To simplify, let C. elegans have 3 chromosomes each with various combinations of A and T. Each chromosome has a top (Crick) strand and a bottom (Watson) strand. If we count the numbers of As and Ts in a top strand we might find more As than Ts (A>T). Because of the Watson-Crick base pairing rules, if we count the bases in the corresponding bottom strand we would find more Ts than As (A<T). In the top strand there is an excess of As, and in the bottom strand there is an equivalent excess of Ts. This is consistent with Chargaff’s first parity rule (PR1) – for duplex DNA, %A = %T.
But there is also Chargaff’s second parity rule (PR2) – for single strands of DNA, %A ≈ %T (Forsdyke, 2006). Thus, differences between the numbers of As and Ts in single strands are small. In a hypothetical case, for the first chromosome the top strand difference (A – T) might be +100 (i.e. A > T); for the second chromosome the top strand difference might be –25 (i.e. A < T); for the third chromosome the top strand difference might be –75 (i.e. A < T). Now, here’s the point. The total difference for three chromosomes, in this case, would be zero (100 – 25 – 75 = 0). The chromosomes would seem to balance their differences out. They would seem to “account.”
However, if the original sequencers had assigned the third chromosome the opposite orientation (i.e. had turned it through 180°), the total difference for the three chromosomes would be 100 – 25 + 75 (= 150). For any set of chromosomes there are alternative orientations to explore. For three chromosomes there are 23 (= 8) alternatives. If we designate the assigned orientation as TTT, then there are 8/2 alternatives (TTT, TBT, TTB, and TBB, where T means top and B means bottom). But there is also a complementary set of 8/2 alternatives for the bottom strand (BBB, BTB, BBT, and BTT). From a cell’s viewpoint these two sets would seem identical.
Thus, for six chromosomes, there are 26/2 = 32 alternatives. If random, there would be only a 1/32 chance that the assigners of chromosome orientation had chosen one that minimized the AT difference (P = 0.03). But this is what we find for the six C. elegans chromosomes. The GC difference is not so clear cut (Forsdyke et al., 2010). My enquiries to date indicate that the original assignments were indeed random.
References
Forsdyke DR. (2006). Evolutionary Bioinformatics. Springer, New York.
Forsdyke DR, Zhang C, and Wei J-F. (2010). Chromosomes as interdependent accounting units. The assigned orientation of C. elegans chromosomes minimizes the total W-base Chargaff difference. J. Biol. Sys. (in press) 
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Error in your editing at end of first paragraph, which should read:
” … more Ts than As (A<T). In the top strand there is an excess of As, and in the bottom strand there is an equivalent excess of Ts. This is consistent with Chargaff’s first parity rule (PR1) – for duplex DNA, %A = %T."
Whoops. More editorial errors:
Third paragraph should read: “2 [superscript] 3 (= 8) alternatives” not “23 (=8) alternatives”.
Fourth paragraph should read: “2 [superscript] 6/2 = 32 alternatives” not “26/2 = 32 alternatives”
Thanks for your keen eye, and my apologies for the editing SNAFUs. We’ve corrected these and put in place checks for the next issue.
— eds.