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TOPICS IN QUANTITATIVE AND POPULATION GENETICS GENETICS 202 Jon Bernstein Department of Pediatrics October 6, 2015 Session Goals Understand and demonstrate how the Hardy-Weinberg distribution can be used to estimate carrier frequencies from disease incidence for autosomal recessive conditions. Recognize that different individuals have different perceptions of quantitatively similar risks. Understand how and to what extent consanguinity can increase the risk of autosomal recessive conditions. Session Goals Understand how Bayes Theorem can be used to revise a risk estimate based on new clinical information ◦ (short video available) Lecture Outline Clinical Case ◦ Hardy-Weinberg Distribution ◦ Consanguinity – Identity by descent ◦ Bayes Theorem Updating a risk estimate using new information Clinical case: A father in a first cousin marriage is a carrier for ARPKD. They have had had 1 healthy child. ◦ How does the presence of consanguinity and the health child affect the risk estimate for the mother to be a carrier? Aa ?? ? ?? Autosomal Recessive Polycystic Kidney Disease Incidence ◦ Approximately 1 in 15,000 Mutations in PKHD1 on chromosome 6p Classically features multiple bilateral kidney cysts at a young age and hepatic fibrosis Avni, 2002, Pediatric Radiology, Hereditary polycystic kidney diseases in children: changing sonographic appearances through childhood, PMID: 12164348 Autosomal Recessive – Recurrence Risk – The generic problem What is the chance that each parent is a carrier and passes on a mutation? Aa (_ x _) x( _ x _) Aa ? aa (1 x 1/2) x( 1 x 1/2)=1/4 ?? Autosomal Recessive – Recurrence Risk What if one parent is known to be a carrier and the genetic status of the other is unknown? Aa ?? ? (_ x _) x ( _ x _) (1 x 1/2) x( ? x 1/2)=? ?? Population Frequency and Risk to be a Carrier If the incidence of a recessive disease is 1 in 5,000; is the risk for the circled family member to be a carrier higher or lower than if the incidence were 1 in 15,000? Aa ?? ? ?? How can we estimate the risk that an individual is a carrier? In general, each person has two copies (alleles) of each autosomal gene. A=dominant allele a=recessive allele Possible genotypes at a given locus AA Aa aA aa These can be considered together How can we estimate the risk that an individual is a carrier? The population of people has a population of alleles – the gene pool. Under certain conditions one can assume that the two alleles at a locus in an individual are selected at random from the gene pool. AA a a a A A The Gene Pool How can we estimate the risk that an individual is a carrier? If we select one allele at random the chance that it will be “A” or “a” is dependent on the relative frequency of these alleles in the gene pool. _ AA a a a A A The Gene Pool How can we estimate the risk that an individual is a carrier? What if we are selecting two alleles from the gene pool? There are four possible combinations (3 genotypes) AA Aa Can be considered together aA aa __ AA a a a A A The Gene Pool How can we estimate the risk that an individual is a carrier? What are the chances of having each of these genotypes? AA Aa aA aa Chance of AA is: Probability of drawing A twice in a row. If “p” is the probability of drawing an “A” then the chance of drawing two “A”’s in a row (AA) is p x p=p2 How can we estimate the risk that an individual is a carrier? What are the chances of having each of these genotypes? AA Aa aA aa Chance of aa is: Probability of drawing “a” twice in a row. If q is the probability of drawing an “a” then the chance of drawing two “a”’s in a row (aa) is q x q=q2 How can we estimate the risk that an individual is a carrier? What are the chances of having each of these genotypes? AA Aa aA aa Can be considered together Chance of drawing Aa or aA is: Probability of drawing “A” and then “a” OR “a” and then “A” Chances of drawing Aa is p x q=pq Chances of drawing aA is q x p=pq Chances of Aa or aA is pq + pq=2pq How can we estimate the risk that an individual is a carrier? At a given locus, in the case of two alleles ◦ p=frequency the dominant allele, A ◦ q=frequency of the recessive allele, a p2 is the frequency of homozygotes, AA q2 is the frequency of homozygotes, aa 2pq=frequency of heterozygotes, Aa or aA, (collectively referred to as Aa) The Hardy-Weinberg Distribution ◦ p2+2pq+q2 ◦ (p+q)(p+q) Autosomal Recessive – Recurrence Risk Carrier frequency in population = 2pq ◦ This is also the risk for someone with no family history of the condition to be a carrier If (p+q)(p+q)=p2+2pq+q2=1 and the recessive condition is rare 2pq 2 X 1 x 𝐼𝑛𝑐𝑖𝑑𝑒𝑛𝑐𝑒 𝑜𝑓 𝑡ℎ𝑒 𝑐𝑜𝑛𝑑𝑖𝑡𝑖𝑜𝑛 Autosomal Recessive – Recurrence Risk What is the risk to the next child if one parent is known to be a carrier and the genetic status of the other is unknown? (_ x _) x ( _ x _) (1 x 1/2) x( ? x 1/2)=? Aa ?? ? ?? Carrier frequency for ARPKD Incidence = 1 in 15,000 2pq2(1)( 1 / 15,000 )=1/61 Autosomal Recessive – Recurrence Risk - ARPKD What is the risk to the next child if one parent is known to be a carrier and the genetic status of the other is unknown? Aa (_ x _) x ( _ x _) (1 x 1/2) x( 1/61 x 1/2)=1/244 ?? ? ?? 1 in 5,000 versus 1 in 15,000 If the population frequency of a recessive disease is 1 in 5,000 ◦ 2pq 2(1)( 1/5,000)= 1/35 = Carrier Frequency If the population frequency of a recessive disease is 1 in 10,000 ◦ 2pq 2(1)( 1/15,000)=1/61 = Carrier Frequency We advise the couple that they have a risk of 1/244 to have a child with ARPKD How might they react? What factors might influence this reaction? We advise the couple that they have a risk of 1/244 to have a child with ARPKD What factors might influence this reaction? ◦ Severity of the disease Individual’s impression of the severity and impact of the disease ◦ Prior notion of risk was higher or lower ◦ Patient’s understanding of probability and how the information is delivered 1/244 risk of affected, 0.4% risk (Negative framing) 243/244 chance of unaffected, 99.6% chance (Positive framing) What if there are three alleles (A, B and O) There are nine possible combinations (6 genotypes) AA AO Can be considered together OA OO AB Can be considered together BA BB BO Can be considered together OB __ AB A O O B A The Gene Pool What if there are three alleles? The allele frequencies can be considered as p, q and r for A, B and O respectively __ A OA B O B A The distribution of genotypes will be p2+2pq+2pr+2qr+q2+r2 (p+q+r)(p+q+r) BB AA AB AO BO OO The Gene Pool What are assumptions in using the Hardy-Weinberg distribution? The populations of alleles is stable, the population is very large and mating is random ◦ ◦ ◦ ◦ Assortative (random) mating No migration No new mutations No selection Why is it ok to assume no selection for autosomal recessive traits? Fraction of alleles 2(q2)/(1(2pq)+2(q2)) = q Derivations of fraction of alleles exposed to selection q/1(2pq)+2(q2)+q = 1/2p+2q+1= 1/3 New Clinical Genetics 2e Andrew Read and Dian Donnai ISBN: 9781904842804 © Scion Publishing Ltd, 2011 When can selection operate effectively on a recessive trait? Heterozygote advantage ◦ When heterozygous carriers of a recessive allele have greater relative fitness, than either class of homozygotes Relative Fitness The probability that individuals with a particular genotype will successfully reproduce relative to those of another genotype Autosomal Recessive – Recurrence Risk with Consanguinity If the parents are related does that increase the risk their child will be affected by a recessive disease? (_ x _) x ( _ x _) (1 x 1/2) x( ? x 1/2)=? Aa ?? ? ?? Consanguinity • Clinical definition of consanguinity • The union of individuals who are related to each other as second cousins (5th degree relatives) or closer. http://www.consang.net First, Second and Third Degree Relatives 2 2 2 Grandmother Grandfather 2 Aunt 3 First cousin 2 Grandfather 1 1 Father Mother 1 Patient 2 Uncle 1 Brother Sister 2 1 Niece Son 1 Daughter Grandmother 2 Nephew 3 First cousin Consanguinity – and the risk of recessive disease The more closely related the parents, the greater the increase in risk ◦ The risk decreases the more distantly related the parents are ◦ The risk is proportional to the fraction of genes which are shared by the parents Identity by descent Autosomal Recessive – Recurrence Risk If the parents are related does that increase the risk their child will be affected? (_ x _) x ( _ x _) (1 x 1/2) x( ? x 1/2)=? How closely related are the parents? Aa ?? ? ?? If the parents are first cousins…. (1 x 1/2)x(1/8 x 1/2)=1/32 1/2 1/2 1/2 Compare risk of affected child to that estimated by H-W ?? Aa ? ?? Identity by Descent and the Coefficient of inbreeding Identity by descent ◦ When both copies of an allele in an individual are identical because they were inherited from a common ancestor Coefficient of inbreeding ◦ The fraction of a person’s genes that are identical by descent ◦ The probability that a person is identical by descent at a particular locus Cc ?? ? cc Identity by Descent – First Cousins The chance that allele A1 will be inherited by individual IV1 in a homozygous state is: ◦ (1/2x1/2x1/2)x(1/2x1/2x1/2)=1/64 The chance that allele A1, A2, A3 or A4 will be inherited by individual IV-1 in a homozygous state is 4x1/64=1/16 1/2 1/2 1/2 1/2 1/2 1/2 Thompson and Thompson, Genetics in Medicine, 7th Edition, 2007 If the parents are second cousins…. (1 x 1/2)x(1/32 x 1/2)=1/128 1/2 1/2 1/2 1/2 1/2 ?? Aa ? ?? Fraction of genes shared in consanguineous unions Thompson and Thompson, Genetics in Medicine, 7th Edition, 2007 What if there is more than one consanguineous union in the family? New Clinical Genetics 2e Andrew Read and Dian Donnai ISBN: 9781904842804 © Scion Publishing Ltd, 2011 Clinical case: A father in a first cousin marriage is a carrier for ARPKD. They have had had 1 healthy child. ◦ How does the presence of consanguinity and the healthy child affect the risk estimate for the mother to be a carrier? Aa ?? ? ?? Modifying a risk using additional information Given a risk of a particular event, this risk can be modified using Bayes’ theorem to take into account additional information ◦ Risk of developing a disease can be modified by Family History Test Results Age (with age dependent penetrance) Autosomal Recessive Recurrence Risk What is the chance that the patient’s unaffected brother is a carrier? At conception there are four possible pairs of alleles he could inherit with equal probability. ◦ AA, Aa, aA, aa But we know that he is not aa (conditional probability of zero), so 2/3rds risk of being a carrier Unaffected AA Aa Aa Aa ? aa ?? Affected aA aa = brother is a carrier Given genotype what is the chance of no symptoms in first child, and given no symptoms what is chance of each genotype No symptoms Genotype 1/4 0 1 aa 1 Aa 0/4 aa and symptoms 1/4 Aa and no symptoms 2/4 Aa and symptoms 0/4 AA and no symptoms 1/4 AA and symptoms 0/4 0 1/2 1/4 aa and no symptoms 1 AA 0 Incorporating additional information into risk estimates If the couple (first cousin example) has had 1 healthy child. ◦ Does this increase or decrease the probability that the mother is a carrier? Aa ?? ? ?? Given mother a carrier, chance for first child to be unaffected Affected 1/32 Mother carrier and child affected 1/4 C 3/4 1/8 7/8 NC 3/32 Mother carrier and child unaffected Not affected Affected 0/32 Mother not a carrier and child affected 0 1 Not affected 28/32 Mother not a carrier and child unaffected Given first child unaffected, chance for mother to be a carrier Affected 1/32 Mother carrier and child affected 1/4 C 3/4 1/8 7/8 NC 3/32 Mother carrier and child unaffected Not affected Affected 0/32 Mother not a carrier and child affected 0 1 Not affected 28/32 Mother not a carrier and child unaffected Example of Risk Adjustment Using Bayes Method Mother is a carrier Mother is not a carrier Prior Probability 1/8=.125 7/8=.875 Conditional Probability 3/4 1/1 Joint Probability 3/32 7/8=28/32 Final/Posterior Probability (3/32) / (3/32+28/32) = 3/31 =.1 (28/32)/(3/32+28/32) = 28/31=.90 Aa ?? ? aa Students will not be asked to perform this calculation on the final exam ?? Visual Depiction of Conditional Probabilities Mother is a Carrier Mother is Not a Carrier A U U U U U U U U U U U U U U U U U U U U U U U U U U U U U U U + 1/4 - 3/4 Aa ?? ? aa ?? = brother is a unaffected ¾ of children will be unaffected - 28/28 All children will be unaffected Incorporating additional information into risk estimates If the couple has had 2 healthy children. Aa ◦ Does this increase or decrease the probability that the mother is a carrier? ?? ? aa ?? What if there are two unaffected children? Mother is a carrier Mother is not a carrier Prior Probability 1/8=.125 7/8=.875 Conditional Probability 3/4 1/1 Conditional Probability #2 3/4 1/1 Joint Probability 9/128 7/8=112/128 Final/Posterior Probability (9/128) / (9/128+112/128) = 9/121 =.07 (112/128)/(9/1+112/128)= 112/121=.93 Aa ?? ? aa Students will not be asked to perform this calculation on the final exam ?? Lecture Summary The Hardy-Weinberg distribution can be used to estimate carrier frequencies for autosomal recessive conditions given the incidence of the condition Consanguinity can increase the risk of autosomal recessive conditions ◦ The degree of increased risk is related to the degree of consanguinity of the couple ◦ The increase in risk falls off quickly for relationships more distant than second cousins Lecture Summary Bayes’ Theorem can be used to update risk estimates to take into account new clinical information ◦ Outcomes that are unlikely generally require stronger evidence than those that are likely Review Question A severe autosomal recessive condition is less subject to selection than an equally severe autosomal dominant condition because ◦ A) Dominant conditions are more severe than recessive conditions ◦ B) Both parents need to be carriers for a recessive condition to occur ◦ C) Most of the recessive alleles are present in healthy carriers ◦ D) There can be variable expressivity in a dominant condition Review Question The fraction of parental alleles inherited in common by siblings is ◦ ◦ ◦ ◦ A) 0 B)1/4 C)1/2 D)3/4 Synthesis Question If a patient is suspected to have a genetic disease and undergoes a DNA test that is negative. The chances that he is affected by the disease are? ◦ A) 100% ◦ B) 25% ◦ C) 0% ◦ D) It depends on the characteristics of the test ◦ E) It depends on the characteristics of the test and the probability that the patient might have the disease before having the test