遗传咨询与风险评估-医学遗传学中的伦理问题(英文版)

Genetic

Counseling

and

Risk

AssessmentEthical

Issues

in

Medical

GeneticsThe

process

of

genetic

counselingDetermining

recurrence

risksEthical

issues

in

medical

geneticsOutlineTO

MASTER:Common

indications

for

genetic

counseling·Risks

estimation

by

use

of

conditional

probabilitywhen

alternative

genotypes

are

possibleMajor

ethical

policy

in

medical

geneticsWhat

is

genetic

counseling?A

process

of

exploration

and

communication

toprovide

information

and

support

to

families

atrisk

for

having,

or

who

already

have,

memberswith

birth

defects

or

genetic

disease.In

common

disorders

→

given

by

the

family

doctor,the

pediatrician

or

the

obstetrician.RECENTLY

,by

Genetic

Counselor.Genetic

Counselor

(GC)

is

a

health

care

professionalwith

a

master’s

degree

in

human

genetics

andcounseling.

This

training

enables

GCs

to

discusstechnical

genetic

information

in

practical,

useful

termWho

provide

genetic

counseling?Common

indicationsPrevious

child

with

genetic

disorderFamily

history

of

a

hereditary

conditionPrenatal

diagnosisConsanguinityTeratogen

exposureRepeated

pregnancy

loss

or

infertilityNewly

diagnosed

genetic

conditionbefore

undertaking

genetic

testing

and

afterreceiving

resultsAs

follow-up

for

a

positive

resultGenetic

counseling

casemanagementCollection

of

informationFamily

historyMedical

historyTest

and/or

additionalassessmentAssessmentPhysical

examinationValidation

orestablishment

ofdiagnosisCounselingNature

and

consequence

ofdisorderRecurrence

riskAvailability

of

further

or

futuretestingDecision

makingReferral

to

other

specialists,health

agencies,

supportgroupsFollow-upContinuing

clinical

assessmentPsychosocial

supportDetermining

recurrence

risks-----

a

central

concern

in

genetic

counselingRisk

estimation

when

genotypes

arefully

known

by

use

of

Mendel’s

laws3Autosomal

dominantWhat

is

the

probability

that

individual

3

will

beaffected?12AaaaAutosomal

recessiveWhat

is

the

probability

that

individual

5

will

beaffected?1345AaAa2X-linked

recessiveWhat

is

theprobability

that

thewoman

will

havean

affected

son?Risk

estimation

by

use

conditional

probabilitywhen

alternative

genotypes

are

possibleBayesian

analysisA

method

that

takes

advantage

ofphenotypic

information

in

a

pedigree

toassess

the

relative

probability

of

two

ormore

alternative

genotypic

possibilitiesand

condition

the

risk

on

the

basis

of

thatinformation.Conditional

probabilityFamily

Acarrier

risk50%IIIIII121212345Family

Bcarrier

risk3%IIIIII12123412345How

to

calculate

actual

risk

usingBayes’

theoremI123412345IIIIII123412345IIIIII123

412345A:

II-2

is

a

carrier,

but

theconsultand

did

not

inherited

themutant

alleleB:

II-2

is

a

carrier,

and

theconsultand

did

inherited

the

mutantalleleC:

II-2

is

not

a

carrier

and

theconsultandcould

not

haveinherited

the

mutant

allelePriorprobabilitiesIIIIIConditionalprobabilitiesJointprobabilitiesPosteriorprobabilities1/21/21/2(1/2)41/2(1/2)41/21411/2×(1/2)4

×1/2=1/641/2×14

×1=1/21/2×(1/2)4

×1/2=1/641/641/64+1/64+1/2=1/341/641/64+1/64+1/2=1/34≈3%1/21/64+1/64+1/2=16/17Prior

probability

that

a

Female

in

the

populationis

a

Carrier

of

an

X-linked

lethal

disorderH:

the

population

frequency

of

female

carriersAssume:H

is

constant

from

generation

to

generation

The

mutation

rateinany

one

gamete=μ,

and

it

is

the

same

inmales

and

females.Three

ways

that

any

female

could

be

a

carrier:A

mutant

allele

from

a

carrier

mother=1/2HA

newly

mutant

allele

on

the

X

from

her

mother=μA

newly

mutant

allele

on

the

X

from

her

father=μ·

H=(1/2×H)+μ+μH=4μConditional

probability

in

X-linkedlethal

disordersA:

III-1

is

a

new

mutationB:

I-1

is

not

a

carrier,

II-1

must

bethe

product

of

a

maternal

orpaternal

new

mutationC:

I-1

and

II-2

are

both

carriersPrioruJointprobabilitiesu4u×1/2×1/2=u2u×1/2=uIIIprobabilitiesIIIConditionalprobabilities121212IIIIII2u1

21212IIIIII1/21212124u1/21/2Posteriorprobabilitiesu/3u=1/3

u/3u=1/3

u/3u=1/3The

risk

that

II-1

is

a

carrier:

1/3+1/3=2/3The

risk

that

III-2

is

a

carrier:

2/3×1/2=1/3The

penetrance

is

70%.What

is

the

probabilitythat

individual

3

will

havean

affected

child?123Disorder

with

incomplete

penetranceAaaaPriorprobability1/21/2Conditionalprobability1－7/10＝3/101Jointprobability1/2×3/10=3/201/2Posteriorprobability(3/20)

/（3/20+1/2）=

3/13(1/2)

/（3/20+1/2）=10/13The

probability

is

3/13×1/2×7/10

≈

8%.Individual

3

is

30year-old.What

is

the

probabilitythat

individual

3

willbe

affected?123Disorder

with

late

age

at

onsetopulation

(%)10090807060504030201001020304050607080Age

(yr)AaaaPriorprobability1/21/2Conditionalprobability1－1/3＝2/31Jointprobability1/2×2/3=1/31/2Posteriorprobability(2/6)

/（2/6+3/6）=

2/5(1/2)

/（2/6+3/6）=

3/5The

probability

is

40%.AaaaSIn

this

pedigree,

what

is

the

probability

thatindividual

S

is

affected?ⅠⅡⅢⅣⅢ-1

is

a

carrierⅢ-1

is

not

a

carrierPriorprobability1/43/4Conditionalprobability1/2×1/2×1/2=1/81Jointprobability1/4×1/8=1/323/4

×1=3/4Posteriorprobability(1/32)/(1/32+24/32)=1/25(24/32)/(1/32+24/32)=24/25She

will

have

an

affected

son:1/25×1/2=1/50X-linked

recessiveWhat

is

theprobability

that

thewoman

M

will

havean

affected

son?MThe

mother

is

a

carrierThe

mother

is

not

acarrierPri