Dihybrid Crosses and the Law of Independent Assortment

A dihybrid cross follows the inheritance of two genes at the same time, and Mendel’s law of independent assortment states that alleles of different genes are sorted into gametes independently of each other. In brief: crossing two individuals that are heterozygous for both genes (RrYy × RrYy) gives a phenotype ratio of 9 : 3 : 3 : 1 in the offspring, which comes from a 16-box Punnett square. The law holds when the genes are on different chromosomes or far apart on the same chromosome.

Dihybrid crosses look complicated, but a systematic method makes them simple. This guide explains the law of independent assortment and its basis in meiosis, shows how to find gametes and build a 16-box Punnett square, derives the 9:3:3:1 ratio and a shortcut with probabilities, works through two examples and corrects frequent mistakes.

What is independent assortment?

Mendel crossed pea plants that differed in two traits: seed shape (round R, dominant; wrinkled r) and seed color (yellow Y, dominant; green y). He crossed pure-breeding round yellow plants (RRYY) with pure-breeding wrinkled green plants (rryy). All F1 plants were round and yellow (RrYy). When he crossed these F1 plants with each other, the F2 generation showed four phenotypes in a ratio of about 9 round yellow : 3 round green : 3 wrinkled yellow : 1 wrinkled green. In his data there were 315 round yellow, 108 round green, 101 wrinkled yellow and 32 wrinkled green seeds, a total of 556.

This ratio showed that the allele for shape and the allele for color were passed on independently of each other. This is the law of independent assortment. The cellular basis is meiosis: during metaphase I, the pairs of homologous chromosomes line up randomly, so the way in which the pair carrying one gene separates does not influence the way in which the pair carrying another gene separates.

How do you find the gametes?

Each gamete receives one allele of each gene. List all combinations of one allele from each gene. For RrYy, the first gene gives R or r, and the second gives Y or y. The possible gametes are:

• RY
• Ry
• rY
• ry

Each type occurs with a probability of 1/4. A quick check: a heterozygous individual for n independent genes makes 2ⁿ types of gamete. For two genes, 2² = 4. For a homozygous parent such as RRYY, there is just one type, RY.

A common method for listing gametes is the “FOIL” approach, borrowed from algebra: take the first alleles of each gene, then the outer, the inner and the last (R with Y, R with y, r with Y, r with y).

How do you build the Punnett square?

Write the four gametes of one parent across the top and the four gametes of the other parent down the side. There are 16 boxes. Each box shows the genotype of the offspring.

Where does the 9:3:3:1 ratio come from?

Count the phenotypes in the 16 boxes. Dominant for both traits (round and yellow) requires at least one R and at least one Y.

The total is 9 + 3 + 3 + 1 = 16. In the square, the genotype RrYy appears 4 times, RRYy twice, RrYY twice, Rryy twice, rrYy twice, and RRYY, RRyy, rrYY and rryy once each, giving 4 + 2 + 2 + 2 + 2 + 4 = 16.

Is there a faster method?

Because the genes assort independently, you can use the product rule of probability and treat each gene as a separate monohybrid cross. For Rr × Rr, the probability of a round seed is 3/4 and of a wrinkled seed is 1/4. For Yy × Yy, it is 3/4 yellow and 1/4 green. Then multiply:

• Round and yellow: 3/4 × 3/4 = 9/16.
• Round and green: 3/4 × 1/4 = 3/16.
• Wrinkled and yellow: 1/4 × 3/4 = 3/16.
• Wrinkled and green: 1/4 × 1/4 = 1/16.

This method is quicker for crosses with three or more genes, for which a Punnett square would be too big. For three genes, a Punnett square would have 64 boxes.

Worked example: a dihybrid cross in guinea pigs

In guinea pigs, black coat (B) is dominant over brown (b), and short hair (S) is dominant over long hair (s). A black, short-haired guinea pig of genotype BbSs is crossed with a brown, long-haired guinea pig (bbss). This is a dihybrid test cross.

1. Gametes: BbSs produces BS, Bs, bS and bs with 1/4 each. bbss produces only bs.
2. Offspring: BbSs (black short), Bbss (black long), bbSs (brown short) and bbss (brown long), each with a probability of 1/4.
3. Ratio: 1 : 1 : 1 : 1. In a litter-based sample of 80 offspring, about 20 of each type are expected.

Now suppose a cross of BbSs × BbSs produces 160 offspring. The expected numbers are 9/16 × 160 = 90 black short, 3/16 × 160 = 30 black long, 30 brown short and 1/16 × 160 = 10 brown long. If the observed numbers are 88, 32, 29 and 11, they are close to the expected values, and a chi-squared test could confirm that the difference is due to chance. The test is calculated as the sum of (observed − expected)² ÷ expected: (88 − 90)² ÷ 90 = 0.044; (32 − 30)² ÷ 30 = 0.133; (29 − 30)² ÷ 30 = 0.033; (11 − 10)² ÷ 10 = 0.100. The total χ² ≈ 0.31, which is far below the critical value of 7.81 for 3 degrees of freedom at the 5 percent level, so the data fit the 9:3:3:1 ratio.

When does independent assortment not apply?

Genes that are close together on the same chromosome are inherited together, and this is called linkage. They do not assort independently, because they are rarely separated by crossing over. A cross of two linked genes produces mostly parental combinations, and the ratio differs from 9:3:3:1. Gene linkage is covered in a later article. Interaction between genes (epistasis) also changes the ratios, for example to 9:3:4 or 9:7.

How was it discovered?

Mendel presented his dihybrid results in 1865. When the importance of chromosomes became clear in the early twentieth century, Thomas Hunt Morgan and his students found cases in fruit flies in which genes did not assort independently, which led to the idea of linkage and to the construction of the first genetic maps by Alfred Sturtevant in 1913.

Which mistakes are common?

Mistake 1: Writing gametes with two alleles of the same gene

A gamete carries one allele of each gene. RY is correct. RR is not.

Mistake 2: Missing combinations of gametes

A heterozygote for two genes makes four gamete types. Check that there are four, and that each combination appears once.

Mistake 3: Treating 9:3:3:1 as a rule for any cross

It applies to a cross of two double heterozygotes with independent assortment and complete dominance.

Mistake 4: Forgetting the multiplication rule

For independent genes, multiply probabilities, which can save a lot of time.