Greg Rousseau named AFC Defensive Player of the Week

The rookie defensive end was a standout against the Chiefs on Sunday night
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Buffalo Bills defensive end Greg Rousseau has been named the AFC's Defensive Player of the Week for his Week 5 performance against the Kansas City Chiefs on Sunday night at Arrowhead Stadium.

The rookie pass rusher was all over the field, collecting his first career interception to go along with five total tackles, including one for a loss, a sack, a quarterback hurry, and a pass breakup in the team's 38-20 win.

Rousseau becomes the first Bills rookie defensive player to win the weekly NFL honor since linebacker Cornelius Bennett did so in 1987.

So far through five games with the Bills, the 21-year-old has already outperformed his early expectations, registering 18 total tackles, four tackles for loss, three quarterback hits, three sacks and his interception.

The Bills' first round pick (30th overall) in the 2021 NFL Draft joins linebacker Tremaine Edmunds as players to earn AFC Defensive Player of the Week honors this season. Edmunds earned his AFC Defensive Player of the Week honor in Buffalo's Week 4 win over the Houston Texans.

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Rousseau has already become a big piece to the Buffalo defense through the first five games this season. The Bills currently have the top-ranked defense in the NFL, averaging a league-low 251.8 yards-against per-game and only averaging an impressive 12.8 points-against per-game.

Buffalo also leads the league this year with 15 takeaways through five games, which includes nine interceptions and six fumble recoveries.

Rousseau is also the third Bills player to already earn AFC Player of the Week honors this season. Along with Edmunds, Josh Allen earned AFC Offensive Player of the Week honors for his performance in Buffalo's Week 3 win over the Washington Football Team.

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