AP BIO LAB. I have the first part, I just need the second As question says that

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AP BIO LAB. I have the first part, I just need the second

As question says that there are total 100 beans in a bad. In which 75 are light colored (FF) and 25 are dark colored (ff). If the gene pool which is produced by mating between homozygous (FF) and and heterozygous (Ff) determine then we can determine the number of dominant alleles in the population and the number of reces let we picked FF-28 Ff-19 f-3 Then we can calculate the allele frequency We can calculate dominant allele number of offspring with genotype FF (two light beans) 28 × 2-56 F alleles number of offspring with genotype Ff(one dark one light) 19 × 19 F allcles Total F alleles = 75 p- Total number of F alleles Total number of alleles in population (Total number of bcans you used) p= 75/100-0.75 We can calculate recessive allele by using the same formula number of offspring with genotype ff (two dark beans) 3-2-6 F alleles number of offspring with genotype Ff (one dark one light) 19 x 1 -25 F alleles Total f alleles = 25 q Total number of F alleles Total number of alleles in population (Total number of beans you used) q25/100-0.25 Predictions for dominant allele frequency Generation Start sive alleles in the ulation Number 75 28(FF) 19(Ff) 32(FF) 1(Ff) Frequency 075 percentage 75 28x2-56 19x1-19 32x2-64 11x1-11 34x2-68 07x1=7 30x2-60 15x1=15 27x2-54 21x1-21 75/100 0.75 75/100-0.75 75/100-0.75 6(FF) 23(Ff) 75 /1 00-0.75 75/100-0.75 31(Ff) Predictions for recessive allele frequency Generation Start umbe 25 3(ff) 19(Ff) Percentage 25 3(ff x2-6 Frequency 0.25 19x1-19 25/100-0.25 7(ff) 11(Ff) 9(ff) 7(Ff) S(ff) 15(Ff) 2(ff) 1(F0 7x2-14 11x1-11 9x2-18 7x1-7 5x2=10 15x1=15 2x2-4 21x1-21 25/100-0.25 25/100-0.25 25/100 0.25 25/100-0.25

Explanation / Answer

This case is same as in case 1 the 75 FF and 25 ff of beans. The only difference is that the human condition sickle-cell anemia which is caused by a mutation on one allele, in which a homozygous recessive does not survive to reproduce. We will assume that the homozygous recessive (ff) individuals never survive while heterozygous (Ff) and homozygous dominant (FF) individuals always survive. If every time offspring ff is skipped then it affects the allele f frequency whereas does not affects allele F frequency.

number of offspring with genotype FF = 28× 2 = 56 F alleles
number of offspring with genotype Ff = 19 × 1 = 19 F alleles
Total F alleles = 75
p = Total number of F alleles / Total number of alleles in population (Total number used)

75/100=0.75

We can calculate allele frequency of recessive allele

number of offspring with genotype ff = 0× 2 = 0 f alleles
number of offspring with genotype Ff = 19 × 1 = 19 f alleles
Total f alleles = 19
q = Total number of F alleles / Total number of alleles in population (Total number used)

q= 25/100=0.19

For allele F

28(FF)

19(Ff)

28x2=56

19x1=19

32(FF)

15(Ff)

32x2=64

15x1=15

30(FF)

15(Ff)

30x2=60

15x1=15

34(FF)

07(Ff)

34x2=68

7x1=7

27(FF)

21(Ff)

27x2=54

21x1=21

For allele f

3 (ff)

19(Ff)

0x2=0

19x1=19

7(ff)

11(Ff)

0x2=0

11x1=11

5(ff)

15(Ff)

0x2=0

15x1=15

9(ff)

7(Ff)

0x2=0

7x1=7

2(ff)

21(Ff)

0x2=0

21x1=21

As the data shows that the frequency of allele f is decreased if the recessive genotype ff is not reproduce.

Generation number percentage frequency start 75 75 0.75 1

28(FF)

19(Ff)

28x2=56

19x1=19

75/100=0.75 2

32(FF)

15(Ff)

32x2=64

15x1=15

75/100/=0.75 3

30(FF)

15(Ff)

30x2=60

15x1=15

75/100=0.75 4

34(FF)

07(Ff)

34x2=68

7x1=7

75/100=0.75 5

27(FF)

21(Ff)

27x2=54

21x1=21

75/100=0.75

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