Subject: Risk Assessment - part 4

In part 3 the following equations were developed:

(a) Psire = Pa(sire) +.5 x Pc(sire), probability of getting a defective gene from the sire
(b) Pdam = Pa(dam) + .5 x Pc(dam), probability of getting a defective gene from the dam

(1) Pa(offspring) = Psire x Pdam, probability that the offspring is affected
(2) Pc(offspring) = Psire x (1- Pdam) + Pdam x (1 - Psire), probability that the offspring is a carrier

For dogs with no known ancestors (1) and (2) are replaced by:

(1a) Pa = average breed rate of affecteds
(2a) Pc = average breed rate of carriers

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Returning to the pedigree example of part 1:

                                        Old Joe (P)
                       Harry
                                        Jacqueline (C)
	(Puppy)
                                        Old Sam
                       Tabatha (U)
                                        Jacqueline(C)

For a 1% breed average rate of affecteds: Pa =.01, Pc = .18

Starting at the right end of the pedigree with Old Joe with no known ancestors, we would get

     Pa' = .01
     Pc' = .18

but Old Joe (P) has produced affected offspring, so he must be either affected or a carrier, and applying this condition

     Pa = .01/ (.01 + .18) = .0526
     Pc = .18/ (.01 + .18) = .947

Moving to Jacqueline who is a designated carrier (which makes it easy)

     Pa = 0
     Pc = 1

For Old Sam we use the Hardy-Weinberg stats

     Pa = .01
     Pc = .18

which completes the grandparents. Moving next to Harry and using equation (a) and (b)

     Psire = .0526 + .5 x .947 = .526
     Pdam = 0 + .5 x 1 = .5

now using (1) an (2)

     Pa = Psire x Pdam = .263      Pc = (1-Psire) x Pdam + Psire x (1-Pdam) = .5

Tabatha is designated not affected, so starting again using (a) and (b)

     Psire = .01 + .5 x .18 = .1
     Pdam = 0 + .5 x 1 = .5

the probability that Tabatha is clear (she got NO defective genes from her parents) is

     Pclear = (1- Psire) x (1 - Pdam) = .45

using (2) we would expect

     Pc' = .9 x .5 + .1 x .5 = .5

but Tabatha cannot be affected so this result must be conditioned,

     Pc = .5 / (Pc' + Pclear) = .5 / (.5 + .45) = .526
     Pa = 0

Finally, moving to a puppy that would be produced by breeding Harry x Tabatha. First, using (a) an (b)

     Psire = .263 + .5 x .5 = .513
     Pdam = 0 + .5 x .526 = .263

using (1) and (2)

     Pa = .135 or 13.5% risk of being affected
     Pc = (1- .513) x .263 + .513 x (1 - .263) = .506 or 50.6% risk of being a carrier

For those who followed the steps -- now you know why computers are useful! Check it out with Jim Andrew's genetic risk calculator, available free at:

http://azdogs.com/pedrisk.htm