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#1




Calculating the "P" in HardyWeinberg Equation
here is question about PKU relating to following equation
p2 + 2pq + q2 = 1 prevalence of PKU is 1 in 10,000 live births WHAT IS CARRIER FREQUENCY here q2= 1/10,000 we want to calculate 2pq q= 1/100 =0.01 HOW WE CAN CALCULATE p?? prof says p would be 1q how p would be 1q following above equation? 
#2




Second part to the equation is P+q=1
so thts how u get p=1q 
#3




can u explain plz how p+q=1?
i didnt find second part of equation in book 


#4




P+q=1 is as follows
p=frequency of normal allele q= frequency of disease allele So if we add these both we get total allele frequency irrespective of diseased or not which represents entire population. reference http://anthro.palomar.edu/synthetic/synth_2.htm 
The above post was thanked by:  
INCOGNITO (05162011) 
#5




Quote:
great idea its deduction of equation from main stem equation which is p+q=1 taking square of it p2+2pq+q2=1 thanx a lot 
#6




Quote:
Is the answer 1/25 ? 
#7





#8




yeah my bad
From goljan Prevalence rate = [carrier rate/2] raised to 2 
#9




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then it does not fit this scenerio... how to explain according to goljan? 
#10




yeah
RR path goljan page 82 third editon He discuss how to calculate prevalence if carrier rate is given .i just posted as i saw prevalence and carrier rate,so posted ..........so it cant be used here or is this forumula only for AD ? Last edited by noothan; 05162011 at 10:20 AM. Reason: error 
#11




Quote:
q^ = 1/10,000 => q = 1/100 then, P^2 + 2P*(1/100) + 1/10,000 = 1 in order to get 1, since p and p^2 are too small, P must approach to 1... finally: CARRIER FREQUENCY = 2Pq = 2*1*(1/100) = 2/100 = 1/50 I hope this help Last edited by Claus_CU; 05282011 at 10:29 AM. 
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#12




The way I remember this for the exam is: The carrier frequency = 2 x the square root of the prevalence of the disease.
The assumption that this makes (and is generally true) is that the prevalence of the nondiseased approximates 1. That is why 2pq is generally just 2 p since the q can be dropped as it approximates 1. 
#13




So... we first multiply (P*q) and then we multiply that result by 2??? Is that correct?
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"Disease is very old, and nothing about it has changed. It is we who change as we learn to recognize what was formerly imperceptible." JMC 
#14




If q is 1/1,000,000 or whatever than p is 999,999/1,000,000 no matter how low is q, bare in mind that p is always 1q. So carrier frequency will be 2x(1q)xq you can choose the closest answer option to your value. Practice this and then when u see the logic u can assume the p is 1 or very close to it when q is extremely low.
But in cas of CF let's say in Norway, q will be around 1/5...
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The above post was thanked by:  
Dr. Mexito (03252012) 
Tags 
BiostatisticsEpidemiology, Genetics 
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