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Genetic Diagnosis

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Chapter 5

Lecture structure

Genetic diagnosis
Applications of genetic diagnosis

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Genetic diagnosis

Once a gene is identified, the associated genetic disease in at-risk individuals can be diagnosed.
The goal of genetic diagnosis is to determine whether an at-risk individual has inherited a disease-causing gene.

Genetic diagnosis can be distinguished into 2 types:

Direct diagnosis: the mutation itself is examined
Indirect diagnosis: linked markers are used to infer whether the individual has inherited the chromosome segment containing the disease-causing mutation

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Direct Diagnosis

PCR and allele-specific oligonucleotide (ASO) probes

ASO probes are short nucleotide sequences that bind specifically to a single allele of the gene.
For example, the most common mutation causing hemochromatosis is the C282Y mutation that results from a G to A substitution in codon 282.

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Direct Diagnosis

PCR and allele-specific oligonucleotide (ASO) probes

The 2 ASOs could be used to probe the PCR-amplified material on a dot blot.

ASO Probes in hemochromatosis

The results show that
Individual 1 is homozygous for the normal HFE allele.
Individual 2 is heterozygous for the normal and C282Y alleles.
Individual 3 is homozygous for the C282Y allele.

Only individual 3 would be expected to have symptoms.
Note that this test merely determines genotype, and many considerations must be taken into account before predictions about phenotype could be made.
Hemochromatosis has only about 15% penetrance, and in those who do have symptoms, variable expression is seen.

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Direct Diagnosis

DNA chips

This approach involves embedding thousands of different oligonucleotides, representing various mutations and normal sequences, on a silicone chip.
Patient DNA from specific regions is amplified by PCR, tagged with a fluorescent label, and exposed to the oligonucleotides on the chip.
The sites of hybridization on the chip are recorded by a computer.

This approach has the advantages of ready computerization and miniaturization (hundreds of thousands of oligonucleotides can be embedded on a single 2-cm² chip).

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Direct Diagnosis

Restriction fragment length polymorphism (RFLP) analysis of PCR

products (RFLP-PCR)

Occasionally a mutation that creates a disease-producing allele also destroys (or creates in some instances) a restriction enzyme site, as illustrated by the following case:

A 14-year-old girl has been diagnosed with Gaucher disease (glucocerebrosidase A deficiency), an autosomal recessive disorder of sphingolipid catabolism. The mutation, T1448C, in this family also affects an HphI restriction site. PCR amplification of the area containing the mutation yields a 150-bp product.

The PCR product from the normal allele of the gene is not cut by HphI. The PCR product of the mutant allele T1448C is cut by HphI to yield 114- and 36-bp fragments.

The PCR product(s) is visualized directly

by gel electrophoresis. Based on the results shown in the Figure using this assay on DNA samples from this family, what is the most likely conclusion about sibling 2? Sibling 2 is also affected

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Direct Diagnosis

RFLP diagnosis of myotonic dystrophy

RFLP analysis is also useful in a few cases in which polymorphisms are too large to conveniently amplify with a PCR.
One such case is myotonic dystrophy, in which the expanded sequence is within the gene region itself (a CTG in the 3′ untranslated region).
This disease shows anticipation, and family members with a severe form of myotonic dystrophy may have several thousand copies of this repeat.

As shown in the figure, when EcoRI digests are analyzed by Southern blotting, a probe reveals 9- to 10-kb fragments in unaffected individuals.
The size of the fragment can reach 20 kb in severely affected individuals.

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Direct Diagnosis

Direct DNA sequencing

Sequencing of the entire gene (or at least the exons and the intron–exon boundaries) is time-consuming and expensive.
However, it is sometimes necessary if no specific set of mutations is responsible for most cases of a disease (e.g., familial breast cancer caused by any of several hundred mutations of the BRCA1 or BRCA2 genes).
DNA sequencing is typically done using automated sequencing machines.

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Indirect Diagnosis

If the mutation causing a disease in a family is not known, indirect genetic analysis can be used to infer whether a parent has transmitted the mutation to his or her offspring.
Indirect genetic analysis uses genetic markers that are closely linked (showing <1% recombination) to the disease locus.

The markers are the same ones used in genetic mapping studies:

Restriction fragment length polymorphisms (RFLPs),
Short tandem repeat polymorphisms (STRPs), and
Single nucleotide polymorphisms (SNPs).

Because STRPs can have multiple alleles (with each allele representing a different number of repeats), they are often informative markers to use.

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Indirect Diagnosis

Indirect genetic diagnosis using STRPs

Suppose there is a 3-generation family in which Marfan syndrome is being transmitted.
Each family member has been typed for a 4-allele STRP that is closely linked to the disease locus.
The affected father in generation I transmitted the disease-causing mutation to his daughter, and he also transmitted allele 3 of the marker.
This allows us to establish linkage phase in this family.

The genotype of a closely linked marker locus is shown below each individual.

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Indirect Diagnosis

Indirect genetic diagnosis using STRPs

Because of the close linkage between the marker and the disease locus, we can predict accurately that the offspring in generation III who receive allele 3 from their mother will also receive the disease-causing mutation.
Thus, the risk for each child, instead of being the standard 50% recurrence risk for an autosomal dominant disease, is much more definitive: nearly 100% or nearly 0%.

The genotype of a closely linked marker locus is shown below each individual.

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Indirect Diagnosis

Indirect genetic diagnosis using STRPs

Recurrence risks may have to take into account the small chance of recombination between the marker allele and the disease-causing gene.
If the STR and the disease-causing gene used in this case show 1% recombination, then the recurrence risk for a fetus in generation III whose marker genotype is 2,2 would be 1% rather than 0%.
If a fetus in generation III had the marker genotype 2,3, the recurrence risk for that child would be 99%.

The genotype of a closely linked marker locus is shown below each individual.

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Indirect Diagnosis

Indirect genetic testing using RFLPs

A man and a woman seek genetic counseling because the woman is 8 weeks

pregnant, and they had a previous child who died in the perinatal period. A

retrospective diagnosis of long-chain acyl-CoA dehydrogenase (LCAD)

deficiency was made based on the results of mass spectrometry performed

on a blood sample. The couple also has an unaffected 4-year-old daughter

with a normal level of LCAD activity consistent with homozygosity for the

normal LCAD allele.

The parents wish to know whether the current pregnancy will result in a child with the same rare condition as the previous child who died.

DNA samples from both parents and their unaffected 4-year-old daughter are tested for mutations in the LCAD gene. All test negative for the common mutations. The family is then tested for polymorphism at a BamII site within exon 3 of the LCAD gene by using a probe for the relevant region of this exon.

The RFLP marker proves informative. Fetal DNA obtained by amniocentesis is also tested in the same way. The results of the Southern blot are shown below. What is the best conclusion about the fetus?

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Indirect genetic testing using RFLPs

If an RFLP is used as a marker for a disease-causing gene, the data may be analyzed by using Southern blotting and a probe for the gene region.
Although RFLP analysis can be used as both an indirect test and a direct test, there is a significant difference between the two situations.
In the direct test, the mutation causing the disease is the same as the one that alters the restriction site. There is no distance separating the mutations and no chance for recombination to occur, which might lead to an incorrect conclusion.
In the indirect assay, the mutation in the restriction site (a marker) has occurred independently of the mutation causing the disease. Because the mutations are close together on the chromosome, the RFLP can be used as a surrogate marker for the disease-producing mutation.
Linkage phase in each family must be established.
Because the RFLP and the locus of the disease-producing mutation are some distance apart, there is a small chance for recombination and incorrect conclusions.

Fetus is homozygous for LCAD mutation and should be clinically affected

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RFLP analysis for an X-linked disease

Individual II-2 in the family shown below has Lesch Nyhan disease.
His sister, II-4, is pregnant and wants to know the likelihood that her child will be affected.
The mutation in this family is uncharacterized, but is mapped to within 0.05 cM of an EcoR1 site that is informative in this family.
DNA from all family members is obtained. Fetal DNA is obtained by chorionic villus sampling. What is the best conclusion about the fetus?

Indirect Diagnosis

Answer: Fetus (a girl) will not be affected; nor will she be a carrier because her mother, II-4, is not a carrier

RFLP Analysis of HGPRT Deficiency in a Family

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Features of Indirect and Direct Genetic Diagnosis

Direct versus Indirect Genetic Diagnosis

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Applications of genetic diagnosis

Genetic diagnosis is used in a variety of settings, including the ones listed below.

• Carrier diagnosis in recessive diseases

• Presymptomatic diagnosis for late-onset diseases

• Asymptomatic diagnosis for diseases with reduced penetrance

• Prenatal diagnosis

• Preimplantation testing

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Applications of genetic diagnosis

Prenatal Genetic Diagnosis

Prenatal diagnosis is one of the most common applications of genetic diagnosis.
Diagnosis of a genetic disease in a fetus may assist parents in making an informed decision regarding pregnancy termination and in preparing them emotionally and medically for the birth of an affected child.
There are various types of prenatal diagnosis.
Amniocentesis
With amniocentesis, a small sample of amniotic fluid (10–20 mL) is collected at approximately 16 weeks’ gestation.

Fetal cells are present in the amniotic fluid and can be used to diagnose single-gene disorders, chromosome abnormalities, and some biochemical disorders.
Elevated α-fetoprotein levels indicate a fetus with a neural tube defect.
The risk of fetal demise due to amniocentesis is estimated to be approximately 1/200.

Amniocentesis

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Applications of genetic diagnosis

Chorionic villus sampling
This technique, typically performed at 10–12 weeks’ gestation, involves the removal of a small sample of chorionic villus material (either a transcervical or a transabdominal approach may be used).
The villi are of fetal origin and thus provide a large sample of actively dividing fetal cells for diagnosis.
This technique has the advantage of providing a diagnosis earlier in the pregnancy.
There is a small possibility of diagnostic error because of placental mosaicism (i.e., multiple cell types in the villi).
The risk of fetal demise is higher than with amniocentesis (about 1/100).

Chorionic villus sampling

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Applications of genetic diagnosis

Preimplantation diagnosis
Embryos derived from in vitro fertilization can be diagnosed by removing a single cell, typically from the eight-cell stage (this does not harm the embryo).
DNA is PCR amplified and is used to make a genetic diagnosis.
The advantage of this technique is that pregnancy termination need not be considered: only embryos without the mutation are implanted.
There is a possibility of diagnostic error as a result of PCR amplification from a single cell.

Preimplantation diagnosis

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Chapter 5: Genetic Diagnosis

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Review Questions

Select the ONE best answer.

1. The pedigree below shows a family in which hemophilia A, an X-linked

disorder, is segregating. PCR products for each member of the family are

also shown for a short tandem repeat polymorphism located within an

intron of the factor VIII gene. What is the best explanation for the phenotype of individual II-1?

A. Heterozygous for the disease-producing allele

B. Homozygous for the disease-producing allele

C. Homozygous for the normal allele

D. Incomplete penetrance

E. Manifesting heterozygote

1. Answer: A. The female II-1 in this family is heterozygous for the marker

(from the gel) and also has an unaffected father. Her mother is a carrier

and the bottom band in the mother’s pattern is associated with the diseaseproducing

allele of the factor VIII gene. All observations are consistent

with II-1 being heterozygous (Xx) for the factor VIII gene. She has no

symptoms, so she is not a manifesting heterozygote (choice E). She cannot

be homozygous for the disease-producing allele (choice B) because her

father is unaffected. Homozygosity for the normal allele (choice C) is

inconsistent with the results shown on the gel. She has inherited the chromosome

from her mother (bottom band) that carries the mutant factor

VIII allele, but from her father she has received a chromosome carrying

the normal allele. Note that her father is not affected, and the bottom band

in his pattern is in linkage phase with the normal allele of the gene. This is

a case where linkage phase is different in the mother and the father.

Incomplete penetrance (choice D) is not a good choice because the female

(II-1) does not have the disease-producing genotype. She is heterozygous

for the recessive and (dominant) normal allele. One would expect from

her genotype that she would be unaffected.

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Chapter 5: Genetic Diagnosis

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Review Questions

Select the ONE best answer.

2. A 22-year-old woman with Marfan syndrome, a dominant genetic disorder,

is referred to a prenatal genetics clinic during her tenth week of pregnancy.

Her family pedigree is shown below (the arrow indicates the pregnant

woman). PCR amplification of a short tandem repeat (STR) located in an

intron of the fibrillin gene is carried out on DNA from each family member.

What is the best conclusion about the fetus (III-1)?

A. Has a 25% change of having Marfan syndrome

B. Has a 50% chance of having Marfan syndrome

C. Will develop Marfan syndrome

D. Will not develop Marfan syndrome

E. Will not develop Marfan syndrome, but will be a carrier of the disease allele

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Chapter 5: Genetic Diagnosis

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Review Questions

Select the ONE best answer.

3. The pedigree below represents a family in which phenylketonuria (PKU),

an autosomal recessive disease, is segregating. Southern blots for each family member are also shown for an RFLP that maps 10 million bp upstream

from the phenylalanine hydroxylase gene. What is the most likely explanation

for the phenotype of II-3?

A. A large percentage of her cells have the paternal X chromosome carrying the PKU allele active

B. Heteroplasmy

C. Male I-2 is not the biologic father

D. PKU shows incomplete penetrance

E. Recombination has occurred

3. Answer: E. Although II-3 has an RFLP pattern consistent with heterozygosity

for the PKU allele, she has PKU. The best explanation offered is

that recombination has occurred, and although she is heterozygous for

the restriction site generating the RFLP pattern, she is homozygous for

the mutation causing PKU. The restriction site is 10 million bp upstream

from the phenylalanine hydroxylase gene so there is a minimum chance

of recombination of 10%. Although this is small, it is the most likely of

the options listed. The phenylalanine hydroxylase gene is not on the X

chromosome (choice A). Heteroplasmy (choice B) is associated with

mitochondrial pedigrees, and the phenylalanine hydroxylase gene is a

nuclear one. The RFLP pattern is quite consistent with I-2 being the biologic

father (choice C), and he is a known carrier of the PKU mutation

because he has another affected child (II-1). If II-3’s RFLP pattern showed

homozygosity for the marker (identical to II-1), and she had no symptoms,

incomplete penetrance (choice D) would be a good choice.

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Select the ONE best answer.

4. A 14-year-old boy has Becker muscular dystrophy (BMD), an X-linked

recessive disease. A maternal uncle is also affected. His sisters, aged 20 and

18, wish to know their genetic status with respect to the BMD. Neither the

boy nor his affected uncle has any of the known mutations in the dystrophin

gene associated with BMD. Family members are typed for a HindII restriction site polymorphism that maps to the 5′ end of intron 12 of the dystrophin gene. The region around the restriction site is amplified with a PCR. The amplified product is treated with the restriction enzyme HindII and the fragments separated by agarose gel electrophoresis. The results are shown below. What is the most likely status of individual III-2?

A. Carrier of the disease-producing allele

B. Hemizygous for the disease-producing allele

C. Homozygous for the normal allele

D. Homozygous for the disease-producing allele

E. Manifesting heterozygote

4. Answer: C. The disease-producing allele of the gene is associated with

the presence of the HindII site. Notice that both affected males show two

smaller bands (75 and 40 bp). II-3, a carrier female, also has these two

smaller bands in her pattern, in addition to a larger PCR product (115 bp),

representing the absence of the HindII site on her normal chromosome.

III-2 has only the larger PCR product (notice the density because both

chromosomes yielded this product). She is homozygous for the normal

allele. Choice A, carrier, would be correct if her pattern had looked like

those of II-3 and III-1. All the males shown are hemizygous (choice B)

for the dystrophin gene because they have only one copy. II-1 and III-3

are hemizygous for the disease-producing allele, and II-2 is hemizygous

for the normal allele. No one in the family is homozygous for the disease producing

allele (choice D). In an X-linked pattern, this would be

characteristic

of a female with two copies of the disease-producing allele

and is very rarely seen. III-2 is not a manifesting heterozygote (choice E)

because she has no symptoms and is not a heterozygote.

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Chapter 5: Genetic Diagnosis

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Review Questions

Select the ONE best answer.

5. Two phenotypically normal second cousins marry and would like to have

a child. They are aware that one ancestor (great-grandfather) had PKU and

are concerned about having an affected offspring. They request ASO testing

and get the following results. What is the probability that their child will be affected?

A. 1.0

B. 0.75

C. 0.67

D. 0.50

E. 0.25

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Chapter 5: Genetic Diagnosis

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Review Questions

Select the ONE best answer.

6. A 66-year-old man (I-2) has recently been diagnosed with Huntington disease, a late-onset, autosomal dominant condition. His granddaughter (III-1) wishes to know whether she has inherited the disease-producing allele, but

her 48-year-old father (II-1) does not wish to be tested or to have his status

known. The grandfather, his unaffected wife, the granddaughter, and her

mother (II-2) are tested for alleles of a marker closely linked to the huntingtin gene on 4p16.3. The pedigree and the results of testing are shown below. What is the best information that can be given to the granddaughter (III-1) about her risk for developing Huntington disease?

A. 50%

B. 25%

C. Marker is not informative

D. Nearly 100%

E. Nearly 0%

6. Answer: A. The affected grandfather has marker alleles DS2 and DS3.

There is no information about which one is in linkage phase with his disease producing

huntingtin allele. On the basis of the pedigree alone, the

daughter has a 25% change of inheriting the grandfather’s disease-producing

huntingtin allele (choice B); however, she would like more information.

Because her father (II-1) does not wish to be tested or know about his

genetic status with respect to Huntington’s, it is unethical to test the

daughter for the triplet repeat expansion. The results would necessarily

reveal the status of her father also. By doing an indirect genetic test, one

can see the daughter has inherited one of her marker alleles (DS2) from

the grandfather via her father. This means she has a 50% chance of developing

Huntington’s because there is a 50% chance that DS2 is a marker for

the disease-producing huntingtin allele in the grandfather and a 50%

chance it is not (and DS3 is). Notice the result does not reveal additional

information about her father (II-1). Before her testing, he had a 50%

chance of having the disease-producing huntingtin allele. His risk is still

50% with the information from the daughter’s test. However, if the father

(II-1) does develop Huntington’s in the future, that will mean that the

daughter has a 100% chance of having the disease also (choice D).

If her marker status had been DS1/DS1, her chances of developing

Huntington’s

would have been near 0 (choice E) because she did not

inherit these alleles from her grandfather. One came from her grandmother

(via her father) and one from her mother. This result still would

not reveal additional relevant information about her father (II-1), whose

risk would remain 50%.