Download NAME_________KEY____________________________ Page 2

NAME_________KEY____________________________ Page 2
EXAM#_______
1.
(15 points) Next to each term in the left-hand column place the number from the right-hand column that
best corresponds:
Apoprotein E 18
1) their rate of evolution equals their mutation rate
2) survival of the fittest
neutral mutations 1
3) the time inverse of DNA replication
4) the population of potential gametes of a deme
genetic variation within 13
demes
5) authors of a study showing that the environment can strongly influence IQ score
evolution 30
7) an inherited enzyme deficiency that interacts with diet to cause mental retardation
6) an example of a bottleneck effect
8) the proportion of the phenotypic variance explained by the additive genotypic
deviations
gene pool 4
9) allows allele frequencies to be predicted from genotype frequencies
10) found on low-density lipoprotein (LDL)
5-α-steroid reductase 21
deficiency
11) the response to natural selection is determined by this quantity for the
phenotype of fitness
12) authors of a study showing that genes, not the environment, is the major
determinant of the IQ scores of adopted children
PKU 7
13) decreased by drift
14) a quantitative template ligation assay
broad-sense heritability 26
15) the effect of distance measured in centiMorgans on linkage disequilibrium
16) an inherited enzyme deficiency causing mental retardation
QTL 19
17) non-random association between alleles at different loci due to linkage
18) can bind to the LDL-receptor protein
Skodak and Skeels 5
19) a genomic region associated with phenotypic variation by linkage to markers
20) the genetic fusion of two populations by interbreeding
Hardy-Weinberg Law 29
21) increased by a founder effect in one human population
22) non-random association between alleles at different loci in the gene pool
coalescence 3
23) mutations that do not lead to evolutionary change
24) a model of gene flow that fits well to the human data
average excess 11
25) the population of all genotypes
26) the proportion of the phenotypic variance explained by genotypic variation
isolation by distance 24
27) the average phenotype of a genotype minus the population average
28) decreased by gene flow
linkage disequilibrium 22
29) allows genotype frequencies to be predicted from allele frequencies
30) a change in gamete frequency
NAME_________KEY____________________________ Page 3
2.
EXAM#_______
(6 points) List the three basic components of the phenotype of fitness in a given environment.
1. Viability
2. Mating Success
3. Fecundity/fertility
3. (4 points) Haplotypes were determined in an autosomal DNA region for eight individuals by using special
molecular haplotyping techniques for four SNPs in this region, with the following results:
1. GCTA
CACT
2. GACT
CCTA
3. CCTT
GACA
4. CACA
GCTT
5. GCCT
CATA
6. CCCA
GATT
7. GCCA
CATT
8. GATA
CCCT
Group these individuals into the distinct categories that could be distinguished with a standard automated DNA
sequencer.
All are indistinguishable and yield identical patterns with an automated DNA sequencer.
NAME_________KEY____________________________ Page 4
EXAM#_______
4. (8 points) Calculate all the allele frequencies in the following populations, given an autosomal locus:
a. (2 points)
Genotype:
Frequency:
AA
0.1
Aa
0.4
aa
0.5
Freq. A = 0.1 +1 /2 (0.4) = 0.3
Freq. a = 0.5+1 /2 (0.4) = 0.7 or Freq. a = 1-Freq. A = 0.7
b. (2 points)
Genotype:
Frequency:
AA
0.4
Aa
0.6
aa
0.0
Freq. A = 0.4 +1 /2 (0.6) = 0.7
Freq. a = 0.0+1 /2 (0.6) = 0.3 or Freq. a = 1-Freq. A = 0.3
c. (3 points)
Genotype:
Frequency:
AA
0.2
AB
0.2
BB
0.3
AC
0.1
BC
0.1
CC
0.1
Freq. A = 0.2+1 /2 (0.2)+ 1/2 (0.1) = 0.35
Freq. B = 0.3+1 /2 (0.2)+ 1/2 (0.1) =0.45
Freq. C = 0.1+1 /2 (0.1)+ 1/2 (0.1) = 0.2
d. (2 points) Assuming Hardy-Weinberg, what is the allele frequency of A given the following dominant and
recessive phenotype frequencies:
Phenotype:
A- (dominant)
aa (recessive)
Frequency:
0.36
0.64
q2 =0.64 which implies q=0.8. So Freq. A = p =1-q = 0.2
NAME_________KEY____________________________ Page 5
EXAM#_______
5. (6 points). The allele frequencies of a population are p = 0.6 for allele A and q = 0.4 for allele a. The observed
numbers of genotypes are:
Genotypes:
Numbers:
AA
40
Aa
52
aa
8
Test the hypothesis that this population is in Hardy-Weinberg equilibrium with the given allele frequencies.
1. Calculate the expected values. Given p =0.6 and HW, the expected frequencies and numbers are
Genotypes
Expected Freq.
Exp. No. =Exp. Freq.×100
AA
0.36
36
Aa
0.48
48
aa
0.16
16
2. Now calculate the chi-square statistic:
Genotypes
Obs.
Expected
(O-E)2 /E
AA
40
36
0.44
Aa
52
48
0.33
aa
8
16
4.00
Sum
100
100
4.78 = χ2
3. Determine degrees of freedom and statistical significance.
There are 2 degrees of freedom because we start with three categories, subtract one, then substract the
number of parameters estimated from the data. The only parameter needed to estimated the HW
frequencies is the freq. of the A allele, but that was given, not estimated. Using the chi-square table on
the front of the exam, a chi-square of 4.78 with 2 degrees of freedom is NOT significant at the 0.05 level,
the standard cut-off for statistical significance. Therefore, we fail to reject the null hypothesis of HW
with p=0.6 for these data.
NAME_________KEY____________________________ Page 6
EXAM#_______
6. (8 points) Four populations are independently isolated from a large ancestral population with p = 0.25 for the A
allele at an autosomal locus. The table below gives the frequencies of the A allele in the original founders and in
the offspring produced by the founders:
Population
1
2
3
4
Frequency of A in Founders
0.24
0.36
0.12
0.21
Frequency of A in Founder’s Offspring
0.38
0.35
0.06
0.52
a. (4 points) Rank these populations according to their most probable founder sizes, starting with the largest
founder population size and ending with the smallest.
Under genetic drift, the expected magnitude of the change in allele frequency is inversely proportional
to the population size. Therefore, the founder population sizes should be inversely proportion to the
magnitude of the change in allele frequency in the founders versus the ancestral freq. of 0.25. The
changes in allele frequency follow:
Population
1
2
3
4
Freq. of A in Founders – Ancestral Freq. of A
0.24-0.25=-0.01
0.36-0.25=0.11
0.12-0.25=-0.13
0.21-0.25=-0.04
Ranking the populations inversely by the magnitude of the change in allele frequency yields:
Pop. 1 > Pop. 4 > Pop. 2 > Pop. 3
b. (4 points) Rank these populations according to their most probable sizes for the offspring generation,
starting with the largest number of offspring and ending with the smallest.
The offspring population sizes should be inversely proportion to the magnitude of the change in allele
frequency in the offspring versus the founders. The changes in allele frequency follow:
Population
1
2
3
4
Freq. of A in Offspring – Freq. of A in Founders
0.38-0.24=0.14
0.35-0.36=-0.01
0.06-0.12=-0.06
0.52-0.21=0.31
Ranking the populations inversely by the magnitude of the change in allele frequency yields:
Pop. 2 > Pop. 3 > Pop. 1 > Pop. 4
NAME_________KEY____________________________ Page 7
EXAM#_______
7. (8 points) Two "races", Race 1 and Race 2, originally lived on different continents, but some individuals from
both races came to live together on a third continent and interbred to create a hybrid population. Although all
individuals of this hybrid population are regarded as being of the same “race”, there is much variation in the
amount of individual admixture (measured by percent of genes from Race 1 in the individuals from the hybrid
population). Race 1 has a higher mean phenotypic value than Race 2 for two traits, with Trait 1 having a high
heritability in both races, and Trait 2 having no heritability in either race. The degree of admixture was then
measured from molecular data in a sample of 1000 individuals from the hybrid population and the individual
phenotypic values for the two traits were then plotted against the degree of admixture with the following results:
High
High
Trait
Value
Trait
Value
Low
Low
0%
35%
Percent of Genes From Race 1
0%
Percent of Genes From Race1
35%
a) (2 points) What does the information about heritabilities tell you about the genetic basis for the mean
differences in traits 1 and 2 between the “races”?
NOTHING. h2 is not defined for means nor for between populations. Therefore, h2 tells you
nothing about the genetic basis for the mean differences in traits 1 and 2 between the “races.”
b) (4 points) What do the two graphs given above tell you about the genetic basis for the mean differences
in traits 1 and 2 between the “races”?
As in the Scarr et al study, the flat slope (no correlation) indicates that no genetic differences exist
between the groups for the mean differences in Trait 1.
As in the Scarr et al study, the positive slope (positive correlation) indicates that genetic
differences do exist between the groups for the mean differences in Trait 2.
c) (2 points) What do the graphs combined with the heritabilities tell you about the genetic basis for the
mean differences in traits 1 and 2 between the “races”?
For the same reasons given in a), h2 is completely non-informative, so adding it tells you nothing
more than already learned in part b).
NAME_________KEY____________________________ Page 8
EXAM#_______
8. (14 points)
a). (3 points) The genotypic values for genotypes AA, Aa and aa are 20, 40, and 60 respectively in a
population with a mean phenotype of 50 and a phenotypic variance of 225. What are the genotypic deviations?
Genotypic Deviation = Genotypic Value - Mean
Genotypes
Genotypic Deviations
AA
20-50=-30
Aa
40-50=-10
aa
60-50=10
b). (4 points) For genotypic deviations of -2, 1 and 4 respectively for AA, Aa and aa; a mean phenotype of
100; a frequency of the A allele of 0.75; and a random mating population, what are the average excesses for the A
and a alleles?
Gametes
Average Excess
A
(0.75)(-2)+(0.25)(1)= -1.25
a
(0.75)(1)+(0.25)(4)= 1.75
c) (3 points) If the average excesses of two alleles are 10 for the A allele and -5 for the a allele in a
randomly mating population whose mean phenotype is 50 and if the frequency of the A allele is 0.7, what are the
additive genotypic deviations of the three genotypes?
Genotypes
Add. Genotypic
Deviations
AA
10+10=20
Aa
10-5=5
aa
-5-5= -10
d) (4 points) The phenotypic variance of a trait is 100, the correlation between parent and offspring is 0.1,
and the correlation between sibs is 0.2. What are the additive genetic variance and the heritability of the trait?
1
/2 σ2 a /σ2 P = /2 σ2 a /100 = Corr.(Parent-Offspring)=0.1, so σ2 a = 2(0.1)100 = 20 and h2 = 2(0.1)=0.2
NAME_________KEY____________________________ Page 9
EXAM#_______
9. (6 points) Consider the following gene pool as characterized at two loci, each with two alleles (A and a at one
locus, B and b at the second locus):
Gamete
AB
aB
Ab
ab
Gamete
Frequency
0.6
0.1
0.2
0.1
a) (2 points) Calculate the linkage disequilibrium between these two loci.
D=(0.6)(0.1)-(0.1)(0.2)= 0.06-0.02 = 0.04
b) (4 points) Calculate the gamete frequencies at linkage equilibrium
First, calculate the allele frequencies at each locus:
Freq. A = Freq.(AB)+Freq.(Ab) = 0.6+0.2 = 0.8
Freq. a = Freq.(aB)+Freq.(ab) = 0.1+0.1 = 0.2
Freq. B = Freq.(AB)+Freq.(aB) = 0.6+0.1 = 0.7
Freq. b = Freq.(Ab)+Freq.(ab) = 0.2+0.1 = 0.3
Second, multiply the allele frequencies to get the two-locus gamete frequencies:
Freq.(AB)=Freq.(A)×Freq.(B) = (0.8)(0.7) = 0.56
Freq.(aB)=Freq.(a)×Freq.(B) = (0.2)(0.7) = 0.14
Freq.(Ab)=Freq.(A)×Freq.(b) = (0.8)(0.3) = 0.24
Freq.(ab)=Freq.(a)×Freq.(b) = (0.2)(0.3) = 0.06
NAME_________KEY____________________________ Page 10
EXAM#_______
10. (9 points) One chimpanzee male and five human males were sequenced for a small region on the Xchromosome showing no recombination to yield the following haplotypes (only variable sites are shown):
Nucleotide Position:
1 3 5 6 7 9
Chimp:
ATTCCG
Human 1:
GCGATC
Human 2:
GCTACG
Human 3:
GTTACG
Human 4:
GCGATG
Human 5:
GCTATG
a) (4 points) What is the haplotype tree that minimizes the number of mutational changes (maximum parsimony)?
Be sure to indicate the root of the human part of the tree.
Chimp:A T T C C G
→ →
Human 1:G C G A T C
↑
Human 4:G C G A T G
↑
Human 5:G C T A T G
↑
Human 2:G C T A C G
↑
Human 3:G T T A C G (the Root of the Human Tree)
b). (3 points) This region is a candidate locus for trait X, and the mutations that occurred in human evolution at
nucleotide positions 1 and 7 interact to increase the value of trait X by 10% in their male bearers (without both
mutations, there is no phenotypic effect). Assuming that all other mutations in this DNA region have no functional
effect, which haplotypes are expected to increase trait X by 10% in human males?
As the tree shows, all human haplotypes have the site 1 mutation, so the increased value in trait X
appears when the site 7 mutation occurs. The tree above shows it occurs between Human2 and
Human5. Therefore, Human5 and its descendants, Human4 and Human1, all increase the trait value.
c) (2 points) Suppose each of the nucleotide positions were analyzed one-by-one in humans for associations with
elevated levels of trait X. Which nucleotide positions would show such an association?
Because all humans have the mutation at site 1, it cannot be a cause of variation. The SNP at site 7 will
show phenotypic associations since it is the causative, variable site. Because of no recombination and
descent from the ancestral haplotype being SNP 7, SNPs 5 and 9 will also show phenotypic associations
through linkage disequilibrium.
NAME_________KEY____________________________ Page 11
EXAM#_______
11. (6 points). A protein coding gene and an associated pseudogene are sequenced in several different species to
determine their rates of molecular evolution.
a). (2 points) The coding gene is observed to evolve more rapidly than the pseudogene. Is this compatible
with the neutral theory of molecular evolution?
Under the neutral theory, a pseudogene is less selectively constrained than a functional gene, so it
should have a faster rate of evolution. Therefore, this observation is NOT COMPATIBLE.
b). (2 points) Within the coding gene, the first two nucleotide positions in a codon evolve more rapidly than
the third. Is this compatible with the neutral theory of molecular evolution?
Under the neutral theory, third positions tend to be silent and therefore less selectively constrained than
the first two codon positions, so third positions should have a faster rate of evolution. Therefore, this
observation is NOT COMPATIBLE.
c) (2 points) Within the pseudogene, all nucleotide positions in the former coding region are evolving at the
same rate. Is this compatible with the neutral theory of molecular evolution?
Under the neutral theory, all positions in a pseudogene are without functional significance and
therefore should have no selective constraint and thus the same rate of evolution. Therefore, this
observation is COMPATIBLE.
NAME_________KEY____________________________ Page 12
EXAM#_______
12 (5 points) Answer the following questions in reference to the “PCR and Genetic Variation in Humans”
experiment we did in lab:
a) (2 points) Diagram one round (cycle) of PCR. In addition, briefly note in words what happens at each step.
step 1: template DNA is denatured
step 2: primers anneal to template DNA
step 3: DNA polymerization/synthesis
b) (1 point) How do D1S80 alleles differ from each other?
they differ by the number of 16 bp repeats that they have
c) (2 points) How does the technique of PCR allow one to detect the differences in D1S80 alleles?
The template sequence will vary in length according to the number of 16 bp repeat sequences it has.
Since PCR just copies the template sequence, the larger template will result in a longer
amplification product, while the shorter template sequence will result in a shorter amplification
product. Difference in amplification product length can be determined with gel electrophoresis.
NAME_________KEY____________________________ Page 13
EXAM#_______
13.
(5 points) Assume for this question that trichome number in Brassica rapa is heritable, and that trichome
number is determined by five genes (A, B, C, D and E) with two alleles each.
a) Diagram an example of a parent offspring regression graph, where mother trichome number is regressed
against offspring trichome number. Include a regression line and be sure to label axes.
regression line must show positive slope
b) Describe how the regression line can be used to calculate heritability.
h2 = 2 x (slope of the regression line)
note to grader: students can include the standard deviation of the mother in the numerator and the
standard deviation of the offspring in the denominator, but that is not necessary for full credit
c) Assume you continued the selection for high trichome number for five more generations. Would you
expect the mean trichome number of the offspring generation to increase each generation? Would you expect
there ever to be an upper limit in trichome number, such that selection had no additional affect? Explain your
reasoning (using the gene symbols as described above).
You would expect trichome number to increase for five more generations, because the question
stipulated that trichome number is heritable. Eventually an upper limit would be reached however,
when all plants are fixed for the alleles determining a high number of trichomes. Using the symbols
A, B, C, D, E, where a capitol letter denotes a high trichome number allele and a lower case letter
denotes a low trichome number allele, after many generations of selection all plants in the
population would be AABBCCDDEE.