AIMS OF WORKSHOP

1. AIMS OF WORKSHOP • Maths isn’t everyone’s favourite subject! • Cannot be avoided – risk analysis integral part of Clinical Scientist role • Many of you have a strong background in Genetics so what we cover may be familiar • Intend to refresh and remind you of some of the concepts of risk analysis • Give you confidence to tackle common problems • Promote problem sharing and collaboration

2. Topics Covered • Application of Hardy-Weinberg Equilibrium • Basic Bayes Analysis • Linkage Analysis Advanced workshop (21st May 2007, or next year) • Further Bayes Analysis (X-linked disease) • LOD scores

3. Timetable

4. 5 Black 6 White 3 Black 4 White 6 Black 3 White 9 Black 5 White WEIRD PROBABILITIES!! 5/11 > 3/7 0.455 > 0.429 Choose Red 6/9 > 9/14 0.666 > 0.643 Choose Red

5. 5 Black 6 White 11 Black 9 White 3 Black 4 White 12 Black 9 White 6 Black 3 White 9 Black 5 White WEIRD PROBABILITIES!!

6. 11 Black 9 White 12 Black 9 White WEIRD PROBABILITIES!! 11/20 < 12/21 0.55 < 0.571 Choose Blue!!! SIMPSON’S PARADOX

7. ? THE MONTY HALL PROBLEM- YOUR CHOICE? We're easily deceived! You should always change your mind – it doubles your chance of winning the car! Is the best strategy to (i) change your mind (ii) stick with original door (iii) doesn’t matter

8. Hardy-Weinberg Equilibrium What is it? Describes the relationship between allele, genotype and phenotype frequencies within a stable population

9. Hardy-Weinberg Equilibrium Imagine a founder population: 30 individuals - genotype ‘AA’ 70 individuals - genotype ‘aa’ Allele frequencies: A = 0.3 a = 0.7 Genotype frequencies: AA = 0.3 Aa = 0.0 aa = 0.7 After a period of breeding, the numbers of heterozygotes will increase, and the homozygotes decrease. After many generations, the numbers of individuals with these genotypes will stabilise. This will be at Hardy-Weinberg equilibrium - if certain assumptions are made...

10. Hardy-Weinberg Equilibrium Assumptions for Hardy-Weinberg equilibrium: 1. Random mating 2. No mutation 3. No selection 4. No Genetic drift 5. No migration

11. Hardy-Weinberg Equilibrium 1. Random Mating: Assortative mating observed for most visible characteristics (eg stature) Random mating expected for ‘invisible’ characteristics (eg blood group)

12. Hardy-Weinberg Equilibrium 2. New mutations: A high new-mutation rate will introduce new mutant alleles 3. Selection Clearly, for disease genes, selection exists against disease status But there may be selection for carrier status (eg sickle cell carriers have resistance to malaria)

13. Hardy-Weinberg Equilibrium 4. Genetic drift: Allele and genotype frequencies can alter by chance - ‘genetic drift’ The probability of a neutral allele reaching fixation in a population = its frequency in the population The time to reach fixation depends on the size of the population - fast in small population, slow in large population The problem of genetic drift can be ignored in large populations

14. Hardy-Weinberg Equilibrium At equilibrium 2 allele system: Allele: A a Frequency: p q p + q = 1 Genotype: AA Aa aa Frequency: p2 2pq q2 p2 + 2pq + q2 = 1

15. Hardy-Weinberg Equilibrium Back to our previous founder population: p = 0.3 q = 0.7 At equilibrium: p2 = 0.09 2pq = 0.42 q2 = 0.49 Despite the limitations (assumptions) of Hardy-Weinberg, we can use this knowledge to estimate relationships between alleles and genotypes for human genetic diseases

16. Hardy-Weinberg Equilibrium What use is it in Human Genetics? It is used mainly in Autosomal Recessive diseases for • Estimation of gene frequencies • Estimation of carrier frequencies

17. Hardy-Weinberg Equilibrium- an example • Congenital Adrenal Hyperplasia is an autosomal recessive disease with an incidence of 1 in 10,000 • A CAH affected man and his wife with no family history wish to know the risk that their current pregnancy is affected by CAH

18. Hardy-Weinberg Equilibrium- an example • We already know that the man is affected, so they can have an affected child if the woman is a carrier of CAH • From the incidence of CAH we can calculate the gene frequency of CAH mutations • The incidence of CAH (1 in 10,000) = q2 • Thus q = SQR 1 in 10,000 = 0.01 • Then as p + q = 1 the frequency of the normal alleles must be = 0.99

19. Hardy-Weinberg Equilibrium- an example • Next we can calculate the carrier frequency of CAH from the 2pq term in the Hardy-Weinberg formula • For CAH • 2pq = 2 x 0.99 x 0.01 = 0.0198 ~ 1 in 50 • So our affected man’s wife is at 1 in 50 risk of being a CAH carrier

20. Hardy-Weinberg Equilibrium- an example • the probability that they are at risk of having an affected child is 1 in 50 • BUT EVEN IF she is a carrier the risk to the current pregnancy is 1 in 2 • OVERALL the risk to the CURRENT pregnancy is 1 in 100

21. Bayes Theorem Allows the calculation of the likelihood of an event happening in the future – given that the event hasn’t happened yet!! By combining information from different sources to modify the probability of an outcome

22. 2 1 Bayes Theorem & Common Sense ? You are allowed to draw three cards from one pile to help you guess Obviously if one of the three is a spade then you have identified which is the pile with spades But you draw three cards from pile no 2 – none are spades – what does this tell you? It is obvious that pile 2 (from which you drew three none spades) is more likely to be the one without spades This is Bayes theorem in action – although the outcomes are not always so predictable! Using what you learn today – try to calculate what the relative probabilities of the two piles of cards are Imagine 2 piles of 12 cards One pile 3 cards from each suit     Other has no  but 4 cards from each of the other suits    You don’t know which one is which

23. Bayes Theorem LATE BUS PROBLEM • You’re on time and waiting for a bus - some mornings the bus doesn’t turn up (the next one is not for another hour) • you wait for 10 minutes, but the bus is late • should you walk? • what is the likelihood of the bus turning up if you wait longer?

24. Bayes Theorem • With a bit more information you can work out the probability that the bus will still come - even though it is already 10 minutes late! • 10% of buses don’t turn up • Of buses that do turn up 80% are no more than 10 minutes late Intuitively it doesn’t seem worth waiting for the bus – BUT we can calculate the probability

25. A x M A x M + B x N B x N A x M + B x N Bayes Theorem –A Bayes Table A B M N A x M B X N

26. 18/100 18/100 + 10/100 10/100 18/100 + 10/100 Bayes Theorem – Late Bus Problem 9/10 1/10 2/10 1 18/100 10/100 18/28 ~ 2/3 So it is worth waiting! (anyway, chances are - the bus will have come before you’ve finished working this out).

27. Modifying Risks using Bayes • Work out a prior risk • - population risk or increased risk due to family history? • • Ask what additional information is available (from tests/pedigree) • - +ve/-ve test results? • unaffected children/siblings? • other indicators – e.g. age in late onset disease  Set out in a table and calculate the jointand final risks

28. 1/200 1/200 + 190/200 190/200 1/200 + 190/100 Bayes Theorem – a CF example Calculate the CF carrier risk of someone with no family history of the disease but who has tested negative for 29 common CF mutations (CF29m test)? Assumptions No Family History of disease Population Carrier Frequency 1 in 20 CF29m test detects 90% of mutations 1/20 19/20 1/10 1 1/200 190/200 1/191

29. Some Things to Remember! • You cannot use any piece of information twice! • - take care with pedigree information

30. Some Things to Remember! If you ask the wrong question you will get the wrong answer!! - think carefully about the column headings and conditional risks you put in the table BAYES = FUN!!!!!

31. Linkage Analysis • Linkage tests are becoming rarer in diagnostic labs – direct mutation analysis • Remains an important skill • Experience is an essential requirement for registration • Interpret with care – sources of error are • - Recombination • - Genetic heterogeneity • - Mosaicism • - Non-paternity • - Null-alleles • - And others!

32. 1,2 1,1 1,2 1,2 1,2 1,2 Linkage Analysis • Testing by association – an allele by itself means nothing – an allele inherited along with disease does • Important concept is setting phase (coupling) Phase unknown Phase known

33. 2,2 1,3 Phase inferred 1,2 Phase known Setting Linkage Phase • Phase is only set in certain individuals in a pedigree! • Phase is inferred in the father • Probability of phase in father with allele 3 = recombination rate (q)

34. A: 1,3 B: 2,2 C:(1,2) A: 2,4 B: 1,1 C:(1,2) A: 1,4 B: 2,1 C:(1,2) Paternal Maternal Building Haplotypes • A haplotype is a series of alleles at linked loci inherited together – need to build haplotypes in linkage analysis • Phase is known for markers A & B:A-1;B-2 paternal haplotype; A-4;B-1 maternal haplotype • Phase is unknown for marker C – cannot include in haplotype

35. 2 1,3 1 2,3 1 • New mutation (mosaicism) • Recombination Unusual result (1) An X-linked recessive pedigree Possible causes?

36. 3,3 1,2 2,3 1,3 2,2 How could you distinguish between the possibilities? • Non paternity • Null allele • Primer SNP • Deletion / rearrangement Unusual result (2) An autosomal recessive pedigree Possible causes?

37. AIMS OF WORKSHOP • Give you confidence to tackle common problems • Promote problem sharing and collaboration • Stimulate and provoke thought! • Challenges can be rewarding & fun! • Practice regularly – ask for problems • Only an introduction