Exam 5 – BZ508 – November 21, 2005
Name: ____________________
Take home exam: Please return to the Biology Office by noon Wednesday Nov 23.
Undergraduates: Please answer 4 of the 6 questions. Graduates: Please answer 5 of the 6
questions.
1. Suppose that scientists at UK obtain the sequence of a protein-coding gene from individuals of
two different species of prairie dog: white-footed vs black-footed. The gene is 1000 nucleotides
long and 900 of the nueleotide sites are invariant in all sequences that were obtained. Of the
variable nueleotide sites, 40 are polymorphic within species; of these, 10 cause an amino acid
change in the translated protein. For the remaining 60 sites, the two species are fixed for different
nucleotides, and 5 of these sites cause an amino acid change.
1a. Are the data consistent with the neutral theory…..why or why not……explain in full. How
can you test this hypothesis?
Knowledge of nonsynonymous and synonmous variation both within and between
populations can be used to test the neutral theory using the McDonald-Kreitman test.
We need to set up a 2-by-2 table that counts the number of synonymous and
nonsynonymous within-population substitutions (polymorphisms, P) and betweenpopulation substitutions (fixed differences, D). The data tell us that there are forty
polymorphic sites, of which 10 are non-synonymous. There are 60 sites with fixed
differences, of which 5 are non-synonymous. We can set up the following table and
calculate the NS/S ratio for each type of variation.
P
D
NS
10
5
S
30
55
NS/S 0.33 0.09
If the mutation rate in the two populations is the same and the sequences are neutral
or affected only by functional constraints that are common to both populations, we
expect NS/S to be the same within and between populations. If the sequences are
affected by unique selection pressures in each population, NS/S should be different
within and between populations. NS/S is useful because it is generally not affected by
differences in the mutation rate or in drift between populations, which affect both
synonymous and nonsynonymous sites equally.
1b. What do the results suggest about the role of selection on this sequence?
The fact that S is greater than NS in both columns shows that the sequences are not
absolutely neutral. But this finding is consistent with the neutral theory, which says
that functional constraints will reduce the frequency of many nonsynymous
substitutions.
The key comparison is between the two entries in the table’s bottom row: there is
considerably more nonsynonymous variation within populations than between them. This
finding departs from the neutral expectation. Nonsynonymous substitutions occurred
faster within populations than between them. This suggests that selection may be
driving populations to develop more internal diversity than would occur under
neutrality. This pattern is known to occur, for instance, with immune system genes
that are involved in recognizing diverse antigens. It also occurs for genes in viruses
and parasitic organisms, where generating more diversity within a population helps the
parasite to evade the host’s immune responses.
In general the McDonald-Kreitman test is not absolutely decisive in rejecting neutrality
or near-neutrality, because it is conceivable that population sizes could have affected
the behavior of weakly selected alleles differently between and within populations. For
instance, if there a bottleneck in each species soon after they diverged from each
other (as is thought to often occur at speciation), then sites that were polymorphic in
the original population will be rapidly fixed for different alleles in each population,
leading to a greater between-population (D) than within-population variability (P).
2. Brian Storz presented a guest lecture in which he described variation in developmental rate
within and between different species of spadefoot toad. Species diverge from one another in
growth rate only after they reach a specific stage of development. Up until this developmental
stage, different species exhibit similar rates of development. Interestingly, individuals of species
that alter developmental rate in response to environmental factors only do so after reaching this
same developmental stage. What do you think is the significance of this finding for understanding
the evolution of developmental rate variation in spadefoot toads? Please embellish liberally and
be sure to include the words “canalization”, “phenotypic plasticity”, and any other words you
feel are important.
The data suggest that selection can only modify developmental rate variation after a
critical developmental stage has been reached. This suggests that developmental rate
is canalized until this developmental stage is reached. After an individual reaches this
developmental stage during ontogeny, it can exhibit phenotypic plasticity in
developmental rate in response to environmental factors. For example, a tadpole may
speed up development after reaching the critical developmental stage in response to a
drying pond environment.
Brian’s data also suggests that variation in developmental rate within populations may
depend upon the same mechanism that explains developmental rate variation between
populations.
3. Why does coalescence time depend upon population size? In your book (p. 244), it is
estimated that the world’s human population may have descended from an effective population
size of only 4600 people. How was this estimate obtained? What additional parameters and
parameter values are needed to obtain this estimate?
a. Coalescence time depends upon population size because genetic drift has a stronger
effect on allele frequencies in a smaller population. In a small population, drift is stronger,
so existing alleles must trace back to a more recent common ancestor than would alleles in a
large population.
b. The estimate was obtained by assuming a relationship between effective population size
(Ne), nucleotide diversity (p), and the mutation rate (m). Ne =  / 4 
c. Estimates of nucleotide diversity and mutation rate are needed.
4. Assume that all loci except one affecting body size in a population are fixed in homozygous
form. At the one variable locus there are two alleles, B1 and B2. B1 tends to increase body size
and B2 tends to decrease body size by the same relative amount: the difference between the two
homozygotes is 6 units of body size. The heterozygote, B1B2, has the same value as B1B1. The
frequencies of two alleles are p = 0.8 and q = 0.2.
a. Compute frequencies for the three genotypes in the population.
b. Compute the mean for each genotype.
c. Compute the mean for the population.
Genotype
Value
Freq
Freq x Value
B1B1
3
.64
1.92
B1B2
3
.32
.96
B2B2
-3
.04
-.12
Pop mean allele freq method = a(p-q) + 2dpq = 3(.8-.2) + 2(3)(.8)(.2) = 2.76
Pop mean value method = sum of Freq x values = 1.92 + .96 – 1.6 = 2.76
d. Compute the average effect of B1 and B2 in the population.
1 = q[a+d(q-p)] = .2[3+3[.2-.8)]=.24
2 = -p[a+d(q-p)]= -.8[3+3(.2-.8)]=-.96
e. Compute the breeding value of each genotype.
B1B1=21= 2(.24) = .48
B1B2=2+ 2 = .24 - .96 = -.72
B2B2 = 22= 2(-.96) = -1.92
5. Defend or refute the following statements:
5a. “It is a fallacy to think of a characteristic as being “genetic” to a certain extent and
“environmental “ to another extent”.
5b. “If a trait has a high heritability, it will not be strongly affected by the environment.”
5a. In one sense the statement is correct: it is possible to partition the genetic and
environmental components of variation for a given, well-planned and well-executed
study. However the components of variation are a property of the study and
conclusions about the relative magnitudes of the variance components cannot be
extrapolated to other studies. In much the same sense, the genetic component of
variation for a characteristic is a property of the population from which it is measured
and the environmental conditions under which it is measured. When another population
is examined or the character is measured under different conditions, the variance
components are likely to change.
5b. If a trait has a high heritability (I’m assuming narrowsense hear), this means that
variance attributable to the additive effects of alleles (V A) is contributing greatly to
the overall phenotypic variance V(P). If it contributed all of the total phenotypic
variance, then there would be no estimated environmental effect. However, it is
important to note that heritability estimates depend upon the degree to which
environmental variation is controlled; the estimate for heritability may change if the
environments in which characters are measured change.
6. The degree of antisocial behavior in human males is a quantitative trait. The figure below
shows antisocial behavior among men with different levels of MAOA activity as a function of the
way they were treated during their childhoods. MAOA is a brain enzyme that breaks down a
variety of neurotransmitters that nerve cells use to communicate with each other. Childhood
treatment was categorized as no maltreatment, probable maltreatment, or severe maltreatment.
Answer the following questions:
1
0.75
Index of
Antisocial
Behavior
0.25
0
MAOA
High Activity
-0.25
Low Activity
-0.5
None
Probable
Severe
Childhood Treatment
6a. Is variation among men in antisocial behavior at least partially due to differences in genotype
and/or environment? Explain.
Yes. Variation in antisocial behavior appears to depend upon an interaction between
genotype and environment. This is shown by the crossing lines: A high activity genotype
has a higher antisocial index score in a normal environment, but a lower antisocial index
score in the other environments characterized by abuse.
6b. Is antisocial behavior heritable? Explain.
In a broad sense, it does look like there is a phenotypic difference attributable to the
inheritance of different MAOA alleles. So there appears to be a genetic difference
but from this study it is impossible to accurately measure the effect or to measure
narrow-sense heritability.
6c. Do these data influence your opinion about how men who exhibit antisocial behavior should
be treated and/or punished?
I guess it makes you wonder if there should be genetic tests administered to antisocial
behavior individuals that break the law before administering punishment.