Pittsburgh Steelers defensive end Stephon Tuitt had a great game Sunday against the Cleveland Browns and the former second-round draft pick out of Notre Dame has been named the AFC Defensive Player of the Week for his performance.
Tuitt recorded 6 total tackles against the Browns and that included 2.5 sacks and 5 total quarterback hits.
With fellow defensive end Cameron Heyward lost for the remainder of the season last week with a pectoral injury, Tuitt was moved around quite a bit on the Steelers defensive line against the Browns and he responded in a big way.
Tuitt now enters Week 12 with 33 total tackles on the season, three and a half sacks, nine quarterback pressures, and two forced fumbles. He is the first Steelers’ player to be named AFC Defensive Player of the Week since safety Troy Polamalu won the award in Week 12 of 2013 season.
Tuitt is now in his third season in the league and being as next year will be the final year of his rookie contract, he might receive a lucrative extension at some point prior to the start of the regular season.
