Rayshawn Jenkins told reporters he had never recorded a pick-six at any level of football until he did so for Sunday’s walk-off, overtime victory over the Cowboys.
That play is part of why he’s now earned something else he never has before — a player of the week award.
Jenkins is the AFC defensive player of the week after he recorded a whopping 18 tackles and two interceptions in the win against Dallas.
According to the league, Jenkins’ 18 tackles were the most for a player in a game with multiple interceptions since 1991.
Jenkins is also the first Jacksonville defensive player to earn a player of the week award since Week Nine of the 2021 season.
In his second season with the Jaguars, Jenkins has 95 total tackles, 11 passes defensed, two forced fumbles, and three interceptions.
The Jaguars will play again on Thursday night in a critical matchup against the Jets.
