Download MCDB 1041 Activity 8: Genetic testing Part I. Using Restriction

MCDB 1041 Activity 8: Genetic testing Part I. Using Restriction Enzymes to identify disease alleles You work in a clinic doing prenatal testing and genetic counseling. You use PCR analysis combined with
restriction enzyme digests to determine whether fetuses are affected by cystic fibrosis, caused by a mutation on
both copies of chromosome 7, in the cystic fibrosis (CF) gene. Below is a region of DNA (from the middle of
the CF gene). Sequence of normal CF gene:
5’ ACGCCGCTACGT TAGACTTCGCTACAAGACGG 3’
3’ TGCGGCG ATGCAATCTGAAGCGATGTTCTGCC 5’
Sequence of mutant CF gene
5’ ACGCCGCTACGT TAGAATTCGCTACAAGACGG 3’
3’ TGCGGCG ATGCAATCTTAAGCGATGTTCTGCC 5’
1. Circle the mutation in the sequence above.
One of the restriction enzymes (R.E.) you can use cuts at the following sequence, at the stars: G*AATTC
CTTAA*G
2. Mark on the sequence(s) above where this R.E. will cut the DNA.
You have been charged with doing DNA analysis on three recently born babies. You determine that one child
is normal (no mutations in the cystic fibrosis gene on chromosome 7), one is a carrier (one mutation), and one
has cystic fibrosis (a mutation on each chromosome)—see the diagram below.
The line in the diagram indicates the site of the mutation within the CF gene that you just found in questions 8
and 9 above. The CF gene DNA is isolated from cells from the fetus’ cells by PCR, and is 10 KB in length.
The restriction enzyme shown above cuts the mutant CF gene into two pieces (7 kb and 3 kb). You cut your
samples of DNA from each fetus, and then run the DNA on the gel.
Ch. 7
Ch. 7
Ch. 7
normal
Ch. 7
Ch. 7
Cystic fibrosis
Ch. 7
carrier
3. What would you expect to see when you run the gel? Draw the bands each child has. Make sure to indicate
the correct intensity of each band (ie, more DNA, darker band).
Child A is normal.
Child B has cystic fibrosis
Child C is a carrier
A
B
C
12 KB
11 KB
10 KB
9 KB
6 KB
3 KB
1 KB
4. Below is a pedigree showing inheritance of an autosomal recessive disease. Carriers are marked. The gene
being analyzed is 15 KB in length normally. The mutation is a deletion of 2 KB. For each individual, imagine
you have done PCR to amplify just this gene sequence from their DNA. For each numbered individual, draw in
the band(s) of DNA on the gel that would result from the PCR.
1
1
2
__
2
3
4
5
ladder
20
15
3
4
5
10
8
+
6
5. Let’s say you are testing newborns to see if they have or are carriers of an X-linked recessive disease,
hemophilia (due to a mutation in “coagulation factor VIII” –also called “F8”-- which results in failure of blood
to clot properly).
a. Using the white boards: make up a pedigree for three generations that shows unaffected, affected and
carrier individuals in a pedigree for hemophilia. You can copy your pedigree into the space below, so you can
refer back to it later.
b. Say that the F8 gene is 15 KB in length. If you wanted to analyze whether individuals in a family had
or carried hemophilia before they showed any symptoms, what kind of mutations would you be able to assay
using just PCR and gel electrophoresis. What kind of mutations could you assay using PCR, restriction enzyme
digests and gel electrophoresis?
c. Imagine that the mutation you are assaying that leads to hemophilia is a single base change that
removes a restriction enzyme site. If that site is normally at the 5 KB point of the gene, indicate what results
would be expected for unaffected, affected and carrier individuals by drawing a gel and drawing the pattern on
the gel expected for your analysis of each individual in the pedigree.
Part II: Using STRs for finding genes associated with traits
Objectives:
• Interpret and draw conclusions from STR data, including predicting relatedness and likelihood of
disease inheritance.
• Visualize how multiple genes can lead to a distribution of phenotypes
Microsatellites (STRs) are short sequences of repetitive DNA. Here is an STR DNA sequence for Mary. This
STR is called MB1041 and is located on chromosome 4. Mary’s MB1041 STR consists of 4 repeats of CAAT.
Primer #1
GCGCTACAATCAATCAATCAATTCGAGC
CGCGATGTTAGTTAGTTAGTTAAGCTCG
Primer #2
The number of repetitive sequences in an STR varies in individuals. For example, Bob has 8 CAAT repeats for
the MB1041 STR.
Primer#1
GCGCTACAATCAATCAATCAATCAATCAATCAATCAATTCGAGC
CGCGATGTTAGTTAGTTAGTTAGTTAGTTAGTTAGTTAAGCTCG
Primer #2
To look at the variability in microsatellite length, PCR primers #1 and #2 are designed using the unique
sequences on either side of (flanking) the STR sequences. During the PCR reaction primers #1 and #2 bind to
these unique sequences and the STR sequence is amplified. A separate PCR reaction is done on the DNA for
each individual, and their samples then loaded into separate lanes of the same gel. Mary’s MB1041 STR DNA
is the same length on her maternally and paternally derived chromosome 4. Similarly Bob’s, MB1041 STR is
the same length on his maternally and paternally derived chromosome 4.
Here is a gel that shows the results of the PCR.
1. Draw the approximate location of Bob’s MB1041
band on the gel. Explain how you know
approximately where it will run on this gel.
2. DNA from Mary and Bob’s son Jacob was also
amplified with primers #1 and #2 and run on the same
gel above. Draw in Jacob’s bands, and explain how
you know their approximate positions.
Wells where
DNA from the
PCR reaction
is loaded
Mary’s
lane
MaryÕ
s
Bob’s
lanes
BobÕ
Lane
Lane
Jacob’s
lanes
JacobÕ
Lane
3. Is the thickness of the bands in Jacob’s lane the same as in Mary and Bob’s lanes? Why or why not?
4. When you run out the DNA from a PCR amplification of an individual’s STR DNA, what is the maximum
number of bands you can see on a gel? Explain
Why use an STR sequence as opposed to PCR or restriction digests of a gene known to cause disease?
Remember we have discussed how a mutation could cause a change in the sequence of a gene such that a
restriction enzyme may not longer cut it (or may cut it when before it did not). Of course this will not always be
the case! So STR analysis is just ANOTHER way to provide additional genotypic information when there is a
limited amount of information in a pedigree. STRs are also especially useful if scientist don’t yet know the
exact location of a gene that causes an illness.
Because there are many STRs, it is relatively easy to find one that is close to a gene of interest. When an
STR sequence is very close to a gene, the likelihood of recombination is low. Thus, detecting a variation (or
allele) of an STR can be used to detect an allele of a gene. For all the questions on this activity, assume there is
no recombination between the STR and the gene of interest. Bottom line: if a specific STR allele is always
present when someone has a disease, this shows scientists that the disease-related allele is located close to
that STR, on that specific chromosome.
The gene that causes cardiac valvular dysplasia when mutant has not yet been identified, but it is known to be
linked to an STR called INT3. Examples of this region of DNA are shown below.
INT3 Allele A
---------gene of interest-------------CATACATACATACATACATACATACATACATACATACATA--------INT3 Allele D
---------gene of interest------------- CATACATACATACATACATA----------I.
Individual II-4 in the pedigree on the right comes into
the clinic and you determine that he has problems with
his aortic valve. His symptoms are consistent with a
II.
disorder called cardiac valvular dysplasia. Beneath the
1
pedigree is a gel with the results of PCR amplification
of INT3 for members of this family. There are 4
different alleles of INT3 called A, B, C, and D. Note:
There could be carriers in this pedigree (they are
I-1 I-2
INT3
unknown, and so not marked)
5. If you just had this family pedigree and no STR
data, what modes of inheritance would be plausible?
Could you rule out any modes of inheritance from the
pedigree alone?
1
2
2
3
II-1 II-2 II-3
4
5
6
II-4 II-5 II-6 II-7
A.
B.
C.
D.
6. Using the STR data and the pedigree, what is the mode of inheritance for cardiac valvular dysplasia?
Explain your reasoning.
7
7. The mutation that causes cardiac valvular dysplasia is on a chromosome with which allele of INT3?
8. The two bands for individual II-3 indicate that he has what chromosomal abnormality?
9. In which parent did the nondisjunction event occur (resulting in the chromosomal abnormality above) and did
it occur during meiosis I or II? Explain.
10. Which members of this family are carriers for cardiac valvular dysplasia?
11. II-5 just had a normal XX baby girl and her DNA is tested. Assume meiosis was
normal in both production of egg and sperm. The gel to the right shows the results of the
PCR amplification of microsatellite INT3 for the daughter. II-5 seemed especially
nervous about having her daughter’s DNA tested. Why might she be nervous?
INT3
A.
B.
C.
D.
Patients with palmoplantar keratoderma (PPK) have yellowish thickening of the skin on their hands and feet.
PPK can be caused by a mutation in either of two different keratin genes, keratin 1 or keratin 9.
Here is the pedigree of a family that has
PPK. Filled symbols indicate individuals that have
yellowish thickening of the skin. Carriers are not marked,
but could exist in the pedigree.
I.
1
II.
1
Beneath the pedigree are two gels with results of PCR
amplification of two STRs. D12S3 is located near the
keratin 1 gene and D17S5 is located near the keratin 9 gene.
There are 5 alleles of D12S3 (A-E) and 4 alleles of D17S5
(A-D).
12. In this family, do individuals with yellowish thickening
of the skin likely have a mutation in the keratin 1, keratin 9,
or both genes and why? From the data, which STR allele is
associated with the disease?
2
D12S3
2
3
I-1 I-2 II-1 II-2 II-3
Keratin 1
A.
B.
C.
D.
E.
D17S5
A.
B.
C.
D.
I-1 I-2 II-1 II-2 II-3
Keratin 9
13. If you just had the family pedigree and no STR data, what modes of inheritance would be plausible? Could
you rule out any modes of inheritance from the pedigree information alone?
14. Using the STR data and the pedigree, what is the mode of inheritance for PPK in this family? Explain your
reasoning.
I-1 and I-2 want to have another baby, but they are having problems conceiving another child so they use in
vitro fertilization. They also use preimplantation genetic diagnosis (using one cell from an 8-cell embryo) to
determine whether embryos that are being selected for implantation will develop PPK.
PCR amplification of D12S3 and D17S5 was done with DNA isolated from 4
embryos.
15. Using the original data and the data shown
to the right, which of the following embryos
(labeled E1-E4) are likely to develop PPK?
Explain.
D12S3
E1 E2 E3 E4
1
1
B.
D12S3
2
I-1 I-2 II-1 II-2 II-3
A.
D.
B.
E.
C.
D17S5
2
II.
A.
C.
16. What would be a possible explanation for a
situation in which an individual inherited a
disease, but from the STR data, did not have the
STR allele associated with the disease allele?
(hint: this is an unlikely event!)
I.
E1 E2 E3 E4
D.
A.
E.
B.
D17S5
C.
A.
D.
B.
I-1 I-2 II-1 II-2 II-3
C.
D.
17. Turn your answer (as a group) to this question in to your LA on a piece of paper.
Why, in general, can you predict the likelihood of inheritance of a gene by looking at the STR alleles that an
individual has? Explain the relationship between an STR allele of interest and the gene of interest.
3
Practice Questions
1. Mutations in the Retinitis Pigmentosa 2 (RP2) gene cause Retinitis Pigmentosa, a disorder that leads to
constriction of visual fields and night blindness. An STR marker, DXS255 (with five alleles, A,B,C,D, and E) is
tightly linked to the RP2 gene. Below is a pedigree of a family that is afflicted by Retinitis Pigmentosa and a gel
that shows the results of PCR amplification of DXS255.
Wells where DNA from the PCR
reaction is loaded
The mutation causing Retinitis Pigmentosa is on the chromosome with which allele of DXS255 (A, B, C, D, or
E)?
a. Allele A
b. Allele B
c. Allele C
d. Allele D
e. Allele E
2. STR marker DXS255 consists of repeats of the sequences ACT. Which allele of the DXS225 marker has the
most ACT repeats?
a. Allele A
b. Allele B
c. Allele C
d. Allele D
e. Allele E
3. Using all of the data presented above, what is the mode of inheritance for Retinitis Pigmentosa?
a. Autosomal dominant
b. Autosomal recessive
c. X-linked dominant
d. X-linked recessive
e. Mitochondrial
f. More than one of the above is possible
4. II-4 and II-5 have a baby boy. What is the chance that he has Retinitis Pigmentosa?
a. 0%
b. 25%
c. 50%
d. 75%
e. 100%
5. The magic gene is closely linked to a microsatellite called Lumos525. There are 4 alleles of Lumos525
called A, B, C, and D. Below is a pedigree of their family and a gel showing the results of PCR amplification
of Lumos525. In the pedigree, the symbols of the people who can do magic are filled in. For this data, the
intensity of the bands is meaningful (darker, thicker bands on the gel indicate that a person has 2 copies of the
same allele).
I-1 I-2 II-1 II-2 II-3 II-4 II-5 III-1 III-2 IV-1 IV-2
A
B
C
D
I-1 and I-2 adopted one child. Which of their children in generation II is adopted?
a. II-1
b. II-2
c. II-3
d. II-4
e. II-5
6. III-1 and III-2 just had twins: IV-1 is a girl and IV-2 is a boy. They have not undergone genetic testing yet,
but which alleles would you predict each twin has?
a. IV-1 (girl) AA, IV-2 (boy) A
b. IV-1 (girl) AC, IV-2 (boy) A
c. IV-1 (girl) AC, IV-2 (boy) C
d. IV-1 (girl) CC, IV-2 (boy) C
e. more than one answer above is possible
7. You are a scientist at a genetic testing company. Explain how you would take one approach to testing
relatedness between two individuals, and a different approach to determine an individuals susceptibility to
diabetes.