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DNA METHODS FOR HLA TYPING
A WORKBOOK FOR BEGINNERS
New in version 8: next generation sequencing and allele frequencies
Carolyn Katovich Hurley, Ph.D., D(ABHI)
C. W. Bill Young Marrow Donor Recruitment and Research Program
Department of Oncology, Georgetown University School of Medicine,
Washington, DC 20057
Version 8
Copyright (c) 1993, 1998, 2004, 2008, 2015 by Georgetown University
PURPOSE OF THE MANUAL
The purpose of this manual is to provide the reader with the basic background
needed to understand the HLA system and the molecular biology methods used to identify
HLA "types". It is assumed that the reader has a college education and has taken courses
in basic biology and biochemistry. The best results will be obtained if the reader starts at
the beginning of the manual and follows all of the instructions in the manual. The manual
is meant to be a workbook and spaces for answers are provided in the text. Answers are
provided at the end of the manual. Readers are encouraged to discuss their questions with
their laboratory director.
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TABLE OF CONTENTS
Transplantation and HLA Typing ..................................................................................... 3
1. General Concepts in Molecular Biology .................................................................... 4
2. HLA Class I Proteins and Genes ............................................................................ 17
3. HLA Class II Proteins and Genes ........................................................................... 27
4. HLA Alleles and Inheritance .................................................................................... 30
5. HLA Types Defined by Serology ............................................................................. 43
6. Gene Amplification Using the Polymerase Chain Reaction .................................... 50
7. Use of Oligonucleotide Probes to Detect Specific DNA Sequences ....................... 61
8. Sanger-Based DNA Sequencing of HLA Genes ..................................................... 68
9. Next Generation DNA Sequencing for HLA ............................................................ 75
10. Other Molecular Biology Techniques for HLA Typing............................................. 80
11. Interpretation of DNA Typing Results ..................................................................... 84
Answers to the Questions ............................................................................................. 88
REFERENCE WEB SITES
Web Address
Content
http://hla.alleles.org
Nomenclature
http://igdawg.org/cwd.html
Common and well documented alleles
http://www.ebi.ac.uk/ipd/imgt/hla/
Allele history, alignments, serologic links
https://bioinformatics.bethematchclinical.org/
Allele/haplotype frequencies in US, haplotype
tools, multi-allele codes
World-wide allele distribution maps
http://www.pypop.org/popdata/
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TRANSPLANTATION AND HLA TYPING
Hematopoietic stem cells found in the bone marrow differentiate to become blood
cells. These cells play important roles in fighting infection (white blood cells or
lymphocytes), in transporting oxygen to the tissues (red blood cells), and in causing blood
clotting (platelets). Certain types of cancer affect these stem cells and can be fatal to the
patient. One therapy for the treatment of these blood diseases like leukemia and aplastic
anemia is bone marrow (or hematopoietic stem cell) transplantation. In this procedure, the
abnormal bone marrow of the patient is destroyed by irradiation and chemotherapy. The
patient (recipient) is then transfused with stem cells from another individual (donor). The
transfused cells travel to the cavities in the bones and grow there forming new blood cells.
In order for hematopoietic stem cell transplantation to be successful, the patient and
the donor of the stem cells must be matched for a group of proteins called HLA molecules
(or HLA antigens). This manual describes HLA molecules and discusses the techniques
that are used to determine the HLA "types" of the patient and potential donors. If the
patient and a potential donor have the same HLA type, the transplant has a good chance
of being successful. If a patient and a donor are HLA mismatched, the donor’s stem cells
may be destroyed by the patient's immune system (graft rejection) or the immune system
cells in the donor’s stem cell preparation may attempt to destroy the patient's own cells
(graft vs. host disease). These events can result in major complications or death of the
patient. Transplantation of organs like the kidney face similar histocompatibility barriers.
This manual is designed to be used by readers who will participate in using
molecular biology techniques to identify HLA types. It begins with a description of some
basic molecular biology concepts.
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CHAPTER 1
GENERAL CONCEPTS IN MOLECULAR BIOLOGY
The purpose of this chapter is to describe basic principles regarding DNA and to describe
the methods used to study DNA.
I.
Structure of DNA and RNA
A.
DNA (deoxyribonucleic acid) is composed of a phosphate and sugar
(deoxyribose) backbone attached to bases (Adenine, Thymine, Cytosine,
Guanine) [Figure 1-1]. The combination of a base, a sugar, and a phosphate
group is called a nucleotide (e.g., dATP is an abbreviation of
deoxyadenosine triphosphate; dNTP is an abbreviation which means any
nucleotide). Nucleotides are the basic building blocks of DNA.
B.
Nucleotides are linked together to form a linear chain of nucleotides. One
end of this single DNA strand is called 5'; the other end is called 3'. The two
ends have different structures. It is common practice to put the 5' nucleotide
on the left side of the page when writing down a sequence of nucleotides
(e.g., 5' TAAGGCT 3').
Figure 1-1
Structure of DNA
Nucleotide
5’
3’
P
S
P
S
P
S
P
G
A
T
C
C
T
A
G
S
P
S
P
S
P
3’
S
S
P
5’
dGTP, deoxyguanosine triphosphate
C.
Two strands of DNA form a ladder (or double helix) by base pairing (G-C and
A-T). The bases are paired through hydrogen bonds. G-C pairs form 3
hydrogen bonds while A-T pairs form 2 hydrogen bonds. This difference
means that it is harder to break the bonds that hold together a G-C pair than
the bonds holding together an A-T pair.
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II.
D.
The two strands of DNA that form a double helix run in opposite directions.
This means that the 3’ end of one strand is next to the 5’ end of the second
strand.
E.
The length of double-stranded DNA is measured in base pairs. 1000 base
pairs is called a kilobase (1 kb).
F.
The two strands of DNA are complementary to one another; if you are given
the sequence of one strand, you automatically know the sequence of the
second strand. To save space, the sequence of only one strand of the DNA,
the coding strand, is reported in the literature. The sequence of the coding
strand is the same sequence as found in the mRNA that is transcribed from
that gene and is always written with the 5' end on the left (like a sentence).
[More on mRNA in Chapter 2.]
G.
RNA (ribonucleic acid) is composed of phosphate, ribose, and bases (A,C,G,
Uracil). RNA containing the coding information for a protein (messenger
RNA or mRNA) is single stranded.
Restriction endonucleases (RE)
A.
Why are restriction enzymes important to understand? We don’t use RE for
HLA typing any more but the concept of the frequency of specific sequences
in the DNA will be important for other methods of DNA typing. Also, next
generation sequencing ligates DNA fragments together in library creation and
this is a topic covered in this section.
Restriction endonucleases (RE) are enzymes that cut double-stranded DNA
at a specific sequence of base pairs. These sequences are palindromes,
that is, they read the same forwards (coding strand) as backwards
(noncoding strand). For example, the restriction enzyme EcoRI cleaves DNA
at the sequence:
---GAATTC--- (coding strand)
---CTTAAG--- (noncoding strand)
B.
Depending on the restriction enzyme, the ends generated after cleavage are
blunt (or flat) or have a 5' or 3' single strand protrusion (cohesive or sticky
end).
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Blunt ends are generated by the RE SmaI:
5' ----CCCGGG---- 3' ===>
3' ----GGGCCC---- 5'
5' ---CCC 3'
3' ---GGG 5'
5' GGG--- 3'
3' CCC--- 5'
A 5' single strand protrusion is formed by the RE EcoRI:
5' ----GAATTC---- 3' --->
3' ----CTTAAG---- 5'
5' ----G 3'
5' AATTC---- 3'
3' ----CTTAA 5'
3' G---- 5'
═══════════════════════════════════════════════════════════════════
QUESTION 1: Look up the recognition sequence for the restriction enzyme PstI on the
internet. Write out the sequence of a double stranded DNA with the restriction enzyme site
in it. Label the 5' and 3' ends. Draw the fragments generated after cleavage with the 5'
and 3' ends labeled. What kind of protrusion is generated?
═══════════════════════════════════════════════════════════════════
C.
Cohesive-ended fragments can be ligated (or linked) to one another only if
their protrusions are compatible (have complementary sequences). Any
DNA fragments which have blunt ends can be ligated together.
D.
The number of nucleotides defining the cleavage site of a restriction enzyme
can vary. RE with shorter recognition sequences cut DNA more frequently
than those with longer recognition sequences. Assuming that a piece of
DNA has an equal content of each base (A, C, G, and T), a RE with a 4-base
recognition sequence will cleave the DNA, on average, every 4 4 (256) bases
compared to every 46 (4096) bases for a RE with a 6-base recognition
sequence.
Note: The probability of cleavage for a RE that cuts GGCT is calculated as
the probability of finding that sequence in the DNA. Probability of a specific
base, e.g., G, occurring is 1/4 (based on 4 bases: A,C,G,T) so probability of
the specific sequence GGCT is 1/4x1/4x1/4x1/4 = 1/256.
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E.
RE are endonucleases. If a restriction site appears at the end of a DNA
fragment, the restriction enzyme will not cut the DNA.
5' GAATTCGACTGCCATA 3'
3' CTTAAGCTGACGGTAT 5' will not be cut by EcoRI
═══════════════════════════════════════════════════════════════════
QUESTION 2: Write down the DNA sequence cleaved by the RE NotI. Will NotI cleave
genomic DNA more or less frequently than EcoRI? [Hint: You will need to look up the
recognition sequence of this enzyme before you can answer the question.]
═══════════════════════════════════════════════════════════════════
III.
Denaturation/Hybridization
A.
Denaturation disrupts the hydrogen bonds which hold the bases and, hence,
the double stranded DNA together. Denaturation can be accomplished by
heating the DNA or by treating the DNA with alkali (NaOH) or polar solvents
(e.g., dimethyl sulfoxide (DMSO), formamide) which break hydrogen bonds.
B.
The melting temperature (Tm) is defined as the temperature at which 50% of
the DNA is hybridized (i.e., found in a double stranded form) and 50% is
denatured. The Tm for a short piece of DNA can be estimated by [4 x G+C
pairs] + [2 x A+T pairs]. The Tm is influenced by the base composition and
the length of the double stranded DNA. Heating DNA at 94 oC or higher will
usually denature all double stranded DNA regardless of the length or base
composition. For example, the Tm of the double stranded DNA sequence:
AATGCGGAT
TTACGCCTA
is (4x4)+(2x5)=26oC.
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═══════════════════════════════════════════════════════════════════
QUESTION 3: Write out the sequence of any piece of DNA that is 18 base pairs in length
and determine its approximate melting temperature. Is the melting temperature higher or
lower than the example shown above? What would be the melting temperature of the 18
base pair sequence if it was made up of only G-C pairs? Only A-T pairs?
═══════════════════════════════════════════════════════════════════
C.
Hybridization or reannealing of DNA regenerates the base pairs and yields
double stranded DNA. The efficiency of hybridization depends on:
1.
Concentration of DNA: For example, if we denature the DNA found in
a human cell, it will take a long time for one strand of an HLA gene to
find its complement among all the other genes present. In contrast,
the strand can rapidly find its complement in a solution that contains
only many copies of that HLA gene.
Figure 1-2
DNA Denaturation / Annealing
Stringent
Non-Stringent
DNA
G
C
A
T
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D.
2.
Time that the hybridization is given to take place: The more complex
the mixture of DNA, the more time a single stranded piece of DNA
needs to find its match. For example, one strand of an HLA gene will
have a hard time finding its complement in a mixture containing all the
genes found in a human cell. The longer the hybridization time, the
more likely the strands are to find one another and anneal.
3.
Base composition of the DNA: Stretches of GC pairs tend to anneal
more rapidly than stretches of AT pairs to form double stranded DNA.
This probably happens because the formation of 3 hydrogen bonds
between a single GC pair stabilizes the hybridized strands to a greater
extent than the 2 hydrogen bonds formed by an AT pair. TMAC
(tetramethylammonium chloride) is a chemical which causes DNA to
reanneal at a rate related only to its length so that the time of
reannealing does not depend on the numbers of GC pairs.
Single strands of DNA will bind to imperfectly matched single strands under
conditions of low stringency [Figure 1-2]. Under high stringency conditions,
only perfect matches are found. Usually, salt concentration and temperature
are used to control the stringency.
1.
Temperature of the reaction: High temperatures cause the DNA
strands to move more rapidly. Imperfectly matched DNA hybrids will
dissociate more rapidly at high temperatures. Therefore, higher
temperatures favor the generation of perfectly matched hybrids.
Obviously, if the temperature is too high, the two strands will not
reanneal.
2.
Salt concentration of the reaction mixture: High concentrations of salt
allow imperfectly matched hybrids to be formed. At low salt
concentrations, only perfectly matched hybrids will form.
3.
Low salt and high temperature create high stringency conditions. The
stringency of the match is controlled during the incubation of single
stranded DNA (hybridization) or during the wash following the
hybridization reaction. The temperature is the easiest parameter to
adjust to obtain perfect matches. If using a short piece of single
stranded DNA for hybridization, the final high stringency wash is often
carried out 3-50C below the melting temperature to keep the perfectly
matched strands hybridized but to eliminate imperfectly matched
hybrids [Figure 1-3].
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Figure 1-3
Denaturation of DNA
+ Single stranded synthetic DNA (oligo)
Control Stringency
Hybridization
Wash to remove
imperfectly matched
hybrids
E.
Probes are used to detect specific DNA sequences by hybridization. For
HLA typing, these probes are synthetic single stranded DNA
(oligonucleotides) which are 12-26 nucleotides in length. Probes are often
called sequence specific oligonucleotide probes (SSOP).
1.
F.
IV.
Because many copies of these probes can be added to create a high
concentration, hybridization takes place rapidly.
One strand of the DNA is often attached to a solid support (e.g., membrane
or bead) to increase the speed of hybridization and to aid in the detection of
a successful hybridization reaction.
Detection of hybridization
A.
If the probe is labeled, hybridization can be detected.
1.
Probes can be labeled using radioactive phosphate (P32) or by
adding a modified or unusual nucleotide to a probe. The label is
added to the probe either during or after synthesis of the probe.
Methods to do this are described later.
2.
Binding can be detected by autoradiography (to detect P32) or
through detection of color, chemiluminescence (light emitted by a
chemical reaction) or fluorescence. Detectors may be X-ray film,
ELISA plate readers, fluorescence or chemiluminescence detectors,
or the human eye. Figure 1-4 illustrates a detection system.
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Figure 1-4
Detection of DNA Hybridized to Probe
Probe
Y
Enzyme
Antibody
to tag
tag
Substrate
DNA-1
DNA-2
DNA-3
DNA-4
Solid support--membrane
Probe hybridizes to DNA-2
V.
Electrophoresis
A.
Electrophoresis through agarose or polyacrylamide gels is one method to
separate, identify, and purify DNA fragments [Figure 1-5]. DNA is negatively
charged (due to the phosphate backbone) and moves away from the
negative pole (cathode) and toward the positive pole (anode) in an electric
field. Pieces of DNA are identified by size. Small pieces of DNA move more
rapidly than large pieces of DNA. DNA is visualized on the gel by UV light if
the DNA is stained with ethidium bromide or other less hazardous dyes or by
X-ray film if the DNA is labeled with a radioactive isotope.
Figure 1-5
Electrophoresis of DNA
DNA – Negative Phosphate Backbone
-
1.5-2% Agarose Gel
+
Tris/Acetate/EDTA or Tris/Borate/EDTA buffer
Power
Supply
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B.
The gel itself is a matrix through which the pieces of DNA travel.
Polyacrylamide-like gels are used to separate small fragments of DNA (5-500
base pairs) and are used for Sanger-based DNA sequencing or
oligonucleotide purification. Agarose gels are used for DNA from 200 base
pairs to 50,000 base pairs (50 kilobases). Two DNA fragments which are 50
bp and 55 bp will likely migrate at different rates and will be identified on an
acrylamide gel but will migrate together (and not be resolved) on an agarose
gel.
C.
The percent composition of acrylamide or agarose determines how easily
pieces of DNA of different sizes can travel through the matrix; therefore, the
percent composition of the gel will determine the range over which DNA
fragments are resolved (separated from one another).
═══════════════════════════════════════════════════════════════════
QUESTION 4: Figure 1-6 is a picture of a stained agarose gel. The lane labeled M
contains commercially purchased marker DNA which contains DNA fragments that are
multiples of 100 base pairs in length. The other two lanes contain pieces of DNA that have
been electrophoresed in the gel. Label the positive and negative poles of the gel and
indicate the direction of DNA migration. What are the approximate sizes of the two DNA
fragments in the lanes?
Figure 1-6
Agarose Gel
M
BP
700
400
100
═══════════════════════════════════════════════════════════════════
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VI.
Synthesis of DNA
A.
Enzymes called DNA polymerases synthesize single strands of DNA [Figure
1-7]. The enzymes require:
1.
A single stranded DNA template to copy. The polymerases
synthesize a product whose sequence is complementary to the
template. Thermostable DNA polymerase (Taq DNA polymerase)
and similar enzymes require a DNA template. Denatured genomic
DNA can served as this template. When starting with a mRNA
template, the enzyme reverse transcriptase copies mRNA to make
complementary DNA (cDNA).
Figure 1-7
DNA Synthesis
5'
5'
3'
3'
3'
5'
 Single stranded DNA
 DNA polymerase (enzyme)
 Single stranded oligo primer
 Nucleotides (dNTP: N=A, C, G, T)
dNTP, 2’ deoxynucleoside 5’ triphosphate
2.
A single stranded DNA primer (a synthetic oligonucleotide) that is
hybridized to the template. Polymerases usually add nucleotides on
to the 3' end of a primer and extend the newly synthesized strand
from 5' to 3'.
3.
Nucleotides (dNTP) to form the new DNA strand.
B.
If modified nucleotides are added during synthesis or if a modified (e.g.,
biotinylated) primer is used for synthesis, the newly synthesized DNA
becomes labeled.
C.
Single stranded DNA of a defined sequence (oligonucleotide) can be made
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in the laboratory using an automated DNA synthesizer and phosphoramidite
chemistry. These pieces of DNA can be used as primers for DNA synthesis
or as probes in hybridization reactions. Modified nucleotides might be used
instead of A,C,G,T to provide more stability or to enhance hybridization to
certain sequences.
D.
DNA ligases (e.g., T4 DNA ligase) catalyze the formation of the phosphate
backbone in double stranded DNA and can be used to join up DNA
fragments [Figure 1-8]. For example, a piece of DNA cut with the RE EcoRI
into two fragments can be put together again by incubating with a ligase.
Figure 1-8
Restriction Enzyme (EcoRI)
G
CTTAA
AATTC
G
GTATTC
CATAAG
G
CTTAA
AATTC
G
GTATTC
CATAAG
Ligase
Kinase
GTATTC
*P- GTATTC
TDT (Terminal transferase)
TGCTTAC
VII.
TGCTTACUUUU
E.
Kinases (e.g., T4 polynucleotide kinase) add phosphates on to the 5' ends of
DNA fragments [Figure 1-8]. This enables DNA to be ligated or attached to
another piece of DNA.
F.
Terminal transferase adds nucleotides to the 3' ends of DNA molecules to
make a single stranded tail [Figure 1-8]. Terminal transferase can be used to
create homopolymer tails (tails of all one nucleotide) for cloning or for
labeling the DNA.
Extraction of DNA from cells
A.
Any cell with a nucleus can be used as a source of DNA [Figure 1-9]. Red
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blood cells do not contain nuclei; however, other cells in the blood like white
blood cells are a good source of DNA. Cell lines like transformed B cells are
also a good source of DNA. Because transformed cells can be grown in
culture in the laboratory, they provide an inexhaustible supply of DNA.
B.
Many different protocols can be used to isolate DNA from cells. One
protocol uses Triton X-100 (a detergent) to lyse the cell membrane releasing
the nuclei. If the starting material was whole blood, these nuclei must then
be washed extensively to wash away any hemoglobin released by the red
blood cells. The heme portion of hemoglobin interfers with the gene
amplification reaction used to determine HLA types.
Figure 1-9
DNA & Its Preparation
 More stable than RNA
 Found in all nucleated cells
Nucleus
 Lyse cell / nuclear membranes
DNA
 Remove proteins bound to
Protein
DNA
 Isolate DNA away from other
cell components
Cell
C.
The nuclei are lysed using another detergent, Tween-20, and the DNA freed
from the proteins bound to it by treatment with Proteinase K, an enzyme
which destroys proteins.
D.
After the proteins are destroyed, the Proteinase K is also destroyed by
incubation of the DNA at high temperatures (90oC). It is important to destroy
the Proteinase K because it can degrade the enzyme used in the HLA typing
reaction.
E.
Other methods may employ a solid phase to bind DNA for isolation.
F.
A number of vendors sell kits to prepare DNA.
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Reference:
Green and Sambrook. Molecular Cloning, A Laboratory Manual, 4th edition. Cold Spring
Harbor Laboratory Press
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CHAPTER 2
HLA CLASS I PROTEINS AND GENES
The purpose of this chapter is to describe an important group of HLA molecules, the class I
molecules. Matching for bone marrow transplantation involves identification of these class
I molecules, HLA-A, HLA-B and HLA-C. This chapter also discusses the general
properties of proteins and genes.
I.
HLA is an abbreviation for human leukocyte antigen. These protein molecules are
expressed on the surfaces of our cells and play an important role in transplantation.
Because of this, the HLA molecules of graft donor and recipient must be identified
to find the best match. These molecules are often called histocompatibility
molecules because of their role in compatibility of tissue.
II.
Class I HLA molecules are histocompatibility molecules which are identified during
HLA typing.
A.
The class I molecules are found on the surface of essentially all nucleated
cells in the body.
Figure 2-1
Class I Protein Structure
NH2
Alpha 1 domain
Alpha 2 domain
NH2
Alpha 3 domain
Beta 2 microglobulin
Transmembrane region
Cytoplasmic region
COOH
B.
At least three different class I molecules, HLA-A, HLA-B, and HLA-C, are
expressed on each cell of an individual. These molecules are very similar to
one another. There are approximately 0.5-1 million class I molecules on the
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surface of a cell.
C.
Class I molecules are comprised of a cell membrane glycopolypeptide (44K
MW, alpha or heavy chain) associated with a second polypeptide, beta-2
microglobulin (12K MW) [Figure 2-1]. Each chain is a linear string of amino
acids. Each class I protein has a specific sequence of amino acids that
differs from other proteins. The sequence of an HLA-A molecule is different
from the sequence of an HLA-B molecule. [Note: Class II molecules also
have alpha chains but they are different in amino acid sequence from class I
alpha chains. Likewise, the HLA-A,B,C alpha chains differ from one another
in protein sequence.]
Table 2-1. One- and Three-Letter Codes for Amino Acids
Alanine
Ala
A
Leucine
Leu
L
Arginine
Arg
R
Lysine
Lys
K
Asparagine
Asn
N
Methionine
Met
M
Aspartic Acid
Asp
D
Phenylalanine
Phe
F
Cysteine
Cys
C
Proline
Pro
P
Glutamic Acid
Glu
E
Serine
Ser
S
Glutamine
Gln
Q
Threonine
Thr
T
Glycine
Gly
G
Trytophan
Trp
W
Histidine
His
H
Tyrosine
Tyr
Y
Isoleucine
Ile
I
Valine
Val
V
There are 20 amino acids (e.g., lysine, serine, leucine) [Table 2-1]. To save
space, publications sometimes use a single letter to refer to an amino acid.
For example, S is used for serine, L for leucine, and K for lysine.
One end of the polypeptide chain is called the amino-terminus (or Nterminus) and the other end is called the carboxy-terminus (or C-terminus).
D.
Each class I alpha (or heavy) chain can be divided into regions: three
extracellular domains (each ~100 amino acids in length), transmembrane
region and cytoplasmic tail. Each alpha chain is initially synthesized with a
short signal peptide on its amino terminus. This peptide is removed from the
polypeptide during transport to the cell surface and is not found in the mature
protein.
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III.
Class I genes
A.
Some of the information specifying the amino acid sequences of the HLA
class I molecules is located on human chromosome 6 in a region called the
major histocompatibility complex (MHC) [Figure 2-2]. The sequence of base
pairs containing the genetic information that specifies a protein sequence
defines a gene. The genes that encode the three class I 44K MW alpha
chains are located next to one another in the MHC. Beta-2 microglobulin is
encoded on another chromosome.
Figure 2-2
Human MHC
HLA Class II Region
DP
B1
A1
DQ
B1
HLA Class I Region
DR
A1
B1
B3
B
C
A
A
B4
B5
B.
The genes that encode the alpha chains of the class I proteins are located
next to one another. This cluster of genes is part of the major
histocompatibility gene complex (MHC). This complex encompasses
approximately 3500 kb (3,500,000 bases) of DNA. Beta2 microglobulin is
encoded on another chromosome. When the location of a gene on a
chromosome is known, that gene can also be called a locus.
C.
The World Health Organization (W.H.O.) has a committee that assigns
names to HLA loci. For example, HLA-A is the name of the locus that
encodes the HLA-A alpha chain.
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D.
The information for a single gene is not found in a single stretch of base
pairs but is found in multiple segments of double stranded DNA called exons
[Figure 2-3]. These segments are separated by intervening segments of
DNA called introns. The entire set of exons containing the coding information
for an entire polypeptide chain is called a "gene" [Figure 2-3].
Figure 2-3
Class I Alpha Chain Locus
DNA (exons/introns)
L
1
A1
A2
A3
TM
2
3
4
5
Cyt
6
7
8
mRNA (cDNA/CDS)
1
2
3
4
5
6
1
2
3
4
5
6
7
8
Polypeptide
IV.
Transfer of information from DNA to RNA to protein.
A.
The DNA encoding the HLA molecule is copied (transcribed) into mRNA.
After processing, the mRNA contains only the exon sequences.
═══════════════════════════════════════════════════════════════════
QUESTION 1: Figure 2-4 lists the genomic or DNA sequence of an HLA-A gene. Figure 25 lists the coding sequence (CDS) (=mRNA sequence) sequence from the same HLA-A
gene. Use the information in Figure 2-5 to identify and box the exons which encode the
HLA-A gene in the genomic DNA sequence in Figure 2-4. Remember that genomic DNA
contains exons and introns. Hint: The vertical lines in Figure 2-4 and the shading added to
Figure 2-5 will help you find the exons.
═══════════════════════════════════════════════════════════════════
C.W. Bill Young Marrow Donor Recruitment and Research Program
20
Figure 2-4.
Full length genomic sequence A*01:01:01:01
A*01:01:01:01
-291
-281
-271
-261
-251
-241
-231
-221
-211
-201
CAGGAGCAGA GGGGTCAGGG CGAAGTCCCA GGGCCCCAGG CGTGGCTCTC AGGGTCTCAG GCCCCGAAGG CGGTGTATGG ATTGGGGAGT CCCAGCCTTG
A*01:01:01:01
-191
-181
-171
-161
-151
-141
-131
-121
-111
-101
GGGATTCCCC AACTCCGCAG TTTCTTTTCT CCCTCTCCCA ACCTACGTAG GGTCCTTCAT CCTGGATACT CACGACGCGG ACCCAGTTCT CACTCCCATT
A*01:01:01:01
-91
-81
-71
-61
-51
-41
-31
-21
-11
-1
GGGTGTCGGG TTTCCAGAGA AGCCAATCAG TGTCGTCGCG GTCGCTGTTC TAAAGTCCGC ACGCACCCAC CGGGACTCAG ATTCTCCCCA GACGCCGAGG
A*01:01:01:01
10
20
30
40
50
60
70
80
90
100
|ATGGCCGTCA TGGCGCCCCG AACCCTCCTC CTGCTACTCT CGGGGGCCCT GGCCCTGACC CAGACCTGGG CGG|GTGAGTG CGGGGTCGGG AGGGAAACCG
A*01:01:01:01
110
120
130
140
150
160
170
180
190
200
CCTCTGCGGG GAGAAGCAAG GGGCCCTCCT GGCGGGGGCG CAGGACCGGG GGAGCCGCGC CGGGAGGAGG GTCGGGCAGG TCTCAGCCAC TGCTCGCCCC
A*01:01:01:01
210
220
230
240
250
260
270
280
290
300
CAG|GCTCCCA CTCCATGAGG TATTTCTTCA CATCCGTGTC CCGGCCCGGC CGCGGGGAGC CCCGCTTCAT CGCCGTGGGC TACGTGGACG ACACGCAGTT
A*01:01:01:01
310
320
330
340
350
360
370
380
390
400
CGTGCGGTTC GACAGCGACG CCGCGAGCCA GAAGATGGAG CCGCGGGCGC CGTGGATAGA GCAGGAGGGG CCGGAGTATT GGGACCAGGA GACACGGAAT
A*01:01:01:01
410
420
430
440
450
460
470
480
490
500
ATGAAGGCCC ACTCACAGAC TGACCGAGCG AACCTGGGGA CCCTGCGCGG CTACTACAAC CAGAGCGAGG ACG|GTGAGTG ACCCCGGCCC GGGGCGCAGG
A*01:01:01:0
510
520
530
540
550
560
570
580
590
600
TCACGACCCC TCATCCCCCA CGGACGGGCC AGGTCGCCCA CAGTCTCCGG GTCCGAGATC CACCCCGAAG CCGCGGGACT CCGAGACCCT TGTCCCGGGA
A*01:01:01:01
610
620
630
640
650
660
670
680
690
700
GAGGCCCAGG CGCCTTTACC CGGTTTCATT TTCAGTTTAG GCCAAAAATC CCCCCGGGTT GGTCGGGGCG GGGCGGGGCT CGGGGGACTG GGCTGACCGC
A*01:01:01:01
710
720
730
740
750
760
770
780
790
800
GGGGTCGGGG CCAG|GTTCTC ACACCATCCA GATAATGTAT GGCTGCGACG TGGGGCCGGA CGGGCGCTTC CTCCGCGGGT ACCGGCAGGA CGCCTACGAC
A*01:01:01:01
810
820
830
840
850
860
870
880
890
900
GGCAAGGATT ACATCGCCCT GAACGAGGAC CTGCGCTCTT GGACCGCGGC GGACATGGCA GCTCAGATCA CCAAGCGCAA GTGGGAGGCG GTCCATGCGG
A*01:01:01:01
910
920
930
940
950
960
970
980
990
1000
CGGAGCAGCG GAGAGTCTAC CTGGAGGGCC GGTGCGTGGA CGGGCTCCGC AGATACCTGG AGAACGGGAA GGAGACGCTG CAGCGCACGG |GTACCAGGGG
A*01:01:01:01
1010
1020
1030
1040
1050
1060
1070
1080
1090
1100
CCACGGGGCG CCTCCCTGAT CGCCTATAGA TCTCCCGGGC TGGCCTCCCA CAAGGAGGGG AGACAATTGG GACCAACACT AGAATATCAC CCTCCCTCTG
A*01:01:01:01
1110
1120
1130
1140
1150
1160
1170
1180
1190
1200
GTCCTGAGGG AGAGGAATCC TCCTGGGTTT CCAGATCCTG TACCAGAGAG TGACTCTGAG GTTCCGCCCT GCTCTCTGAC ACAATTAAGG GATAAAATCT
A*01:01:01:01
1210
1220
1230
1240
1250
1260
1270
1280
1290
1300
CTGAAGGAGT GACGGGAAGA CGATCCCTCG AATACTGATG AGTGGTTCCC TTTGACACCG GCAGCAGCCT TGGGCCCGTG ACTTTTCCTC TCAGGCCTTG
A*01:01:01:01
1310
1320
1330
1340
1350
1360
1370
1380
1390
1400
TTCTCTGCTT CACACTCAAT GTGTGTGGGG GTCTGAGTCC AGCACTTCTG AGTCTCTCAG CCTCCACTCA GGTCAGGACC AGAAGTCGCT GTTCCCTTCT
A*01:01:01:01
1410
1420
1430
1440
1450
1460
1470
1480
1490
1500
CAGGGAATAG AAGATTATCC CAGGTGCCTG TGTCCAGGCT GGTGTCTGGG TTCTGTGCTC TCTTCCCCAT CCCGGGTGTC CTGTCCATTC TCAAGATGGC
A*01:01:01:01
1510
1520
1530
1540
1550
1560
1570
1580
1590
1600
CACATGCGTG CTGGTGGAGT GTCCCATGAC AGATGCAAAA TGCCTGAATT TTCTGACTCT TCCCGTCAG|A CCCCCCCAAG ACACATATGA CCCACCACCC
A*01:01:01:01
1610
1620
1630
1640
1650
1660
1670
1680
1690
1700
CATCTCTGAC CATGAGGCCA CCCTGAGGTG CTGGGCCCTG GGCTTCTACC CTGCGGAGAT CACACTGACC TGGCAGCGGG ATGGGGAGGA CCAGACCCAG
A*01:01:01:01
1710
1720
1730
1740
1750
1760
1770
1780
1790
1800
GACACGGAGC TCGTGGAGAC CAGGCCTGCA GGGGATGGAA CCTTCCAGAA GTGGGCGGCT GTGGTGGTGC CTTCTGGAGA GGAGCAGAGA TACACCTGCC
A*01:01:01:01
1810
1820
1830
1840
1850
1860
1870
1880
1890
1900
ATGTGCAGCA TGAGGGTCTG CCCAAGCCCC TCACCCTGAG ATGGG|GTAAG GAGGGAGATG GGGGTGTCAT GTCTCTTAGG GAAAGCAGGA GCCTCTCTGG
1910
1920
1930
1940
1950
1960
1970
C.W. Bill Young Marrow Donor Recruitment and Research Program
1980
1990
2000
21
A*01:01:01:01
AGACCTTTAG CAGGGTCAGG GCCCCTCACC TTCCCCTCTT TTCCCAG|AGC TGTCTTCCCA GCCCACCATC CCCATCGTGG GCATCATTGC TGGCCTGGTT
A*01:01:01:01
2010
2020
2030
2040
2050
2060
2070
2080
2090
2100
CTCCTTGGAG CTGTGATCAC TGGAGCTGTG GTCGCTGCCG TGATGTGGAG GAGGAAGAGC TCAG|GTGGAG AAGGGGTGAA GGGTGGGGTC TGAGATTTCT
A*01:01:01:01
2110
2120
2130
2140
2150
2160
2170
2180
2190
2200
TGTCTCACTG AGGGTTCCAA GCCCCAGCTA GAAATGTGCC CTGTCTCATT ACTGGGAAGC ACCTTCCACA ATCATGGGCC GACCCAGCCT GGGCCCTGTG
A*01:01:01:01
2210
2220
2230
2240
2250
2260
2270
2280
2290
2300
TGCCAGCACT TACTCTTTTG TAAAGCACCT GTTAAAATGA AGGACAGATT TATCACCTTG ATTACGGCGG TGATGGGACC TGATCCCAGC AGTCACAAGT
A*01:01:01:01
2310
2320
2330
2340
2350
2360
2370
2380
2390
2400
CACAGGGGAA GGTCCCTGAG GACAGACCTC AGGAGGGCTA TTGGTCCAGG ACCCACACCT GCTTTCTTCA TGTTTCCTGA TCCCGCCCTG GGTCTGCAGT
A*01:01:01:01
2410
2420
2430
2440
2450
2460
2470
2480
2490
2500
CACACATTTC TGGAAACTTC TCTGGGGTCC AAGACTAGGA GGTTCCTCTA GGACCTTAAG GCCCTGGCTC CTTTCTGGTA TCTCACAGGA CATTTTCTTC
A*01:01:01:01
2510
2520
2530
2540
2550
2560
2570
2580
2590
2600
CCACAG|ATAG AAAAGGAGGG AGTTACACTC AGGCTGCAA|G TAAGTATGAA GGAGGCTGAT GCCTGAGGTC CTTGGGATAT TGTGTTTGGG AGCCCATGGG
A*01:01:01:01
2610
2620
2630
2640
2650
2660
2670
2680
2690
2700
GGAGCTCACC CACCCCACAA TTCCTCCTCT AGCCACATCT TCTGTGGGAT CTGACCAGGT TCTGTTTTTG TTCTACCCCA G|GCAGTGACA GTGCCCAGGG
A*01:01:01:01
2710
2720
2730
2740
2750
2760
2770
2780
2790
2800
CTCTGATGTG TCTCTCACAG CTTGTAAAG|G TGAGAGCTTG GAGGGCCTGA TGTGTGTTGG GTGTTGGGTG GAACAGTGGA CACAGCTGTG CTATGGGGTT
A*01:01:01:01
2810
2820
2830
2840
2850
2860
2870
2880
2890
2900
TCTTTGCGTT GGATGTATTG AGCATGCGAT GGGCTGTTTA AGGTGTGACC CCTCACTGTG ATGGATATGA ATTTGTTCAT GAATATTTTT TTCTATAG|TG
A*01:01:01:01
2910
2920
2930
2940
2950
2960
2970
2980
2990
3000
TGA|GACAGCT GCCTTGTGTG GGACTGAGAG GCAAGAGTTG TTCCTGCCCT TCCCTTTGTG ACTTGAAGAA CCCTGACTTT GTTTCTGCAA AGGCACCTGC
A*01:01:01:01
3010
3020
3030
3040
3050
3060
3070
3080
3090
3100
ATGTGTCTGT GTTCGTGTAG GCATAATGTG AGGAGGTGGG GAGAGCACCC CACCCCCATG TCCACCATGA CCCTCTTCCC ACGCTGACCT GTGCTCCCTC
A*01:01:01:01
3110
3120
3130
3140
3150
3160
3170
3180
3190
3200
CCCAATCATC TTTCCTGTTC CAGAGAGGTG GGGCTGAGGT GTCTCCATCT CTGTCTCAAC TTCATGGTGC ACTGAGCTGT AACTTCTTCC TTCCCTATTA
C.W. Bill Young Marrow Donor Recruitment and Research Program
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Figure 2-5.
Nucleotide CDS (coding sequence) A*01:01:01:01
-20
-15
-10
-5
1
ATG GCC GTC ATG GCG CCC CGA ACC CTC CTC CTG CTA CTC TCG GGG GCC CTG GCC CTG ACC CAG ACC TGG GCG G|GC
A*01:01:01:01
A*01:01:01:01
5
10
15
20
25
TCC CAC TCC ATG AGG TAT TTC TTC ACA TCC GTG TCC CGG CCC GGC CGC GGG GAG CCC CGC TTC ATC GCC GTG GGC
A*01:01:01:01
30
35
40
45
50
TAC GTG GAC GAC ACG CAG TTC GTG CGG TTC GAC AGC GAC GCC GCG AGC CAG AAG ATG GAG CCG CGG GCG CCG TGG
A*01:01:01:01
55
60
65
70
75
ATA GAG CAG GAG GGG CCG GAG TAT TGG GAC CAG GAG ACA CGG AAT ATG AAG GCC CAC TCA CAG ACT GAC CGA GCG
A*01:01:01:01
80
85
90
95
100
AAC CTG GGG ACC CTG CGC GGC TAC TAC AAC CAG AGC GAG GAC G|GT TCT CAC ACC ATC CAG ATA ATG TAT GGC TGC
A*01:01:01:01
105
110
115
120
125
GAC GTG GGG CCG GAC GGG CGC TTC CTC CGC GGG TAC CGG CAG GAC GCC TAC GAC GGC AAG GAT TAC ATC GCC CTG
A*01:01:01:01
130
135
140
145
150
AAC GAG GAC CTG CGC TCT TGG ACC GCG GCG GAC ATG GCA GCT CAG ATC ACC AAG CGC AAG TGG GAG GCG GTC CAT
A*01:01:01:01
155
160
165
170
175
GCG GCG GAG CAG CGG AGA GTC TAC CTG GAG GGC CGG TGC GTG GAC GGG CTC CGC AGA TAC CTG GAG AAC GGG AAG
A*01:01:01:01
180
185
190
195
200
GAG ACG CTG CAG CGC ACG G|AC CCC CCC AAG ACA CAT ATG ACC CAC CAC CCC ATC TCT GAC CAT GAG GCC ACC CTG
A*01:01:01:01
205
210
215
220
225
AGG TGC TGG GCC CTG GGC TTC TAC CCT GCG GAG ATC ACA CTG ACC TGG CAG CGG GAT GGG GAG GAC CAG ACC CAG
A*01:01:01:01
230
235
240
245
250
GAC ACG GAG CTC GTG GAG ACC AGG CCT GCA GGG GAT GGA ACC TTC CAG AAG TGG GCG GCT GTG GTG GTG CCT TCT
A*01:01:01:01
255
260
265
270
275
GGA GAG GAG CAG AGA TAC ACC TGC CAT GTG CAG CAT GAG GGT CTG CCC AAG CCC CTC ACC CTG AGA TGG G|AG CTG
A*01:01:01:01
280
285
290
295
300
TCT TCC CAG CCC ACC ATC CCC ATC GTG GGC ATC ATT GCT GGC CTG GTT CTC CTT GGA GCT GTG ATC ACT GGA GCT
A*01:01:01:01
305
310
315
320
325
GTG GTC GCT GCC GTG ATG TGG AGG AGG AAG AGC TCA G|AT AGA AAA GGA GGG AGT TAC ACT CAG GCT GCA A|GC AGT
A*01:01:01:01
330
335
340
GAC AGT GCC CAG GGC TCT GAT GTG TCT CTC ACA GCT TGT AAA G|TG TGA
C.W. Bill Young Marrow Donor Recruitment and Research Program
23
B.
The mRNA is translated into protein by the ribosomes. The ribosome reads
the genetic code to convert RNA sequences into a protein sequence.
The genetic code consists of all possible triplet combinations of RNA bases.
Each triplet (codon) specifies one amino acid. For example, the codon UCU
(TCT in DNA) specifies the amino acid serine. An amino acid can be
specified by more than one triplet. For example, UCC, UCA, and UCG also
specify serine.
The class I protein sequences start with a methionine at the amino-terminus
of the signal peptide encoded by an AUG codon in the mRNA. Some triplets
are called "stop codons" and identify the end of the protein sequence
(carboxy-terminus). UGA is a stop codon.
C.
The class I signal peptide and part of the first amino acid of the alpha-1
domain are encoded in exon 1 [Figure 2-3]. The rest of the first domain is
encoded in exon 2. Exon 3 encodes the second domain and exon 4
encodes the third domain. The remainder of the exons encode the
transmembrane and cytoplasmic regions.
D.
Intron 1 follows exon 1, intron 2 follows exon 2 and so forth.
E.
If the gene is characterized by a sequence analysis of a DNA copy of the
mRNA, that sequence is called a cDNA (complementary DNA) sequence.
The cDNA sequence reported in the literature has the same sequence as the
mRNA and is always written 5' (on the left) to 3'.
═══════════════════════════════════════════════════════════════════
QUESTION 2: Use Figure 2-5 and Table 2-2 to translate the cDNA (mRNA) sequence for
HLA-A into a protein sequence. The codon encoding the first amino acid in the leader
sequence is indicated. Circle the stop codon.
═══════════════════════════════════════════════════════════════════
V.
The polypeptide specified by an alpha gene associates with the polypeptide
specified by the beta-2 microglobulin gene to form a class I protein [Figure 2-1].
Polypeptide is a term used to indicate that the alpha and beta chains are not usually
found alone but are found in an complex (also called a heterodimer).
C.W. Bill Young Marrow Donor Recruitment and Research Program
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Table 2-2. The Alphabet of the Genetic Code 1
1
2
3
UUU2
Phe
UCU
Ser
UAU
Tyr
UGU
Cys
UUC
Phe
UCC
Ser
UAC
Tyr
UGC
Cys
UUA
Leu
UCA
Ser
UAA
Term3
UGA
Term
UUG
Leu
UCG
Ser
UAG
Term
UGG
Trp
CUU
Leu
CCU
Pro
CAU
His
CGU
Arg
CUC
Leu
CCC
Pro
CAC
His
CGC
Arg
CUA
Leu
CCA
Pro
CAA
Gln
CGA
Arg
CUG
Leu
CCG
Pro
CAG
Gln
CGG
Arg
AUU
Ile
ACU
Thr
AAU
Asn
AGU
Ser
AUC
Ile
ACC
Thr
AAC
Asn
AGC
Ser
AUA
Ile
ACA
Thr
AAA
Lys
AGA
Arg
AUG
Met
ACG
Thr
AAG
Lys
AGG
Arg
GUU
Val
GCU
Ala
GAU
Asp
GGU
Gly
GUC
Val
GCC
Ala
GAC
Asp
GGC
Gly
GUA
Val
GCA
Ala
GAA
Glu
GGA
Gly
GUG
Val
GCG
Ala
GAG
Glu
GGG
Gly
mRNA codons which specify a particular amino acid.
U is found in mRNA; T is found in DNA.
Term indicates a termination codon which halts protein synthesis.
C.W. Bill Young Marrow Donor Recruitment and Research Program
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References:
Nelson and Cox. Lehninger’s Principles of Biochemistry. 2012.
Alberts et al. Molecular Biology of the Cell. 2007.
Murphy. Janeway’s Immunobiology. 2011.
C.W. Bill Young Marrow Donor Recruitment and Research Program
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CHAPTER 3
HLA CLASS II PROTEINS AND GENES
The purpose of this chapter is to describe a second important group of HLA molecules, the
class II molecules. Matching of donor and recipient for bone marrow transplantation
involves identification of at least one of the class II molecules, HLA-DR. Many of the
characteristics of the class II molecules are similar to the class I molecules discussed in
Chapter 2.
I.
Class II HLA molecules
A.
These molecules are found on the surface of cells of the immune system like
B cells and dendritic cells although other cell types may express these
molecules under certain conditions (e.g., under the influence of cytokines).
Figure 3-1
Class II Protein Structure
Amino-termini
Alpha 1 domain
Beta 1 domain
Alpha 2 domain
Beta 2 domain
Transmembrane region
Cytoplasmic region
Carboxy-termini
B.
Three different class II molecules, HLA-DR, HLA-DQ, and HLA-DP, are
expressed on immune system cells by an individual. These molecules are
very similar to one another. There are approximately 0.5-1 million class II
molecules on the cell surface.
C.W. Bill Young Marrow Donor Recruitment and Research Program
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II.
C.
The class II molecules are cell membrane glycoproteins consisting of an
alpha polypeptide chain (~34K MW) and a beta polypeptide chain (~28K
MW) [Figure 3-1]. Each protein has a specific sequence of amino acids that
differs from other proteins. The sequence of an HLA-DR molecule is
different from the sequence of an HLA-DP molecule.
D.
Each class II polypeptide chain can be divided into regions: two extracellular
domains (each ~100 amino acids in length), a transmembrane region, and a
cytoplasmic tail. Each polypeptide is initially synthesized with a short signal
peptide on its amino terminus. This peptide is removed from the polypeptide
during transport to the cell surface after synthesis and is not found in the
mature protein.
Class II genes
A.
The information specifying the amino acid sequences of the HLA class II
molecules is located on human chromosome 6 [Figure 3-2]. The sequence
of base pairs containing the genetic information that specifies a protein
sequence defines a gene.
Figure 3-2
Human MHC
HLA Class II Region
DP
B1
A1
DQ
B1
HLA Class I Region
DR
A1
B1
B3
B
C
A
A
B4
B5
B.
The genes that encode the alpha and beta chains of the class II proteins are
located next to one another in the MHC.
C.W. Bill Young Marrow Donor Recruitment and Research Program
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C.
DQB1 is the name of the locus that encodes the DQ beta chain and DQA1 is
the name of the locus that encodes the DQ alpha chain. The numbers (e.g.,
DQA1) were added because there are other class II-like genes in the MHC.
D. The information for a single gene is found in multiple segments of double
stranded DNA called exons [Figure 3-3].
═══════════════════════════════════════════════════════════════════
QUESTION 1: Go to the website http://hla.alleles.org/ and find a copy of the current listing
of the W.H.O. nomenclature for HLA alleles. What is the name of the locus that encodes
the DR alpha chain?
═══════════════════════════════════════════════════════════════════
Figure 3-3
DR Beta Chain Locus
DNA (exons/introns)
L
1
B1
B2
2
3
Tm
4
Cyt
5
6
mRNA (cDNA/CDS)
1
2
3
4
5
6
1
2
3
4
5
6
Polypeptide
C.W. Bill Young Marrow Donor Recruitment and Research Program
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CHAPTER 4
HLA ALLELES AND INHERITANCE
The purpose of this chapter is to describe the HLA diversity in the human population. It is
this diversity that makes it is so difficult to find an unrelated donor with the same HLA
alleles as a patient.
I.
HLA genes are highly polymorphic (many forms). This means that, if we look at an
HLA gene in unrelated individuals in a large population of humans, we will see that
many people have a different DNA sequence for that gene. Each different DNA
sequence is called an allele. These different gene sequences give rise to many
different class I and class II allelic products (polypeptides).
A.
Some loci have many alleles. For example, the DRB1 locus has over 1500
alleles. Other loci have only a few alleles. For example, DRA has seven
alleles.
B.
It is unlikely that two unrelated individuals carry the same class II alleles at
DR, DQ and DP loci. A ballpark estimate for common HLA alleles is that
only 1 person in 20,000 people from the same ethnic group will carry the
same HLA alleles as another person in that group.
C.
The two alleles carried by an individual is called their genotype.
Figure 4-1
C.W. Bill Young Marrow Donor Recruitment and Research Program
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II.
In the class II alleles, the second exons which encode the first domain of each
polypeptide chain contain the majority of the allelic differences. Allelic differences
can also be found in other exons. HLA class II typing usually focuses on defining
the differences in the second exon [Figure 4-1].
═══════════════════════════════════════════════════════════════════
QUESTION 1: Go to the web, http://www.ebi.ac.uk/ipd/imgt/hla/, and find a copy of the
current list of the CDS (coding region) sequences of the DRB1 alleles. Practice using the
alignment tool.
═══════════════════════════════════════════════════════════════════
III.
The W.H.O. assigns names to HLA alleles. For example, alleles at the DRB1 locus
are called: DRB1*01:01:01, DRB1*01:02:01, DRB1*04:02:01.
A.
Each HLA allele is designated by the name of the gene or locus followed by
an asterisk and two or more fields separated by colons [Figure 4-2]. For
example, DPB1*02:01:02 is an allele of the HLA-DP B1 gene;
DQB1*03:01:01:01 is an allele of the HLA-DQ B1 gene; DQA1*01:01:01 is
an allele of the HLA-DQ A1 gene. The first field following the asterisk is a
numerical designation often based on the serologic type of the resultant
protein (explained in Chapter 5) and/or the similarity to other alleles in that
group or family. The next field in the allele designation refers to the order in
which that allele was discovered. The allele name should be viewed as
simply a unique name for the allele and does not necessarily imply anything
about its relationship to other alleles with similar names.
═══════════════════════════════════════════════════════════════════
QUESTION 2: Use http://hla.alleles.org/ to list a few alleles of the DQB1 locus.
═══════════════════════════════════════════════════════════════════
B.
Sometimes shorter names are used such as DRB1*01:01. This designation
means that the typing does not distinguish among alleles whose names all
start with DRB1*01:01.
These alleles include DRB1*01:01:01,
DRB1*01:01:02 and DRB1*01:01:03.
C.
Bone marrow donor registries may use letter codes to designate subsets of
HLA alleles. For example, the letters AF in DRB1*14:AF mean *14:01 or
*14:09. For example, DRB1*04:ABC means DRB1*04:03 or *04:04 or *04:06
or *04:07 or*04:08 or *04:10 or *04:11 or *04:17 or *04:19 or *04:20 or
*04:23. The letter codes can be found at https://bioinformatics.bethematch
clinical.org/.
D.
The number of people carrying a specific HLA allele varies. These frequency
C.W. Bill Young Marrow Donor Recruitment and Research Program
31
differences mean that some alleles are considered common and others may
have been only observed in one individual so far. A list of common alleles
can be found at http://igdawg.org/cwd.html. Maps showing the world-wide
distribution
of
some
HLA
alleles
can
be
found
at
http://www.pypop.org/popdata/.
═══════════════════════════════════════════════════════════════════
QUESTION 3: The DRB1*01:01 allele is found in 1.9% of individuals in a population. The
DRB1*03:01 allele is found in 7%. What is the probability of finding an individual from that
population who carries both alleles?
═══════════════════════════════════════════════════════════════════
Figure 4-2
IV.
DNA sequence differences between alleles can result in differences in the protein
sequence or can be silent at the protein level.
A.
An example of alleles that do not differ at the protein sequence level is
DRB1*11:01:01 and DRB1*11:01:02. Both alleles have the same protein
sequence but differ in the codons used to specify that sequence (silent or
synonymous substitutions). The W.H.O committee indicates silent changes
by adding a third field to the allele name. Since the immune system detects
C.W. Bill Young Marrow Donor Recruitment and Research Program
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B.
differences in the HLA proteins, silent differences that do not change the
protein sequence are not considered important in selecting an HLA matched
donor.
N is added to indicated a null or nonexpressed allele (e.g.,
DRB4*01:03:01:02N). Since the N is optional, this allele is also correctly
named DRB4*01:03:01:02. Null alleles are important to consider in the
selection of an HLA matched donor. The fourth field of this allele name
indicates that it differs from DRB4*01:03:01:01 in the intron or 5’ or 3’ regions
of the gene (that is, the exon sequences are identical).
═══════════════════════════════════════════════════════════════════
QUESTION 4: Using http://www.ebi.ac.uk/ipd/imgt/hla/, translate the first 20 codons of
DRB1*01:01:01 and DRB1*13:02:01 into amino acids using Table 2-2. [The web site will
also convert DNA sequence to protein sequence but give it a try using the table!]
═══════════════════════════════════════════════════════════════════
═══════════════════════════════════════════════════════════════════
QUESTION 5: Write down a different DNA sequence that would encode the same
polypeptide.
ACT GGT TAC TTC
Thr Gly
Tyr
Phe
T
G
Y
F
GAG
Glu
E
═══════════════════════════════════════════════════════════════════
V.
Each person carries two copies of each HLA-DR, -DQ and -DP alpha and beta
gene, one inherited from their mother and one from their father [Figures 4-3 and 44]. Furthermore, since all of the HLA genes are found on a single chromosome,
each person has inherited one copy of chromosome 6 carrying one HLA gene
complex from their mother and one copy carrying a second HLA gene complex from
their father.
C.W. Bill Young Marrow Donor Recruitment and Research Program
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A.
The HLA genes are codominantly expressed, that is, both copies encode
proteins that are expressed on a single cell.
B.
A person can have two identical alleles of a single gene (homozygous) or
may have two different alleles of a single gene (heterozygous).
C.
A parent and child share one chromosome or haplotype (haploidentical). A
sibling (brother or sister) has a 1 in 4 chance of receiving the same two
copies of chromosome 6 from their parents as another sibling and becoming
HLA identical. Thus, the class II genes, DR, DQ, DP, are inherited as a
package. Traditionally, chromosomes from the father are labeled "a" and "b"
and chromosomes from the mother are labeled "c" and "d".
Figure 4-3
Inheritance of HLA-DRB1 Locus Alleles
in a Family
DRB1*01:01
DRB1*03:01
DRB1*01:01
DRB1*03:01
DRB1*01:01
DRB1*15:01
DRB1*03:01
DRB1*15:01
DRB1*03:01
DRB1*03:01
DRB1*03:01
DRB1*15:01
 4 possible genotypes in children
 Heterozygous vs homozygous
 1 in 4 chance that two sibs inherit same two alleles
D.
Sometimes the two copies of chromosome 6, which carry the class II genes,
exchange gene segments, a process called reciprocal recombination [Figure
4-4]. This exchange reshuffles the DR, DQ, DP combinations. If this
happens in the germ cells (egg and sperm), that person's offspring may
inherit the new combination.
C.W. Bill Young Marrow Donor Recruitment and Research Program
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═══════════════════════════════════════════════════════════════════
QUESTION
6:
Father
is
DRB1*04:02,DRB1*11:03
and
mother
is
DRB1*01:01,DRB1*03:01. What are the possible DRB1 allele combinations inherited by
their children?
Can two children be DR identical (i.e., share DRB1 alleles)? Will any of the children be
homozygous?
═══════════════════════════════════════════════════════════════════
Figure 4-4
Segregation of
Haplotypes in
Families
Sib 1
A*01:01,B*07:02,DRB1*04:01
A*03:01,B*15:01,DRB1*11:04
ac
Sib 2
Father
ab
A*01:01,B*07:02,DRB1*04:01
A*02:01,B*08:01,DRB1*01:01
A*02:01,B*08:01,DRB1*01:01
A*03:01,B*15:01,DRB1*11:04
Sib 3
A*01:01,B*07:02,DRB1*04:01
A*02:01,B*53:01,DRB1*11:01
cd
A*03:01,B*15:01,DRB1*11:04
A*02:01,B*53:01,DRB1*11:01
Mother
Haplotype = combination of
alleles on chromosome
Haploidentical = share 1 haplotype
bc
ad
Sib 4
A*02:01,B*08:01,DRB1*01:01
A*02:01,B*53:01,DRB1*11:01
A*02:01,B*08:01,DRB1*01:01
A*03:01
A*03:01,B*53:01,DRB1*11:01
bd
Recombinant
═══════════════════════════════════════════════════════════════════
QUESTION 7: Father is a: DRB1*04:03,DPB1*02:01; b: DRB1*11:01,DPB1*02:01 and
mother is c: DRB1*01:01,DPB1*02:01; d: DRB1*03:02,DPB1*01:01. What are the
possible DR,DP allele combinations inherited by their children?
Could a child be DRB1*04:03,DPB1*02:01; DRB1*01:01,DPB1*01:01?
═══════════════════════════════════════════════════════════════════
C.W. Bill Young Marrow Donor Recruitment and Research Program
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VI.
The number of genes in the MHC may vary among different individuals carrying
different haplotypes or different copies of chromosome 6.
A.
Some chromosomes carry only one expressed DR beta gene [Figure 4-5].
An expressed gene is one that encodes a polypeptide. The DR beta locus is
called DRB1.
Figure 4-5
DR Genes and DR Proteins
DRB1 Gene
DRA Gene
DR protein
DRB1 Gene
DR protein
VII.
DRB4 Gene
DRA Gene
DR protein
B.
Other copies of chromosome 6 carry two expressed DR beta genes. One
locus is DRB1. The second beta chain locus is called DRB3 or DRB4 or
DRB5. Individuals who carry haplotypes containing two DR beta loci express
two different DR molecules encoded by that haplotype [Figure 4-5].
C.
Each chromosome carries a single DQA1, DQB1, DPA1, and DPB1 gene
and, thus, encodes a single DQ and a single DP molecule.
D.
Some class II genes in the MHC are pseudogenes (e.g., DRB2) and are
defective in some way. Others (e.g. DM,DO) specify proteins that are not
considered important for transplantation matching.
Particular HLA-DR alleles are often found together on chromosomes that carry two
C.W. Bill Young Marrow Donor Recruitment and Research Program
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DR beta loci.
A.
DRB1 and DRB4. Alleles whose names start with DRB1*04, DRB1*07, and
DRB1*09 are usually found on the same chromosome as alleles whose
names begin with DRB4 [Figure 4-6].
B.
DRB1 and DRB3. Alleles whose names start with DRB1*03, DRB1*11,
DRB1*12, DRB1*13, and DRB1*14, are usually found on the same
chromosome as alleles whose names begin with DRB3 [Figure 4-6].
C.
DRB1 and DRB5. Alleles whose names start with DRB1*15 or DRB1*16 are
usually found on the same chromosome as alleles whose names begin with
DRB5 [Figure 4-6].
D.
These associations are common but not always found. For example,
DRB1*15:01 can be found on a chromosome without a DRB5 allele and
DRB5 can be found on a chromosome without a DRB1*15:01 allele.
Figure 4-6
DR Genes and DR Proteins
DRB1 Gene
DRB1*04:02
DRB3 Gene
DRB3*01:01
DRA Gene
DRA*01:01
DR protein
DR52 protein
DRB1 Gene
DRB1*04:02
DRB4 Gene
DRB4*01:01
DR protein
DRB1 Gene
DRB1*16:02
DR protein
E.
DRA Gene
DRA*01:01
DR53 protein
DRB5 Gene
DRB5*02:01
DRA Gene
DRA*01:02
DR51 protein
The names of the DRB loci (DRB1, DRB3, DRB4, DRB5) are derived from
the hypothesized evolutionary origin of the DRB genes [Figure 4-7]. These
genes are thought to have arisen from duplication, deletion, and
C.W. Bill Young Marrow Donor Recruitment and Research Program
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diversification over evolutionary time.
Figure 4-7
Gene
Duplication
Generates the
DR Subregion
DRB
Duplication of gene
DRB1
DRB2
Duplication of genes
DRB1
DRB2
DRB1’
DRB2’
Diversification of genes
DRB1
DRB2
DRB3
DRB4
Deletion of gene
Deletion of gene
DRB1
DRB2
DRB4
DRB1
DRB2
DRB3
═══════════════════════════════════════════════════════════════════
Question 8: Draw the class II loci present on the two copies of chromosome 6 from a
person who carries DRB1*13:04, DRB1*04:02, DRB3*02:01, DRB4*01:01, DQA1*02:01,
DQB1*02:01, DPA1*01:04, DPB1*04:01 alleles.
Is the person homozygous or heterozygous for DQ and DP alleles?
═══════════════════════════════════════════════════════════════════
VIII.
The class II genes are closely related to one another and share many segments of
sequence. This sharing may result in difficulties in distinguishing specific HLA
alleles.
A.
Some alleles fall into families based on similarities in their DNA sequences.
These families are designated by their similar nomenclature. For example,
C.W. Bill Young Marrow Donor Recruitment and Research Program
38
alleles, DRB1*08:01:01, DRB1*08:02:03, DRB1*08:03:02, DRB1*08:04:04,
are very similar in sequence to one another as denoted by the digits in the
first field of the allele name, 08.
B.
Some segments of each class II gene sequence are shared by all alleles.
C.
Other segments of the gene sequences are polymorphic or vary among
alleles. Alleles from different allele families may share these polymorphic
segments of sequence.
═══════════════════════════════════════════════════════════════════
QUESTION 9: Compare the DNA sequences of DRB1*01:02:01 and DRB5*02:02 and
identify sequences that are shared. Some sequences are common to many DRB alleles;
other sequence segments are found only in a few alleles. Identify the shared sequences
in these two alleles. [Hint: sequences are usually compared to one sequence (e.g.,
DRB1*01:01) and nucleotides that are identical are indicated by a dash (-).]
Compare the DNA sequences of DRB1*01:03 and DRB1*13:01:01 and identify sequences
that are shared.
Observe the nucleotides that differ between alleles.
═══════════════════════════════════════════════════════════════════
IX.
Class I loci are also highly polymorphic.
A.
In the class I alleles, the second and third exons which encode the first and
second domains of the alpha chain contain the majority of the allelic
differences.
B.
The class I loci are very similar to one another. The sequences that identify
an A locus allele from a B or C locus allele are located in the first exon
(encoding the signal peptide) or in the 3' exons (encoding the
transmembrane and cytoplasmic regions) or in the introns separating the
exons.
C.
Alleles are designated by numbers.
For example, A*02:02 and
A*03:01:01:01 are alleles of the HLA-A locus. B*27:02 is an allele of the
HLA-B locus [Figure 4-8].
C.W. Bill Young Marrow Donor Recruitment and Research Program
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D.
Null or nonexpressed alleles may be indicated by an “N” in the allele name
e.g., A*24:11N. An “L” indicates an allele which exhibits decreased protein
expression. An “S” indicates that the HLA protein product is secreted. While
null alleles are important to consider in matching of patient and donor, the
importance of low vs. normal expression levels or secreted vs cell surface
molecule is not yet known. “Q” means that the level of expression is not
known but expected to be low or absent.
Figure 4-8
═══════════════════════════════════════════════════════════════════
QUESTION 10: Use the W.H.O. nomenclature report to list a few alleles of the C locus.
═══════════════════════════════════════════════════════════════════
E.
The HLA genes are closely related to one another and share many segments
of sequence. This sharing may result in difficulties in distinguishing specific
HLA class I alleles or molecules.
C.W. Bill Young Marrow Donor Recruitment and Research Program
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═══════════════════════════════════════════════════════════════════
QUESTION 11: Go to the web page: http://www.ebi.ac.uk/ipd/imgt/hla/ and find a copy of
the current list of the CDS sequences of the class I alleles. Compare the nucleotide
sequences of B*07:02:01, B*08:01, and B*42:01 and identify sequences that are shared
between alleles.
═══════════════════════════════════════════════════════════════════
X.
Each person carries two copies of each HLA-A, -B, and -C locus, one inherited from
their mother and one from their father.
A.
These genes are codominantly expressed, that is, both copies encode
proteins that are expressed on a single cell [Figure 4-9]. If this is a cell of the
immune system, it will also express all the class II alleles the cell carries.
Figure 4-9
B.
A parent and child share one chromosome or haplotype (haploidentical). A
sibling has a 1 in 4 chance of receiving the same two chromosomes (HLA
identical) as another sibling.
C.W. Bill Young Marrow Donor Recruitment and Research Program
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C.
XI.
Sometimes reciprocal recombination alters these allele combinations.
Other class I molecules (HLA-E, HLA-G) are expressed in specific tissues at
specific times. For example, HLA-G is expressed at the maternal/fetal interface.
These molecules are less polymorphic than HLA-A, -B, and -C and are not
considered important for HLA matching for transplantation. Other class I-like genes
(e.g., HLA-Y) are pseudogenes and are defective, not expressing a protein.
References:
Fernandez Vina MA, Hollenbach JA, Lyke KE, Sztein MB, Maiers M, Klitz W, Cano P,
Mack S, Single R, Brautbar C, Israel S, Raimondi E, Khoriaty E, Inati A, Andreani M, Testi
M, Moraes ME, Thomson G, Stastny P, Cao K. 2012. Tracking human migrations by the
analysis of the distribution of HLA alleles, lineages and haplotypes in closed and open
populations. Philos Trans R Soc Lond B Biol Sci. 367(1590):820-9.
Mack SJ, Cano P, Hollenbach JA, He J, Hurley CK, Middleton D, Moraes ME, Pereira SE,
Kempenich JH, Reed EF, Setterholm M, Smith AG, Tilanus MG, Torres M, Varney MD,
Voorter CE, Fischer GF, Fleischhauer K, Goodridge D, Klitz W, Little AM, Maiers M, et al.
2013. Common and well-documented HLA alleles: 2012 update to the CWD catalogue.
Tissue Antigens 81(4):194-203.
Parham, P., Lomen, C.E., Lawlor, D.A., Ways, J.P., Holmes, N., Coppin, H.L., Salter, R.D.,
Wan, A.M., Ennis, P.D. 1988. Nature of polymorphism in HLA-A, -B, -C molecules. Proc.
Natl. Acad. Sci. USA 85:4005-4009.
Parham, P. and Ohta, T. Population biology of antigen presentation by MHC Class I
molecules. Science 272:67-74, 1996.
Solberg OD, Mack SJ, Lancaster AK, Single RM, Tsai Y, Sanchez-Mazas A, Thomson G.
2008. Balancing selection and heterogeneity across the classical human leukocyte antigen
loci: a meta-analytic review of 497 population studies. Hum Immunol. 69(7):443-64.
C.W. Bill Young Marrow Donor Recruitment and Research Program
42
CHAPTER 5
HLA TYPES DEFINED BY SEROLOGY
The purpose of this chapter is to describe the terms used by HLA serologists to describe
the many forms of the HLA molecules found in the human population.
I.
Serology is used to identify the HLA proteins on the surfaces of cells.
A.
The different forms of the HLA proteins found in the human population may
be detected serologically using antibodies (human alloantisera or monoclonal
antibodies) in a test called a microcytotoxicity assay [Figure 5-1].
Figure 5-1
HLA Typing -- Microcytotoxicity Assay
Y
Antibody
HLA Molecule
Cell
Complement
Y
B.
The alloantisera utilized in this assay are obtained from humans who have
been sensitized to foreign HLA molecules by pregnancy or previous
transplant. These antibodies are used as reagents to identify serologic
determinants or specificities (or HLA types) by reacting with the HLA
molecules present on the cell surfaces.
C.
Because humans mount an immune response to foreign HLA molecules, the
HLA molecules are often called antigens. An antigen is any substance that
an antibody can bind.
C.W. Bill Young Marrow Donor Recruitment and Research Program
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II.
Each serologic specificity (or HLA type) is designated by a letter indicating the kind
of HLA antigen (A, B, C, DR, DQ, DP) and a number. The number indicates the
order in which the type was originally discovered or its relationship to other defined
antigens. For examples, DQ7 is a serologic specificity localized on an HLA-DQ
antigen, DR4 is a serologic specificity localized on an HLA-DR antigen, and B8 is a
serologic specificity localized on an HLA-B antigen.
═══════════════════════════════════════════════════════════════════
QUESTION 1: Go to the web http://hla.alleles.org/ and find a list of the accepted serologic
specificities.
═══════════════════════════════════════════════════════════════════
III.
Broad antigens and splits
A.
Some antibodies define clusters of HLA proteins (broad antigens) that are
similar and that are indistinguishable using a particular antibody preparation
[Figure 5-2]. For example, the A2 specificity is likely found on over 200
different HLA-A molecules.
B.
Other more specific antibodies may allow the definition of subdivisions or
"splits" of broad antigens. A nontechnical example of splits which may be
clearer is that a stranger may recognize you as a member of the Jones
family (broad specificity) but a friend may recognize you as Mary Jones (a
split).
Figure 5-2
C.W. Bill Young Marrow Donor Recruitment and Research Program
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═══════════════════════════════════════════════════════════════════
EXAMPLE: Antibodies define A9, a broad antigen [Figure 5-2]. A9 was the 9th HLA-A
specificity described. The splits of A9 have been labeled A23 and A24. Thus, the
specificity A23 is often called a split of the broad specificity A9 and can be also written as
A23(9). Because these serologic specificities were named as they were discovered, the
splits of A9 were designated as A23 and A24 because they were the 23rd and 24th HLA-A
antigens discovered. An individual who types as A24 using A24-specific antibodies would
also type as A9 using A9-specific antibodies but they would not type as A23.
═══════════════════════════════════════════════════════════════════
C.
Different HLA alleles defined by DNA typing can specify HLA proteins which
are indistinguishable using serologic typing. For example, an individual
carrying the DRB1*04:01:01 allele would have the same serologic type (DR4)
as an individual carrying the DRB1*04:12 allele. Thus, DRB1*04:01:01 and
DRB1*04:12 are splits of the broad specificity DR4. These splits are
identified by DNA typing but, because we do not have serologic reagents
specific enough to define this split, we can not serologically distinguish
between the antigens specified by the two alleles, DRB1*04:01:01 and
DRB1*04:12.
═══════════════════════════════════════════════════════════════════
QUESTION 2: Go to http://www.ebi.ac.uk/ipd/imgt/hla/ under HLA dictionary. Look up the
tables that list the serologic specificities of HLA-A,-B,-DR and links them to the alleles.
═══════════════════════════════════════════════════════════════════
D.
Although HLA types have been defined using serology for many years, the
available antibodies lack the resolution required to identify all of the specific
products of the HLA alleles [Figures 5-3]. For example, serology can not
determine whether a donor and recipient who are typed as DR4 carry the
same alleles of DR4. This is one reason that, in some situations such as
typing for bone marrow transplantation, that serology has been replaced by
DNA-based typing methods.
═══════════════════════════════════════════════════════════════════
QUESTION 3: Do the donor and recipient below carry the same DR alleles?
Donor: DR8,DR3
Recipient: DR8,DR3
═══════════════════════════════════════════════════════════════════
C.W. Bill Young Marrow Donor Recruitment and Research Program
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Figure 5-3
═══════════════════════════════════════════════════════════════════
QUESTION 4: Father is DR4, DR11 and mother is DR1, DR3. What are the possible DR
types inherited by their children?
What are one possible set of DR alleles that might be found in the family?
═══════════════════════════════════════════════════════════════════
E.
IV.
Cellular assays such as the mixed lymphocyte culture (MLC) measure the
differences in class II proteins between individuals. A cellular assay is more
sensitive in detecting HLA differences than serologic typing since even a
single amino acid differences can cause stimulation of white blood cells.
Because of the difficulty in generating and maintaining cellular reagents and
because of difficulties in interpreting experimental results, these types of
assays have been replaced by DNA-based typing methods.
The HLA genes encode specific HLA antigens.
C.W. Bill Young Marrow Donor Recruitment and Research Program
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A.
The DR molecules encoded by the DRA and DRB1 loci carry the DR1,
DR15, DR16, DR3, DR4, etc. serologic specificities [Figure 5-4].
Figure 5-4
B.
Molecules encoded by the DRA and DRB3 loci carry the DR52 serological
specificity Molecules encoded by the DRA and DRB4 loci carry the DR53
serological specificity. Molecules encoded by the DRA and DRB5 loci carry
the DR51 serological specificity.
C.
Molecules encoded by the DQA1 and DQB1 loci carry DQ serological
specificities. Molecules encoded by the DPA1 and DPB1 loci carry DP
serological specificities. [Note: There are very few alloantisera that detect
DP types.]
D.
Molecules encoded by the A, B, and C loci carry HLA-A, HLA-B, and HLA-C
serologic specificities. HLA-C is poorly defined by serologic reagents.
C.W. Bill Young Marrow Donor Recruitment and Research Program
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E.
Individuals who carry haplotypes (or chromosomes) with two DR beta loci
express two different DR molecules specified by that haplotype and, since
the allele associations are usually fixed, the combination of DR molecules
can usually be predicted [Figure 5-4]. [Review Chapter 4.]
1.
Individuals who express a DR4 molecule usually also express a DR53
molecule and carry a DR4 beta chain allele (e.g., DRB1*04:02) and a
DR53 beta chain allele (e.g., DRB4*01:01:01:01) [Figure 5-4]. This is
also true for DR7 and DR9.
2.
Individuals who express a DR3 molecule usually also express a DR52
molecule and carry a DR3 beta chain allele (e.g., DRB1*03:01:01)
and a DR52 beta chain allele (e.g., DRB3*01:01:02:01). This is also
true for DR11, DR12, DR13, and DR14.
3.
Individuals who express a DR15 molecule usually also express a
DR51 molecule and carry a DR15 beta chain allele (e.g.,
DRB1*15:01:01) and a DR51 beta chain allele (e.g., DRB5*01:01:01).
This is also true for DR16.
═══════════════════════════════════════════════════════════════════
QUESTION 5: Draw the DR loci present in the MHC of a person who is serologically typed
as DR11, DR52.
List one set of possible alleles. [Hint: There are more than one possible allele for each
locus, just pick one.]
═══════════════════════════════════════════════════════════════════
F.
There are exceptions to these associations. For example, some DR7
haplotypes carry a DRB4 allele but it is not expressed as a DR53 molecule
on the cell surface because of a defect in the gene (e.g., a termination codon
halting protein synthesis). The allele specifying this null allele is sometimes
labeled with an “N” (e.g., DRB4*01:03:01:02N).
═══════════════════════════════════════════════════════════════════
QUESTION 6: Father expresses HLA-A2,A3,B27,B53 and mother expresses HLAA2,A11,B51,B71. What are the possible HLA-A and -B types that their children might
express?
C.W. Bill Young Marrow Donor Recruitment and Research Program
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One of the children expresses HLA-A2,B27,B71. What are the most likely haplotypes
carried by the parents?
═══════════════════════════════════════════════════════════════════
V.
Complications.
A.
Unfortunately, the serologic types associated with some HLA alleles are not
yet known. For example, the serologic type specified by B*08:08 is not
known.
B.
In some cases, the name of the allele does not reflect its serologic type. For
example, the HLA molecule specified by B*50:02 is serologically typed as
B45.
═══════════════════════════════════════════════════════════════════
QUESTION 7: Using the DNA dictionary, what is the serologic type assigned to the
following alleles: A*68:01:01, B*13:01, B*18:04, DRB1*03:01:01, DRB1*11:22?
═══════════════════════════════════════════════════════════════════
═══════════════════════════════════════════════════════════════════
QUESTION 8: A patient carries A*02:01:01:01 and A*24:09N alleles. What is the
serologic type of this patient? Would a donor serologically typed as A2, A24 be a
match for this patient?
═══════════════════════════════════════════════════════════════════
References:
HLA websites with serology information:
http://www.ebi.ac.uk/ipd/imgt/hla/
http://hla.alleles.org/
C.W. Bill Young Marrow Donor Recruitment and Research Program
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CHAPTER 6
GENE AMPLIFICATION USING THE POLYMERASE CHAIN REACTION
The polymerase chain reaction (PCR) is a rapid way of isolating large quantities of specific
HLA genes for HLA typing [Figure 6-1]. Using this method, we can generate millions of
copies of a specific gene.
Figure 6-1
PCR
Amplification for
Identification of
HLA Alleles
After
HLA Gene
Other Genes
Before
HLA Gene
I.
PCR is a method of DNA synthesis [Figure 1-7] using:
A.
A DNA polymerase that is not destroyed when heated (thermostable): Taq
polymerase. This enzyme comes from bacteria that live in hot springs.
B.
DNA template: Crude cell lysates containing heat denatured DNA can be
used although better results are obtained with DNA that has been more
extensively purified.
C.
Nucleotides.
D.
Primers.
1.
One of the advantages of PCR is that it uses two primers so that both
C.W. Bill Young Marrow Donor Recruitment and Research Program
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DNA strands are copied. The two single stranded primers must flank
the region to be amplified. Since most of the polymorphism of the
class II genes is found in the second exon, primers usually are
designed to flank this region [Figure 6-2]. Primers for class I alleles
usually flank the polymorphism found in exons 2 and 3.
Figure 6-2
PCR Primers Amplify One Part of an
HLA Gene
Class II Gene
L
1
B1
B2
Tm
2
3
4
Cyt
5
6
2.
The primers hybridize to opposite strands of the DNA ladder. One
primer is called a forward (or 5' or sense or coding) primer and the
other primer is called a reverse (or 3' or antisense or noncoding)
primer. Remember the discussion in Chapter 1 about DNA synthesis.
3.
The primer sequences must complement (match) their target
sequence and be sufficiently long (20-30 nucleotides) to bind to only
the HLA gene that you want to amplify.
4.
A primer should not contain any stretches of sequence that would
anneal to the other primer (form primer dimers). For example, 5'
AGCACTTTT and 5' TCCATAAAA would not be a good choice
because the stretches of Ts and As would anneal. The polymerase
would add complementary nucleotides to make a short double strand
DNA called a primer dimer.
5’
AGCACTTTT 3’
3’
AAAATACCT 5’
===> 5’AGCACTTTTATGGA
TCGTGAAAATACCT
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II.
DNA is amplified by using a three step procedure:
A. DNA denaturation (94-96oC) to generate a single stranded template [Figure 6-3].
Figure 6-3
B.
Annealing of the primers (45-65oC) using hybridization conditions that
guarantee that the primers will bind to perfectly matched sequences (target
sequence) and not to sequences that are not matched [Figure 6-4]. The
temperature of annealing controls, in part, the specificity of the amplification
[Hint: Remember Chapter 1]. The higher the temperature, the more specific
the amplification until you get to the melting temperature of the primer. At
that temperature, amplification does not work very well, if at all.
C.W. Bill Young Marrow Donor Recruitment and Research Program
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Figure 6-4
PCR – Anneal Primers at 5065ºC
5’GGGGTGCCCCCCCCTTTTTTTGAAAAA3’
3’CTTTTT5’
5’GGGGT3’
3’CCCCACGGGGGGGGAAAAAAACTTTTT5’
Annealing temperature depends on length, GC content of primer ie melting temperature
Figure 6-5
PCR – Extension at 72ºC
5’GGGGTGCCCCCCCCTTTTTTTGAAAAA3’
3’CTTTTT5’
5’GGGGT3’
3’CCCCACGGGGGGGGAAAAAAACTTTTT5’
Use temperature optimal for polymerase
C.W. Bill Young Marrow Donor Recruitment and Research Program
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o
C.
Extension (synthesis of DNA) (around 72 C) [Figure 6-5].
D.
At the end of the first cycle, amount of target DNA is doubled [Figure 6-6].
Figure 6-6
PCR – End of 1st Cycle
5’GGGGTGCCCCCCCCTTTTTTTGAAAAA3’
3’CCCCACGGGGGGGGAAAAAAACTTTTT5’
3’CCCCACGGGGGGGGAAAAAAACTTTTT5’
5’GGGGTGCCCCCCCCTTTTTTTGAAAAA3’
Note: primers become part of the newly synthesized DNA
E.
The three steps are repeated over and over by simply changing the
temperature of the reaction mix using an instrument called a thermal cycler.
The newly synthesized strands serve as templates for synthesis in the next
cycle (another advantage of PCR). Usually 25-30 cycles of amplification are
carried out to yield millions of copies of the gene of interest.
F.
Figure 6-7 illustrates how the reaction yields an amplicon of defined length
after many cycles. Question 2 gives you a chance to demonstrate this to
yourself.
C.W. Bill Young Marrow Donor Recruitment and Research Program
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Figure 6-7.
═══════════════════════════════════════════════════════════════════
QUESTION 1: What would happen if you used a primer set that had the following
sequences to amplify a gene from human DNA:
5' CC
5' TA
Would primers that were 20 bases long be more or less specific in priming the amplification
reaction? Why? [Hint: Remember the discussion about the specificity of restriction
enzymes. What is the probability that you will find a two nucleotide-long sequence in the
DNA compared to the probability of finding a 20 nucleotide-long sequence?]
═══════════════════════════════════════════════════════════════════
═══════════════════════════════════════════════════════════════════
QUESTION 2: Draw out a PCR reaction amplifying the DNA listed below using primers
listed. What happens at each step of the amplification (denaturation, annealing,
extension)? What do the amplified products look like after the second cycle of
C.W. Bill Young Marrow Donor Recruitment and Research Program
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amplification? After the third cycle?
5' AATAATAATAATAATAATTATGGCGGCTATCGGCGGCGGCTTTATTATTATTATTCCCC 3'
3' TTATTATTATTATTATTAATACCGCCGATAGCCGCCGCCGAAATAATAATAATAAGGGG 5'
Primers: 5'TTATG and 5'AAAGC
How many base pairs in length is the amplified fragment?
If you started with one piece of DNA, after 3 cycles, how many copies do you have? After
10 cycles?
═══════════════════════════════════════════════════════════════════
═══════════════════════════════════════════════════════════════════
QUESTION 3: The sequences of the two primers that are used to amplify HLA-A alleles
are:
Forward: 5’ CCC AGA CGC CGA GGA TGG CCG 3’
Hint: 5’UTR-exon 1 boundary
Reverse: 5’ GCA GGG CGG AAC CTC AGA GTC ACT CTC T 3’
Hint: Intron 3; remember 5’ to 3’ and complementary DNA strands running in opposite
directions
Use the HLA-A sequence in Figure 2-4 to locate these sequences in the DNA. Remember
that DNA synthesis proceeds 5' to 3' and that we are only interested in the polymorphic
regions (second and third exon).
Why can't you find the primer sequences in the CDS sequence of HLA-A in Figure 2-5?
═══════════════════════════════════════════════════════════════════
═══════════════════════════════════════════════════════════════════
QUESTION 4: Locate the C locus primers used to amplify exons 2 and 3:
AGCGAGG(GT)GCCCGCCCGGCGA and GGAGATGGGGAAGGCTCCCCACT. [Hint:
C.W. Bill Young Marrow Donor Recruitment and Research Program
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(GT) means that two primers are synthesized, one with G at this position and one with T in
this position. Hint: Look at the genomic sequences because the primers anneal in the
introns.
Note: Primers described in Cereb, N. et al. 1996. Nucleotide sequences of MHC class I
introns 1,2, and 3 in humans and intron 2 in nonhuman primates. Tissue Antigens 47:498511. Correction in Tissue Antigens 48:235-236, 1996.
═══════════════════════════════════════════════════════════════════
═══════════════════════════════════════════════════════════════════
QUESTION 5: Design primers to amplify the second exon of the DQA1 alleles. You will
need to find a publication listing the sequences of all the DQA1 alleles to do this [Hint:
HLA/IMGT web site.].
═══════════════════════════════════════════════════════════════════
III.
To determine if DNA has been amplified following the PCR reaction, the PCR
product is often analyzed by gel electrophoresis to identify a fragment of a specific
size. For example, amplification of exon 2 of the DRB genes will yield a fragment of
DNA approximately 270 base pairs in length and amplification of exons 2 and 3 of
an HLA-B gene will yield a fragment of DNA approximately 1000 base pairs in
length.
A.
IV.
V.
Note that a primer dimer may also detected on the gel during
electrophoresis. It will be smaller that the HLA amplicon so it will move
farther down the gel.
Advantages of PCR
A.
Large quantities of the HLA gene of interest will allow the rapid detection of
the HLA type.
B.
Amplification of a specific HLA gene means that detection of an HLA type will
not be influenced by the presence of other HLA genes or non-HLA genes
during probe hybridization or sequencing.
Contamination with previously amplified DNA is the biggest potential problem. It is
relatively easy to contaminate the work area with amplified DNA because there are
so many copies of an HLA gene following amplification.
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═══════════════════════════════════════════════════════════════════
QUESTION 6: Assume that a PCR amplification generates 27 million (27 x 10 6) copies of
a piece of DNA in a 100 microliter reaction volume. How many copies will be in a 1
microliter drop that was sucked into the barrel of a pipetter?
═══════════════════════════════════════════════════════════════════
VI.
Allele or group-specific PCR or sequence-specific primer (SSP) typing. A
modification of this approach is called ARMS (amplification refractory mutation
system).
A.
In this technique, PCR primers are designed to anneal only to a specific set
of alleles or to a single allele [Figure 6-8]. One or both primers include
sequences unique to the allele(s). These unique sequences should be
located at the 3' end of the primer for maximum specificity in the annealing
step. Remember, that to get efficient amplification, both PCR primers must
anneal to the DNA.
Figure 6-8
Sequence Specific PCR
DRB1*01, DRB1*15
DRB1*01
primer
DRB1*03
primer
DRB1*15
primer
DRB1*01
DRB1*15
MW
Sample
marker
Gel
B.
MW
Sample
marker
Gel
MW
Sample
marker
Gel
The primer set can be designed to give an amplified fragment of a specific
C.W. Bill Young Marrow Donor Recruitment and Research Program
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size which can be detected by gel electrophoresis.
C.
Failure to use the appropriate amplification conditions can cause
amplification that will give either false positives (wrong alleles amplified) or
false negatives (correct allele not amplified).
D.
Primers used in the ARMS system may have an additional mismatch
incorporated in their sequence. This makes the primers mismatched to all
alleles; however, the allele(s) which amplify are less mismatched than the
alleles which should not amplify.
E.
The 3’ end of the primer is most important in controlling the specificity of
annealing and amplification. Differences between the primer and template
near the 5’ end of the primer are likely not to affect annealing of the primer
and will give rise to false positives (i.e. amplification occurs even though
mismatched).
═══════════════════════════════════════════════════════════════════
QUESTION 7: Design a forward PCR primer that will amplify all the DRB1*04 alleles and
not most of the other DRB1 alleles. You can use 5’ CCG CTG CAC TGT GAA GCT CT as
a reverse primer (end of exon 2) but you need to design a forward primer. You will have to
look at the DRB1 allele sequences that you looked at in Chapter 4.
Can you design a forward PCR primer that will amplify only the DRB1*11 alleles?
Can you design a primer set that will amplify only DRB1*11:01 alleles?
═══════════════════════════════════════════════════════════════════
VII.
The amplified DNA is used to identify HLA types as described in the next chapters.
References:
Green and Sambrook. Molecular Cloning, A Laboratory Manual, 4th edition. Cold Spring
Harbor Laboratory Press
Bunce, M. O’Neill, C.M., Barnardo, M.C., Krausa, P., Browning, M.J., Morris, P.J. & Welsh,
C.W. Bill Young Marrow Donor Recruitment and Research Program
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K.I. 1995. Phototyping: comprehensive DNA typing for HLA-A,B,C DRB1, DRB3,
DRB4,DRB5 & DQB1 by PCR with 144 primer mixes utilizing sequence-specific primers
(PCR-SSP). Tissue Antigens 46:355-367.
Cereb, N., Kong, Y., Lee, S., Maye, P., Yang, S.Y. 1996. Nucleotide sequences of MHC
class I introns 1,2, and 3 in humans and intron 2 in nonhuman primates. Tissue Antigens
47:498-511. Correction in Tissue Antigens 48:235-236, 1996.
Krausa, P., Bodmer, J.G., and Browning, M.J. 1993. Defining the common subtypes of
HLA A9, A10, A28 and A19 by use of ARMS/PCR. Tissue Antigens 42:91-99.
Mullis, K. The unusual origin of the polymerase chain reaction. Scientific American, April
1990.
Olerup, O. and Zetterquist, H. 1992. HLA-DR typing by PCR amplification with sequence
specific primers (PCR-SSP) in 2 hours: An alternative to serological DR typing in clincial
practice including donor-recipient matching in cadaveric transplantations. Tissue Antigens
39:225-235.
Saiki, R.K., Gelfand, D.H., Stoffel, S., Scharf, S.J., Higuchi, R., Horn, G.T., Mullis, K.B.,
and Erlich, H.A. 1988. Primer-directed enzymatic amplification of DNA with a thermostable
DNA polymerase. Science 239: 487-91.
C.W. Bill Young Marrow Donor Recruitment and Research Program
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CHAPTER 7
USE OF OLIGONUCLEOTIDE PROBES TO DETECT SPECIFIC DNA SEQUENCES
The amplified DNA generated using the PCR is used to identify HLA types. This chapter
discusses the use of oligonucleotide probe hybridization (Figure 7-1) to detect specific HLA
alleles.
Figure 7-1
SSOPH – Standard Format
Probe 2
Oligo Probe #1
Probe 3
Probe 4
Label
Different
genomic
DNA
samples on
membrane
I.
Multiple membranes,
One for each probe
Selection and synthesis of probes (oligonucleotides) to detect specific alleles
(SSOP=sequence specific oligonucleotide probes) (also called SSO) by
hybridization (SSOPH).
A.
Oligonucleotides ("oligos") (single stranded DNAs) must complement their
target sequence and be sufficiently long (~18 nucleotides) to allow the use of
hybridization conditions that guarantee discrimination between the target
sequence and other closely related sequences [Figure 7-2]. The more
differences between the matched and mismatched sequences, the easier it
is to establish specific hybridization conditions. If the sequences are
mismatched for only one nucleotide, the mismatched base should be placed
in the middle of the oligonucleotide probe. Stringent washes are more likely
to remove single base mismatches in this position; it is more destabilizing
C.W. Bill Young Marrow Donor Recruitment and Research Program
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than a mismatch at one end of the oligonucleotide.
Figure 7-2
B.
The oligo should be approximately 50% G+C (if possible) and should not
contain any complementary sequences that might cause the oligo to anneal
to
itself
(form
a
hairpin-like
structure).
For
example,
AAAAATGCCGCTATTTTT would not be a good choice.
C.
The oligo can be complementary to either strand of the DNA. Most people
synthesize oligos which are identical in sequence to the coding strand (CDS,
the sequence published in the literature) so that they are easier to find in the
HLA allele sequences.
II.
Labeling of oligo probes to detect binding. Probes are usually end-labeled, tailed, or
contain modified nucleotides. If probes are attached to beads, the bead can be
fluorescently tagged.
III.
Hybridization of probes to DNA samples.
A.
Amplified DNA is linked to a solid support, denatured, and then hybridized to
a labeled SSOP [Figures 7-1, 7-3]. Support can be a membrane, plastic
C.W. Bill Young Marrow Donor Recruitment and Research Program
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plate or a bead.
B.
The conditions of the hybridization and/or wash to remove nonspecifically
bound probe are very important in controlling specificity of hybridization
[Figure 7-4]. Most people hybridize using nonstringent conditions and utilize
stringent conditions for washing.
Figure 7-3
Probe Hybridization
Synthetic
oligonucleotide
(oligo)
Solid support
DNA
Figure 7-4
Denaturation of DNA
+ Single stranded synthetic DNA (oligo)
Hybridization
Wash to remove
imperfectly matched
hybrids
Control Stringency
PCR
Probes
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C.
In one form of this technique, samples of DNA from many people are applied
to a single membrane and hybridized with a single probe [Figure 7-1]. Each
assay requires a separate membrane for each probe.
D.
Alternatively, the probes may be linked to the solid support (like a 24 well
plate) and then hybridized to labeled, denatured, amplified DNA (reverse dot
blot) [Figure 7-5]. Support can be a membrane, plastic plate or a bead.
E.
Another variation on the SSOP technique is the “chip” technology. In this
methodology, short overlapping oligonucleotides representing the entire
sequences of alleles are attached to a solid support about the size of a dime
(a “chip”)]. Labeled denatured amplified DNA is incubated with this chip and
the binding of the DNA to individual oligos is detected by a excitation of the
label with a laser].
Figure 7-5
SSOPH--Reverse Format
1 Sample / Membrane
Amplified HLA Gene
Label
Oligo
Multiple probes on solid support
═══════════════════════════════════════════════════════════════════
COMMENT: Both probes and primers are single stranded pieces of synthetic DNA. When
the oligonucleotide is used to prime DNA synthesis, it is called a primer. When the
oligonucleotide is used for hybridization and that hybridization event is used to define an
HLA type, the oligonucleotide is called a probe.
═══════════════════════════════════════════════════════════════════
IV.
Interpretation of hybridization results
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A.
V.
When a sample is tested with a panel of oligo probes to type DRB1, for
example, some probes will hybridize and some will not, based on the
sequences of the two DRB1 alleles present. A software program then takes
these data and compares it to the DNA sequences of all the known DRB1
alleles and determines which alleles might be present. See Chapter 11.
Difficulty with this system:
A.
Alleles often share nucleotide sequences so probes usually detect more than
one allele. One may need to use multiple probes for typing a single allele.
One may also need to carry out group specific amplification [Figures 7-6].
Identification of individual alleles is called allele-level typing. Typing which
narrows down the allele possibilities but still includes more than one possible
allele is termed intermediate resolution testing. An example is a sample
typed as (DRB1*11:01 or DRB1*11:04) and (DRB1*03:02 or DRB1*03:03).
B.
Since oligo probes identify only specific sequences within the gene,
differences outside of these regions or at the edges of the probe will go
unnoticed. Thus new alleles might be missed.
C.
Contamination with other DNAs may give false positives. For example, if
DNA amplified in a previous assay contaminates a pipettor, that DNA may be
transferred into a DNA sample in another assay.
═══════════════════════════════════════════════════════════════════
QUESTION 1: The two sequences below differ by a single nucleotide. Design a 10 base
long oligo probe to detect sequence #1.
#1 5' ATACAGAGGTACTACGCCTAATATGGCGCTA
#2 5' ATACAGAGGTACTACACCTAATATGGCGCTA
What is the G+C content of the oligo that you have designed? What is its approximate
melting temperature?
If you carry out the hybridization wash at 50oC, what will happen?
What would happen if you put the discriminating nucleotide at the 3’ or 5’ end of the probe?
═══════════════════════════════════════════════════════════════════
═══════════════════════════════════════════════════════════════════
QUESTION 2: Using the nomenclature website that lists all of the DRB1 allele sequences,
design an oligonucleotide to specifically detect DRB1*11 alleles. This oligo will be
equivalent to a DR11-specific antibody used in serologic HLA typing. If we use the
oligonucleotide probes to define "serologic types", this is called antigen level or low
C.W. Bill Young Marrow Donor Recruitment and Research Program
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resolution typing.
[Note that the probe will also hybridize to a few non-DRB1*11 alleles like DRB1*04:15 and
DRB1*03:08.]
Design an oligonucleotide to detect all DRB1 alleles.
Design an oligonucleotide to detect only the DRB1*10:01 alleles. If we use the
oligonucleotide probes to define alleles, this is called allele level or high resolution typing.
═══════════════════════════════════════════════════════════════════
Figure 7-6
═══════════════════════════════════════════════════════════════════
QUESTION 3: How would you design a protocol to identify a DRB1*01:02 allele from a
DRB1*01:01 allele?
═══════════════════════════════════════════════════════════════════
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References (a few examples):
Bugawan, T. L., Apple, R., and Erlich, H. A. 1994. A method for typing polymorphism at
the HLA-A locus using PCR amplification and immobilized oligonucleotide probes. Tissue
Antigens 44:137-147.
Bugawan, T. and Erlich. 1991. Rapid typing of HLA-DQB1 DNA polymorphism using
nonradioactive oligonucleotide probes and amplified DNA. Immunogenetics 33:163.
Middleton, D., Williams, F., Hamill, M. A., et. al. 2000. Frequency of HLA-B alleles in a
Caucasoid population determined by a two-stage PCR-SSOP typing stategy. Human
Immunol 61:1285-1297.
Fernandez-Vina, M., Lazaro, A.M., Sun, Y., Miller, S., Forero, L., and Stastny, P. 1995.
Population diversity of B-locus alleles observed by high-resolution DNA typing. Tissue
Antigens 45:153-168.
Gao, X., Fernandez-Vina, M., Shumway, W., and Stastny, P. 1990. DNA typing for class II
HLA antigens with allele-specific or group-specific amplification: typing for subsets of HLADR4. Human Immunol. 27:40-50.
Fulton, R.J., R.L. McDade, P.L. Smith, L.J. Kienker, and J.R. Kettman Jr. 1997. Advance
multiplexed analysis with the FlowMetrix TM system. Clinical Chemistry 43: 1749-1756.
Saiki, R.K., Walsh, P.S., Levenson, C.H., and Erlich, H.A. 1989. Genetic analysis of
amplified DNA with immobilized sequence-specific oligonucleotide probes. Proc. Natl.
Acad. Sci. USA 86:6230-6234.
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CHAPTER 8
SANGER-BASED DNA SEQUENCING OF HLA GENES
It is not yet know how many HLA alleles exist in the human population. It is possible that
there will be so many HLA alleles that SSOP typing will require huge panels of
oligonucleotide probes and SSP will require huge primer panels to identify alleles. It may
be more practical and informative to use DNA sequencing to determine if a specific donor
and recipient carry the same HLA alleles.
I.
DNA sequencing can be used to determine the exact base sequence of the DNA or
mRNA (cDNA) encoding an HLA molecule.
II.
Method.
A.
The DNA of the gene must be amplifed by PCR [Figure 8-1].
1.
PCR is more commonly used today to obtain enough copies of a gene
to sequence (Chapter 6). The primers used for PCR amplification
determine how many alleles are coamplified. For example, if primers
that amplify all HLA-B alleles are used, then the amplified product will
contain both HLA-B alleles expressed by an individual.
Figure 8-1
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2.
If a sequence-specific primer set is used for PCR, each of the two
alleles may be isolated for sequencing
Cloning is more work and would likely not be used routinely for HLA
typing. To make a DNA library or to "clone" a piece of DNA, pieces of
DNA or DNA copies of the mRNA (cDNA) are inserted into a vector.
Vectors are pieces of DNA that can replicate like a plasmid or virus.
Each vector containing an inserted piece of DNA is propagated in
bacteria and, as the bacteria replicate, millions of copies of the vector
and inserted DNA are made.
Cloning can be used to isolate one HLA allele for characterization
from a heterozygous individual. PCR amplified DNA (e.g., HLA-B
PCR product) is cloned into a vector and individual bacterial colonies
isolated. Each colony carries one vector and, hence, one of the two
HLA alleles.
DNA/cDNA libraries can be made that contain most of the pieces of
DNA (genomic library) carried by a cell or most of the mRNAs (cDNA
library) expressed by a cell. From these libraries, the genes or
mRNAs (cDNAs) that encode the HLA molecules can be isolated.
═══════════════════════════════════════════════════════════════════
QUESTION 1: If a person was typed as DRB1*01:01 and DRB1*07:01, how would you
obtain only the DRB1*07:01 allele for sequence analysis?
═══════════════════════════════════════════════════════════════════
B.
Sanger-based DNA sequencing is a method of DNA synthesis (Chapter 1)
requiring:
1.
A single stranded template containing the PCR-amplified copies of the
HLA gene being sequenced.
2.
One primer that anneals to the template (called the sequencing
primer).
3.
A DNA polymerase like Taq polymerase.
4.
Nucleotides.
a.
dATP, dTTP, dGTP, dCTP.
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b.
5.
C.
Dideoxynucleotides (ddATP,ddTTP,ddCTP,ddGTP). When
the polymerase incorporates a dideoxynucleotide into the
growing DNA strand, synthesis stops at that point.
Some method of labeling is used to identify the newly synthesized
DNA strand.
a.
The dideoxynucleotide could be labeled with a fluorescent dye
b.
Or the primer could be labeled with a fluorescent dye
The method used for sequencing is called the Sanger chain termination
sequencing method [Figure 8-2]:
1.
The amplified DNA is denatured and the primer is annealed.
2.
The DNA is divided into 4 aliquots. A different chain-terminating
nucleotide is added to each aliquot in addition to all four of the normal
nucleotides. For example, ddCTP + dATP, dTTP, dGTP, dCTP are
added to one tube [Figure 8-2].
Figure 8-2
Labeled Terminator Sequencing
3’GGGGTGCCCCCCCCTTTTTTTGAAAAA5’
5’CCCCA
C
G
A
T
C-d ideoxy
3’GGGGTGCCCCCCCCTTTTTTTGAAAAA5’
5’CCCCACd
3’GGGGTGCCCCCCCCTTTTTTTGAAAAA5’
5’CCCCACGGGGGGGGAAAAAAACd
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As the polymerase is synthesizing a complementary DNA strand (the
sequence shown in Figure 8-2), it has a choice of nucleotides for the
incorporation. If it uses a normal nucleotide, synthesis proceeds. If
the polymerase incorporates a dideoxynucleotide, synthesis halts.
Remember that many identical strands of DNA are being synthesized
at the same time.
4.
Each reaction generates populations of labeled oligonucleotides of
different lengths that begin from a fixed point (the primer) and
terminate randomly at the residue represented by the ddNTP in that
aliquot [Figure 8-3].
5.
The populations of oligonucleotides of different lengths are resolved
by electrophoresis on a polyacrylamide like gel. If different colored
fluorescent dyes are used to label each dideoxy aliquot, the four
aliquots can be run in the same lane. A laser reads each color as the
fragments pass by a detector (automated sequencer).
6.
The “read” from a single sequencing primer is usually more than the
length of an HLA exon but doesn’t usually include all of the exons in
the amplicon. Usually multiple sequencing primers are used in
different reactions to produce fragments covering all of the exons
included in the amplicon. Sense and antisense primers are used to
obtain the sequence of both strands of the DNA being sequenced.
Figure 8-3
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═══════════════════════════════════════════════════════════════════
QUESTION 2: Draw out the sequencing reaction for the following piece of single stranded
DNA using the primer listed below. What would the sequencing gel look like?
3' AAAAAATGCCGAATCCGATACGTCGGGCATT 5'
Primer: 5' TTTTTTA 3'
═══════════════════════════════════════════════════════════════════
D.
In sequence based typing (SBT), amplified HLA alleles are sequenced
directly (i.e., without cloning) to identify the alleles carried by the individual.
1.
Depending on the PCR primers used, the sequence may contain
either a single allele or two alleles mixed together [Figure 8-4].
Figure 8-4
Sequence-Based Typing
Heterozygote
T G C
C
A
T
G
C
A
[M]
a.
When two alleles are sequenced simultaneously, positions
where the two alleles differ in sequence will show two
nucleotides. For example, in Figure 8-4, the fourth position
shows both a C and an A. This is labeled as M meaning both
C.W. Bill Young Marrow Donor Recruitment and Research Program
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C and A. Table 8-1 lists the codes used for multiple
nucleotides.
b.
2.
A sequence-specific sequencing primer may be used to
produce the sequence of a single allele from a PCR reaction
containing both alleles at a locus.
A software program is used to identify the alleles based on their
sequence.
Table 8-1
IUB Codes for Multiple Nucleotides
R
A/G (puRine)
Y
C/T (pYrimidine)
K
G/T (Keto)
M
A/C (aMino)
S
G/C (Strong 3H)
W
A/T (Weak 2H
N
A/C/T/G aNy base
IUB, International Union of Biochemistry Nomenclature Committee
III.
Sanger sequencing is a great method for identifying the HLA alleles carried by an
individual but, without isolating each of the two alleles, doesn’t tell us the phase of
the nucleotides across the entire region sequenced. Phase identifies which
polymorphic residues are carried on the maternal versus the paternal chromosome
6 [Figure 8-5]. Without phasing, it may not be possible to tell which genotype
(combination of 2 alleles) an individual carries.
C.W. Bill Young Marrow Donor Recruitment and Research Program
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Phase of Polymorphisms
Paternal chrom 6
Maternal chrom 6
DRB1*010101
1
5
10
15
20
CTG GCT TTG GCT GGG GAC ACC CGA C|CA CGT TTC TTG TGG CAG CTT AAG TTT GAA TGT CAT TTC TTC AAT GGG ACG
DRB1*040101
--- --- --- --- --- --- --- --- -|-- --- --- --- GA- --- G-- --A CA- --G --- --- --- --- --C --- ---
DRB1*010101
25
30
35
40
45
GAG CGG GTG CGG TTG CTG GAA AGA TGC ATC TAT AAC CAA GAG GAG TCC GTG CGC TTC GAC AGC GAC GTG GGG GAG
DRB1*040101
--- --- --- --- --C --- --C --- -A- T-- --- C-- --- --- --- -A- --- --- --- --- --- --- --- --- ---
DRB1*010101
DRB1*040101
50
55
60
65
70
TAC CGG GCG GTG ACG GAG CTG GGG CGG CCT GAT GCC GAG TAC TGG AAC AGC CAG AAG GAC CTC CTG GAG CAG AGG
--- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- -A-
DRB1*010101
75
80
85
90
95
CGG GCC GCG GTG GAC ACC TAC TGC AGA CAC AAC TAC GGG GTT GGT GAG AGC TTC ACA GTG CAG CGG CGA G|TT GAG
DRB1*040101
--- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- -|-C T-T
References:
Sanger, F., Nicklen, S., and Coulson, A.R. 1977. DNA sequencing with chain-terminating
inhibitors. Proc. Natl. Acad. Sci. USA 74:5463-5467.
Kotsch, K., J. Wehling, S. Kohler, and R. Blasczyk. 1997. Sequencing of HLA class I
genes based on the conserved diversity of the noncoding regions: sequencing-based
typing of the HLA-A gene. Tissue Antigens 50:178-191.
Versluis, L.F., Rozemuller, E., Tonks, S., Marsh, S. G. E., Bouwens, A. G. M., Bodmer, J.
G., and Tilanus, M. G. J. 1993. High-resolution HLA-DPB typing based upon
computerized analysis of data obtained by fluorescent sequencing of the amplified
polymorphic exon 2. Human Immunology 38:277-283.
McGinnis, M.D., Conrad, M.P., Bouwens, A.G.M., Tilanus, M.G.J., and Kronick, M.N. 1995.
Automated, solid-phase sequencing of DRB region genes using T7 sequencing chemistry
and dye-labeled primers. Tissue Antigens 46:173-179.
Petersdorf, E.W. and Hansen, J.A. 1995. A comprehensive approach for typing alleles of
the HLA-B locus by automated sequencing. Tissue Antigens 46:73-85.
C.W. Bill Young Marrow Donor Recruitment and Research Program
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CHAPTER 9
NEXT GENERATION DNA SEQUENCING OF HLA
The purpose of this chapter is to describe new methods of DNA sequencing which are
replacing the Sanger method. The basic sequencing chemistries share similarities with
the Sanger method so it will be useful to read Chapter 8. One advantage is that these new
methods can determine and report the sequence of a single DNA double helix and so allow
the phase of polymorphic nucleotides to be determined.
I.
Next generation sequencing (called NGS for short) is often used to sequence entire
human genomes. It is thought that characterization of the sequence of all of the
genetic information in an individual will help, for example, determine disease
susceptibility, identify mutations that might result in cancer, and aid in disease
treatment decisions. Thus, NGS is providing the technology for “personalized
medicine.”
Although whole genome sequencing should result in sequencing of the HLA genes
carried by an individual, in practice this doesn’t work very well. Therefore, the use
of NGS to identify HLA alleles usually begins with amplification of the HLA genes by
PCR.
II.
There are several platforms for NGS using different instruments and somewhat
different sequencing and detection chemistries. Below is a simplified version of how
all of these methods work. For the details of each platform, see the references for
this chapter.
A.
One advantage of NGS is that it has the capability of sequencing long
stretches of DNA. Thus, some strategies for HLA sequencing include
amplification of the entire HLA gene, exons and introns, with generic or
locus-specific primers.
B.
Because NGS results in the sequences of individual DNA fragments, it is
possible to combine different HLA amplicons for sequencing. So, for
example, it is possible in one sequencing run to obtain the DNA sequences
of all of the HLA genes from an individual simultaneously. It is also possible
to combine the samples from many individuals and sequence all their HLA
genes simultaneously.
C.
Once the HLA genes are amplified by PCR, the amplicons may need to be
broken into smaller fragments for DNA sequencing [Figure 9-1]. For
example, an NGS sequencing chemistry might be limited to determining the
sequence of only 500 base pairs. So, if an HLA amplicon is 6,000 base pairs
long, for example, it must be fragmented into smaller pieces for DNA
sequencing. Any piece of DNA over 500 base pairs will not be completely
C.W. Bill Young Marrow Donor Recruitment and Research Program
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sequenced. One method for fragmenting the DNA is by sonication.
D.
Each long DNA amplicon (remember this includes thousands of copies of the
region amplified) will be randomly broken into pieces of varying sizes. This
will result in a series of overlapping DNA fragments covering the length of the
amplicon.
Figure 9-1
Amplicons and Reads
Thousands of copies of
HLA-B amplicon
containing the entire
gene (5 kb)
Random fragments of
HLA-B amplicon
generated by sonication
Each fragments binds to
solid support and is
sequenced separately
II.
Before sequencing the amplified DNA, the laboratory will need to prepare a DNA
library. The methods used will depend on the platform used.
A.
Library preparation will usually involve ligating short pieces of DNA to both 5’
and 3’ ends of each DNA fragment [Figure 9-2]. These short DNA
sequences, called “adaptors”, will help the fragments anneal to the surface of
the sequencing reaction chamber and will provide sites where PCR and
sequencing primers can anneal.
B.
The ligation will also attach unique DNA sequences called “indices”. These
short sequences act as barcodes, labeling the fragments as coming from one
individual. The purpose of the indices is to allow the laboratory to combine
the samples from multiple individuals into one sequencing run. The software
will later sort out the information for each individual based on these
barcodes.
C.W. Bill Young Marrow Donor Recruitment and Research Program
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Figure 9-2
DNA Library
Genomic
DNA
or PCR
amplicon
Randomly
fragment DNA
Attach PCR &
sequencing
primer annealing
sites to ends of
each DNA
fragment
Hybridize single
stranded DNA
fragments to
solid support
III.
Once the library is created from each individual sample, samples from many
individuals are combined and DNA sequencing commences. There are several
different NGS sequencing chemistries that can be used. All have some
resemblance to the Sanger sequencing method.
A.
IV.
The unique feature of next generation sequencing is that each DNA fragment
is sequenced separately in what is call massive parallel sequencing. Thus,
the sequences of millions of DNA fragments are determined simultaneously.
Analysis of the huge amount of data collected is then carried out by software
programs.
A.
Once sequencing is complete, a computer program separates all the
sequences (called “reads”) based on their indices into sets of reads deriving
from one individual (called “demultiplexing”).
B.
Then a computer program aligns the reads to each locus based on the
sequences of reference genes [Figure 9-3]. So, for example, the computer
C.W. Bill Young Marrow Donor Recruitment and Research Program
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will identify all the reads that match to a reference HLA-A sequence and will
assemble them to form the sequence of an HLA-A gene.
Figure 9-3
Software Must Align Each Read to
Reference Set of Genes
Class I sequence may yield
thousands of reads
Ref HLA-A
Ref HLA-B
Ref HLA-C
Ref HLA-Y
Genes are very similar, sharing regions of sequence. This is challenging for
programs to address.
C.
The computer program will then separate the reads matching to HLA-A into
the sequences of the two alleles present at that locus by identifying which
polymorphic residues are carried on the same DNA fragments i.e., determine
the phase of polymorphisms [Figure 9-4].
C.W. Bill Young Marrow Donor Recruitment and Research Program
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Figure 9-4
Fragments Are Phased, Separated into
Two Alleles
TGCAATACGCG
GCAGTGCAA
AATACGCGTAACT
GCAGTGCAATACGCGTAACT
GCATTGCATTAC
GCATTACGCGTAGCT
Phased
GCATTGCATTACGCGTAGCT
long distance, loss of phase
Phased
or
D.
The end result is the complete sequences of both alleles at each HLA locus
for an individual. This method is a high volume procedure so that several
hundred individuals might be typed simultaneously.
References:
Gabriel C, Furst D, Fae I, Wenda S, Zollikofer C, Mytilineos J, Fischer GF. 2014. HLA
typing by next-generation sequencing - getting closer to reality. Tissue Antigens
83(2):65-75.
De Santis D, Dinauer D, Duke J, Erlich HA, Holcomb CL, Lind C, Mackiewicz K, Monos
D, Moudgil A, Norman P, Parham P, Sasson A, Allcock RJ. 2013. 16(th) IHIW : review
of HLA typing by NGS. Int J Immunogenet 40(1):72-6.
Erlich H. HLA DNA typing: past, present, and future. 2012. Tissue Antigens 80(1):1-11.
http://en.wikipedia.org/wiki/DNA_sequencing
Platforms (these web sites often have educational videos ):
 Illumina MiSeq: http://www.illumina.com/
 Pacific Biosystems: http://www.pacificbiosciences.com/
 Life Technologies Ion Torrent: http://ioncommunity.lifetechnologies.com/welcome
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CHAPTER 10
OTHER MOLECULAR BIOLOGY TECHNIQUES FOR HLA TYPING
While PCR/SSOP and SSP typing are the most common techniques used for HLA typing,
other techniques have been described.
I.
PCR/Restriction Fragment Length Polymorphism (RFLP)/AFLP [Figure 10-1].
A.
This method can be used if two alleles differ by the presence of a restriction
enzyme site. Following PCR amplification, the amplified DNA is cleaved with
the restriction enzyme and the alleles are identified by the fragmentation
pattern upon gel electrophoresis.
Figure 10-1
PCR / RFLP / AFLP
Allele 1
GAATTC
CTTAAG
Allele 2
GTATTC
CATAAG
Restriction
Enzyme digest
G
CTTAAG
AATTC
CTTAAG
GTATTC
CATAAG
Gel
Allele:
B.
II.
1/1
2/2
1/2
Two problems with the technique are a failure to get complete cleavage
which may make a homozygote look like a heterozygote and difficulty in
finding appropriate restriction sites in all of the alleles at a locus.
Denaturing gradient gel electrophoresis.
A.
As amplified double stranded DNA is electrophoresed through a denaturing
gradient gel, it reaches a point in the gradient where it denatures. This point
is determined by the sequence of the DNA.
B.
Not widely used because the results may be difficult to interpret.
C.W. Bill Young Marrow Donor Recruitment and Research Program
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III.
IV.
Single stranded conformational polymorphism (SSCP).
A.
PCR-amplified DNA is denatured and electrophoresed on a polyacrylamide
gel. Each single strand moves at a position related to its conformation as
determined by its sequence.
B.
Not widely used because the results may be difficult to interpret, this method
could be useful in comparing the alleles of two individuals.
Heteroduplex formation [Figure 10-2].
A.
Amplified DNA is denatured and allowed to reanneal under nonstringent
conditions. If DNA strands are present which do not perfectly match (e.g., in
a person heterozygous for the gene amplified), these will form
heteroduplexes in addition to homoduplexes. These heteroduplexes will have
an altered conformation compared to the homoduplexes and will migrate
differently in an electric field.
B.
Sometimes a labeled reference DNA is added. This DNA is designed to
anneal to the gene of interest creating additional heteroduplexes. In this
case, only the labeled heteroduplexes are detected. This variation is called
reference strand mediated conformational analysis (RSCA).
C.
This method could be used to compare the alleles of two individuals or to
determine an HLA type if compared to known alleles.
Figure 10-2
1
2
Heteroduplex
1 2
3
Denature +
reanneal
nonstringent
+
Gel

Lane 1/2: homozygotes

Lane 3: heterozygote
Homo Hetero Hetero Homo
Duplex Duplex Duplex Duplex
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V. Exonuclease-released fluorescence [Figure 10-3].
A.
An sequence specific oligonucleotide labeled with reporter and quencher
dyes is hybridized to target DNA. Addition of Taq polymerase and a locusspecific primer located 5' of the probe causes the reporter dye to be released
during PCR. Once the reporter dye is separated from the quencher dye,
fluorescence is produced indicating that the probe had hybridized to the
DNA.
Figure 10-3
Exonuclease-Released Fluorescence
Taq
Reporter
Quencher
References:
RFLP:
Maeda, M., Uryu, N., Murayama, N., Ishii, H., Ota, M. Tsuji, K., and Inoko, H. 1990. A
simple and rapid method for HLA-DP genotyping by digestion of PCR-amplified DNA with
allele-specific restriction endonucleases. Human Immunology 27:111-121.
Olerup, O. 1990. HLA class II typing by digestion of PCR-amplified DNA with allele-specific
restriction endonucleases will fail to unequivocally identify the genotypes of many
homozygous and heterozygous individuals. Tissue Antigens 36:83-87.
SSCP:
Orita, M., Iwahana, H., Kanazawa, H., Hayashi, K., and Sekiya, T. 1989. Detection of
polymorphisms of human DNA by gel electrophoresis as single-strand conformation
polymorphisms. Proc. Natl. Acad. Sci. USA 86:2766-2770.
C.W. Bill Young Marrow Donor Recruitment and Research Program
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Lo, Y.M.D., Patel, P., Mehal, W.Z., Fleming, K.A., Bell, J.I., and Wainscoat, J.S. 1992.
Analysis of complex genetic systems by ARMS-SSCP: application to HLA genotyping.
Nucleic Acids Res. 20-1005-1009.
Heteroduplex/RSCA:
Savage et. al. 1996. A rapid HLA-DRB1*04 subtyping method using PCR and DNA
heteroduplex generators. Tissue Antigens 47:284-292.
Summers, C., Morling, F., Taylor, M., Yin, J. L., and Stevens, R. 1994. Donor-recipient
HLA class I bone marrow transplant matching by multilocus heteroduplex analysis.
Transplantation 58: 628-629.
Arguello, J.R., Little, A-M, Bohan, E. et. al. 1998. High resolution HLA class typing by
reference strand mediated conformation analysis (RSCA). Tissue Antigens 52: 57-66.
Exonuclease:
Faas et. al. 1996. Sequence specific priming and exonuclease-released fluorescence
detection of HLA-DQB1 alleles. Tissue Antigens 48:97-112.
Slateva, K., Albis-Camps, M., Blasczyk, et.al. 1998. Fluorotyping of HLA-A by sequence
specific priming and fluorogenic probing. Tissue Antigens 52: 462-72.
C.W. Bill Young Marrow Donor Recruitment and Research Program
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CHAPTER 11
INTERPRETATION OF DNA TYPING RESULTS
The hybridization results obtained with sequence specific oligonucleotide probes or the
amplifications obtained with sequence specific primers are used to identify HLA types.
Likewise the DNA sequences obtained by either Sanger or NGS are used to assign alleles.
I.
Most HLA alleles do not have a unique DNA sequence that characterizes that allele
and that is found in no other allele (Chapters 6 and 7).
II.
Detection of HLA types might use a panel of SSOP (30-40 probes) to obtain low
resolution or serologic typing level resolution results for a single locus e.g., DRB1 or
HLA-A. The use of more probes will produce an intermediate level of typing
resolution.
A.
HLA types are obtained by comparing the positive and negative probe
hybridizations (or the positive and negative sequence specific amplifications)
to the known list of alleles (Figure 11-1).
Figure 11-1
DR
allele
0101
0301
0401
0701
1101
1301
Probe
1
Probe
2
Probe
3
Probe
4
Probe
5
Probe
6
═══════════════════════════════════════════════════════════════════
QUESTION 1:Based on Figure 11-1, what would be the typing if probe 1 and probe 3 were
positive and the remainder negative?
What would the typing be if probe 2, probe 4, probe 5 were positive and the rest negative?
═══════════════════════════════════════════════════════════════════
C.W. Bill Young Marrow Donor Recruitment and Research Program
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═══════════════════════════════════════════════════════════════════
QUESTION 2: Align the sequences of the DRB1 alleles.
If probe DR1001
(5'TGGCAGCTTAAGTTTGAA (codons 9-13)) is positive, one of the DRB1*01 alleles is
present. If probe DR7007 (5'ACATCCTGGAAGACGAGC (codons 66-72)) is also positive,
the DRB1*01:03 allele may be present. What would be your interpretation of the DR type
for this sample if the probe DR1004 (5' GAGCAGGTTAAACATGAG (codons 9-14)) is also
positive in addition to the probes listed above?.
How would your interpretation of DRB1*01:03 change?
═══════════════════════════════════════════════════════════════════
VI.
IV.
DNA sequence interpretation
A.
The use of Sanger sequencing for typing allows identification of the complete
sequence of the HLA region amplified. The testing must be supplemented
with a strategy to determine the phase of the polymorphic nucleotides in
order to identify the genotype(s) present.
B.
NGS sequencing of an entire HLA gene with phasing should result in the
identification of a single genotype at the allele level.
Interpretation of typing results provides us with a genotype for an individual. Often
the typing result yields more than one possible genotype and more testing will be
required to determine which genotype is the correct one for that individual. Figure
11-2 is an example of a sequencing result which identified two possible genotypes:
DRB1*01:01:01, 03:01:02 or DRB1*01:04,03:14.
Figure 11-2
C.W. Bill Young Marrow Donor Recruitment and Research Program
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Alternative Genotypes or Ambiguous
Combinations
DRB1*01:01:01
DRB1*01:04
DRB1*03:01:02
DRB1*03:14
DRB1*01:01:01
DRB1*01:04
DRB1*03:01:02
DRB1*03:14
DRB1*01:01:01
DRB1*01:04
DRB1*03:01:02
DRB1*03:14
DRB1*01:01:01
DRB1*01:04
DRB1*03::01:02
DRB1*0314
1
5
CTG GCT TTG GCT GGG GAC ACC CGA C|CA CGT TTC TTG
*** *** *** *** --- --- --- --- -|-- --- --- ----- --- --- --- --- --- --- A-- -|-- --- --- --*** *** *** *** *** *** *** *** *|-- --- --- ---
10
15
20
TGG CAG CTT AAG TTT GAA TGT CAT TTC TTC AAT GGG ACG
--- --- --- --- --- --- --- --- --- --- --- --- --GA- T-C TC- -C- -C- --G --- --- --- --- --- --- --GA- T-C TC- -C- -C- --G --- --- --- --- --- --- ---
GAG
-------
CGG
-------
GTG
-------
25
CGG
-------
TTG
---AC
-AC
CTG
-------
GAA
----C
--C
AGA
-------
30
TGC
---A-A-
ATC
--T-T--
TAT
--C-C--
AAC
-------
CAA
----G
--G
35
GAG
-------
GAG
-------
TCC
--AAAA-
GTG
-------
CGC
-------
40
TTC
-------
GAC
-------
AGC
-------
GAC
-------
GTG
-------
45
GGG
-------
GAG
-------
TAC
---T-T-
CGG
-------
GCG
-------
50
GTG
-------
ACG
-------
GAG
-------
CTG
-------
GGG
-------
55
CGG
-------
CCT
-------
GAT
-------
GCC
-------
GAG
-------
60
TAC
-------
TGG
-------
AAC
-------
AGC
-------
CAG
-------
65
AAG
-------
GAC
-------
CTC
-------
CTG
-------
GAG
-------
70
CAG
-------
AGG
---A-A-
95
G|TT
-|--|-C
-|**
GAG
--C-T
***
CGG
-------
GCC
---G-G-
GCG
--CGCG-
75
GTG
-------
GAC
-------
ACC
-AT
-AT
---
TAC
-------
TGC
-------
80
AGA
-------
CAC
-------
AAC
-------
TAC
-------
GGG
-------
85
GTT
-------
GGT
-TG
-TG
---
GAG
-------
AGC
-------
TTC
-------
90
ACA
-------
GTG
-------
CAG
-------
CGG
-------
CGA
-------
Alternative allele combinations possible:
DRB1*01:01:01,*03:01:02 vs DRB1*01:04,*03:14
═══════════════════════════════════════════════════════════════════
QUESTION 3: Why can’t these two alternative genotypes be distinguished by sequencing
of both alleles of the DRB1 locus together?
How would you determine which genotype is the correct one?
═══════════════════════════════════════════════════════════════════
V.
New HLA alleles continue to be described. If the same sample is typed with the
same set of probes each year, the interpretation of the hybridization results might
change over time. For example, if DNA from a sample was positive with probe
DR1001AS in 1990, the result would be interpreted as DRB1*01:01 or DRB1*01:02
or DRB1*01:03. If DNA from the same sample was typed again a few years later, a
positive hybridization result with probe DR1001AS would be interpreted as
DRB1*01:01 or DRB1*01:02 or DRB1*01:03 or DRB1*01:04. DRB1*01:04 is a
more recently described allele which carries the sequence detected by DR1001AS.
Note that the hybridization result does not change; only the interpretation of that
result. This means that one must always review the interpretation of a typing result
if the typing occurred some time in the past (e.g., over a year ago). This requires
the knowledge of the sequences of the probes and primers used in the typing and
the positive and negative hybridization results obtained with those probes and
primers.
C.W. Bill Young Marrow Donor Recruitment and Research Program
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═══════════════════════════════════════════════════════════════════
QUESTION 4: A donor on the unrelated bone marrow registry was typed as DRB1*11
using the probe 5703 (5' GCCTGATGAGGAGTACTG (codons 55-61)). What would be
this person’s DRB1 type based on all of the DRB1 alleles that have been identified to date?
(Hint: Look at the list of current DRB1 alleles.)
═══════════════════════════════════════════════════════════════════
References:
Hurley, C.K. 1997. Acquisition and use of DNA-based HLA typing data in bone marrow
registries. 1997. Tissue Antigens 49:323-328.
C.W. Bill Young Marrow Donor Recruitment and Research Program
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ANSWERS TO THE QUESTIONS
Chapter 1:
Question 1:
PstI recognition site
5'
3'
CTGCA|G
This generates a 3' protrusion.
example:
5'
3'
GTAC TGCA|GTC
cut--->
CATG|ACGT CAG
3'
5'
5'
3'
GTACTGCA
CATG
3'
5'
5'
and
3'
GTC
ACGTCAG
3'
5'
Question 2:
NotI recognition site:
5'
3'
GC|GGCCGC = 8 base pairs= 48= one cut every 65,536 bp
EcoRI recognition site:
5'
3'
G|AATTC = 6 base pairs= 46= one cut every 4,096 bp
NotI has a longer recognition sequence than EcoRI; therefore, NotI cuts less frequently.
Question 3:
Example:
ATGCCTTAGGCATCCGTT
TACGGAATCCGTAGGCAA
Tm=[4x9] + [2x9] = 36 + 18 = 54C
This example temperature is higher than the example provided in the manual. If the
sequence is 18 bp of GC, the Tm= [4x18]+[2x0]= 72oC. If the sequence is 18 bp of AT, the
Tm= [4x0] + [2x18]= 36oC
Question 4:
The DNA fragment in lane 1 is approx. 500 bp long. The DNA fragment in lane 2 is
approx. 330 bp long. The migration is from negative (top of gel) to positive (bottom).
Chapter 2:
Question 1:
A*01:01:01:01
-291
-281
-271
-261
-251
-241
-231
-221
-211
-201
CAGGAGCAGA GGGGTCAGGG CGAAGTCCCA GGGCCCCAGG CGTGGCTCTC AGGGTCTCAG GCCCCGAAGG CGGTGTATGG ATTGGGGAGT CCCAGCCTTG
A*01:01:01:01
-191
-181
-171
-161
-151
-141
-131
-121
-111
-101
GGGATTCCCC AACTCCGCAG TTTCTTTTCT CCCTCTCCCA ACCTACGTAG GGTCCTTCAT CCTGGATACT CACGACGCGG ACCCAGTTCT CACTCCCATT
A*01:01:01:01
-91
-81
-71
-61
-51
-41
-31
-21
-11
-1
GGGTGTCGGG TTTCCAGAGA AGCCAATCAG TGTCGTCGCG GTCGCTGTTC TAAAGTCCGC ACGCACCCAC CGGGACTCAG ATTCTCCCCA GACGCCGAGG
A*01:01:01:01
10
20
30
40
50
60
70
80
90
100
|ATGGCCGTCA TGGCGCCCCG AACCCTCCTC CTGCTACTCT CGGGGGCCCT GGCCCTGACC CAGACCTGGG CGG|GTGAGTG CGGGGTCGGG AGGGAAACCG
C.W. Bill Young Marrow Donor Recruitment and Research Program
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A*01:01:01:01
110
120
130
140
150
160
170
180
190
200
CCTCTGCGGG GAGAAGCAAG GGGCCCTCCT GGCGGGGGCG CAGGACCGGG GGAGCCGCGC CGGGAGGAGG GTCGGGCAGG TCTCAGCCAC TGCTCGCCCC
A*01:01:01:01
210
220
230
240
250
260
270
280
290
300
CAG|GCTCCCA CTCCATGAGG TATTTCTTCA CATCCGTGTC CCGGCCCGGC CGCGGGGAGC CCCGCTTCAT CGCCGTGGGC TACGTGGACG ACACGCAGTT
A*01:01:01:01
310
320
330
340
350
360
370
380
390
400
CGTGCGGTTC GACAGCGACG CCGCGAGCCA GAAGATGGAG CCGCGGGCGC CGTGGATAGA GCAGGAGGGG CCGGAGTATT GGGACCAGGA GACACGGAAT
A*01:01:01:01
410
420
430
440
450
460
470
480
490
500
ATGAAGGCCC ACTCACAGAC TGACCGAGCG AACCTGGGGA CCCTGCGCGG CTACTACAAC CAGAGCGAGG ACG|GTGAGTG ACCCCGGCCC GGGGCGCAGG
A*01:01:01:01
510
520
530
540
550
560
570
580
590
600
TCACGACCCC TCATCCCCCA CGGACGGGCC AGGTCGCCCA CAGTCTCCGG GTCCGAGATC CACCCCGAAG CCGCGGGACT CCGAGACCCT TGTCCCGGGA
A*01:01:01:01
610
620
630
640
650
660
670
680
690
700
GAGGCCCAGG CGCCTTTACC CGGTTTCATT TTCAGTTTAG GCCAAAAATC CCCCCGGGTT GGTCGGGGCG GGGCGGGGCT CGGGGGACTG GGCTGACCGC
A*01:01:01:01
710
720
730
740
750
760
770
780
790
800
GGGGTCGGGG CCAG|GTTCTC ACACCATCCA GATAATGTAT GGCTGCGACG TGGGGCCGGA CGGGCGCTTC CTCCGCGGGT ACCGGCAGGA CGCCTACGAC
A*01:01:01:01
810
820
830
840
850
860
870
880
890
900
GGCAAGGATT ACATCGCCCT GAACGAGGAC CTGCGCTCTT GGACCGCGGC GGACATGGCA GCTCAGATCA CCAAGCGCAA GTGGGAGGCG GTCCATGCGG
A*01:01:01:01
910
920
930
940
950
960
970
980
990
1000
CGGAGCAGCG GAGAGTCTAC CTGGAGGGCC GGTGCGTGGA CGGGCTCCGC AGATACCTGG AGAACGGGAA GGAGACGCTG CAGCGCACGG |GTACCAGGGG
A*01:01:01:01
CCTCCCTCTG
1010
1020
1030
1040
1050
1060
1070
1080
1090
CCACGGGGCG CCTCCCTGAT CGCCTATAGA TCTCCCGGGC TGGCCTCCCA CAAGGAGGGG AGACAATTGG GACCAACACT AGAATATCAC
1100
A*01:01:01:01
1110
1120
1130
1140
1150
1160
1170
1180
1190
1200
GTCCTGAGGG AGAGGAATCC TCCTGGGTTT CCAGATCCTG TACCAGAGAG TGACTCTGAG GTTCCGCCCT GCTCTCTGAC ACAATTAAGG GATAAAATCT
A*01:01:01:01
1210
1220
1230
1240
1250
1260
1270
1280
1290
1300
CTGAAGGAGT GACGGGAAGA CGATCCCTCG AATACTGATG AGTGGTTCCC TTTGACACCG GCAGCAGCCT TGGGCCCGTG ACTTTTCCTC TCAGGCCTTG
A*01:01:01:01
1310
1320
1330
1340
1350
1360
1370
1380
1390
1400
TTCTCTGCTT CACACTCAAT GTGTGTGGGG GTCTGAGTCC AGCACTTCTG AGTCTCTCAG CCTCCACTCA GGTCAGGACC AGAAGTCGCT GTTCCCTTCT
A*01:01:01:01
1410
1420
1430
1440
1450
1460
1470
1480
1490
1500
CAGGGAATAG AAGATTATCC CAGGTGCCTG TGTCCAGGCT GGTGTCTGGG TTCTGTGCTC TCTTCCCCAT CCCGGGTGTC CTGTCCATTC TCAAGATGGC
A*01:01:01:01
1510
1520
1530
1540
1550
1560
1570
1580
1590
1600
CACATGCGTG CTGGTGGAGT GTCCCATGAC AGATGCAAAA TGCCTGAATT TTCTGACTCT TCCCGTCAG|A CCCCCCCAAG ACACATATGA CCCACCACCC
A*01:01:01:01
1610
1620
1630
1640
1650
1660
1670
1680
1690
1700
CATCTCTGAC CATGAGGCCA CCCTGAGGTG CTGGGCCCTG GGCTTCTACC CTGCGGAGAT CACACTGACC TGGCAGCGGG ATGGGGAGGA CCAGACCCAG
A*01:01:01:01
1710
1720
1730
1740
1750
1760
1770
1780
1790
1800
GACACGGAGC TCGTGGAGAC CAGGCCTGCA GGGGATGGAA CCTTCCAGAA GTGGGCGGCT GTGGTGGTGC CTTCTGGAGA GGAGCAGAGA TACACCTGCC
A*01:01:01:01
1810
1820
1830
1840
1850
1860
1870
1880
1890
1900
ATGTGCAGCA TGAGGGTCTG CCCAAGCCCC TCACCCTGAG ATGGG|GTAAG GAGGGAGATG GGGGTGTCAT GTCTCTTAGG GAAAGCAGGA GCCTCTCTGG
A*01:01:01:01
1910
1920
1930
1940
1950
1960
1970
1980
1990
2000
AGACCTTTAG CAGGGTCAGG GCCCCTCACC TTCCCCTCTT TTCCCAG|AGC TGTCTTCCCA GCCCACCATC CCCATCGTGG GCATCATTGC TGGCCTGGTT
A*01:01:01:01
2010
2020
2030
2040
2050
2060
2070
2080
2090
2100
CTCCTTGGAG CTGTGATCAC TGGAGCTGTG GTCGCTGCCG TGATGTGGAG GAGGAAGAGC TCAG|GTGGAG AAGGGGTGAA GGGTGGGGTC TGAGATTTCT
A*01:01:01:01
2110
2120
2130
2140
2150
2160
2170
2180
2190
2200
TGTCTCACTG AGGGTTCCAA GCCCCAGCTA GAAATGTGCC CTGTCTCATT ACTGGGAAGC ACCTTCCACA ATCATGGGCC GACCCAGCCT GGGCCCTGTG
A*01:01:01:01
2210
2220
2230
2240
2250
2260
2270
2280
2290
2300
TGCCAGCACT TACTCTTTTG TAAAGCACCT GTTAAAATGA AGGACAGATT TATCACCTTG ATTACGGCGG TGATGGGACC TGATCCCAGC AGTCACAAGT
A*01:01:01:01
2310
2320
2330
2340
2350
2360
2370
2380
2390
2400
CACAGGGGAA GGTCCCTGAG GACAGACCTC AGGAGGGCTA TTGGTCCAGG ACCCACACCT GCTTTCTTCA TGTTTCCTGA TCCCGCCCTG GGTCTGCAGT
A*01:01:01:01
2410
2420
2430
2440
2450
2460
2470
2480
2490
2500
CACACATTTC TGGAAACTTC TCTGGGGTCC AAGACTAGGA GGTTCCTCTA GGACCTTAAG GCCCTGGCTC CTTTCTGGTA TCTCACAGGA CATTTTCTTC
A*01:01:01:01
2510
2520
2530
2540
2550
2560
2570
2580
2590
2600
CCACAG|ATAG AAAAGGAGGG AGTTACACTC AGGCTGCAA|G TAAGTATGAA GGAGGCTGAT GCCTGAGGTC CTTGGGATAT TGTGTTTGGG AGCCCATGGG
A*01:01:01:01
2610
2620
2630
2640
2650
2660
2670
2680
2690
2700
GGAGCTCACC CACCCCACAA TTCCTCCTCT AGCCACATCT TCTGTGGGAT CTGACCAGGT TCTGTTTTTG TTCTACCCCA G|GCAGTGACA GTGCCCAGGG
A*01:01:01:01
2710
2720
2730
2740
2750
2760
2770
2780
2790
2800
CTCTGATGTG TCTCTCACAG CTTGTAAAG|G TGAGAGCTTG GAGGGCCTGA TGTGTGTTGG GTGTTGGGTG GAACAGTGGA CACAGCTGTG CTATGGGGTT
A*01:01:01:01
2810
2820
2830
2840
2850
2860
2870
2880
2890
2900
TCTTTGCGTT GGATGTATTG AGCATGCGAT GGGCTGTTTA AGGTGTGACC CCTCACTGTG ATGGATATGA ATTTGTTCAT GAATATTTTT TTCTATAG|TG
A*01:01:01:01
2910
2920
2930
2940
2950
2960
2970
2980
2990
3000
TGA|GACAGCT GCCTTGTGTG GGACTGAGAG GCAAGAGTTG TTCCTGCCCT TCCCTTTGTG ACTTGAAGAA CCCTGACTTT GTTTCTGCAA AGGCACCTGC
A*01:01:01:01
3010
3020
3030
3040
3050
3060
3070
3080
3090
3100
ATGTGTCTGT GTTCGTGTAG GCATAATGTG AGGAGGTGGG GAGAGCACCC CACCCCCATG TCCACCATGA CCCTCTTCCC ACGCTGACCT GTGCTCCCTC
A*01:01:01:01
3110
3120
3130
3140
3150
3160
3170
3180
3190
3200
CCCAATCATC TTTCCTGTTC CAGAGAGGTG GGGCTGAGGT GTCTCCATCT CTGTCTCAAC TTCATGGTGC ACTGAGCTGT AACTTCTTCC TTCCCTATTA
C.W. Bill Young Marrow Donor Recruitment and Research Program
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Question 2:
A*01:01:01:01
-21
-11
-1
10
20
30
40
50
60
70
MAVM APRTLLLLLS GALALTQTWA GSHSMRYFFT SVSRPGRGEP RFIAVGYVDD TQFVRFDSDA ASQKMEPRAP WIEQEGPEYW DQETRNMKAH
A*01:01:01:01
80
90
100
110
120
130
140
150
160
170
SQTDRANLGT LRGYYNQSED GSHTIQIMYG CDVGPDGRFL RGYRQDAYDG KDYIALNEDL RSWTAADMAA QITKRKWEAV HAAEQRRVYL EGRCVDGLRR
A*01:01:01:01
180
190
200
210
220
230
240
250
260
270
YLENGKETLQ RTDPPKTHMT HHPISDHEAT LRCWALGFYP AEITLTWQRD GEDQTQDTEL VETRPAGDGT FQKWAAVVVP SGEEQRYTCH VQHEGLPKPL
A*01:01:01:01
280
290
300
310
320
330
340
TLRWELSSQP TIPIVGIIAG LVLLGAVITG AVVAAVMWRR KSSDRKGGSY TQAASSDSAQ GSDVSLTACK V
350
See the answer to question 2 for the start codon and leader sequence.
Chapter 3:
Question1:
DRA
Chapter 4:
Question 1:
No question, look up reference.
Question 2:
For example: DQB1*05:01, DQB1*02:01
Question 3:
(1.9/100) x (7/100) = 13.3/10,000 = 0.133/100 = 0.133%
Question 4:
DRB1*01:01:
GGG GAC ACC CGA CCA CGT TTC TTG TGG CAG CTT AAG TTT GAA
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Gly Asp Thr Arg Pro Arg Phe Leu Trp Gln Leu Lys Phe Glu
TGT CAT TTC TTC AAT GGG
Cys His Phe Phe Asn Gly
DRB1*13:02:
GGG GAC ACC
Gly Asp Thr
TGT CAT TTC
Cys His Phe
AGA
Arg
TTC
Phe
Question 5:
Example:
ACC GGA TAT
Thr Gly Tyr
CCA
Pro
AAT
Asn
TTT
Phe
CGT TTC TTG GAG TAC TCT ACG TCT GAG
Arg Phe Leu Glu Tyr Ser Thr Ser Glu
GGG
Gly
GAA
Glu
Question 6:
♂ DRB1*04:01, DRB1*11:03
♀ DRB1*01:01, DRB1*03:01
Possible DRB1 combinations of children:
DRB1*04:02, *03:01
DRB1*04:02, *01:01
DRB1*11:03, *03:01
DRB1*11:03, *01:01
Yes, it is possible for two children to be DR identical. No children are homozygous; all are
heterozygous (i.e., carry 2 different alleles).
Question 7:
♂
DRB1*04:03, DPB1*02:01
DRB1*11:01, DPB1*02:01
♀
DRB1*01:01, DPB1*02:01
DRB1*03:02, DPB1*01:01
Possible DR, DP combinations of children:
DRB1*04:03, DPB1*02:01; DRB1*01:01, DPB1*02:01
DRB1*04:03, DPB1*02:01; DRB1*03:02, DPB1*01:01
DRB1*11:01, DPB1*02:01; DRB1*01:01, DPB1*02:01
DRB1*11:01, DPB1*02:01; DRB1*03:02, DPB1*01:01
and many more possibilities by recombination, for example:
DRB1*04:03, DPB1*02:01; DRB1*03:02, DPB1*02:01
Yes, DRB1*04:03, DPB1*02:01; DRB1*01:01, DPB1*01:01 is possible by recombination.
DRB1*01:01
DPB1*02:01
X
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DRB1*03:02

DRB1*01:01
DPB1*01:01
DRB1*03:02
DPB1*02:01
DPB1*01:01
Question 8:
DRB1*13:04 DRB3*02:01 DQA1*02:01 DQB1*02:01 DPA1*01:04 DPB1*04:01
DRB1*04:02 DRB4*01:01 DQA1*02:01 DQB1*02:01 DPA1*01:04 DPB1*04:01
Note the predicted association of DRB1*13:04 with DRB3*02:01 and DRB1*04:02 with
DRB4*01:01.
The person is homozygous for DQ and DP.
Question 9:
See next two pages.
Question 10:
For example: C*01:01, C*02:01, C*02:02, C*08:01
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Chapter 4-Question 9:
Part One:
DRB1*0101
DRB1*0102
DRB5*0202
1
10
GGG GAC ACC CGA CCA CGT TTC TTG TGG CAG CTT AAG TTT GAA TGT CAT TTC TTC AAT
--- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- ----- --- --- --- --- T-- --- --- CA- --- GA- --- -A- --G --- --- --- --- --C
20
GGG ACG GAG CGG GTG CGG TTG CTG GAA AGA
--- --- --- --- --- --- --- --- --- ----- --- --- --- --- --- --C --- C-C ---
DRB1*0101
DRB1*0102
DRB5*0202
40
50
GAG GAG TCC GTG CGC TTC GAC AGC GAC GTG GGG GAG TAC CGG GCG GTG ACG GAG CTG GGG CGG CCT GAT GCC GAG
--- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- ----- --- AA- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --C --T ---
DRB1*0101
DRB1*0102
DRB5*0202
70
80
90
100
GAG CAG AGG CGG GCC GCG GTG GAC ACC TAC TGC AGA CAC AAC TAC GGG GTT GGT GAG AGC TTC ACA GTG CAG CGG CGA GTT GAG CCT AAG GTG ACT GTG TAT
--- --- --- --- --- --- --- --- --- --T --- --- --- --- --- --- -C- -TG --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- ----- --- GC- --- --- --- --- --- --- --- --- --- --- --- --- --- -C- -TG --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- ---
DRB1*0101
DRB1*0102
DRB5*0202
110
120
130
CCT TCA AAG ACC CAG CCC CTG CAG CAC CAC AAC CTC CTG GTC TGC TCT GTG AGT GGT TTC TAT CCA GGC AGC ATT GAA GTC AGG TGG TTC CGG AAC GGC CAG
--- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- ----- G-- -G- --- --- A-- --- --- --- --- --- --- --- --- --- --- --- -A- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- ---
DRB1*0101
DRB1*0102
DRB5*0202
140
GAA GAG AAG GCT GGG GTG GTG TCC ACA GGC CTG ATC CAG
--- --- --- --- --- --- --- --- --- --- --- --- ----- --- --- --- --- --- --- --- --- --- --- --T ---
DRB1*0101
DRB1*0102
DRB5*0202
180
TAC ACC TGC CAA GTG GAG CAC CCA AGT GTG ACG AGC CCT CTC
--- --- --- --- --- --- --- --- --- --- --- --- --- ----- --- --- --- --- --- --- --- --C --- --- --- --- ---
DRB1*0101
DRB1*0102
DRB5*0202
210
220
GGC TTC GTG CTG GGC CTG CTC TTC CTT GGG GCC GGG CTG TTC ATC TAC TTC AGG AAT CAG AAA GGA CAC TCT GGA
--- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- ----- --T --- --- --- --- --- --- --- --- --- --- --A --- --- --- --- -A- --- --- --- --G --- --- ---
150
AAT GGA GAT TGG ACC TTC CAG ACC CTG GTG
--- --- --- --- --- --- --- --- --- ----- --- --C- --- --- --- --- -TT --- ---
30
TGC ATC TAT AAC CAA
--- --- --- --- --G-- --- --- --- ---
60
TAC TGG AAC AGC CAG AAG GAC CTC CTG
--- --- --- --- --- --- --- --- ----- --- --- --- --- --- --- A-- ---
160
ATG CTG GAA ACA GTT CCT CGG AGT GGA GAG
--- --- --- --- --- --- --- --- --- ----- --- --- --- --- --- --- --- --- ---
170
GTT
-----
190
200
ACA GTG GAA TGG AGA GCA CGG TCT GAA TCT GCA CAG AGC AAG ATG CTG AGT GGA GTC GGG
--- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- ----- --- --- --- --- --- -A- --- --- --- --- --- --- --- --- --- --- --- A-- ---
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230
CTT CAG CCA ACA GGA TTC CTG AGC TGA
--- --- --- --- --- --- --- --- ----- --C --- --- --- C-- G-- --- ---
part two:
DRB1*0101
DRB1*0103
DRB1*1301
1
10
GGG GAC ACC CGA CCA CGT TTC TTG TGG CAG CTT AAG TTT GAA TGT CAT TTC TTC AAT
--- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- ----- --- --- A-- --- --- --- --- GA- T-C TC- -C- -C- --G --- --- --- --- ---
20
GGG ACG GAG CGG GTG CGG TTG CTG GAA AGA
--- --- --- --- --- --- --- --- --- ----- --- --- --- --- --- --C --- --C ---
DRB1*0101
DRB1*0103
DRB1*1301
40
50
GAG GAG TCC GTG CGC TTC GAC AGC GAC GTG GGG GAG TAC CGG GCG GTG ACG GAG CTG GGG CGG CCT GAT GCC GAG
--- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- ----- --- AA- --- --- --- --- --- --- --- --- --- -T- --- --- --- --- --- --- --- --- --- --- --- ---
DRB1*0101
DRB1*0103
DRB1*1301
70
80
90
100
GAG CAG AGG CGG GCC GCG GTG GAC ACC TAC TGC AGA CAC AAC TAC GGG GTT GGT GAG AGC TTC ACA GTG CAG CGG CGA GTT GAG CCT AAG GTG ACT GTG TAT
--A G-C GA- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- ----A G-C GA- --- --- --- --- --- --- --- --- --- --- --- --- --- --- -TG --- --- --- --- --- --- --- --- --C C-T --- --- --- --- --- ---
DRB1*0101
DRB1*0103
DRB1*1301
110
120
130
CCT TCA AAG ACC CAG CCC CTG CAG CAC CAC AAC CTC CTG GTC TGC TCT GTG AGT GGT TTC TAT CCA GGC AGC ATT GAA GTC AGG TGG TTC CGG AAC GGC CAG
--- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- ----- --- --- --- --- --- --- --- --- --- --- --- --- --- --T --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --T --- ---
DRB1*0101
DRB1*0103
DRB1*1301
140
GAA GAG AAG GCT GGG GTG GTG TCC ACA GGC CTG ATC CAG
--- --- --- --- --- --- --- --- --- --- --- --- ----- --- --- A-- --- --- --- --- --- --- --- --- --C
DRB1*0101
DRB1*0103
DRB1*1301
180
TAC ACC TGC CAA GTG GAG CAC CCA AGT GTG ACG AGC CCT CTC
--- --- --- --- --- --- --- --- --- --- --- --- --- ----- --- --- --- --- --- --- --- --C --- --A --- --- ---
DRB1*0101
DRB1*0103
DRB1*1301
210
220
GGC TTC GTG CTG GGC CTG CTC TTC CTT GGG GCC GGG CTG TTC ATC TAC TTC AGG AAT CAG AAA GGA CAC TCT GGA
--- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- ----- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- ---
150
AAT GGA GAT TGG ACC TTC CAG ACC CTG GTG
--- --- --- --- --- --- --- --- --- ----- --- --C --- --- --- --- --- --- ---
30
TGC ATC TAT AAC CAA
--- --- --- --- ---A- T-- C-- --- --G
60
TAC TGG AAC AGC CAG AAG GAC CTC CTG
--- --- --- --- --- --- --- A-- ----- --- --- --- --- --- --- A-- ---
160
ATG CTG GAA ACA GTT CCT CGG AGT GGA GAG
--- --- --- --- --- --- --- --- --- ----- --- --- --- --- --- --- --- --- ---
170
GTT
-----
190
200
ACA GTG GAA TGG AGA GCA CGG TCT GAA TCT GCA CAG AGC AAG ATG CTG AGT GGA GTC GGG
--- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- ----- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- ---
C.W. Bill Young Marrow Donor Recruitment and Research Program
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230
CTT CAG CCA ACA GGA TTC CTG AGC TGA
--- --- --- --- --- --- --- --- ----- --- --- -G- --- --- --- --- ---
Chapter 4, Question 11:
1
ConsensusATGCGGGTCA TGGCGCCCCG
b-07021 ----t----- ---------b-0801 ----t----- ---------b-4201 ----t----- ----------
AACCCTCCTC
----g--------g--------g-----
CTGCTGCTCT
----------------------------
CGGGGGCCCT
---c--------c--------c------
GGCCCTGACC
----------------------------
GAGACCTGGG
----------------------------
CCGGCTCCCA
----------------------------
CTCCATGAGG
----------------------------
100
TATTTCTACA
---------------g------------
Consensus
b-07021
b-0801
b-4201
101
CCGCCGTGTC
--t-----------a-----t-------
CCGGCCCGGC
----------------------------
CGCGGGGAGC
----------------------------
CCCGCTTCAT
----------------------------
CGCAGTGGGC
-t--------t--------t--------
TACGTGGACG
----------------------------
ACACGCAGTT
----c-----------------c-----
CGTGAGGTTC
----------------------------
GACAGCGACG
----------------------------
200
CCGCGAGTCC
----------------------------
Consensus
b-07021
b-0801
b-4201
201
GAGGATGGAG
---aga------aga------aga----
CCGCGGGCGC
----------------------------
CGTGGATAGA
----------------------------
GCAGGAGGGG
----------------------------
CCGGAGTATT
----------------------------
GGGACCGGGA
--------a--------a--------a-
GACACAGATC
c--------c--------c---------
TTCAAGACCA
-a----g--c
----------a----g--c
ACACACAGAC
-gg----------------gg-------
300
TGACCGAGAG
----------------------------
Consensus
b-07021
b-0801
b-4201
301
AGCCTGCGGA
----------------------------
ACCTGCGCGG
----------------------------
CTACTACAAC
----------------------------
CAGAGCGAGG
----------------------------
CCGGGTCTCA
----------------------------
CACCCTCCAG
----------------------------
AGGATGTATG
--c-----c--c-----c--c-----c-
GCTGCGACGT
----------------------------
GGGGCCGGAC
----------------------------
400
GGGCGCCTCC
----------------------------
Consensus
b-07021
b-0801
b-4201
401
TCCGCGGGTA
--------c--------c--------c-
TGACCAGTAC
----------a--------a--------
GCCTACGACG
----------------------------
GCAAGGATTA
----------------------------
CATCGCCCTG
----------------------------
AACGAGGACC
----------------------------
TGCGCTCCTG
----------------------------
GACCGCGGCG
------c---------------------
GACACGGCGG
--------------c--------c----
500
CTCAGATCAC
----------------------------
Consensus
b-07021
b-0801
b-4201
501
CCAGCGCAAG
----------------------------
TGGGAGGCGG
----------------------------
CCCGTGTGGC
------a---------------------
GGAGCAGCTG
--------g-------gac
-------gac
AGAGCCTACC
----------------------------
TGGAGGGCAC
--------ga
-------------------
GTGCGTGGAG
----------------------------
TGGCTCCGCA
----------------------------
GATACCTGGA
----------------------------
600
GAACGGGAAG
----------------------------
Consensus
b-07021
b-0801
b-4201
601
GAGACGCTGC
--c-a----g
--c------g
--c------g
AGCGCGCGGA
-------t--------------------
CCCCCCAAAG
----------------------------
ACACATGTGA
-----c--------c--------c----
CCCACCACCC
----------------------------
CATCTCTGAC
----------------------------
CATGAGGCCA
----------------------------
CCCTGAGGTG
----------------------------
CTGGGCCCTG
----------------------------
700
GGCTTCTACC
--t-------------------------
701
800
Consensus CTGCGGAGAT CACACTGACC TGGCAGCGGG ATGGCGAGGA CCAAACTCAG GACACCGAGC TTGTGGAGAC CAGACCAGCA GGAGATAGAA CCTTCCAGAA
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b-07021
b-0801
b-4201
---------- ---------- ---------- ---------- ---------- -----t---- ---------- ---------- ---------- ------------------- ---------- ---------- ---------- ---------- -----t---- ---------- ---------- ---------- ------------------- ---------- ---------- ---------- ---------- -----t---- ---------- ---------- ---------- ----------
Consensus
b-07021
b-0801
b-4201
801
GTGGGCAGCT
----------------------------
GTGGTGGTGC
----------------------------
CTTCTGGAGA
----------------------------
AGAGCAGAGA
----------------------------
TACACATGCC
----------------------------
ATGTACAGCA
----------------------------
TGAGGGGCTG
----------------------------
CCGAAGCCCC
----------------------------
TCACCCTGAG
----------------------------
900
ATGGGAGCCA
---------g
---------g
---------g
Consensus
b-07021
b-0801
b-4201
901
TCTTCCCAGT
----------------------------
CCACCATCCC
-----g--------g--------g----
CATCGTGGGC
----------------------------
ATTGTTGCTG
----------------------------
GCCTGGCTGT
----------------------------
CCTAGCAGTT
----------------------------
CTAGTGGTCA
...------...------...-------
TCGGAGCTGT
----------------------------
GGTCGCTGCT
----------------------------
1000
GTGATGTGTA
----------------------------
Consensus
b-07021
b-0801
b-4201
1001
GGAGGAAGAG
----------------------------
CTCAGGTGGA
t---------------------------
AAAGGAGGGA
----------------------------
GCTACTCTCA
----------------------------
GGCTGCGTCC
--------g--------g--------g-
AGCGACAGTG
----------------------------
CCCAGGGCTC
----------------------------
TGATGTGTCT
----------------------------
CTCACAGCTT
----------------------------
1100
GAAAAGTGTG
--~~~~~~~~
--~~~~~~~~
--~~~~~~~
Chapter 5
Question 1: No question.
Question 2: No question.
Question 3:
Not necessarily. There are many alleles of DR3 and DR8. Serologic testing cannot distinguish between the alleles.
For example, the donor can be DRB1*08:03, DRB1*03:01 and the recipient can be DRB1*08:04, DRB1*03:02.
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Question 4:
A)♂
DR4/DR11
♀
DR1/DR3
children:
DR4/DR1; DR4/DR3; DR11/DR1; DR11/DR3
B) ♂
DRB1*04:02/DRB1*11:03
♀
DRB1*01:01/DRB1*03:01
children:
DRB1*04:02/DRB1*01:01; DRB1*040:2/DRB1*03:01
DRB1*11:03/DRB1*01:01; DRB1*11:03/DRB1*03:01
Question 5:
Loci: DRB1
DRB3
DRA
DRB1*11:02 DRB3*02:02 DRA*01:01
Question 6:
♂
A2,A3,B27,B53
♀
A2,A11,B51,B71
Children:
A2,B27;A2,B51 -or- A2,B53;A2,B51 -or- A3,B27;A2,B51 -or- A3,B53;A2,B51
A2,B27;A2,B71 -or- A2,B53;A2,B71 -or- A3,B27;A2,B71 -or- A3,B53;A2,B71
A2,B27;A11,B71 -or- A2,B53;A11,B71 -or- A3,B27;A11,B71 -or- A3,B53;A11,B71
A2,B27;A11,B51 -or- A2,B53;A11,B51 -or- A3,B27;A11,B51 -or- A3,B53;A11,B51
parent one: A2,B27 and A3,B53
parent two: A2,B71 and A11,B51
Question 7:
A*68:01=A68(28), B*13:01=B13, B*18:04=serologic type not known, DRB1*03:01=DR17(3),
DRB1*11:22=serologic type not known
Question 8:
Patient exhibits serologic type A2 only; the A*2409N allele does not specify an HLA-A antigen.
An A2,A24 donor would not be a good choice; a A2 donor would be better.
C.W. Bill Young Marrow Donor Recruitment and Research Program
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Chapter 6:
Question 1:
It would amplify many sequences. If 20 bp long, then the primers would be more specific
because the probability of finding a 20 bp sequence is less than finding a 2 bp sequence. A 2
bp sequence would be fairly common and the amplification would not be specific.
1/(42) vs. 1/(420)
1/16
vs. 1/1,099,500,000,000
Question 2:
Denaturation (dsDNA becomes ssDNA):
5' AATAA--------------------------------CCCC
3' TTATT--------------------------------GGGG
Primers Anneal:
5' AATAA-----------TTATGGCGG-------GGCGGCTTTA-------CCCC
|||||
CGAAA 5'
5'TTATG
||||
3 'TTATT------------AATACCGCC-------CCGCCGAAAT-------GGGG
Extension- nucleotides add to the 3' end of the primer and form 2 separate dsDNA molecules
at the end of the first cycle:
5'AATAA----------TTATGGCGG----------GGCGCTTT----------CCCC
◄◄◄◄◄◄◄◄◄◄◄◄◄◄◄◄◄◄◄◄◄◄◄◄◄◄◄◄◄◄CCGCGAAA 5'
5'
TTATGGCGG►►►►►►►►►►►►►►►►►►►►►►►►►►►
3'TTATT----------AATACCGCC----------CCGCGAAA----------GGGG
After second cycle:
5'AATAA--------------------------------GCTTT----------CCCC 3'
3'TTATT--------------------------------CGAAA 5'
5'TTATG---------------------------------CCCC 3'
3'TTATT---------AATAC---------------------------------GGGG 3'
5'TTATG------------------GCTTT 3'
3'TTATT---------AATAC------------------CGAAA
5'TTATG------------------GCTTT----------CCCC 3'
3'AATAC------------------CGAAA 5'
=4 dsDNA molecules
C.W. Bill Young Marrow Donor Recruitment and Research Program
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o
o
The polymerase copies the DNA until the temperature is changed (e.g., 72 C to 94 C). In the
initial cycles of the PCR, the original DNA strands yield long pieces of DNA when copied
by Taq. As the cycles proceed, a major DNA product appears that is a fixed size, bounded by
the primers.
After 3rd cycle:
8 dsDNA molecules
There are 26 bp in the amplified fragment.
after 3 cycles= 23 = 8 copies
after 10 cycles= 210 = 1024 copies
Question 3:
Find the answer in the answer for Chapter 2 question 1, the primer annealing sites are
underlined. The primer sequences cannot be found entirely in the cDNA sequence because
the primers are partially (forward) or completely positioned in the introns.
Question 4:
C*010201
C*04010101
C*050101
C*06020101
C*0802
C*12030101
C*1701
GGTTCTAGAG
-------------------------------------------------------
C*010201
C*04010101
C*050101
C*06020101
C*0802
C*12030101
C*1701
20
30
40
50
60
70
80
90
100
110
TGGCGCCCCG AACCCTCATC CTGCTGCTCT CGGGAGCCCT GGCCCTGACC GAGACCTGGG CCT|GTGAGTG CGGGGTTGGG AGGGAAACGG CCTCTGCGGA
---------- ---------- ---------- ---------- ---------- ---------- --G|------- ---------- ---------- ------G------------ ---------- ---------- ---------- ---------- ---------- ---|------- --A------- ---------- ------------------- ---------- ---------- ---------- ---------- ---------- ---|------- ---------- ---------- ------------------- ---------- ---------- ---------- ---------- ---------- ---|------- --A------- ---------- ------------------- ---------- ---------- ---------- ---------- ---------- ---|------- ---------- ---------- ------------------A -G-----C-- ---------- ---------- --------T- ---------- --G|------- ---------- ---------- ----------
C*010201
C*04010101
C*050101
C*06020101
C*0802
C*12030101
C*1701
120
130
140
150
160
170
180
190
200
210
GAGGAACGAG GTGCCCGCCC GGCGAGGGCG CAGGACCCGG GGAGCCGCGC AGGGAGGAGG GTCGGGCGGG TCTCAGCCCC TCCTCGCCCC CAG|GCTCCCA
-----G---- -G-------- ---------- ---------- ---------- ---------- ---------- --------A- ------T--- ---|-----------G---- -G-------- ---------- ---------- ---------- ---------- ---------- ---------- ---------- ---|-----------G---- -G-------- ---------- ---------- ---------- -------T-- ---------- ---------- ---------- ---|-----------G---- -G-------- ---------- ---------- ---------- ---------- ---------- ---------- ---------- ---|-----------G---- -G-------- ---------- ---------- ---------- ---------- ---------- ---------- ---------- ---|-----------G---- -G-------- ---------- ---------- ---------- ---------- ---------- ---------- ---------- ---|-------
AAGCCAATCA
-------------------------------------------------------
GCGTCTCCGC
---------------------------------------------------A---
AGTCCCGGTT
-------------------------------------------------------
CTAAAGTCCC
-----------------------------------------------G-----..
CAGTCACCCA
---------------------------------------------.....-----
CCCGGACTCA
------------------G
---------G
---------G
---------G
----------
GATTCTCCCC
-------------------------------------------------------
AGACGCCGAG
-------------------------------------------------------
|ATGCGGGTCA
|---------|---------|---------|---------|---------|----------
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C*010201
C*04010101
C*050101
C*06020101
C*0802
C*12030101
C*1701
CAGCGGAGAG
---------------------T-----------------T---------T-----
C*010201
C*04010101
C*050101
C*06020101
C*0802
C*12030101
C*1701
1020
1030
1040
1050
1060
1070
1080
1090
1100
1110
GGGAGCCTTC CCCATCTCCC GTAGATCTCC CGGCATGGCC TCCCACGAGG AGGGGAGGAA AATGGGATCA GCGCTAGAAT ATCGCCCTCC CTTGAATGGA
---------- ---------- ---------- ---G------ ---------- ---------- ---------- ---------- ---------- ------------------- ---------T ---------- ---G------ ---------- ---------- ---------- -----G---- ---------- ------------------- ---------T ---------- ---G------ ---------- ---------- ---------- ---------- ---------- ------------------- ---------T ---------- ---G------ ---------- ---------- ---------- -----G---- ---------- ------------------- ---------T ---------- ---G------ ---------- ---------- ---------- ---------- ---------- ------------------- ---------T A--------- ---G------ ---------- ---------- ---------- ---------- ---------- ----------
CCTACCTGGA
-------------------------------------------------------
GGGCACGTGC
-------------------------------------------------GA----
GTGGAGTGGC
-------------------------------------------------------
Question 5:
For example:
Forward:
5'
3'
GGT GTA AAC TTG TAC CAG
TCCGCAGATA
--------------------------------------------------G----
CCTGGAGAAC
-------------------------------------------------------
GGGAAGGAGA
---------------A-----------------A---------------------
CGCTGCAGCG
-------------------------------------------------------
CGCGG|GTACC
-----|---------|---------|---------|---------|---------|-----
AGGGGCAGTG
-------------------------------------------------------
Reverse:
5'
3'
CAA CTC TAC CGC TGC TAC
Question 6:
270,000 copies
Question 7:
Part one: 5' CTT GGA GCA GGT TAA ACA 3' for example (codon 7-13). Place the mismatch
on the 3' end of a PCR primer.
Part two: Can’t do this. The best choice is: 5' CTG GGG CGG CCT GAT GAG 3' (codon 5258) which will amplify all DRB1*11 alleles plus a few other alleles like DRB1*14:15.
Part three: No, it can not be done!!
Chapter 7:
Question 1:
3' GATGCGGATT 5'. If the DNA is double stranded, the complementary sequence of
5' CTACGCCTAA 3' can also be used.
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50% G+C. Approximate melting temperature is 30([4X5]+[2x5]=30). No hybridization will
occur at 50o, because it is above the melting temperature. If the probe is positioned so the
variation between the alleles is at the 3’ or 5’ end of the probe, it will be very difficult to achieve
specific hybridization; the probe will likely bind well to both alleles.
Question 2:
Part one: For example: 5' CGG CCT GAT GAG GAG TAC (codon 55-60). Place the mismatch
in the center of the oligo for a probe.
Part two: For example: 5' TGG AAC AGC CAG AAG GAC (codon 61-66) although
DRB1*04:11 differs in this region.
Part three: 5' GAG GAG GTT AAG TTT GAG (codon 9 -14).
Question 3:
First, amplify only the DRB1*01alleles with PCR primers designed around codons 25-31 and
DRBAMPB (for example). Then use a probe to distinguish between DRB1*01:01 and
DRB1*01:02, designed around codons 83-89. Use probe with sequence: 5' C GGG GCT GTG
GAG AGC TT (for example to detect DRB1*01:02 and not DRB1*01:01, *01:03, *01:04).
Chapter 8:
Question 1:
create PCR primers that would amplify DRB1*07:01
Question 2:
T
G
C
A
C.W. Bill Young Marrow Donor Recruitment and Research Program
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Chapter 9:
No questions
Chapter 10:
No questions
Chapter 11:
Question 1:
DRB1*01:01, *04:01
DRB1*11:01, *07:01
Question 2:
Sample also carries one of the DRB1*04 alleles. Since one of the DRB1*04 alleles,
DRB1*04:02, also carries the sequence detected by probe DR7007, it is not known if
DRB1*01:03 is present. As a result, the sample would be typed as [DRB1*01:03 and
DRB1*04:01 or *04:03 or *04:04 or ... *04:10] OR [DRB1*01:01 or *01:02 or *01:03 and
DRB1*04:02]. To determine if DBR1*01:03 is present, you would need to amplify DRB1*01
using a group-specific amplification.
Question 3: These two allele combinations would have exactly the same sequence. Positions
of mixed bases (i.e., two bases, a polymorphic residue) are the same. One must isolate
individual alleles to identify which combination is present. Look at the sequence of the two
alternative DRB1*03 alleles and see how knowing whether the DRB1*03 specific sequence at
codons 9-14 is on the same strand of DNA (i.e., in the same allele) as the sequence at codons
77 and 86 will identify which allele, DRB1*03:01:02 or DRB1*03:14, is present.
Question 4: The person could have any one of the DRB1*11 alleles (now numbering over 30)
or DRB1*03:08 or DRB1*04:15 or DRB1*12:04 or DRB1*14:11. The GAG codon at 58 used to
be unique for the DRB1*11 alleles but this is no longer the case.
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