We think that dwarfism in river birch might be a simple Mendelian trait. We have taken a pure breeding normal plant and crossed it with a a pure breeding dwarf plant. The resulting F1 plants were normal height. The F1 plants were selfed and the F2 data is presented below: Normal River Birch 811 Dwarf River Birch 261 Perform a chi square analysis on this data using the following hypothesis: The mode of inheritance for height in river birch is simple Mendelian. What is the calculated chi-square value? Round to three decimal places.

Initially a pure breeding normal plant (AA) was crossed with a pure breeding dwarf plant (aa). Pure breeding means that the individual is homozygous for all the genes being studied.

P AA x aa

F1 Aa

100% of the F1 has the A alelle, therefore normal height.

Then, they selfed the F1 plants:

F1 x F1 Aa x Aa

The resulting plants had the following phenotypes:

Normal height = A_ = 811

Dwarf = aa = 261

Total = 811 + 261 = 1072

These are the OBSERVED number of individuals.

From Mendel's law of segregation, when analyzing a single gene with two traits and after a cross between pure breeds expressing different traits we expect a homogeneous F1 expressing the dominant trait, and a F2 expressing the dominant and recessive trait in a 3:1 ratio.

To calculate the EXPECTED number of individuals, we have to multiply the expected ratio of the phenotypic trait with the total number of individuals obtained in the offspring.

In this problem:

Normal height = [tex]\frac{3}{4}[/tex] x 1072 = 804

Dwarf = Normal height = [tex]\frac{1}{4}[/tex] x 1072 = 268

These are the EXPECTED number of individuals.

Finally, we calculate the chi-square value.

The formula is [tex]X^{2}[/tex] = ∑ [tex]\frac{(O - E)^{2}}{E}[/tex]

Where O is Observed number of individuals and E is expected number of individuals.

Replacing the data in the formula:

[tex]X^{2}[/tex] = [tex]\frac{(811 - 804)^{2}}{804}[/tex] + [tex]\frac{(261 - 268)^{2}}{268}[/tex]

[tex]X^{2}[/tex] = 0.244