Kollin Moore had a big weekend in B1G action.
The conference honored the redshirt junior with the second Big Ten Wrestler of the Week award, the second weekly honor of his career. Ranked No. 2 in the country at 197 pounds, Moore picked up a pair of victories over top-12 opponents to help Ohio State defeat Nebraska and Purdue in the final weekend of the Big Ten dual-meet season.
The conference split the award between Moore and Iowa's Alex Marinelli, a graduate of the famed Graham wrestling program in St. Paris, Ohio.
Moore kicked off the weekend by defeating No. 9 Christian Brunner at Purdue by a 12-5 decision on Friday night, and followed that up with a 7-5 decision over No. 11 Eric Schultz of Nebraska on Sunday in St. John Arena.
In the Nebraska match he hit a big takedown less than 10 seconds into sudden victory to win the bout.
No. 2 Kollin Moore with the takedown in Sudden Victory for a 7-5 decision!!@wrestlingbucks takes an 18-12 lead!#GoBucks pic.twitter.com/bglJLHQlgq
— Ohio State on BTN (@OhioStateOnBTN) February 17, 2019
The Big Ten Network will feature his match against Schultz on the network's Wrestling in 60 program Tuesday at 9 p.m.
With those wins Moore is now 15-1 on the season, with his only loss to Penn State's Bo Nickal, the No. 1 wrestler in the class and a favorite to win the Hodge Trophy.
Moore is the third Buckeye to earn Big Ten Wrestler of the Week honors this season, joining Myles Martin (Dec. 4) and Joey McKenna (Nov. 6).
