Nov. 28, 2012
NEW YORK, N.Y. - Former Maryland linebacker D'Qwell Jackson, currently of the Cleveland Browns, was named the AFC Player of the Week for his play in the Browns' 20-14 win over Pittsburgh last week, the NFL announced Wednesday.
Jackson posted a team-high nine tackles, one forced fumble, one fumble recovery and one pass defensed. His play helped limit the Steelers to 242 total yards (193 passing, 49 rushing).
Jackson was part of a defensive effort that forced eight turnovers (five fumble recoveries, three interceptions), the most by an NFL team since the New Orleans Saints registered eight takeaways on October 28, 2001. The Browns scored 17 points off turnovers.
He also helped hold Pittsburgh to an 11 percent (one of nine) conversion rate on third downs. The win marked Cleveland's first defeat of the Steelers since December 10, 2009.
In his seventh season, this is Jackson's first career Defensive Player of the Week Award. He is the first Brown to win Defensive Player of the Week since David Bowens in 2010 (Week 7).
At Maryland, Jackson was a two-time All-American (2004, 2005) and three-time All-ACC selection (2003-05). He ranks fourth in school history with 447 career tackles.