What phenotypic ratio would one would expect from crossing two heterozygous smooth kernel corn plants that show mendelian inheritance?

The phenotypic ratio from crossing 2 heterozygous smooth kernel corn plants that show distinct dominance/recessiveness would be 3 smooth to 1 rough kernel corn plant.

Showing Mendelian inheritance means that the trait displays distinct dominance and recessiveness.

Thus, assuming the smooth kernel is represented by A allele, the rough kernel would be a.

Crossing 2 heterozygous:

                  Aa     x     Aa

               AA   Aa   Aa   aa

AA, Aa, Aa = smooth kernel

aa = rough kernel

The phenotypic ratio of smooth to rough kernel offspring would, therefore, be 3:1

More on phenotypic ratios can be found here: link

Smooth kernel corn plant : wrinkled  kernel corn plant

Explanation:

Mendelian inheritance follows the laws of segregation and independent assortment. Thus, there exists neither  co-dominance nor incomplete dominance.

Let the allele for smooth kernel trait of corn plants be represented by "S"

and the allele for wrinkled kernel trait of corn plants be represented by "s"

The genotype of  heterozygous smooth kernel corn plants  will be "Ss"

The offspring produced by crossing two heterozygous smooth kernel corn plants would be -

Ss * Ss

SS, Ss, Ss, ss

Out of four offspring , one offspring is  homozygous smooth kernel corn plant

Two offspring are heterozygous smooth kernel corn plants

and the remaining last offspring is  homozygous wrinkled  kernel corn plant

The phenotypic ration is

Smooth kernel corn plant : wrinkled  kernel corn plant

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