For each of the following tasks you\'ll use the \'chado\' database. You need onl

Question

For each of the following tasks you'll use the 'chado' database. You need only provide the query or command you used within MySQL for each.

mysql> show tables
-> ;
+--------------------------+
| Tables in chado |
+--------------------------+
| analysis |
| analysisfeature |
| analysisprop |
| cm_blast |
| cm_cluster_members |
| cm_clusters |
| cm_cvterms |
| cm_gene_structure |
| cm_genes |
| cm_proteins |
| cv |
| cvterm |
| cvterm_dbxref |
| cvterm_relationship |
| cvtermpath |
| cvtermprop |
| cvtermsynonym |
| db |
| dbxref |
| dbxrefprop |
| feature |
| feature_cvterm |
| feature_cvterm_dbxref |
| feature_cvterm_pub |
| feature_cvtermprop |
| feature_dbxref |
| feature_pub |
| feature_relationship |
| feature_relationship_pub |
| feature_relationshipprop |
| feature_relprop_pub |
| feature_synonym |
| featureloc |
| featureprop |
| featureprop_pub |
| organism |
| organism_dbxref |
| organismprop |
| project |
| pub |
| pub_dbxref |
| pub_relationship |
| pubauthor |
| pubprop |
| synonym |
| tableinfo |



mysql> describe cv;
+------------+--------------+------+-----+---------+-------+
| Field | Type | Null | Key | Default | Extra |
+------------+--------------+------+-----+---------+-------+
| cv_id | int(11) | NO | PRI | NULL | |
| name | varchar(255) | NO | UNI | NULL | |
| definition | varchar(255) | YES | | NULL | |


-> describe cvterm;
+---------------------+---------------+------+-----+---------+-------+
| Field | Type | Null | Key | Default | Extra |
+---------------------+---------------+------+-----+---------+-------+
| cvterm_id | int(11) | NO | PRI | NULL | |
| cv_id | int(11) | NO | MUL | NULL | |
| name | varchar(255) | NO | MUL | NULL | |
| definition | varchar(1200) | YES | | NULL | |
| dbxref_id | int(11) | YES | MUL | NULL | |
| is_obsolete | int(11) | NO | | NULL | |
| is_relationshiptype | tinyint(1) | NO | | NULL | |
+---------------------+---------------+------+-----+---------+-------+
7 rows in set (0.01 sec)

select * from cv;
+-------+--------------------------------+------------+
| cv_id | name | definition |
+-------+--------------------------------+------------+
| 1 | relationship | NULL |
| 2 | synonym_types | NULL |
| 3 | annotation_attributes.ontology | NULL |
| 4 | ANNFLG | NULL |
| 5 | TIGR_role | NULL |
| 6 | component.ontology | NULL |
| 7 | EC | NULL |
| 8 | evidence_code | NULL |
| 9 | evidence_code2.obo | NULL |
| 10 | GFF3 | NULL |
| 11 | GO | NULL |
| 12 | biological_process | NULL |
| 13 | molecular_function | NULL |
| 14 | cellular_component | NULL |
| 15 | go_properties | NULL |
| 16 | output.ontology | NULL |
| 17 | SO | NULL |
| 18 | ARD | NULL |
+-------+---------------------------

Explanation / Answer

As per the schema......by look at the cv table.....Go and go_properties are linked controlled vocabulay terms to that cv table.....so it count is 2.
Coming most terms linked are that two only Go and go_properties ......so GO ontology was linked most....

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