Compute probabilities of alleles being shared between two individuals PROVIDE AL

Question

Compute probabilities of alleles being shared between two individuals PROVIDE ALL ANSWERS ON THIS PAGE TO THE NEAREST 0.0001 5. RELATEDNESS. Depicted to the right is a pedigree (family tree) just like the one in class, representing males (squares) and females (circles) that mate to produce offspring Males : :0 Females : Compute the Probabilities of alleles being shared between two individuals due to the ancestry depicted and write these in the boxes in the table below. (1 pt each) The probability you will calculate is the probability of an allele in the individual listed in the leftmost column being in the individual listed in the 2nd column. For the 3ed column (first column of boxes) consider allele on an autosome. 58 For the 4 column (second column of boxes) consider allele also in if allele on if allele on X if allele on Y alleles on the X in indiv. indiv. autosome chromosome chromosome For the 5th column (second column of boxes) consider alleles on the Y 0.3125 -= 0.2500 = 16 32 0.1563 = 0.3125 0.2500 -= 0, 5000 :0, 1563 = 0.5000 2

Explanation / Answer

This is a question of conditional probablity whereinwe need to find the chances of presences of a particular alleles in an individual when it is know that the other individual carries it.

Now let us take case of B and C

Clearly C is son of B (male).

Now three cases exist for B when it is known that a particular alleles is present in C

alleles on autosome : clearly this is possible in C, if it comes to him eaither from his mom or dad. Hence this probablity is 1/2.

Allelese on X chromosomes

If C has alleles on X chromosomes then it must have come from mom since females carry 2 X chromosomes.

Hence chances of it coming from dad is 0.

Alleles from Y chromosomes

This comes in C from B since females dont carry Y chromosomes.. hence tihs probablity is 1.

The similar situations occur when we look at its ocnverse. that is calculating the prbablity of particular alleles in C.given B. This justifies the prabablity calculations as shown in table.

Probablity for A given C

1. when it comes from X or Y chromosomes

Clearly this will come into a child only from his parents .If we look at A , it can come from his mother /father. His mom is wife of B and so the X/Y chromosome Alleles in A cant come in him from B as A is not the offspring of B.

Hence probablity in each case is 0.

2. when it comes from autosome (autosomes are non sex determining chromosomes)

So when we need to find the realtinoship prbablity then we need to find the transverse of the prbablity of autosomes.

It is known that C has autosomes. Then if this is true then we need to find the probablity of A having the alleles .

Now C got it either from his mom or dad (B) who in turn get it from his parents who in turn get it from thier parents.

Focus on A

Where does he gets his gene pools form?

His parents who get it from thier parents who get it from thier parents.(3 generations)

So lets from 1st generation.

The prbablity of autosome alleles =1/2 ( may be ,may be ont)

Probablity of non autosome alleles =(1-1/2) =1/2

Probablity of autosome alleles transfering to kid (let say P2) =1/2*1/2=1/4 (= probablity from mother*probablity from dad)

This is the probalbity of autosome .Thus we need to ifnd probalbity of sexual non auotsome alleges =1-P2

=1-1/2*1/2 =3/4 =Probalbity in 2nd generation father

Probalbity of non autosome in non realted mother =1-1/2 =1/2

Thus probalbity in 3rd generation kid P3=p(mother).p(father) =1/2*3/4 =3/8

Probablity of non autosome =1-P3 =1-3/8 =5/8 =Probablity in father of A

Now C has it. This means it might have come in him from B (his father )

Thus mother's probalbity =1/2*1/2=1/4

Thus P(A) =P(father)*p(mother) =5/8*1/4 =5/32 .

Converse B given A will also be the same.

A given B

Clearly for Y choromsome alleles it has to be zero since it comes ot A from his father only.

When X chromosomes alleles, then can come from his father or mother hence 1/2*1/2 =1/4

When it is autosome

Previously we have ofund that A's father has on autosome alleles probalbity =5/8.

A's mother probablity of carry alleles will be a simple usual 1/2 since it is known that B is carrying it .Hence B's wife (who is A's omther) doesnt have it.

Thus, Probalbity (A) =P(father)*P(mother)=5/8*1/2 =5/16

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