Question

1. in garden peas, inflated pod (i) is dominant and constricted pod (i) is recessive. if a plant homozygous for inflated pod is crossed with a plant homozygous for constricted pod, what ratio of phenotypes would you expect to find in the f2 generation? (1 point) round, yellow pea inflated, green pod constricted, yellow pod tall plant purple, terminal flower white, axial flower short plant wrinkled, green pea all inflated all constricted 1:1 inflated to constricted 3:1 constricted to inflated 3:1 inflated to constricted

Explanation:

Step1: Determine parental genotypes

Let the allele for inflated pod be \(I\) (dominant) and for constricted pod be \(i\) (recessive). The homozygous inflated - pod parent has genotype \(II\) and the homozygous constricted - pod parent has genotype \(ii\).

Step2: Determine F1 generation genotype

When \(II\times ii\), all F1 offspring have the genotype \(Ii\) (heterozygous).

Step3: Determine F2 generation genotypes and phenotypes

When F1 individuals (\(Ii\)) self - cross (\(Ii\times Ii\)), using the Punnett - square method, the genotypes and their ratios are \(II:Ii:ii = 1:2:1\). The phenotypes are determined by the genotypes. Since \(I\) is dominant, \(II\) and \(Ii\) individuals have inflated pods and \(ii\) individuals have constricted pods. So the phenotypic ratio of inflated pods to constricted pods is \(3:1\) (3 inflated : 1 constricted).

Answer:

3:1 inflated to constricted