Answer:
B
Step-by-step explanation:
The two loci did not assort independently.
In order to check for independent assortment or otherwise of the result from the cross, Chi-square is used to see if the result conforms with that of Mendelian standard of 9:3:3:1.
Phenotype Observed f Expected f
=

AB/ab 391 9/16 x 1584 = 891
= 280.58
ab/ab 401 3/16 x 1584 = 297
= 36.42
aB/ab 406 3/16 x 1584 = 297
= 40.00
Ab/ab 386 1/16 x 1584 = 99
= 832.01
Total
= 280.58 + 36.42 + 40.00 + 832.01 = 1,189.01
Degree of freedom = 4 - 1 = 3
Tabulated
at degree 3 freedom and 95% level = 7.815
The calculated
value is more than the tabulated value. Therefore, we conclude that the outcome of the cross is not in agreement with Mendelian standard and hence, the A and B loci did not assort independently.
The correct option is B.