Iowa Will Play Host To 2008 Big Ten Men's Tennis Championship :: Ohio State is top seed in this year's event

April 23, 2008

Iowa will host the 2008 Big Ten Men's Tennis Championship this weekend with matches beginning on Thursday, April 24, and continuing through Sunday, April
29.

The opening day of the tournament will feature six schools with No. 8 Iowa and No. 9 Purdue leading off at 9 a.m. CT. No. 6 Indiana will then battle No. 11 Northwestern at Noon CT. To conclude the opening round, No. 7 Minnesota will face No. 10 Michigan State at 3 p.m. CT.

The five top seeds all have first-round byes and will begin their portion of the championship on Friday. No. 1 Ohio State, No. 2 Wisconsin and No. 3 Illinois will play the winners of Thursday's matches. No. 4 Michigan and No. 5 Penn State will face off in the other quarterfinal match.

Ohio State, ranked No. 2 nationally by the ITA, is looking for its third-consecutive Big Ten Championship this weekend. The Buckeyes were a perfect 10-0 in conference
play and finished the regular season with an overall record of 29-1. Individually, OSU's Bryan Koniecko is ranked 17th, and teammate Justin Kronauge is ranked
22nd in the nation by ITA.

In last year's event, Illinois fell to Ohio State in the championship match. The No. 3-seed Fighting Illini will be led this year by Ryan Rowe, who is currently ranked No. 19 in singles play by the ITA. In all, 13 Big Ten athletes are ranked in the ITA top-125 singles, 12 pairs rank in the top 60 doubles and six teams fall in the top 50.

This year's event will be broadcast nationally by the Big Ten Network. The network will air the championship match on tape delay on Wednesday, May 7, at 8 p.m. ET.