The number of possible global alignments between two sequences of length N is _____(a) \(\frac{2^N}{\sqrt{πN}}\)(b) \(\frac{2^{2N}}{\sqrt{πN}}\)(c) \(\frac{2^{(N-1)}}{\sqrt{πN}}\)(d) \(\frac{2^{2N}}{\sqrt{N}}\)I have been asked this question during an interview for a job.My doubt is from Needleman topic in chapter Multiple Sequence Alignment of Bioinformatics

The number of possible global alignments between two sequences of leng

1.	The number of possible global alignments between two sequences of length N is _____(a) \(\frac{2^N}{\sqrt{πN}}\)(b) \(\frac{2^{2N}}{\sqrt{πN}}\)(c) \(\frac{2^{(N-1)}}{\sqrt{πN}}\)(d) \(\frac{2^{2N}}{\sqrt{N}}\)I have been asked this question during an interview for a job.My doubt is from Needleman topic in chapter Multiple Sequence Alignment of Bioinformatics
Answer» Right choice is (b) \(\frac{2^{2N}}{\sqrt{πN}}\) For EXPLANATION I would SAY: By the total number of permutations and combinations option b gives the accurate number of possible GLOBAL ALIGNMENTS between two sequences of length N. For two sequences of 250 residues this is 10^149.

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