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Bud Dupree named AFC Defensive Player of the Week

Dupree has been one of the biggest surprises of the 2019 season, and it should pay off with a big contract this coming offseason.

Indianapolis Colts v Pittsburgh Steelers Photo by Justin K. Aller/Getty Images

Steelers linebacker and former Kentucky Wildcats star Bud Dupree has been named the AFC Defensive Player of the Week.

The fifth-year veteran played a major part in the Steelers’ Week 9 upset of the Colts, as Dupree racked up two sacks, a forced fumble and a fumble recovery in the 26-24 win.

Dupree now has five sacks over his last five games. The Steelers have gone 4-1 during that stretch after starting the season 0-3 and seemingly out of the playoff mix before the calendar even turned to October.

“Bud is balling,” said Steelers linebacker T.J. Watt after the game via ESPN. “I said it last week, and I see the way that he approaches watching film and practice, and you just have a feeling when you’re playing. It seems like every week for me and Bud and this whole defense, we all just have feelings that it’s going to be a great game.”

Dupree, who is in a contract year, leads the Steelers in tackles for loss (8) and is second in sacks (6), QB hits (9) and forced fumbles (2) to go with 32 total tackles. His previous career-high for sacks in a single season was six in 2017. and he’s already hit that mark in just eight games.

“It’s opportunities and also just preparation,” said Dupree after Sunday’s win via “I took my game to a different level. It’s just been different moments where we come together as a unit and make stuff happen on the field and off the field.”

The former first-round pick is putting himself in line to sign a big contract extension this offseason.

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