Mendel's Second Law (Law of Independent Assortment)

What the law states

The law of independent assortment states that, when two or more genes are located on different chromosomes, the alleles of one gene separate into gametes independently of how the alleles of another gene separate. Which allele of gene A ends up in a gamete does not affect which allele of gene B ends up in that same gamete.

Why it happens: the link to meiosis

During meiosis I, each homologous chromosome pair lines up and separates independently of every other pair. Because genes on different chromosomes are carried on different homologous pairs, the way one pair separates does not influence the way another pair separates, so all combinations of alleles are equally possible in the gametes.

Producing four gamete types

Consider a dihybrid individual TtYy, where T/t and Y/y are on different chromosomes. Independent assortment means T can combine with either Y or y, and t can also combine with either Y or y, giving four equally likely gamete types.

• TY
• Ty
• tY
• ty

What it means for a dihybrid cross

Because a TtYy individual produces four gamete types in equal proportion, a dihybrid Punnett square needs a 4 x 4 grid, and crossing two TtYy individuals gives the classic 9:3:3:1 phenotype ratio.

Worked problem 1 (easy): a dihybrid test cross

In pea plants, round seed shape (R) is dominant over wrinkled (r), and yellow seed colour (Y) is dominant over green (y), with the two genes on different chromosomes. A dihybrid plant (RrYy) is crossed with a fully recessive plant (rryy).

Work out the phenotype ratio of the offspring.

By independent assortment the RrYy plant forms four gamete types, RY, Ry, rY and ry, in equal numbers, while the rryy plant forms only ry gametes. Each offspring therefore takes one of the four gamete types from the RrYy parent and pairs it with ry, giving a 1 round yellow : 1 round green : 1 wrinkled yellow : 1 wrinkled green ratio.

Worked problem 2 (medium): the full dihybrid cross

Two dihybrid pea plants (RrYy) are crossed. Work out the phenotype ratio of the offspring.

Each parent forms four gamete types, RY, Ry, rY and ry, so a 4 x 4 Punnett square gives sixteen equally likely combinations. Counting phenotypes gives 9 with both dominant traits (round and yellow), 3 round and green, 3 wrinkled and yellow, and 1 wrinkled and green, the classic 9:3:3:1 dihybrid ratio.

Worked problem 3 (SPM-level): one gene heterozygous, one homozygous

A pea plant heterozygous for both seed shape and seed colour (RrYy) is crossed with a plant heterozygous for shape but true-breeding green (Rryy). Predict the phenotype ratio of the offspring, using independent assortment.

The RrYy parent forms four gamete types (RY, Ry, rY, ry), but the Rryy parent forms only two (Ry and ry), so the Punnett square is 4 x 2 with eight combinations. Colour behaves like a 1:1 test cross (Yy x yy) and shape behaves like a 3:1 cross (Rr x Rr); combined, the offspring show 3 round yellow : 3 round green : 1 wrinkled yellow : 1 wrinkled green, a 3:3:1:1 ratio.

Practice problems

1. In pea plants, tall (T) is dominant over short (t), and green pod (G) is dominant over yellow pod (g). A TtGg plant is crossed with another TtGg plant. State the phenotype ratio expected among the offspring.
2. List the four types of gamete produced by a TtGg plant.
3. A TtGg plant is crossed with a ttgg plant. State the expected phenotype ratio of the offspring and name this type of cross.

Common mistakes

Source:SRC-DSKP-EN

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