Q.58 A rare genetic disorder resulting from homozygosity for a recessive allele (r) occurs in 2
out of every 10,000 individuals in a population. Assuming that (i) the disorder is not
lethal, (ii) the disorder does not impact reproductive success, (iii) no new mutations are
introduced in the population, and (iv) the population follows Hardy–Weinberg
equilibrium, the percentage (%) of the carriers in the population that pass the r allele to
offspring is_________. (rounded off to 1 decimal)

Problem Breakdown

The disorder frequency is q² = 2/10000 = 0.0002, where q is the recessive allele (r) frequency. Thus, q = √0.0002 ≈ 0.01414, and dominant allele frequency p = 1 – q ≈ 0.98586. Carriers (heterozygotes, genotype Rr) have frequency 2pq ≈ 0.0279 or 2.79% of the population.

Key Calculation

Each carrier transmits the r allele with probability 0.5 to any gamete or offspring, regardless of mate’s genotype. Assumptions (non-lethal, no fitness impact, no mutation, equilibrium) ensure allele frequencies remain stable, so transmission stays at 50%. The question targets this fixed Mendelian ratio for carriers specifically.

Why 50.0%?

No options exist; it’s a fill-in numerical answer. Population carrier frequency informs q estimation but does not alter per-carrier transmission (always 50%). Rounding to one decimal gives 50.0. Alternative misreads (like overall population transmission) yield ~1.4%, but query specifies “carriers…that pass the r allele.”

Introduction to Hardy-Weinberg Equilibrium in Genetic Disorders

Hardy-Weinberg equilibrium recessive disorder analysis predicts genotype frequencies from allele frequencies in stable populations. For rare recessive disorders like the one with 2 out of 10,000 affected (homozygous rr), calculate carrier frequency and r allele transmission precisely. This CSIR NET-level problem tests equilibrium assumptions: no lethality, full reproductive success, no mutations, random mating.

Step-by-Step Solution: Recessive Allele Frequency

Start with affected frequency: q² = 2/10000 = 0.0002.
Take square root: q = √0.0002 = 0.01414 (r allele frequency).
Then p = 1 – q = 0.98586 (R allele frequency).
Carrier (Rr) frequency: 2pq = 2 × 0.98586 × 0.01414 ≈ 0.0279 or 2.79%.

Carrier Transmission of Recessive Allele

Heterozygous carriers (Rr) produce gametes: 50% R, 50% r. Thus, percentage passing r allele to offspring is exactly 50.0% (0.5 × 100, rounded to 1 decimal). Equilibrium ensures no selection bias; each carrier transmits independently. This holds even for rare alleles where p ≈ 1 simplifies to ~2q carriers.

Common Exam Mistakes and Clarifications

• Confusing carrier frequency (2.8%) with transmission: Question asks transmission from carriers, not carrier proportion.
• Ignoring per-carrier basis: Population-wide r transmission is ~1.4%, but carriers specifically give 50%.
• Approximation errors: Use exact √0.0002; p ≠ 1 exactly but irrelevant for transmission. No options to debate—direct numerical answer.

Applications in CSIR NET Life Sciences

Such problems integrate quantitative genetics, population biology for exams. Practice with cystic fibrosis (1/2500 affected → carriers ~5%) or sickle cell analogs. Tools like Perinatology calculator verify: input q²=0.0002 yields carriers 2pq≈2.79%.