Q.91 In a population of 1000 wild dogs in a grassland, 360 and 480 dogs had black
body colour with genotypes BB and Bb, respectively. In the same population,
remaining dogs were white in colour with a genotype of bb. Based on this data,
the frequency of allele “b” in the population is ______ (round off to one decimal
place).
The frequency of allele “b” in this wild dog population is 0.4. This calculation uses the Hardy-Weinberg principle to count recessive alleles from genotype data. Total “b” alleles total 800 out of 2000 alleles in the population of 1000 dogs.
Step-by-Step Calculation
First, identify genotypes: 360 BB (black), 480 Bb (black), and 160 bb (white, as 1000 – 360 – 480 = 160).
Total alleles equal 1000 dogs × 2 = 2000.
Count “b” alleles:
- From bb: 160 × 2 = 320
- From Bb: 480 × 1 = 480
- From BB: 360 × 0 = 0
Total “b” = 320 + 480 + 0 = 800. Frequency = 800 / 2000 = 0.4 (rounded to one decimal).
Hardy-Weinberg Context
In Hardy-Weinberg equilibrium, allele frequency q for recessive “b” derives directly from genotype counts without assuming equilibrium, as shown. Verify: q² = 160/1000 = 0.16, so q = √0.16 = 0.4, matching the count.
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Introduction: Mastering Allele Frequency in Wild Dogs Population
Determining the frequency of allele b wild dogs population is a core skill for CSIR NET Life Sciences aspirants tackling Hardy-Weinberg problems. This analysis of 1000 grassland wild dogs—360 BB (black), 480 Bb (black), 160 bb (white)—reveals precise genetic insights.
Detailed Solution Breakdown
Population totals 1000 dogs with 2000 alleles. Recessive “b” alleles come from bb (320) and half of Bb (480), totaling 800 “b” alleles. Thus, frequency of allele b = 800/2000 = 0.4.
Alternative: q = √(bb/N) = √0.16 = 0.4.
Why No Options to Explain?
Query lacks multiple-choice options, so focus remains on direct computation. Common pitfalls include forgetting heterozygote contribution or total alleles—always double-count “b” sources.
Applications in Population Genetics
Such calculations track evolution, breeding patterns in wild dogs, and disease alleles. For CSIR NET, practice verifies equilibrium via p + q = 1 (here, p=0.6).
