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 (R_Y_)
• 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.

Worked example

A pea breeder crosses two dihybrid plants (RrYy times RrYy). The genes for seed shape (R/r) and seed colour (Y/y) are on different chromosomes, so they assort independently. From 800 offspring, she expects:

• Round yellow (R_Y_): 9/16 × 800 = 450.
• Round green (R_yy): 3/16 × 800 = 150.
• Wrinkled yellow (rrY_): 3/16 × 800 = 150.
• Wrinkled green (rryy): 1/16 × 800 = 50.

She actually gets 452, 144, 156 and 48: very close to 9:3:3:1, confirming independent assortment.

A different breeder works with two genes she suspects are linked. She test-crosses an AaBb plant with aabb and gets 410:425:78:87. Far from 1:1:1:1, so the genes are linked. The parental classes (AaBb and aabb, the largest) tell her the AaBb parent had the cis configuration AB / ab. Recombination frequency = (78 + 87) / 1000 = 16.5%, so the two loci are 16.5 cM apart.

Common traps

Confusing 9:3:3:1 with 1:1:1:1. 9:3:3:1 is the F2 of a dihybrid self-cross (RrYy × RrYy). 1:1:1:1 is the test cross (RrYy × rryy) result for independent assortment.

Calling a 9:3:3:1 result "linkage". It is the opposite: 9:3:3:1 is the signature of independent assortment.

Forgetting that distant linked genes look unlinked. Two genes on the same chromosome but very far apart can show RF approaching 50%, indistinguishable from independent assortment in a single cross.

Saying crossing over always happens between linked genes. It happens with a probability proportional to distance. Very close genes rarely recombine.

Calling a deviation from 9:3:3:1 "always linkage". Other causes include sample size variation, lethal allele combinations, or epistasis. Linkage shows up as parental classes much larger than recombinant classes.

Forgetting cis vs trans. The same set of alleles can be in different parental arrangements; this changes which offspring are parental.

In one sentence

Two genes that assort independently (on different chromosomes or very far apart on the same chromosome) give a 9:3:3:1 phenotype ratio in a dihybrid self-cross and 1:1:1:1 in a test cross, while two genes linked on the same chromosome produce mostly parental gametes (matching the cis or trans arrangement in the parent) with a smaller fraction of recombinant gametes (whose frequency, the recombination frequency, equals the genetic distance in centimorgans and tops out at 50% for distant or unlinked loci).