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HRR Block 1 > Mendelian Genetics II & Hardy Weinberg > Flashcards

Mendelian Genetics II & Hardy Weinberg Flashcards

1
Q

 Baby boy is born with a ventricular septal defect … Can be part of a genetic syndrome or not
 Genetic Counseling impacts what two things?

A

Genetic Counseling Example
 Baby boy is born with a ventricular septal defect … Can be part of a genetic syndrome or not
 Impacts medical management & recurrence risk
1. Features of baby
2. Family history & pedigree
3. Can both be explained by a syndrome?

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2
Q

22q11.2 Deletion Syndrome
 Velocardiofacial (VCF) syndrome or DiGeorge syndrome
 Contiguous gene deletion syndrome on chromosome 22q11.2

what type of inheritance?

Penetrance is complete but effects range from mild to severe =? Variable Expression

A

22q11.2 Deletion Syndrome
 Velocardiofacial (VCF) syndrome or DiGeorge syndrome
 Contiguous gene deletion syndrome on chromosome 22q11.2  AD inheritance
 Penetrance is complete but effects range from mild to severe  Variable Expression
 74% have CHD; 69% have palatal abnormalities; ~70% learning disabilities; ~77% immune deficiency
 Pleiotropy

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3
Q

after confirming 22q11.2 Deletion Syndrome diagnosis with genetic testing….

A

Next steps
 Confirm dx with genetic testing
 Baby at risk for other multisystem problems (growth, immunodeficiency, VPI, feeding, psych)
 Explains mom and uncle’s histories; test and coordinate care
 Future children of mom or uncle at 50% risk
 Future children of baby at 50% risk  May be more or less severe
 Prenatal diagnosis available
 Help with adaptation, resources, referrals and support

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4
Q

note

A

note

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5
Q
  1. Marfan syndrome is a dominant genetic syndrome only caused by FBN1 gene mutations. All who have the gene mutation show some symptoms. Symptoms involve ocular, skeletal and/or cardiovascular systems.
    Which of the following factors are observed?

A. Pleiotropy & Variable expressivity
B. Genetic heterogeneity & Pleiotropy
C. Incomplete penetrance & Genetic heterogeneity
D. A&C

A

A

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6
Q
  1. Marfan syndrome is a dominant genetic syndrome only caused by FBN1 gene mutations. All who have the gene mutation show some symptoms. Symptoms involve ocular, skeletal and/or cardiovascular systems.
    Which of the following factors are observed?

Does the 4yo patient have Marfan syndrome?

     A. No B. Yes C. More information is needed to confirm or exclude a diagnosis
A

More information is needed to confirm or exclude a diagnosis

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7
Q

Example and Clicker Questions:
2. John and Amy have a baby who is diagnosed with Cystic Fibrosis. Neither of them have any family history of cystic fibrosis. Neither of them have had genetic testing.

We expect that:
A. Either John or Amy must have subclinical CF
B. The baby must have two spontaneous CFTR mutations
C. Both parents are carriers of CF

A

c : recessive conditions parents are obligate carriers

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8
Q

blank note

A

We expect that:
A. Either John or Amy must have subclinical CF
B. The baby must have two spontaneous CFTR mutations
C. Both parents are carriers of CF

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9
Q

John and Amy have a baby who is diagnosed with Cystic Fibrosis. Neither of them have any family history of cystic fibrosis. Neither of them have had genetic testing.

John and Amy have carrier testing and both are carriers. What is the risk for their next child to have CF?

                                                A. 1 in 4 (25%) B. 1 in 3 (33%) C. 1 in 2 (50%) D. 3 in 4 (75%)
A

A. 1 in 4 (25%)

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10
Q

Advanced Mendelian Genetics Outline
 2/3 carrier risk for unaffected sibs in AR disease
 New mutation in X‐LR lethal disorders (the second 2/3
rule)
 Carrier risk ‐ when to use the pedigree, and what about when the family history is “negative?”
 Hardy‐Weinberg Equilibrium
• Assumptions
• Forces that alter HWE
• Genotype calculations in AR disease
 Examples

A

Advanced Mendelian Genetics Outline
 2/3 carrier risk for unaffected sibs in AR disease
 New mutation in X‐LR lethal disorders (the second 2/3
rule)
 Carrier risk ‐ when to use the pedigree, and what about when the family history is “negative?”
 Hardy‐Weinberg Equilibrium
• Assumptions
• Forces that alter HWE
• Genotype calculations in AR disease
 Examples

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11
Q

explain Autosomal Recessive ____Rule

A

Autosomal Recessive 2/3 Rule

 Unaffected sibling of someone with AR disease has a 2/3 carrier risk

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12
Q

X‐Linked Recessive _____ Rule

A

X‐Linked Recessive 2/3 Rule
 In a population at equilibrium for a sex‐linked lethal trait (reproductively limiting), 1/3 of the mutations must arise anew each generation
 1/3 of cases occur due to new mutations and 2/3 are inherited
**Assign 2/3 carrier risk to the mother
of the proband, if the only case in the family.

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13
Q

Bayesian Derived Rule

A

Bayesian Derived Rule
 AKA Conditional Probability
 Uses phenotypic information in pedigree to assess relative probability of two or more alternative genotypic possibilities (carrier vs. non‐carrier)
 μ= new mutation rate  Extremely small
 Female Carrier Rate (Rare X‐linked recessive)
 2μ + 1⁄2(2μ) + 1⁄4(2μ) + 1/8(2μ) + 1/16(2μ) +… = 4μ
 This accounts for chance that the female had a new mutation on either X chromosome plus the chance that she inherited a mutation from anyone in the female lineage (as females are unaffected carriers)
 A priori chance to be a non‐carrier, given negative family history = 1‐ 4μ= ~1
 General population (non)carrier risk
 Changed by having affected son

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14
Q

 μ=

A

 μ= new mutation rate  Extremely small

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15
Q

General population (non)carrier risk

A

General population (non)carrier risk

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16
Q 
Prior probability (Female at birth): carrier vs Non-Carrier Conditional Probability (Son is affected): carrier vs Non-Carrier 
Joint Probability: carrier vs Non-Carrier

Posterior Probability: carrier vs Non-Carrier

A 
Prior probability (Female at birth): carrier vs Non-Carrier 
4μ
1-(4μ ) = 1
Conditional Probability (Son is affected)
1/2
μ (new mutation in son)
Joint Probability
2μ
μ
Posterior Probability
2/3
1/3
17
Q

XLR scenarios where mom IS a carrier

A

XLR scenarios where mom IS a carrier

???

18
Q

Pedigree vs. Population Data

A

Pedigree vs. Population Data
• Use pedigree data when informative to find genotype probabilities
• e.g. to find carrier risk when positive family history
• Use population genetics rules or given information to find genotype probabilities when pedigree information is not informative (negative).
• e.g. to find carrier risk when there’s no family history of the disorder

19
Q

Hardy‐Weinberg Equilibrium

A

Hardy‐Weinberg Equilibrium  Population genetics law
 Independently formulated by English mathematician (Hardy) & German physician (Weinberg) in 1908
 Apply to Recessive disorders to:
 Calculate genotype frequencies from allele frequencies
 Estimate heterozygote frequency (carrier rate) in a population
based on disease incidence
 Estimate a person’s carrier risk or probability of disease occurrence when family history is negative

20
Q

Hardy‐Weinberg Law- assumptions

A

Hardy‐Weinberg Law
The Hardy‐Weinberg Law states that under conditions assumed to exist in large human populations …
The frequencies of homozygous dominant AA, heterozygous Aa and homozygous recessive aa genotypes remain stable from generation to generation
Where p represents the allele frequency of A, q represents the allele frequency of a and

***p+q=1

21
Q

Assumptions of the Hardy‐Weinberg Law For such “genetic equilibrium” to exist

A

Assumptions of the Hardy‐Weinberg Law For such “genetic equilibrium” to exist
 Mating must be random i.e., mate selection should not be influenced by genotype
 New mutation should be rare enough to ignore
 All genotypes should be equally fit to contribute to the gene pool (reproduce)

22
Q

Forces that Alter the Equilibrium of Alleles

A

Forces that Alter the Equilibrium of Alleles (so i think we assume for hardy that none of these exists)

● Preferential mating: mate selection on the basis of particular phenotypic attributes
● Mutations
● Mortality
● Infertility
● Immigration and emigration
● Genetic drift: random fluctuation of frequencies in small populations
● Founder effect (e.g. Amish, Ashkenazi Jewish)
● Heterozygote Advantage (e.g. sickle cell trait and malaria)

23
Q

HWE Genotype & Phenotype Frequencies

A

HWE Genotype & Phenotype Frequencies
• The frequency of the three genotypes AA, Aa and aa is given by the terms of the binomial expansion of (p + q)2
• (p+q)x(p+q)=(p2)+(2pq)+(q2)
• The sum of all alleles in the population = 100%, and sum of all genotypes in the population = 100%
• p+q=1
• (p2)+(2pq)+(q2)=1

*q2 is what you can see & count; affected, observable phenotype
(q2 = disease incidence)
Genotype vs HWE term
AA: p^2 
Aa :2pq 
aa: q^2
24
Q

Hardy Weinberg Equilibrium

p + q = 1 (“the sum of all alleles in the population =

A

Hardy Weinberg Equilibrium
p + q = 1 (“the sum of all alleles in the population = 100%”)
p2 + 2pq + q2 = 1 (“the sum of all genotypes in the population = 100%”)
 Remember, a genotype is the combination of 2 alleles  Using HWE
1. Identify which term the question asks you to solve for (e.g. a question asking for a carrier rate is asking you to solve for the 2pq term using given information)
2. Identify which terms are given (e.g. a disease incidence gives you q2)
3. Identify which terms you can solve for based on given information (e.g. you can find q if you know q2, you can find p if you know q…)

25
Q

HWE: Genotypes

A 
HWE
Genotypes
p2 +2pq+q2 =1
 p2 = homozygous dominant AA
 2pq= heterozygous Aa = carrier rate
 q2 = homozygous recessive aa = disease incidenc
26
Q

HWE: Alleles

A 
Alleles
p+q=1
 p = frequency of dominant
allele in population
 q = frequency of recessive (disease) allele(s) in population
27
Q

How we estimate carrier frequencies using the Hardy‐Weinberg Law
Phenylketonuria (PKU) as an example
PKU alleles are distributed in the population as follows
Normal allele (A) has frequency of p
Mutant allele (a) has frequency of q

Incidence of PKU genotype (i.e., aa with frequency q2) is 1/10,000

Frequency of PKU carriers is :

A

How we estimate carrier frequencies using the Hardy‐Weinberg Law
Phenylketonuria (PKU) as an example
PKU alleles are distributed in the population as follows
Normal allele (A) has frequency of p
Mutant allele (a) has frequency of q
the sum of p + q = 100% of PKU alleles in the population
Assuming genetic equilibrium, various genotype frequencies will be AA=p2 + Aa=2pq + aa=q2
Incidence of PKU genotype (i.e., aa with frequency q2) is 1/10,000
Therefore q = the square root of 1/10,000 = 1/100 p = 1‐ (1/100) = 99/100 or approximately 1
Frequency of PKU carriers is 2pq = 2 x 1 x (1/100) = 1/50

28
Q

Example steps and tricks for HWE in rare disease
 Incidence is given = q2

*if q=1/100then 2pq=?

A

Example steps and tricks for HWE in rare disease
 Incidence is given = q2  Solve for q
• Square root of q2  Solve for p
• 1–q
• Simplify to “1” (disorder is sufficiently rare)  Find carrier rate (2pq)
• 2xpxq=2x~1xq  Simplifies to 2 x q
 Tip: keep q in fraction form for easiest use in pedigree problems
o e.g.ifq=1/100then2pq=2x1x1/100=2/100reduceto1/50
• Alternate trick: you can divide the denominator of q by 2 to get the same answer for 2pq
• e.g. where q = 1/100, 2pq = 1/50

29
Q

Putting It Together
 Some reproductive risk problems require you to use HWE AND pedigree data
 Pedigree data informs for
 Use pedigree data to determine
 If someone does not have a family history of a condition in their bloodline,
 Use __ and ____ to assess risk to offspring

A

Putting It Together
 Some reproductive risk problems require you to use HWE AND pedigree data
 Pedigree data informs for one partner (affected or family history) and population data for the other (negative family history)
 Use pedigree data to determine genotypes or genotype probabilities, where possible
 If someone does not have a family history of a condition in their bloodline, use given information or HWE to figure out their genotype probabilities
 Use genotypes of parents and pattern of inheritance to assess risk to offspring

30
Q

Examples ‐ #1
1. A couple is considering having children. The wife has PKU, a recessive disease, and the husband has no known family history of PKU. The carrier rate of PKU is 1/50. What is the chance they will have a child with PKU?

A

Examples ‐ #1
1. A couple is considering having children. The wife has PKU, a recessive disease, and the husband has no known family history of PKU. The carrier rate of PKU is 1/50. What is the chance they will have a child with PKU?
 Q= chance for child with PKU?
 PKU is recessive (will have if inherit mutation from both parents)
 Background carrier rate is given (dad)
 Mom’s genotype is given (affected)

31
Q 
#1 continued
 A couple is considering having children. The wife has PKU, a recessive disease, and the husband has no known family history of PKU. The carrier rate of PKU is 1/50. What is the chance they will have a child with PKU?
A 
#1 continued
 A couple is considering having children. The wife has PKU, a recessive disease, and the husband has no known family history of PKU. The carrier rate of PKU is 1/50. What is the chance they will have a child with PKU?
1/50 chance Aa
aa ? = chance of aa
Wife’s risk to pass on mutation= 100% Husband’s risk to be a carrier= 1/50 (pop. risk)
• If carrier, 50% chance to pass on mutation (Female’s chance to have mutation x chance to pass it on) x
(Male’s chance to have mutation x chance to pass it on) (1x1)x(1/50x1⁄2)= 1/100
32
Q

Examples ‐ #2

  1. A healthy man has a brother with CF and his wife has no family history of CF. The incidence of CF is 1/2500.
    a) What is the chance that the man is a carrier?
A

Examples ‐ #2
2. A healthy man has a brother with CF and his wife has no family history of CF. The incidence of CF is 1/2500.
a) What is the chance that the man is a carrier?
 2/3
P
 Note AR unaffected sib 2/3 rule informed by family history (pedigree data)

33
Q
  1. A healthy man has a brother with CF and his wife has no family history of CF. The incidence of CF is 1/2500.
b) What is the chance his wife is a carrier?  Recall incidence of CF is given: 1/2500
 Wife is “general population” – no fhx
Use HWE:
p2+2pq+q2 =1
q2= incidence= 1/2500 q= 1/50
p+q=1
p=~1
2pq= (2 x 1 x 1/50) = 1/25
A 
b) What is the chance his wife is a carrier?  Recall incidence of CF is given: 1/2500
 Wife is “general population” – no fhx
Use HWE:
p2+2pq+q2 =1
q2= incidence= 1/2500 q= 1/50
p+q=1
p=~1
2pq= (2 x 1 x 1/50) = 1/25
34
Q
  1. A healthy man has a brother with CF and his wife has no family history of CF. The incidence of CF is 1/2500.
    c) What is the chance they have a child with CF?
A

c) What is the chance they have a child with CF?
 We know CF is AR
(2/3 x 1⁄2) x (1/25 x 1⁄2) = 1/150
Or 2/3 x 1/25 x 1⁄4 = 1/150

i.e. dad’s carrier chance x chance to pass on mutation x mom’s carrier chance x chance to pass on mutation
The 1⁄4 in AR “both parents carriers” problems encompasses chance for both parents to pass on mutation

35
Q

 In this type of problem there was informative pedigree data for one member of the couple’s carrier risk.
 You used HWE and given population data for the other partner.
 You used patterns of inheritance and allele tracking to find risk to their offspring.

A

 In this type of problem there was informative pedigree data for one member of the couple’s carrier risk.
 You used HWE and given population data for the other partner.
 You used patterns of inheritance and allele tracking to find risk to their offspring.

36
Q

A woman had a brother who died from DMD (X‐linked recessive). She and her husband want to know what the risk is for their baby to have DMD as well.
What is the chance that the woman is a carrier?

A

A woman had a brother who died from DMD (X‐linked recessive). She and her husband want to know what the risk is for their baby to have DMD as well.
What is the chance that the woman is a carrier?
1/3
 remember 1/3 cases are new mutations
 2/3 carrier risk to mother of single case
 Use X‐LR inheritance and allele tracking to find carrier risk for woman in question (1/2 her mother’s ris

2/3 * 1/2

37
Q

What is the chance that her child will be affected with DMD?

A

What is the chance that her child will be affected with DMD?
1/3x1/2 x1/2= 1/12
Her Chance carrier she
risk passes on
mutation (versus normal X)
Chance child is male!
Read and think carefully – only males affected in X‐LR.

38
Q

3 FAQs

 Q: But what about the baby’s dad?
 Remember risk for X‐LR affected child depends on mom’s
carrier status
 All boys get their dad’s Y – no male to male transmission
 Q: The gender part confused me.
 In early reproductive risk assessment gender is 50‐50
chance, hence the extra 1⁄2 term when only M affected
 If problem stated “ultrasound showed baby is a boy – what
is the chance that HE is affected” then answer would be 1/6.

A

3 FAQs

 Q: But what about the baby’s dad?
 Remember risk for X‐LR affected child depends on mom’s
carrier status
 All boys get their dad’s Y – no male to male transmission
 Q: The gender part confused me.
 In early reproductive risk assessment gender is 50‐50
chance, hence the extra 1⁄2 term when only M affected
 If problem stated “ultrasound showed baby is a boy – what
is the chance that HE is affected” then answer would be 1/6.

39
Q
  • Use pedigree data when :

* Use population genetics rules or given information to find genotype probabilities when :

A
  • Use pedigree data when informative to find genotype probabilities
  • e.g. to find carrier risk when positive family history
  • Use population genetics rules or given information to find genotype probabilities when pedigree information is not informative (negative).
  • e.g. to find carrier risk when there’s no family history of the disorder

HRR Block 1 (6 decks)

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