In Drosophila psuedoobscura, the alleles P and p determine red (dom.) versus pin
ID: 15055 • Letter: I
Question
In Drosophila psuedoobscura, the alleles P and p determine red (dom.) versus pink eyes, and N and n determine wide (dom.) versus narrow wings. A dihybrid cross was carried out to produce flies homozygous for both p and n. The following phenotypes were obtained in the F2 generation:red eyes, wide wings 554
red eyes, narrow wings 184
pink eyes, wide wings 191
pink eyes, narrow wings 71
Test these data to see if they follow the expected dihybrid ratios expected under the hypothesis that the two pairs of alleles undergo independent assortment.
Reply to this with your answer (including chi-square value, acceptance or rejection of null hypothesis and what this means).
Explanation / Answer
The null hypothesis states that these genes are not linked; to reject the null hypothesis, the chi-squared value must correspond with a probability of 5% or less. A dihybrid cross would mean that the F1 parents were both heterozygous; the cross was PpNn x PpNn; from this cross, under the hypothesis of independent assortment, the expected phenotypic ratios would be 9:3:3:1, where: red eyes, wide wings ~9/16 --> 562.5 --> ~563 expected red eyes, narrow wings ~3/16 --> 187.5 --> ~188 expected pink eyes, wide wings ~3/16 --> 187.5 --> ~188 expected pink eyes, narrow wings ~1/16 --> 62.5 --> ~63 expected So if you put the F2 data into a chi-squared test, the formula for each phenotype is ((observed - expected)^2)/expected. The result: red eyes, wide wings = 0.1439 red eyes, narrow wings = 0.0851 pink eyes, wide wings =0.04787 pink eyes, narrow wings = 1.0159 Total sum = 1.2927 Because there are 4 phenotypes, there are 3 degrees of freedom on the chi-squared table. You find degrees of freedom by number of phenotypes minus one. On the 3 degrees of freedom lane, the total sum is greater than the 0.80 but smaller than the 0.70 column values. The greatest individual chi-squared result for a single phenotype is also greater than the 0.80 column value but smaller than the 0.70 value; therefore, the chances of these phenotype results in the observed F2 generation happening randomly in nature is between 70% to 80%, proving the null hypothesis since it is greater than 5% chance - so we accept the null hypothesis that these genes are NOT linked. Hope this helps!