Example I

See Genetics 15: pg 95

Cross AABB x aabb F₁ AaBb. Self-fertilize AaBb to develop F₂ family. The following data was observed:

This excess of A_B_ and aabb classes indicates that the loci are in coupling linkage. For coupling we use the ratio st/ru to use the table and find p. In this example:

We can use Table 2 (pg. 88) to find p by looking up under the coupling column between values 0.01419 and 0.01586. The crossover value is between 0.095 and 0.10.

We can then add 0.095 + 0.0048 = 0.0998 = p

We had to interpolate between 0.095 and 0.10 because st/ru = 0.01579 which was between 0.01419 and 0.01586. The associated standard error of p is also found from table 2.

The standard error is between 0.2097/(N)^0.5 and 0.2153/(N)^0.5.

160 x 0.0056 = 0.0053
167

0.0053 + 0.2097 = 0.2150/(N)^0.5 is the standard error of p by interpolation.