Is a t-test applicable here in comparing two proportions? I have 3 categories of tissues examined by microscopy (A , B and C, 12 of each). I want to test whether the proportion of tissues staining by a particular dye, differs in the three categories. Each  of the 36 tissue specimens is stained and examined under the microscope. The data is collected in the form of proportion of cells staining positive for the dye in each tissue specimen.
Now, I understand that the proportions have a binomial distribution and t-test is probably not appropriate because the variance of proportions is known. But I don't know how to apply the z-test to detect the difference in proportion here, since the data collected itself is in the form of "proportion" for each specimen. Do I calculate the mean proportion ($p_A,p_B,p_C$ for each category) by averaging the recorded proportions under each observation and then apply the z-test?
As an alternative thought, am I not estimating the proportion from my study population by averaging the proportions themselves. Thus, since I am actually estimating the true population proportion from sample itself, would the assumptions of z-test hold here?

(Not thirty minutes ago was I reading a paper that made me think about this same topic!)
The t-test is okay for testing proportions. However, proportions are a rare time where we $100\%$ know the population distribution. Sure, there are other times where we do not know the population distribution and rely on the decent robustness of the t-test to violations of the normality assumption, but with proportions, we know that each observation is Bernoulli. Consequently, there is no statistical reason not to use a test for Bernoulli data (with the caveat that doing so is one of the smaller errors that one can make, due to the robustness of the t-test, so it probably is not that big of a deal).
If you have the data to t-test proportions, you have the data to z-test proportions. To do the t-test, you have to have the mean (the proportion) and the sample size (to get the degrees of freedom in the t-distribution). From these values, you can determine how many instances there are of your $0$-case and your $1$-case, and then you can fill in the variables in all of the equations for z-testing (or other proportion tests, like Fisher’s exact test or the chi-squared test).
