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Calculate the allele frequencies in this population of palm trees. A model consisting of a box containing fourteen, randomly arranged colored circles, and a key to the right of the box. The box is labeled Time 1 and contains four red circles, six purple circles, and four blue circles. The key indicates that red represents big F big F, purple represents big F little f, and blue represents little f little f. Calculate the allele frequencies in this population of palm trees. A model consisting of a box containing fourteen, randomly arranged colored circles, and a key to the right of the box. The box is labeled Time 1 and contains four red circles, six purple circles, and four blue circles. The key indicates that red represents big F big F, purple represents big F little f, and blue represents little f little f. 4 FF, 6 Ff, 4 ff 0.3 FF, 0.4 Ff, 0.3 ff 0.5 F, 0.5 f 0.3 F, 0.7 f 0.3 F, 0.3f

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Answer:

The correct answer is

0.5 F, 0.5 f

Step-by-step explanation:

We note the following

Number of colored circles in the box = 14

Number of red circles in the box = 4

Number of purple circles in the box = 6

Number of blue circles in the box = 4

The allele frequency are as follows

Where the frequency is given as

Genotype            Frequency                         Relative frequency        

FF = Red                    4                                       4/14 (0.29≈0.3)                  

Ff = Purple                 6                                       6/14 (0.43≈0.4)

ff = Blue                     4                                       4/14(0.29≈0.3)

Within this population,

We however  have 4 FF = 8 F

                                6 Ff =   6 F + 6 f  and

                                4 ff = 8 f

Total allele = 8+6+6+8 = 28

Relative frequency of F = (8+6)/28 = 14/28 = 0.5

relative frequency of f = (8+6)/28 = 0.5

Therefore the allele frequencies in the palm tree population is

0.5 F, 0.5 f

When in equilibrium we have

However the FF has the product of F×F which is = F² = 0.29 so the frequency of F = √(0.29) = 0.535≈ 0.5

The frequency of Ff is Ff or fF =  0.43 since there is equal number of each allele in Ff we have fF or Ff = Ff = 0.43

Which hives 0.43/2 = F =f ≈ 0.2

To  

and ff = 0.29 so that f = 0.535 ≈ 0.5

Therefore f = F  = 0.5 + 0.2 = 0.7

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