Colts’ DeForest Buckner named AFC Defensive Player of the Week


Indianapolis Colts defensive tackle DeForest Buckner (99) celebrates a turnover during an NFL football game against the Houston Texans on Sunday, Dec. 20, 2020, in Indianapolis. (AP Photo/Zach Bolinger)

INDIANAPOLIS — The NFL has named DeForest Buckner, defensive tackle for the Indianapolis Colts, as the AFC Defensive Player of the Week for Week 15.

He’s being recognized for his work on the field in the Colts’ 27-20 win against the Houston Texans.

During that game, Buckner totaled four tackles, 2.0 tackles for loss , a career-high 3.0 sacks and one forced tumble.

In a release from the Colts, the organization said Buckner is one of just four players in the NFL this season to register at least four tackles, 2.0 tackles for loss, 3.0 sacks and one forced fumble in a single game.

When it comes to defensive tackles across the league, Buckner is tied third for tackles, tied second for solo tackles, second in sacks and tied for third in forced fumbles.

Buckner is the fifth Indianapolis player to earn AFC Player of the Week honors this season. It is the first time in his career he has received this recognition from the NFL.

You can follow Mike Chappell on Twitter at @mchappell51.

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