Consider the genealogy of six haploid individuals in the left upper corner of the figure. Substitution events are depicted as circles for a total of five loci. Which of the five site-frequency-spectra best corresponds to the data?
E
Which of the following is not a reason that an effective population size can be smaller than the actual population size?
A) There are different numbers of males and females in the population
B) Every individual produces exactly two offspring
C) Generations overlap
D) There is natural selection (a high variance in the number of offspring)
E) Population size fluctuates
Every individual produced exactly two offspring
You genotype a bi-allelic locus in a very small population of only two chromosomes (N=2) and find that both alleles are present. What is the probability that in two generations, both alleles are still present?
A) 0
B) 1/2
C) 1/4
D) 1/16
E) 1/8
Since each parent has a 50% chance of passing on either allele, there's a 1/2 * 1/2 = 1/4 chance of each of the homozygous scenarios (AA or aa) and a 2 * (1/2 * 1/2) = 1/2 chance of the heterozygous scenario (Aa).
So, there is a 1/2 chance that the first generation is Aa.
Therefore, given that the first generation is Aa, the probability that the second generation is also Aa (and thus has both alleles) is 1/2.
Answer: C) 1/4
Which of the following statements concerning the following equation is correct?
A) Heterozygosity increases with time.
B) Selection reduces heterozygosity over time.
C) Reduction in heterozygosity over time depends on population size.
D) Change in heterozygosity depends on population size, but is independent of time.
E) This equation is incorrect. 2N should be in the numerator.
-Reduction in heterozygosity over time depends on population size.
In a world of no inbreeding, how many ancestors would you have when Shakespeare was
presumably born (*1564) assuming a human generation time of possible 20 years. Approximately
Step 1: Determine the number of generations
• Generation time: 20 years
• Time from 1564 to 2024: 460 years
• Number of generations:
Ancestors at n = 2^n
8 x 10^6
Which statement concerning the ancestral recombination graph is incorrect?
A) Recombination introduces a change in the genealogy.
B ) If recombination rate exceeds mutation rate inference on the undertying genealogy is no longer possible well. ~
C) Genealogies change abruptly at recombination breakpoints.
D) The ancestral recombination graph applies to all sexually reproducing, outcrossing species
E) Identity by descent can no longer be establishe
E)
Consider two populations of equal size with the following genotype frequencies AA, Aa and aa.
Pop 1: 5, 10, 5.
Pop 2: 3, 10, 7.
FIS (inbreeding coefficient)
FST is the proportion of the total genetic variance contained in a subpopulation (the S subscript) relative to the total genetic variance (the T subscript)
What is FIS for population 1 and 2 and FST (FIS_1/FIS_2/FST, rounded to two digits)?
Population 1:
Genotype frequencies: AA = 5, Aa = 10, aa = 5. Total individuals = 20
Allele frequencies:
p = (2 * AA + Aa) / (2 * Total) = (2 * 5 + 10) / (2 * 20) = 20 / 40 = 0.5
q = (2 * aa + Aa) / (2 * Total) = (2 * 5 + 10) / (2 * 20) = 20 / 40 = 0.5
Observed heterozygosity (Ho): Ho = Number of Aa / Total = 10 / 20 = 0.5
Expected heterozygosity (He): He = 2pq = 2 * 0.5 * 0.5 = 0.5
FIS: FIS = (He - Ho) / He = (0.5 - 0.5) / 0.5 = 0
Population 2:
Genotype frequencies: AA = 3, Aa = 10, aa = 7. Total individuals = 20
p = (2 * 3 + 10) / (2 * 20) = 16 / 40 = 0.4
q = (2 * 7 + 10) / (2 * 20) = 24 / 40 = 0.6
Observed heterozygosity (Ho): Ho = 10 / 20 = 0.5
Expected heterozygosity (He): He = 2 * 0.4 * 0.6 = 0.48
FIS: FIS = (0.48 - 0.5) / 0.48 = -0.04166... ≈ -0.04 (rounded to two decimal places)
FST Calculation:
To calculate FST, we need the total allele frequencies and the average expected heterozygosity within subpopulations (Hs).
Total allele frequencies:
pT = (5 + 3 + 10) / (20 + 20) = 18 / 40 = 0.45
qT = (5 + 7 + 10) / (20 + 20) = 22 / 40 = 0.55
Total expected heterozygosity (HT): HT = 2 * 0.45 * 0.55 = 0.495
Average expected heterozygosity within subpopulations (HS): HS = (He1 + He2) / 2 = (0.5 + 0.48) / 2 = 0.49
FST: FST = (HT - HS) / HT = (0.495 - 0.49) / 0.495 = 0.005 / 0.495 ≈ 0.0101 ≈ 0.01 (rounded to two decimal places)
Final Answer: 0.00, -0.04, 0.01
Which of the following is not a component of fitness according to evolutionary biology?
mating success
fertilization success
viability
mutation
Fecundity
muatation
Inversions often stably maintain allelic combinations of genes in so-called supergenes. In a large population, two inversion types (A and B) segregate. Individuals heterozygous for the inversions (AB) have a survival advantage, while individuals homozygous for either type (AA or BB) have reduced survival.
In a population with the following genotype frequencies among adult individuals:
AA (n = 10)
AB (n = 190)
BB (n = 200)
What are the selection coefficients for the respective inversion types A and B (rounded to two decimal places)?
Let the relative fitnesses be:
Fitness of AA = 1 - s_A
Fitness of AB = 1 (reference)
Fitness of BB = 1 - s_B
Total population size: N = 10 + 190 + 200 = 400
Observed genotype frequencies:
f(AA) = 10 / 400 = 0.025
f(AB) = 190 / 400 = 0.475
f(BB) = 200 / 400 = 0.50
p (frequency of A) = f(AA) + (0.5 × f(AB)) = 0.025 + (0.5 × 0.475) = 0.2625
q (frequency of B) = 1 - p = 1 - 0.2625 = 0.7375
Expected frequency of AA = p² = (0.2625)² = 0.069
Expected frequency of AB = 2pq = 2 × (0.2625) × (0.7375) = 0.387
Expected frequency of BB = q² = (0.7375)² = 0.544
The selection coefficient for each inversion type is given by:
1 - s_A = (observed frequency of AA) ÷ (expected frequency of AA) = 0.025 ÷ 0.069 = 0.30
1 - s_B = (observed frequency of BB) ÷ (expected frequency of BB) = 0.50 ÷ 0.544 = 0.75
Solving for s_A and s_B:
s_A = 1 - 0.30 = 0.70
s_B = 1 - 0.75 = 0.25
s_A = 0.70, s_B = 0.25
The following plot shows the change in allele frequency of a beneficial allele (A) over several generations under three different genetic architectures.
Which fitness architectures (wAA, wAa, waa) correspond to the three curves: black (B), dark grey (DG), and light grey (LG)?
Answer:
B (Black): wAA>wAa>waa (Dominant Advantage) → Fastest fixation
DG (Dark Grey): wAA=wAa>waa (Additive Selection) → Medium-speed fixation
LG (Light Grey): wAA>wAa=waa (Recessive Advantage) → Slowest fixation
Explanation:
Dominant: Aa individuals benefit immediately, leading to rapid spread.
Additive: Gradual increase, as Aa has intermediate fitness.
Recessive: Initially slow, as Aa doesn’t benefit; fixation accelerates once AA is common
B: wAA>wAa>waa
DG: wAA=wAa>waa
LG: wAA>wAa=waa
Have a look at the three adaptive landscapes (FL used to visualize the relationship between genotypes and reproductive success) below showing the fitness of the homozygotes of two diallelic loci.
We assume that the ancestral genotype is aacc.
Which of the following statements is correct is we assume that drift does not play a role?
A) Only FL 1 can evolve to the highest fitness peak
B) FL 1 shows no epistatsis
C) Genotype aaCC can only be reached in FL 1 and 3
D) FL 3 will reach the hightest fitness peak despite epistatsis
E) FL 2 cannot reach the highest fitness peak because of epistatis
Epistasis is when the effect of one gene depends on the presence of other genes. In these landscapes, we can infer epistasis by looking at whether the combined fitness of homozygous genotypes is simply the product of their individual fitnesses. If it isn't, there is epistasis.
When comparing recombination rates, we always observe …
Uniform recombination rates across different chromosomes.
Higher recombination rates in females compared to males.
A direct relationship between physical and genetic map length.
None of the answers is correct
Thomas Hunt Morgan and colleagues spent many years crossing wildtype and mutant fruit flies.
If Morgan crossed a female fruit fly homozygous recessive at two linked loci controlling ( not transmitted independently) color and wing length (i.e., bb / vgvg), to a male fly heterozygous at both these loci (i.e., +b / +vg), of the options below, what distribution of offspring genotypes is possible.
bb/vgvg:576 +b/+vg:692 +b/vgvg:0 bb/+vg:0
No recombination since linkage
In the figure, the star represents a recent mutation at a gene locus. The letters represent other loci. Which other locus is most likely to be in linkage disequilibrium with the mutation (all else being equal; assume a sexually reproducing organism)?
C
In a population with two diallelic loci (A, with alleles = A1 or A2; and B, with alleles = B1 or B2) that are in linkage disequilibrium
(combinations of alleles at different loci (genes or positions on a chromosome) occur more or less frequently than would be expected if the loci were independently assorting)
If the frequency of allele A1 = pA and the frequency of allele B1 = pB the frequency of haplotype A1B2 could be determined as:
The term pA * (1 - pB) represents the expected frequency of the A1B2 haplotype if the A and B loci were independently assorting (i.e., in linkage equilibrium)
The probability of the A1 allele is pA.
The probability of the B2 allele is (1 - pB) (since it's the alternative to B1, which has a frequency of pB).
However, because the loci are in linkage disequilibrium, we need to account for the LD coefficient (D).
pA * (1 - pB) - D
Which of the following statements about quantitative traits is not correct?
Quantitative traits are polygenic, i.e., they are controlled by many genes, and there are environmental effects that alter the phenotypic state of each individual
A) Quantitative traits vary continuously
B) Quantitative traits are normally determined by multiple alleles at multiple loci
C) Quantitative traits are not affected by the environment
D) Quantitative traits can be affected by drift
E) Quantitative traits can be affected by natural selection
Why are permutation tests commonly employed in QTL mapping studies? Choose the best answer.
QTLs are associated with traits with continuous variance
QTL mapping is a statistical analysis to identify which molecular markers lead to a quantitative change of a particular trait
A) To estimate effect sizes of identified QTLs accurately.
B) To determine the statistical significance of QTLs as it allows us to correct for multiple testing.
C) To identify candidate genes associated with QTLs.
D) To directly manipulate the phenotypes of individuals in the mapping population.
E) To improve the resolution of QTL mapping by introducing additional genetic variation.
B)
Dobzhansky-Muller incompatibilities (DMIs) ... (choose the best answer)
explain how reproductive isolation, and ultimately speciation, can arise between populations.
A) can explain how assortative mating can evolve in the face of gene flow
B) can explain how hybrid sterility can arise
C) will reduce population fitness
D) rely on chromosomal rearrangements
E) explain why male hybrids are more often infertile than female hybrids
How it Works:
Independent Evolution: Imagine two populations of the same species that become geographically separated. Over time, each population accumulates different mutations. Crucially, some of these mutations may become fixed (common) within each population because they are beneficial in that specific genetic background.
Epistasis: Epistasis is when the effect of one gene depends on the presence of other genes. In the context of DMIs, this means that the effect of a particular allele at one locus depends on which allele is present at another locus.
Incompatibility: The problem arises when individuals from these two populations hybridize. The hybrid offspring inherit a mix of alleles from both parents. The combinations of alleles that were beneficial in their original populations may now clash in the hybrid. For example, an allele at locus A from population 1 might work fine with the allele at locus B also from population 1. But when that same allele at locus A is paired with the allele at locus B from population 2, it might result in a harmful interaction, leading to hybrid sterility or inviability.
What mechanisms may contribute to maintaining associations between traits involved in prezygotic isolation and those under divergent selection?
prezygotic isolation =before the gametes fertilize to create a zygote
Choose the best answer.
A) Random mating and genetic drift
B) Physical linkage, inversions, or pleiotropy
C) Hardy-Weinberg equilibrium
D) Homologous recombination
E) Phenotypic plasticity
The correct answer is B) Physical linkage, inversions, or pleiotropy.
Here's why:
The question asks about mechanisms that maintain associations between traits involved in prezygotic isolation and those under divergent selection. Let's break down the key concepts:
Prezygotic Isolation: Reproductive isolation that occurs before the formation of a zygote (fertilized egg). This prevents mating or fertilization between different groups. Examples include differences in mating rituals, timing of reproduction, or physical incompatibility of reproductive organs.
Divergent Selection: When natural selection favors different traits or alleles in different populations of a species. This can lead to the evolution of distinct characteristics in each population.
Maintaining Associations: The question is asking how traits related to prezygotic isolation can remain associated with traits under divergent selection. In other words, how can the traits that help with prezygotic isolation stay linked to the traits that are being favored by divergent selection?
When comparing human and mouse, most genes have:
A) dN/dS < 0
B) dN/dS = 0
C) 0 < dN/dS < 1
D) dN/dS = 1
E) dN/dS > 1
dN/dS is the ratio of nonsynonymous (dN) to synonymous (dS) substitutions in protein-coding genes.
dN/dS < 1 suggests purifying selection, where most amino acid changes are harmful and removed by natural selection. This is the most common pattern for most genes as it indicates that the protein sequence is under functional constraint.
dN/dS = 1 suggests neutral evolution, where amino acid changes have no significant effect on fitness.
dN/dS > 1 suggests positive selection, where amino acid changes are advantageous and favored by natural selection. This is less common and usually seen in genes involved in adaptation or immune response.
A negative value of Tajima's D indicates an excess of:
A) Common variants
B) Rare variants
C) Synonymous variants
D) Nonsynonymous variants
E) Transversions
Correct Answer: B) Rare variants
Tajima's D is a statistic used to test for deviations from the neutral model of evolution in a population. It compares the number of segregating sites (polymorphisms) in a DNA sequence to the average pairwise difference between sequences.
A negative Tajima's D value indicates an excess of rare variants in the population. This pattern can be caused by:
Population expansion: After a population bottleneck or rapid expansion, there is an increase in the number of new mutations, which are initially rare.
Positive (directional) selection: When a beneficial mutation sweeps through a population, linked neutral or slightly deleterious variants can "hitchhike" along with it. This can also lead to an excess of rare variants.
A positive Tajima's D value suggests an excess of intermediate-frequency variants, which can be a sign of balancing selection or population subdivision
Which of the following can explain the observed positive correlation between DNA polymorphism and recombination rate?
A) Dominance
B) Overdominance
C) Genetic hitchhiking
D) Genetic drift
E) Balancing selection
Answer: C) Genetic hitchhiking
Regions of high recombination tend to have higher levels of genetic diversity (polymorphism).
Genetic hitchhiking explains this: When a beneficial mutation arises, it often carries along linked neutral or slightly deleterious variants. In regions of high recombination, these linked variants are more readily separated from the beneficial mutation, reducing the chance that they are lost due to negative selection or random drift. This allows for higher overall diversity in areas with high recombination.
Which model of human evolution involves extensive gene flow among populations?
A) Candelabra model
B) Multiregional model
C) Replacement model
D) Adaptive model 1
E) Hybridization model
Answer: B) Multiregional model
The multiregional evolution model proposes that modern humans evolved from Homo erectus in multiple regions of the world, with continuous gene flow between these populations. This gene flow is crucial in this model, as it prevents complete reproductive isolation and allows for the sharing of beneficial mutations.
UPGMA is an example of a __________ method.
A) Bayesian
B) Parsimony
C) Maximum likelihood
D) Maximum branching
E) Distance
Answer: E) Distance
UPGMA (Unweighted Pair Group Method with Arithmetic Mean) is a distance-based method for constructing phylogenetic trees. It uses a matrix of pairwise distances between taxa (e.g., genetic distance, sequence divergence) to cluster the most similar taxa together. Other methods like Maximum Likelihood and Bayesian approaches use sequence data directly and model evolutionary changes explicitly. Parsimony aims to find the tree that requires the fewest evolutionary changes.
The mutations responsible for lactase persistence in humans occur in:
A) The 5' UTR of the LCT gene
B) The coding sequence of the LCT gene
C) The 3' UTR of the LCT gene
D) An intron of the LCT gene
E) An intron of a gene neighboring the LCT gene
E) An intron of a gene neighboring the LCT gene is also a possible location for lactase persistence mutations.
The homeodomain is a protein domain of __________ amino acids:
A) 20
B) 40
C) 60
D) 120
E) 180
C) 60: The homeodomain is a conserved 60-amino-acid DNA-binding domain.
The Drosophila homolog of the mammalian Pax6 gene is known as:
A) Dax
B) eyeless
C) white
D) yellow
E) Fmr1
B) eyeless: Eyeless is the Drosophila homolog of the Pax6 gene, both crucial for eye development.
In Drosophila, the X chromosome has __________ genes as the Y chromosome:
A) The same number of
B) One-half as many
C) One-third as many
D) Two times as many
E) Over ten times as many
E) Over ten times as many: The X chromosome in Drosophila has far more genes than the Y
Which type of site is expected to show the least divergence between species?
A) non-degenerate
B) 2-fold degenerate
C) 3-fold degenerate
D) 4-fold degenerate
E) 5-fold degenerate
A) non-degenerate: Non-degenerate sites are under stronger selective pressure, thus less prone to change.
Mexican and South American dog breeds:
A) Were domesticated in the Americas from American wolves
B) Were brought to the Americas by Europeans after 1492
C) Were brought to the Americas by Vikings before 1492
D) Were derived from domesticated dogs brought to the Americas from Asia
E) Are hybrids between modern Asian and European dog breeds
D) Were derived from domesticated dogs brought to the Americas from Asia: Genetic evidence supports an Asian origin for American dog breeds via the Bering Land Bridge.
Which of the following statements is true of background selection?
A) It has the greatest effect in regions of low recombination
B) It has the greatest effect in regions of high recombination
C) It has the greatest effect when there is strong positive selection
D) It does not affect levels of genetic variation
E) It relies solely on genetic drift
A) It has the greatest effect in regions of low recombination: Background selection is more effective in low recombination areas because linked neutral variants are more likely to be eliminated along with deleterious mutations.
What type of genetic variation is associated with number of children in the Icelandic population?
A) Gene copy number
B) Synonymous SNP
C) Nonsynonymous SNP
D) Microsatellite repeat number
E) Chromosomal inversion
E) Chromosomal inversion is a more directly established association with the number of children in the Icelandic population.
What type of species has the largest known genome?
A) Bacteria
B) Amoeba
C) Diatom
D) Insect
E) Vertebrate
Answer: B) Amoeba
While vertebrates have large genomes, some amoebas have far larger genomes, sometimes hundreds of times greater than the human genome. This is largely due to vast amounts of repetitive DNA.
The defining feature of a cis-regulatory variant is that it:
A) Alters the tissue in which a gene is expressed
B) Only has an effect in heterozygotes
C) Affects expression differently in males and females
D) Is genetically linked to the gene it regulates
E) Leads to the creation of a new splicing site
Answer: D) Is genetically linked to the gene it regulates
Cis-regulatory elements are on the same DNA molecule as the gene they regulate. They affect the expression of nearby genes on the same chromosome.
Which pair of species/taxa are male-heterogametic? (heterogametic sex is the one in which the sex chromosomes differ.)
A) Drosophila and honeybee
B) Drosophila and human
C) Honeybee and chicken
D) Human and chicken
E) Anolis lizard and sea turtle
While humans are male-heterogametic (XY), Drosophila also has an XY sex-determination system, making males heterogametic. Honeybees, on the other hand, have a haplodiploid system where males are haploid (one set of chromosomes) and females are diploid (two sets).
In sea turtles, an increase in the proportion of females is associated with:
A) Degeneration of the Y chromosome
B) A lethal, male-specific virus
C) Higher temperature
D) Adaptive evolution of the X chromosome
E) Whole genome duplication
Answer: C) Higher temperature
Sea turtle sex determination is temperature-dependent. Higher temperatures during incubation tend to produce more females.
Both selective sweeps and background selection have been proposed to explain:
A) The positive correlation between recombination rate and nucleotide polymorphism
B) The positive correlation between genetic divergence and time (in years)
C) The positive correlation between genetic divergence and time (in generations)
D) The negative correlation between dN and dS
E) The negative correlation between pi and theta
Answer: A) The positive correlation between recombination rate and nucleotide polymorphism
Both selective sweeps (beneficial mutations rapidly spreading) and background selection (removal of deleterious mutations) can lead to reduced genetic diversity around the selected site. However, in regions of high recombination, the effects of these processes are more localized, leading to the observed positive correlation because linked neutral variants are more readily separated from the selected site.
What type of proteins are encoded by Hox genes?
A) DNA polymerases
B) RNA polymerases
C) Elongation factors
D) Transcription factors
E) Nuclear transferases
Answer: D) Transcription factors
Hox genes are a family of transcription factors that play a crucial role in regulating development, particularly in establishing the body plan of animals. They control the expression of other genes involved in determining segment identity and body axis formation.
Last changed10 days ago
