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04 November 2015


The National Football League announced today that Tampa Bay Buccaneers linebacker Kwon Alexander has been named NFC Defensive Player of the Week for Week 8.
Alexander, earning his first career Player of the Week Award, set a career-high with 11 tackles in Tampa Bay’s overtime win at Atlanta (11/1), while adding a pass defensed, an interception, a forced fumble and a fumble recovery. He is the first rookie linebacker to record an interception, a forced fumble and a fumble recovery in the same game since Demeco Ryans (2006) and the first Buccaneers player to accomplish the feat since John Lynch (1996). His two takeaways led to 10 Buccaneers points, while his 11 tackles were tied for the eighth-most in the NFL in Week 8.
Drafted in the fourth round (124th overall), Alexander has started all seven games for Tampa Bay this season, tallying 49 tackles, two tackles for loss, seven passes defensed, two interceptions (15 return yards), one sack, one forced fumble and one fumble recovery. He leads all NFL linebackers in passes defensed, with seven and is tied for the league led in interceptions by a linebacker. Alexander is the only player in the league to have at least 45 tackles, two interceptions, one sack, one forced fumble and one fumble recovery.
Alexander is the first Buccaneers rookie to earn Player of the Week honors since RB Doug Martin in 2012 (Week 9) and is only the third Tampa Bay rookie to be named Defensive Player of the Week, joining Santana Dotson (Week 4, 1992) and Paul Tripoli (Week 4, 1987). Alexander, who earned Pepsi Rookie of the Week honors following Tampa Bay’s Week 3 contest at Houston, joins Martin as the only rookies in team history to earn both Player of the Week and Rookie of the Week honors during their first campaign.
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