THE DRAFT GENOME OF Ruditapes philippinarum (THE MANILA CLAM)

Ruditapes philippinarum (Bivalvia Veneridae) Synonyms: Venerupis philippinarum, Tapes philippinarum, Venerupis japonica, Tapes japonica. Common Names: Japanese littleneck, Manila clam C-value = 1.97 pg (Genome Size = 1.93 Gb)

Sequencing

1. PacBio RSII P6-C4 Chemistry, 10-50 Kb BluePippin size selected libraries, 54 SMRTcells. Coverage ~30x.

2. Illumina HiSeq 2500, 2x250 bp, 1 short insert library (450-500bp) + 1 long insert library (1500-1700bp). Coverage ~40x.

UCI Genomics University of California,

Irvine

Why R. philippinarum?

1. BIVALVIA is a highly successful and ancient Class (20,000+ known species). Important group for EVOLUTIONARY and BIODIVERSITY studies.

2. FOOD: Bivalve aquaculture production: 13.2+ million metric tons (20% of the global aquaculture production). Clams are first in production (5 million tons), followed by oysters (4.7 million tons), mussels (1.8 million tons), and scallops (1.6 million tons).

3. GOOD MODEL FOR STUDYING ADAPTATION: anoxia and hypoxia (intertidal zones, buried in sediment); salinity and temperature (same species, different environments).

3. BIOINDICATORS: they filter water to get food and oxygen accumulating heavy metals and pollutants.

4. EXCEPTIONAL MODEL FOR STUDYING MITOCHONDRIAL INHERITANCE AND EVOLUTION: THE DOUBLY UNIPARENTAL INHERITANCE (DUI)

UPC Genome Core University of Southern California, Los Angeles

fabrizio.ghiselli@unibo.it www.mozoolab.unibo.it

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Fabrizio Ghiselli, Aleksey Komissarov, Liliana Milani, Joseph P. Dunham, Sophie Breton, Sergey V. Nuzhdin, Marco Passamonti

Manila Clams: Hatchery and Nursery Methods

Innovative Aquaculture Products Ltd. WWW.InnovativeAqua.com46

Develop your ownstandards to evaluate yourclam sets. Keep accuraterecords.

screen (500 µ to remove any larger material) onto a130 µ. screen. The total volume is measured in agraduated cylinder. Three samples are taken thencounted on a prepared slide (being sure to include onlylive animals as a part of the final count). The totalvolume is divided by the sample volume and thenmultiplied by an average of the samples

Total larvae added to system = 250000 animalsTotal volume of the post set clams = 12 ml.Sample #1 = 0.01 ml = 160 animalsSample #2 = 0.01 ml = 155 animalsSample #3 = 0.01 ml = 171 animals

To find the total live animals in the total volume ofpost set clams :Add sub samples and average 160+155+171 / 3 =162 animals per 0.01 mlDivide the total volume by 0.01, then multiply thatnumber by the 162 animals in the sample.12ml/0.01 = 1200 1200 X 162 = 194400 liveanimalsTo find the percent survival or "success" :Divide total live animals by the total number initiallyadded to the system and multiply by 100 %194400/250000 X 100 = 77.8%

a. Natural heteroplasmy b. Coevolution, conflicts c. Mitochondria under selection for male

functions d. Mitochondrial inheritance and bottleneck e. Role of mitochondria in germ line

development f. Mitochondrial recombination

R. philippinarum Complete 99 40% 1,967 Mb 30X + 40X R. philippinarum Partial 185 75%

[Modified from: Murgarella et al. 2016 PLoS ONE 15;11(3):e0151561]

De novo assembly

We have tried to assembly the data with all available hybrid or PacBio-only assembly pipelines. The best results were obtained with PacBio reads assembled with Canu. Major problems: 1. High heterozygosity 2. Large number of tandem repeats 3. Many repeats larger than PacBio reads 4. Correction and scaffolding of PacBio reads

Now what?

Right now we are working on PacBio correction, and on scaffolding of PacBio reads using Illumina super-reads.