Solutions For An Introduction to Genetic Analysis Chapter 2 Problem 49P

When we carefully observe the results of the crosses in the table, we can interpret them as follows:

Winged trait is dominant and wingless is recessive.

Winged is A/- homozygous dominant or heterozygous dominant.

Wingless is homozygous recessive a/a.

In cross 1, the winged is selfed and the progeny is all winged except one. This is due to non-genetic reason. This is due to environmental factors, which affects the development of plants (Scarcity of water or so). So, the parents are A/A homozygous dominant to get all the progeny dominant.

In cross 2, the cross is between winged (selfed). The progeny is 3:1, for winged to wingless. So, we infer that the parents must be heterozygous, for the dominant phenotype (A/a).

In cross 3, the cross is between wingless (selfed). Wingless is homozygous recessive. So, it is a/a X a/a -- Almost all the progeny is wingless except 4 winged. This exception may be due to human error or contamination and has non-genetic explanation.

In cross 4, it is between winged and wingless. And the progeny is all winged. This shows that the winged parent is homozygous dominant A/A. So, A/A X a/a - All A/a (heterozygous winged).

In cross 5, it is between winged and wingless. The progeny is half winged and half wingless. To get 1:1 ratio, the parents should be A/a X a/a-- A/a, A/a, a/a, a/a. So, half A/a winged, half a/a wingless.

In cross 6, the cross is between winged and wingless. The progeny is all winged. This shows that the winged parents should be A/A X a/a- All A/a winged.

In cross 7, the cross is between winged and winged. The progeny is all winged. This shows that the parents should be A/A X A/- -- All winged A/-

In cross 8, the cross is between winged and winged. The progeny is all winged. The parents must be A/A X A/- -- All winged A/-.

The crossed are tabulated as follows: