Denovo solution

Q1. Illumina

Q1A. discarded contains reads that are too short, pair1 and pair2 files contain the read pairs were both passed trimming and singleton are reads were one of the two pairs were discarded.

Q2. Around 84

Q3. N = (M*L)/(L-K+1) = (84*99)/(99-15+1) = 97.84 Genome_size = T/N = (213721367+212523694)/97.84 = 4.35Mb

Q4. Mean = 259 ; SD = 11

Q5. It is lower, this means that the actual kmer peak we found (unless you found one higher than 84) is higher (this would give a lower genome size).

Q6. 10 of 195 contigs were scaffolded into scaffolds, this is quite few - normally it is much higher. A reason for this could be that our insert size is quite low (~250 bp) and the repeats in the genome are larger than this.

Q7. Repeat regions

Q8. Contaminations

Q9. Because we use the reference genome as the truth it may be hard to distinguish what is a misassembly and what is true variation from the reference genome.

Q10. This is of course just visual, but it seems that most part of the reference genome is covered by our assembly, so yes.

Q11. Yes, a couple of the small contigs does not map at all, and the C1097 only maps partially. This could be sequence in our strain, but not in the reference genome.

Q12. This is a region with a lot of repeats, this is also why we cant really assemble it. It is used by V. cholerae to integrate new genes into its genome.

Q13. The Nanopore assembly only has 2 contigs and pacbio 1!