Foye Oluokun Named AFC Defensive Player of the Week

The Jacksonville Jaguars achieved a notable victory over the Cincinnati Bengals last Sunday, with linebacker Foye Oluokun playing a crucial role in the team's success.

In the game, Oluokun recorded seven tackles, an interception, a forced fumble, and a quarterback hit, contributing to the Jaguars' 22-17 win. His interception, which occurred at the Jaguars' 10-yard-line in the fourth quarter, was pivotal in maintaining a 16-10 lead as the game approached its conclusion.

On Wednesday, the NFL announced Oluokun's selection as the AFC defensive player of the week, marking the fourth time he has received this honor. He previously earned two such awards while playing for the Atlanta Falcons, with the latest two coming during his tenure with the Jaguars.