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

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

the answer is thin because within context, the stanza concludes the and frail tone the people have toward each other, and the fact that it would be thin, makes the most sense compared to "pliable" or "not hard or firm to the touch". hope this , good luck.

step-by-step explanation:

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thats not a system of equations, that is a solution