Rams’ Donald to miss first game due to injury

NFL

THOUSAND OAKS, Calif. — Los Angeles Rams defensive tackle Aaron Donald will not play on Sunday against the Seattle Seahawks because of a high ankle sprain, missing a game due to injury for the first time in his NFL career.

Rams head coach Sean McVay said Wednesday that Donald’s ankle injury will not require surgery, but the team will take it week by week with his availability. Donald injured his ankle early on against the Kansas City Chiefs in Week 12 but finished the game for Los Angeles.

“That’s what makes Aaron, Aaron,” McVay said. “He’s unbelievable. I mean, to be able to play through that. The toughness. Everything that he embodies is what you love about Aaron Donald.”

Donald has missed two games in his career, both during the 2017 season, but one was because his holdout ended the day before the Rams’ opener and the other because McVay rested several players in the last game of the regular season.

In addition to Donald, the Rams are expecting to play Sunday without quarterback Matthew Stafford, who is dealing with a neck injury and is in the concussion protocol. McVay said Stafford would not practice on Wednesday and said it’s “safe to say” he would not play against Seattle.

Los Angeles will also be without wide receiver Cooper Kupp, who has a high ankle sprain, and wide receiver Allen Robinson, who has a stress fracture in his foot.

In 11 games this season, Donald has five sacks, 10 tackles for a loss, 11 quarterback hits and a forced fumble.

Products You May Like

Articles You May Like

Ravens ‘haven’t lost any confidence’ in Tucker
Cowboys lose Pro Bowl TE Ferguson to concussion
Our guide to every Week 12 NFL game: Matchup previews, predictions, picks and nuggets
Torn PCL could sideline Packers’ Alexander longer
Mavs’ Doncic out at least week with wrist strain

Leave a Reply

Your email address will not be published. Required fields are marked *