Calculation of risk for genetic disorder, using probabilities

1. Risk Calculation
Ethical and Legal
Issues
MPH
6-4-15

2. Learning Objectives
• By the end of this lecture……….we all should
be able to:
– Appreciate the role of Clinical Genetics
– Explain the role of Genetics in Clinical Practice
– Calculate the probabilities of simple genetic
problems
– Calculate risks of genetic diseases in given cases.
– Discuss the ethical and Legal issues related to
Genetics

3. Clinical Genetics
Clinical Genetics is a medical specialty which
provides a diagnostic services and genetic
counseling for individuals or families with, or
at risk of, conditions which may have a genetic
basis.

4. Who is referred to Clinical Geneticisit’s
Clinic?
• A person with a known genetic condition in the family, wanting to
know the risks.
• Parents of a child with physical and/or learning difficulties which
may be due to a genetic condition, referred to see if a diagnosis can
be made
• Person with a strong family history of cancer, wanting to know if
they are at increased risk, and if they are, what options they have.
• A person with a known genetic condition wanting specialist advice
about the condition
• A person with a possible genetic condition in the family wanting to
know if a diagnosis can be made, and if so, their risks and options
• A pregnant couple told that a test has given an abnormal result,
wanting to talk about what the result means, and what options are
available.

5. Genetic Screening
Family Screening:
• Screening of individuals and couples known to be at
significant or high risk because of a positive family history -
sometimes referred to as targeted, or family, screening
because it focuses on those most likely to benefit.
– Carriers
– Heterozygote screening.
– Pre-symptomatic testing.

6. Genetic screening
Objectives
General screening
• Low risk population.
• Primary objective is to
enhance autonomy by
enabling individuals to be
better informed about genetic
risks and reproductive options.
• Secondary goal is the
prevention of morbidity due to
genetic disease and alleviation
of the suffering that this would
impose.

7. Presymptomatic Diagnostic Methods
• Clinical Examination
• Specialist Investigation
• Biochemical Tests
• Linked DNA Markers

8. Presymptomatic Diagnosis of
Neurofibromatosis I
Cinical Examination
• Autosomal dominant disorder
neurofibromatosis type I (NF I
) can have a number of
different clinical features .
• Unaffected relative of NF I
pateitns can be examined for
signs such as cafe-au-lait spots
or cutaneous neurofibromas.
NF I is a relatively rare
example of a dominantly
inherited disorder that is
virtually l 00% penetrant by
the age of 5 or 6 years, with
visible external features.

9. Specialist Examiniation
• Similarly, assessment for
Marfan syndrome involves
ophthalmic examination for
evidence of ectopia lentis,
echocardiography for
measurement of the aortic
root diameter, and magnetic
resonance imaging of the
lumbar spine to look for
evidence of dural ectasia - all
of these features count as
major criteria in the
disorder.

10. Biochemical Tests
• Serum Cholesterol Level for the
hypercholesterolemia

11. Linked DNA Markers
• The demonstration of linkage between a DNA
sequence variant and a disease locus
overcomes the need to identify a biochemical
defect or protein marker and the necessity for
it to be expressed in accessible tissues.
• In addition, use of markers at the DNA level
also overcomes the difficulties that occur in
carrier detection due to X-inactivation for
women at risk for X-linked disorders.

12. Diseases where
Linked DNA Markers
can be used

13. Risk Calculation
Anatomy of a Pedigree Chart

14. Problem-1
In Pedigree, which of the following females is least likely to be a heterozygote for the rare X-
linked recessive gene,III-1, III-3, or III-5?

15. Problem-2
• A patient of yours is getting married and comes to
you for counseling. She has a brother with a rare
X-linked recessive disease. Her mother's father
also had the disease. She wants to know the
probability of her being a carrier of the disease
and the probability that she will pass the disease
to her children.
What is your advice?

16. Answer
• Her mother was a carrier
• She has a 50% of being a carrier, depending upon
which of her mother's X chromosomes she
inherited.
• If she is a carrier she will pass the affected X to
her son 50% of the time.
• Her daughters will not be affected because they
will always get a normal X from their father.

17. Recurrence
risk
• One of the most important aspects of genetic
counseling is the provision of a risk figure, called
Recurrence risk.
• Estimation of the recurrence risk usually requires
careful consideration and takes into account:
– The diagnosis and its mode of inheritance
– Analysis of the family pedigree
– The results of tests that can include linkage studies.

18. Probablity
• The probability of an outcome can be defined as the
proportion of times it occurs in a large series of events.
• Conventionally, probability is indicated as a proportion
of 1.
– A probability of 0 implies that an outcome will never be
observed.
– A probability of 1 implies that it will always be observed.
– Therefore, a probability of 0.25 indicates that, on average,
a particular outcome or event will be observed on 1 in 4
occasions, or 25%.
– The probability that the outcome will not occur is 0.75,
which can also be expressed as 3 chances out of 4, or 75%.

19. Let’s practice

20. Prior/Anterior Probability
• The initial probability of each event is known
as its prior probability, and is based on
ancestral or anterior information.

21. Anterior and Posterior Probabilities
• Anterior probability of getting heads on a single toss of a
penny is 1/2, since there are two equal possibilities, H or T.
• Then, given two pennies tossed at random, HH, HT, TH, and
TT are all equally likely and the anterior probability of
getting at least one head is 3/4.
• The anterior probability that any combination with at least
one head will have two tails (HT or TH vs HH) is 2/3.
• If one penny is tossed and it shows H, What is the
probability that the other is also H? The a posterior
probability is 1/2 : given the knowledge that one coin is H,
the other is H or T with equal probability.
• In analyzing the results of any particular experiment, the
added information changes probabilities a posteriori.

22. Mutually Exclusive Events
– Law of Addition:
If the events are mutually exclusive, then the probability that
either one or the other will occur equals the sum of their
individual probabilities. The probability that the baby will be
either a boy or a girl equals1 (1/2 + 1/2)
– Law of Multiplication:
If two or more events or outcomes are independent, then the
probability that both the first and the second will occur equals
the product of their individual probabilities. If the mother is
found on ultrasonography to be carrying twins who are non-
identical, then the probability that both the first and the
second twin will be boys equals 1/4 (1/2 x 1/2).

23. Bayes’ Theorem
Applications
• Devised by the Reverend Thomas Bayes.
• Provides a valuable method for determining
the overall probability of an event or outcome.
• It considers all initial possibilities (e.g. carrier
or non-carrier) and then modifies these by
incorporating information, such as test results.
• It combines the probability that an event will
occur with the probability that it will not
occur.

24. Concepts
Conditional Probability
The observations that modify these prior
probabilities allow conditional probabilities to
be determined.

25. Concepts
Posterior Information
In genetic counseling these are usually based
on numbers of offspring and/ or the results of
tests.

26. Concepts
Joint Probability
The resulting probability for each event or
outcome is known as its joint probability.

27. Concepts
Posterior/ Relative probability
The final probability for each event is known as
its posterior or relative probability and is
obtained by dividing the joint probability for
that event by the sum of all the joint
probabilities.

28. Worked Example
Consider a pedigree
where the consultand
II1 wishes to know
the probability that
her male fetus will
have X-linked
Haemophilia

29. Worked Example
History
1. She has an
unaffected son
2. She had a brother
and two maternal
uncles with a
clinical diagnosis
but no molecular
testing is available

30. Worked Example
Calculations
1. On the basis of her mother being obligate carrier, she gets a probability of ½ of being a
carrier
2. Her anterior probability of not being a carrier is also ½.
3. Since they are mutually exclusive events, there sum equals to 1
4. Her unaffected son decreases the chance of her being a career.
5. The conditional probability that a career would have an unaffected son is ½P
6. The conditional probability that a non-carrier will have an unaffected son is 1
7. The posterior probability that she is a carrier is 1/3
8. The posterior probability that she is not a carrier is 2/3
9. Since it is an X-linked condition, there is a 50% chance that she would transmit the disease
to her son if she is a carrier
10. So the chance that the fetus will be affected is 1/3 x 1/2 = 1/6

31. Genetics
Ethical & Legal Issues

32. Thank you!