DeSanto Honored By Big Ten

Iowa's Austin DeSanto (left) was named the Big Ten's wrestler of the week on Tuesday. (Darren Miller/hawkeyesports.com)
John Bohnenkamp

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.

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.

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