EARTH CITY, Mo. -- The NFL announced their weekly awards Wednesday morning and Rams defensive end Robert Quinn was prominently involved.
Quinn earned NFC Defensive Player of the Week honors for a thoroughly dominant performance against Arizona in which he finished with three sacks and a pair of forced fumbles. The fumble he caused in the fourth quarter led to the game-tying field goal.
In the coaches' review of the game, Quinn had two tackles, two forced fumbles, three sacks, three quarterback pressures and three quarterback hits. He also drew a holding penalty against clearly overmatched Arizona tackle Levi Brown.
The folks over at Pro Football Focus certainly believed Quinn was deserving of the honor. In their Insider piece rating the best and worst performance of week one, Quinn drew a grade of +8.2, the highest score in the league and the only one that surpassed the 8.0 mark.
Apparently playing against Arizona is good not only for Quinn, who registered three sacks against the Cardinals in the teams' meeting in St. Louis last year, but for winning the defensive honors all together. Rams cornerback Janoris Jenkins was the most recent Ram to win the award, claiming it after coming up with two pick-sixes against, you guessed it, the Cardinals in Week 12 last year.
The award is the second weekly honor of Quinn's career but his first for his work on defense. He won the special-teams award in 2011 for blocking a punt in a win against New Orleans.
The Rams believe Quinn has many more performances like he had on Sunday in him. He's regularly shown he can win matchups that he should but the next step will be regularly beating some of the elite tackles he's going to see against the likes of San Francisco, Seattle and Houston. There's little doubt the ability is there for him to do just that and claim the accolades that go with it along the way.