Answer:
Null Hypothesis -
The observed frequency is approximately equal to the expected frequency of phenotype.
Explanation:
Pure Breeding Cross - TTww x ttWW
Genotype of offspring in F2 generation - TtWw
Null Hypothesis -
The observed frequency is approximately equal to the expected frequency of phenotype.
The chi square analysis is attached
The degree of freedom for this question is 3
The p value for X^2 estimated through chi square test is 0.5
Hence the null hypothesis is accepted.
B. sulfur tetrafluoride + water → hydrogen fluoride + sulfur dioxide