Stem length, seed shape and flower colour in pea plants are determined by three independently assorting genes. Each trait has a dominant and a recessive phenotype. If a plant that is heterozygous for the three traits is allowed to self-fertilise, you would expect 3/8 of the offspring to be
A) homozygous for all three traits.
B) homozygous for the three recessive traits.
C) homozygous for one and heterozygous for two.
D) heterozygous for all three traits.
E) homozygous for two traits and homozygous for one.

Question

Quizplus · Accepted Answer

For each trait, the heterozygous parent (Aa) can produce two types of gametes (A and a) with equal probability. When self-fertilizing, the probability for each trait to be heterozygous in the offspring is 1/2 (Aa x Aa = 1/4 AA, 1/2 Aa, 1/4 aa). Since the traits assort independently, the probability that all three traits are heterozygous in the offspring is (1/2)^3 = 1/8, not 3/8 as stated in the question. Therefore, none of the options perfectly match the expected outcome based on Mendelian genetics principles. However, the closest option in the context given, considering a possible misunderstanding in the question's phrasing, would be D, as it directly refers to being heterozygous for all three traits, which aligns with the scenario of a heterozygous parent self-fertilizing.

Stem Length, Seed Shape and Flower Colour in Pea Plants