Principles of Inheritance and Variation · Lesson 4 of 12
Dihybrid Crosses and Independent Assortment
“Predicting combined inheritance of two characters and establishing when 9:3:3:1 applies.”
• Identify gametes from a dihybrid genotype. • Construct and interpret a 4×4 Punnett square. • Derive the 9:3:3:1 phenotype ratio. • State the conditions limiting independent assortment.
From one trait to two
A dihybrid cross tracks inheritance of two characters simultaneously. Mendel investigated pea seed shape and colour using parental lines with round yellow seeds (RRYY) and wrinkled green seeds (rryy). The corresponding alleles are R/r for round/wrinkled shape and Y/y for yellow/green colour. The RRYY parent provides only RY gametes; rryy provides only ry. Every F1 offspring becomes RrYy and shows round yellow seeds when R and Y are dominant. The interesting question is whether the two allele pairs are transmitted together as permanent combinations or can form new combinations.
Deriving four F1 gamete classes
In a RrYy plant, segregation first requires that each gamete receive either R or r and either Y or y. If the two loci assort independently, the choice of R versus r does not affect the choice of Y versus y. Therefore P(RY) = P(R) × P(Y) = 1/2 × 1/2 = 1/4. By the same reasoning, Ry, rY and ry each occur with expected frequency 1/4. Four equal gamete classes are the source of 16 equally likely fertilisation cells in a RrYy × RrYy Punnett square. These combinations represent new allele combinations even though neither locus has mutated.
Producing a 9:3:3:1 phenotype ratio
At the shape locus, Rr × Rr produces round phenotype with probability 3/4 and wrinkled with 1/4. At the colour locus, Yy × Yy produces yellow with 3/4 and green with 1/4. Under independence, multiply probabilities for combined traits: round yellow = 3/4 × 3/4 = 9/16; round green = 3/4 × 1/4 = 3/16; wrinkled yellow = 1/4 × 3/4 = 3/16; wrinkled green = 1/4 × 1/4 = 1/16. Thus the phenotypic ratio is 9:3:3:1. The ratio does not mean that 16 offspring must appear in precisely those numbers.
Genotypes are not the same as phenotype categories
A 4×4 square contains 16 fertilisation cells but only nine distinct two-locus genotype categories: RRYY, RRYy, RRyy, RrYY, RrYy, Rryy, rrYY, rrYy and rryy. Different genotypes can produce the same visible phenotype under complete dominance. For example, round yellow seeds may have RRYY, RRYy, RrYY or RrYy. To count genotypes, multiply the possible one-locus genotype counts (3 × 3 = 9). To obtain a combined genotype's probability, multiply the relevant one-locus probabilities when independence holds.
When multiplication and addition are appropriate
Multiplication applies to the joint occurrence of independent events, such as the inheritance of R and Y at independently assorting loci. Addition applies when mutually exclusive routes lead to the same target category. For Rr × Rr, an offspring is heterozygous through either R egg + r pollen or r egg + R pollen; probabilities are 1/4 + 1/4 = 1/2. In a two-locus cross, a phenotype such as round yellow groups several genotypes, so we add the probabilities of its mutually exclusive genotype routes. These rules give each arithmetic operation a biological interpretation.
Independent assortment has a physical condition
Allele pairs on different chromosomes generally assort independently because different chromosome pairs orient independently in meiosis I. Loci far apart on the same chromosome may also have approximately independent recombinant outcomes. Closely linked genes on one chromosome do not normally produce four equal gamete classes. Consequently, a 9:3:3:1 ratio requires independent assortment, complete dominance at each studied locus, broadly equal gamete viability, random fertilisation, and no strong survival effects that distort counted phenotypes. A deviation is evidence to investigate, not automatically proof of linkage.
A mixed-genotype example
Consider TtYy × Ttyy with independently assorting height and seed-colour loci. For height, Tt × Tt gives P(tall) = 3/4 and P(dwarf) = 1/4. For colour, Yy × yy gives P(yellow) = 1/2 and P(green) = 1/2. Multiply the independent character probabilities: tall green = 3/4 × 1/2 = 3/8; dwarf green = 1/4 × 1/2 = 1/8. This approach is faster than a full square once the genetic model is established.
Quick Quiz
Quiz
How many equally likely gamete types does RrYy form under independent assortment?
What is the expected probability of round green in RrYy × RrYy?
What does 9:3:3:1 describe in the standard dihybrid model?
How many distinct two-locus genotype combinations occur in RrYy selfing?
When does independent assortment commonly fail?
In TtYy × Ttyy, what is P(dwarf and green) under independence?
Practice Problems
- Derive the expected four F2 phenotypes in RrYy selfing.
- List gametes from AaBbCcDd when all loci assort independently, and calculate the number of classes.
- Calculate tall-green and dwarf-green proportions for TtYy × Ttyy.
- Explain why a pair of loci producing mostly parental combinations suggests linkage.
For each locus, dominant phenotype probability is 3/4 and recessive phenotype probability 1/4. Multiply across independent loci: R_Y_ = 9/16; R_yy = 3/16; rrY_ = 3/16; rryy = 1/16. The combined ratio is 9:3:3:1.
Key Takeaways
• A dihybrid cross follows two characters at once. • Independent assortment gives four RrYy gamete types: RY, Ry, rY and ry. • The four gamete types each have expected frequency 1/4 under independence. • Complete dominance plus independent assortment yields the 9:3:3:1 F2 phenotype ratio. • The 16 square cells represent nine distinct two-locus genotype classes. • Multiply probabilities for independent events and add mutually exclusive routes. • Physical linkage and biological selection can disrupt a simple 9:3:3:1 expectation.