Download NAME_______________________________ EXAM

NAME_______________________________
EXAM#_______
1. (15 points) Next to each unnumbered item in the left column place the number from the right column/bottom that best corresponds:
9 Lemba
1) additive genotypic deviation
2) broad-sense heritability
5 collared lizards
3) genotypes collectively shared by the individuals of a deme
4) together describe a normal distribution.
6 dn = do(1-2m)n
5) apprehended using dental floss
6) approaches zero over many generations
19 coronary artery disease (CAD)
7) effectively treated by angioplasty regardless of patients' genotypes
8) average value of 2pq measured within subpopulations of the total
14 neutrality theory
population
9) Y-chromosomal haplotypes and mitochondrial-DNA haplotypes
15 environmental variance
coalesce to different geographic populations.
10) describes the genetic divergence between populations caused by
30 HT
genetic drift
11) This method permits measurement of epistatic variance involving two
11 candidate-locus approach
or more loci.
12) the proportion of an individual's phenotype attributable to
28 isolation by distance
environmental factors
13) explains why long-term rates of protein evolution are proportional to
4 mean and variance
population sizes
14) explains why the pseudogene for beta hemoglobin evolves more
26 norm of reaction
rapidly than the active gene
15) the proportion of populational phenotypic variation not explained by
20 Fisher's fundamental theorem
of natural selection
16 phenylketonuria
modeled genetic variation for a quantitative trait
16) illustrates inheritance of a response to dietary environment
17) population of gene copies collectively shared by the individuals of a
deme
17 gene pool
18) maintain geographically widespread populations connected by
gene flow throughout forests of the Missouri Ozarks
1 breeding value
19) No single genetic or environmental factor is both necessary and
sufficient for producing a particular quantitative phenotype.
20) Rate of evolution by selection equals the additive genetic variance in fitness of a population.
21) a range of phenotypes collectively expressed by all genotypes at a locus in a population
22) a founder population used to find the chromosomal locus for Huntington's disease.
23) Rate of evolution by natural selection is influenced by a population's system of mating. Occurs
24) illustrates heritability of mental retardation in human populations
25) A high heritability indicates close correspondence between parents and offspring for both parameters.
26) a range of phenotypes associated with a particular genotype in interaction with diverse environmental and genetic backgrounds
27) a method using genomic scans to identify chromosomal segments affecting a quantitative trait
28) Geographic distance among populations is the best predictor of their amount of genetic divergence.
29) Geographically distant populations do not exchange genes and form separate evolutionary lineages.
30) probability that a pair of gametes sampled at random from a total population contains two different alleles at a particular locus.
1
NAME_______________________________
EXAM#_______
2. (16 points). Two human demes of equal size are surveyed for MN blood types with the following
results:
M
MN
N
Total
Deme 1
38
324
638
1000
Deme 2
245
510
245
1000
a. (2 points) Calculate the frequencies of the M and N alleles in demes 1 and 2.
Deme 1: freq(M) = (38 + 162)/1000 = 0.2; freq (N) = 1 - 0.2 = 0.8
Deme 2: freq (M) = (245 + 255)/1000 = 0.5; freq (N) = 1 - 0.5 = 0.5
b. (6 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. (Critical Chi-square values at alpha = 0.05 for df 1 through 5: 1 df = 3.84, 2 df = 5.99, 3 df =
7.81, 4 df = 9.49, 5 df = 11.07)
Deme 1: expected M = (0.2)2(1000) = 40; expected MN = 2(0.2)(0.8)(1000) = 320;
expected (N) = (0.8)2(1000) = 640.
Chi-square = (38-40)2/40 + (324-320)2/320 + (638-640)2/640 = 0.1 + 0.05 + 0.006
= 0.156 HW not rejected.
Deme 2: expected M = (0.5)2(1000) = 250; expected MN = 2(0.5)(0.5)(1000) = 500;
expected (N) = (0.5)2(1000) = 250.
Chi-square = (245-250)2/250 + (510-500)2/500 + (245-250)2/250 = 0.1 + 0.2 + 0.1
= 0.4 HW not rejected.
c. (4 points) Calculate the FST value for these 2 populations.
Hs = [2(0.2)(0.8) + 2(0.5)(0.5)]/2 = 0.41; HT = 2(0.35)(0.65) = 0.455
FST = (0.455 - 0.41)/0.455 = 0.099
d. (2 points) Based on your answer to part c, are the populations compared more or less similar genetically than expected for
two randomly-chosen human demes?
Average value for comparisons of human populations = 0.15, so these populations are slightly more similar than
expected on average.
e. (2 points) Suppose that these demes begin to exchange migrants symmetrically so that 10% of each population moves to the
other one each generation. What will be the difference in frequency of the M allele between the populations after one
generation of gene flow?
dn = do(1-2m)n = 0.3[1-2(.1)] = 0.24
2
NAME_______________________________
3.
EXAM#_______
3
(5 points) After getting the sequencing results from the T. saxatilus experiment, you calculated FST values for burned and
unburned glade areas. What are these two resulting FST values? What do they indicate about the balance between genetic drift
and gene flow acting on these populations and the effect of the experimental treatment on these evolutionary forces?
FST is 0.039 for the glades in the burned area and 0.075 for glades in the unburned areas. Both values indicate that
gene flow greatly dominates the genetic structures of these populations regardless of burning. The burning appears
to increase the prevalence of gene flow over genetic drift.
4.
(6 points) Name and describe (including using formulas if appropriate) the two methods we used for calculating heritability in
our artificial selection experiment with Brassica rapa. Be sure to define any symbols you use.
response to selection:
mean trichome # in G2 - mean trichome # of G1
mean trichome # of parents- mean trichome # of G1
parent offspring regression:
need to find the slope of the regression line when regressing mother trichome # (x axis) versus offspring trichome # (y
axis)
heritability = 2 x slope x standard dev. of parents
standard dev. of offspring
5.
(6 points) Based on our Evolve simulations, describe one difference and one similarity in the results when selection favors a
recessive allele versus a dominant allele. Assume in both cases that the favored allele starts at a very low frequency in the
population, and that only two alleles are possible at the hypothetical locus.
similarity: In both cases (selection for recessive or dominant allele), the allele with the selective advantage becomes fixed in
the population.
differences: (1) With selection for a dominant allele, the time to reach an allele frequency of 0.5 (or fixation) was faster
(fewer generations). or (2) With selection for a dominant, the time to reach allele frequency 0.5 was the same in every
trial (selection was so strong we saw no effect of genetic drift).
NAME_______________________________
EXAM#_______
6. (14 points) For each of the following discoveries regarding human population structure and history, indicate how the conclusions
were inferred from analyses of molecular genetic variation using (1) FST values and/or (2) haplotype trees, as appropriate.
a.
The balance between gene flow and genetic drift among human populations on a worldwide basis is comparable on average
to that measured among Kenyan impala populations.
FST values both ~ 0.15.
b. Isolation by distance is a prevalent feature of human evolution both currently and in the past.
FST
How- Isolation by distance (IBD) shows a pattern of FST values between population pairs show positive correlation
with geographic distance between the pairs (i.e) plot FST vs distance
Haplotype trees
How- IBD shows a pattern of older haplotypes having a larger geographic distributions than younger haplotypes.
This is observed in many nuclear and mtDNA markers covering different scales of evolutionary time.
c. Three major expansions of populations out of Africa into Europe and Asia occurred over the past 1.7 million years.
Haplotype trees
How- Observe younger widespread haplotypes in Europe and Asia coalesce to ancestral haplotypes found in Africa
at three different points in human evolutionary history (FST values are not relevant to this conclusion.)
d. A recent range expansion from northeastern Asia established native American populations, which initially remained isolated
from other human populations.
Haplotypes trees
How- Observe young widespread haplotypes in native Americans coalesce to older less widespread haplotypes in
northeastern Asia indicating a recent range expansion from Asia into the Americas with no subsequent gene flow
prior to recent times. (FST values cannot distinguish recent range expansion followed by isolation from ongoing gene
flow)
e. A traditional hypothesis that humans form three distinct biological races is refuted by comparative genetic data.
FST
How- has overall low value; Shows IBD pattern from part b; No discrete breaks in genetic variation among
populations that correspond to “races” (i.e. relative dissimilarity of Melanesians and Africans from FST distance tree)
Haplotype trees
How- They show IBD pattern from part b; None of the lineages of haplotypes (evolutionary lineages) in the tree from
the root correspond to “race” categories.
4
NAME_______________________________
EXAM#_______
7. (10 points) For each of the following statements regarding linkage disequilibrium, indicate whether the statement is true or false.
For false statements, explain why the statement is false.
a. In large, randomly-mating populations, linkage disequilibria are unlikely to occur between loci located more than 1 cM apart
on a linkage map.
True
b. Genome scans for chromosomal locations of quantitative-trait loci using SNP markers rely upon occurrence of linkage
disequilibria between quantitative trait loci and genetically linked SNP loci.
True
c. Populations that have undergone a recent bottleneck in size, including founder events, often have higher average levels of
linkage disequilibrium than their larger source populations because of the consequences of genetic drift for haplotype
frequencies.
True
d. Haplotype trees are a crucial means of analysis for explaining linkage disequilibria occurring among sites within a small
chromosomal region in which crossing-over is rare or nonexistent.
True
e. Variable genetic markers that are closely linked (within 1 cM of each other) almost always show high linkage
disequilibrium.
False - A figure shown in lecture depicting levels of linkage disequilibria between pairs of SNP markers from a Costa Rican
founder population showed that only a fraction of the closely linked markers showed high linkage disequilibrium. Also, the
variable sites within the apoE gene region are tightly linked but not all pairs of sites showed high linkage disequilibrium.
5
NAME_______________________________
EXAM#_______
8. (18 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 frequencies
at the ApoE locus are: allele 2 (0.1), allele 3(0.8) and allele 4(0.1).
Genotype: 2/2
2/3
2/4
3/3
3/4
4/4
H-W freq. .01
.16
.02
.64
.16
.01
LDL-chol . 76
84
90
100
100
100 (mean = 97)
a. (4 points) Calculate the genotypic deviations for each genotype. What are the units of
measurement ?
Genotype:
2/2
2/3
-21
-13
2/4
-7
3/3
3/4
4/4
3
3
3 all in mg/dl
b. (2 points) Indicate the genetic variance for LDL cholesterol in this population, including units of measurement (answer
may be left in polynomial form).
.01(-21)2 + . 16(-13)2 + .02(-7)2 + .64(3)2+ .16(3)2+ .01(3)2 =
4.41 + 27.04 + 0.98 + 5.76 + 1.44 + 0.09 = 39.72 mg2/dl2
c. (4 points) Calculate the average excesses for alleles 2, 3 and 4. Include units of measurement.
Allele 2 = .1(-21) + .8(-13) + .1(-7) = -2.1 -10.4 -.7 = -13.2 mg/dl
Allele 3 = .1(-13) + .8(3) + .1(3) = -1.3 + 2.4 + .3 = 1.4 mg/dl
Allele 4 = .1(-7) + .8(3) + .1(3) = -0.7 + 2.4 + 0.3 = 2.0 mg/dl
d. (4 points) Calculate the additive genotypic deviations for each genotype.
What are the units of measurement?
Genotype:
2/2
2/3
2/4
3/3
3/4
4/4
-26.4
-11.8
-11.2
2.8
3.4
4.0 all in mg/dl
e. (2 points) Indicate the additive genetic variance for the population, including units of
measurement (answer may be left in polynomial form).
.01(-26.4)2 + . 16(-11.8)2 + .02(-11.2)2 + .64(2.8)2+ .01(3.4)2+ .01(4.0)2 =
6.97 + 22.28 + 2.51 + 5.02 + 1.16 + 0.16 = 38.1 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?
Either total phenotypic variance or environmental variance
6
NAME_______________________________
EXAM#_______
9. (10 points) For each of the following statements regarding the hemoglobin beta locus in humans, indicate whether the statement is
true or false. For false statements, indicate why the statement is false.
a. The hemoglobin A and S alleles form a balanced polymorphism in central African populations because the heterozygous
genotype has a higher fitness than the alternative AA and SS genotypes in a Malarial environment.
True
b. In some middle-eastern populations, persistence of fetal hemoglobin interacts epistatically with the homozygous
hemoglobin S genotype to alleviate the anemia associated with that genotype in other genetic backgrounds.
True
c. At a selective equilibrium in central African populations, the hemoglobin A and C alleles occur at frequencies of 0.89 and
0.11 respectively.
False - This equilibrium pertains to the A and S alleles, not C.
d. The expected coalescence time for all copies of homologous DNA for this locus in the human population would be 4N
generations, where N is the number of living individuals.
True
e. Given the following values of average excess for hemoglobin A, C and S alleles in a particular population, only the A allele
is expected to increase in frequency by natural selection in the next generation: aA = -0.001, aS = 0.081, aC = 0.015.
False - The A allele should decrease slightly and the other two alleles should increase in frequency.
7