DeSanto earns Big Ten Wrestler of the Week

Austin DeSanto’s upset of No. 1 Seth Gross earned him the first Big Ten honor for Iowa this season.


Shivansh Ahuja

Iowa’s 133-pound Austin DeSanto wrestles Wisconsin’s Seth Goss during a wrestling match between No.1 Iowa and No. 6 Wisconsin at Carver-Hawkeye Arena on Sunday, Dec. 1, 2019. DeSanto won by decision, 6-2, and the Hawkeyes defeated the Badgers, 32-3.

Anna Kayser, Sports Editor

Iowa junior 133-pounder Austin DeSanto was named the Big Ten Wrestler of the Week after upsetting No. 1 Seth Gross of Wisconsin on Sunday to take over his top-ranked spot.

DeSanto was ranked No. 2 before the dual, which Iowa won 32-3. DeSanto won his match 6-2, with a 3-0 advantage in takedowns.

This season, he wrestled at 141 for Iowa’s season-opener against Tennessee-Chattanooga and won the match by technical fall. He then moved back down to 133 against Iowa State and took the match by a major decision.

DeSanto transferred to Iowa last season and went 23-6. He placed fourth in the Big Ten Championships and fifth at the NCAA Championships. When bonus points were in play, he was 12-0 last season with two pins.



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