Numerical/Problem-solving
1. TITLE
Hardy-Weinberg numerical — The method that works every time
2. PROBLEM TYPE INTRO
This is a Population Genetics numerical based on the Hardy-Weinberg Principle. You will need the algebraic equations $p + q = 1$ and $p^2 + 2pq + q^2 = 1$. The most typical mistake students make is confusing the genotype frequency ($q^2$) with the allele frequency ($q$), or forgetting to take the square root before calculating the dominant allele.
3. MCQ
Question:
In a random mating population of 2,000 individuals, 180 individuals exhibit the recessive trait for albinism. Assuming the population is in Hardy-Weinberg equilibrium, calculate the frequency of the dominant allele and the number of heterozygous individuals (carriers).
Options:
A) Dominant allele = 0.3; Heterozygotes = 420
B) Dominant allele = 0.91; Heterozygotes = 1638
C) Dominant allele = 0.7; Heterozygotes = 840
D) Dominant allele = 0.7; Heterozygotes = 420
4. STEP-BY-STEP SOLUTION
SOLUTION
Given:
* Total Population ($N$) = 2,000
* Number of Recessive Individuals (Genotype $aa$) = 180
To Find:
1. Frequency of the dominant allele ($p$)
2. Number of heterozygous individuals ($2pq \times N$)
Formula:
1. $Frequency = \frac{\text{Count}}{\text{Total Population}}$
2. $p + q = 1$ (Sum of allele frequencies is constant)
3. $p^2 + 2pq + q^2 = 1$ (Binomial expansion for genotype frequencies)
Use these when the population is in genetic equilibrium.
Step 1: Calculate the frequency of the recessive genotype ($q^2$)
The recessive phenotype represents the homozygous recessive genotype ($q^2$).
Step 2: Calculate the frequency of the recessive allele ($q$)
Take the square root of the genotype frequency.
Step 3: Calculate the frequency of the dominant allele ($p$)
Using the formula $p + q = 1$:
Step 4: Calculate the number of heterozygous individuals ($2pq$)
First, find the frequency of heterozygotes ($2pq$):
Now, multiply by the total population to find the number of individuals:
Answer: Option (C) — Dominant allele = 0.7; Heterozygotes = 840
5. WHERE STUDENTS GO WRONG
List 2-3 common mistakes:
* ❌ Mistake 1: Calculating $p$ directly from $q^2$ ($p = 1 - 0.09 = 0.91$) without taking the square root first. Always find $q$ first!.
* ❌ Mistake 2: Confusing "Frequency" with "Number". The frequency is 0.42, but the number of individuals is 840.
* ❌ Mistake 3: Assuming the dominant phenotype count (Total - Recessive) equals $p^2$. The dominant phenotype includes both homozygous ($p^2$) and heterozygous ($2pq$) individuals.
6. QUICK FORMULA REFERENCE
Formulas Used:
* $q = \sqrt{\frac{\text{Recessive Count}}{\text{Total}}}$ — Use to find allele frequency from phenotype data.
* $p = 1 - q$ — Use to find the dominant allele frequency.
* $2pq \times \text{Total}$ — Use to find the number of carrier individuals.
7. SIMILAR PROBLEM HINT
"If NEET changes the values, the method stays the same. Practice with: A population of 100 where 36 are recessive. (Hint: $q^2 = 0.36$, so $q = 0.6$)."
8. NCERT REFERENCE
Based on: NCERT Class 12, Chapter 7 (Evolution), Section 7.7 (Hardy-Weinberg Principle),.
9. CTA
More solved numericals → Genetics and Evolution Complete Guide
Practice problems → Mock exams