In a certain plant species, tallness (T) is dominant over dwarfness (t), and red flower color (R) is dominant over white flower color (r). A dihybrid heterozygous tall, red-flowered plant (TtRr) is test-crossed with a homozygous dwarf, white-flowered plant (ttrr). Assuming independent assortment, calculate the expected phenotypic and genotypic frequencies of the offspring. Show all intermediate steps and calculations.

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Accepted Answer

Problem Statement Analysis Cross Type: Test cross between a dihybrid heterozygous individual and a double recessive homozygous individual. Genotype of Parent 1 (Dihybrid Heterozygous): $TtRr$ Genotype of Parent 2 (Double Recessive Tester): $ttrr$ Step 1: Determine Gametes Produced by Parent 1 ($TtRr$)
Assuming independent assortment, the heterozygous parent $TtRr$ will produce four types of gametes in equal proportions (25% or 1/4 each): $TR$ $Tr$ $tR$ $tr$ Step 2: Determine Gametes Produced by Parent 2 ($ttrr$)
The double recessive parent is homozygous for both traits and will produce only one type of gamete: $tr$ (100% or 1) Step 3: Set up the Punnett Square / Cross Combinations
Crossing the 4 gametes of $TtRr$ with the single gamete type ($tr$) of $ttrr$ yields 4 possible offspring genotypes: $TR 	imes tr ightarrow TtRr$ $Tr 	imes tr ightarrow Ttrr$ $tR 	imes tr ightarrow ttRr$ $tr 	imes tr ightarrow ttrr$ Step 4: Calculate Genotypic Frequencies
Since each of the four gametes from the heterozygous parent is produced with an equal probability of $\frac{1}{4}$ (or 25%), the genotypes of the offspring will appear in equal proportions: $TtRr$ (Tall, Red): $\frac{1}{4}$ or 25% $Ttrr$ (Tall, White): $\frac{1}{4}$ or 25% $ttRr$ (Dwarf, Red): $\frac{1}{4}$ or 25% $ttrr$ (Dwarf, White): $\frac{1}{4}$ or 25% Step 5: Calculate Phenotypic Frequencies
Each genotype corresponds to a unique phenotype because of complete dominance: Tall and Red ($TtRr$): 25% or $\frac{1}{4}$ Tall and White ($Ttrr$): 25% or $\frac{1}{4}$ Dwarf and Red ($ttRr$): 25% or $\frac{1}{4}$ Dwarf and White ($ttrr$): 25% or $\frac{1}{4}$ Final Ratio: The phenotypic and genotypic ratio for a dihybrid test cross under independent assortment is 1 : 1 : 1 : 1.

Problem Statement Analysis

Step 1: Determine Gametes Produced by Parent 1 ($TtRr$)\nAssuming independent assortment, the heterozygous parent $TtRr$ will produce four types of gametes in equal proportions (25% or 1/4 each):

Step 2: Determine Gametes Produced by Parent 2 ($ttrr$)\nThe double recessive parent is homozygous for both traits and will produce only one type of gamete:

Step 3: Set up the Punnett Square / Cross Combinations\nCrossing the 4 gametes of $TtRr$ with the single gamete type ($tr$) of $ttrr$ yields 4 possible offspring genotypes:

Step 4: Calculate Genotypic Frequencies\nSince each of the four gametes from the heterozygous parent is produced with an equal probability of $\frac{1}{4}$ (or 25%), the genotypes of the offspring will appear in equal proportions:

Step 5: Calculate Phenotypic Frequencies\nEach genotype corresponds to a unique phenotype because of complete dominance: