If 200 animals were tested, and two cases were found, we find the percentage of affected animals by dividing 2 by 200.

2 / 200 = .01 = 1%

Definitions:
P = gene for the normal (dominant)
p = gene for PRA (recessive)
A = the frequency of the P gene
B = the frequency of the p gene

Since all the genes at each locus must add up to 100% (or 1.0), A + B = 1
therefore AP + Bp = 1
But each dog carries 2 genes at the locus concerned, so we have 3 possible genotypes,

PP = genotype of genetically clear animals
Pp = genotype of carrier animals
pp = genotype of affected animals

When a male with genes P and p is mated to a female with P and p, according to the familiar quadratic formula,  the probable outcome is:    
A² PP + 2AB Pp + B² pp

If the frequency of the genotype of affected animals is found to be 1% (.01)
(here you substitute the frequency you found in the population)
then B² = 0.01 and B = 0.1 (the square root of 0.01)

but since B + A = 1,  then A = 0.9

Substituting, we get: 0.9² PP + 2(0.9 x 0.1) Pp + 0.1² pp
or 0.81 PP + 0.18 Pp + 0.01 pp

Thus the incidence of:
the clear genotype is 0.81 (81%)
the carrier genotype is 0.18 (18%)
the affected genotype is 0.01 (1%)