In: Biology
When a monohybrid heterozygote is test crossed, the following phenotypes are produced:
Phenotype I = 213
Phenotype II = 87
Determine if this cross follows Mendelian rules of segregation. Be careful to state your hypothesis and conclusion, degrees of freedom, and p value (range is ok), and SHOW YOUR WORK.
Null Hypothesis: The phenotype follows the Mendelian rule of
segregation.
Alternate Hypothesis: The phenotype does not follow the Mendelian
rule of segregation.
Based on this hypothesis, the test cross Aa x aa should produce progeny of the two phenotypes in an equal proportion, and therefore the number of individuals of each phenotype should be the same. Given that 300 individuals were observed, the expected number of individuals for each phenotype is 150.
Phenotype | Expected (E) |
Observed (O) |
E - O | (E - O)2 | (E - O)2/E |
Phenotype I | 150 | 213 | -63 | 3969 | 26.46 |
Phenotype II | 150 | 87 | 63 | 3969 | 26.46 |
= 52.92 |
Here, the degrees of freedom is 1, as we can calculate the number of individuals with Phenotype II just by knowing the total number of individuals counted and the number of individuals with Phenotype I.
Now, the critical Chi-squared value at 1 degrees of freedom is 3.841. This is the value, at which, the probability of the results obtained here being due to chance alone is 5%. However, the Chi-squared value obtained here is 52.92, which means that the probability of these results occurring due to chance alone is less than 0.1% and therefore the null hypothesis is rejected.