Calculate allele and genotype frequencies — Hinglish notes

5.5.2 · Biology › Population Genetics & Speciation

1. Core vocabulary

HUM kya count kar rahe hain? Do alag-alag cheezein:

Genotype frequency individuals count karta hai (unka count $N$ hai).
Allele frequency allele copies count karta hai (ek diploid mein $2 N$ hain, kyunki har individual 2 copies hold karta hai).

2. Formulas ko scratch se derive karna

Setup. Maano $N$ diploid individuals ki ek population mein:

$n_{AA}$ individuals of genotype $AA$
$n_{A a}$ individuals of genotype $A a$
$n_{aa}$ individuals of genotype $aa$

with $n_{AA} + n_{A a} + n_{aa} = N$ .

Genotype frequencies (aasan — bas $N$ se divide karo)

$f (AA) = \frac{n _{AA}}{N}, f (A a) = \frac{n _{A a}}{N}, f (aa) = \frac{n _{aa}}{N}$

Yeh step kyun? Genotypes individuals ko label karte hain, isliye total pool individuals ki sankhya $N$ hai. Yeh teeno 1 tak add hone chahiye kyunki har individual ka exactly ek genotype hota hai.

Allele frequencies (copies count karo)

Total allele copies $= 2 N$ (har ek $N$ insaan 2 carries karta hai).

Kitne $A$ copies hain?

Har $AA$ insaan 2 $A$ 's contribute karta hai → $2 n_{AA}$
Har $A a$ insaan 1 $A$ contribute karta hai → $n_{A a}$
Har $aa$ insaan 0 $A$ 's contribute karta hai → $0$

$p = f (A) = \frac{2 n _{AA} + n _{A a}}{2 N}$

2's kyun? Kyunki hum copies count kar rahe hain, log nahi. Ek $AA$ homozygote bag mein do $A$ tokens deta hai.

Fraction ko split karo ek khoobsurat shortcut dikhane ke liye:

$p = \frac{2 n _{AA}}{2 N} + \frac{n _{A a}}{2 N} = \frac{n _{AA}}{N} + \frac{1}{2} \cdot \frac{n _{A a}}{N}$

Kyunki sirf do alleles hain, har copy ya toh $A$ hai ya $a$ , isliye:

$p + q = 1$

Kyun? $p + q = f (AA) + f (A a) + f (aa) = 1$ (teeno genotype frequencies 1 tak sum hoti hain). Yeh koi assumption nahi hai — yeh counting se khud nikal ke aata hai.

4. Recall checkpoints

Recall Answers chhupa lo — kya tum formulas khud bana sakte ho?

Hum allele counts ko $N$ ki jagah $2 N$ se kyun divide karte hain? → Kyunki diploids 2 copies each carry karte hain; hum copies count karte hain, log nahi.
Heterozygotes par $\frac{1}{2}$ factor kyun? → Woh allele of interest ki sirf ek copy carry karte hain.
$p + q$ kya equal hona chahiye aur kyun? → 1, kyunki har copy do mein se ek allele hai.
$f (aa) = 0.16$ aur $f (A a) = 0.48$ diya hai, $q$ nikalo. → $q = 0.16 + 0.24 = 0.40$ .

Recall Feynman: ek 12-saal ke bacche ko explain karo

Socho ek jar socks se bhara hai, aur har baccha exactly do socks uthata hai. Red socks allele $A$ hain, blue socks allele $a$ hain. Do reds wala baccha " $AA$ " hai, do blues wala " $aa$ " hai, ek-ek wala " $A a$ " hai. Yeh jaanne ke liye ki poori class kitni "red" hai,