Can someone answer 5-9? C) Homozygosity B. Balance Selection In a population s,
ID: 47569 • Letter: C
Question
Can someone answer 5-9? C) Homozygosity B. Balance Selection In a population s, allele causes normal appearance (as opposed to dented of sandworms appearence) on fouth body segment, and 'n' allele is recessi and'n'allele is recessive to 'N'. In a population of 300 25 sandworms, 252 had normal fourth segemnts. The frequency of A. 0.46 denl homozygous dominant sandworms for this character is INnn E, 0.6 D. 0.4 otes? Nn D. 0.36 (c) 0.36 C 0.36 B. 0.06 he above population, what is the frequency of heterozygotes? What is the frequency of homozygous recessive sandworms in the above population? nn What is the PHENOTYPIC FREQUENCY of the normal appearance in above population? 6. In In the above E. 0.16 A. 0.84 B) 0.48 C. 0.7 7 E) 0.16 D. 0.36 C. 0.04 A. 0.84 B. 0.36 0.34 8. E) 0.84 D. 0.75 A. 0.49 B. 0.16 C. 0.51 What is the frequency of dominant allele in the above population? A.) 0.6 9. E. 0.8 C. 0.48 D. 0.7 B. 0.3 A 0.6Explanation / Answer
5.
From the data,
48 are homozygous recessive.
Frequency of homozygous recessive q2 = 48/300 = 0.16
Therefore, q = root of 0.16 = 0.4
p = 1- q
= 1- 0.4 = 0.6
Therefore, the frequency of the homozygous dominant p2 = 0.6 x 0.6 = 0.36 (Option C)
6.
Frequency of heterozygotes = 2pq = 2 x 0.4 x 0.6 = 0.48 (option B).
7.
Frequency of homozygous recessive q2 = 0.4 x 0.4 = 0.16 (option E)
8.
Phenotypic frequency of normal appearance = 252/300 = 0.84 (option E)
9.
Frequency of dominant allele
p = 1- q
= 1- 0.4
= 0.6 (option A)