Question

1. In a population of 1000 fruit flies, 640 have red eyes while 360 have sepia eyes. Red eyes are dominant to sepia. How many individuals would you expect to be homozygous for red eye color?

a. 160

b. 360

c. 480

d. 640

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2. In a small population, 30 frogs were AA, 60 were Aa, and 30 were aa. After several generations genotype testing revealed 20 AA, 50 Aa, and 50 aa frogs. What are the initial allelic frequencies and final frequencies for this population of frogs?

Initial: p= _________ q= _________

Final: p= _________ q= _________

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3. Does this population appear to be in equilibrium? Why or why not?

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Please help, and please answer all questions! I don't understand this stuff. Thank you so much!

Answer 1

1) Hardy-Weinberg equation is used to calculate the frequency of different genotypes

p² + 2pq + q² = 1

p² = homozygous dominant

2pq = heterozygous

q² = Homozygous recessive

Here RR denotes homozygous dominant red, SS denotes Homozygous recessive sepia and RS is heterozygous

Sepia eyes (SS) have recessive gene that is both recessive alleles should be present for the phenotype

360 individuals have sepia eyes which means the genotype frequency is 360/1000 = 0.36

That is q² = 0.36, q = 0.6

p + q = 1

so p = 0.4

p² = 0.16

p² + 2pq + q² = 0.16 + 2x0.4x0.6 + 0.36 = 1

So the p² is homozygous dominant (RR) = 0.16 x 1000 = 160

Out of 1000 individuals, 160 are homozygous dominant (RR)

2) Initial 30 AA, 60 Aa, 30 aa

q² = 30/120 = 0.25

q = 0.5

p+q = 1, so p =0.5

Initial frequency of p = 0.5 and q = 0.5

Final frequency 20AA, 50 Aa, 50 aa

q² = 50/120 = 0.41

q = 0.64

p+q =1

p = 0.36

Final frequency of p =0.36 and q =0.64

3) The population (Frog) is not in equilibrium because the initial and final frequencies for p and q are not the same