Bayes’ Theorem - az9194.vo.msecnd.netaz9194.vo.msecnd.net/pdfs/130401/12.pdf · Disclosure(s) In...

The Association for Molecular Pathology Education. Innovation and Improved Patient Care. Advocacy.

www.amp.org

Bayes’ Theorem

Sherri J. Bale, PhD, FACMG Managing Director, GeneDx, Inc.

April 2013

Disclosure(s)

In accordance with ACCME guidelines, any individual in a position to influence and/or control the content of this ASCP CME activity has disclosed all relevant financial relationships within the past 12 months with commercial interests that provide products and/or services related to the content of this CME activity.

The individual below has responded that he/she has no relevant financial relationship(s) with commercial interest(s) to disclose:

Sherri J. Bale, PhD, FACMG

2

Bayes Theorem

Pr {Hypothesis | Data} =

P( | ) ● P ( )

=

P( | ) ● P ( ) + P( | ) ● P ( )

Introduces subjectivity

Pr { H|D }

D H H

D H H D H H

Example: An X-linked recessive trait Emily asks,

“What is my chance of having an affected son?

Emily

Doris

= carrier

(a) Without evoking Bayes theorem:

Chance Emily is a carrier:

Chance a son will be affected:



Emily

Doris

= carrier

(a) Without evoking Bayes theorem:

Chance Emily is a carrier:

Chance a son will be affected:

XX 1 2

1 2

1 2

1 2

1 4

Chance she’s a carrier

Chance she has a son

1 4

1 2

1 8



Emily

Doris

(b) Evoking Bayes theorem:

Define 2 Hypotheses:

H: ________ is a carrier

Alt H: ________ is NOT carrier

Doris Doris

State of Nature Table

H _________________

Alt H ____________________

Prior Probability

P(H) =

P(Alt H) =

Conditional Probability

P(D|H) =

P(D|Alt H) =

Joint Probability (Prior x Conditional)

P(D|H) ● P(H) =

P(D|Alt H) ● P(Alt H) =

Posterior Probability

A B

A B

B

A B

A

State of Nature Table H

_________________ Alt H

____________________

Prior Probability

P(H) =

P(Alt H) =

Conditional Probability [2 unaffected sons]

P(D|H) =

P(D|Alt H) =


P(D|H) ● P(H) =



A B

A B

B

A B

A

Doris is a Carrier Doris is NOT a Carrier

= 1

1 2

1 2 x 1 1 x

1 2

1 2

1 2

x x

1 8

= 3

1 2 ( )

1 2

4 8

1 2

1 2

x x 1 1

1 2 =

4 8

4 8

1 8

1 8

1 8

1 5

4 5 = 1

Now we can answer:

(1) What is the chance Emily is a carrier?

1 2

Pr (Doris is a carrier)

x 1 2

=

1 5 x

1 2

= 1 10

(2) What is the chance Emily’s son will be affected? Pr (Emily is a carrier)

x ? =

1 10

x 1 2 =

1 20

with Bayes

without Bayes 1 8



Emily

Doris

Without Baye’s

1 5 With Baye’s

1 8

1 20

Working Some Problems I-1 was affected with Hemophilia (F VIII def). What is

the probability that the pregnancy (IV-4) will be an affected boy?

I

II

III

IV

1 2

1 2

1 2

1 2 3 4

P


H _________________

Alt H ____________________

Prior Probability

P(H) =

P(Alt H) =


P(D|H) =

P(D|Alt H) =


P(D|H) ● P(H) =



A B

A B

B

A B

A

I

II

III

IV

1 2

1 2

1 2

1 2 3 4

P

III-1 is a Carrier III-1 is NOT a Carrier

1 2

1 2

[3 unaffected boys]

3 1 2 ( ) = 1

8 1 x x 1 1

1 2 x

1 8

= 1 16

1 2 x 1

= 1 2

4 1

2 ( ) 4

1

2 ( ) 1 2

1/16

1/16 + 8/16 = 1 9

8 9

= 1

The QUESTION: What is the probability that IV-4 is an affected boy?

I

II

III

IV

1 2

1 2

1 2

1 2 3 4

P

= Prob III-1 is a carrier

Given III-1 is a carrier, there are 4 possible pregnancy outcomes:

Prob: 1 4

1 4

1 4

1 4

1 9

1 9

1 4

x = 1 36

What if we already knew she was pregnant with a SON?

Prob: 1 2

1 2

X

1 9

1 2

x = 1 18

WARNING: 1. Read the Question. 2. Read the Question again!

A woman has a brother and maternal uncle who died with severe hydrocephalus due to x-linked aqueductal stenosis. She has 2 healthy sons and is pregnant again. The probability that her fetus is severely affected with hydrocephalus is?

P

w


H _________________

Alt H ____________________

Prior Probability

P(H) =

P(Alt H) =


P(D|H) =

P(D|Alt H) =


P(D|H) ● P(H) =



A B

A B

B

A B

A

W is a Carrier W is NOT a Carrier

1 2

1 2

[2 unaffected sons] =

1 4

1 x = 1 1

1 2 x

1 4

= 1 8

1 2 x 1

= 1 2

1

8

1

8 4 8

= 1 5

4 5

= 1

P

w

1 2

x 1 2

= 4 8

We are answering: What is the probability that her fetus is affected with hydrocephalus?

= Prob W is a carrier

If W is a carrier, 4 outcomes of pregnancy:

Prob: 1 4

1 4

1 4

1 4

1 5

1 5

1 4

x = 1 20

Mutation in ______ gene causes X-linked hydrocephalus?

4 5

1 5

to

4 : 1

Odds in favor of being a non-carrier to a carrier?

LICAM

A healthy 60 y.o. male presents with history of father with Huntington Disease (HD). What is the probability this man has inherited his father’s HD gene?

Data given: The probability being affected with HD by age 60, if one has the gene, is 75%.

Charlie age 60

45 60 75

50

100

Pr. affected


_________________ Alt H

____________________

Prior Probability

P(H) =

P(Alt H) =


P(D|H) =

P(D|Alt H) =


P(D|H) ● P(H) =



A B

A B

B

A B

A

Charlie inherited HD gene Charlie missed the bullet

= 1

1 4 1

1 4

1 2

x

1 8 =

1 2

4 8

1 2

1 2

x 1

1 2 =

4 8

4 8

1 8

1 8

1 8

1 5

4 5 = 1

[unaffected at age 60]

Charlie has a daughter, age 45, who is not affected. How does this added info change his probability of having inherited the HD gene?

Given: The probability being affected with HD by age 45 is 50%.

Charlie age 60

45 60 75

50

100

Pr. affected

Dawn age 45


_________________ Alt H

____________________

Prior Probability

P(H) =

P(Alt H) =


P(D|H) =

P(D|Alt H) =


P(D|H) ● P(H) =



A B

A B

B

A B

A

Charlie inherited HD gene Charlie missed the bullet

= 1

1 4 1

{¼ x (½ + [ ½ x ½ ])}

3 32 =

1 2

16 32

1 2

1 2

x 1

1 2 =

3 32

3 19

[Charlie: unaff. @ 60]

[Dawn: unaff. @ 45] AND

H <--- OR ---> Alt H No HD Gene HD Gene, Unaff. @ 45

½ + ½ (1 – 50%)

● 1

2 ¼ x (2/4 + ¼) ¼ x (3/4)

3/16

3 32

= 16% 84%

x 1

A phenotypically female child is found to have XY karyotype. A mutation in the X-linked AR gene is found. She has a phenotypically male twin. Mom asks: Could they actually by MZ twins?

xy xy

You do some genotyping and got: Locus Dad Mom Twin 1 Twin 2

ABO

D12S14

D17S800

A1B O A1 A1

1,2 1,2 1,1 1,1

1,7 2,9 1,9 1,9

Circle the best answer: a. .03 b. .97

c. .94 d. 1.00


_________________ Alt H

____________________

Prior Probability

P(H) =

P(Alt H) =


P(D|H) =

P(D|Alt H) =


P(D|H) ● P(H) =



A B

A B

B

A B

A

Twins are DZ Twins are MZ

= 1

1 4 1

2

3 x

1 96

=

x

1 3 =

1 33

=

ABO

D12S14 D17S800

Gender

1/2

1/4 1/4

1/2

2 3

1 3

1 1 1 1

(½x¼x¼x½ ) 1 3

(1x1x1x1 )

= 32

96

1/96

1/96 32/96

32/96

1/96 32/96

32 33

0.97

Note: If you don’t include GENDER in your calculations, final answer is: 0.94

Example: Autosomal Recessive

I

II

III

1 2

2 1

1

3

P

CF

Given: Incidence of CF = 1/2500. No genetic testing has been done.

Question: What is the chance II-2 and II-3 will have an affected child?

Any conditional info available?

None…just calculate directly.

Prob II-2 is a carrier =

Prob II-3 is a carrier =

Chance III-1 is affected? Use standard info about carrier frequency and segregation:

q2 = 1/2500

2pq = 1/25

2/3 x 1/25 x 1/4 = = 2/300 = 1/150

P(II-2 carrier) x P (II-3 carrier) x segregation =

2/3

1/25

Can we modify the risk estimate by doing CF carrier testing on II-3?

II-3 negative on standard 23 mutation panel. (She is Caucasian, non-Hispanic). So, sensitivity of that test for her is 90%.

Let’s do some genetic testing on II-3. Can we modify the risk for the couple? II-3 tested for 23 mutation panel. She is Caucasian,

non-Hispanic. She is negative 90% sensitive test.

H _________________

Alt H ____________________

Prior Probability

2pq =

1 – 2pq =



P(D|H) ● P(H) =



A B

A B

B

A B

A

II-3 is a Carrier II-3 is NOT a Carrier

1

1 250

=

x

1 241

[Negative Carrier Test]

24 25

= 240

250

1/250

1/250 240/250

240 241

1 – Sensitivity of negative test

1 25

1 10

1 25

1 10

x 1 24 25

240/250

240/250 1/250

= =

Question: Chance II-2 and II-3 will have child with CF?

P(II-2 is carrier) x P (II-3 is carrier) x segregation ratio =

2/3 x 1/241 x 1/4 =

1/1446 ( vs 1/150 )

Considering Mutation

Assumptions:

(2)

(1) m m m = =

There is a prior prob of 4m

that any is a carrier of a particular X-linked is a carrier of a particular X-linked recessive, male lethal, disorder (Fitness = 0 for lethal)

Joanne has a brother with an X-linked disorder. No family history otherwise.

Question: What is the chance Joanne is a carrier? Tina

Joanne

H _________________

Alt H ____________________

Prior Probability

P(H) =

P(Alt H) =


P(D|H) =

P(D|Alt H) =


P(D|H) ● P(H) =



A B

A B

B

A B

A

Tina is a Carrier Tina is NOT a Carrier

m

2 3

[Affected Son]

2m

2m m

1 3

1 2

= =

4m 1

2m m

Tina is carrier

So, probability Joanne is a carrier: = 2/3 x 1/2 = 1/3

Question: What is the chance Emily is a carrier? Kathy

Emily

H _________________

Alt H ____________________

Prior Probability

P(H) =

P(Alt H) =


P(D|H) =

P(D|Alt H) =


P(D|H) ● P(H) =



A B

A B

B

A B

A

Kathy is a Carrier Kathy is NOT a Carrier

m

1 3

a) affected son b) 2 unaffected sons

m/2

m/2 m

2 3

1 2

= =

4m 1

4m x ½ x (½)2 =

m

Kathy is carrier

3 1 2 ( )

2 1 2 ( ) x

x 1 1

½ m

So, probability Emily is a carrier:

= 1/3 x 1/2 = 1/6

Association for Molecular Pathology 9650 Rockville Pike

Bethesda, MD 20814

[email protected]

www.amp.org

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