A test cross of PQR over pqr with pqr over pqr gives 100 offspring in eight classes…

Step 1. (a) PqR differs from PQR in the gene Q only. So Q is in the middle, and the order is P, Q, R.

Step 2. (b) The classes Pqr and pQR are single crossovers between P and Q ($5 + 5 = 10$). Add the double crossovers ($1 + 1 = 2$). Total $12$ of $100$, so $12$ map units.

Step 3. (c) PQr and pqR are single crossovers between Q and R ($2 + 2 = 4$). Add the double crossovers ($2$). Total $6$ of $100$, so $6$ map units.

Step 4. (d) The map distance is $12 + 6 = 18$ map units. By direct count, P and R show only $10 + 4 = 14$ recombinants, because the double crossovers cancel.