Houda Bakkali Awarded at American Illustration 38 for “Transgression”

Houda Bakkali was awarded at American Illustration 38 for her work “Transgression,” from her “Beautiful African Woman” series. The piece celebrates the power, emancipation, and visibility of women, with the artist’s own mother serving as its muse.

American Illustration 38 was celebrated in New York on November 7, 2019, with “The Party,” a gala honoring that year’s award-winning artists and bringing together some of the world’s leading illustrators.The selected works were chosen by a prestigious jury of art directors from Crown Publishing, Smithsonian Magazine, BuzzFeed News, The New York Times, Politico, Art & Mechanical, and National Geographic.

Women’s strength and empowerment are among the central themes running throughout Bakkali’s work.