Experiment 1 (401)
Evolution of the Vertebrate Genome
Background Information
The theory of organic evolution is based upon the belief that present-day
organisms have descended with modifications from forms of life existing in the
past. Phylogeny is the evolutionary history of a species, genus, or larger group.
Taxonomy is the science of classification of organisms according to the degree of
kinship and evolutionary relationships. The doctrine of organic evolution is one of
the most important generalizations in science. It is supported by evidence drawn
from genetics, paleontology, geographical distribution of species, and comparative
anatomy and embryology. Results from the molecular biology laboratory have
also provided strong evidence for the theory of evolution and have suggested
possible mechanisms by which evolutionary changes occur.
A central aim of evolutionary biology is to reconstruct the phylogenetic tree by
describing its branching pattern and, if possible, by dating each branch point.
A molecular approach has been applied to this problem during the past two
decades and the results have usually complemented traditional studies based
on comparative morphology, embryology and the fossil record. The molecular
approach to phylogeny is based on the idea that the structure of macromolecules
in living organisms, like the morphological characteristics of the organisms
themselves, have undergone progressive modifications as species evolved into
different forms. As a result of these changes, the structures of macromolecules
in organisms with different ancestral backgrounds are less similar than those
in closely related species. The macromolecules most frequently compared in
evolutionary studies are the proteins and DNA.
Protein Evolution
A comparison of the amino acid sequence of the same protein in different organisms
has provided a way to study molecular evolution. A comparison of the amino acid
sequence of cytochrome C from over 80 species has revealed that the amino acid
sequence of this protein from different species varies and the degree of variation
corresponds to the distance that separates two species on the evolutionary tree.
That is, the greater the taxonomic difference, the more the cytochromes are likely
to differ in their order of amino acid residues. For example, the cytochrome C
molecules in men and chimpanzees contain 104 amino acid residues and their
order of amino acids residues are exactly the same. In contrast, the cytochrome
C in man differs from the cytochrome C found in yeast in 44 out of the 104
amino acid residues. The number of amino acid replacements in cytochrome C
of 12 species are compared in Table 1. This type of information has led to the
construction of family trees of organisms that agree well with those obtained from
the classical anatomical record. (See Figure 1-1 ).
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Table 1. Variations in the Amino Acid Sequence of Cytochrome C
Cytochromes Compared toNumber of Variant Amino
Human Cytochrome
Acid Residues
Chimpanzee0
Rhesus monkey
1
Kangaroo10
Dog11
Horse12
Chicken13
Rattlesnake14
Tuna fish
21
Dogfish23
Moth31
Wheat35
Neurospora43
Yeast44
Figure 1-1. A Family Tree of Organisms Deduced
from Their Cytochrome C.
The end of each branch of the tree represents a different species. The total length
of the branch or branches connecting two species is proportional to the number of
amino acid residues in cytochrome C that are different.
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Evolution or DNA
A rich source of information for evolutionary relationships is locked in the
sequences of nucleotides in DNA. DNA hybridization procedures have permitted
investigators to begin to unlock these secrets because the techniques can be used
to determine the degree of similarities of base sequences in two preparations of
DNA. With these techniques, it has been possible to compare species in terms
of their total nucleotide sequences and the nucleotide sequence along specific
genes. In general, these studies have supported comparative morphological
analysis and family trees derived from DNA studies look remarkably similar to
those obtained by traditional procedures. For example, human DNA hybridizes
extensively with DNA from gorilla, orangutan and baboon but much less so with
DNA from non-primate mammals. Similarly, the restriction maps of the entire
chromosomal regions coding for the ἀ-globin chains of hemoglobin in man,
orangutan and chimpanzee are essentially identical and are distinct from those
in nonprimates. On a number of occasions, comparative DNA studies have even
been used to clarify and expand on phylogenetic relationships. For example,
although biologists have long disagreed about the taxonomic placement of the
giant panda, recent studies using DNA hybridization techniques strongly suggest
that this animal is more closely related to bears than to raccoons. Another longstanding debate concerns the taxonomic position of the flamingo. Flamingos are
similar to water fowl (ducks, geese, swans) in their bill structures and webbed
feet and some investigators have suggested that flamingos and water fowl share a
recent common ancestry. However, their long legs, necks and other morphological
characteristics have been interpreted by others as evidence that flamingos are
most closely related to ibises and storks. Recent studies by Sibley and coworkers
using DNA comparisons show that the flamingo line shares a recent common
ancestry with the ibises and storks and that the divergence between these groups
and the waterfowl occurred much earlier on the evolutionary time table. Thus, the
morphological similarities between flamingos and waterfowl is likely to be due to
convergent evolution, and not recent common ancestry.
In this exercise, you will use a “Dot” hybridization procedure to compare the
DNA sequences in salmon, turkey,chicken and cow. In the experiment, single­
stranded DNA from these four species will be immobilized on a membrane filter
made of nylon. The immobilization is accomplished simply by pipeting the DNA
onto the filter. The filters will then be incubated with biotinylated DNA from
cow and chicken under conditions which favor hybrid formation. Following the
reaction, the biotinylated probes that have hybridized to the immobilized DNA
will be detected by the color-producing peroxidase reaction. If you have not as
yet read the preceding section of this manual titled “Hybridization Analysis”, you
should do so at this time.
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Objective
To compare the nucleotide sequences in cow, chicken, turkey, and salmon DNA
by filter hybridization analysis.
Materials Provided
DNA
DNA Gel Stain (methylene blue)
Cow DNA (1mg/ml) - from calf thymus
Chicken DNA (1mg/ml)- from chicken erythrocytes
Turkey DNA (1mg/ml) - from turkey erythrocytes Salmon DNA (1mg/ml) from salmon sperm Biotinylated Chicken DNA
Biotinylated Cow DNA
The two biotinylated DNAs were prepared by incorporating
photobiotin into chicken and cow DNA.
Hybridization
*Tris-Buffer Saline (TBS)
*Hybridization Buffer- The buffer should be warmed to 37° C before use.
*1/4 X Hybridization Buffer
*Nylon Membrane Strips, 1cm x 6cm (16) Small Plastic Bags (16)
Foam Rack
Opaque Plastic Trays with Lids (4)
Color Development Reaction
*Gelatin-TBS
*Tris-Buffer Saline (TBS)
*TBS+NP-40
Avidin-Peroxidase
*Color Development Buffer
*Color Development Solution
(This solution is unstable and should be prepared immediately before the
color development step- see below).
* Prepared as described in Appendix 2 of the Instructor Manual.
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Materials Not Provided
A water bath, with lid, that will maintain a constant temperature of 60-65° C
and 37° C. A bath capable of gentle shaking is preferred but not absolutely
necessary.
Rulers (8)
Paper towels
One large (4 liter) beaker
Needles ( 18-22 gauge) or similar sharp objects
Microliter dispensers (8)
Graduated cylinders or beakers (8-25ml) Razor blades
Forceps (8)
Distilled or deionized water
Gloves (optional)
Small glass (7-10ml) test tubes Pasteur or transfer pipets with bulbs Black
construction paper
Small paper clips (around 2.5 em long)
Beaker with ice chips
Paper towels
Procedure
The experiment was designed for eight students working individually, or sixteen
students working in pairs. Each student (or student pair) will use two strips of
nylon membrane. The exercise requires approximately two 2-3 hour laboratory
periods.
A. Hybridization - Day 1
l.
Prior to hybridization, the DNA samples that will be used in this
experiment must be denatured. This operation can be accom­plished with
the aid of the gray foam tube rack. In a large beaker (4 liter), bring to
boil sufficient water (about 500 ml) to immerse but not completely cover
the rack. Pierce the top of the six tubes containing DNA with a needle or
similar sharp object to provide a vent for escaping steam during boiling.
Once the vent has been formed, make sure not to invert the tubes nor
to completely submerge them or their labels during boiling. When the
water is boiling vigorously, place the tubes in the rack in boiling water
for a period of five minutes. Note the positions of each tube in the rack
because the tube labels may come off during boiling. After boiling, place
the tubes directly into a beaker containing ice chips.
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2. Obtain two strips of nylon membrane. With a sharp pencil, draw a line
across each strip I em from one end as shown in the figure below. Place
your initials in the 1cm2 area created by this line and the letter A on one
strip and Bon the other. Note: Gloves should be worn when handling
nylon to prevent transfer of proteins from your hands to the membrane.
If gloves are not available, use forceps. Touch only the edges of the
membranes with gloves or forceps.
3. The nylon membranes must be wet before DNA application. Float each
membrane in deionized or distilled water in the tray, then submerge and
wet thoroughly. Pour off the water and replace with 5 ml of Tris-BufferSaline (TBS).
4. Place the membranes on a damp paper towel next to a ruler.
5. Remove excess moisture from the membranes by blotting with a paper
towel and pipet 5μι) each of cow DNA, chicken DNA, turkey DNA and
salmon DNA onto each strip. These DNAs should be carefully pi petted
onto the filters to form individual spots or “dots” at 2 em, 3 em, 4 em, and
5 em, respectively, from the end of the strip with your initials as shown
below.
Sample Application
6. Allow 15-20 minutes for the DNA to be adsorbed onto the mem­brane.
Then, rinse the membranes by dipping them in the tray containing TBS.
Exchange one strip with a classmate so that you have either two strips
labeled A or two strips labeled B. One membrane strip from the entire
class (one of the sixteen strips) should be placed in DNA gel stain for
ten minutes. Following DNA staining, each member of the class should
observe this strip. The four DNA samples should appear as circles or dots
of equal intensities.
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7. Place the two membranes in a hybridization bag and push one end of each
strip to the bottom of the bag using a gloved finger or forceps. Add 3ml
of hybridization buffer. Seal the bags by pressing along the length of the
zippers between your thumb and fingers. Fold over the top 2.5 em of the
bag and place a paper clip on each side of the bag such that the long axis
of the clips is parallel to the bag’s zipper as shown in the figure below.
The clips will help to maintain the seal during hybridization.
The Hybridization Bag
8. Lay the bags flat in the bottom of the tray (2 bags per tray), place the
lid on the tray and float the container in a water bath maintained at
60-65° C for about 15 minutes. This prehybridization step serves to
equilibrate the membranes with hybridization buffer and to block sites
on the membranes which can bind to free probe nonspecifically during
hybridization. Powdered milk is present in the hybridization solution
and the milk proteins will bind to, and prevent, binding of hybridization
probes to membrane sites not occupied by DNA molecules. The
hybridization buffer also contains the detergent sodium dodecyl sulfate
(SDS) which will reduce nonspecific probe binding and prevent nuclease
activity.
9. After the 15 minute prehybridization period, remove the bags from the
incubator. Straighten the paper clips before opening the bag and then
open the bag and discard the clips and hybridization buffer. If the bag
becomes punctured with a clip during this operation, discard it and
transfer the membranes to an unused bag. Press gently on the bags to
squeeze out any air or buffer that may be trapped within.
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10. Obtain one small tube, and add 3.5 ml of hybridization buffer.
11.
If your strips are labeled A, add 50μι of biotinylated calf DNA to the
tube. If your strips are labeled B, add 50μι of biotinylated chicken DNA
to the tube.
12. Using a transfer or pasteur pipet, transfer the solution to the plastic bag
containing the membranes. During the transfer, try to avoid introducing
air bubbles into the bag. If air is trapped in the bag, remove it by gently
stroking the bag with your index finger to squeeze the air, but not the
buffer, out of the open end of the bag.
13. Reseal the bags with clips as described in section A, step 7, place them in
the membrane-holding trays and transfer the closed trays to the 60-65° C
incubator.
14. Incubate the bags at this temperature for 18-24 hours.
B. Detection of the DNA-DNA Hybrids - Day 2
1. Remove the bags from the incubator after 18-24 hours of hybridi­zation.
The bags may be stored in the refrigerator for a few days or, preferably,
the membranes will be processed immediately as described below.
2. Remove the membranes from the bags and place them in a clean tray
containing 10 ml of 114 X hybridization buffer. Forceps may be needed
to remove the membranes from the bag. Note that the tray should be
rinsed thoroughly with tap water and then rinsed once with distilled or
deionized water between each of the steps listed below.
3. Place the tray in the water bath at 60-65° C. After 20 minutes, replace
the 114 X hybridization buffer with the same amount of fresh 1/4 X
hybridization buffer. These washes remove probe molecules that are not
hybridized to DNA on the membrane.
4. Place the two membranes into the tray with about 10 ml of TBS and
incubate at 37° C. After 5 minutes, replace the TBS with 15 ml of TBSgelatin solution and incubate at 37° C for an additional 10 minutes. After
this incubation, the membranes may be stored in the gelatin solution
for a few days in the refrigerator or may be processed immediately as
described below. This step serves to equilibrate the membrane with the
avidin-peroxidase buffer and the gelatin blocks sites on the membrane
which can bind to the avidin-peroxidase complex.
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5. Place 3 ml of TBS-gelatin into a small tube and add 25μι of the avidinperoxidase.
6. Place the two membrane strips into a single bag and transfer the avidinperoxidase solution to the bag. Remove air that may be trapped in the
bag as described above and seal the bag with clips as described in section
A, step 7.
7. Place the bags in the trays (2 bags/tray) and incubate at 37° C for
40-50 minutes. During this incubation, the avidin portion of the complex
should bind to the biotinylated probe on the membranes.
8. Remove the membranes from the bag and wash them in the tray by
incubating at room temperature in the following solutions for the times
indicated:
A. 10 ml
TBS 5 min.
B. 15 ml TBS-NP40 5 min.
C. 15 ml Color Development Buffer
5 min.
9. While the membranes are washing, the instructor or one member of the
class should prepare the color development solution. (See Appendix 2,
Instructor Manual) This solution contains 4-Chloro­napthol and hydrogen
peroxide dissolved in color development buffer. This solution is required
for color development. (See the section in this manual on hybridization
techniques).
10. Pour the color development buffer off of the membranes and add about
10 ml of the color development solution using a pasteur pipet, transfer
pipet or a graduated cylinder. Gently rock the tray containing the
membranes until purple spots appear. Maximal spot intensities should be
observed in 15-25 minutes. Rinse the membranes in water.
11. Record your results and the results of your classmates, noting the position
of the purple dots on the membranes and their relative intensities. The
membranes may be stored protected from heat and light (between two
sheets of black construction paper, for ex­ample).
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Study Questions and Analysis
1. Compare the hybridization signals from the various species. Which
organism shows the greatest similarity to chicken with respect to
nucleotide sequence?
2. How does this analysis compare with the traditional taxonomic
relationships reported for these animals?
3. Describe how your results would have been affected if biotinylated
salmon DNA had been used in the analysis.
4. Evolutionary relationships at the DNA level are frequently studied by
measuring the difference in melting temperatures of native DNA and
hybrid DNA after reassociation. For example, the Tm of human DNA
is 86° C and the Tm of green monkey DNA is 85° C, while the Tm
of human - Green monkey hybrid is 77° C. Explain the rationale for
measuring evolutionary relationships in this manner.
5. Why is it important to transfer the tubes containing DNA directly into an
ice bath after boiling?
Suggested Reading and References
1.
The Ancestry of the Giant Panda, Stephen J. O’Brien. Scientific American
256:102, 1987.
2.
Molecules and Morphology in Evolution. Colin Patterson, Ed. Cambridge
University Press, New York, 1987. There are many excellent chapters in
this book. Bird relationships using DNA-DNA hybridization techniques
are evaluated in the chapter by Sibley and Ahlquist.
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