l6 b terminology

8/10/2019 L6 B Terminology

1/42


2/42

Terminology

Pleiotropy

Locus heterogeneityAllelic heterogeneity

Variable expressivity

Reduced penetrance


3/42


4/42

Hemophilia:two different genes on X chromosome, encoding

factor F8 and factor F9 (hemophilia A andhemophilia B)

Retinitis pigmentosa , a common cause of visualimpairment due to photoreceptor degeneration

14 autosomal dominant forms24 autosomal recessive forms

5 X-linked forms

1 Y-linked form

Locus (gene) heterogeneity

Mutations in different genes may cause the

same clinical phenotype, e.g.


5/42

Cystic fibrosis (CF)

> 1400 disease causing mutation in the CFTR genePhenylketonuria

> 400 disease causing mutations in the PAH gene

Allelic heterogeneity

The existence of many different diseasecausing alleles at a given locus resulting in

the same disease


6/42

Locus heterogeneity :

different genes the same disease

Allelic heterogeneity:

different mutations in the same gene the same disease

Locus heterogeneityversus

allelic heterogeneity


7/42

The extent to which a genetic defect is expressed.

If a disease shows variable expression, the symptoms can

vary from mild to severe, but are never completely

unexpressed in individuals who have the appropriate

genotype.

For example, neurofibromatosis type 1 (NF1) can be

manifested just by several cafe-au-lait spots in mild

cases, but thousands of neurofibromas in severe cases

Variable expressivity


8/42

Neurofibromatosis type 1

cafe-au-lait spots

Multiple neurofibromas


9/42

Penetrance is

the fraction of individuals with a genotype known to

cause a disease that actually have any signs or

symptoms of the disease.

Penetrance = 80%

among individuals with mutant genotype 80% reveal

disease (are affected) while 20% remains unaffected

Reduced penetrance means that not all individuals

who have mutant genotype have also the disease

Penetrance


10/42

Ectrodactyly: lobster claw syndrome; claw-like; split hand/split foot malformation

Ectrodactyly on both feet

of a one-year old child
http://upload.wikimedia.org/wikipedia/commons/0/04/Deux_pieds_1_an.jpghttp://upload.wikimedia.org/wikipedia/commons/0/04/Deux_pieds_1_an.jpghttp://upload.wikimedia.org/wikipedia/commons/0/04/Deux_pieds_1_an.jpg


11/42

Pedigree of autosomal dominant split-handdeformity demonstrating failure of penetrance in

the mother of the consultand (arrow).


12/42

Variable expressivity versus

reduced penertance

Variable expressivity: disease

Reduced penetrance: no disease


13/42

Risk calculation for autosomal dominantdiseases with reduced penetrance

The husband is affected with autosomaldominant disorder which is characterized by

reduced penetrance (80%) . His spouse is healthy.What is the risk for their offspring to have thedisease?

Risk of getting mutation from father = 0.5 (50%)

Risk of having disease = Risk of mutation x penetrance=0.5 x 0.8 = 0.40 (40%)

One affected parent


14/42

Risk calculation for autosomal dominantdiseases with reduced penetrance

Two individuals who both have autosomaldominant disorder mate. The disease is

characterized by reduced penetrance (80%).What is the risk for their offspring to have the

disease?

Risk of getting mutant genotype= 0.75 (75%)

Risk of having disease = Risk of mutation x penetrance=0.75 x 0.8 = 0.60 (60%)

Both parents are affected


15/42

Neurofibromatosistype 1(NF1)


16/42

Autosomal dominantMultiple caf-au-lait spots

Neurofibromas

Lish nodules

Neurofibromatosis type 1 (NF1)

Disease is caused by mutation in NF1 gene, which is atumor suppressor geneNo common mutationMost mutations are unique to a particular familyNF1 is a large gene (>350 Kb) with 60 exons


17/42


18/42

Lisch nodules. Clinical photograph of the iris in a patientwith NF1 shows yellow and brown nodules (arrows) thatare elevated above the iris surface and affect the

entire area of the iris.

Lisch nodules
http://radiographics.rsna.org/content/25/2/455/F3.large.jpg


19/42

Neurofibromas


20/42


21/42

Mutations in the NF1gene

NF1 is a tumor suppressor gene

More than 1,000 NF1mutations that causeneurofibromatosis type 1 have been identified. Mostof these mutations are unique to a particular family.


22/42

Exercises


23/42

In assessing a patient with osteogenesis imperfecta, a

history of bone fractures, as well as blue sclerae, isnoted. These findings are an example of which of the

following?

A. Allelic heterogeneity

B. Gain-of-function mutation

C. Locus heterogeneity

D. Multiple mutations

E. Pleiotropy

Q1


24/42

In assessing a patient with osteogenesis imperfecta, a

history of bone fractures, as well as blue sclerae, isnoted. These findings are an example of which of the

following?

A. Allelic heterogeneity

B. Gain-of-function mutation

C. Locus heterogeneity

D. Multiple mutations

E. Pleiotropy

A1


25/42

A 40-year-old man with neurofibromatosis type 1 hasa plexiform neurofibroma of his leg that has caused

significant disfigurement, and hundreds of cutaneousneurofibromas. His 43-year-old sister has cafe-au-laitspots and Lisch nodules of the iris, but noneurofibromas. His 19-year-old son has cafe-au-laitspots, a dozen cutaneous neurofibromas, and scoliosis(curvature of the spine). Choose one of the followingterms that best matches a clinical vignette.

A. allelic heterogeneity

B. incomplete penetranceC. locus heterogeneity

D. variable expressivity

Q2


26/42

A 40-year-old man with neurofibromatosis type 1 hasa plexiform neurofibroma of his leg that has caused

significant disfigurement, and hundreds of cutaneousneurofibromas. His 43-year-old sister has cafe-au-laitspots and Lisch nodules of the iris, but noneurofibromas. His 19-year-old son has cafe-au-laitspots, a dozen cutaneous neurofibromas, and scoliosis(curvature of the spine). Choose one of the followingterms that best matches a clinical vignette.

A. allelic heterogeneity

B. incomplete penetranceC. locus heterogeneity

D. variable expressivity

A2


27/42

Ectrodactyly causes missing middle fingers (lobsterclaw malformation) and exhibits genetic heterogeneity

with autosomal dominant and recessive forms. One typeof ectrodactyly (split hand-foot malformation) is anautosomal dominant. A grandfather and grandson havethis form of ecrodactyly, but the intervening father hasnormal hands by X-ray. Which of the following terms

applies to this family?(A)new mutation

(B) incomplete penetrance

(C) variable expressivity

(D) germ-line mosaicism

(E)anticipation

Q3


28/42

Ectrodactyly causes missing middle fingers (lobsterclaw malformation) and exhibits genetic heterogeneity

with autosomal dominant and recessive forms. One typeof ectrodactyly (split hand-foot malformation) is anautosomal dominant. A grandfather and grandson havethis form of ecrodactyly, but the intervening father hasnormal hands by X-ray. Which of the following terms

applies to this family?(A)new mutation

(B) incomplete penetrance

(C) variable expressivity

(D) germ-line mosaicism

(E)anticipation

A3


29/42

Hemophilia A and hemophilia B havenearly identical phenotypes, but they

result from mutations in different geneson the X chromosome. This is an exampleof:

A. variable expressivityB. compound heterozygosityC. locus heterogeneityD. germ-line mosaicism

E. allelic heterogeneity

Q4


30/42

Hemophilia A and hemophilia B havenearly identical phenotypes, but they

result from mutations in different geneson the X chromosome. This is an exampleof:

A. variable expressivityB. compound heterozygosityC. locus heterogeneityD. germ-line mosaicism

E. allelic heterogeneity

A4


31/42

Molecular analysis of the DMD gene is carried outon two unrelated boys with Duchenne musculardystrophy. One is found to have a large deletion inthe 5' end of the gene; the other has a nonsensemutation in exon 5 of the gene. Choose one of the

following terms that best matches a clinicalvignette:

A. anticipationB. allelic heterogeneity

C. incomplete penetranceD. locus heterogeneityE. variable expressivity

Q5


32/42

Molecular analysis of the DMD gene is carried outon two unrelated boys with Duchenne musculardystrophy. One is found to have a large deletion inthe 5' end of the gene; the other has a nonsensemutation in exon 5 of the gene. Choose one of the

following terms that best matches a clinicalvignette:

A. anticipationB. allelic heterogeneity

C. incomplete penetranceD. locus heterogeneityE. variable expressivity

A5

Q


33/42

Two parents are both affected with albinism, an

autosomal recessive disease, and have a normalchild. Which of the following terms best

describes the situation?

A. variable expressivity

B. allelic heterogeneity

C. locus heterogeneityD. incomplete penetrance

E. new mutation

Q

Albinism in a child of Afro-

Caribbean origin

A


34/42

Two parents are both affected with albinism, an

autosomal recessive disease, and have a normalchild. Which of the following terms best

describes the situation?

A. variable expressivity

B. allelic heterogeneity

C. locus heterogeneityD. incomplete penetrance

E. new mutation

A

Albinism can be cased by anyof at least three geneslocated on different autosomes

Q


35/42

In the family shown below, individuals affected withprofound deafness are represented by a shaded

symbol. The phenotypes of individuals in the fourthgeneration can best be explained by

Q

A. PleiotropyB. Variable expressivity

C. Incomplete penetrance

D. Allelic heterogeneity

E. Locus heterogeneity

A


36/42

In the family shown below, individuals affected withprofound deafness are represented by a shaded

symbol. The phenotypes of individuals in the fourthgeneration can best be explained by

A

A. PleiotropyB. Variable expressivity

C. Incomplete penetrance

D. Allelic heterogeneity

E. Locus heterogeneity


37/42

A woman is affected with an autosomaldominant disease. Her husband is healthy and

there is no history of this disease among hisrelatives. The disease is characterized byreduced penetrance 70%. What is therecurrence risk for offspring to be affected

with the disease?

(A) 75%

(B) 50%(C)35%(D)25%(E)15%

Q6

A6


38/42

Answer = 0.5 x 0.7 = 0.35 or 35%

A woman is affected with an autosomal dominantdisease. Her husband is healthy and there is no history

of this disease among his relatives. The disease ischaracterized by reduced penetrance 70%. What is therecurrence risk for offspring to be affected with thedisease?

(A) 75%(B) 50%(C)35%(D)25%

(E)15%

A6


39/42

A woman is affected with an autosomaldominant disease. Her husband is healthy and

there is no history of this disease among hisrelatives. The disease is characterized byreduced penetrance 70%. What is theprobability that they will produce phenotypically

normal offspring?

(A) 75%(B) 65%(C) 50%(D) 35%(E) 25%

Q7


40/42

A woman is affected with an autosomal dominantdisease. Her husband is healthy and there is no history

of this disease among his relatives. The disease ischaracterized by reduced penetrance 70%. What is theprobability that they will produce phenotypically normaloffspring?

(A) 75%(B) 65%(C) 50%(D) 35%(E) 25%

A7

Answer: there are two ways

I:Risk to be affected = 0.5 x 0.7 =0.35Probability to be healthy = 1 0.35 = 0.65

II:

Probability to have normal genotype andtherefore be healthy = 0.5Probability to be healthy with mutantgenotype = 0.5 x 0.3= 0.15Total probability = 0.5+0.15=0.65


41/42

A man and woman are both affected by an autosomal

dominant disorder that has 80% penetrance in allaffected individuals. They are both heterozygotes for

the disease-causing mutation. What is the probability

that they will produce phenotypically normal offspring?A. 20%

B. 25%

C. 40%D. 60%

E. 80%

Q8

A8


42/42

A man and woman are both affected by an autosomal dominant

disorder that has 80% penetrance in all affected individuals.

They are both heterozygotes for the disease-causingmutation. What is the probability that they will produce

phenotypically normal offspring?

A. 20%

B. 25%

C. 40%

D. 60%

E. 80%

A8

Answer: there are two ways

I:Risk to be affected = 0.75 x 0.8 =0.6Probability to be healthy = 1 0.6 = 0.4

II:

Probability to have normal genotype andtherefore be healthy = 0.25Probability to be healthy with mutantgenotype = 0.75 x 0.2= 0.15Total probability = 0.25+0.15=0.4

l6 b terminology

Documents