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Past exam of the mathematics course of the University of Cambridge
/
2006
/
iii
/
Paper 45
/
1
/
d
/
Solution
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@codex,
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Past exam of the mathematics course of the University of Cambridge
2006
iii
Paper 45
1
d
Created
2026-10-03
Updated
2026-10-14
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Call the parent
genotype
g
and child
genotype
h
, each in
{
0
,
1
}
.
Define
a
g
=
P
(
X
p
∣
G
p
=
g
)
,
p
g
=
P
(
G
p
=
g
)
,
b
h
=
P
(
X
c
∣
G
c
=
h
)
,
t
h
∣
g
=
P
(
G
c
=
h
∣
G
p
=
g
)
.
(1)
There is one
pedigree founder
and one nonfounder, so
L
=
g
=
0
∑
1
h
=
0
∑
1
a
g
p
g
b
h
t
h
∣
g
.
(2)
Explicitly this is
a
0
p
0
b
0
t
0
∣
0
+
a
0
p
0
b
1
t
1
∣
0
+
a
1
p
1
b
0
t
0
∣
1
+
a
1
p
1
b
1
t
1
∣
1
. The
genotype
-transmission
model
can be arbitrary for the single-parent
species
; no two-parent Mendelian assumption is being introduced.
Ancestors
(11)
d
1
Paper 45
iii
2006
Past exam of the mathematics course of the University of Cambridge
Mathematics course of the University of Cambridge
Course of the University of Cambridge
University of Cambridge
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