Chapter 11 Chapter 11 Recurrence risk of singleRecurrence risk of singleRecurrence risk of singleRecurrence risk of single--

gene disordersgene disordersgene disordersgene disorders

Recurrence risk is the probability of a Recurrence risk is the probability of a child being affected with a diseasechild being affected with a disease given agiven achild being affected with a disease child being affected with a disease given a given a positive family historypositive family history of that disease.of that disease.

The recurrence risk of singleThe recurrence risk of single--gene gene The recurrence risk of singleThe recurrence risk of single gene gene disordersdisorders

Prevention Prevention ＞＞ treatmenttreatment

RRRR＞＞ 10% high risk;10% high risk;11～～10% 10% moderatemoderate risk;risk;＜＜1% low risk.1% low risk.

O tliO tliOutlineOutline

Direct estimationDirect estimation

Estimating using Bayes RuleEstimating using Bayes Rule

When the genotypes of the couple are When the genotypes of the couple are When the genotypes of the couple are When the genotypes of the couple are determinate determinate ……

AD AD -- Completely dominantCompletely dominant–– (most of the affected in the population are (most of the affected in the population are ( p p( p p

heterozygotes)heterozygotes)

①① Unaffected Unaffected ×× unaffectedunaffected aa aa ×× aaaa–– commonly unaffected (or the percentage of onset = commonly unaffected (or the percentage of onset =

d f l )d f l )incidence of population)incidence of population)②② AffectedAffected×× unaffectedunaffected Aa Aa ×× aaaa–– ½½ of offspring will be affectedof offspring will be affected③③ Affected Affected ×× affectedaffected Aa Aa ××AaAa–– ¾¾ of offspring will be affectedof offspring will be affected

AD AD -- Incompletely dominantIncompletely dominant–– According to the onset and severity of the individuals According to the onset and severity of the individuals g yg y

Unaffected Unaffected ×× unaffected unaffected aaaa××aaaa–– commonly unaffected (or the percentage of onset = incidence of commonly unaffected (or the percentage of onset = incidence of

population)population)Slightly affectedSlightly affected××unaffected unaffected AaAa××aaaaSlightly affectedSlightly affected××unaffected unaffected AaAa××aaaa–– ½½ of offspring will be affectedof offspring will be affected

Slightly affectedSlightly affected×× slightly affected slightly affected AaAa××AaAag yg y g yg y–– ¾¾ of offspring will be affectedof offspring will be affected

Severely affectedSeverely affected×× unaffected unaffected AAAA××aaaall f ff i ill b li h l ff dll f ff i ill b li h l ff d–– all of offspring will be slightly affectedall of offspring will be slightly affected

ARAR

1. 1. UnaffectedUnaffected××unaffectedunaffected

① AA × AA Commonly normal

② AA × Aa

？

C l l② AA × Aa？

Commonly normal

③ Aa × Aa¼ affected？？

2. 2. UnaffectedUnaffected××affectedaffected

？

AA × aa All carriers？

Aa × aa？

½ affected？

3. 3. Affected Affected ×× affectedaffected

aa ×aa

?Pay attention to geneticPay attention to genetic heterogeneity

×bbaa ×bb?

XDXD

1 Normal female 1 Normal female ×× normal male normal male XXaaXXa a ×× XXaaYY1. Normal female 1. Normal female ×× normal male normal male XX XX ×× XX YY–– Commonly normalCommonly normal

2 Affected female 2 Affected female ×× normal male normal male XXAAXXa a ×× XXaaYY2. Affected female 2. Affected female ×× normal male normal male XXAAXXa a ×× XXaaYY–– ½½ sons & sons & ½½ daughters are affecteddaughters are affected

3 Aff t d l 3 Aff t d l ×× l f l l f l XXAAY Y ×× XXaaXXaa3. Affected male 3. Affected male ×× normal female normal female XXAAY Y ×× XXaaXXaa

–– All daughters & none of sons are affected; All daughters & none of sons are affected;

XRXRXRXR

1. Normal female 1. Normal female ××normal male normal male

XAXA × XAY

C l l

？

Commonly normal

XAXa × XAY

？Female: ½ normal, ½ carrier;

Male: ½ normal, ½ affected,

2 Normal female 2 Normal female ×× affected male affected male 2. Normal female 2. Normal female ×× affected male affected male

XXAAXXA A ×× XXaaYY–– Commonly unaffected (daughters are carriers)Commonly unaffected (daughters are carriers)

XXAAXXa a ×× XXaaYYXX XX ×× XX YY–– ½½ sons & sons & ½½ daughters are affecteddaughters are affected

？？

3. Normal male 3. Normal male ×× affected female affected female ––

AAXXAAY Y ×× XXaaXXaa

–– Sons are affected, daughters are carriersSons are affected, daughters are carriers

？？

Example Example Example Example

Hyperdactylia AD penetrance 90%Hyperdactylia AD penetrance 90%Hyperdactylia AD penetrance 90%Hyperdactylia AD penetrance 90%

ⅠAa

Ⅰ

Ⅱ

1 2Aa Aa or aa

？ ？

Ⅱ

Ⅲ

42 3

1

1

21 2

Q: How about the risk of Ⅲ ？Q: How about the risk of Ⅲ1？

A: The risk of Ⅲ1：1/2×90%=45%

Q: How about the risk of Ⅲ2？

When either of the genotypes of the When either of the genotypes of the couple is indeterminate couple is indeterminate ……couple is indeterminate couple is indeterminate ……

Bayes Rule Bayes Rule (Bayesian analysis)(Bayesian analysis)

Thomas BayesThomas BayesThomas BayesThomas BayesMathematician who first used probabilityinductively and established a mathematicalinductively and established a mathematical basis for probability inference (a means of calculating, from the number of times an event has not occurred the probabilityevent has not occurred, the probability that it will occur in future trials). He set down his findings on probability in "Essay Towards Solving a Problem inEssay Towards Solving a Problem in the Doctrine of Chances" (1763), published posthumously in the Philosophical Transactions of the Royal Society of London. Bayes'of the Royal Society of London. Bayes contributions are immortalized by naming a fundamental proposition in probability, called Bayes Rule, after him.

Born: 1702 in London, EnglandDied: 7 April 1761 in Tunbridge Wells, Kent, England

y ,

D fi itiD fi itiDefinitionsDefinitions

Prior probabilityPrior probability -- the probability of an event the probability of an event before new information (finding) is acquired; before new information (finding) is acquired; before new information (finding) is acquired; before new information (finding) is acquired; pretest probability or risk pretest probability or risk 先验概率先验概率

C di i l b biliC di i l b bili i f i i f i Conditional probabilityConditional probability -- new information new information (finding) (finding) Joint probabilityJoint probability -- the probability of both events the probability of both events occur togetheroccur togetherPosterior probabilityPosterior probability -- the probability of an the probability of an event after new information (finding) is acquired; event after new information (finding) is acquired; ( g) q ;( g) q ;posttest probability or riskposttest probability or risk

前概率前概率 按有关遗传理论或遗传方式 列出个体按有关遗传理论或遗传方式 列出个体前概率前概率————按有关遗传理论或遗传方式，列出个体按有关遗传理论或遗传方式，列出个体可能的基因型以及产生这种基因型的分离率。可能的基因型以及产生这种基因型的分离率。

条件概率条件概率————指在某种特定条件下产生这种特定情指在某种特定条件下产生这种特定情况的概率。况的概率。

联合概率联合概率————指在某一基因型前提下，前概率和条指在某一基因型前提下，前概率和条件概率所说明的两个事件同时出现的概率件概率所说明的两个事件同时出现的概率件概率所说明的两个事件同时出现的概率。件概率所说明的两个事件同时出现的概率。

某事件的联合概率

各事件的联合概率之和=后概率后概率

各事件的联合概率之和

Example 1Example 1Example 1Example 1

HyperdactyliaHyperdactylia ADAD penetrancepenetrance 90%90%HyperdactyliaHyperdactylia AD AD penetrancepenetrance 90%90%

ⅠAa

Ⅰ

Ⅱ

1 2Aa Aa or aa

？ ？

Ⅱ

Ⅲ

42 3

1

1

21 2

Q: How about the risk of Ⅲ2？

Ⅱ3’s genotype Aa aaⅡ3 s ge o ype Aa aa

PP 1/2 1/2

CP 1-90％ 1

JP ½×（ ％） ½×1JP ½×（1-90％） ½×1

Posterior P1/20 1/2

Posterior P1/20＋1/2 1/20＋1/2

＝1/11 ＝10/11

For Ⅱ3: probability of a carrier is 1/11，

＝

Recurrence risk of Ⅱ3’s offspring :

1/11×1/2×90％＝9/2201/11×1/2×90％＝9/220

E l 2 E l 2 Example 2: Example 2: A phenotypicaly normal couple, the husband with blood A phenotypicaly normal couple, the husband with blood type A while the wife with blood type O, had two children type A while the wife with blood type O, had two children type A while the wife with blood type O, had two children type A while the wife with blood type O, had two children who died for who died for hemolysishemolysis..Hypothesize that the frequencies in the population of Hypothesize that the frequencies in the population of IIAAIIAAand and IIAAii are respectively 6.35% and 32.5%. What will be the are respectively 6.35% and 32.5%. What will be the onset probability of the next progeny of this couple?onset probability of the next progeny of this couple?例例22 一对表型正常的夫妇，丈夫血型为一对表型正常的夫妇，丈夫血型为AA型，妻子为型，妻子为OO型，婚后有两个孩子型，婚后有两个孩子例例2 2 一对表型正常的夫妇，丈夫血型为一对表型正常的夫妇，丈夫血型为AA型，妻子为型，妻子为OO型，婚后有两个孩子型，婚后有两个孩子死于新生儿溶血症。假设群体中死于新生儿溶血症。假设群体中IIAAIIAA和和IIAAii的基因型个体的频率分别为的基因型个体的频率分别为6.356.35％和％和32.532.5％，这对夫妇的下个孩子出现新生儿溶血的概率是多少？％，这对夫妇的下个孩子出现新生儿溶血的概率是多少？

Ⅰ1 2

Ⅱ21 33

h h b d IAIA IAiThe husband’s genotype IAIA IAi

6 35％ 32 5％PP 6.35％ 32.5％

6.35％＋32.5％ 6.35％＋32.5％

＝ ＝

CP 12 （½）2

＝0.16 ＝0.84

（ ）

JP 0.16 0.21

Posterior P 0.16

0 16＋0 21

0.21

0 16＋0 210.16＋0.21 0.16＋0.21

＝0.43 ＝0.57＝0.43 ＝0.57

The father’s genotype would be IAIA (the probability is 0.43) or IAi (the probability isprobability is 0.43) or I i (the probability is 0.57). The probability that the next child will be type A is ：be type A is ：

0.43×1＋0.57×1/2＝0.715

Example 3: Example 3: Example 3: Example 3:

H ti t Di H ti t Di Huntington Disease Huntington Disease ----Autosomal delayed Autosomal delayed d i t d i t

Ⅰ

Aa aadominant dominant The percentage of the The percentage of the i di id l tt k d i di id l tt k d

1

1

2

2Ⅰ

Ⅱ aa43

individuals attacked individuals attacked before 20 is 10%, and that before 20 is 10%, and that

f th b f 43 i 70% f th b f 43 i 70% 1

1

2

Ⅲ

aaAa

aa

20of those before 43 is 70%. of those before 43 is 70%. In this kindred, what is In this kindred, what is th i k f th i k f ⅢⅢ t b t b aa

Aathe risk of the risk of ⅢⅢ11 to be to be attacked?attacked?

Ⅱ2’s genotype Aa aa

CP （1-70%） 1

2 g ypPP 1/2 1/2CP （1 70%） 1JP 3/20 1/2

P t i P ＝Posterior P 10/13=0.773/13＝0.23

Ⅲ1’s genotype Aa aa1 g ypPP 0.23×1/2 0.77+0.23×1/2CP （1 10%） 1CP （1-10%） 1JP 0.1035 0.885

0.105Posterior P

The recurrence risk of Ⅲ is 10 5%．The recurrence risk of Ⅲ1 is 10.5%．

OrOr

Ⅱ2 genotype Aa aa2 g yp

Ⅲ1 genotype Aa aa aa

PP 1/2×1/2 1/21/2×1/2

CCP (1-70%)× (1-10%) 1(1-70%)× 1

JP 27/400 30/100 1

Posterior P 0.105

JP 27/400 30/100 1

The recurrence risk of Ⅲ1 is 10.5%．

Example 4Example 4、、酪氨酸血症酪氨酸血症Example 4Example 4、、酪氨酸血症酪氨酸血症(AR)(AR)

1Ⅰ

A

aa

Ⅱ1

Aa

AaAaHow about the How about the risk of risk of ⅤⅤ33??

Ⅲ1 2 ３ ４

ⅣⅣ AA

AAAa

1 2ⅣⅣ AA

Aaaa

1 2 ３ⅤⅤ

Ⅲ３genotype Aa AA

Ⅳ２genotype AA Aa AA

PP 1/2×1/2 1/2×11/2×1/2

CP 1212 (½)2

JP 1/4 1/16 1/2

CP 11 (½)

Posterior P 4/13 8/131/13

ⅣⅣ２２: the probability of a carrier is : the probability of a carrier is 1/131/13．．

For the offspring of For the offspring of ⅣⅣ and and ⅣⅣ the recurrence risk is : the recurrence risk is :For the offspring of For the offspring of ⅣⅣ1 1 and and ⅣⅣ22 , the recurrence risk is :, the recurrence risk is :1/131/13××1/2=1/261/2=1/26

1Ⅰ

Ⅱ1

Ⅲ1 2 ３ ４

1 2ⅣⅣ

1 2 ３ⅤⅤ

Example 5: Hemophilia (XR)Example 5: Hemophilia (XR)

Ⅰ1 2

XAXaXAYXAXA

Ⅱ21 4 53

X Y

XAXa

1 2 3 4Ⅲ

XaY

1 2 3 4

Ⅳ1

XAXA

XAXa

How about the risk of Ⅲ2’s male progeny？

Ⅱ4 genotype XAXa XAXA

PP 1/21/2

CP 12(½)2

JP 1/8 1/2

CP 12(½)2

Posterior P 4/51/5

For Ⅱ4: the probability of a carrier is 1/5For Ⅲ2: the probability of a carrier is 1/5×1/2=1/10

The recurrence risk of Ⅲ2’s son is 1/10×1/2＝1/20.

For Ⅲ2: the probability of a carrier is 1/5 1/2 1/10

IfⅡ4 have born 4 normal sons，then the answer is？

Ⅱ4 genotype XAXa XAXA

IfⅡ4 have born 4 normal sons，then the answer is？

Ⅱ4 genotype X X X X

PP 1/21/2PP 1/21/2

CP 14(½)4

JP 1/32 1/2

( )

Posterior P 16/171/17

For Ⅱ4: the probability of a carrier is 1/17For Ⅲ2: the probability of a carrier is 1/17×1/2=1/34

The recurrence risk of Ⅲ2’s son is 1/34×1/2＝1/68.

Ⅰ1 2

XAXa

Ⅱ21 4 53

1 2 3 4Ⅲ

XAXAⅣ1XAXA

XAXa

Ⅲ × ＝The recurrence risk of Ⅲ2’s progeny is 1/2×1/4＝1/8