TR2005 1978

Relational Taxonomy Treeand BioDataBaseby Huhn-Kie Lee: 

Relational Taxonomy Tree and BioDataBase by Huhn-Kie Lee

Part I.Relational Taxonomy Tree: 

Part I. Relational Taxonomy Tree

Relational Taxonomy Tree (RTT): 

Relational Taxonomy Tree (RTT) Taxonomic hierachy Kingdom, phylum, class, order, family, genus, species Lower level inherits higher level’s property: Properties may be stored “redundantly” Siblings differ by some properties: Properties are “disparate,” so we need different relation schemes

Relational Taxonomy Tree (RTT): 

Relational Taxonomy Tree (RTT) Carnivore herbivore animal Dalmatian cats dogs Chihuahua Russian cat Italian cat

Relational Taxonomy Tree (RTT): 

Relational Taxonomy Tree (RTT) Carnivore (prey, hunting method) Herbivore (feeding plant, chewing method) animal Dalmatian cats dogs Chihuahua Russian cat Italian cat

Relational Taxonomy Tree (RTT): 

Relational Taxonomy Tree (RTT) Carnivore Herbivore animal Dalmatian Cats (meowing sound, whiskers size) Dogs (barking sound, snout size) Chihuahua Russian cat Italian cat

Relational Taxonomy Tree (RTT): 

Relational Taxonomy Tree (RTT) Carnivore Herbivore animal Cats (meowing sound, whiskers size) Dogs (barking sound, snout size)

Relational Taxonomy Tree (RTT): 

Relational Taxonomy Tree (RTT) Carnivore (prey, preying method) Herbivore animal

Relational Taxonomy Tree (RTT): 

Relational Taxonomy Tree (RTT) Vertical query: join a relation with its ancestor relation “Find hunting method of a dog which barks “bow-bow” “ (See relations in slide 6, 5) SELECT Carn.hunting_method FROM Dogs D, Carnivore Carn WHERE D.speciesID = Carn.speciesID AND D.barking_sound = “bow-bow”

Relational Taxonomy Tree (RTT): 

Relational Taxonomy Tree (RTT) Horizontal query: join any two relations (may not in same level) “Find (barking sound, meowing sound) pair of dogs and cats which prey on the same animal (See relations in slide 6, 5) SELECT D.barking_sound, C.meowing_sound FROM Dogs D, Carnivore Carn1,Carn2, Cats C WHERE D.speciesID = Carn1.speciesID AND C.speciesID = Carn2.speciesID AND Carn1.prey = Carn2.prey

Multiple Inheritance from same-level parents : 

Multiple Inheritance from same-level parents Carnivore (prey, hunting method) Herbivore (feeding plant, chewing method) animal bear Black bear Grizzly bear Omnivore(prey, hunt, plant, chew)

Multiple Inheritance from diff-level parents: 

Multiple Inheritance from diff-level parents Carnivore (prey, hunting method) Herbivore (feeding plant, chewing method) animal Cats(meowing sound, whiskers size) dogs Pseudo-cat(meow,whisker,plant,chew)

Multi-Inherit Rules: 

Multi-Inherit Rules AB CD MN Add a taxon whose attribute set is MNABCD AB CD MN ABCD

Multi-Inherit Rules: 

Multi-Inherit Rules AB CD MN Add a taxon whose attribute set is MNCDEF AB CD MN EF EF EF

Multi-Inherit Rules: 

Multi-Inherit Rules AB CD MN Add a taxon whose attribute set is MNBC AB CD MN BC

Multi-Inherit Rules: 

Multi-Inherit Rules AB CD MN Add a taxon whose attribute set is KL AB CD MN KL

Multi-Inherit Rules: 

Multi-Inherit Rules AB CD MN Add a taxon whose attribute set is MK AB CD N K M

RTT is skewed: 

RTT is skewed karyote virus Genorg Multi-cellular prokaryote eukaryote Gram+bact1,2… bacteria archaea Virus1, virus2…. Gram+ bact gram - bact Archaea1,archaea2… Gram-bact1,2… mono-cellular

Terminal Relation: 

Terminal Relation karyote virus Genorg Multi-cellular eukaryote mono-cellular

Non-terminal Relation: 

Non-terminal Relation karyote virus Genorg Multi-cellular eukaryote mono-cellular -Save general trend in each subtaxon.

Non-terminal Relation: 

Non-terminal Relation animal plant Sexual eukaryote -Save common values of each subtaxon. -Terminal relation would be redundant. Asexual eukaryote eukaryote

Part II.BioDataBase: 

Part II. BioDataBase

BioDataBase (BDB): 

BioDataBase (BDB) Want to store all the information about all the living organisms on the planet Too many data! Solution: partition database into “Domains” Each domain has its own database that stores relevant biological infomation Want to find correlation between different domains’ information

BioDataBase (BDB): 

BioDataBase (BDB) Consider 3 domains and their relevant info: Genomics: genes of each species Ecology: population distribution of species Environment: a location’s humidity, temperature

BioDataBase (BDB): 

BioDataBase (BDB) Genomics: Species/gene is many-to-many relation Hence, (species, gene) relation lion zebra geneB geneA geneC

BioDataBase (BDB): 

BioDataBase (BDB) Ecology: Want to store species_A lives in location_B and the number of them is population_C PRIMARY KEY: (speciesID, locationID)

BioDataBase (BDB): 

BioDataBase (BDB) Environment: Want to store environmental factors that affect living organisms

BioDataBase (BDB): 

BioDataBase (BDB) Want to answer a query that spans all 3 domains: simply join relations from 3 domains! “Find genes that are common to (genomics) all species that live in the area (ecology) where humidity is low (environment)”

BioDataBase (BDB): 

BioDataBase (BDB) “Find genes that are common to all species that live in the area where the humidity is low“ (see relations in 14,15,16) (SELECT G.geneID, G.speciesID FROM Genomic G, Ecology Eco, Environment Env WHERE G.speciesID = Eco.speciesID AND Eco.location = Env.location AND Env.humidity = low ) DIVIDE (SELECT Eco.speciesID FROM Ecology Eco, Environment Env WHERE Eco.location = Env.location AND Env.humidity = low )

Part III.Conclusion & cs632 Project: 

Part III. Conclusion & cs632 Project

Conclusion: 

Conclusion Relational Taxonomy Tree solves Redundancy problem: diff. species have common attributes. Disparity problem: diff. species have diff. attributes RTT and BDB can serve as the prototype for the infrastructure of the Library of Life Project.

Tentative Project Suggestion: 

Tentative Project Suggestion There are four of us: Helgi, Yoni, Shobhi, mi. Two of us work on implementation of mini-Relational Taxonomy Tree The other two of us work implement mini-BioDataBase All of us implement a program that can process SQL queries on RTT & BDB

So what do you say?: 

So what do you say?