Biology Reference
In-Depth Information
Alleles from tested man
Alleles from mother
C
D
A
A,C
A,D
B
B
,
C
B,D
If the tested man
is
the biological father then only one of the four possible com-
binations match the genotype of the child, therefore we have a probability of 0.25
that the mother and tested man would have a child with the genotype (
B,C
) (shown
in bold).
If the tested man
is not
the biological father then the mother must pass on allele
B
, which she will do with a probability of 0.5; the probability that a male other than
the tested man is the biological father is dependent upon the frequency of allele
C
(
P
c
) in the relevant population. The combined probability that this woman and a
'random man' would have a child with the genotype
B,C
is 0.5
×
P
c
.Thisgivesa
LR of:
1
2
P
c
The same process can be used for any of the possible combinations. Consider the
version where the mother is heterozygous (
A,B
), the child is heterozygous (
A,C
)and
the tested man is homozygous (
C,C
).
0
.
25
PI
=
5
P
c
=
0
.
Alleles from tested man
Alleles from mother
C
C
A
A
,
C
A
,
C
B
B,C
B,C
If the tested man
is
the biological father then there would be a 0.5 probability that
the child's genotype would be (
A,C
), as two of the four combinations are (
A,C
).
If the tested man
is not
the biological father then the mother must pass on allele
A
, which she will do with a probability of 0.5; the probability that a male other than
the tested man is the biological father is dependent upon the frequency of allele
C
(
P
c
) in the relevant population. The combined probability that this woman and a
'random man' would have a child with the genotype
A,C
is 0.5
×
P
c
.Thisgivesa
LR of:
1
P
c
If we consider a final case when the mother is (
A,B
), the child is (
A,B
) and the tested
man is (
A,C
).
5
0
.
5
P
c
=
0
.
PI
=
Alleles from tested man
Alleles from mother
A
C
A
A,A
A,C
B
A
,
B
B,C