Jarrett named AFC Defensive Player of the Week

Florham Park N.J.- Brooklyn native Jaiquawn Jarrett was rewarded for his outstanding play in the Jets 20-13 win against the Pittsburgh Steelers by being named AFC Defensive Player of the Week.

Jarrett was the star of Sunday’s win with ten tackles (seven solo, three assists) a fumble recovery, a sack, and two interceptions.

Jarrett’s three takeaways equaled the amount the Jets defense forced through the first nine games of the season.

“After the performance he had, I’m sure he’ll probably be, if not, he should be, the (AFC) Defensive Player of the Week. It’s hard for me to fathom somebody had a better game than he had” said head coach Rex Ryan.

This is the first time Jarrett has won the Defensive Player of the Week award and is the first Jet to do so since to win the award since Dee Milliner won it for his two interception game against the Miami Dolphins in week 17 of last year.


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