Denovo solution: Difference between revisions
(Created page with "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 g...") |
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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. | 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 | Q10. This is just visual, but it seems that most 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. | 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. | ||
Revision as of 13:04, 28 November 2024
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 just visual, but it seems that most 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!