The cross AB/ab X ab/ab produces the following progeny: AB/ab 391, ab/ab 401, aB/ab 406, Ab/ab 386. From these data, one can conclude that the A and B loci assort independently. A. True
B. False

B

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.