### An Introduction to Risk Analysis in Inherited X-Linked Recessive Disorders

Introduction

Risk assessment is an essential part of genetic counselling and much of this is based upon Bayesian risk analysis. Bayesian analysis is a method of combining probabilities and which is used to calculate the probability of having or not having a disease causing mutation after a negative test is obtained. Bayesian analysis is derived from ‘An Essay Toward Solving a Problem in the Doctrine of Chances' by the Reverend Thomas Bayes and published posthumously in 1763.

In X-linked disorders, risk analysis is generally undertaken for 2 reasons:

An affected male with haemophilia exists within a family and the female members in that kindred wish to establish whether they are haemophilia carriers or not.

A female with no family history of a bleeding disorder gives birth to a boy with haemophilia and wishes to know if she is a carrier and therefore, at risk of having similarly affected children. Other females within that family may also wish to know if there are carriers.

Risk assessment is undertaken by an analysis of:

- the Pedigree

- the Phenotype

- the Genotype

- and by combining these it may be possible to establish the probability of carriership in an 'at risk' female. The aim is to establish definitively whether an 'at-risk' female is a carrier or not. However, as you will see this is not always possible.

### Principles

Pedigree Analysis

An accurate family pedigree forms an essential part of the initial investigation of any family with a bleeding disorder. On the basis of the family pedigree a female may be ‘classified’ as an obligate carrier, a potential carrier or 'normal.' Risk analysis has to assume correct paternity. Historically the risk of non-paternity has been estimated to be as high as 10% but more recent data suggests that it is in the region of 1.9%.

For Guidelines on pedigree drawing and recommendations for standardised Human pedigree nomenclature - see References. Various packages are available for drawing family pedigrees - see References

### Obligate Carriers of X-Linked Recessive Disorders

In haemophilia A or B, the four pedigrees below illustrate the concept of obligate carriership i.e. a female must be a carrier of the abnormal *F8* or *F9* gene from the pedigree. This assumes, of course, correct paternity.

Pedigree | Interpretation |
---|---|

Pedigree 1 |
The daughters of an affected male are obligate carriers (in this case II:3) as they must inherit the affected X-chromosome from their biological father. In this pedigree (and others): Solid square = affected male Open square = unaffected male Open circle = unaffected female Circle with a dot in the centre = a female carrier of an X-linked disorder Circle with a ? in the centre = a possible female carrier of an X-linked disorder |

Pedigree 2 |
A woman (I:2) with two affected sons. A single affected son does not establish the mother as an obligate carrier as the affected child may represent a new, sporadic mutation. |

Pedigree 3 |
A mother with a single affected son but from a family in which there is a history of an X-linked disorder such as Haemophilia. In this case II:2 is a carrier as she has a son with haemophilia and a brother with haemophilia. |

Pedigree 4 |
The mother of two daughters each of whom has an affected male son. II:2 and II:4 are obligate carriers as they are sisters each with an affected male son. In this pedigree I:2 is likely to be a carrier - assuming that I:1 was normal i.e. not a male with haemophilia. |

### Bayesian Analysis Using Pedigree Data:

Pedigree 5

Let us now the consider the pedigree shown below [Pedigree 5]:

I:2 has 2 sons with (in this pedigree) severe Haemophilia A (VIII:C<1 IU/dL) - and she is, therefore, an obligate carrier of severe Haemophilia A. She has a daughter II:3 who wishes to know if she is a carrier or not.

We can use Bayesian risk analysis to establish the risk that II:3 is or is not a carrier. We are assuming that *F8* mutational analysis is not available or that this is one of the families, approximately 3-4% of cases, in which no mutation can be identified within the *F8* gene.

So - Bayes theorem or law states:

1. If the prior probability of an event occurring C is denoted as P(C) **and**

2. the prior probability an event C not occurring is denoted as P(NC) **and**

3. the conditional probability of an observation O occurring if C occurs is P(O/C) **and**

4. the conditional probability of an observation O occurring if C does not occur is P(O/NC)

5. **then** the overall probability of event C given O, equals:

6. and the overall probability of event C not occurring equals:

We can use this information in Pedigree 5 to establish the risk that II-3 is or is not a carrier of haemophilia. Bayesian analysis usually starts with mutually exclusive hypotheses – in this case whether II:3 is a carrier of severe haemophilia A or not. Each hypothesis is set out as a column in a table.

This is summarised in the table below:

Probability | II:3 is a carrier[Event occurs] |
II:3 is not a carrier[Event does not occur] |
---|---|---|

Prior probability | P(C) | P(NC) |

Conditional probability that O occurs | P(O/C) | P(O/NC) |

Joint probability | PC x P(O/C) | P(NC) x P(O/NC) |

Final [or posterior] probability |

The event (C) in this case is whether or not II:3 is a carrier of severe haemophilia and so:

P(C): P(NC): |
The probability that an event (C) occurs The probability that an event (C) does not occur |

P(O/C): P(O/NC): |
The conditional probability of observation (O) occurring if event C occurs The conditional probability observation (O) occurring if event C does not occur |

P(C) x P(O/C): P(NC) x P(O/NC): |
The joint probability of observation (O) occurring if event C occurs The joint probability of observation (O) occurring if event C does not occur |

The final or posterior probability of observation (O) occurring if event C occurs The final or posterior probability of observation (O) occurring if event C does not occur |

It is simpler to see how this works[!!] if we make the substitutions in the table taking into account the pedigree data from Pedigree 5: So -

Probability II:3 is a carrier

[Event occurs]II:3 is not a carrier

[Event does not occur]Prior probability 1/2 1/2 Conditional probability that O occurs 1 ~1 Joint probability 1/2 1/2 Final [or posterior] probability 1/2

--------------

1/2 + 1/2

= 1/21/2

--------------

1/2 + 1/2

= 1/2

**Prior Probability:**

1. The prior probability of II:3 being a carrier is 1/2 as she has a 1/2 chance of inheriting the abnormal X-chromosome from her mother that carries the *F8* gene mutation.

2. The risk of II:3 not being a carrier is also 1/2 as she has a 1/2 chance of inheriting the normal X-chromosome from her mother that does not carry the *F8* gene mutation.

**Conditional Probability:**

1. The conditional probability that II:3 is a carrier if she inherits the X-chromosome from her mother that carries the abnormal *F8* gene is 1 i.e. she must be a carrier if she inherits the abnormal *F8* gene.

2. The conditional probability that II:3 is a not carrier if she does not inherit the X-chromosome from her mother that carries the abnormal *F8* gene is almost 1 i.e. she cannot be a carrier if she does not inherit the abnormal *F8* gene but she could develop a new mutation in her *F8* gene.
If II:3 is not a carrier the conditional probability that a son would be normal is [1-µ] where µ is the *F8* gene mutation rate in females. This is an extremely small risk [in the region of 1/10,000] and to simplify the calculations we have ignored this and assumed the probability is ~1.

**Joint Probability:**

1. The joint probability of observation (O) occurring if event C occurs takes into account both the prior probability and the conditional probability and is derived by multiplying the two events e.g. 1/2 x 1 = 1/2.

2. Similarly the joint probability of observation (O) occurring if event C does not occur takes into account both the prior probability and the conditional probability and is derived by multiplying the two events e.g. 1/2 x 1 = 1/2.

**Final [or posterior] Probability**

1. The final or posterior probability of observation (O) occurring if event C occurs is derived from the joint probability (1/2) divided by the sum of the joint probabilities (1/2 + 1/2 = 1) = 1/2.

2. Similarly the final or posterior probability of observation (O) occurring if event C does not occur is derived from the joint probability (1/2) divided by the sum of the joint probabilities (1/2 + 1/2 = 1) = 1/2.

[**Note:** The sum of the final of observation (O) occurring if event C occurs and the final or posterior probability of observation (O) occurring if event C does not occur is always 1.]

So in this case - Bayesian analysis provides us with no additional information on the risk that II:3 is or is not a carrier. Our calculations show that the risk of II:3 being a carrier is 1/2 and of not being a carrier is also 1/2.

### Bayesian Analysis Using Pedigrees with Additional Data:

Pedigree 6a

Let us now the consider the pedigree (Pedigree 6A) shown below in which there are several affected males with severe haemophilia A [VIII:C <1 IU/dL]. This is similar to Pedigree 5 but with additional family members:

II:2 has 2 sons with (in this pedigree) severe Haemophilia A (VIII:C<1 IU/dl) and a brother with haemophilia - she is, therefore, an obligate carrier of severe Haemophilia A (see definitions of obligate carriers above). She has a daughter III:3 who wishes to know if she is a carrier of severe Haemophilia A or not.

III:3 already has two sons who do not have haemophilia. We can use this additional data i.e. the fact that III:3 has two normal sons without haemophilia in a Bayesian risk analysis to establish the risks that III:3 is or is not a carrier. So...

Probability III:3 is a carrier

[Event occurs]III:3 is not a carrier

[Event does not occur]Prior probability 1/2 1/2 Conditional probability that O occurs 1/2 x 1/2 = 1/4 ~1 Joint probability 1/2 x 1/4 = 1/8 1/2 x 1 = 1/2 Final [or posterior] probability 1/8

----------

1/8 + 1/2

= 1/51/2

----------

1/8 + 1/2

= 4/5

**1. Prior Probability:** So – in the pedigree above (and the table) – the prior probability [P(C)] that III:3 is a carrier of severe haemophilia A is 1/2 and the prior probability that she is not a carrier [P(NC)] of severe haemophilia A is also 1/2. [III:3 has a 1/2 chance of inheriting or not inheriting the abnormal X-chromosome from her mother that carries the *F8* gene mutation depending on which of the two X-chromosomes she inherits.]

**2. Conditional Probability:** III:3 has 2 normal sons.

If III:3 is a carrier then the conditional probability that both would be normal i.e. do not have severe haemophilia A, is 1/2 x 1/2 = 1/4. [Each son of III:3 has a 1/2 chance, if their mother is a carrier, of being an affected male.]

If she is not a carrier then the conditional probability that both sons would be normal is almost 1. It is not exactly 1 because there is a risk that a son could be affected because of a new mutation.

**3. Joint Probability:** The joint probability that III:3 is a carrier [P(C) x P(O/C)] is: 1/2 x 1/4 = 1/8 and the joint probability that III:3 is not a carrier [P(NC) x P(O/NC)] is: 1/2 x 1 = 1/2.

Any observation that modifies this prior probability allows conditional probabilities to be established for each hypothesis using posterior information such as the results of carrier detection studies or in the pedigree described – the finding that III:3 has two normal boys. The conditional probabilities are the probabilities that III:3 would have 2 unaffected sons if she was a carrier or not a carrier. The resulting probability for each is known as the joint probability and is derived from multiplying the prior probability by the conditional probability.

**4. The Final [or Posterior] Probability:** The overall final probability for each event is known as the posterior or relative probability and is obtained by dividing the joint probability by the sum of all the joint probabilities. This ensures that the sum of all the posterior probabilities will equal 1.

So in this pedigree - the risk the III:3 is a carrier of severe haemophilia A is 1/2 based solely on the finding that her mother is an obligate carrier but this probability changes to 1/5 that she is a carrier and 4/5 that she is not a carrier if we take into account the fact that she has had two unaffected sons.

### Pedigree 6B

If III:3 was to have another son without haemophilia - this would further reduce the risk that she is a carrier.

Probability III:3 is a carrier

[Event occurs]III:3 is not a carrier

[Event does not occur]Prior probability 1/2 1/2 Conditional probability that O occurs 1/2 x 1/2 x 1/2 = 1/8 ~1 Joint probability 1/2 x 1/8 = 1/16 1/2 x 1 = 1/2 Final [or posterior] probability 1/16

----------

1/16 + 1/2

= 1/91/2

----------

1/16 + 1/2

= 8/9

The risk, therefore, that III:3 is a carrier if she has three unaffected boys is 1/9 and the risk that she is not a carrier is 8/9. So you can see that the pedigree data can significantly alter the probability that someone is or is not a carrier. The data interpretation section will provide you with additional pedigrees to analyse.

### Bayesian Analysis in families there is only a single affected male:

Pedigree 7a

Bayesian analysis can be used to assess carrier risk in families in which there is only a single affected male. In the pedigree below, I-2 has a son (II:1) with severe haemophilia B and a daughter (II:2). II-2 wishes to establish her carrier status.

The probability that any female is a carrier can be derived from the formula:

- where *µ* is the mutation rate in females, *v* is the mutation rate in males and *f* is the biological fitness i.e. the ability to reproduce and have healthy children. *
f* can vary between 0 - a lethal mutation (as is the case with some X-linked disorders such as Duchenne Muscular Dystrophy) and 1 - the normal allele. A biological fitness of 0.5 implies that these males have only half as many children as their unaffected males siblings or healthy males in the general population.

In haemophilia the biological fitness of affected males (*f*) has been estimated to lie between 0.3 – 0.7 (but may be greater in the 21st century because of effective replacement therapy).

The mutation rate in male gametes (*v**)* is estimated at about 10X that of female gametes (*µ)*. This is because of the greater number of cell divisions that cells undergo during meiosis and the formation of males gametes compared to female gametes ]

Therefore – if we assume for the above pedigree that *f* = 0.5 and *v *= 10, then the prior probability of a female being a carrier is:

- where *µ* is the mutation rate in females. So there is a prior probability of 46µ that any female is a carrier. Remember - this assumes that the biological fitness is not zero (0) and that there is a difference in mutation rates between females and males. If the biological fitness is zero and there is no difference in the mutation rates between male and female gametes then the formula simplifies to 4µ.

In order to derive the risk that II:2 is a carrier in Pedigree 7a, the risk for I:2 must be calculated:

Probability | I:2 is a carrier[Event occurs] |
I:2 is not a carrier[Event does not occur] |
---|---|---|

Prior | 46µ | 1-46µ |

Conditional probability that O occurs | 1/2 | µ |

Joint probability | 46µ x 1/2 = 23µ | 1 x µ = µ |

Final [or posterior] probability | 23µ ---------- 23µ + µ = 23/24 |
23µ ---------- 23µ + µ = 1/24 |

So - from the pedigree and taking into account both the biological fitness and the difference in mutation rates between male and female gametes, I:2 is highly likely to be a carrier of haemophilia A [23/23 = 95%). This is confirmed by a number of studies that have shown that in families with a single male with haemophilia, the mother is a carrier in about 85-90% cases.

We can now use this data to establish the risk that II:3 is a carrier or not. The risk that she is a carrier of severe haemophilia A is 23/24 x 1/2 =22/48.

### Pedigree 7b

Let us now consider the case of Pedigree 7b. This is similar to Pedigree 7a but with additional family members:

In the pedigree below, I-2 has a son (II:1) with severe haemophilia A, a boys without haemophilia and a daughter (II:3). II-3 has two boys without haemophilia but wishes to establish her carrier status. We start again but establishing the probability I:2 is or is not a carrier:

Probability | I:2 is a carrier[Event occurs] |
I:2 is not a carrier[Event does not occur] |
---|---|---|

Prior | 46µ | 1-46µ |

Conditional probability that O occurs 1 son with Haemophilia A I unaffected son |
1/2 1/2 |
µ 1 |

Joint probability | 46µ x 1/2 x 1/2 = 46µ/4 | 1 x µ x 1= µ |

Final [or posterior] probability |

So we now use this data to establish the risk that II:3 is a carrier taking into account that she also has 2 sons without haemophilia. So -

Probability | I:2 is a carrier |
I:2 is not a carrier |
||
---|---|---|---|---|

II:3 | Carrier |
Not a carrier |
Carrier |
Not a carrier |

Prior | 46/100 |
46/100 |
4/100 |
4/100 |

Conditional probability 2 unaffected sons |
1/2 x 1/2 = 1/4 |
1 x 1 = 1 |
1/2 x 1/2 = 1/4 |
1 x 1 = 1 |

Joint probability | 46/400 |
46/100 |
4/400 |
4/100 |

Final [or posterior] probability | 46/230 |
176/230 |
1/5 |
4/5 |

So - If I:2 is a carrier, the probability that II:2 is not a carrier is 176/230 = 76.5%.

If I:2 is not a carrier, the probability that II:2 is not a carrier is 4/5 = 80%.

### Incorporating Phenotypic and Genotypic Data into Bayesian Risk Calculations

We can increase the complexity of these calculations [considerably] by adding in additional phenotypic data [e.g. the results of FVIII & VWF:Ag assays] and genotypic data [e.g. the results of genetic linkage studies.]

**A. Phenotypic (coagulation) Data:** In the following pedigree [Pedigree 6A] with severe Haemophilia A, II:3 wishes to establish her carrier status and her risk of having children with severe haemophilia A.

If we have no other data available other than the pedigree, then the risks that III:3 is or is not a carrier is shown below:

Probability II:3 is a carrier

[Event occurs]II:3 is not a carrier

[Event does not occur]Prior probability 1/2 1/2 Conditional probability that O occurs 1/2 x 1/2 = 1/4 ~1 Joint probability 1/2 x 1/4 = 1/8 1/2 x 1 = 1/2 Final [or posterior] probability 1/8

----------

1/8 + 1/2

= 1/51/2

----------

1/8 + 1/2

= 4/5

If we now obtain the results of factor assays [VIII:C and VWF:Ag data] and these show a VIII:C/VWF:Ag ratio of 0.63. From previous measurements of VIII:C and VWF:Ag and a calculation of the VIII:C/VWF:Ag ratio in a cohort of obligate carriers of Haemophilia A and of normal women, 3/4 obligate carriers were found to have an VIII:C/VWF:Ag ratio <0.65 but only 1/16 normal women.

If we incorporate this additional phenotypic data data into our risk analysis table, then:

Probability II:3 is a carrier

[Event occurs]II:3 is not a carrier

[Event does not occur]Prior probability 1/2 1/2 Conditional probability that O occurs

2 normal sons

VIII:C/VWF:Ag ratio = 0.63

1/2 x 1/2 = 1/4

3/4

~1

1/16Joint probability 1/2 x 1/4 x 3/4 = 3/32 1/2 x 1 x 1/16 =1/32 Final [or posterior] probability 3/32

---------------

3/32 + 1/32

= 3/41/32

---------------

3/32 + 1/32

= 1/4

So we can see that the phenotypic data has completely changed the probability that II:3 is a carrier [from 1/5 to 3/4]. Additional data such as genotypic data might further alter this risk.

To take this a step further - what happens if the phenotypic data is reversed i.e. the VIII:C/VWF:Ag Ratio = 0.72.

Then if we make the substitutions in the risk table:

Probability II:3 is a carrier

[Event occurs]II:3 is not a carrier

[Event does not occur]Prior probability 1/2 1/2 Conditional probability that O occurs

2 normal sons

VIII:C/VWF:Ag ratio = 0.63

1/2 x 1/2 = 1/4

1/16

~1

3/4Joint probability 1/2 x 1/4 x 1/6 = 1/128 1/2 x 1 x 3/4 =3/8 Final [or posterior] probability 1/128

---------------

1/128 + 3/8

= 1/493/8

---------------

1/128 + 3/8

= 48/49

So - in this case the phenotypic data has significantly changed the chances that II:3 is a carrier.

**B****. Genotypic Data:** In the following pedigree [Pedigree 8] with severe haemophilia A, linkage analysis was used to track the abnormal *F8* gene within a family. Mutation data was not available. The family was informative for only a single, extragenic polymorphism [Bgl II] which is closely linked to the *F8* gene locus but not part of it. There is a 5% chance of recombination taking place using this linked marker.

The section on Linkage Analysis explains this in more detail.

Analysis of the pedigree shows that the haemophilic *F8* gene is linked to the 5.8kb allele and that III:3 has inherited this from her mother II:1. However as this is a linked marker - there is a 5% chance of recombination taking place and therefore a 5% chance that our assumption that she [II:3] is a carrier is incorrect. Measuring her FVIII:C/vWF:Ag ratio may be of value but in this case was 0.78 i.e. normal.

The recombination fraction (θ) is an estimate of the likelihood of recombination taking place. In this case - the recombination fraction (θ) is 5% (0.05.) So - if we construct our risk calculation table as before:

Probability II:3 is a carrier

[Event occurs]II:3 is not a carrier

[Event does not occur]Prior probability 1/2 1/2 Conditional probability that O occurs

2 normal sons

Bgl II linkage data

1/2 x 1/2 = 1/4

1-θ

~1

θJoint probability 1/2 x 1/4 x (1-θ) = (1-θ)/8 1/2 x 1 x θ = θ/2 Final [or posterior] probability

Remember θ is the recombination fraction and using this linked extragenic marker, the risk of recombination is 5% (0.05.) So - if we substitute 0.05 for θ then the risk of II:3 being a carrier is:

and the risk of II:3 not being a carrier is:

This family is interesting in that II:3 has inherited the same allele (5.8kb) as the males that have haemophilia and which strongly suggests that she is a carrier. However, Bayesian risk analysis takes into account both the 5% chance of recombination using this linked extragenic marker and that the fact that II:3 has had two boys without haemophilia.

In practice due to the risk of recombination using linked extragenic markers, assignment of carriership should use intragenic markers and if these are not available a number [i.e. at least 2] of extragenic makers closely linked to the gene under study should be used. .

### What Test Next

The next tests would include:

a. Factor assays

b. Some form of genetic testing which is commonly mutation analysis but may include linkage analysis in some families in which no mutation can be identified (approximately 4% of families) or if mutation analysis is not available.

### Data Interpretation

Click HERE to go to the Data Interpretation Exercises.