Can I apply a one-way ANOVA and Tukey test to this data? The experiment:
Effect of pH on yeast respiration. I tested 5 different pH levels and measured the amount of CO2 produced. The data shows a trend similar to upside down parabola. Would it make sense to apply ANOVA and Tukey, just ANOVA, or is it not really applicable?
 A: With the clarification in your comment, and with the understanding that the measurements on different pH levels are independent, then you can treat this as a one-way ANOVA. The null hypothesis would be that
all pH levels have the same population mean level of $\text{CO}_2,$
That is $H_0: \mu_1 = \mu_2 = \mu_3 = \mu_4.$ The alternative hypothesis is that not all five of the $\mu_i$ are equal. For a standard one-way ANOVA computation we assume that the carbon dioxide measurements are normally distributed and that all five pH groups have the same population variance. 
If $H_0$ is rejected, then we suppose there are some significant differences among the $\mu_i$ and we would like to know the pattern of those differences. Then the Tukey HSD procedure would be one
of several ways to judge which of the $\mu_i$ are different from which other(s).
I don't have the actual data at hand, so here is an illustration
using Minitab software on fake data.
Here are group means and standard deviations of the fake data I used.
 Level  N    Mean  StDev 
 pH1    5  60.818  2.375 
 pH2    5  52.098  1.354 
 pH3    5  50.712  2.984 
 pH4    5  60.926  2.793 
 pH5    5  57.926  2.801 


From the plots it does seem that there may be significant
differences among the five groups. Here is the ANOVA table.
 Source  DF      SS      MS      F      P
 Factor   4  465.76  116.44  18.20  0.000
 Error   20  127.94    6.40
 Total   24  593.70

The P-value of the test of $H_0$ is printed as 0.000,
which means less than 0.0005. So there are there are some
significant differences at any reasonable level of significance.
Because $H_0$ is rejected, it makes sense to use the Tukey method to see where the
significant differences may lie. (If $H_0$ is not rejected
it is not appropriate to use the Tukey method.)
A Tukey HSD procedure gave the following diagram:
 Grouping Information Using Tukey Method

      N    Mean  Grouping
 pH4  5  60.926  A
 pH1  5  60.818  A
 pH5  5  57.926  A
 pH2  5  52.098    B
 pH3  5  50.712    B

This means that levels pH1, pH4, and pH5 are not deemed significantly different among themselves. Similarly levels pH2 and pH3 are not
deemed significantly different. However the three levels marked A
all deemed significantly different from either of the levels marked B. 
Note: Because these are simulated data the actual pattern of population
means is known: $\mu_1 = 60,\, \mu_2 = 54,\, \mu_3 = 50,\, \mu_4 = 53,\, \mu_5 = 58$ and the population standard deviation of each group
is $\sigma = 3.$ Notice that because of random variation, the
order of the $sample$ means is not quite the same as for the known
$population$ means. Also, there are some differences in population
means that were not detected by the Tukey method. With only
five replications per group, no statistical analysis can reliably reclaim the exact pattern of the population means. (In practice
of course, the true population means are not going to be known.)
