Evaluation of DNA mixtures involving two pairs of relatives

Abstract

This paper considers the statistical evaluation of DNA mixtures in the following situations: (1) two unknown contributors are related respectively to two typed persons, (2) two of the unknown or untyped contributors are related and the third unknown contributor is related to a typed person, or (3) there are two pairs of related unknown contributors to the DNA mixture. The corresponding formulas for evaluating the likelihood ratios on the strength of DNA evidence are derived and the kinship coefficients for the related persons are incorporated into the calculations. Two examples are analyzed for illustration.

Appendix

Lemma 1

If the kinship coefficients of two individuals X₁ and X₂ are (k₀,2k₁,k₂), then for any set of alleles D

$$P(\chi _{1} \subset D,\chi _{2} \subset D) = k_{0} {\mathop {{\left( {{\sum\limits_{i \in D} }\,p_{i} } \right)}}\nolimits^4 } + 2k_{1} {\mathop {{\left( {{\sum\limits_{i \in D} }\,p_{i} } \right)}}\nolimits^3 } + k_{2} {\mathop {{\left( {{\sum\limits_{i \in D} }\,p_{i} } \right)}}\nolimits^2 },$$

(10)

$$P(\chi _{1} \subset D|X_{2} = x_{{21}} x_{{22}} ) = k_{0} {\mathop {{\left( {{\sum\limits_{i \in D} }\,p_{i} } \right)}}\nolimits^2 } + k_{1} (I_{D} (x_{{21}} ) + I_{D} (x_{{22}} )){\sum\limits_{i \in D} }\,p_{i} + k_{2} I_{D} (x_{{21}} )I_{D} (x_{{22}} ),$$

(11)

where χ₁ and χ₂ are the genetic profiles of X₁ and X₂ respectively.

Proof

They are just Eqs. 4 and 5 in Hu and Fung (2003).

Lemma 2

For any U⊂M, and any pairwise distinct alleles a,b,c,d,∈U,

$$Q(n,U \setminus \{ a\} ) = {\sum\limits_{M \setminus U \subset D \subset M} }\,I_{D} (a)( - 1)^{{|M \setminus D|}} {\mathop {{\left( {{\sum\limits_{i \in D} }\,p_{i} } \right)}}\nolimits^n },$$

(12)

$$Q(n,U \setminus \{ a,b\} ) = {\sum\limits_{M \setminus U \subset D \subset M} }\,I_{D} (a)I_{D} (b)( - 1)^{{|M \setminus D|}} {\mathop {{\left( {{\sum\limits_{i \in D} }\,p_{i} } \right)}}\nolimits^n },$$

(13)

$$Q(n,U \setminus \{ a,b,c\} ) = {\sum\limits_{M \setminus U \subset D \subset M} }\,I_{D} (a)I_{D} (b)I_{D} (c)( - 1)^{{|M \setminus D|}} {\mathop {{\left( {{\sum\limits_{i \in D} }\,p_{i} } \right)}}\nolimits^n },$$

(14)

$$Q(n,U \setminus \{ a,b,c,d\} ) = {\sum\limits_{M \setminus U \subset D \subset M} }\,I_{D} (a)I_{D} (b)I_{D} (c)I_{D} (d)( - 1)^{{|M \setminus D|}} {\mathop {{\left( {{\sum\limits_{i \in D} }\,p_{i} } \right)}}\nolimits^n }.$$

(15)

Proof

It is sufficient to note that $I_{D} (a) = 1 \Leftrightarrow a \in D \Leftrightarrow (M \setminus U) \cup \{ a\} \subset D \subset M \Leftrightarrow M \setminus (U \setminus \{ a\} ) \subset D \subset M$, so

$${\sum\limits_{M \setminus U \subset D \subset M} }\,I_{D} (a)( - 1)^{{|M \setminus D|}} {\mathop {{\left( {{\sum\limits_{i \in D} }\,p_{i} } \right)}}\nolimits^n } = {\sum\limits_{M \setminus (U \setminus \{ a\} ) \subset D \subset M} }\,( - 1)^{{|M \setminus D|}} {\mathop {{\left( {{\sum\limits_{i \in D} }\,p_{i} } \right)}}\nolimits^n }.$$

Thus Eq. 12 follows immediately from Eq. 3.

Similarly, $I_{D} (a)I_{D} (b) = 1 \Leftrightarrow a,b \in D \Leftrightarrow (M \setminus U) \cup \{ a,b\} \subset D \subset M \Leftrightarrow M \setminus (U \setminus \{ a,b\} ) \subset D \subset M$. So Eq. 13 holds by Eq. 3.$I_{D} (a)I_{D} (b)I_{D} (c) = 1 \Leftrightarrow a,b,c \in D \Leftrightarrow (M \setminus U) \cup \{ a,b,c\} \subset D \subset M \Leftrightarrow M \setminus (U \setminus \{ a,b,c\} ) \subset D \subset M$, so Eq. 14 holds by Eq. 3.$I_{D} (a)I_{D} (b)I_{D} (c)I_{D} (d) = 1 \Leftrightarrow a,b,c,d \in D \Leftrightarrow (M \setminus U) \cup \{ a,b,c,d\} \subset D \subset M \Leftrightarrow M \setminus (U \setminus \{ a,b,c,d\} ) \subset D \subset M$, so Eq. 15 holds by Eq. 3.

Lemma 3

For any U⊂M, and any alleles a,b,c,d, we have

$${\sum\limits_{M \setminus U \subset D \subset M} }\,I_{D} (a)( - 1)^{{|M \setminus D|}} {\mathop {{\left( {{\sum\limits_{i \in D} }\,p_{i} } \right)}}\nolimits^n } = I_{M} (a)Q(n,U \setminus \{ a\} ),$$

(16)

$${\sum\limits_{M \setminus U \subset D \subset M} }\,I_{D} (a)I_{D} (b)( - 1)^{{|M \setminus D|}} {\mathop {{\left( {{\sum\limits_{i \in D} }\,p_{i} } \right)}}\nolimits^n } = I_{M} (a)I_{M} (b)Q(n,U \setminus \{ a\} \cup \{ b\} )$$

(17)

$${\sum\limits_{M \setminus U \subset D \subset M} }\,I_{D} (a)I_{D} (b)I_{D} (c)( - 1)^{{|M \setminus D|}} {\mathop {{\left( {{\sum\limits_{i \in D} }\,p_{i} } \right)}}\nolimits^n } = I_{M} (a)I_{M} (b)I_{M} (c)Q(n,U \setminus \{ a\} \cup \{ b\} \cup \{ c\} )$$

(18)

$$\begin{aligned} & {\sum\limits_{M \setminus U \subset D \subset M} }\,I_{D} (a)I_{D} (b)I_{D} (c)I_{D} (d)( - 1)^{{|M \setminus D|}} {\mathop {{\left( {{\sum\limits_{i \in D} }\,p_{i} } \right)}}\nolimits^n } \\ & = I_{M} (a)I_{M} (b)I_{M} (c)I_{M} (d)Q(n,U \setminus \{ a\} \cup \{ b\} \cup \{ c\} \cup \{ d\} ). \\ \end{aligned}$$

(19)

Proof

If a∉M, then both sides of Eq. 16 are zero; if a∈M∖U, then Eq. 16 is just Eq. 3; if a∈U, then Eq. 16 is just Eq. 12. Thus, Eq. 16 holds.

If a=b, then Eq. 17 is just Eq. 16. Next, we consider a≠b. If a or b is not in M, then both sides of Eq. 17 are zero; if a,b, ∈U, then Eq. 17 is just Eq. 13; if a∈U and b∈M∖U, then Eq. 17 is just Eq. 16; if a,b∈M∖U, then Eq. 17 is just Eq. 3. Thus, we have Eq. 17 by the symmetry of a and b.

If two of a,b,c, are identical, then Eq. 18 is just Eq. 17. So we consider that a,b and c are pairwise distinct. If one of a,b, and c is not in M, then both sides of Eq. 18 are zero; if a,b,c∈U, then Eq. 18 is just Eq. 14; if a,b,∈U and c∈M∖U, then Eq. 18 is just Eq. 17; if a ∈U and b,c∈M∖U, then Eq. 18 is just Eq. 16; if a,b,c ∈M∖U, then Eq. 18 is just Eq. 3. By the symmetry of a,b and c, Eq. 18 is hence proved.

If two of a,b,c and d are identical, then Eq. 19 is just Eq. 18. So we assume a,b,c and d are pairwise distinct. If one of a,b,c and d is not in M, then both sides of Eq. 19 are zero; if a,b,c,d∈U, then Eq. 19 is just Eq. 15; if a,b,c,∈U and d∈M∖U, then Eq. 19 is just Eq. 18; if a,b,∈U and c,d∈M∖U, then Eq. 19 is just Eq. 17; if a∈U and b,c,d∈M∖U, then Eq. 19 is just Eq. 16; if a,b,c,d∈M∖U, then Eq. 19 is just Eq. 3. Thus we prove Eq. 19 by the symmetry of a, b, c and d.

Proof of Eq. 5

Under the hypothesis declared in Eq. 4, the only related individuals are X₁ and T₁, and X₂ and T₂. Let χ₁, χ₂, and χ₀ be the genetic profiles of X₁, X₂, and the other x−2 unknown contributors, respectively, then we have χ=χ₁⋃χ₂⋃χ₀ and further by Eq. 2

$$W(D) = P({\mathop \chi \nolimits_1 } \subset D,{\mathop \chi \nolimits_2 } \subset D,{\mathop \chi \nolimits_0 } \subset D|K) = P({\mathop \chi \nolimits_1 } \subset D|T_{1} )P({\mathop \chi \nolimits_2 } \subset D|T_{2} )P({\mathop \chi \nolimits_0 } \subset D).$$

(20)

Using Eq. 11 for the first two items in Eq. 20, $P({\mathop \chi \nolimits_0 } \subset D) = ({\sum\nolimits_{i \in D} }\,p_{i} )^{{2(x - 2)}}$, and the results of Lemma 3, we can have Eq. 5 from Eq. 1 after some matrix manipulation.

Proof of Eq. 7

Under hypothesis declared in Eq. 6, the only related individuals are X₁ and T₁, and X₂ and X₃. Let χ₁, χ₂, χ₃, and χ₀ be the genetic profiles of X₁, X₂, X₃, and the other x−3 unknown contributors, respectively, then we have χ=χ₁⋃χ₂⋃χ₃⋃χ₀ and further by Eq. 2

$$W(D) = P({\mathop \chi \nolimits_1 } \subset D|T_{1} )P({\mathop \chi \nolimits_2 } \subset D,{\mathop \chi \nolimits_3 } \subset D)P({\mathop \chi \nolimits_0 } \subset D),$$

(21)

Substituting Eqs. 10, 11, and $P({\mathop \chi \nolimits_0 } \subset D) = ({\sum\nolimits_{i \in D} }\,p_{i} )^{{2(x - 3)}}$ into the expression of W(D) in Eq. 21 and then using Lemma 3, we can have Eq. 7 from Eq. 1 after some matrix manipulation.

Proof of Eq. 9

Under the hypothesis declared in Eq. 8, the only related individuals are X₁ and X₂, and X₃ and X₄. Let χ₁, χ₂, χ₃, χ₄, and χ₀ be the genetic profiles of X₁, X₂, X₃, X₄, and the other x−4 unknown contributors, respectively, then we have χ=χ₁⋃χ₂⋃χ₃⋃χ₄⋃χ₀ and further by Eq. 2

$$W(D) = P({\mathop \chi \nolimits_1 } \subset D,{\mathop \chi \nolimits_2 } \subset D)P({\mathop \chi \nolimits_3 } \subset D,{\mathop \chi \nolimits_4 } \subset D)P({\mathop \chi \nolimits_0 } \subset D),$$

(22)

Substituting Eq. 10 and $P({\mathop \chi \nolimits_0 } \subset D) = ({\sum\nolimits_{i \in D} }\,p_{i} )^{{2(x - 4)}}$ into Eq. 22 and then using Lemma 3, we have Eq. 9 from Eq. 1 after some matrix manipulation.