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CHAPTER 27
Application and Experimental Questions
E1.
Two populations of snakes are separated by a river. The snakes cross the river
only on rare occasions. The snakes in the two populations look very similar to each other,
except that the members of the population on the eastern bank of the river have a yellow
spot on the top of their head, while the members of the western population have an
orange spot on the top of their head. Discuss two experimental methods that you might
follow to determine whether the two populations are members of the same species or
members of different species.
Answer: Some possible experimental approaches are listed here:
1.
You could karyotype some members of each population. Different species
can often be distinguished by changes in chromosome structure and number.
2.
You could see if eastern and western snakes could mate with each other in
captivity. If they cannot, this would suggest reproductive isolation. If they can,
they still might be different species if mating never occurs in nature.
3.
You could sit in a blind (a box where someone cannot be seen by animals)
and watch both populations to see if eastern and western snakes ever mate with
each other in nature.
E2.
Sympatric speciation by allotetraploidy has been proposed as a common
mechanism for speciation. Let’s suppose you were interested in the origin of certain grass
species in Southern California. Experimentally, how would you go about determining if
some of the grass species are the result of allotetraploidy?
Answer: Perhaps the easiest way to determine allotetraploidy is by the chromosomal
examination of closely related species. A researcher could karyotype the chromosomes
from many different species and look for homologous chromosomes that have similar
banding patterns. This may enable them to identify allotetraploids that contain a diploid
set of chromosomes from two different species.
E3.
Two diploid species of closely related frogs, which we will call species A and
species B, were analyzed with regard to genes that encode an enzyme called hexokinase.
Species A has two distinct copies of this gene: A1 and A2. In other words, this diploid
species is A1A1 A2A2. The other species has three copies of the hexokinase gene, which
we will call B1, B2, and B3. A diploid individual of species B would be B1B1 B2B2
B3B3. These hexokinase genes from the two species were subjected to DNA sequencing,
and the percentage of sequence identity was compared among these genes. The results are
shown here:
A1
A2
B1
B2
Percentage of DNA Sequence Identity
A1
A2
B1
B2
B3
100
62
54
94
53
62
100
91
49
92
54
91
100
67
90
94
49
67
100
64
B3
53
92
90
64
100
If we assume that hexokinase genes were never lost in the evolution of these frog
species, how many distinct hexokinase genes do you think there were in the most recent
ancestor that preceded the divergence of these two species? Explain your answer. Also
explain why species B has three distinct copies of this gene while species A only has two.
Answer: If we compare the percentages of sequence identity among the five genes, we
see that A1 is very similar to B2, and A2 is very similar to B1 and B3. This suggests that
the most recent ancestor had two copies of the gene prior to the formation of these two
species. After the formation of species A and B, one of the B genes must have duplicated
again in species B to form B1 and B3.
E4.
A researcher sequenced a portion of a bacterial gene and obtained the following
sequence, beginning with the start codon, which is underlined:
ATG CCG GAT TAC CCG GTC CCA AAC AAA ATG ATC GGC CGC CGA ATC
TAT CCC
The bacterial strain that contained this gene has been maintained in the laboratory
and grown serially for many generations. Recently, another person working in the
laboratory isolated DNA from the bacterial strain and sequenced the same region. The
following results were obtained.
ATG CCG GAT TAT CCG GTC CCA AAT AAA ATG ATC GGC CGC CGA ATC
TAC CCC
Explain why these sequencing differences may have occurred.
Answer: The mutations that have occurred in this sequence are neutral mutations. In all
cases, the wobble base has changed, and this change would not affect the amino acid
sequence of the encoded polypeptide. Therefore, a reasonable explanation is that the gene
has accumulated random neutral mutations over the course of many generations. This
observation would be consistent with the neutral theory of evolution. A second
explanation would be that one of these two researchers made a few experimental
mistakes when determining the sequence of this region.
E5.
F1 hybrids between two species of cotton, Gossypium barbadense and G.
hirsutum, are very vigorous plants. However, F1 crosses produce many seeds that do not
germinate and a low percentage of very weak F2 offspring. Suggest two reasons for these
observations.
Answer: Two closely related species oftentimes carry the same kinds of genes and a
similar total genetic complement of DNA. Therefore, interspecies F1 hybrids are
frequently viable. However, due to changes in the arrangement of the genetic material
within the chromosomes, the F1 offspring may not be fertile. For example, the two
species may have different numbers and types of chromosomes. If this is the case, the
chromosomes of the F1 offspring will not be able to pair properly during meiosis, and
this will create highly aneuploid gametes. When these gametes participate in fertilization,
the embryos will either not develop, or they will result in offspring that are not very
healthy. Similarly, the chromosomes of closely related species may differ with regard to
inversions and translocations. As discussed in Chapter 8, these changes in chromosome
structure are likely to produce gametes that have imbalances in their genetic material.
This will also result in embryos that do not survive or have detrimental consequences.
E6.
A species of antelope contains 20 chromosomes per set. The species is divided by
a mountain range into two separate populations, which we will call the eastern and
western population. When comparing the karyotypes between these two populations, it
was discovered that the members of the eastern population are homozygous for a large
inversion within chromosome 14. How would this inversion affect the interbreeding
between the two populations? Could such an inversion play an important role in
speciation?
Answer: Inversions do not affect the total amount of genetic material. Usually, inversions
do not affect the phenotype of the organism. Therefore, if members of the two
populations were to interbreed, the offspring would probably be viable because they
would have inherited a normal amount of genetic material from each parent. However,
such offspring would be inversion heterozygotes. As described in Chapter 8 (see Figure
8.10), crossing over during meiosis may create chromosomes that have too much or too
little genetic material. If these unbalanced chromosomes are passed to the next generation
of offspring, the offspring may not survive. For this reason, inversion heterozygotes (that
are phenotypically normal) may not be very fertile because many of their offspring will
die. Because inversion heterozygotes are less fertile, this would tend to keep the eastern
and western populations reproductively isolated. Over time, this would aid in the
independent evolution of the two populations and would ultimately promote the evolution
of the two populations into separate species.
E7.
Explain why molecular techniques were needed as a way to provide evidence for
the neutral theory of evolution.
Answer: If a mutation is neutral, it will have no effect on the phenotype of the organism.
Therefore, you cannot look at the traits of an organism and conclude that it carries neutral
mutations. However, molecular techniques can determine if a group of individuals have
genetic variation due to the accumulation of neutral mutations. The best method to detect
genetic variation (both neutral and nonneutral) is DNA sequencing. By analyzing the
DNA sequences from different individuals, it is possible to identify changes that are
likely to be phenotypically neutral. For example, changes in the wobble base that do not
affect the polypeptide sequence would be expected to be neutral.
E8.
In Chapter 21, we learned about a technique called in situ hybridization, during
which a cloned piece of DNA is hybridized to a set of chromosomes. Let’s suppose that
we cloned a piece of DNA from G. pubescens (see Figure 27.4) and used it as a probe for
in situ hybridization. What would you expect to happen if we hybridized it to the G.
speciosa, the natural G. tetrahit, or the artificial G. tetrahit strains? Draw your expected
results.
Answer: We would expect the probe to hybridize to the natural G. tetrahit and also the
artificial G. tetrahit, because both of these strains contain two sets of chromosomes from
G. pubescens. We would expect two bright spots in the in situ experiment. Depending on
how closely related G. pubescens and G. speciosa are, the probe may also hybridize to
two sites in the G. speciosa genome, but this is difficult to predict a priori. If so, the G.
tetrahit species would show four spots.
E9.
Discuss how the principle of parsimony can be used in a cladistics approach of
constructing a phylogenetic tree.
Answer: The principle of parsimony chooses a phylogenetic tree that requires the fewest
number of evolutionary changes. When using molecular data, researchers can use
computer programs that compare DNA sequences from homologous genes of different
species and construct a tree that requires the fewest numbers of mutations. Such a tree is
the most likely pathway for the evolution of such species.
E10. A homologous DNA region, which was 20,000 bp in length, was sequenced
among four different species. The following number of nucleotide differences were
obtained:
Species A
Species B
Species C
Species D
Species A
0
443
719
465
Species B
443
0
744
423
Species C
719
744
0
723
Species D
465
423
723
0
Construct a phylogenetic tree that describes the evolutionary relationships among
these four species using the UPGMA method. Your tree should include values that show
the percentage of nucleotide substitutions.
Answer: We begin by identifying the pair that is the most closely related. In this case, it
is species B and D, because they have the fewest nucleotide differences. We divide 423
by 2 to yield the average number of substitutions, which equals 211.5. The percentage of
nucleotide differences between species B and D is:
211.5
20,000
x 100
= 1.06
Species A is the next most closely related species to B and D. The percentage of
nucleotide difference between species A compared to B and D is computed as follows:
There are 465 differences between A and D, and 443 differences between A and
B. So the average number of differences is:
443 + 465
2
= 454
We divide 454 by 2, which equals 227 to yield the number of nucleotide differences that
occur in species A and B or D. The average percentage of nucleotide differences
between species A compared to B and D is:
227 x 100
20,000
= 1.14
Finally, we consider species C. The percentage of nucleotide difference between species
C compared to A, B and D is computed as follows. There are 719 differences between C
and A, 744 differences between C and B, and 723 differences between C and D. So the
average number of differences is:
719 + 744 + 723
3
= 729
We divide 729 by 2, which equals 364.5, to yield the number of nucleotide differences
that occurred in C or A, B, or D. The average percentage of nucleotide differences
between species C compared to A, B and D is:
364.5
20,000
x 100
= 1.82
With these data, we can construct the phylogenetic tree shown here.
E11. As discussed in this chapter and Chapter 25, genes are sometimes transferred
between different species via horizontal gene transfer. Discuss how horizontal gene
transfer might give misleading results when constructing a phylogenetic tree. How could
you overcome this problem?
Answer: If you constructed a phylogenetic tree using a gene sequence that was the result
of horizontal gene transfer, the results might be misleading. For example, if a bacterial
gene was transferred to a protist, and you used that gene to construct your tree, the results
would suggest that bacterium and protist are very closely related, which they are not.
Evolutionary biologists can overcome this problem by choosing several different genes in
the construction of their phylogenetic trees, assuming that most of them are not
transferred horizontally.
E12. If a researcher used genetic engineering techniques to express the Drosophila
eyeless gene in the embryo at the region that will become the tip of the mouse's tail, what
results would you expect in the resulting offspring?
Answer: Possibly, the mouse would have an eye at the tip of its tail!