I met this guy who keeps saying he is a Q-dawg(as he puts it), but he does not have the brand and when he said he was the "dean" of the line, I found out he was trying to get into the fraternity as a graduate. How can I find out if he is really a Q or is he just pretending?
2007-02-11 11:52:14 · 5 answers · asked by angeefive 1 in Society & Culture ➔ Cultures & Groups ➔ Other - Cultures & Groups
Well the simple fact that youre are questioning his membership is an indicator in itself. If he is a member of my beloved Omega Psi Phi Fraternity Incorporated then he will have many friends who are also bruhs. And he will know of events and parties that the bruhs are having. If you never see him around Ques then hes prob not one. If all else fails ...let me speak to him ..I will let u know for sure...LMAO
And no All bruhs dont get branded. It is not mandatory in any way shape or form.
Just ask him to bring you around his frat brothers
2007-02-13 07:43:54 · answer #1 · answered by acedog2006 1 · 0⤊ 0⤋
Omega Psi Phi Branding
2016-09-29 05:07:00 · answer #2 · answered by ? 4 · 0⤊ 0⤋
This Site Might Help You.
RE:
Do all Omega Psi Phi members have get the brand?
I met this guy who keeps saying he is a Q-dawg(as he puts it), but he does not have the brand and when he said he was the "dean" of the line, I found out he was trying to get into the fraternity as a graduate. How can I find out if he is really a Q or is he just pretending?
2015-08-18 10:56:17 · answer #3 · answered by Dela 1 · 0⤊ 0⤋
nope not at all
2017-03-04 20:32:09 · answer #4 · answered by Jomekia 1 · 0⤊ 0⤋
Branding is not required for ANY fraternity....its a choice..my boyfriend had one and only a few others.
Find out who was his line brothers.....that will tell it all !
2007-02-11 12:12:41 · answer #5 · answered by Diamond in the Rough 6 · 0⤊ 1⤋
Go to his frat and find out. Ask the president of the frat!
2007-02-11 16:16:42 · answer #6 · answered by Melissa Breazeale 2 · 0⤊ 0⤋
ahmad white? yeah he is
2016-03-17 07:30:21 · answer #7 · answered by Anonymous · 0⤊ 0⤋