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Jadeveon Clowney named NFC Defensive Player of the Week

SANTA CLARA, CALIFORNIA – NOVEMBER 11: Defensive end Jadeveon Clowney #90 of the Seattle Seahawks recovers a fumble to score a touchdown over the San Francisco 49ers during the second quarter at Levi’s Stadium on November 11, 2019 in Santa Clara, California. (Photo by Thearon W. Henderson/Getty Images)

SEATTLE — Seahawks defensive end Jadeveon Clowney’s has been named NFC Defensive Player of the Week after his standout performance in Monday’s big win over the 49ers.

Clowney had his biggest impact since joining Seattle this season. He returned a fumble 10 yards for the Seahawks’ first touchdown, set up another with a strip sack and had five quarterback hits on the day.

The Seahawks sacked 49ers quarterback Jimmy Garoppolo five times in Santa Clara Monday night. Although Clowney only made one of those sacks, “he was the dominant force that got it all going,” Seahawks reporter John Boyle notes.

Clowney is the second Seahawks player to get the Player of the Week honor this season. Quarterback Russell Wilson has been named NFC Offensive Player of the Week twice this season, once in week 2 and again in week 9 after the Hawks beat the Tampa Bay Buccaneers.

Coach Pete Carroll described the nail-biting overtime win against San Francisco as a “breakout game” for Clowney.

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