In a dihybrid cross RrYy × RrYy, what is the probability that an offspring will have round seeds (R) and yellow seed colour (Y)?
Answer
9/16
The correct answer is 9/16 — Thus, 9 out of 16 possible offspring will be round and yellow.
Step-by-step solution
- Apply the law of independent assortment: the two genes segregate independently.
- Probability of obtaining at least one dominant allele R in a Rr × Rr cross is 3/4.
- Probability of obtaining at least one dominant allele Y in a Yy × Yy cross is also 3/4.
- Since the two traits assort independently, multiply the individual probabilities: (3/4) × (3/4) = 9/16.
- Thus, 9 out of 16 possible offspring will be round and yellow.
Common mistake: Adding the probabilities (3/4 + 3/4) instead of multiplying them.
Concept tested: Independent assortment in dihybrid cross
Published · How this solution was prepared and checked
Related questions
- 1 Genes A and B are located on the same chromosome and are 20 % recombination apart. A plant heterozygous for both genes in coupling phase (genotype AB/ab) is crossed with a plant homozygous recessive (ab/ab). What is the probability that an offspring will inherit the recombinant genotype Ab?
- 2 In pea plants, the allele for round seed shape (R) is dominant over the allele for wrinkled seed shape (r). If a plant heterozygous for seed shape (Rr) is crossed with a plant homozygous recessive (rr), what percentage of the progeny will have round seeds?
- 3 A colour‑blindness (X‑linked recessive) gene is represented by ‘c’. If a normal‑vision male (XY) marries a carrier female (XcX), what is the expected proportion of colour‑blind sons?
- 4 A colour‑blind male (genotype XcY) marries a woman who is a carrier for colour‑blindness (genotype XCXc). What proportion of their daughters will be colour‑blind?
- 5 According to Mendel’s law of dominance, when a pure‑bred tall pea plant (TT) is crossed with a pure‑bred dwarf plant (tt), the F1 generation shows which phenotype?
- 6 Hemophilia is an X‑linked recessive disorder represented by allele h (recessive) and H (normal). A carrier mother (XHXh) marries a normal‑vision male (XHY). What percentage of their sons will be affected and what percentage of their daughters will be carriers?