The question asks to compute P-value and E-value If there are 2000 random sequen

Question

The question asks to compute P-value and E-value

If there are 2000 random sequences and each of the sequence have the length of 1086 nucleotides, we will have the following result:

1997 sequences which have a score smaller than 72

2 sequneces which have a score larger than 72

1 sequence which have a score of 72

Based on the previous count, estimate the probability of a local alignment score great than or equal to 72. If there had been 6,062 sequences, based on this probability how many alignments would you expect to have a score of at least 72?

Please show all your steps for the question. Thank you!

Explanation / Answer

as there are 3 sequence out of 2000 have  score great than or equal to 72

hence probability =3/2000 =0.0015

hence exoected number in 6062 sequences =np =6062*0.0015=9.093

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.