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.

= 0.244

Explanation:

We're analyzing a simple Mendelian trait: one gene with two alelles.

The notation I will use is the following:

Dominant homozygous (AA) or heterozygous (Aa):

A_ = normal plant

Recessive homozygous:

aa = dwarf plant

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

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.

In this problem:

Normal height = x 1072 = 804

Dwarf = Normal height = x 1072 = 268

These are the EXPECTED number of individuals.

The formula is   = ∑

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

Replacing the data in the formula:

= +

= 0.244

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