Final answer:
When crossing true breeding inflated pod plants with true breeding constricted pod plants, the F2 phenotype outcome should follow a phenotypic ratio of 3:1 for inflated to constricted pods according to Mendelian genetics. If 650 offspring in the F2 generation have inflated pods, you would expect approximately 217 constricted-pod plants and in a sample of 100 offspring, about 25 would be constricted.
Step-by-step explanation:
In garden peas, inflated pod is dominant and constricted pod is recessive. When performing a cross between a true breeding inflated pod and a true breeding constricted pod plant, the first generation (F1) will all have inflated pods since the inflated pod trait is dominant. For the F2 generation, you would conduct a self-cross of the F1 heterozygotes, and according to Mendelian genetics, you should expect a phenotypic ratio of 3:1, meaning three inflated pods to one constricted pod.
Having this ratio, if you obtained 650 inflated-pod plants, the number of constricted-pod plants in the F2 generation is calculated as follows:
- Combine the number of both phenotypes which follows the 3:1 ratio (3 inflated + 1 constricted = 4).
- To find out how many parts of the ratio represents 650 inflated-pod plants, divide 650 by 3, which gives approximately 216.67.
- Multiply this number by 1 (representing the constricted pod part of the ratio) to predict the number of constricted-pod plants. Therefore, you would expect approximately 217 constricted-pod plants out of 867 (650+217) total plants.
If we extrapolate this to calculate how many constricted pod plants are found in 100 offspring, we use a simple proportion:
- If 867 plants have 217 constricted plants, then 100 plants would have about 25 constricted plants (100/867 x 217 ≈ 25).