In zebrafish the allele for dark stripes is dominant to the

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

a) True-bred zebrafishes, one having homozygous dominant traits as dark striped-long tailed and other having recessive trait as faint stripe-short tail, when mated will produce heterozygous F1 progeny, exhibiting dark stripe and long tail.

D = Dark stripe

d = faint stripe

T = long tail

t = short tail

Parent cross

DDTT X ddtt

F1 generation: All DdTt

When the F1 progeny are interbred, we get a dihybrid cross.

Hypothesis: When true bred dihybrid zebrafish are crossed and the F1 progeny are interbred, the resulting F2 generation offsprings should follow Mendel\'s law and exhibit phenotypic ratio as 9:3:3:1.

Expected ratio (phenotype): 9:3:3:1.

b) Total number of F2 progeny =320.

The observed offsprings with dark stripes and a long tail = 201

The observed offsprings with faint stripes and a long tail = 44

The observed offsprings with dark stripes and a short tail = 61

The observed offsprings with faint stripes and a short tail = 14

If the hypothesis holds true, then the ratio of offsprings with different phenotype should be 9:3:3:1

Expected offsprings with dark stripes and a long tail = (9/16) x 320 = 180

Expected offsprings with faint stripes and a long tail = (3/16) x 320 = 60

Expected offsprings with dark stripes and a short tail = (3/16) x 320 = 60

Expected offsprings with faint stripes and a short tail = (1/16) x 320 = 20

Table (for chi square calculation)

c) Chi square value X² =Sum of all d² /e values = 8.54

d)The degree of freedom (df) = 1

Using a p-value of 5%, critical value of X² = 3.84 (from table)

Test value of chi square, 8.54 > Table value, 3.84.

Therefore we reject the hypothesis.

e) The chi-square analysis does not support the hypothesis. Therefore the dihybrid cross among the zebrafish of F1 generation does not follow mendelian genetics, hence does not give phenotypic ratio 9:3:3:1.