Construct simple genetic linkage maps

3.2.11 · Biology › Extensions of Mendelian Genetics

Ruler ko first principles se derive karna

KYA chahiye: offspring count ko distance mein convert karna.

KAISE (step by step):

Ek testcross lo: ek doubly heterozygous parent ( $A a B b$ ) × ek homozygous recessive ( $aabb$ ). Testcross kyun? Recessive parent sirf $ab$ gametes contribute karta hai, isliye har offspring ka phenotype seedha reveal karta hai ki heterozygous parent ne kaun sa gamete banaya. Koi hidden dominance nahi.
Agar genes unlinked hain, to independent assortment 4 gamete types equal $25% : 25% : 25% : 25%$ mein deta hai → RF $= 50%$ .
Agar genes linked hain, to do parental gamete types dominate karte hain, aur do recombinant types rare hote hain. Recombinants tabhi bante hain jab loci ke beech crossover hota hai.
Define karo:

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

Map banana: additivity rule

KAISE teen genes (A, B, C) ko order karo:

Saare teen pairwise RFs calculate karo.
Sabse bada RF outermost pair hai → teesra gene unke beech mein hai.
Chhote pairwise distances bade walon ko thoda undercount kar sakte hain double crossovers ki wajah se (do swaps cancel out ho jaate hain aur parental lagte hain), isliye door waale pairs less accurately measure hote hain.

Figure — Construct simple genetic linkage maps

Worked Example 1 — do genes

Ek Drosophila testcross 1000 offspring deta hai:

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

Step 1: Recombinants identify karo. Kyun? Recombinants minority classes hain (parental combos se alag): $105 + 95 = 200$ .

Step 2: Formula apply karo. $RF = \frac{200}{1000} \times 100 = 20%$ Yeh step kyun? RF hi map distance hai.

Answer: Do genes 20 cM apart hain.

Worked Example 2 — teen genes order karo

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

Step 1: Sabse bada RF dhoondo. Kyun? Sabse badi distance = do outer (flanking) genes → A–C = 18% sabse bada hai, isliye A aur C ends hain, aur B beech mein hai.

Step 2: Chhoti distances use karke B ko place karo. A–B = 5, B–C = 15.

Step 3: Additivity check karo: $5 + 15 = 20$ , lekin observed A–C = 18 hai. Gap kyun? A aur C ke beech double crossovers A–C class mein count nahi hote → seedha measure kiya gaya outer distance thoda underestimate hota hai. Chhote intervals ka sum (20 cM) zyada accurate hai.

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

Worked Example 3 — counts back-calculate karo

Do genes 8 cM apart hain; ek testcross 500 offspring deta hai. Kitne recombinants?

$# recomb = \frac{8}{100} \times 500 = 40$ Kyun? cM = expected % recombinant hai, isliye fraction ko total se multiply karo. ~40 recombinants expect karo (do recombinant classes mein ~20 + 20 split), ~460 parentals.

Common mistake Common errors ko steel-man karna

Error 1: "RF pane ke liye saari offspring classes add karo." Kyun sahi lagta hai: RF "saare" data ke baare mein hai. Fix: RF = recombinants ÷ total. Parentals majority hain aur numerator mein nahi aate.

Error 2: "Do sabse badi classes recombinants hain." Kyun sahi lagta hai: bade numbers important lagte hain. Fix: sabse badi classes PARENTAL hain (linked genes saath rehte hain). Recombinants minority hain.

Error 3: "RF of 60% ka matlab 60 cM hai." Kyun sahi lagta hai: formula hamesha scale hota raha lagta hai. Fix: RF 50% se zyada nahi ho sakta; ~50% dekhna matlab genes effectively unlinked hain. Real distances >50 cM ko intervals add karke map karte hain, ek RF se nahi.

Error 4: Double crossovers outer distance ko shrink karte hain yeh bhool jaana. Fix: ek bade measured RF ke bajaye chhote intervals ke sum par trust karo.

Recall Ek 12-saal ke bachche ko samjhao

Socho do dost ek lambi bus mein baith rahe hain. Agar woh ekdum saath baithein, toh woh almost hamesha ek hi stop par saath utarte hain. Agar woh door baithe hain, to kabhi kabhi ek seat change kar leta hai aur woh alag ho jaate hain. Yeh count karke ki dost kitni baar "alag" hote hain, tum andaaza laga sakte ho ki woh kitni door baithe the. Chromosome par genes wahi dost hain, aur "alag hona" crossing over hai. Paas wale genes kam split hote hain; door waale zyada split hote hain — isi se hum map banate hain.

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 iska exception hai.
Crossing Over and Chiasmata in Meiosis I — woh physical event jo recombinants generate karta hai.
Chromosome Theory of Inheritance — genes chromosomes par physical loci hain.
Testcross and Dihybrid Ratios — woh experimental setup jisse hum count karte hain.
Centimorgan vs Physical Distance (base pairs) — kyun cM ≠ exact bp.

Kya map kiya ja raha hai?