Calculate allele and genotype frequencies — Biology notes

1. The core vocabulary

WHAT are we counting? Two different things:

Genotype frequency counts individuals (there are $N$ of them).
Allele frequency counts allele copies (there are $2N$ of them in a diploid, because each individual holds 2 copies).

2. Deriving the formulas from scratch

Setup. Let a population of $N$ diploid individuals have:

$n_{AA}$ individuals of genotype $AA$
$n_{Aa}$ individuals of genotype $Aa$
$n_{aa}$ individuals of genotype $aa$

with $n_{AA} + n_{Aa} + n_{aa} = N$ .

Genotype frequencies (easy — just divide by $N$ )

$f(AA)=\frac{n_{AA}}{N},\quad f(Aa)=\frac{n_{Aa}}{N},\quad f(aa)=\frac{n_{aa}}{N}$

Why this step? Genotypes label individuals, so the total pool is the number of individuals $N$ . These three must sum to 1 because every individual has exactly one genotype.

Allele frequencies (count the copies)

Total allele copies $= 2N$ (each of $N$ people carries 2).

How many $A$ copies are there?

Each $AA$ person contributes 2 $A$ 's → $2n_{AA}$
Each $Aa$ person contributes 1 $A$ → $n_{Aa}$
Each $aa$ person contributes 0 $A$ 's → $0$

$p = f(A) = \frac{2n_{AA} + n_{Aa}}{2N}$

Why the 2's? Because we're counting copies, not people. An $AA$ homozygote donates two $A$ tokens to the bag.

Split the fraction to reveal a beautiful shortcut:

$p = \frac{2n_{AA}}{2N}+\frac{n_{Aa}}{2N}=\frac{n_{AA}}{N}+\frac12\cdot\frac{n_{Aa}}{N}$

Because there are only two alleles, every copy is either $A$ or $a$ , so:

$\boxed{p + q = 1}$

Why? $p+q = f(AA)+f(Aa)+f(aa) = 1$ (the three genotype frequencies sum to 1). This is not an assumption — it falls out of the counting.

3. Worked examples

Figure — Calculate allele and genotype frequencies

Worked example Example 1 — Counting from raw numbers

In 1000 pea plants: $640$ are $AA$ , $320$ are $Aa$ , $40$ are $aa$ .

Step 1 — Genotype frequencies. $f(AA)=640/1000=0.64$ , $f(Aa)=320/1000=0.32$ , $f(aa)=40/1000=0.04$ . Why? Divide individuals by total individuals $N=1000$ . Check: $0.64+0.32+0.04=1$ . ✓

Step 2 — Allele copies. Total copies $=2\times1000=2000$ . $A$ copies $=2(640)+320 = 1600$ . $a$ copies $=2(40)+320=400$ . Why the 2's? Each $AA$ gives two $A$ 's; each $Aa$ gives one.

Step 3 — Allele frequencies. $p=\frac{1600}{2000}=0.8,\qquad q=\frac{400}{2000}=0.2$ Check with the shortcut: $p = 0.64 + \tfrac12(0.32)=0.64+0.16=0.80$ ✓ and $p+q=1$ ✓.

4. Recall checkpoints

Recall Cover the answers — can you rebuild the formulas?

Why do we divide allele counts by $2N$ instead of $N$ ? → Because diploids carry 2 copies each; we count copies, not people.
Why the factor $\tfrac12$ on heterozygotes? → They carry only one copy of the allele in question.
What must $p+q$ equal and why? → 1, because every copy is one of the two alleles.
Given $f(aa)=0.16$ and $f(Aa)=0.48$ , find $q$ . → $q=0.16+0.24=0.40$ .

Recall Feynman: explain to a 12-year-old

Imagine a jar full of socks, and every kid grabs exactly two socks. Red socks are allele $A$ , blue socks are allele $a$ . A kid with two reds is " $AA$ ", two blues is " $aa$ ", one of each is " $Aa$ ". To find out how "red" the whole class is, you don't count kids — you count socks. Add up every red sock and divide by all the socks. The mixed kids only own one red sock each, so they only count "half" toward redness. That "half" is the whole trick!

5. Flashcards

What is an allele frequency?

The proportion of all allele copies in the gene pool that are a particular allele (e.g.

p

for

A

What is a genotype frequency?

The proportion of individuals in the population having a particular genotype.

In a diploid population of N individuals, how many allele copies exist at one locus?

2N

, because each individual carries 2 copies.

Formula for p in terms of allele counts

p = \dfrac{2n_{AA} + n_{Aa}}{2N}

Shortcut for p from genotype frequencies

p = f(AA) + \tfrac12 f(Aa)

("homozygote plus half the heterozygotes").

Why does the heterozygote get a factor of 1/2?

It carries only one copy of the allele of interest, not two.

What does p + q equal and why?

1, because every allele copy is one of the two alleles; genotype frequencies sum to 1.

If 640 AA, 320 Aa, 40 aa, what is p (freq of A)?

p = 0.64 + 0.5(0.32) = 0.80.

For an X-linked gene, how do males contribute to copy counts?

Only 1 copy each (hemizygous), so total copies = 2(females) + 1(males), not 2N.

Given f(aa)=0.09 and f(Aa)=0.42, find q.

q = 0.09 + 0.21 = 0.30.

6. Connections

Hardy-Weinberg Equilibrium — uses $p$ and $q$ to predict genotype frequencies $p^2, 2pq, q^2$ .
Genetic Drift — random change in these allele frequencies in small populations.
Natural Selection & Fitness — differential survival shifts $p$ and $q$ over generations.
Gene Pool — the total collection of alleles we are sampling from.
Speciation — accumulated frequency changes between isolated populations.
Mendelian Genetics — Punnett Squares — the individual-level logic behind heterozygotes carrying one copy.

Concept Map

Hinglish (regional understanding)

Intuition Hinglish mein samjho

Dekho, population genetics ka sabse pehla step hai yeh samajhna ki ek population basically ek bag of alleles hai — jise gene pool kehte hain. Har allele gene ka ek version hai (jaise $A$ ya $a$ ). Ab do cheezein alag hain: genotype frequency matlab kitne individuals ka particular pair hai ( $AA$ , $Aa$ , $aa$ ), aur allele frequency matlab total allele copies mein se kitni copies ek particular allele ki hain. Diploid organism mein har banda 2 copies rakhta hai, isliye total copies $2N$ hoti hain, sirf $N$ nahi.

Formula ratne ki zaroorat nahi — bas ginti karo. $A$ ki total copies nikaalne ke liye: har $AA$ banda 2 $A$ deta hai, har $Aa$ banda sirf 1 $A$ deta hai, aur $aa$ banda 0. Toh $p = \frac{2n_{AA} + n_{Aa}}{2N}$ . Ise tod ke likho toh shortcut milta hai: $p = f(AA) + \tfrac12 f(Aa)$ — yaani "homo full, hetero half". Heterozygote ko aadha count isliye karte hain kyunki uske paas us allele ki sirf ek hi copy hai.

Sabse common galti: log $p = f(AA)+f(Aa)$ likh dete hain — yeh galat hai kyunki $Aa$ waale ko poora count kar diya, jabki uske paas ek hi $A$ hai. Hamesha yaad rakho: hetero ko half do. Aur ek dhyaan ki baat — $2N$ wala rule sirf diploid, autosomal gene ke liye hai. X-linked gene mein males ke paas ek hi copy hoti hai, toh copies alag se ginni padti hain.

Yeh skill kyun important hai? Kyunki iske upar hi Hardy-Weinberg, selection, genetic drift aur aakhir mein speciation sab khada hai. Pehle allele frequency nikalna aata hai, tabhi tum bata paoge ki population evolve ho rahi hai ya nahi.

Test yourself — Population Genetics & Speciation

Connections