Question

Difficulty: EasyMendel's Second Law and Dihybrid Inheritance

In sweet pea plants (*Lathyrus odoratus*), purple flower color (PP) is dominant over red flower color (pp), and long pollen grain (LL) is dominant over round pollen grain (ll). If a heterozygous dihybrid plant with the genotype PpLlPpLl is self-pollinated and yields a total of 160160 F2 seeds, how many of these seeds are expected to produce plants with the double recessive phenotype of red flowers and round pollen grains?

Answer: 10 seeds

Answer

The expected number of seeds producing plants with red flowers and round pollen grains is 10.
In a dihybrid cross involving two heterozygous parents (PpLl×PpLlPpLl \times PpLl), independent assortment results in a 9:3:3:1 phenotypic ratio in the F2 generation. The double recessive phenotype (red flowers and round pollen grains, genotype ppllppll) accounts for 1 out of 16 total offspring. Multiplying this fraction (1/16) by the total yield of 160 seeds produces an expected value of 10 seeds.

Step-by-Step Solution

1
Determine the dihybrid F2 phenotypic ratio
The phenotypic ratio for a cross between two heterozygous dihybrid parents (PpLl×PpLlPpLl \times PpLl) is 9:3:3:1.
According to Mendel's Law of Independent Assortment, the alleles for flower color and pollen shape segregate independently during gamete formation.
2
Calculate the proportion of double recessive offspring
The fraction of offspring displaying both recessive traits (red flowers and round pollen grains, genotype ppllppll) is 1/16.
Out of 16 equal Punnett square combinations, exactly 1 combination represents the homozygous double recessive phenotype.
3
Compute the expected number of double recessive seeds
(1 / 16) * 160 = 10 seeds.
Multiplying the phenotypic probability (1/16) by the total seed population (160) yields the expected quantity.

Key Concept

Mendel's Law of Independent Assortment and Dihybrid Phenotypic Ratios
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