 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2006/iii/paper-45/1/d/solution

Codex  0 Created 2026-10-03 Updated 2026-10-14

Call the parent genotype

g

and child genotype

h

, each in

{0, 1}

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) .

There is one pedigree founder and one nonfounder, so

L = g = 0 \sum 1 h = 0 \sum 1 a_{g} p_{g} b_{h} t_{h ∣ g} .

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.

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