Answers to Practice Problems:

 

1. Given the information, these variables have the following values:

 

            m = 0.005       PI = 0.01         PC = 0.25        s = 0.8

 

            Formula = ˆ PI = PC x [m ÷ (s – m)]

 

                        = 0.25 [0.005/(0.8 – 0.005)]

 

                        = 0.0016

 

2. Frequency of the dullard allele = square root of 40/1000 = 0.2

 

            Fitness of the dullard allele = 0.8, so s = 0.2

 

            µ = 0.005

 

            ˆq = square root of 0.005/0.2

 

            = 0.158

 

3. Nm = 30; Nf = 100

 

            Ne = 4(30) x 100/30 + 100

 

                        = 92.3

 

4.         BB = 25 – 1/26  = 0.48           

 

Bb = 2/26  = 0.03       

 

bb = 25 – 1/26 = 0.48

 

 

5. (a) R = h2(S)

 

            = 0.75(50-47)

 

            = 2.25 cm, so average horn size in next generation = 47 + 2.25 = 49.25 cm

 

    (b) h2 = R/S

 

            R = 48.2 – 47 = 1.2 cm; S = 3.0

 

            h2 = 1.2/3 = 0.40

 

6. (a) Frequency of stone allele = square root of 9/100 = 0.3; this is q0, qn = 0.01

 

            n = (1/0.01) – (1/0.3) = 96.7 generations

 

(b) if w = 0.6, then s = 0.4; from above, q = 0.3, so p = 0.7; thus

 

            ∆q = -spq2/(1-sq2)

 

                        = -0.4(0.7)(0.09)/[1 – (0.4)(0.09)]

 

                        = -0.026

 

7. (a) Frequency of b = square root of 60/1000 = 0.245; thus B = 0.755

 

            Using Hardy-Weinberg, Frequency of each genotype is:

 

            BB = p2 or (.775)(.775) = 0.57

 

            Bb = 2pq or 2(.775)(.245) = 0.37

 

            Bb = q2 or (.245)(.245) = 0.06

 

            Determination of H-W equilibrium and answer to part (b):

 

           

 

BB

Bb

bb

Observed

650

275

75

Expected

570

370

60

Ratio (Obs./Exp)

1.14

0.743

1.25

Fitness (w)

0.91

0.594

1.0

Selection (s)

0.09

0.406

0.0

 

Deviation of observed from expected indicates population is not in H-W equilibrium.  Expected numbers obtained by multiplying total population size (1000) by each genotype frequency.

 

(c) Answer is the same as in question 4.

 

8. (a) Frequency of F = [2(130) + 410]/2000 = 0.335; f = [2(460) + 410]/2000 = 0.665

 

Using Hardy-Weinberg, Frequency of each genotype is:

 

            FF = p2 or (.335)(.335) = 0.112

 

            Ff = 2pq or 2(.665)(.335) = 0.446

 

            ff = q2 or (.665)(.665) = 0.442

 

            Determination of H-W equilibrium and answer to part (b):

 

           

 

BB

Bb

bb

Observed

130

410

460

Expected

112

446

442

Ratio (Obs./Exp)

1.16

0.92

1.04

Fitness (w)

1.0

0.79

0.90

Selection (s)

0.0

0.21

0.10

 

As in question 7, expected numbers obtained by multiplying genotype frequencies by population size (1000).  Population is close to H-W equilibrium.

 

c. s = 0.10; just as in question 6(b), use formula: ∆q = -spq2/(1-sq2)

 

            ∆q = -0.10(0.335)(0.665)2/1 – (0.10)(0.665)2

 

            ∆q = -0.015

 

            q1 = 0.665 – 0.015 = 0.65