Colts’ DeForest Buckner named AFC Defensive Player of the Week

Kevin Hickey
·1 min read

Indianapolis Colts defensive tackle DeForest Buckner was named the AFC Defensive Player of the Week for his efforts in the 27-20 win over the Houston Texans.

Despite being limited to a season-low 51% of snaps, Buckner found several ways to ruin the day for the Texans. He finished with 3.0 sacks, four quarterback hits and two tackles for loss while leading the defense with five total pressures.

Buckner’s presence has been incredible this season, and he’s been the biggest reason the entire defense has taken a leap into the elite category. Despite this, he was snubbed for a Pro Bowl selection.

Bucker is currently second among all interior defenders with 7.5 sacks and 24 quarterback hits while he’s eighth in pressures.

Buckner has been enjoying a monster season, one that could consider him for an All-Pro spot at the end of the season given both his direct and indirect impact on the Colts this season.

Buckner is the fifth Colts player to earn a weekly award this season. Cornerbacks Xavier Rhodes (Week 3) and Kenny Moore (Week 14) were the Defensive Players of the Week while linebacker E.J. Speed (Week 10) and kicker Rodrigo Blankenship (Week 11) earned AFC Special Teams Player of the Week.

Buckner is finally getting some of the national recognition he deserves during this incredible season, and it’s starting in thanks to his big performance against a divisional foe in Week 15.