Doctor Who confirms fan theory about new character's family connections

David Tennant and Catherine Tate in Doctor Who
(Image credit: BBC)

Doctor Who is keeping it in the family. New character Rose, played by Heartstopper's Yasmin Finney, who will be joining the show for the upcoming 60th anniversary specials, is now confirmed to be the daughter of previous companion Donna Noble. 

"Welcome to the Doctor Who family, Rose Noble," the show's official account tweeted. "Wishing a very happy birthday to Yasmin Finney, whose character Rose is the daughter of companion Donna Noble (played by Catherine Tate) and Shaun Temple (Karl Collins)."

A follow-up tweet reads: "So, the question is, what happens when Rose meets one of her mum’s oldest friends? It is a mystery for now, but what we do know is that just like her mother, Rose stumbles across something alien – and from that point her seemingly ordinary family is never quite the same…"

Finney's casting was announced back in May, but we only knew that her first name was Rose. This is the first time her surname has been revealed, but fans have been speculating about her family history after clips saw Rose and Tate's Donna together.

Donna was the companion to David Tennant's Doctor in season 4, which aired in 2008. In the season finale, she saves the universe, but loses her memories of the Doctor in the process. She made another appearance in the 2009 Christmas special, along with her new husband Shaun. 

The Doctor Who 60th anniversary specials are set to air this November on BBC One in the UK and Disney Plus internationally. In the meantime, fill out your watch list with our picks of the best shows on Disney Plus to watch right now.

Entertainment Writer

I’m an Entertainment Writer here at GamesRadar+, covering everything film and TV-related across the Total Film and SFX sections. I help bring you all the latest news and also the occasional feature too. I’ve previously written for publications like HuffPost and i-D after getting my NCTJ Diploma in Multimedia Journalism.