Question

Hardy-Weinberg Equilibrium Problems:

A population of beetles have either long or short antennae, where short antennae are dominant over long antennae. There are 653 individuals in the population. 104 beetles have long antennae and 549 have short antennae.

a. Determine the frequency of the dominant and recessive alleles and the frequency of individuals with dominant, heterozygous, and recessive genotypes. The next generation of beetles in this population has 560 individuals with short antennae and 840 individuals with long antennae.

b. Is the population in Hardy-Weinberg Equilibrium? Solve for p and q.

Answer 1

ANSWER:

Dominant allele is mentioned as the beetles with short antenna. Recessive allele are the ones with long antenna.

Out of 653 individuals in the population. 104 beetles have long antennae and 549 have short antennae.

The allelic frequencies are p (dominant) and q (recessive), and the total number of individuals counted is 104 + 549 = 653. So, the recessive individuals are (q²) and 104/653 = 0.1592. Therefore, q (the square root of q²) is 0.399.

• frequency of recessive allele found as 0.399 ; ie q = 0.399
• therefore, frequency of dominant allele p = 0.601 (since p+q = 1)

Formula

p² + 2 pq + q² = 1 ; where,

based on the expected genotype frequencies under random mating,

• p² = dominant homozygous frequency (AA) = 0.36
• 2 pq = heterozygous frequency (Aa) = 0.4795
• and q²  = recessive homozygous frequency (aa) = 0.159

b) When a population is in Hardy-Weinberg equilibrium, in population genetics, it states that allele and genotype frequencies in a population will remain constant from generation to generation in the absence of other evolutionary influences.

Given that,the next generation of beetles in this population has 560 individuals with short antennae and 840 individuals with long antennae.

Lets calculate the frequencies for same.

The allelic frequencies are p and q, and the total number of individuals counted is 560 + 840 = 1400. So, the recessive individuals are (q²) and 840/1400 = 0.6. Therefore, q (the square root of q²) is 0.7745. Since p + q = 1, then p must equal 1 - 0.7745 = 0.2254.

• frequency of recessive allele found as 0.7745 ; ie q = 0.7745
• therefore, frequency of dominant allele p = 0.2254 (since p+q = 1)
• p² = dominant homozygous frequency (AA) = 0.05
• 2 pq = heterozygous frequency (Aa) = 0.349
• and q²  = recessive homozygous frequency (aa) = 0.6

Since the allele and genotype frequencies are not the same for the next generation (as the previous one), therefore the population is not in Hardy Weinberg equilibrium.