predicted genetic outcomes for two genes that are either linked or assort independently (unlinked)

What this dot point is asking

VCAA wants the dihybrid cross for two genes that assort independently (unlinked), the expected 9:3:3:1 ratio, and how that ratio breaks down when the two genes are linked on the same chromosome. You should also know how crossing over produces a small fraction of recombinant offspring even from linked genes.

The answer

Independent assortment: the 9:3:3:1 ratio

Mendel's Second Law (the Law of Independent Assortment) says that alleles of one gene segregate into gametes independently of alleles of another gene. This holds when the two genes are on different chromosomes or are far apart on the same chromosome.

Consider two genes, R/r and Y/y, on different chromosomes:

• A homozygous parent RRYY (round yellow) crossed with rryy (wrinkled green) gives an F1 all RrYy (round yellow, dihybrid).
• The F1 dihybrid (RrYy) produces four kinds of gametes in equal proportion: RY, Ry, rY, ry (each at 25%) because at metaphase I, the orientation of the two homologous pairs is independent.
• Crossing two F1 dihybrids (RrYy × RrYy) gives a 4-by-4 Punnett square with 16 cells.

The F2 phenotype ratio:

• 9 round yellow (RY)
• 3 round green (R_yy)
• 3 wrinkled yellow (rrY_)
• 1 wrinkled green (rryy)

This is the famous 9:3:3:1 ratio. It is the dihybrid version of Mendel's 3:1.

A dihybrid test cross (RrYy × rryy) gives offspring in a 1:1:1:1 ratio of all four phenotypes, confirming independent assortment.

Why independent assortment works

During metaphase I of meiosis, each homologous chromosome pair lines up at the equator independently of every other pair. For two pairs, there are two equally likely orientations, producing four equally likely gametes:

• Maternal R-chromosome + maternal Y-chromosome: gamete RY.
• Maternal R + paternal y: gamete Ry.
• Paternal r + maternal Y: gamete rY.
• Paternal r + paternal y: gamete ry.

For more pairs, the number of combinations doubles each time. This is the meiotic basis of independent assortment.

Linkage: when 9:3:3:1 breaks down

Two genes on the same chromosome are linked: they tend to be inherited together because they travel as one unit through meiosis.

If linkage were absolute, an RrYy parent with R and Y on one homologue and r and y on the other (the cis or "coupling" arrangement) would produce only two kinds of gametes:

• RY (parental)
• ry (parental)

A test cross would give a 1:1 ratio of only two phenotypes (round yellow and wrinkled green), no Ry or rY. Wildly different from 1:1:1:1.

But linkage is not absolute, because of crossing over.

Crossing over produces recombinants

During prophase I, homologous chromosomes pair up and non-sister chromatids exchange segments at chiasmata. If a crossover happens between two linked loci, it shuffles the alleles:

• Parental (non-recombinant) gametes: RY and ry.
• Recombinant gametes (created by the crossover): Ry and rY.

The recombination frequency (RF) between two genes is:

RF = number of recombinant offspring / total offspring.

• If two genes are completely linked (no crossing over ever happens between them): RF = 0%.
• If two genes are very close on the same chromosome: RF is low (a few percent), so most offspring are parental.
• If two genes are far apart on the same chromosome: RF approaches 50%, indistinguishable from independent assortment.
• If two genes are on different chromosomes: RF = 50% (independent assortment).

RF is roughly proportional to the distance between the two loci on the chromosome (1% RF = 1 centimorgan, cM). This is the basis for genetic mapping.

Identifying linkage from data

A test cross is the cleanest way to detect linkage because it strips away the dominance complication.

If a test cross AaBb × aabb gives:

• 1:1:1:1 of all four phenotypes: the two genes are unlinked (independent assortment).
• More parental than recombinant offspring (a ratio close to 1:1 parental, plus a smaller equal number of recombinants): the two genes are linked.

Example data:

• 425 AaBb (parental)
• 410 aabb (parental)
• 90 Aabb (recombinant)
• 75 aaBb (recombinant)

Parental total = 835. Recombinant total = 165. RF = 165 / 1000 = 16.5%. The two genes are linked, about 16.5 cM apart.

Cis and trans

In a dihybrid (AaBb), the dominant alleles can be on the same chromosome (cis, written AB / ab) or on opposite chromosomes (trans, written Ab / aB).

The cis or trans arrangement determines which gametes are parental and which are recombinant. In cis, parental gametes are AB and ab; recombinant gametes are Ab and aB. In trans, parental gametes are Ab and aB; recombinant gametes are AB and ab.

You determine cis or trans from the parents' genotypes (often inferred from a pedigree), and then identify the most common offspring as parental.

Why this matters for natural populations

Independent assortment and recombination together create enormous genetic diversity (covered in the meiosis-and-genetic-diversity dot point). Linkage limits this diversity for closely-positioned genes: they tend to be inherited as haplotypes.

Mapping linked genes was the foundation of classical genetics (Sturtevant's 1913 chromosome map of Drosophila) and remains the basis of modern genome-wide association studies (GWAS) that find disease genes by looking for linked DNA markers.

Examples in context

Example 1. Linkage mapping in Drosophila at University of Melbourne. University of Melbourne genetics labs use Drosophila melanogaster (vinegar fly) to teach linkage. Two genes on the same chromosome (say, body colour and wing length) tend to be inherited together because they are physically linked. Crossing a heterozygous fly to a homozygous recessive should give a 1:1:1:1 ratio if the genes are unlinked, but linked genes give mostly two parental types and few recombinants. The recombination frequency, measured as (recombinant offspring divided by total offspring) times 100, equals the genetic distance in centimorgans. If only 10 percent of offspring show recombinant phenotypes, the two loci are 10 cM apart.

Example 2. Wheat rust resistance at Plant Breeding Institute, Narrabri. Cereal breeders at the University of Sydney's Plant Breeding Institute in Narrabri stack rust-resistance alleles on a single wheat chromosome so they remain physically linked through generations of selection. The advantage: farmers receive a multi-gene resistance package that breaks down more slowly under pathogen evolution. The disadvantage: if a deleterious allele lies near a desirable rust gene, the linkage drag pulls it through too. Modern molecular markers in the breeding pipeline allow targeted recombination to break unwanted linkages, recovering a 90 percent recurrent parent genome with the rust-resistance "donor block" still intact.

Try this

Q1. Distinguish between linked and unlinked genes with respect to their position on chromosomes and inheritance pattern. [2 marks]

• Cue. Linked: on same chromosome, tend to be inherited together. Unlinked: on different chromosomes or far apart, assort independently.

Q2. A dihybrid test cross gives 405 AaBb, 12 Aabb, 17 aaBb and 396 aabb offspring. Calculate the recombination frequency and state whether the two genes are linked. [3 marks]

• Cue. Recombinants = 12 + 17 = 29; total = 830; RF = 29/830 x 100 = 3.5 percent; strongly linked (close together on same chromosome).

Q3. Refer to two genes A and B with recombination frequency 25 percent. (a) State the genetic distance in centimorgans. (b) Predict the four offspring classes and their approximate ratios from an AaBb x aabb test cross. (c) Explain why RF approaches 50 percent for genes far apart on the same chromosome. [2+2+2 marks]

• Cue. (a) 25 cM. (b) Parental: 37.5 percent AaBb, 37.5 percent aabb. Recombinant: 12.5 percent Aabb, 12.5 percent aaBb. (c) Multiple crossovers between them make them effectively unlinked.