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. propose a hypothesis to explain the data and include the expected ratio based on the hypothesis place the data into an appropriate table calculate the chi-square value using a p value of 5% select the appropriate critical value for x^2. (see Table 3.2, p 52 of text, or search on-line) does the chi-square analysis support the hypothesis?

Solution

a). Hypothesis: The values of expected and observed results are significantly different.

Assume that the gene coding for dark stripes is, D and the faint stripes is, d. Similarly, the gene coding for long tail is L, and the gene coding for short tail is l.

The parental cross is, DDLL* ddll --à DdLl (all dark stripes, long tail) -à F1

Cross between F1 will have the progeny with the following genotypes.

DdLl* DdLl -à D_L_ (9/16, dark stripes, long tail), D_ ll (3/16, dark, short tail), dd L_ (3/16, faint stripes, long tail), dd ll (1/16, faint stripes, short tail).

b). The expected ratio of progeny are,

dark stripes, long tail -à 9/16* 320 = 180

dark stripes, short tail -à 3/16* 320 = 60

faint stripes, Long tail -à 3/16* 320 = 60

faint stripes, short tail à 1/16*320 = 20

c). We can calculate the chi-square value from the below formula.

CHI - SQUARE (X²):

X² = (O - E)² / E

Where O = Observed frequency

E = Expected frequency

P

O

E

(O-E)

(O-E)^2

(O-E)^2/E

dark, long

201

180

21

441

2.45

faint, long

44

60

-16

256

4.266667

dark, short

61

60

1

1

0.016667

faint, short

14

20

-6

36

1.8

320

320

8.533333

The calculated chisquare value is = 8.533

d). Degrees of freedom is = n-1 = 4-1 = 3

The P-Value is 0.03619. The result is significant at p < 0.05.

e). The observed and expected results are significantly different. Thus, the chi-square analysis support the hypothesis.

P	O	E	(O-E)	(O-E)^2	(O-E)^2/E
dark, long	201	180	21	441	2.45
faint, long	44	60	-16	256	4.266667
dark, short	61	60	1	1	0.016667
faint, short	14	20	-6	36	1.8
	320	320			8.533333