Question

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

Answer 1

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
`X^2` =
`((O - E)^2)/(E)`

AB/ab 391 9/16 x 1584 = 891
`((391-891)^2)/(891)` = 280.58

ab/ab 401 3/16 x 1584 = 297
`((401-297)^2)/(297)` = 36.42

aB/ab 406 3/16 x 1584 = 297
`((406-297)^2)/(297)` = 40.00

Ab/ab 386 1/16 x 1584 = 99
`((386-99)^2)/(99)` = 832.01

Total
`X^2` = 280.58 + 36.42 + 40.00 + 832.01 = 1,189.01

Degree of freedom = 4 - 1 = 3

Tabulated
`X^2` at degree 3 freedom and 95% level = 7.815

The calculated
`X^2` 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.