Jones earns the honor for the second time in his career after leading the Denver Broncos to a win in London.
The NFL announced on Wednesday morning that former Ohio State defensive tackle Dre’Mont Jones has been named the AFC Defensive Player of the Week for his performance in the Denver Broncos’ 21-17 win over the Jacksonville Jaguars.
Jones, who was a third-round pick in the 2019 NFL Draft, recorded a season-high seven tackles, three tackles for loss and a sack to help the Broncos snap a four-game losing streak with a win over the Jaguars in London on Sunday.
This marks the second time in Jones’ career that he has garnered defensive player of the week honors, as he did so following a five-tackle, 2.5-sack performance in a 27-17 win over the Detroit Lions during his rookie season in 2019.
Jones has recorded 31 tackles, eight tackle for loss, eight quarterback hits, 5.5 sacks and one forced fumble in eight games, which is why general manager George Paton said this week they want him to remain in Denver beyond his rookie contract, which expires at the end of the season.
“He’s one of our core players,” Paton said during his press conference on Tuesday evening. “We want Dre’Mont here a long time, we’ll just leave it at that.”
Jones becomes the second Buckeye to take home a weekly honor this season, joining New York Jets wide receiver Garrett Wilson, who was named the rookie of the week after he caught eight passes for 102 yards and two touchdowns in a 31-30 win over the Cleveland Browns in Week 2.
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