Zebrafish (Danio rerio) is a small (4-5 cm in length) tropical fresh-water fish
ID: 308627 • Letter: Z
Question
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?
Explanation / Answer
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.
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
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