Gronkowski Named AFC Offensive Player of Week

BOSTON (CBS) – Patriots tight end Rob Gronkowski has been named the AFC Offensive Player of the Week, the team announced Wednesday morning.

Gronkowski had two more touchdowns Sunday in New England’s 34-27 win over the Washington Redskins. He finished the day with six receptions for a career-high 160 yards.

Gronkowski’s first touchdown of the afternoon was his 14th of the season, setting a new NFL single-season mark for touchdown receptions by a tight end.

Gronk has 71 receptions for 1,088 yards and 16 total touchdowns this season.

Additionally, Gronkowski became just the second New England tight end to go over 1,000 yards for the season on Sunday. Ben Coates set the franchise mark for a tight end with 1,174 yards in 1994. Gronkowski needs 203 yards in the final three games to set the NFL record for most yards in a season by a tight end, held by Kellen Winslow, who had 1,290 yards in 1980 with San Diego.

Watch: Patriots Annual Wrap-A-Pat

Tune in to the Patriots-Broncos game Sunday on WBZ-TV and 98.5 The Sports Hub at 4:15pm. Pregame coverage begins Sunday morning at 11:30am on WBZ-TV with Patriots Gameday; pregame coverage on the Hub begins at 1pm. After the game, tune in to The Postgame Show on 98.5 and Patriots Fifth Quarter on MyTV38.

Comments

One Comment

  1. Michele says:

    Congrats to Rob Gronkowski! He is an amazing player, and the Patriots are so lucky to have him!!! GO PATS!!!!!!!!!!!!!!!!!!

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