Construct simple genetic linkage maps

Deriving the ruler from first principles

WHAT we want: turn an offspring count into a distance.

HOW (step by step):

Take a testcross: a doubly heterozygous parent ( $AaBb$ ) × a homozygous recessive ( $aabb$ ). Why a testcross? The recessive parent contributes only $ab$ gametes, so the phenotype of each offspring directly reveals which gamete the heterozygous parent made. No hidden dominance.
If genes are unlinked, independent assortment gives 4 gamete types in equal $25\%:25\%:25\%:25\%$ → RF $= 50\%$ .
If genes are linked, the two parental gamete types dominate, and the two recombinant types are rarer. Recombinants arise only when a crossover occurs between the loci.
Define:

$\text{RF} = \frac{\text{number of recombinant offspring}}{\text{total offspring}} \times 100\%$

Building the map: the additivity rule

HOW to order three genes (A, B, C):

Compute all three pairwise RFs.
The largest RF is the outermost pair → the third gene sits between them.
Smaller pairwise distances may slightly undercount large ones because of double crossovers (two swaps cancel out and look parental), which is why distant pairs are measured less accurately.

Figure — Construct simple genetic linkage maps

Worked Example 1 — two genes

A Drosophila testcross gives 1000 offspring:

410 parental type 1, 390 parental type 2
105 recombinant type 1, 95 recombinant type 2

Step 1: Identify recombinants. Why? Recombinants are the minority classes (departing from parental combos): $105 + 95 = 200$ .

Step 2: Apply formula. $\text{RF} = \frac{200}{1000}\times 100 = 20\%$ Why this step? RF is the map distance.

Answer: The two genes are 20 cM apart.

Worked Example 2 — order three genes

Pairwise RFs measured: A–B = 5%, B–C = 15%, A–C = 18%.

Step 1: Find the largest RF. Why? Largest distance = the two outer (flanking) genes → A–C = 18% is biggest, so A and C are the ends, and B is in the middle.

Step 2: Place B using the smaller distances. A–B = 5, B–C = 15.

Step 3: Check additivity: $5 + 15 = 20$ , but observed A–C = 18. Why the gap? Double crossovers between A and C aren't counted in the A–C class → the directly measured outer distance is a slight underestimate. The sum of short intervals (20 cM) is more accurate.

A ----5---- B ---------15--------- C
|<---------- 18 (measured) -------->|

Worked Example 3 — back-calculate counts

Two genes are 8 cM apart; a testcross yields 500 offspring. How many recombinants?

$\#\text{recomb} = \frac{8}{100}\times 500 = 40$ Why? cM = expected % recombinant, so multiply fraction by total. Expect ~40 recombinants (split ~20 + 20 between the two recombinant classes), ~460 parentals.

Common mistake Steel-manned common errors

Error 1: "Add up all offspring classes to get RF." Why it feels right: RF is "about" all the data. Fix: RF = recombinants ÷ total. Parentals are the majority and are not in the numerator.

Error 2: "The two biggest classes are recombinants." Why it feels right: big numbers look important. Fix: the biggest classes are PARENTAL (linked genes stay together). Recombinants are the minority.

Error 3: "RF of 60% means 60 cM." Why it feels right: the formula seems to scale forever. Fix: RF cannot exceed 50%; observing ~50% means the genes are effectively unlinked. Real distances >50 cM are mapped by adding intervals, not from one RF.

Error 4: Forgetting double crossovers shrink the outer distance. Fix: trust the sum of short intervals over a single large measured RF.

Recall Explain it to a 12-year-old

Imagine two friends sitting on a long bus. If they sit right next to each other, they almost always get off at the same stop together. If they sit far apart, sometimes one moves seats and they end up separated. By counting how often friends get "split up," you can guess how far apart they were sitting. Genes on a chromosome are those friends, and "getting split up" is crossing over. Closer genes split less; far genes split more — that's how we draw the map.

What does a recombination frequency of 1% equal in map units?

1 centimorgan (1 cM / 1 map unit).

In a testcross, which offspring classes are the recombinants?

The two minority classes (the allele combinations differing from both parents).

Why is a testcross (×homozygous recessive) used for mapping?

The recessive parent contributes only recessive gametes, so each offspring's phenotype directly reveals the heterozygous parent's gamete. :::

Formula for map distance from offspring counts?

Distance(cM) = (#recombinants / total offspring) × 100.

What is the maximum possible recombination frequency, and what does it mean?

50%; the genes assort as if unlinked.

Of three linked genes, how do you find which one is in the middle?

The pair with the largest RF are the outer genes; the remaining gene lies between them.

Why does the sum of two short intervals exceed the directly measured outer distance?

Double crossovers between the outer genes go uncounted, so the outer RF underestimates the true distance.

If two genes are 6 cM apart and 800 offspring are produced, how many recombinants are expected?

(6/100)×800 = 48 recombinants.

Connections

Mendel's Law of Independent Assortment — linkage is the exception to it.
Crossing Over and Chiasmata in Meiosis I — the physical event generating recombinants.
Chromosome Theory of Inheritance — genes are physical loci on chromosomes.
Testcross and Dihybrid Ratios — the experimental setup we count from.
Centimorgan vs Physical Distance (base pairs) — why cM ≠ exact bp.

Concept Map

Hinglish (regional understanding)

Intuition Hinglish mein samjho

Dekho, agar do genes ek hi chromosome par paas-paas baithe hain, to woh meiosis ke time aksar saath hi rehte hain — isko linkage kehte hain. Lekin meiosis mein homologous chromosomes apne tukde aapas mein swap kar lete hain, jise crossing over bolte hain. Jitna door do genes honge, utni zyada chance ki crossover unke beech mein gire, aur woh alag ho jayein. Yahi alag hone wale offspring ko recombinant kehte hain.

Mapping ka trick simple hai: testcross lagao (heterozygous parent × homozygous recessive). Recessive parent sirf recessive gamete deta hai, isliye har bachche ka phenotype seedha bata deta hai ki heterozygous parent ne kaunsa gamete banaya. Ab recombinant offspring gino, total se divide karo, ×100 — yeh recombination frequency hai, aur yahi map distance in centimorgan (cM) hai. Yaad rakho: majority classes = parental, minority classes = recombinant. Yeh ulta mat samajhna!

Teen genes ko order karna ho to teeno pairwise RF nikaalo. Jiska RF sabse bada ho, woh dono outer (kinare wale) genes hain, aur teesra gene beech mein. Chhote intervals ko jodo (additive), kyunki double crossover ke kaaran bada wala distance thoda kam dikh sakta hai. Ek baat aur — RF kabhi bhi 50% se zyada nahi ho sakti; agar 50% aaye to genes practically unlinked behave karte hain. Bas itna samajh lo, to linkage map banana aasaan ho jaata hai!

What is being mapped?

Deriving the ruler from first principles

Building the map: the additivity rule

Worked Example 1 — two genes

Worked Example 2 — order three genes

Worked Example 3 — back-calculate counts

Connections

Concept Map

Hinglish (regional understanding)

Test yourself — Extensions of Mendelian Genetics

Connections