Data Interpretation: Genetics  Answers
Introduction
This section of PracticalHaemostasis.com is designed to allow you to work thorough a number of clinical genetic scenarios that range from the relatively straightforward to the more taxing.
Question 1  

This is the pedigree of a family with severe Haemophilia A. From the pedigree data shown below, what is the risk that III3 is or is not a carrier? Click here for Part 2
From the pedigree we know that II:2 is an obligate carrier of haemophilia A as she has 2 sons and a brother with severe haemophilia A. Her daughter III:3 has a 1/2 chance that she is a carrier from the pedigree but this probability changes because she has 3 boys without haemophilia:
Prior Probability: When you have derived this risk  if III:3 decided to have another child what is the risk that if this a boy he will have severe haemophilia A? 
Question 2 

This is the pedigree of a family with severe Haemophilia B [FIX:C <1 IU/dL.] From the pedigree data shown below, what is the risk that II3 is or is not a carrier?
Click here for Part 2

Question 3  

This is the pedigree of a family with severe Haemophilia A. 1. From the pedigree data shown below, what is the risk that III3 is or is not a carrier? 2. Why is II2 an obligate carrier of severe haemophilia A? Click HERE for the answers
1. From the pedigree II:2 is an obligate carrier and therefore, there is a 1:2 chance that III:3 is a carrier and a 1:2 chance that she is not a carrier. III:3 has 2 normal boys and this changes the probability that she is or is not a carrier.
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 Xchromosome from her mother that carries the F8 gene mutation depending on which of the two Xchromosomes she inherits.] 2. Conditional Probability: III:3 has 2 normal sons. 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. 
Question 4  

This is the pedigree of a family with severe Haemophilia B. Click HERE for the answers
1. From the pedigree the risk that III:5 is a carrier is 1/2 [her mother II:2 is an obligate carrier as she has a brother with haemophilia and two sons also with haemophilia.
From the pedigree data we can construct a table:
If III:5 is a carrier then there is a 1:4 risk that she will have a boy with severe haemophilia B. So her overall risk is 1/1 x 1/4 = 1/8. 
Question 5 

This is the pedigree of a family with severe Haemophilia A [FVIII:C <1 IU/dL]. 1. From the pedigree data shown below, what is the risk that III4 is or is not a carrier? 2. The additional genetic information is generated using two linked intragenic markers within the F8 gene. The risk of recombination is negligible and can be ignored. The two alleles are 0.8/1.1 [kb] and 4.8/5.2 [kb]. Using this additional data what is the risk the III4 is a carrier?
Click HERE for the answers to Part 1
The risk from the pedigree that III:4 is a carrier is 1/2 as her mother is an obligate carrier of haemophilia.
Analysis of the pedigree shows that the abnormal F8 gene is linked to the 0.8kb/4.8kb haplotype and that III:4 has inherited this from her mother II:3. As these two markers are intragenic SNPs [i.e. located within the F8 gene] the chance of recombination occurring is negligible. Therefore, III:4 is a carrier of severe haemophilia. A. Remember  we do not know what the mutation is only that III:4 has inherited this.
Click here for Part 2
Is there still a role for linkage analysis in the investigation of families with haemophilia and if so why? Would you use intragenic or extragenic [linked] markers for these types of studies 
Question 6 

This is the pedigree of a family with severe Haemophilia A. 1. From the pedigree data shown below, what is the risk that II3 is or is not a carrier? 2. The additional information is generated using two intragenic markers within the F8 gene [0.8/1.1 and 4.8/5.2] and 1 extragenic marker [5.8/2.8]. The risk of recombination for the two intragenic markers is negligible and can be ignored but is 5% for the extragenic marker. Using this additional data what is the risk the III4 is a carrier? Click HERE for the answers
1. From the pedigree data, the risk that II:3 is a carrier of severe haemophilia A is 1/2 as her mother is an obligate carrier of severe haemophilia A.
2. Linkage analysis demonstrates that the F8 gene associated with haemophilia A in this pedigree is marked by the 4.8/0.8/5.8kb haplotype. III:3 must have inherited the 4.8/0.8/5.8kb haplotype from her mother and she is, therefore, a carrier of severe haemophilia A. 
Question 7 

These are electropherograms from a region of the human F9 gene. What mutation is shown.
Click HERE for the answer
The wild type ['normal'] sequence is: GCTCCTCTGCCTTCTGCTTG
The sequence from our patient with haemophilia A is: GCTCCTGCCTTCTGCTTG So there is a 2bp deletion of the nucleotides TC. As this will destroy the correct reading frame [the mutation is within an exon]  this can lead to a variety of possible abnormalities including the creation of a premature stop codon. This will alter the protein structure and therefore, one can predict that this is likely to be associated with a severe phenotype. 
Question 8  

In the following pedigree, II:3 died from neonatal purpura fulminans secondary to homozygous Protein C deficiency. His sister [II:1] seeks genetic counselling to establish what is the risk that her baby [III:1] will have homozygous Protein C deficiency. For the purposes of this exercise, assume the frequency of heterozygous Protein C deficiency in the population is 1:500. What is the risk that III:1 will have homozygous Protein C deficiency?
Click HERE for the answers
1. From the pedigree we can establish that that II:1 has a 1/3 chance that she has not inherited a PC mutation, and a 2/3 chance that she has [we know that she does not have homozygous PC deficiency and so we can remove this from the estimation of risk.]
[Remember  the probability that a healthy sibling of someone with an autosomal recessive disorder is a carrier is 2/3.] Therefore, her risk of having a child with homozygous PC deficiency is 2/3 x 1/500 x 1/4 = 1/3000. The probability that III:1 will be affected is the probabilities of these two independent events [2/3 x 1/500] multiplied by 1/4  the chance that they will both pass on an abnormal gene. 2. The probability changes significantly if her partner III:2 is has undiagnosed heterozygous PC deficiency: 2/3 x 1/2 x 1/2 = 2/12 = 1/6. 2/3 [the risk that II:1 is a carrier] x 1/2 [the risk that II:2 will pass on an abnormal PC gene] = 2/12 = 1/6. The following table illustrates this:
3. You might wish to consider the chance of two unrelated individuals having a child with homozygous PC deficiency assuming a prevalence of heterozygous PC deficiency in the general population of 1/500. Then the risk is: 1/500 x 1/500 x 1/4 = 1/1,000,000 [i.e. small]. 
Question 9 

In the following pedigree, II1 has 2N Von Willebrand Disease. He and his partner consult you as they are keen to know the risk that their child [III:1] will also have 2N VWD. What is the risk that III:1 will have 2N VWD? What is the risk that his sister [II:3] will have 2N VWD. [Assume that the frequency of the 2N heterozygosity in the general population is 1:750.] Would mutational analysis help in this family? Click HERE for the answers
1. Type 2N VWD is inherited as an autosomal recessive disorder and therefore II:1 is homozyous or a compound heterozygote for mutations that lead to defective binding of FVIII by VWF. In some cases a gene deletion and a heterozygous 2N mutation can lead to the same phenotype i.e. pseudohomozygosity but in these cases the VWF levels are usually reduced whereas in classical 2N VWD the VWF levels [immunological and functional] are normal. The frequency of 2N VWD heterozygotes in the general population [we are told] is 1/750. Therefore, the risk that III:1 will have 2N VWD is 1 x 1/750 x 1/2 = 1/1500. [Remember, III:1 has to inherit an abnormal VWF allele from his father II:1 but has a 1/2 chance of inheriting the abnormal allele from his mother II:2 if she is a carrier and the chances of her being a carrier are 1/750.] 
Question 10 

In the following pedigree, I1 has Severe Haemophilia A [FVIII:C <1 IU/dL] and his partner [I2] has 2N Von Willebrand Disease. [Assume that the frequency of the 2N heterozygosity in the general population is 1:750.] What are the risks that I:1 and I:2 will: 1. Will have a son with severe haemophilia A? 2. Will have a son with 2N VWD? 3. Will have a daughter who is a carrier of severe haemophilia A? 4. Will have a daughter with 2N VWD?
Click HERE for the answers
1. They cannot have a son with haemophilia A as any boys will inherit the Y chromosome from their father. However, there is a small chance that a new mutation could arise and so the actual risk is not zero. Remember that all the female offspring of I:1 will be obligate carriers of haemophilia A. 2. The risk is the same as in the previous question [Q9]: The frequency of 2N VWD heterozygotes in the general population is [we are told] is 1/750. Therefore, the risk that I:1 and I:2 will have a child with 2N VWD is 1 x 1/750 x 1/2 = 1/1500. 3. II:1 is an obligate carrier of severe Haemophilia by definition as her father has severe haemophilia A. 4. 2N VWD is not Xlinked and therefore the risks for males and females are the same  so see 2. Remember these are not true incidence rates but have been chosen to simplify the maths. 
Question 11 

You are asked to see a family with a bleeding disorder and obtain the following information: Click HERE for the answers
Shown below is the pedigree that you should have drawn:

Question 12 

A 45yearold man is diagnosed with Type 1 antithrombin deficiency [AT:Act 45 U/dL] and is concerned that he may have passed it onto his three children. His partner has normal functional antithrombin levels. He consults you and asks what is the risk that all 3 of his children will have Type 1 Antithrombin deficiency. Click Here for the Answers.
We can derive the answers for these questions using Pascal's Triangle  derived by expanding (p + n)^{n}.
This is show diagrammatically below [Click HERE for an enlarged image].
p = the probability of an event not occurring [in this case not inheriting the disease] q = the probability of an event occurring [in this case inheriting the disease] n = the total number of disorders [in this case the number of potential or actual offspring] So  if we look at the family with 3 possible children [n = 3] then if read along we see that for all 3 of the children to have Type I antithrombin deficiency the probability is [p]^{3} where p = 1/2 [Remember because this is an autosomally inherited disorders the risk of inheriting the mutant allele from a parent is 1/2 and of not inheriting it is also 1/2]. [1/2]^{3} = 1/8. So the chance that all 3 children will be normal is 1/8 and similarly the risk that all 3 children will be affected [ = [q]^{3}] is also 1/8. To establish if only 1 of the children will be affected and the other two normal then again we read along from n=3 to where we see 3pq^{2} and so 3pq^{2} = 3 x 1/2 x [1/2]^{2} = 3 x 1/2 x 1/4 =3/8. So the risk that they will have one affected child and two normal children is 3/8. [Remember p is the probability of an event not occurring [in this case not inheriting the disease and q is the probability of an event occurring [in this case inheriting the disease]] Finally the risk that they will have two affected children and one normal child is 3p^{2}q = 3 x [1/2]^{2} x 1/2 = 3/8. If they had a further child, then the Sample SIze =4 but the method of working out the various risks is the same  just read along from n=4 to derive the formulae for calculating the risks. 
Question 13 

This is the pedigree of a family with severe Haemophilia A. No mutation data was available and linkage analysis was used to establish the carrier status of II:3. 1. I:2 is an obligate carrier of severe Haemophilia A as she has two boys with severe Haemophilia. 
Question 14 

Look at the following pedigree which is from a family with severe Haemophilia A. III:1 and III:2 are identical twins. Click HERE for the answers
1. The important message here is that the VIII:C in the identical twin may not be the same as the other twin and in this case the twin had a low FVIII:C of 0.05 IU/dl. Although both twins in this case have inherited the mutated F8 gene from their mother [II:3] the discordance in FVIII levels arises from the differing patterns of Xinactivation or Lyonisation which subsequently occurs.
2. II:2 and II:3 are obligate carriers of haemophilia as their father had severe Haemophilia A. 
Click HERE to return to the top of the page