Lamarr Houston Named AFC Defensive Player of the Week

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DE Lamarr Houston forces a fumble that leads to a Raiders recovery and a game-winning field goal in overtime. Photo by Tony Gonzales

Oakland Raiders defensive end Lamarr Houston has earned the AFC Defensive Player of the Week award, the National Football League will announce Wednesday.

Houston, a 6-foot-3, 300-pound veteran in his third year out of Texas, totaled seven tackles (six solo), one sack and a monumental forced fumble in the Raiders' 26-23 overtime win over Jacksonville Sunday. On the third play in overtime, Houston closed in on Cecil Shorts from behind, separating the ball from the receiver. Teammate Joselio Hanson recovered, setting up Sebastian Janikowski's game-winning field goal from 40 yards out.

Sunday marked the second time this season Houston has chased down a ball-carrier from behind to force a fumble and return possession to the Raiders. He also sprinted more than 50 yards to cause a Demaryius Thomas fumble, preventing a touchdown at Denver Sept. 30.

Houston on Sunday made four significant plays over the final four Jacksonville drives, including the overtime fumble. He began the fourth quarter by sacking Chad Henne, shooting at the QB's feet from a sprinter's stance to force Jacksonville to settle for a field goal after a Raiders fumble deep in their own territory, then stopped Rashad Jennings for a 1-yard gain just prior to the two-minute warning in a tie game. He also stuffed Jennings on the first play of overtime.

Earlier against Jacksonville, Houston nearly registered a safety by flooring Jennings for a 4-yard loss to the Jags' 3, and registered a pair of pressures, one to force an incompletion one play after a Raiders interception, and another that came dangerously close to intentional grounding.

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