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Answers to Practice Problems:

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

m = 0.005 P_I = 0.01 P_C = 0.25 s = 0.8

Formula = ˆ P_I = P_C 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, q_n = 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. Note that there was a mistake in the formulation of this problem initially. Check to make sure the problem you are attempting to solve is the current version.

(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