In this case, purple (T) is dominant to yellow (t) and round (R) is dominant to wrinkled (r).
Parents:
Phenotype: purple round purple round
Genotype: TtRr TtRr
Offspring:
From the table, we can see that 9 of the offspring are purple round (TTRR or TTRr or TtRr or TtRR); 3 are purple wrinkled (Ttrr or TTrr); 3 are yellow round (ttRR or ttRr) and 1 is yellow wrinkled (ttrr).
In the long maize cob, the ratio of the phenotypes is approximately 1 (purple round): 1 (purple wrinkled): 1 (yellow round): 1 (yellow wrinkled), which is also a classic dihybrid ratio (cross test).
Purple (T) is dominant to yellow (t) and round (R) is dominant to wrinkled (r).
Parents:
Phenotypes: purple round yellow wrinkled
Genotypes: TtRr ttrr
Gametes: TR Tr tR tr all tr
Offspring genotypes: TtRr Ttrr ttRr ttrr
Phenotypes: purple round purple wrinkled yellow round yellow wrinkled
We can see that each of the 4 different phenotype counts for a quarter of the total number of offspring.
Calculation (CHI-squared test):
Short maize cob
=0.08+0.10285+0.0677+0.554=0.8036
Long maize cob
=0.159+0.00036+0.0032+0.13=0.29256
Conclusion
According to the CHI-squared table, we can see our value of 0.8036 lies between 0.6 and 1.0. From this we can say that the probability that this result is significant is between 80% and 90%. Put another way, it tells us that the probability that the result is due to chance is between 10% and 20%. Therefore we accept the hypothesis.
According to the CHI-squared table, 0.29256 lies between 0.4 and 0.6. From this we can say that the probability that the result is significant is higher than 95% so we accept the hypothesis.
The actual result is not exactly the same as the expected result is because: mutation might have occurred; personal inaccuracy in counting; some offspring might have rotted so could not be counted.