Changes between Version 29 and Version 30 of SOPs/InProgress

Timestamp:

12/07/15 16:07:16 (10 years ago)

Author:

gbell

Comment:

--

SOPs/InProgress

@@ -1 @@
-'''Calculating number of reads needed for certain coverage'''
+== How long reads should you use?  Should they be single or paired-end? ==
 
-* These are some useful references:
-   Ref1 for estimates of the number of reads required for single nucleotide variant calling: [http://www.ncbi.nlm.nih.gov/pubmed/21771779/]
-
-   Ref2 for estimates of the number of reads required for single nucleotide variant calling: [http://www.ncbi.nlm.nih.gov/pubmed/24434847]
-
-   For estimates of the number of reads required for RNA-seq and ChIP-seq experiments: [http://www.ncbi.nlm.nih.gov/pubmed/24434847/]
+ * What is the goal of your experiment?
+   * For typical RNA-seq expression level quantification, a read or read pair gets one count, regardless of the length.  As a result, shorter reads may provide just as good data, as long as they aren't so short that repetitive mapping is a problem.
+   * Longer and/or paired reads are surely beneficial if the experimental goal is 
+       * novel gene discovery: longer reads are much better at identifying novel splice junctions
+   * For variant discovery, coverage is key, whether it's fewer long reads or more shorter reads (as long as the reads are long enough to map uniquely)
 
 
- *  ''Example 1''
+== If you are able to sequence more than one lane, how should the samples be divided? ==
 
- For a  3e+9 nt genome if we want 35x coverage we would need: 
+   * The magnitude of a lane effect is typically small but probably non-zero.
+   * To balance any lane effect, sequence all of your samples on each of your lanes.
+   * Another benefit of barcoding and mixing all samples together is that the samples can be re-sequenced in other lanes in the future (from the same library preparation) without unbalancing the experimental design.
 
- 3e+9 * 35 / 40 = 2.625e+09 = 2.6 billion 40-nt reads[[BR]]
- or[[BR]]
- 3e+9 * 35 / 100 = 1.05e+09 = 1 billion 100-nt reads
+== Calculating number of DNA or RNA reads needed to obtain the desired coverage ==
 
+ * Some useful references:
+   * Sims et al., 2014.  [http://www.ncbi.nlm.nih.gov/pubmed/24434847 Sequencing depth and coverage: key considerations in genomic analyses.]  
+       * Includes methods to estimate the number of reads required for single nucleotide variant calling, and RNA-seq and ChIP-seq experiments
+   * Ajay et al., 2011. [http://www.ncbi.nlm.nih.gov/pubmed/21771779/ Accurate and comprehensive sequencing of personal genomes.]  
+       * Includes methods to estimate the number of reads required for single nucleotide variant calling 
 
- *  '' Example 2''
+ * ''Example 1'' (genome sequencing): For a genome of 3e+9 nt, to get 35x coverage we would need: 
+   * For 40-nt reads:
+      * 3e+9 * 35 / 40 = 2.625e+09 => ~2.6 billion reads
+   * For 100-nt reads:
+      * 3e+9 * 35 / 100 = 1.05e+09 => ~1 billion reads
 
- For an RNA_seq experiment:
- If we have 6 million paired end reads and a genome with ~7000 genes expressed X 5741 bp average gene length = 40,187,000. That is 40 mill nt to cover.
-
- 6M reads x 70 bp (35 pb per paired end reads) = 420 mill bp that we will cover.[[BR]]
- 420 mill nt that we will cover/ 40 mill nt to cover ~ 10 x coverage.
+ *  '' Example 2'' (RNA_seq experiment): 
+   * If we have 
+     * 6 million 35x35-nt paired end reads
+     * a genome with ~7000 genes expressed 
+     * average gene length = 5741 bp
+   * then the total length of the transcriptome is 7000 x 5741 => 38,297,000 nt
+   * and the total length of the reads is 6 million x 70 nt [35 + 35] => 420,000,000 nt
+   * so the average coverage will be 420,000,000 / 38,297,000 => ~11x
+   * but note that coverage will be very irregular to due a wide range of expression levels