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GRADE A TRAINEES ADVANCED RISK ANALYSIS WORKSHOP

GRADE A TRAINEES ADVANCED RISK ANALYSIS WORKSHOP. National Genetics Reference Laboratory (Manchester) 21 st May 2007. Thanks to: CMGS TAB committee Andrew Wallace Assistants Manchester NGRL & NOWGEN. Timetable. 10:30 – 10:50 Tea/Coffee 10:50 – 11:00 Introduction

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GRADE A TRAINEES ADVANCED RISK ANALYSIS WORKSHOP

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  1. GRADE A TRAINEES ADVANCED RISK ANALYSIS WORKSHOP National Genetics Reference Laboratory (Manchester) 21st May 2007 Thanks to: CMGS TAB committee Andrew Wallace Assistants Manchester NGRL & NOWGEN

  2. Timetable • 10:30 – 10:50 Tea/Coffee • 10:50 – 11:00 Introduction • 11:00 – 11:30 X-linked Bayes’ • 11:30 – 12:45 X-linked Bayes’ problems • 12:45 – 13:15 Lunch • 13:15 – 13:45 LOD scores • 13:45 – 15:15 LOD score problems • 15:15 – 15:30 Feedback & Discussion • 15:30 Close

  3. Aims of the Workshop • Build on the Introductory Course • Introduce new elements, more complex calculations • New Elements • X-linked Bayes’ analysis • LOD scores

  4. Why These Topics? • X-linked Bayes’ Calculations – • relevant for clinically important diseases e.g. DMD/BMD • LOD Scores • core method used in Genetics • used for mapping loci • still useful in analysis of heterogeneous disease

  5. Words of Caution! • Be realistic - don’t expect to become an expert in one day • Some calculations can be very complex – best done on a computer • Don’t expect to complete all the problems – some are pretty tough! • To get the most out of the course you will need to go away and work through the specimen answers • Also get practice in the lab on real cases

  6. Words of Justification! • If we do use computers to calculate risks we need to be able to check the answers – garbage in garbage out • Important that we have an understanding of the theory – we are Clinical Scientists after all • It can be satisfying to solve a difficult problem • What a great way of making yourself indispensable (and embarrassing your boss) by becoming the lab expert!

  7. The Two Envelope Paradox • Imagine you are taking part in a Game Show • You are given the choice of opening one of two envelopes • All you know is that one contains twice as much money as the other • You have no indication of the amount

  8. The Two Envelope Paradox • So you open one and it contains a cheque for £40,000 but the host gives you the chance to swap. What should you do?

  9. £40,000 £80,000 £40,000 £20,000 The Two Envelope Paradox -£20,000 +£40,000

  10. The Two Envelope Paradox • Always Stick – will win £40,000 • Swap 0.5 x £80,000 • Swap 0.5 x £20,000 • Therefore if swap average win will be (£20,000 + £80,000)/2 • = £50,000 • It always pays to swap • Can this be correct?!

  11. GRADE A TRAINEES ADVANCED RISK ANALYSIS WORKSHOP Part 1 X-linked Bayes’ analysis

  12. Recap on Bayes’ Theorem Allows the calculation of the likelihood of an event happening in the future – given that the event hasn’t happened yet!! By logically combining evidence from different sources to modify the probability of the event

  13. A x M A x M + B x N B x N A x M + B x N Setting out a Bayes’ Table A B M N A x M B X N

  14. An Example Bayes’ Problem • You have two purses • Each contains two coins • One contains a gold sovereign – the other coins are worthless • You pick a coin out at random, it is plastic • What is the probability that the other coin in the same purse is the sovereign?

  15. An Example Bayes’ Problem 1/2 1/2 1 1/2 1/2 1/4 2/3 1/3

  16. X-linked Bayes’ • OK so what’s so different? • New Mutation events • Up to now e.g. with CF and other ‘basic’ problems we have been able to ignore new mutations as their probability is very small compared with other factors • Cannot ignore new mutations with many X-linked diseases and pedigrees

  17. m n New Mutations - ‘m’ Female germ-line mutation rate Male germ-line mutation rate n & m are often assumed to be equal – if so both referred to as m

  18. Why is this important? • We can use these values for our prior probabilities for females when possibility of new mutation is important • So a randomly selected female is at 4m prior risk of carrying an X-linked recessive disease where f = 0 • Similarly a randomly selected female is at 10m prior risk of carrying an X-linked recessive disease where f = 0.5 • Luckily we don’t need to know the value of m to carry out Bayes’ calculations!

  19. m m An ‘Adam and Eve’ Explanation I Adam Eve 2m

  20. m m 1/2 x 2m +2m = 3m An ‘Adam and Eve’ Explanation II Adam Eve 2m

  21. m m 1/2 x 3m +2m = 3.5m An ‘Adam and Eve’ Explanation III Adam Eve 2m 3m

  22. m m 1/2 x 3.5m +2m = 3.75m An ‘Adam and Eve’ Explanation IV Adam Eve 2m 3m 3.5m limit N = 4m

  23. μ μ μ μ μ μ μ 4μ = 1/4 female population 3μ = 1/2 female population μ μ μ μ μ 5μ = 1/8 female population Proportioncarrier risk of population 1/2 3μ 1/4 4μ 1/8 5μ 1/16 6μ 1/32 7μ etc etc average carrier risk in population = 4μ

  24. Reproductive Fitness - f • f is an important factor in X-linked calculations • f is a measure of affected individual’s ability to pass on their genotype • Value of f is specific to a disease:– 0 <f< 1 • If affected males never survive to reproduce or are unable to then f = 0 e.g. Duchenne Muscular Dystrophy • If there is no effect and affected males reproduce as well as unaffected sibs then f = 1 e.g. colour-blindness • by definition, f = 0 for lethal X-linked diseases – but is above zero for many X-linked diseases, e.g. BMD (f = 0.7)

  25. Co-efficient of m • This is the population frequency of female carriers expressed as a factor of m • Can be calculated using the formula below – provided m and n are equal(generally, a reasonable assumption for DMD) • N = (4 + 2f)/(1 - f) • Where N = co-efficient of m • So the co-efficient of m where f = 0 then N = 4 • and where f = 0.5 • N = (4 + 2 x 0.5)/(1 – 0.5) = 10

  26. Biological fitness: 0 <f< 1 f = 0 for lethal diseases; f = 1 for full fitness At equilibrium: Where f = 0 (e.g. DMD) prior risk = 4μ 2μ 2μ 2μ 2μ μ μ μ 3μ μ 3μ Where f = 0.7 (e.g. BMD) prior risk = 18μ 10μ x f = 7μ 9μ 9μ 9μ 9μ μ μ μ 10μ 8μ 10μ

  27. X – linked Bayes Consider this simple DMD pedigree • What is the mother’s carrier risk? There are two possibilities (i) she is a carrier (ii) she is not a carrier and her son is a new mutation But quantifying these risks is easy using Bayes’!

  28. New Mutations Carrying out a Bayes’ calculation • We can treat the fact that she has had an affected son as a conditional piece of information • And use the population carrier risks as prior carrier risks in a Bayes’ table Carrier = 4m Non-carrier = 1 – 4m ~ 1

  29. 4m 1 - 4m ~ 1 New Mutation Bayes’ Calculation

  30. 4m 1/2 1 - 4m ~ 1 m New Mutation Bayes’ Calculation

  31. m New mutation if a non-carrier

  32. 4m 1/2 2m 1 - 4m ~ 1 m m 2m/3m = 2/3 m/3m = 1/3 New Mutation Calculation

  33. 1 2 Using a single Bayes Table I This is something that is recommended to avoid errors Why? Enables you to take into account ALL the information

  34. CARRIER NON-CARRIER PRIOR 4m 1-4m ~ 1 G’ma C1 Aff son C2 Norm son Dau ghter C NC C NC 1/2 1/2 2m 1-2m ~ 1 C3 Norm son JOINT FINAL A B C D Split the columns!

  35. CARRIER NON-CARRIER PRIOR 4m 1-4m ~ 1 G’ma C1 Aff son 1/2 m C2 Norm son 1/2 1 Dau ghter C NC C NC 1/2 1/2 2m 1-2m ~ 1 C3 Norm son 1/2 1 1/2 1 JOINT 1/4m 1/2m m2 ~ 0 m FINAL 1/7 2/7 0 4/7 A B C D Split the columns! Grandma’s risk = A + B/A + B + C + D = 3/7 Daughter’s risk = A + C/A + B + C + D = 1/7

  36. CARRIER CARRIER NON-CARRIER NON-CARRIER PRIOR PRIOR 4m 1/4 1-4m ~ 1 3/4 Grandma Daughter C3 Normal son C1 Affected son 1/2 1/2 m 1 JOINT C2 Normal son 1/8 1/2 6/8 1 JOINT FINAL m 1/7 6/7 m FINAL 1/2 1/2 Using two Bayes’ Tables Daughter’s risk is still correct but Grandma’s is incorrect! Failed to include daughter’s unaffected son in Grandma’s risk

  37. Where to Start? Always commence your calculation with the most senior ‘relevant’ ancestor!

  38. Final Points • Tried to grade the questions – first ones should be quite easy • Final ones are quite difficult! • First problem involves an obligate carrier female so you don’t need to start the Bayes’ table using m

  39. Problem Session • You will need a scientific calculator, paper and pens • Split into groups of 4/5 • Attempt the problems together • We will be on hand to provide help

  40. GRADE A TRAINEES ADVANCED RISK ANALYSIS WORKSHOP Part 2 LOD Score analysis

  41. LOD Scores What is a LOD score? LOD = Logarithm of Odds ratio for linkage Logarithm • A logarithm is the exponent of 10 required to give a number e.g. expressed as 10n • So 100 expressed as a logarithm = 2; as 102 = 100 Odds ratio • A measure of the relative likelihood of two mutually exclusive events e.g. a 1:1 odds ratio indicates two equally likely possibilities So a 100:1 odds ratio for linkage = LOD of 2 LOD scores first described by Newton Morton to calculate linkage distances

  42. LOD scores A LOD score is calculated as follows • Where  = the recombination rate (fraction) • Hence a LOD score of 3 indicates a relative likelihood of 1000:1 in favour of linkage (103 = 1000)

  43. What are LODs used for?

  44. What are LODs used for? • Lots of things - but we will be • Estimating the significance of genetic linkage • Analysis of heterogeneous disease - using linkage analysis to determine which locus is significant

  45. Categorise Individuals • Think about what information is being gained from each person in a family – not everyone is ‘informative’ • 5 categories • Uninformative (U) • Definite non-recombinant (NR) • Definite recombinant (R) • Probable non-recombinant (PNR) • Probable recombinant (PR)

  46. 1,1 2,2 phase set 3,3 1,2 1,3 An Autosomal Dominant Pedigree I U U NR  = 0.1 Likelihood of result if linked = 1 - = 0.9 Likelihood of result if unlinked = ½ (0.5) LOD = log10 0.9/0.5 = 0.255  = 0 Likelihood of result if linked = 1 -  = 1 Likelihood of result if unlinked = ½ (0.5) LOD = log10 1.0/0.5 = 0.301

  47. 1,1 2,2 phase set 3,3 1,2 1,3 1,3 An Autosomal Dominant Pedigree II U U NR NR  = 0.1 Likelihood of result if linked = (1 - ) x (1 - ) =0.9 x 0.9 = 0.81 Likelihood of result if unlinked = 0.5 x 0.5 = 0.25 LOD = log10 0.81/0.25 = 0.511  = 0 Likelihood of result if linked = (1 - ) x (1 - ) = 1 x 1 = 1 Likelihood of result if unlinked = 0.5 x 0.5 = 0.25 LOD = log10 1.0/0.25 = 0.602

  48. 2,2 1,1 phase set 3,3 1,2 1,3 1,3 An Autosomal Dominant Pedigree III U U 2,3 NR NR R  = 0.1 Likelihood of result if linked = (1 - ) x  x (1 - ) =0.9 x 0.1 x 0.9 = 0.081 Likelihood of result if unlinked = 0.5 x 0.5 x 0.5 = 0.125 LOD = log10 0.081/0.125 = -0.188  = 0 Likelihood of result if linked = (1 - ) x  x (1 - )= 1 x 0 x 1 = 0 Likelihood of result if unlinked = 0. 5 x 0.5 x 0.5 = 0.125 LOD = log10 0/0.125 = -

  49. (1- ) NR x R log10 0.5(NR+R) LOD scores A LOD score is calculated for a pedigree using the following general formula (PROVIDED PHASE OF INHERITANCE IS KNOWN!)

  50. phase unknown An AD Phase Unknown Pedigree U 3,3 1,2 1,3 1,3 PNR 1,3 PNR PNR  = 0 Likelihood of result if linked = ((1 x 1 x 1) + (0 x 0 x 0))/2 = 0.5 Likelihood of result if unlinked = 0. 5 x 0.5 x 0.5 = 0.125 LOD = log10 0.5/0.125 = 0.602  = 0.1 Likelihood of result if linked = ((0.9 x 0.9 x 0.9) +(0.1 x 0.1 x 0.1))/2 = 0.365 Likelihood of result if unlinked = 0.5 x 0.5 x 0.5 = 0.125 LOD = log10 0.365/0.125 = 0.465

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