Question

Geneticists examined the distribution of seed coat color in cultivated amaranth grains, Amaranthus caudatus. Crossing black-seeded and pale-seeded A. caudatus populations gave the following counts of black, brown, and pale seeds in the second generation.

Seed Coat Color	black	brown	pale
Seed Count	308	82	40

According to genetics laws, dominant epistasis should lead to  

12

16

  of all such seeds being black,  

3

16

  brown, and  

1

16

  pale. We want to test this theory at the 5% significance level.

(a)	[2 marks] Find the value of the test statistic.

(b)	[1 mark] Fiind the critical value.

(c)	[1 mark] What is the conclusion?

Answer 1

Part a

Find the value of the test statistic.

Test statistic formula is given as below:

Chi square = ∑[(O – E)^2/E]

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

The required calculation table is given as below:

Seed Coat Color	Prop	O	E	(O - E)^2/E
Black	12/16 = 0.75	308	430*0.75 = 322.5	0.6519379845
Brown	3/16 = 0.1875	82	430*0.1875 = 80.625	0.0234496124
Pale	1/16 = 0.0625	40	430*0.0625 = 26.875	6.4098837209
Total	1	430	430	7.0852713178

Test statistic = Chi square = 7.0852713178

Part b

We are given

N = 3

df = N – 1 = 2

α = 0.05

So, critical value by using Chi square table is given as below:

Critical value = 5.99146455

Part c
The conclusion is that critical value is less than chi-square test statistic.
So, the null hypothesis is rejected at 5% level of significance.