Unbeaten welterweight prospect Brandun Lee will face Samuel Teah on March 10, it was announced Wednesday night.

The 10-round bout will take place at the Mohegan Sun Casino in Uncasville, Connecticut and will headline a three-bout ShoBox telecast (Showtime, 9 pm ET/ 6 pm PT). 

The hard-hitting Lee (21-0, 19 knockouts), who resides in Yuba City, California, last fought in a stay-busy fight on December 19, stopping Dakota Linger in round 3. In his previous fight, which was also part of a ShoBox telecast, Lee dropped Jimmy Williams three times before winning by knockout.

March 10 will mark Lee’s third appearance on the popular series. The 21-year-old debuted on ShoBox March 13, stopping Camilo Prieto in round 3. 

Teah will be Lee’s toughest test on paper and it will be interesting to see if he can take Lee into the later rounds. Lee has only reached round 4 on two occasions.

The 33-year-old Teah (17-3, 7 KOs), who is originally from Monrovia, Liberia and now resides in Philadelphia, Pennsylvania, defeated journeyman Dieumerci Nzau by unanimous decision in his last bout on March 7. In his previous fight in November 2019, Teah won a hard-fought majority decision over Sonny Fredrickson. 

Teah’s fight against Lee will mark his fourth appearance on ShoBox. Teah defeated O’Shaquie Foster, who is currently ranked No. 10 by The Ring at 130 pounds, in November 2015, but lost by majority decision to Montana Love in February 2018. Over nine months after the loss to Love, Teah defeated Kenneth Sims, Jr. by unanimous decision.

Francisco A. Salazar has written for The Ring since October of 2013 and has covered boxing in Southern California and abroad since 2000. Francisco also covers boxing for the Ventura County (Calif.) Star newspaper. He can be reached by email at [email protected] or on Twitter at FSalazarBoxing