Zebrafish Danio rerio is a small 45 cm in length tropical fr

Zebrafish (Danio rerio) is a small (4-5 cm in length) tropical fresh-water fish which has external fertilization, produces large clutches of eggs, has a short generation time, and a rapid embryogenesis which is similar to that of higher vertebrates. As such zebrafish is a widely used model organism for genetic studies in vertebrate development, developmental biology, and a variety of human congenital and genetic diseases.

In zebrafish, the allele for dark stripes is dominant to the allele for faint stripes, and a long tail is dominant to the allele for a short tail. A true breeding dark-striped fish with a long tail was crossed with a true breeding faint striped fish with a short tail. Their progeny were interbred. Of the 320 offspring, 61 had dark stripes and a short tail; 201 had dark stripes and a long tail; 14 had faint stripes and a short tail; and 44 had faint stripes and a long tail.

a) propose a hypothesis to explain the data and include the expected ratio based on the hypothesis

b) place the data into an appropriate table

c) calculate the chi-square value

d) using a p value of 5% select the appropriate critical value for x2.( search on-line)

e) does the chi-square analysis support the hypothesis?

Solution

Let the allele for dark stripes be \'D\' and long tail be \'L\' ; similarly, let faint stripes be \'d\' and short tail be \'l\'

Then, DDLL X ddll

F1 result will be DdLl (dark and long)

F2 cross will be DdLl X DdLl

F2 results are: D_ll = 61; D_L_ = 201; ddll=14; ddL_=44

a) Null hypothesis: Ho: Zebrafish does not obey Mendelian laws, and independent assortment (9:3:3:1 ratio) is not there according to HW equilibrium.

b)

Observed

Expected

(O-E)^2

(O-E)^2/E= chi square value

D_L_

201

144

3249

22.5625

D_ll

61

48

169

3.52083333

ddL_

44

48

16

0.33333333

ddll

14

16

4

0.25

Total

320

26.6666667

Calculated value = 26.67 (Answer to part \'c\')

Degree of freedom = (rows-1) (columns-1) = (4-1)* (4-1) = 9

Tabulated value = taken from chi square table = 16.92

Level of significance (given) = 0.05

Null hypothesis is accepted if calculated value is less than tabulated value.

Here, tabulated value<<<<calculated value, so, null hypothesis should be rejected.

So, the population is not in HW equilibrium; and is not assorting independently according to Mendelian ratio of 9:3:3:1 (part e). Chi-square test does not support the hypothesis.