scala naturale refers to
Aristotele’s hierarchichal organization of organismal life
For two biallelic loci A and B, which one refers to a haplotype?
Ab
BB
Aa, Bb
AB
In female heterogametic species like birds the mutation of the Z chromosome compared to autosomes is expected to be
higher
lower
equal
1/3
two alleles A and a segregate at a single bi-allelic locus. Assuming finite population size, random mating and no selection, the A allele will eventually be
fixed
either lost or fixed
preserved
at equilibrium
Which of the folloewing statements is correct? Linkage disequilibrium
Increase the level of heterozygosity in each of the loci under consideration
cannot occur because of genetic drift
diminishes as a function of time and recombination rate
is independent of gene genealogies
You calculate F-statistics for a set of two subpopulations in a diploid organism. In subpopulation 1 you find FIS to be positive. What does that mean ?
The number of heterozygotes in subpopulation 1 is lower than expected under hardy weinberg equilibrium
What does the Wahlund effect refer to?
The reduction of heterozygosity in a population due to population substructure
Which of the following conditions will most likely lead to an increase in the frequency of altruistic behavior in a population?
Groups that differ in therms of altruistic and selfish genotypes have differential reproduction survival
The altruistic behavior benefits the population
the rate of extinction of the selfish genotype is high
The benificaries of the behavior are related to the individual performing it
In a diploid population with alleles A and a, the aa genotype has a selective disadvantage of 40%. The dominance coefficient h in heterozygotes is 0.3. What is the fitness of the three genotypes ?
wAA = 1, wAa = 0.88, waa = 0.6
wAA =1
wAa = 1 - h * s = 1 - 0.4 * 0.3
Waa = 1 - s = 1 - 0.4 = 0.6
Where h = 0.3 & s = 40% = 0.4
One obligate cross-over per chromosome arm is required for successful meiosis. Consider an organism with 23 metacentric chromosomes. What is the minimum genetic map length of the chromosome?
2300cM
23 * 100
Using a linear regression of phenotypic values of the offspring over the mean of the two parental phenotypic values can estimate:
H^2 = VG / VP, assuming the parents are not related
H^2 = VA / VP, assuming the parents are not related
H^2 = VA / VP, assuming the parents and offspring environments are not correlated
H^2 = VG / VP, assuming the parents and offspring environments are not correlated
H^2 = VA / VP, assuming both isn’t correlated
The biological species concept is useful because
there are no expections
it emphasizes the role of reproduction isolation
it includes asexual species
it is easy to test
Genealogy of 6 haploid individuals. Substitution events are circles -> 5 loci. Which site-frequence-spectra (SFS) fits best?
E !
Y axis, number of mutations at that frequency -> the closer to the root the higher the frequency of the alleles
( A bit unsecure about this shit)
X-linked, biallelic locus in male heterogametic species. X-allele has pf in females and pm in males, x-allele qf and qm respectively. very large population and no selection. What is incorrect?
Female genotype frequencies correspond to hardy-wenberg expectations after one round of mating
Overall frequency p is always equal to 2/3 pf x 1/3 pm, regardless whether pf=pm or not
pm is equal to pf in the preceding generation
pf is not independent of pm
the absolute difference between pf and pm is halfed each generation
Incorrect:
Because:
does not hold immediatly to x-linked loci in a male-heterogamic species
Which of the following is not a reason that an effecctive population size can be smaller than the actual population size
There are different numbers of males and females
Every individual produces exactly 2 offspring
generations overlap
there is natural selection
population size fluctuates
Everey individual produces 2 offspring
You genotype a biallelic locus in a very small population of only 2 chromosomes (N=2) and find that both alleles are present. What is the probability that in two generations, both alleles are still present?
1/2 * 1/2 = 1/4
statement about ancestral recombination graph is incorrect:
recombination introduces a change in the genealogy
if recombination rate exceeds mutation rate inference of the underlying genealogy is no longer possible well.
genealogies change aprubtly at recombination breakpoints
the ancestral recombination graph applies to all sexually reproducing, outcrossing species
identity by descent can no longer be established
identity by descent can no longer be established is incorrect
Even with recombination identity by descent can still be established for segments of DNA that remain unbroken by recombination
which of the answers is not a component of fitness according to evolutionary biology ?
mating success
fertilization success
fecundity
mutation
Viability
ancestral genotype “aacc”, adaptive landscapes (FL) fitness of the homozygote of two diallelic loci.
Which statement is correct assuming drift doesn’t play a role ?
only FL1 can evolve towards high fitness values
FL1 shows no epistasis
genotype aaCC can only be reached by FL1 and FL3
FL3 will reach the highest fitness peak despite epistasis
FL2 cannot reach the highest fitness peak because of epistasis
epistasis here:
value in between is lower than than ancestorial value
The following plot describes the change in allele frequency p of a benefical allele A over several generations for three different genetic architectures. Which fitness architectures (wAA, wAa, waa) underlie the three curves, black (B), dark grey (DG) and light grey (LG)?
B: wAA>wAa>waa /DG: wAA<wAa>waa /LG: wAA<wAa<waa
B: wAA<wAa<waa /DG: wAA=wAa<waa /LG: wAA<wAa=waa
B: wAA>wAa>waa /DG: wAA=wAa>waa /LG: wAA>wAa=waa
B: wAA=wAa>waa /DG: wAA>wAa>waa /LG: wAA>wAa=waa
B: wAA>wAa>waa /DG: wAA>wAa<waa /LG: wAA>wAa=waa
(3rd option)
Consider 2 populations of equal sizewith the following genotype frequencies AA, Aa and aa. Pop1: 5, 10, 5 adn Pop2: 3, 10, 7
What is FIS for population 1 and 2 and FST, rounded by 2 digits?
2 dimensional model of evolution after:
Carolus Linnaeus
Jean-Baptiste Lamarck
CHarles R. Darwin
Assign given different mutatiuons to given NUC changes
a) TAC -> TAT (Tyr -> Tyr)
b) TAC -> TAG. (Tyr -> Stop)
c) TAC -> CAC. (Tyr -> His)
d) TAC -> TCAC. (Tyr -> ???)
missense mutation
frameshift mutation
non-sense mutation
silent mutation
a) silent
b) non-sense
c) missense
d) frameshift
F-statistics for:
p1 = 0.5, p2 = 0.2
Hobs1 = 0.5, Hobs2 = 0.2
Hexp1 = 0.5, Hexp2 = 0.32
…..
Fis1 = 0
Fis2 = 0,375
FSt = -0,17
Expected zygote frequencies
Exp AA = p^2 * N
EXP Aa = 2pq * N
EXP aa = q^2 *N
relative fitness per genotype with expected zygote frequency
observed / expected zygote frequency
compute linkage disequilibrium
D = PAB * Pab - PAb * PaB
Where P__ = #combination / #all
Fitness as a function of offspring genotypes, viability selection and large pop size
Assign:
Allele A1 will eventually get fices in the pop
Allele A2 will eventually get fixed in the pop
Allele A1 and A2 sill be stably maintained as a polymorphism
Allele A1 will be maintained at frequency 1/2
Allele A1 will be a t higher frequency than Allele 2 at equilibrium
Random genetic drift will dominate allele frequency change
A
B
D & C
—- or D
—- -> large pop
dung flies size, mean pop size = 6, selection = 9, offspring = 8
-> narrow sense heritability
calc: offspring - avg pop / selection - avg pop
here: 8-6/9-6 = 2/3
Felsenstein’s one allele vs two allele models
key difference considering genetic changes
why speciation more likely under one-allele
Consider 2 diallelic loci A and B that you genotyped for 7 haploid individuals:
A1B2, A2B1, A1B2, A2B2, A1B1, A2B2, A2B2
1 A1B1, 2 A1B2, 1 A2B1, 3 A2B2
How can you test for evidence of recombination ?
And what is an initial sign of recombination ?
Computing linkage disequilibrium!
Haplotype frequencies
A1B1: 1/7
A1B2: 2/7
A2B1: 1/7
A2B2: 3/7
Compute linkage disequilibrium
D = P(A1B1) * P(A2B2) - P(A1B2) * P(A2B1)
D = 1/7 * 3/7 - 2/7 * 1/7 = 0.0204
-> D != 0 therefore recombination is present
Initial sign: all haplotypes present
What happens with heterozygosity over time ?
decreases
frequencies of next generation with fitness values ?
average fitness = (p^2 * wAA) + (2pq * wAa) + (q^2 * waa)
e.g. freq. wAA:
p^2*wAA / avg. fitness
evolution and time
evolution is measured in number of generations
-> fossil record doesn’t show evolution in continuous pace
directional - stabilizing - disruptive
which of the following cannot be performed using only within-species polymorphism data?
Hudson-Kreitman-Aguade test
Tajima’s D test
Hudson’s haplotype test
haplotype diversity test
haplotype number test
The “Fast-X” effect is expected when benefical mutations are:
recessive
common
overdominant
rare
dominant
Mutations that change one body part into another are
homeotic
In the male germline of Drosphilia melanogaster:
Expression of the X Chromosome is supressed relative to the autosomes
The X chromosome is enriched with mal-biased genes relative to the autosomes
The X chromosome is completely deactivated
X-chromosomal gene expression is upregulated approximatly two-fold
X-chromosomal gene expression is upregulated approximatly four-fold
in the male somatic cells of Drosphilia melanogaster:
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