Practice Questions for Exam 2 (Scroll down for answers)
Know the meaning of the following terms:
Allopatric speciation
Sympatric speciation
Peripheral isolate
Magic trait (with respect to speciation)
Epistasis
Bateson-Dobzhansky-Muller model
Allopolyploidy
Monophyletic group
Paraphyletic group
Synapomorphy
Symplesiomorphy
Maximum likelihood
Maximum parsimony
Gene tree
Species tree
Synonymous substitution
Pseudogene
Exon shuffling
Transposable element
Retroelement
P-element
Horizontal gene transfer
Segregation distortion genes
Transcription factor
Cis-regulatory element
Homeobox
MADS box gene
Semelparity
Iteroparity
Age specific survivorship
Senescence (will be covered in Tuesday's lecture)
1)Explain how whole genome duplication facilitates the process of speciation by hybridization.
2) Explain why disruptive selection alone is not likely to lead to sympatric speciation. What other process would we have to add to disruptive selection in order to make sympatric speciation likely?
3)In constructing a phylogeny of vertebrates, explain why the trait "has lungs" can be used as evidence to unite lizards and mammals, while the trait "not having lungs" can not be used as evidence uniting sharks and tuna.
4)Explain the difference between maximum likelihood and maximum parsimony as ways to choose the best phylogenetic tree. Identify a situation in which one of them is expected to yield the wrong answer.
5)What do the terms KA and KS refer to? Explain why observing that KA/KS << 1 for a particular gene suggests that the gene has been under purifying selection.
6)Consider a locus, A, with two alleles, A1 and A2, each at frequency 0.5. What evidence would suggest that A2 is currently increasing in frequency due to selection while A1 is decreasing in frequency?
7)What properties of retroposon insertions make them particularly useful for reconstructing phylogenies?
8)Consider the following data for four taxa, A, B, C, and D:
| Trait 1 | Trait 2 | Trait 3 | Trait 4 | Trait 5 | Trait 6 | |
| A | 1 | 0 | 1 | 0 | 0 | 1 |
| B | 0 | 0 | 0 | 0 | 1 | 0 |
| C | 1 | 1 | 1 | 1 | 0 | 0 |
| D | 1 | 0 | 1 | 1 | 0 | 0 |
9)Explain the difference between a gene tree and a species tree. What properties of a locus increase the probability that the gene tree for that locus will not match the species tree?
10)You construct a maximum parsimony tree for a set of species and notice that two of the species, which show up as closest relatives on your tree, each show many more autapomorphies than any of the other species. Why might this make you suspicious of your tree?
11)Why might exon shuffling be more likely to produce an adaptive variant protein than would ordinary point mutation?
12)How can a gene duplication event produce two genes that already
have different sequences?
13) (This will be covered on Tuesday) Consider a species with very high and consistent adult age specific survivorship, as indicated by the following lifetable:
x 0 1 2 3 4 5 lx 1 1 .95 .9 .86 .81 mx 0 .2 .2 .2 .2 .2
How would you expect the schedule of reproduction (i.e. the distribution of mx values) to change if a new predator arrived that reduced the px values of adults (age class 1 and up) by a factor of 1/2?
Answers
1)Hybrids between different species, if they survive at all, are sterile. This is because they do not have pairs of homologous chromosomes - meaning that Meiosis does not proceed properly and they produce no viable gametes.
Duplicating the entire genome results in two identical copies of each chromosome, so each chromosome now has a homologue and Meiosis can proceed to produce viable gametes.
2) In a population with random mating, disruptive selection (selection favoring extreme phenotypes and against intermediate phenotypes) tends to cause one or the other extreme phenotype to go to fixation because most of the offspring of parents with the rarer extreme phenotype will be hybrids, and thus have an intermediate phenotype and low fitness. (5 pts). However, if we combine such selection with assortative mating that is correlated with the selected trait, such that individuals with the favored phenotypes preferentially mate with others with the same phenotype, then both of the extreme phenotypes can increase in frequency as hybrids decrease.
3)Lungs are derived within vertebrates, so the fact that Lizards and Mammals both have them is evidence that these two groups are members of a clade within the vertebrates that arose after the appearance of lungs.
The trait "not having lungs" is the ancestral state for all vertebrates (and all life). The fact that sharks and tuna share this trait is thus evidence that each of their lineages split off before the appearance of lungs, but not that they are members of the same clade within the vertebrates.
4)Maximum likelihood chooses the tree that maximizes the probability of the data that we have. Maximum parsimony chooses the tree that requires the smallest number of character state changes. When the rate of evolution is highly variable, such that distant branches have many more character state changes than the intervening branches, parsimony tends to put these "long branches" together.
5)Ka is defined as the number of non-synonymous differences between copies of a gene from two different species, divided by the number of possible non-synonymous mutations in that gene.
Ks is the number of synonymous differences divided by the number of possible synonymous differences.
If Ka/Ks << 1, this implies that the probability that a non-synonymous mutation would get fixed in the population is much lower than the probability that a synonymous mutation would get fixed. This suggests that most mutations that change the gene product are deleterious, so selection is "purifying" in that it is primarily eliminating deleterious mutations, rather than fixing advantageous ones.
6)If A2 is currently increasing in frequency, then we would expect to see little variation at neutral loci that are closely linked to A2 (since all of the A2 alleles are recently derived from a small number of copies).
By contrast, neutral loci linked to A1 should show substantial variation, since A1 was the common allele until recently and there was time for variation to build up at nearby loci.
7)The two principle confounding factors in building a phylogeny are homoplasy (repeated appearance of the same trait) and loss of a trait that had previously appeared. Retroposon insertion sites avoid both of these problems:
1) The number of sites at which a particular retroposon can insert is so vast that the probability of two identical retroposons inserting in the exact same place in two different individuals is so small that we can ignore the possibility. We thus do not expect them to exhibit homoplasy (the same character state appearing more than once in a phylogeny).
2) Once inserted, it is very unlikely that a retroposon would be removed without leaving a trace (or without deleting a larger piece of the chromosome).
8)a)
b)Monophyletic = (BACD) or (ACD) or (CD)
Paraphyletic = (BAC) or (BA) or (AC) or (AD).
9)A gene tree shows the historical relationships between a set of homologous DNA sequences. A species tree shows the historical relationships between a set of reproductively isolated populations. The gene tree for a particular locus may not match the species tree if that locus remains polymorphic for the entire period of time between speciation events. This is most likely if there is selection favoring heterozygotes at the locus and/or the time between speciation events is short.
10)Maximum parsimony can be positively misleading when the rates of evolution are very high on branches that are actually far apart on the tree; parsimony tends to put these "long" branches together. When this happens, the species that are erroneously put together will each tend to have many derived character states that are not seen in any other species (autapomorphies).
11)Proteins are often constructed of different functional domains, each working largely independently of the others. These domains often correspond to individual exons in the gene for the protein. Exon shuffling, either within or between genes, can thus rearrange functional domains without damaging the functionality of any of them. By contrast, non synonymous point mutations will usually just damage one or another of the existing functional domains of a protein.
12)If there are different alleles of a gene present in the population, then unequal crossing over in a heterozygote can place a copy of each of two different alleles on the same chromosome.
13)Reducing adult survivorship by this much would select for concentrating reproduction early in the lifecycle. Jul 8, 2021