Agbortative mating E. Genetic drift A disease-ca disease-causing recessive allel
ID: 47566 • Letter: A
Question
Agbortative mating E. Genetic drift A disease-ca disease-causing recessive allele occurs at a frequency o equilibrium genetic n would be carriers (but not sick) for this allele? E. 0.21 equilibrium, what fraction of the population would be ca f 0.2 in a population. At A. 0.24 4 B 0.32 C. 0.42 D. 0.48 Positive assortative mating is likely to cause more more of (a) gene(s) in a population. A. Mutations E. Heterozygosity D. Balance B. Selection C.Homozygosity S. In a populat a population of sandworms, 'N allele causes normal appearaIn appearence) sandworms, 252 had normal fourth segemnts. of sandworms, "N' allele causes normal appearance (as opposed to dented population on fouth body segment, and .n' allele is recessive to-N. In a population of and'nallele is recessive to N. In a population of 300 segemnts. Na nor ns dent The freq uency of homoz ncy of homozygous dominant sandworms for this character is NNn ygous domina 0.6 D. 0.4 A. 0.46 B. 0.06 0.36 n 6. In t he above population, what is the frequency of heterozygotes? E. 0.16 D. 0.36 B) 0.48 C. 0.7 What is the frequency of homozygous recessive sandworms in the above population? nn A. 0.84 A. 0.84 7. E.) 0.16 7. D. 0.36 B. 0.36 C. 0.04 What is the PHENOTYPIC FREQUENCY of the normal appearance in above population? A. 0.49 8. 0.84 C. 0.51 D. 0.75 B. 0.16 9. What is the frequency of dominant allele in the above population? E. 0.8 D. 0.7 C. 0.48 B. 0.3 A) 0.6Explanation / Answer
3.
Hardy and Weinberg mathematically proved that in a population, all dominant and recessive alleles comprise all alleles for that gene.
This was mathematically represented as p+ q = 1.0
Where,
p = frequency of dominant alleles
q = frequency of recessive alleles.
Given that q = 0.2; hence p = 0.8
Hardy-Weinberg equation p2 + 2pq + q2 = 1.0
Where,
p2 = proportion of homozygous dominant individuals
q2 = proportion of homozygous recessive individuals
2pq = proportion of heterozygotes.
The frequency of carriers = 2pq = 2 * 0.2 * 0.8 = 0.32.
Thus, the correct option is B. 0.32.