Let p^ = population proportion of class of interest, here p^ = 0.52; Za/2 = population distribution for one sided test; and E = maximum error allow, say 0.03. The population proportion sample size is given by:

n = Za/22p^(1 - p^) / e2

If Za/2(0.95) = 1.96; p^ = 0.52 and e = 0.03, then the sample size is:

n = (1.96)2(0.52)(0.48) / (0.03)2

n = 3.84(0.2496) / 0.0009

n = 0.9585 / 0.0009

n = 1065

https://onlinecourses.science.psu.edu/stat414/node/212

Let p^ = population proportion of class of interest, here p^ = 0.52; Za/2 = population distribution for one sided test; and E = maximum error allow, say 0.03. The population proportion sample size is given by:

n = Za/22p^(1 - p^) / e2

If Za/2(0.95) = 1.96; p^ = 0.52 and e = 0.03, then the sample size is:

n = (1.96)2(0.52)(0.48) / (0.03)2

n = 3.84(0.2496) / 0.0009

n = 0.9585 / 0.0009

n = 1065

https://onlinecourses.science.psu.edu/stat414/node/212

I suppose your primary aim is to find out if an association exists between the variation and OP.  Use http://www.openepi.com/SampleSize/SSCohort.htm to calculate your sample size. The following information is needed for sample size calculation:

1. What is the anticipated proportion of the variation in the general population?

2. what is the minimum odds ratio you want to detect? OR What is the expected or anticipated proportion of the variation OP patients.

once you know 1 & 2 above just put in the values in the appropriate tables in the link given and your sample size will be calculated.

Let's suppose the expected or anticipated proportion of the variation in general population is 10% and you expect the variation proportion in the OP patients to be 15% (or if the difference in the two proportions is at least 5% then such a difference would be clinically significant for you). The sample size will be 1380 given that the prevalence of OP is 52% (which I use to calculate the ratio of exposed versus not exposed)

These calculations provided assume that there is no confounding in the study, however I think that Farahnaz may have a number of confounders, as she indicates there are genetic variations that she believes will impact OP. Given that, it may be important to try and run sample size calculations via simulation, plugging in some estimates for the confounders. In Stata, for example, there is a fantastic user-written command called powersim that conducts simulation-based power analysis for linear and generalized linear models.

Based on Dr. Inaamul Haq is answer, I think you need re-set the alpha level to allow for the multiple testing adjustment. It is safe to set a genomic significant level of alpha, that is 10*-7. 

please see the following calculation:

We are planning a study of independent cases and controls with 1 control(s) per case. Prior data indicate that the probability of exposure among controls is .1. If the true odds ratio for disease in exposed subjects relative to unexposed subjects is 2, we will need to study 1243 case patients and 1243 control patients to be able to reject the null hypothesis that this odds ratio equals 1 with probability (power) .8. The Type I error probability associated with this test of this null hypothesis is .0000005. We will use an uncorrected chi-squared statistic to evaluate this null hypothesis.

if the OR is 1.5, the sample size needed for each group increase to  3997.

I'd like to ask what is the distribution of the genetic variables in your population. when you get the answer you might be able to calculate the sample size you need, because your study looks like a case control study where the OP are your cases andnon-OP controls. so its not he proportion of OP you need to calculate the sample size.

this article is very useful for sample size determination

Article Sample size calculation