DeSanto Honored By Big Ten

Junior 133-pounder named conference's wrestler of the week.

Iowa's Austin DeSanto was named the Big Ten's wrestler of the week on Tuesday.

DeSanto, ranked No. 2 at 133 pounds last week, defeated Wisconsin's Seth Gross, the top-ranked wrestler in the class, 6-2 in last Sunday's dual meet at Carver-Hawkeye Arena. DeSanto moved to the No. 1 ranking after the win.

It was the first loss at 133 for Gross since the 2017 NCAA finals.

See why Austin DeSanto was tabbed the Big Ten Wrestler of the Week.

The @Hawks_Wrestling standout at 133 pounds claimed the honor after earning a 6-2 decision over top-ranked Seth Gross. pic.twitter.com/aCRqy8rJrd

— Iowa On BTN (@IowaOnBTN) December 3, 2019

DeSanto scored one takedown in the first period and two more in the second for the win.

It's the first Big Ten honor for the Hawkeyes this season, and the second of DeSanto’s career. He was named co-wrestler of the week on Jan. 22, 2019.

DeSanto and the top-ranked Hawkeyes wrestle Sunday at No. 12 Princeton.