Aresearcher planted 160 seeds from an f2 cross of two hybrid purple flowered plants and got 123 purple flowered progeny and 37 white flowered progeny. what was the chi- square value for this result?
how do we find the expected value for this problem if we are given the observe value?

Chi square value = 0.3

In order to find the expected value, an assumption of purple flower colour dominance over white flower colour has to be made.

Explanation:

First, we will assume that the purple flower trait is dominant over white flower trait, and that the alleles segregate according to Mendelian standard. Hence, the expected F2 phenotypic ratio will be 3:1.

In order to get the Chi square, the expected phenotype frequency will be calculated thus;

Expected frequency (E) for purple flowered plants = 3/4 x 160 = 120

Expected frequency (E) for white flowered plants = 1/4 x 160 = 40

Observed frequency (O) for purple flowered plants = 123

Observed frequency (O) for white flowered plants = 37

Chi square =

For the purple colour flower

, which gives 0.075

For the white colour flower

, which gives 0.225

Total = 0.075 + 0.225, which gives 0.3

step 1 ammonium can be absorbed by plants, but some is converted into nitrates which are better for plants to absorb. this is called nitrification. 2. step 2 the deer eats the plants and uses the nitrogen-containing proteins for cell growth. 3. step 3 bacteria in the soil conduct nitrogen fixation to convert nitrogen gas from the atmosphere into ammonium. 4. step 4 some of the nitrates are absorbed back into plants, but denitrification breaks down the nitrates into nitrogen gas released back into the atmosphere. 5. step 5 nitrogen is found in the urine, which gets broken down into ammonium through the process of ammonification. 6. step 6 plants absorb the ammonium and begin assimilation, the process to add the nitrogen to protein.

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