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
Search WWH ::




Custom Search