Download NAME_______________________________ EXAM

NAME_______________________________
EXAM#_______
1
1. (15 points) Next to each unnumbered item in the left column place the number from the right
column/bottom that best corresponds:
26 additive genotypic deviation
1) a formula for calculating the change of allele frequency due to
natural selection in a population
6 average excess
2) a formula for calculating the change of allele frequency in a
population due to genetic drift
3 Bonellia
3) a larva consumed by a female worm is induced to become a parasitic
male gonad
21 broad-sense heritability
4) an X-linked dominant genetic disease
5) a protein sequence that is highly functionally constrained and
1 Dp = paA/ W
therefore shows very low evolutionary rates
6) a quantitative-genetic parameter that assigns a “phenotype” to a
30 envelope glycoprotein 120 V3
gamete
7) a relatively imprecise quantitative-genetic approach used mainly when
24 environmental variance
correlations among relatives are difficult to measure
8) can be defined genetically as heritable variation in fitness
19 FST among human populations 9) efficiently eliminates from a population alleles for autosomal
recessive genetic diseases
15 isolation by distance
10) equals 1 /2 h2 between parents and offspring and equals 1 /2 h2 +
1
/4 sd 2 /sp 2 between pairs of siblings in Fisher’s quantitative-genetic
23 linkage disequilibrium
model
11) equals the narrow-sense heritability of a trait in a population
25 measured genotype approach 12) equals the sum of the dominance deviations in Fisher’s quantitativegenetic model
8 natural selection
13) expected equilibrium frequency of an allele for a recessive lethal
disease in humans
28 “non-Darwinian evolution” 14) expected equilibrium frequency of the sickle-cell allele in malarial
areas of central Africa
10 phenotypic correlation
15) gene flow occurs mainly between neighboring populations; a new
mutation may spread across a large geographic distance over many
13 0.003
generations by traversing neighboring populations even if no
organisms make large migrations
16) geographically distant populations exchange no genes even over long periods of time
17) has a chromosomal mechanism of sex determination
18) a polymorphism maintained by natural selection
19) is approximately 0.15 on a worldwide basis
20) is much greater than expected for populations of large mammals
21) measures the importance of genetic variation as a contributor to phenotypic variation within a
generation
22) measures only the dominance and epistatic components of variance
23) occurs when the nucleotide state at one polymorphic nucleotide site is preferentially associated with the
nucleotide state at a second polymorphic nucleotide site in a populationOccurs
24) phenotypic variation not explained by genotypic variation in Fisher’s simple additive genetic model
25) quantitative genetic studies using saturated genome-wide linkage mapping, candidate loci, or a
combination of these methods
NAME_______________________________
EXAM#_______
2
26) sum of the average excesses of both alleles borne by a diploid organism in a population
27) tends to be larger between parents and offspring than between siblings
28) the hypothesis that most evolutionary changes in DNA and protein sequences occur by random genetic
drift of selectively neutral alleles
29) the observation that natural selection sometimes produces adaptively suboptimal results
30) undergoes large amounts of evolution by natural selection in HIV populations present in an infected
person
2. (15 points). Two human demes of equal size are surveyed for MN blood types with the following
results:
M
MN
N
Total
Deme 1
158
484
358
1000
Deme 2
12
176
812
1000
a. (2 points) Calculate the frequencies of the M and N alleles in demes 1 and 2.
Deme 1: p(M) = 0.4, q(N) = 0.6
Deme 2: p(M) = 0.1, q(N) = 0.9
b. (7 points) Test the hypotheses that each deme has Hardy-Weinberg genotypic frequencies. Be
sure to give your chi-square values and degrees of freedom, and state whether the hypotheses of
Hardy-Weinberg proportions are statistically rejected.
Deme 1 chi-square = (158-160)2 /160 + (484-480)2 /480 + (358-360)2 /360
= 0.025 + 0.033 + 0.011 = 0.069 (H-W not rejected)
Deme 2 chi-square = (12-10)2 /10 + (176-180)2 /180 + (812-810)2 /810
= 0.4 + 0.089 + 0.005 = 0.494 (H-W not rejected)
c. (4 points) Calculate the FST value for these 2 populations.
HS = 0.5(0.48 + 0.18) = 0.33
HT = 2(0.25)(0.75) = 0.375
FST = (0.375 - 0.33)/0.375 = 0.12
d. (2 points) Assuming that the M and N alleles are selectively equivalent (neutral) in these demes, FST
measures the balance between what two evolutionary forces?
Genetic drift and gene flow
NAME_______________________________
EXAM#_______
3
3. (5 points) Given the following data from a hypothetical artificial selection experiment similar to the
one we did with Brassica rapa, calculate the narrow-sense heritability using the response-to-selection
method. Assume that from the first generation, the five plants with the most trichomes were chosen and
cross pollinated to produce the second generation.
trichome number for individual first generation plants:
0, 0, 0, 0, 1, 1, 4, 4, 7, 8, 9, 12, 15, 16, 16, 19, 21, 30, 31, 45
trichome number for individual second generation plants:
0, 2, 5, 5, 7, 8, 9, 9, 11, 12, 17, 18, 18, 24, 26, 27, 30, 30, 32, 37
Generation1 mean = 11.95
Parental generation mean = 29.2
Generation 2 mean = 16. 35
h2 = (MG2 - MG1)/ (MPG - MG1) = (16.35 - 11.95)/ (29.2 - 11.95) = 0.255
4. (5 points) Describe specifically how alleles at the D1S80 locus differ from one another. In lab, how
did we determine which D1S80 alleles an individual had?
Alleles differ by the number of 16 bp repeat sequences
We amplified the D1S80 region and electrophoresed to determine the size of the
amplification product
5. (5 points) Using the EVOLVE program, we found that in a population where the allele with the selective
advantage started at a very low frequency in the population, selection for a dominant allele resulted in a
rise to a frequency of 0.5 for that dominant allele more quickly than selection for a recessive allele
resulted in a frequency of 0.5 for that recessive allele. Explain those findings.
When selection favors the dominant allele, both homozygous dominant and heterozygous
individuals are selectively favored. When the favored allele is recessive, only
homozygous recessive individuals are selectively favored.
NAME_______________________________
EXAM#_______
4
6. (14 points) Provide short answers to the following questions.
a. (6 points) For the diseases phenylketonuria and scurvy, discuss the importance of genetic and
environmental factors in (1) causing the disease phenotype in individual people and (2) causing
phenotypic variation at the population level. What general principle regarding causation of
variation in populations is illustrated by the contrast between these diseases?
Both diseases occur in individuals homozygous for an enzyme deficiency and who have a
particular dietary environment (high phenylalanine for PKU, low vitamin C for scurvy). An
individual phenotype is an inseparable interaction of genotype and environment. For PKU,
the genetic factor is rare in the population whereas the dietary factor is fixed; populational
variation in occurrence of the disease is therefore associated with genotypic variation. For
scurvy, the dietary factor is rare in the population whereas the genetic factor is fixed;
populational variation in occurrence of the disease is therefore associated with dietary
environmental variation.
b. (8 points) For each of the following statements regarding heritability, indicate whether the
statement is true or false. For false statements, explain why the statement is false.
i. A trait must be inherited to be heritable, but the reverse is not true.
True
ii. A high heritability for a trait indicates that the environment has relatively little importance in
determining the mean value of that trait in the population.
False - Environmental factors may greatly alter mean values in a population
without altering heritability.
iii. If a trait is highly heritable in two different populations, then the difference between
those populations for the trait is largely genetically determined.
False - Heritability is defined only within populations and does not measure
genetic bases of interpopulational variation.
iv. The Tay-Sachs disease is not heritable in the human population because there is no
additive genetic variance for this trait.
True
NAME_______________________________
EXAM#_______
5
7. (7 points) How are the evolutionary rates of the three proteins shown in this figure interpreted as
evidence for the importance of neutral alleles in evolution? Include in your answer reference to the
concepts of functional density of a protein, the molecular clock, and unit evolutionary period (UEP).
Graph not available
Evolutionary rate for neutral alleles is equal to the neutral mutation rate, which is generally
constant for a particular protein (= "molecular clock") but differs among proteins. Functional
density measures the proportion of amino acids in a protein that are required to perform specific
functions and are therefore not free to vary in the form of neutral alleles. Proteins having high
functional density should evolve more slowly than proteins having low functional density. Unit
evolutionary period measures that amount of time needed for 1% amino-acid sequence
divergence to occur between diverging proteins (a calibration of the molecular clock for each
protein shown in Myr on the graph). Fibrinopeptides have low functional density and evolve
rapidly; cytochrome c has high functional density and evolves slowly, with hemoglobin being
intermediate in functional density and evolutionary rate. These are the results predicted for
neutral evolution .
NAME_______________________________
EXAM#_______
6
8. (8 points) For each of the following statements regarding haplotype trees, indicate whether the statement
is true or false. For false statements, explain why the statement is false.
a. Studies using haplotype trees provide evidence that the progression of HIV infection of a patient
requires evolution of viral populations by natural selection.
True
b. Haplotype trees were important for determining that in central African populations occupying malarial
environments, the equilibrium frequencies of the Hb A and Hb S alleles are 0.89 and 0.11,
respectively.
False - These equilibrium frequencies were calculated from average excesses of the Hb A
and Hb S alleles, not haplotype trees.
c. Haplotype trees provide critical information needed to explain patterns of linkage equilibrium versus
linkage disequilibrium observed among SNP markers within a candidate locus, such as ApoE.
True
d. Haplotype trees provide useful information for reconstructing the history of population structure,
such as the relative importance of isolation by distance, recent range expansions, and long-distance
migrations in human evolutionary history.
True
NAME_______________________________
EXAM#_______
7
9. (26 points) The following data are used for an analysis of variance in LDL cholesterol levels (measured
in mg/dl of blood serum) associated with genotypes at the ApoE locus. Allele frequences at the ApoE
locus are: allele 2 (0.1), allele 3(0.7) and allele 4(0.2).
Genotype:
H-W freq.
LDL-chol .
2/2
.01
76
2/3
.14
90
2/4
.04
90
3/3
.49
100
3/4
.28
100
4/4
.04
100 (mean = 98)
a. (6 points) Calculate the genotypic deviations for each genotype. What are the units of
measurement ?
Genotype: 2/2
2/3
2/4
3/3
3/4
4/4
-22
-8
-8
2
2
2
in mg/dl
b. (2 points) Calculate the genetic variance for LDL cholesterol in this population. Include units
of measurement.
0.01(-22)2 + 0.14(-8)2 + 0.04(-8)2 + 0.49(2)2 + 0.28(2)2 + 0.04(2)2 = 19.6 mg2 /dl2
c. (6 points) Calculate the average excesses for alleles 2, 3 and 4. Include units of measurement.
Average excess of allele 2 = 0.1(-22) + 0.7(-8) + 0.2(-8) = -9.4 mg/dl
Average excess of allele 3 = 0.1(-8) + 0.7(2) + 0.2(2) = 1.0 mg/dl
Average excess of allele 4 = 0.1(-8) + 0.7(2) + 0.2(2) = 1.0 mg/dl
d. (6 points) Calculate the additive genotypic deviations (= breeding value) for each genotype.
What are the units of measurement?
2/2
2/3
2/4
3/3
3/4
4/4
-18.8
-8.4
-8.4
2
2
2
in mg/dl
e. (2 points) Calculate the additive genetic variance for the population. Include units of
measurement.
0.01(-18.8)2 + 0.14(-8.4)2 + 0.04(-8.4)2 + 0.49(2)2 + 0.28(2)2 + 0.04(2)2 = 19.47 mg2 /dl2
f. (2 points) What information not provided above is necessary for determining the broad-sense
and narrow-sense heritabilities for cholesterol level in this population?
Total phenotypic variance (or environmental variance)
g. (2 points) Suppose that another population (population 2) is examined and found to have the
same ApoE genotypes as the population described above (population 1), although the average
levels of cholesterol measured for the ApoE genotypes are different:
Genotype: 2/2
2/3
2/4
3/3
3/4
4/4
LDL-chol . 76
90
115
100
100
106 (mean = 102)
Investigation of genetic variation at the LDLR locus shows that population 2 has a high
frequency of LDLR allele 2 whereas this allele is rare in population 1. From this information,
how might you explain the differences in LDL cholesterol phenotypes associated with ApoE
genotypes between these populations?
Epistatic interaction between the LDLR allele 2 and the ApoE locus is responsible
for differences in genotypic deviations of genotypes containing alleles 3 and 4.