AimisT CUmUlity ald 3. Please refer to the pedigree of King Charles Il of Spain
ID: 177728 • Letter: A
Question
AimisT CUmUlity ald 3. Please refer to the pedigree of King Charles Il of Spain for this last question. The ancestry of the Habsburg royal family in Europe is an interesting case of human inbreeding. In order to keep the bloodline "pure", members of the royal family married close relatives. This led to an accumulation of many defects over time, as exemplified by the final member of this family, King Charles Il of Spain. He was deformed, sterile, mentally handicapped, exhibited the "Habsburg lip", and died at an early age. Circle each member of the family who you believe to be inbred then, choose two inbred family members within this pedigree and calculate their inbreeding coefficient, or the probability that individual has of inheriting the same allele twice from a common ancestor.Explanation / Answer
Maria of spain and Maximilian II holy roman emperor were inbred to produce Anne of Austria.
Philip II of Spain and Anne of Austria were inbred to produce Philip III of Spain.
Philip III of Spain and Margarita of Austria to produce Philip IV of Spain.
Philip IV of Spain and Manana of Austria to produce Charles II of Spain.
Charles Il of Austria and Maria Anna of Bavaria to produce Ferdinand II& Margarita of Austria.
Ferdinand II & Maria Anna of Bavaria to produce Ferdinand III.
Ferdinand III & Maria Anna of Spain to produce Mariana of Austria.
William Duke of Bavaria & Renata of Larraine to produce Maria Anna of Bavaria.
Now the inbreeding coefficient of Anne of Austria will be as follows
In Charles V chances of passing the allele is 1/2. From Maria chances are (1/2)^2=1/4
In Ferdinand I, chances are 1/2, Maximilian II chances are (1/2)^2=1/4.
Thus in Anne inbreeding coefficient is (1/4×1/2)+(1/4×1/2)=1/8+1/8=1/4
In Philip III from Anne chances are 1/4×1/2=1/8 & from his father Philip II of Spain 1/8. Thus inbreeding coefficient will be 1/8+1/8=1/4.
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