Question

1A)

Two autosomal genes control horn color in dragons. Pure-breeding gold-horned dragons were mated to pure-breeding silver-horned dragons. All of the F1 were gold. The F1 were intermated and the F2 generation consisted of 148 gold, 17 silver and 96 bronze.

Choose the genetically based phenotypic ratio that should be hypothesized to explain the F2 data.

Select one:

9:3:3:1

9:3:4

12:3:1

9:7

15:1

9:6:1

13:3

B)Conduct a chi-square test to test the appropriate type of epistasis.
In the chi-square test, what is the expected number of gold-horned dragons? Round to two decimal places.

C)Conduct a chi-square test to test the appropriate type of epistasis.
In the chi-square test, what is the expected number of silver-horned dragons? Round to two decimal places.

D)Conduct a chi-square test to test the appropriate type of epistasis.
In the chi-square test, what is the expected number of bronze-horned dragons? Round to two decimal places.

E)Conduct a chi-square test to test the appropriate type of epistasis.
In the chi-square test, what is the chi-square value? Round to three decimal places.

F)

What is the number for the degrees of freedom?

H)What is the critical value? Round to two decimal places.

What is your conclusion from the chi-square test?

Select one:

Fail to reject H_o

Reject H_o

Answer 1

ANSWER 1.A)

The pure breeding gold-horned dragons are 147 in numbers.

The pure breeding silver-horned dragons are 16 in numbers.

Bronze-horned dragons are 93 in numbers.

So, the ratio should be (147:16:93) which corresponds to around ( 9:1:6 ).

1. (147 / 16 = 9.18 ≈ 9)

2. (93 / 16 = 5.81 ≈ 6)

As, there are three different phenotypes, the ratio corresponds to 9:1:6.

ANSWER 1. B / C / D / E)

If taken into consideration the values in terms of ratio, we get the following chi square distribution table :-

	Observed	Expected
Good horned	9.18/16 (147)	9 (144)
Silver horned	1.00/16 (16)	1 (16)
Bronze horned	5.81/16 (93)	6 (96)

Chi square = Σ (Observed - Expected)² / Expected

1. (147-144)² / 144 = 0.0625

2. (16-16)² / 16 = 0

3. (93-96)² / 96 = 0.0937

So, chi-square value is (0.1562). Rounding the value to three decimal places, we get (0.156).

ANSWER F)

The degree of freedom is given by (n-1) which is total observations minus one. So, (n-1) = (3-1=2). So, the degrees of freedom value is 2.

ANSWER H)

The alpha value is taken as 0.05.

The critical value for the chi square corresponds to 5.991. (5.91) rounded to two digit. Taking into consideration the value should be present between (0.900 - 0.950), the chi square value is more than the p-value of 0.90 and hence the Null hypothesis is rejected. As the value is more than the alpha value of 0.05 (0.90 > 0.05).