Sentence examples for mean coverage divided from inspiring English sources

From these sets the mean coverage divided by the coverage of the sequences library results in a redundancy rate of 25%.

For each region we calculated the density, or total normalized coverage divided by region length.

Normalized coverage-distribution plots were calculated as follows: The mean coverage per exon was divided by the overall mean coverage of all exons as normalized coverage (x-axis).

The mean coverage was obtained by dividing the 286,454,112 bp of sequence that maps to the assembled genome by the 258,954,041 mappable bases of the assembled genome.

The copy number, per haploid B. malayi genome, was estimated by dividing the mean coverage of the respective nuwt by the mean coverage of B. malayi (i.e. 131 for paired end reads).

Normalized gene expression was calculated as the coverage of a contig from a given sample divided by the mean coverage of all the contigs in the reference sequence from the sample (Table  1).

Mean coverage depth (total base pairs in aligned reads divided by total base pairs in reference genome) for these strains over all samples was 3.9×, with depth-of-coverage means varying widely by body site from 0.04× (right antecubital fossa) to 11.1× (tongue dorsum) (Supplementary Table 8).

As the coverage is a function of the number of reads obtained for a given sample (see Table  3 for detailed per sample read and coverage statistics), data were normalized by dividing the coverage of each base by the mean coverage for that sample.

Mean coverage of the targeted exons was 697X.

The mean coverage for the leptin gene was ~ 80× [26].

Mean coverage probabilities in per cent.