Probability
Probability definition Probability of an event P(A) denotes the frequency of it s occurrence in a long (infinite) series of repeats P( A) = A Ω Ω A 0 P It s s represented by a number in the range from 0 (event never occurs) to (event occurs every time) Probability 0.25 (/4) indicates that certain event is observed in from 4 or 25% occurrences
Rules of probability Rule of addition If events are mutually exclusive, then the probability of either event occurring is a sum of the individual probabilities of the occurrence of any of the events P of giving birth to a boy or girl by a pregnant woman: / 2 + / 2 = P of rolling 2 or 3 on six-sided die: / 6 + / 6 = 2 / 6 = / 3
Rules of probability Rule of multiplication If two or more events are independent, then their joint probability is a product of their individual probabilities P of rolling a pair of 6 s on two six-sided dice simultaneously : / 6 / 6 = / 36 P of giving birth to two boys by a woman in binovular pregnancy: / 2 / 2 = / 4
Autosomal recessive diseases Albinism
Autosomal recessive pattern of inheritance Parents carriers of recessive trait (e.g. cystic fibrosis): What is P: affected homozygote? healthy carrier (heterozygote)? / 2 / 2 / 2 / 2 / 2 / 2 / 2 / 2 / 4 / 4 / 4 + / 4 = / 2 Please pay attention: is the question about a phenotype or a genotype?!
Autosomal recessive pattern of inheritance Parents carriers of recessive trait: risk of having an affected child : / 2 / 2 = / 4 probability of having an unaffected child: ( / 2 / 2 ) + ( / 2 / 2 ) + ( / 2 / 2 ) = 3 / 4 healthy homozygote heterozygote (sum of two mutually exclusive events)
Autosomal recessive pattern of inheritance Mucoviscidosis A pair of unrelated Caucasians with negative medical history of cystic fibrosis Risk of having an affected child? Frequency of carriers of disease-causing allele in population: / 25 The risk of having an affected child is the product of three (four) independent events : / = / 25 25 / 2 / 2 / 2500
Mr Smith has two children What's the probability that both of them are boys? Two independent events: / 2 / 2 = / 4 or: four possibilities => / 4
Autosomal dominant pattern of inheritance A couple of one healthy person and one affected by a dominant disease (e.g. Marfan syndrome) A * a A * A * aa (2 normal alleles!) (statistically 50% of their children would be affected) Why 50% and not 00% homozygotes often lethal (not in every disease) additionally: a simplification for teaching purposes
Autosomal dominant pattern of inheritance A couple of one healthy person and one affected by a dominant disease (e.g. Marfan syndrome) A * a A * A * aa (2 normal alleles!) (statistically 50% of their children would be affected) What is, in fact, P that all three would be affected? / 2 / 2 / 2 = / 8
Autosomal dominant pattern of inheritance What s the probability that two out of three will be affected? / 2 / 2 / 2 = / 8? P of any of them being affected = P of any of them being healthy
Autosomal dominant pattern of inheritance A couple of one healthy person and one affected by a dominant disease (e.g. Marfan syndrome) A * a A * A * aa (2 normal alleles!) (statistically 50% of their children would be affected) What is P that two out of three would be affected? ( / = 2 / 2 / 2 ) / 3 8 =? 3 / 8 BUT... (rule of addition)
Risk of recurrence One of the most important aspects of genetic counselling is to assess a genetic risk (risk of recurrence) Easy in mendelian diseases Empirical in polygenic diseases and in chromosomal abnormalities
X-linked recessive inheritance pattern e.g. Duchenne muscular dystrophy Risk of fourth baby being affected?
Bayes theorem P ( A B) = P( A) P( B P( B) A) Determination of individual risk of carrier state by combining the data from the pedigree with other data after taking into account all modifying factors
Bayes theorem example
Bayes theorem example Two assumptions (mutually exclusive)
Bayes theorem example Two assumptions II2 is a carrier II2 is not a carrier
Bayes theorem example Probability that II2 is a carrier II2 is not a carrier
Bayes theorem example Probability that II2 is a carrier II2 is not a carrier a priori /2 /2
Bayes theorem example Probability that II2 is a carrier II2 is not a carrier a priori /2 /2 conditional (3 healthy sons) /8 (/2 /2 /2) ()
Bayes theorem example Probability that II2 is a carrier II2 is not a carrier a priori /2 /2 conditional (3 healthy sons) joint odds /8 (/2 /2 /2) /6 (/2 /8) () /2 (/2 )
Twierdzenie Bayesa przykład Probability that II2 is a carrier II2 is not a carrier a priori /2 /2 conditional (3 healthy sons) /8 (/2 /2 /2) () joint odds final risk (joint odds divided by the sum of joint odds) /6 (/2 /8) 6 /2 (/2 ) /9 8/9 6 + 2
Reduced penetrance Penetrance - proportion of individuals carrying dominant allele that also express particular trait (70% or 0.7) Penetrance is said to be reduced or incomplete when some individuals fail to express the trait, even though they carry the disease-causing allele. Risk of inheriting a dominant disease: P = / 2 pen Risk for the child whose parent suffered from retinoblastoma (pen=0,8): / 2 0.8 = 0.4
Reduced penetrance Retinoblastoma pen = 0.8 Risk of a baby: A) being affected? B) being sick?
Bayes theorem reduced penetrance Probability that II2 is a carrier II2 is not a carrier
Bayes theorem reduced penetrance Probability that II2 is a carrier II2 is not a carrier a priori /2 /2
Bayes theorem reduced penetrance Probability that II2 is a carrier II2 is not a carrier a priori /2 /2 conditional (healthy) - pen
Bayes theorem reduced penetrance Probability that II2 is a carrier II2 is not a carrier a priori /2 /2 conditional (healthy) - pen joint odds ½ (-pen) /2
Bayes theorem reduced penetrance Probability that II2 is a carrier II2 is not a carrier a priori /2 /2 conditional (healthy) - pen joint odds ½ (-pen) /2 final risk (joint odds divided by the sum of joint odds) 2 2 ( pen) ( pen) + pen 2 pen = 6 2 pen = 2 pen pen= 0,8
Reduced penetrance Retinoblastoma pen = 0.8 Risk of disease for III while pen=0,8: /6?
Reduced penetrance Retinoblastoma pen = 0.8 Risk of disease for III while pen=0,8: pen 2 6 2 4 5 P II = = 5 /6?
Odds Odds - describe probability as a ratio (proportion) Odd( A) = P( A) P( A) Odds take higher values than probability For example, if probability of getting ill is /5, the odds of getting ill is: odds= 5 5 = 4 5 5 = 4
Odds vs. probability Frequency of blue balls: 4/0 = 2/5 (two out of five balls are blue) Proportion of blue balls vs. yellow balls: 4:6 = 2:3 (2 to 3)
Odds vs. probability Probability of drawing a blue ball: 2/5 (two blue balls in five attempts) Odds of drawing a blue ball: 2:3 (in five attempts two ball will be blue and three will be yellow)