In this second case you will modify the simulation to make it more realistic. In

Question

In this second case you will modify the simulation to make it more realistic. In the natural environment, not all genotypes have the same rate of survival; that is, the environment might favor some genotypes while selecting against others. An example is the human condition sickle-cell anemia. It is a condition caused by a mutation on one allele, in which a homozygous recessive does not survive to reproduce. For this simulation you will assume that the homozygous recessive (ff) individuals never survive while heterozygous (Ff) and homozygous dominant (FF) individuals always survive.

Create a second data chart similar to Chart 2. Start again with your initial genotype and produce your "offspring" as in Case 1. This time, however, there is one important difference. Every time your offspring is ff it does not reproduce. Because we want to maintain a constant population size, the same two parents must try again until they produce two surviving offspring. Repeat the procedure 49 more times. In other words, every time you pull an ff combination, do not record it. Put it back and pull again.

Before you begin, make a prediction about what you expect to observe regarding this population's allele frequency over several generations. Record your hypothesis in your lab report.

Record your results as you did in Case Study 1. Replenish your bag according to the previous draw for the remainder of the four draws.

Now that you have collected data on your own, be sure to share with your partner(s) to compile within the lab report.

Chart 2 (example data chart)

Allele Frequencies

Allele F

Allele f

Generation

Number

Percentage

Frequency

Number

Percentage

Frequency

Start

75

75

0.75

25

25

0.25

1

2

3

4

5

6 (if needed)

Allele Frequencies

Allele F

Allele f

Generation

Number

Percentage

Frequency

Number

Percentage

Frequency

Start

75

75

0.75

25

25

0.25

1

2

3

4

5

6 (if needed)

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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