Gene Mapping & Linked Genes

Discover how crossing over during meiosis creates recombinant chromosomes — and how scientists exploit recombination frequency to map gene positions.

Crossing Over During Meiosis I

In prophase I, homologous chromosomes pair tightly (synapsis). At points called chiasmata, non-sister chromatids exchange segments. This produces chromosomes with new allele combinations — recombinants.

Prophase I — homologous chromosomes align (synapsis) L P Chromosome 1: L — P l p Chromosome 2: l — p chiasma (crossover point) meiosis PARENTAL GAMETES L P LP l p lp RECOMBINANT GAMETES L p Lp ✗ l P lP ✗ Long stem, Purple flower (LP) short stem, white flower (lp) ↑ Crossover between L and P loci → recombinant gametes Lp and lP
Step 1 / 4
Key Insight: The closer two genes are on the same chromosome, the less likely a crossover will occur between them — so they tend to be inherited together. When very far apart (or on different chromosomes), they assort independently.

Distance Controls Recombination Frequency

Slide to see how physical distance between genes affects how often they recombine. Each map unit equals 1% recombination frequency.



A Gene A B Gene B 5 map units RF = 5% → genes are tightly linked
Tightly linked — low recombination

When Mendel's Laws Break Down: Linked Genes

Mendel's Law of Independent Assortment states that genes on different chromosomes assort independently, producing equal ratios of all four phenotype classes in a testcross (1:1:1:1). But genes on the same chromosome violate this law.

Independent Assortment (Mendel)
Genes on different chromosomes: gametes AB, Ab, aB, ab formed in equal proportions → 1:1:1:1 testcross ratio.
Linked Genes (Morgan)
Genes on the same chromosome: parental combinations far more common than recombinants → skewed ratios.

Thomas Hunt Morgan (1910s) studied Drosophila and found pairs of genes that did not assort independently. He realized they must be physically located on the same chromosome — linked. Crossing over partially breaks the linkage, producing recombinants at a rate that reflects the distance between loci.

Crossing over saves partial independence! Without crossing over, linked genes would always travel together and we'd only see two phenotypes in a testcross. Crossing over generates the rare recombinant classes, and their frequency is our measuring stick for distance.

How Linkage Changes Testcross Ratios

Gene RelationshipExpected Testcross RatioExample Counts (n=400)
Unlinked — different chromosomes 1 : 1 : 1 : 1 100 : 100 : 100 : 100
Loosely linked — far apart, same chromosome Slightly skewed 120 : 130 : 80 : 70
Tightly linked — close together Strongly skewed 188 : 194 : 9 : 9
Completely linked — no crossover possible 1 : 1 : 0 : 0 200 : 200 : 0 : 0

Gene Mapping Tutorial

A step-by-step walkthrough of how to calculate recombination frequency and build a chromosome map.

Step 1 — Set Up the Testcross

Gene mapping requires crossing a doubly heterozygous individual with a homozygous recessive individual. The recessive parent contributes only recessive alleles, so each offspring's phenotype directly reveals which gamete the heterozygous parent produced.

Heterozygous parent
Ll Pp
×
Homozygous recessive (tester)
ll pp
Why homozygous recessive? Because ll pp only contributes l and p alleles. Any dominant phenotype in the offspring must come from the heterozygous parent's gamete — there is no masking.

Step 2 — Identify Parental and Recombinant Types

Examine the offspring counts. The two most abundant classes are Parental types — they match the original chromosome configuration (no crossover between the two loci). The two rarest classes are Recombinant types — they arose from a crossover.

PhenotypeCountClassWhy?
Purple long450 Parental Gamete LP — original chromosome 1 intact
White short445 Parental Gamete lp — original chromosome 2 intact
White long18 Recombinant Gamete Lp — crossover between L and P loci
Purple short15 Recombinant Gamete lP — crossover between L and P loci
How to know which are parental? The heterozygous parent originally received its alleles from its own parents. Purple and long came together (LP on one chromosome); white and short came together (lp on the other). Those original combos = parental. New combos (Lp, lP) = recombinant.

Step 3 — Calculate Recombination Frequency (RF)

RF (%) = Recombinant offspringTotal offspring × 100

Using the sample data:

RF = 18 + 15450 + 445 + 18 + 15 × 100 = 33928 × 100 = 3.56 %

Step 4 — Convert to Map Units

1% recombination frequency = 1 map unit

So a RF of 3.56% means the L and P genes are 3.56 map units apart on the chromosome.

L Stem length P Flower color 3.56 map units

Important limit: Even genes very far apart on the same chromosome cannot show more than 50% recombination (because multiple crossovers cancel out). Anything approaching 50 map units looks statistically similar to unlinked genes.

Summary Checklist

  1. Set up a testcross — heterozygous × homozygous recessive
  2. Count offspring in each phenotype class
  3. Identify parental types (most common) and recombinant types (rarest)
  4. Calculate RF = recombinants ÷ total × 100%
  5. State map distance in map units (numerically equal to RF%)
  6. Draw the chromosome map with genes at the correct relative positions

Practice Problems

Work through each problem step by step. Use hints when stuck.

Problem 1
Stem Length & Flower Color
Problem 2
Seed Color & Texture
Problem 3
Petal Color & Leaf Shape

Problem 1 — Stem Length & Flower Color

Gene Information

In a plant species, long stems (L) are dominant over short stems (l), and purple flowers (P) are dominant over white flowers (p). Both genes are located on the same chromosome.

The Cross: A white-flowered, short-stemmed plant (homozygous recessive for both traits) is crossed with a heterozygous purple-flowered, long-stemmed plant. The heterozygous parent carries L and P alleles on the same chromosome.

Parental Genotypes

White short (Parent 1)
ll pp
×
Purple long, heterozygous (Parent 2)
Ll Pp  (LP/lp linked)

Offspring Data

PhenotypeCount
Purple long450
White short445
White long18
Purple short15

Total offspring: 928

Work Through It

Step 1 of 4
Click each phenotype to label it as Parental or Recombinant (click again to cycle).
Purple long
450
(click to classify)
White short
445
(click to classify)
White long
18
(click to classify)
Purple short
15
(click to classify)
The heterozygous parent carries L and P on one chromosome (LP) and l and p on the other (lp). Parental gametes = LP and lp → purple long and white short are parental types (most common). Crossover between the loci → white long (Lp) and purple short (lP) are recombinants.
Step 2 of 4
How many total recombinant offspring were produced?
Add white long (18) + purple short (15).
Step 3 of 4
Calculate the recombination frequency. RF = (recombinants ÷ total) × 100. Round to 2 decimal places.
%
RF = 33 ÷ 928 × 100 = ?
Step 4 of 4
How many map units separate the L and P genes?
map units
1% RF = 1 map unit. Your RF value in % equals your map distance in map units.

Problem 2 — Seed Color & Seed Texture

Gene Information

In pea plants, yellow seeds (Y) are dominant over green seeds (y), and smooth seeds (S) are dominant over wrinkled seeds (s). Both genes are on chromosome 1.

The Cross: A plant homozygous for green wrinkled seeds is crossed with a plant heterozygous for both traits. The heterozygous plant has its dominant alleles (Y and S) on the same chromosome.

Parental Genotypes

Green wrinkled (Parent 1)
yy ss
×
Yellow smooth, heterozygous (Parent 2)
Yy Ss  (YS/ys linked)

Offspring Data

PhenotypeCount
Yellow smooth452
Green wrinkled448
Yellow wrinkled51
Green smooth49

Total offspring: 1000

Work Through It

Step 1 of 4
Click each phenotype to classify it as Parental or Recombinant.
Yellow smooth
452
(click to classify)
Green wrinkled
448
(click to classify)
Yellow wrinkled
51
(click to classify)
Green smooth
49
(click to classify)
The heterozygous parent carries Y and S linked together (YS/ys). Its parental gametes are YS → yellow smooth, and ys → green wrinkled. Crossovers produce Ys (yellow wrinkled) and yS (green smooth).
Step 2 of 4
Total recombinant offspring?
Yellow wrinkled (51) + Green smooth (49).
Step 3 of 4
Calculate the recombination frequency (%). Round to 2 decimal places.
%
RF = 100 ÷ 1000 × 100 = ?
Step 4 of 4
How many map units separate Y and S?
map units
Map distance in map units = RF%.

Problem 3 — Petal Color & Leaf Shape

Gene Information

In a flowering plant, red petals (R) are dominant over white petals (r), and broad leaves (B) are dominant over narrow leaves (b). Both gene loci are on the same chromosome.

The Cross: A plant with white petals and narrow leaves (homozygous recessive for both) is testcrossed against a plant heterozygous for both traits. The heterozygous parent carries the R and B alleles on the same chromosome (cis configuration).

Parental Genotypes

White narrow (Parent 1)
rr bb
×
Red broad, heterozygous (Parent 2)
Rr Bb  (RB/rb linked)

Offspring Data

PhenotypeCount
Red broad330
White narrow324
Red narrow75
White broad71

Total offspring: 800

Work Through It

Step 1 of 4
Click each phenotype to classify it as Parental or Recombinant.
Red broad
330
(click to classify)
White narrow
324
(click to classify)
Red narrow
75
(click to classify)
White broad
71
(click to classify)
The heterozygous parent has R and B on the same chromosome (RB/rb). Its parental gametes are RB → red broad, and rb → white narrow. Crossovers give Rb (red narrow) and rB (white broad).
Step 2 of 4
Total recombinant offspring?
Red narrow (75) + White broad (71) = ?
Step 3 of 4
Calculate recombination frequency (%). Round to 2 decimal places.
%
RF = 146 ÷ 800 × 100 = ?
Step 4 of 4
How many map units separate R and B?
map units
Map distance in map units = RF%.