1.1
a) 13.58
b) 160.6
c) 47.1
1.2
a) 0.43
b) 2.88
c) 288.8
1.3
a) r=0.898
b) r=0.974
c) r=1.0
1.4
a) slope = -1.016, y-intercept = 30.38
b) slope = 0.101, y-intercept = 19.99
c) slope = 0.984, y-intercept = -96.60
1.5
a) 5.748 < 7.815 YES fits distribution
b) 0.505 < 3.841 YES equal number
1.6
a) |-0.121| < 2.056 NO significant difference
b) |-2.115| > 2.080 YES significant increase
2.1 h2 = 0.75
2.2 h2 = 0.67
2.3 h2 = 0.01
2.4 h2 = 0.85
2.5 h2 = 0.90
2.6 h2 = 0.51
3.1
a) 0.44 Directional
b) 0.40 Disruptive
c) 0.45 Stabilizing
3.2
a) 0.81 Disruptive
b) 0.65 Directional
c) 0.52 Stabilizing
4.1
f(A)=0.4525
f(a)=0.5475
Expected Counts:
AA |
Aa |
aa |
204.75624 |
495.48750 |
299.75625 |
0.001 < 3.841 Therefore the population IS in HW Equilibrium at this locus
4.2
f(A)=0.46052
f(a)=0.53947
Expected Counts:
AA |
Aa |
aa |
48.35526 |
113.28947 |
66.35526 |
4.152 > 3.841 Therefore the population is NOT in HW Equilibrium at this locus
4.3
f(A)=0.87500
f(a)=0.12500
Expected Counts:
AA |
Aa |
aa |
76.56250 |
21.87500 |
1.56250 |
9.878 > 3.841 Therefore the population is NOT in HW Equilibrium at this locus
4.4
f(A)=0.08333
f(a)=0.91666
Expected Counts:
AA |
Aa |
aa |
1.79166 |
39.41666 |
216.79166 |
3.239 < 3.841 Therefore the population IS in HW Equilibrium at this locus
5.1
a)
f(AA)=0.510 f(Aa)=0.244 f(aa)=0.244
f(BB)=0.183 f(Aa)=0.408 f(aa)=0.408
Expected Counts:
|
AA |
Aa |
aa |
BB |
22.959 |
11.020 |
11.020 |
Bb |
51.020 |
24.489 |
24.489 |
bb |
51.020 |
24.489 |
24.489 |
4.537 < 9.488 Therefore in linkage equilibrium
b)
f(AA)=0.595 f(Aa)=0.354 f(aa)=0.049
f(BB)=0.248 f(Aa)=0.503 f(aa)=0.248
Expected Counts:
|
AA |
Aa |
aa |
BB |
59.553 |
35.438 |
4.962 |
Bb |
120.893 |
72.032 |
10.07 |
bb |
59.553 |
35.483 |
4.962 |
0.467 < 9.488 Therefore in linkage equilibrium
c)
f(AA)=0.227 f(Aa)=0.485 f(aa)=0.287
f(BB)=0.237 f(Aa)=0.514 f(aa)=0.247
Expected Counts:
|
AA |
Aa |
aa |
BB |
5.465 |
11.643 |
6.891 |
Bb |
11.841 |
25.227 |
14.930 |
bb |
5.693 |
12.128 |
7.178 |
0.245 < 9.488 Therefore in linkage equilibrium
d)
f(AA)=0.447 f(Aa)=0.219 f(aa)=0.333
f(BB)=0.238 f(Aa)=0.476 f(aa)=0.285
Expected Counts:
|
AA |
Aa |
aa |
BB |
11.190 |
5.476 |
8.333 |
Bb |
22.380 |
10.952 |
16.666 |
bb |
13.428 |
6.571 |
10.000 |
50.162 > 9.488 Therefore in linkage disequilibrium
5.2
a)D=-0.04
b)D=-0.11
c)D=+0.00
d)D=+0.07