Download Introduction You are going to investigate a genetic disease that

Introduction
You are going to investigate a genetic disease that segregates in human families.
Although the disease is hypothetical, you will be able to identify the gene(s)
whose mutations cause the disease in the second part of this exercise.
The process of ‘positional cloning’ (defining the genomic position of a disease
locus and determining which gene is linked to a disease), typically includes the
following four stages:
1. Initial genetic mapping of the disease phenotype to a chromosomal region by
classical linkage (segregation) analysis using highly polymorphic markers
2. The use of recombinants to define flanking markers to encompass the
chromosome region containing the gene
3. Detection of genes (either by experimentation or using online databases) in
the critical region and homology searches to aid in finding the function of the
genes
4. Detection of mutations (either by experimentation or using online databases)
in a candidate gene to confirm association with a disease phenotype.
Stages 1 and 2 are part of Part 1 of the practical.
Part 1.
Clinicians from the Sick People Hospital identified several families in which there
were patients suffering from anemia, a very common disease that leads to iron
depletion in many tissues of the body.
The figure below shows you an example of one of the families. (Family1)
Question 1.Using the information of the lecture on segregation patterns,
determine the most likely segregation patternof the trait and motivate your
choice.
The doctors at the hospital were anxious to learn more about the disease in order
to determine
-
the recurrence risk: thelikelihood that a trait or disorder present in one
family member will occur again in other family members in the same or
subsequent generations, and
-
to detect non-penetrant carriers within the families, that is family members
that carry the disease mutation but do not have the disease because of
other factors influence the expression of the disease.
They approached the virtually famous Institute of Human Genetics at the VU
University and requested assistance in defining the genetic cause of the trait.
You, Dr. Watson, are a student in genetics and are applying for a research
position. The director of the Institute, Prof. G. Mendel is impressed by your CV
and offers you a fellowship and small space in his laboratory to carry out this
important investigation. You will have the help of Mrs. Crick, a laboratory
technician, as well.
Within the Institute there are extensive facilities for genome analysis, including
large-scale
automated
DNA
sequencing
and
genotyping,
an
excellent
bioinformatics section, cell biology and biochemistry and transgenic animal
facilities. Prof. Mendel wants you to carry out an analysis to isolate and
characterize the responsible gene for this disease.
Once you have settled in the lab, the clinicians from the Sick People hospital
send you blood samples of members of two different families. You isolate DNA
from these samples, give them a unique DNA number and store this number
together with all clinical data in a database. The starting point for the search of
the gene responsible will be 48 DNA samples available from all the living
members, including some patients. A clinical geneticist was asked to construct
the pedigrees and the individual family members are denoted in the pedigrees.
Family 1 you have seen above,
Family 2 is a smaller family:
Question 2.What is the most likely segregation pattern of the disease in family
2? Motivate your choice.
Linkage studies
Your lab is well equipped to map disease genes. A few years ago your
colleagues constructed a set of 400 highly polymorphic Short Tandem Repeat
(STR) markers covering all 22 autosomes. Remember that each STRs marker
has numerous alleles (i.e. numerous forms in which it can occur), making STRs
highly informative for identifying individuals. In the lab there is also a pipetting
robot and an automated DNA sequence analyzer to separate individual
genotypes.
Although this equipment makes life (relatively) easy the key to success is that
you and Mrs. Crick are exceptionally skilled. At least that is what we assume
here!!
Below is an outline of the whole process starting from the ascertainment of the
families.
Together with Mrs. Crick you have carried out a genome scan in these two
families aiming to determine the chromosomal location of this disease gene. This
genome screen took a few months and about 19,200 genotypes were generated
(400 markers x 48 individuals).
The analysis involves screening each DNA sample with 400 STR markers. Part
of a genotype Image, the output of the automated sequencer, is shown below
with each lane containing DNA from a single person and with several markers in
each lane.
The markers are distinguished from each other by a combination of length and
color by running them on a “gel”. This gel separates the DNA fragments by size.
Smaller fragments travel faster then larger fragments on this gel. At the end of
the gel a laser and a CCD camera detect the passing DNA fragments. This gel
contains several green (5), blue (4), and yellow (3) markers. The red bands are
size standards that are used to determine the exact size (in base pairs) of the
DNA fragments in the gel..
The content of each lane is then analyzed by a "genotyping" program that
defines the exact length of each marker individually and, based on the colors,
determines the genotype for each individual on each marker. See below:
This length of each band corresponds with differences in repeat numbers and
can be scored in a table. Example below:
Note that the smallest allele is listed first because it runs faster then the larger
allele.
Because of small experimental variation (noise) in the length of the fragments
these "raw" genotypes or alleles are then grouped by a "binning" program and
renumbered to single digits. For example allele with length 317.24 is recoded into
allele ‘3’ (the pink cells below), the allele with length 328.10 and 328.38 into allele
‘1’ (green cells below). The resulting "binned" genotypes can be used for
segregation checks and after passing these checks they can be used for linkage
- and haplotype analysis.
After all these checks linkage analysis is carried out for each of the 400 tested
markers marker for the family 1. The small family number 2 is not really strong
enough to provide significant lod scores by itself and will only be used for
confirmation of positive findings.
Family 1 gives positive lod scores for several markers on the long arm of
chromosome 2. You become very excited and cannot sleep until you can confirm
that you found linkage.
You are very curious to see if the smaller family 2 also shows evidence for
linkage to the same region. You will construct haplotypes for potentially
interesting markers aiming to detect the borders of the linkage region. Remember
a haplotype tells you which alleles from different markers lie on the same
ancestral chromosome. For example, let’s say there are two STR markers,
marker 1 with alleles A,B,C,D and marker 2 with alleles E,F,G,H. After
genotyping a certain person you know that that person has alleles A and B for
marker 1 and alleles E and F for marker two, thus the genotypes for this person
are AB and EF. However, you do not know whether the A allele of marker 1 was
transmitted together with the E allele of marker 2, that is whether they came from
the same parent. This information is important for disease mapping, because we
search for disease genes that are transmitted together with markers. If there is a
disease gene that lies on a chromosome between the A and E alleles, than we
expect that an affected person has received both the A and E allele from an
affected parent. It is thus important to determine haplotypes, which cannot be
inferred from genotypes alone, but can sometimes be derived when parental
information is available.
The genotypes for each available family member of family 2 are provided in the
Table below. As you can see things are a bit complicated because none of the
parents were available for genotyping. But with the help of the genotypes of the
children you will be able to reconstruct the haplotype of the parents.
Notes:
-
We have listed genotypes not haplotypes. You have to put the genotypes
in the pedigree to reconstruct the two parental chromosomes of each
individual.
-
The ID numbers correspond to the numbers in the pedigree of family 2
given above.
-
The markers in the table are already in the correct chromosomal positional
order.
ID
1
2
3
4
5
6
7
8
D2S335
D2S2314
D2S2273
D2S117
14
12
24
44
13
23
13
11
12
33
23
33
13
12
14
24
44
44
23
13
Question 3: Using logical deduction, determine the haplotypes for each family
member. Also try to reconstruct the haplotypes of dead or otherwise missing
family members. Hint: first deduce the genotypes of missing family members,
than construct haplotypes. Use the following notation to show haplotype:
1|2
2|1
3|4
4|4
and (1 2) to denote a genotype.
Question 4: Do specific alleles of the markers co-segregate with the disease? If
so, which ones?
Question 5: And if yes, are there recombinations that have helped you to
determine the border of the critical region?
Question 6: What is the area of linkage where you think the disease gene lies?
Now that you have found linkage for your disease you should buy a bottle of
champagne for your lab to celebrate mapping this gene!!!!