----------------------------------Chapter 7 Linkage and Chromosome Mapping -----------------------------------(each task is dedicated by page number and task number in 2009 edition book, or in 2005 edition
book)
p.88/Task 1 (p.99,T1)
Map distance p
0 cM
20 cM
50 cM
genotype
Frequency
Frequency
frequency
AaBb
0.5
0.4
0.25
Aabb
0
0.1
0.25
aaBb
0
0.1
0.25
aabb
0.5
0.4
0.25
AaBb
0
0.1
0.25
Aabb
0.5
0.4
0.25
aaBb
0.5
0.4
0.25
aabb
0
0.1
0.25
p.88/Task 2 (p.99, T2)
Map distance p
0 cM
20 cM
50 cM
Genotype
Frequency
Frequency
Frequency
p.89/Task 3 (p.99,T3)
a)
phenotype
Ab
AB
aB
genotype
AAbb
AaBb
aaBB
ratio
0.25
0.5
0.25
b)
gametes
AB-0.1
Ab-0.4
aB-0.4
ab-0.1
AB- 0.1
AABB- 0.01
AABb-0.04
AaBB-0.04
AaBb-0.01
phenotype
AB
Ab
aB
ab
Ab-0.4
AABb-0.04
AAbb-0.16
AaBb-0.16
Aabb-0.04
frequency
0.51
0.24
0.24
0.01
aB-0.4
AaBB-0.04
AaBb-0.16
aaBB-0.16
aaBb-0.04
ab-0.1
AaBb-0.01
Aabb-0.04
aaBb-0.04
aabb-0.01
calculation
3*0.01+4*0.04+2*0.16
0.16+2*0.04
0.16+2*0.04
0.01
c)
gametes
AB-0.25
Ab-0.25
aB-0.25
ab-0.25
AB-0.25
AABB-0.0625
AABb-0.0625
AaBB-0.0625
AaBb-0.0625
Ab-0.25
AABb-0.0625
AAbb-0.0625
AaBb-0.0625
Aabb-0.0625
aB-0.25
AaBB-0.0625
AaBb-0.0625
aaBB-0.0625
aaBb-0,0625
ab-0.25
AaBb-0.0625
Aabb-0.0625
aaBb-0.0625
aabb-0.0625
phenotype
AB
Ab
aB
ab
frequency
0.5625
0.1875
0.1875
0.0625
calculation
9*0.0625
3*0.0625
3*0.0625
0.0625
p.90/Task 4 (p.100,T4)
Back-cross -> If the frequency 0.1 (ab phenotype) is to be represented by 10 individuals then at least
100 individuals should be tested
Inter-cross -> If the frequency 0.01 (ab phenotype) is to be represented by 10 individuals then at
least 1000 individuals should be tested.
p.90/Task 5 (p.100,T5)
a)
b)
c)
d)
Yes
Trans
Θ= 0.175
Same result
p.91/Task 6 (p.100,T6)
Chromosome 3
p.92/Task 7 (p.102, T7)
XrgY
Xrg/hX
II/4 -> 0.225
X+Y
II/5 -> 0.025
Xrg/hY
p.92/Task 8 (p.102, T8)
a) Yes
Xrg/hY
X
rg/h rg
X
XrgX+
b) I/1 -> R1el/R2el; I/2 -> R1El/rel
c) Yes -> II/7
d) 1/9=11.1 cM
93/Task 9 (p.103,T9)
Maximal value of lod for both families equals θ = 0.15
93/Task 10 (p.104,T10)
a)
b)
c)
d)
e)
f)
AD
Yes
I/1 -> 0 NPS+/0 NPS+; I/2 -> 0 NPS+/B NPSmut
II/5, II/8, II/14; III/3
4/16=0.25
Yes -> recombination present in III/3 originated in meiosis in man II/3
p.105, T11
Family I:
a) II/1 and II/5
b) 2/5
Family II:
a) III/4, III/6, IV/2, IV/3, IV/5, IV/7, IV/9
b) 7/16