.33
0 TOL → DEL 2/3 2/3 0 0 1 0 1/3 0 0 2.66
0 DEL → TOL 0 0 0 0 0 0 0 1/3 0 0.33
0 DEL → DEL 0 0 0 0 0 0 0 2/3 0 0.66
4 TOL → TOL 0 0 1 0 0 2/3 0 0 1 2.66
4 TOL → DEL 0 0 0 0 0 0 0 0 0 0
4 DEL → TOL 0 0 0 0 0 1/3 0 0 0 0.33
4 DEL → DEL 0 0 0 0 0 0 0 0 0 0

Note.—For each position, the SIFT score of the four possible nucleotides is given. The nucleotides present in the alignment are in bold, with the score in italics corresponding to the derived alleles. From this, each polymorphism can be assigned to a degeneracy category (0 or 4) and a delta SIFT score category (TOL →TOL, TOL →DEL, DEL →TOL, DEL →DEL). In the example, SNPs are thus classified as follows: 0-TOL →TOL (pos. 1), 4-TOL →TOL (pos. 3), 0-TOL →DEL (pos. 5), and 0-DEL →TOL (pos. 8). Each position also contributes to the length of the eight possible categories depending on the opportunity of mutations at this site. For example, at position 1, starting from the ancestral nucleotide C (TOL), one possible mutation is TOL →TOL and the two others are TOL →DEL, so this position contributes 1/3 the length of 0-TOL →TOL category and 2/3 to the 0-TOL →DEL category. The contribution of all positions is then summed across the ancestral sequence to obtain the total length of each category.