Question

Cultivated amaranth grains (Amaranthus caudatus) come in three colors: black, brown, and pale. Geneticists asked whether this phenotype could result from a dominant epistatic control. Crossing black-seeded and pale-seeded A. caudatus populations (homozygous lines) gave the following counts of black, brown, and pale seeds in the second generation (F2).

Seed coat color	Black	Brown	Pale
Seed count	344	82	31

In genetics, dominant epistasis should lead to 12/16 of all such seeds being black, 3/16 Brown, and 1/16 pale. Do the experimental data above support the conclusion that seed coat color in cultivated amaranth grains is due to dominant epistasis?

Please find the Ho, Ha, test statistic, df, exact probability of the test statistic, and conclusion relative to the hypothesis. Is this a 1-tailed or 2-tailed test? Use Excel functions and math calculations, please.

I think you Chi-Squared for this, but I am not sure.

Answer 1

Solution:

Chi square test for goodness of fit

H₀: The seed coat color in cultivated amaranth grains is due to dominant epistasis.

H_a: The seed coat color in cultivated amaranth grains is not due to dominant epistasis.

This is a two tailed test.

Test statistic formula is given as below:

Chi square = ∑[(O – E)^2/E]

Where, O is observed frequencies and E is expected frequencies.

Calculation table is given as below:

Seed coat color	Seed count (O) Observed Frequency	Proportion	(E) Expected Frequency
Black	344	12/16 = 0.75	457*0.75 = 342.75
Brown	82	3/16 = 0.1875	85.6875
Pale	31	1/16 = 0.0625	28.5625
Total	457	1	457

No.	O	E	(O - E)^2/E
1	344	342.75	0.004558716
2	82	85.6875	0.158688913
3	31	28.5625	0.208014223
Total	457	457	0.371261853

Chi square = ∑[(O – E)^2/E] = 0.371261853

df = n – 1 = 3 – 1 = 2

P-value = 0.830580085

P-value > α = 0.05

So, we do not reject H₀

There is sufficient evidence to conclude that the seed coat color in cultivated amaranth grains is due to dominant epistasis.