Featherweights set to meet on July 7 in Las Vegas
A title challenger and a recent lightweight will meet in a featherweight bout on the UFC’s big Independence Day weekend card.
Chad Mendes, the last challenger to Jose Aldo‘s featherweight title, will face Cody McKenzie, who fought just last week as a lightweight, at UFC 148 in July. UFC officials announced the fight over the weekend.
UFC 148 takes place July 7 at the MGM Grand Garden Arena in Las Vegas. The card features a main event middleweight title fight between Anderson Silva and Chael Sonnen, plus a co-main event interim bantamweight title fight between Mendes training partner Urijah Faber and Renan Barao.
Mendes (11-1, 2-1 UFC) in January dropped a featherweight title fight to Aldo at UFC 142 in Rio de Janeiro, Brazil. Aldo connected with a knee to the head and follow-up punches to get the knockout with just one second left in the first round. That loss was the Team Alpha Male product’s first of his professional career.
Mendes started his career 5-0 before getting a call form the WEC, then went 4-0 there, including wins over Erik Koch and Cub Swanson, before moving to the UFC in the promotions’ merger. He picked up decision wins over Michihiro Omigawa and Rani Yahya last year before getting his shot at Aldo.
McKenzie (13-2, 2-2 UFC) is coming off a first-round guillotine choke submission win over Marcus LeVesseur at UFC on Fuel TV 3 last week in Fairfax, Va. That victory snapped a two-fight losing streak that saw him drop submissions to Yves Edwards and Vagner Rocha in 2011.
McKenzie, a product of Season 12 of “The Ultimate Fighter,” amazingly started his career with 12 straight submission wins – 10 by a modified guillotine choke he calls the “McKenzietine.” His TUF 12 run ended with a TKO loss to Nam Phan in the quarterfinals – though that doesn’t count as an official loss on his record. But before that, he won two fights in the house with a pair of McKenzietine chokes, and his win over Aaron Wilkinson at the TUF 12 Finale was also with the special guillotine.
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