Q.105 In a population which is in Hardy-Weinberg equilibrium, the frequency of occurrence of a
disorder caused by recessive allele (q) is 1 in 1100. The frequency of heterozygotes in the
population will be ______________. (Give the answer to three decimal places).

Answer: Frequency of heterozygotes = 0.058

Under Hardy–Weinberg equilibrium, the frequency of heterozygotes (carriers) in the population is
0.058.

The recessive disorder occurs with a frequency of:

q² = 1 / 1100 ≈ 0.000909

---

Step-by-Step Solution

1. Calculate the recessive allele frequency (q)

Since affected individuals are homozygous recessive:

q = √(1 / 1100)

q ≈ 0.030151

2. Calculate the dominant allele frequency (p)

p = 1 − q = 1 − 0.030151 ≈ 0.969849

3. Calculate heterozygote frequency (2pq)

2pq = 2 × 0.969849 × 0.030151

2pq ≈ 0.058

Thus, the frequency of heterozygous carriers in the population is:

0.058

---

Hardy–Weinberg Principles

The Hardy–Weinberg equilibrium is expressed as:

p² + 2pq + q² = 1

This model assumes:

• Random mating
• No mutation
• No migration
• No natural selection
• Large population size (no genetic drift)

For recessive disorders:

• q² = frequency of affected individuals
• 2pq = frequency of heterozygous carriers

---

Introduction to Hardy–Weinberg Equilibrium and Recessive Allele Frequency

Hardy–Weinberg equilibrium calculations are essential in population genetics and
are frequently tested in CSIR NET Life Sciences.
When the frequency of a recessive disorder is known (for example, 1 in 1100),
allele frequencies and carrier proportions can be directly estimated.

---

Detailed Calculation Summary (1 in 1100 Disorder)

• q² = 1 / 1100 ≈ 0.000909
• q ≈ 0.03015
• p ≈ 0.96985
• 2pq ≈ 0.058

This value matches the expected heterozygote frequency in a CSIR NET–style question.

---

Applications in Genetics and Exam Preparation

• Used in carrier screening for recessive disorders
• Helps estimate hidden genetic load in populations
• Foundation for understanding deviations due to selection or drift

Practice Tip:
If q² = 1 / 10,000, then q = 0.01 and 2pq ≈ 0.02.

---

Key Takeaways

• Recessive phenotype frequency directly gives q²
• When q is small, 2pq ≈ 2q (since p ≈ 1)
• Always confirm that p + q = 1
• Genotype frequencies must sum to 1