Answer:
The genotype of both parents will be heterozygous for the trait, Ii.
Step-by-step explanation:
Let us assume that the shape of the seeds is controlled by a single diallelic gene.
- Dominant allele I expresses the inflated phenotype
- Recessive allele i expresses the constricted phenotype
- The dominant phenotype is inflated seeds, expresses by genotype II and Ii. The dominant allele hides the expression of the recessive allele, so if the genotype has at least one dominant allele, the phenotype will be inflated.
- The recessive phenotype is constricted seeds, expressed by the genotype ii. In this genotype there is no dominant allele, so the recessive allele can be expressed.
The individual with constricted seeds must have two recessive alleles i in its genotype ⇒ ii.
One of the i alleles was inherited from one of the parental plants, while the other parental plant provided the other i allele. So both parental plants must carry the recessive allele in their genotype.
We know that both parental plants have inflated seeds, which is the dominant phenotype.
So, if both parental plants have inflated seeds and also carry the recessive allele, then the parentals´ genotype must be heterozygous, Ii.
Both plants exhibit the dominant inflated phenotype but carry a recessive allele which will be provided to some of the descendants. These last individuals will have constricted seeds.