Spider-Man '94 will be a four-issue limited series, written by J.M. DeMatteis and illustrated by Jim Towe. Fans don't exactly ...

There's no Mary-Jane Watson - as we've known her - in the MCU ... yet. Some Spider-Man fans believe this could change, and Georgie Henley could be involved.

Peter Parker and Mary Jane Watson are among the most beloved couples in not just Spider-Man comics, but comic books in ...

Tony-nominated actress and singer Jennifer Damiano shared her new EP, I'm So Sorry, Mary Jane. The collection of songs, ...

Mary Jane Watson has been mostly known for decades as Spider-Man's girlfriend/wife. When does she get to be her own person? ... One could argue she was the original MPDG (Manic Pixie Dream Girl).

Mary Jane Watson is a reporter and Peter Parker’s girlfriend. Despite her lack of superpowers, she has proven to be a valuable ally to Peter due to her courage and investigative prowess.

While longtime Spider-Man fans weren’t all that shocked to meet the Ultimate Universe’s version of Anna May Watson (Mary Jane’s Aunt and neighbor to Peter’s Aunt May in the main Marvel ...

Mary Jane Watson will officially take up the mantle of Jackpot in Amazing Spider-Man #31, ... The original Jackpot was Sara Ehret, a scientist who developed a serum that gave her super strength.

Mary Jane Watson's sister, who made some minor, if prominent appearances way back when in the 616. Created by Tom DeFalco and Ron Frenz for Amazing Spider-Man #259 in 1984, she's back for her ...

Mary Jane Watson, A.K.A. JACKPOT, gets her first solo super story since her debut in ASM #31!" ... The original Jackpot was Sara Ehret, a scientist who was imbued with super-strength.

Twitter definitely likes the idea of Zendaya as Mary Jane Watson in next summer’s Spider-Man: Homecoming.. Social media went nuts Thursday night when industry website The Wrap reported that ...

Reddit user u/uptowndrunk7 responded to a thread about dream pairings in the MCU. During their message, they posited Georgie Henley as Mary Jane Watson in the MCU. As one user pointed out, however ...