1. a. 1+ q = 2 - q                                         b. 1 - 2q = 2p - 1

-1 + q = -p                                                        2 - 2q = 2p

-1 = -p - q                                                         1 - q = p

1 = p + q                                                         1 = p + q

c. p - p2 = q - q2

p - q = -q2 + p2

(p - q) = (p - q)(p + q)

1 = p + q

e. 1 - p/q = 2 - 1/p

p(1 - p/q) = (2 - 1/p)p

p - p2/q = 2p - 1

q(p - p2/q) = (2p - 1)q

pq - p2 = 2pq - q

0 = p2 + pq - q

cannot be factored further

2. The choreic allele is dominant, so all individuals homozygous (CC, with frequency p^2) and heterozygous
(Cc, with frequency 2pq), will develop the disease. Since this is 4 of every 10,000 individuals, the remaining 9996 individuals must be homozygous for the recessive allele (cc, with frequency q^2).

q2 = 0.9996

q = 0.9998

1 - q = p

1 - 0.9998 = p

0.0002 = p (frequency of the choreic allele)

3.

Fallele S = 120/192 = 0.625 = p              Remember that your population is 96, not 100!

Fallele T = 72/192 = 0.375 = q

expected frequencies:

p^2 = 0.390625

2pq = 0.46875

q^2 = 0.140625

expected number of individuals (multiply expected frequencies by 96 individuals):

p^2 = 37.5

2pq = 45

q^2 = 13.5

X2 = (45 - 37.5)^2 + (30 - 45)^2 + (21 - 13.5)^2
                37.5                    45                  13.5

X2 = 1.5 + 5 + 4.167

X2 = 10.667 df = 3 - 1 - 1 = 1

X2 at 1 df and a 0.05 significance level is 4. Since this value is smaller than that of the calculated c ^2 of 10.667, we can reject the hypothesis that the population described by the 2- allele locus is in Hardy-Weinberg equilibrium.

4. There will be 3 homozygous and 3 heterozygous genotypes for this scenario:

Poo = r^2 = (0.20)^2 = 0.04

Pss = p^2 = (0.50)^2 = 0.25

Ptt = q^2 = (0.30)^2 = 0.09

Pso = 2pr = 2(0.5)(0.2) = 0.20

Pto = 2qr = 2(0.3)(0.2) = 0.12

Pts = 2qp = 2(0.3)(0.5) = 0.30

So, the expected phenotypic frequencies would be:

ss = (0.25)(100) = 25

so = (0.20)(100) = 20 total s phenotype individs: 45

tt = (0.09)(100) = 9

to = (0.12)(100) = 12 total t phenotype individs: 21

oo = (0.04)(100) = 4 total o/not detectable individs: 4

ts = (0.30)(100) = 30 total st phenotype individs: 30

In this case, you use the full population of 100 individuals, because while your assay
can't detect the oo genotype per se, you can still figure out that they are not any of the other
proteins, so they MUST be oo.

c ^2 = (45 - 45)^2 + (21 - 21)^2 + (30 - 30)^2 + (4 - 4)^2
               45                     21                  30                     4

df = 4 - 1 - 0 = 3

c^ 2 at 3 df = 0. Accept the hypothesis that with three alleles at the locus, the population is in Hardy-Weinberg equilibrium. The silent alleles serve to mask the other observable alleles, underestimating the allelic frequencies in the population. Clearly, all alleles must be considered in tests on Hardy-Weinberg equilibrium.

5. Let P11 represent the gamete L1M1

Let P12 represent the gamete L1M2

Let P21 represent the gamete L2M1

Let P22 represent the gamete L2M2

When genotypes of L and M combinations are examined, they yield gamete frequencies in the following frequencies for the next generation:

P11’ = P11 - rD

P12’ = P12 + rD

P21’ = P21 + rD

P22’ = P22 - rD

When you plug these P’ values into the equation

D’ = P11’P22’ - P12’P21’

you get:

D’ = (P11 - rD)(P22 - rD) - (P12 + rD)(P21 + rD)

D’ = [P11P22 - P22rD - P11rD + (rD)^2] - [P12P21 + P21rD + P12rD + (rD)^2]

D’ = P11P22 - P22rD - P11rD + (rD)^2 - P12P21 - P21rD - P12rD -(rD)^2

D’ = P11P22 - P22rD - P11rD - P12P21 - P21rD - P12rD

D’ = (P11P22 - P12P21) - rD(P22 + P21 + P12 + P11)

D’ = D - rD(1)