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Code 111 Code 111 Code 111 Circle the best single letter choice for each of the following questions before transferring your answers to your computer sheet within the time limit. Note: Questions may have 3, 4 or 5 choices. Section A – Multiple choice ( 36 questions; 1 mark each; plan 90 minutes total )
Point mutations at the human EPAS1 gene can result in the creation of a mutant allele ( EPAS1o ) that improves physiological function in oxygen-poor environments (such as mountaintops and other high-altitude regions). Humans in oxygen-poor environments, such as the mountains of Nepal and Peru, have much higher frequencies of the EPAS1o allele than populations in oxygen-rich environments. What is the most likely explanation for the distribution of the EPAS1o allele? A. Mutations that change the wild-type allele to the EPAS1o allele are more likely to occur in oxygen-poor environments than in oxygen-rich environments. B. Oxygen-poor environments are associated with higher rates of mutation in general, including mutations that convert the wild-type allele to the EPAS1o allele. C. The EPAS1o allele is favoured by selection in oxygen-poor environments, but not in oxygen- rich environments. D. The EPAS1o allele is maintained by heterozygote advantage.
The locus controlling shell strength in a population of turtles is called T. There are two alleles, T 1 and T 2. The effects of genetic drift are minimal since there are hundreds of thousands of turtles in this population. The starting frequencies of alleles T 1 and T 2 are 0. and 0.3, respectively. Under which of the following scenarios will T 1 eventually reach a frequency of 1 and T 2 eventually disappear completely?
wT1T1 > wT1T2 > wT2T
wT1T1 > wT1T2 < wT2T
wT1T1 > wT1T2 = wT2T
wT1T1 < wT1T2 > wT2T A. 1, 2 & 3 only B. 1 & 3 only C. 2 & 4 only D. 4 only E. All of 1, 2, 3 and 4 are correct.
In a hypothetical population of porcupines, needle length is governed by a single locus with two alleles ( N1 and N2 ). In a sample of 500 porcupines, a biologist counts 20 with long needles (genotype N1N1 ), 160 with medium-length needles (genotype N1N2 ), and 320 with short needles (genotype N2N2 ). Which of the following statements is most likely to be correct? A. Allele N 1 is completely recessive and allele N2 is completely dominant. B. Allele N 1 is completely dominant and allele N2 is completely recessive. C. All three genotypes have the same average fitness. D. Selection favours allele N 2.
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In Skills Session 5A, your team worked to bring together a wide range of information in a proposal for a Marine Protected Area to conserve wolffish populations. Which of the following statements about wolffish is correct?
Wolffish protect kelp beds by preying on sea urchins.
Wolffish are negatively impacted by low levels of dissolved oxygen in seawater.
Wolffish habitat is degraded by fishing practices that disturb the ocean floor.
Wolffish do not migrate over long distances in a given season. A. 1, 2 and 3 B. 1 and 3 C. 2 and 4 D. 4 only E. All of 1, 2, 3 and 4 are correct.
Over the course of Skills Modules dealing with studies of genetic variation in rattlesnakes, zebra mussels and wolffish, we encountered the concept of “structured” vs. “unstructured” populations. What does it mean for a population to be “structured”? A. Structured means that a population shows the expected p^2 + 2pq + q^2 distribution of AA, Aa & aa genotypes, respectively. B. Structured means that a population is spread over diverse habitats. C. Structured means that a population is responding to external disturbances. D. Structured means that a population is composed of distinct sub-populations.
Which of the following hypothetical populations is most likely to be mating assortatively at the M locus? Numbers refer to observed percentages of each genotype. A. 81 MM , 18 Mm , 1 mm B. 25 MM , 50 Mm , 25 mm C. 20 MM , 20 Mm , 60 mm D. 20 MM , 60 Mm , 20 mm
Imagine a team of students studying the distribution of genotypes in a population using chi- square analysis. They have calculated the difference between the observed and expected genotype frequencies for a given locus, giving rise to a chi-square statistic of 18.6. If the critical value is 5.99 at 2 degrees of freedom and p value of 0.05, what decision should this team make with respect to their null hypothesis? A. Accept the null hypothesis, and conclude that the population is not evolving. B. Reject the null hypothesis, and conclude that the frequencies are not in Hardy Weinberg equilibrium. C. Reject the experiment as flawed, since the alternate hypothesis was not tested. D. Accept the null hypothesis, since the chi-square value is greater than the critical value.
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- Consider the cotton plant that is tetraploid (4n = 104) and makes spores that are diploid (2n = 52 ). Which of the following comparisons identifies a pair of metaphase cells containing the same number of chromatids? A. Mitosis in sporophyte: Meiosis II in sporophyte. B. Mitosis in gametophyte: Meiosis II in sporophyte. C. Mitosis in zygote: Mitosis in sporophyte. D. Both B and C are correct.
- A company called 23andMe provides genetic testing for a wide range of human traits for $199. Assume that you and your partner receive the data below for three traits: MN blood group : autosomal gene, two codominant alleles, M and N. Ear wax : autosomal gene, two alleles, E1 codes for dominant wet wax; E2 is recessive dry wax. Androgen insensitivity syndrome: sex-linked trait, recessive ais allele results in affected XY males developing as females. Homozygous and heterozygous XX females ( ais ais; ais AIS ) are not affected and develop normally. MN Genotype Ear Wax Androgen Insensitivity Female Partner MN E1E2 1 ais allele Male Partner MN E1E2 0 ais allele Assuming that all genes assort independently, what is the likelihood that your first born child will develop as a boy and have blood type MN and have wet ear wax? A. 1/ B. 1/ C. 3/ D. 1/ E. 3/
- About 25% of humans exhibit the autosomal dominant phototic sneeze reflex (PSR) that causes them to sneeze a given number of times (often twice) when suddenly facing directly into full sun. (You know who you are.) Imagine that you and your partner are both sun sneezers but your mothers are not. Given the poor outcomes of summer photos of affected individuals, you are concerned that some of your children will be sun sneezers. What is the likelihood that your first two children will be sun sneezers? A. (3/4)^2 B. 3/4 x 2 C. ¾ + ¾ D. 2 3/
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This Figure represents a typical gel after separation of the various forms of MDH-1 and MDH-2 enzymes by cellulose acetate gel electrophoresis. Which of the following properties determines the separation of the two forms of MDH on the gel?
Binding affinity to cellulose acetate
Size
Enzyme activity
Net charge A. 1, 2 and 3 B. 1 and 3 C. 2 and 4 D. 4 only E. All of 1, 2, 3 and 4 are correct. 1 4. The diagram below appeared on the November Term Test where it supposedly showed the last replication bubble at the end of a baboon chromosome. However, more careful attention to the structure of this bubble revealed that the diagram is fundamentally flawed. In which of the following ways is this diagram incorrect? A. The leading strand synthesis should be toward the fork; the lagging strand synthesis should be away from the fork. B. The “bottom” blue strand should have the Okazaki fragments, not the “top” blue strand. C. The blue strands should be labelled “New strands” instead of “Parent DNA”. D. There should be Okazaki fragments on both the top strand and the bottom strand.
While snorkeling in the Thames River one summer, you are excited to discover a new species of fish. For males in this species, body size does not influence fitness (as estimated from survival and reproductive success). However, you notice that smaller females have very low fitness, and larger females have very high fitness. According to the size-advantage principle, what type of reproductive pattern is most likely to be favoured in these fish? A. Asexual reproduction B. Changing sex from female to male (protogyny) C. Changing sex from male to female (protandry) D. Simultaneous hermaphroditism
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In many species of beetle, once mating and egg-laying are completed, neither parent provides any care to the offspring. However, similar to species that do show parental care, male beetles compete for access to females, and females are very choosy when selecting mates. How can we explain this sex difference in behaviour, in a species with no parental care? A. Eggs are more expensive to produce than sperm, so even without parental care, females invest more in offspring than males do. B. Potential fitness is probably higher for females than for males. C. Average fitness is probably higher for males than for females. D. By definition, males are the competitive sex and females are the choosy sex.
Why is reproducing sexually, rather than asexually, thought to be so widespread?
Reproducing sexually reduces the risk that mutation will create harmful alleles.
The earliest forms of life reproduced sexually; sexual reproduction is the ancestral state.
Reproducing sexually increases the chance that mutation will create helpful alleles.
Reproducing sexually increases genetic variation in one’s offspring: this is valuable if the environment is changing or if the population is engaged in evolutionary arms races. A. 1, 2 and 3 B. 1 and 3 C. 2 and 4 D. 4 only E. 1, 2, 3 and 4 are all correct
Which of the following actions would reduce your direct fitness and increase your indirect fitness? Assume that you are related to all your brothers by r=0.5, and to your children by r=0.5.
Sacrifice one of your children, to save the lives of three of your brothers.
Sacrifice your own life to save the life of one of your brothers.
Sacrifice your own life to save the lives of three of your brothers.
Sacrifice three of your brothers, to save the life of one of your children. A. 1, 2 and 3 B. 1 and 3 C. 2 and 4 D. 4 only E. 1, 2, 3 and 4 are all correct
With reference to the previous question, which of the actions described above would increase your inclusive fitness? A. 1, 2 and 3 B. 1 and 3 C. 2 and 4 D. 4 only E. 1, 2, 3 and 4 are all correct
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Beyonce and Solange are full-siblings ( r=0.5 ). Beyonce has the opportunity to provide Solange with some help. Helping her sister will reduce Beyonce’s direct fitness by 8 arbitrary units ( c = 8 ), but will increase Solange’s direct fitness by some amount ( b , expressed in the same units). According to the principle of kin selection, under what circumstances (what range(s) of b ) should these two individuals likely agree about whether or not Beyonce should help Solange?
b < 4
4 < b < 8
b > 16
8 < b < 16 A. 1, 2 and 3 B. 1 and 3 C. 2 and 4 D. 4 only E. 1, 2, 3 and 4 are all correct
The Prisoner’s Dilemma is a two-player game in which each player independently decides to cooperate with their partner or to cheat. If you will be playing a single round of this game, which combination of choices will cause you to receive the BEST possible payoff? A. You cooperate; your partner cooperates. B. You cheat; your partner cooperates. C. You cooperate; your partner cheats. D. You cheat; your partner cooperates. E. You and your partner both make the same choice (that is, both players cooperate, OR both players cheat).
Which of the following statements about evolutionary arms races is correct?
Evolutionary arms races usually place the same strength of selection on both species.
Evolutionary arms races only occur between species that interact antagonistically.
Evolutionary arms races usually result in both species improving their fitness and success.
Evolutionary arms races usually result in both species evolving adaptations that help them to interact with the other species. A. 1, 2 and 3 B. 1 and 3 C. 2 and 4 D. 4 only E. 1, 2, 3 and 4 are all correct
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Which of the phylogenetic trees shown below is consistent with the following sequence alignments? (_ indicates that the base is missing in that species.) Species A: AACTCGGTCTAAGAAGAATTTAAACG Species B: ATCTCGGTGTAAGAAGAATTTAATCG Species C: ATCTCGGTGTAAGAAGAATTTAAACG Species D: AACTCGGTCTAAGTTGAATTTAAACG Species E: AACTGCC_CGG_GAAGAAATT_AACG 3 1. Darwin’s finches (Geospizini) are a group of about 15 species of tanagers living in the Galapagos Islands. Although all species of Darwin’s finch are closely related to one another, they differ tremendously in the shape of their beaks. Some species of Darwin’s finch have beaks that resemble those of warblers more than they do those of other Darwin’s finch species. Why should you NOT use beak shape to build a phylogeny of Darwin’s finches and their relatives? A. Beak shape is probably subject to homoplasy. B. Beak shape is not probably subject to natural selection. C. Bird beaks do not fossilize well, so we do not know what shape beak the common ancestor of Darwin’s finch would have had. D. Similarities in beak shape reflect evolutionary relatedness, so this trait provides little additional information about past selection pressures.
Which of the following principles best characterizes the field of evolutionary medicine? A. When trying to explain and understand human illness, it is best to consider both ultimate and proximate explanations. B. Human health was generally better in hunter-gatherer times than it is today. C. If we discover that a particular medical condition (for example, intense anxiety) first evolved as a defence against danger, the condition should not be treated medically. D. Traits that increase survival or reproductive success late in life are favoured by selection, even if they reduce survival or reproductive success early in life.
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Allopatric speciation involves one ancestral population becoming subdivided into two daughter populations. In some cases, these populations may later come back into secondary contact. At what stage is selection likely to directly favour the evolution of prezygotic isolating mechanisms between the two daughter populations? A. As soon as the populations become physically isolated from each other. B. After divergence has started, but before the populations come back into secondary contact. C. After the populations come back into secondary contact, but only if they have already become at least partly postzygotically isolated. D. As soon as the populations become postzygotically isolated, regardless of whether they have come back into secondary contact.
The Table below shows the distribution of traits found in six hypothetical species of bird (species A, B, C, D, E, F) as well as an outgroup species of crocodile (X). Species Eye placement Teeth Feathers Tail X (outgroup) Front facing Yes No Short A Side facing Yes Yes Short B Front facing No Yes^ Short C Front facing No Yes^ Long D Front facing No^ Yes^ Short E Front facing No^ Yes^ Short F Front facing No^ Yes^ Long Which of the following traits are likely to be phylogenetically informative?
Side-facing eyes
Having teeth
Having feathers
Long tail A. 1, 2 and 3 B. 1 and 3 C. 2 and 4 D. 4 only E. All of 1, 2, 3 and 4 are correct.
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Section B : Case Study: (8 Questions; 25 marks total; plan about 50 minutes total )
The following scientific article is real but highly edited. It speaks to various physiological, genetic and evolutionary aspects of host parasite interactions. You should look briefly at the questions and then read the article to find needed information.
Resistance of song sparrows to local versus non-local malarial parasites
Yanina Sarquis-Adamson and Elizabeth A. MacDougall-Shackleton Department of Biology, University of Western Ontario, London, Ontario, Canada Introduction Parasites and their hosts are key features of one another’s environments, resulting in evolutionary arms races. Selection on parasites favours the ability to exploit locally common host genotypes, which may result in parasites being better able to infect hosts from their local area (“local”) than hosts from another area (“nonlocal”). At the same time, however, selection on hosts favours resistance to locally common parasite strains, which may result in hosts being more resistant to local parasites than to nonlocal parasites. The outcome of these arms races, i.e. whether parasites are better able to infect hosts local hosts than n o n l o c a l hosts (interpreted as parasites “winning” the arms race in their local area), or whether hosts are better able to resist infection by l o c a l p a r a s i t e s t h a n b y nonlocal parasites (interpreted as hosts “winning the arms race in their local area), can have important consequences. For example, if hosts are more resistant to nonlocal than local parasites, selection on hosts may favour individuals that move away from where they were born, in order to escape the local parasites. Conversely, if hosts are more resistant to local than nonlocal parasites, then selection should favour hosts that remain close to where they were born, in order to escape nonlocal parasites. Because of a warming global climate, the geographic ranges of many parasites (and the insects that transmit the parasites) are changing rapidly. This makes it important to predict the relative susceptibility of host populations to local and nonlocal parasites. Although the outcome of an evolutionary arms race is not completely predictable, parasites usually have an evolutionary advantage over their hosts. Thus, parasites usually have a “home-field advantage” and are better able to infect hosts from the local area than hosts from elsewhere. In birds, rates of malaria infection have tripled over the past 70 years. This disease has already caused the extinction of several bird species, making the interactions between birds and their malarial parasites particularly compelling from a conservation perspective. Avian malaria is caused by a single-celled eukaryotic parasite ( Plasmodium ) that spends the asexual (haploid) portion of its lifecycle in the bloodstream of a bird, and the sexual (diploid) portion of its lifecycle in the gut and salivary glands of a mosquito. The parasite is transmitted between the two types of host when mosquitos feed on the blood of an infected bird, and when infected mosquitos inject saliva into the bird bloodstream. We conducted an experiment to test whether song sparrows have an advantage in defending against their local malarial parasites, or conversely whether parasites in this system have an advantage in infecting their local hosts. We captured sparrows from two different breeding sites, identified local Plasmodium strains in each, and assessed resistance to local versus nonlocal strains.
Student Number: ______________ Code 111 Code 111 Code 111 Material andmethods Study system:hosts andparasites We captured 65 song sparrows from each of two breeding locations: an eastern site at Newboro, Ontario, and a western site at London, Ontario. The most commonly observed blood-borne parasites in these birds are Plasmodium, which causes avian malaria. Because asexual haploid reproduction of the parasite occurs within circulating host blood cells, Plasmodium infections can be transmitted to new hosts through inoculation with infected blood. We identified birds that were already infected with Plasmodium at the time of capture by examining their red blood cells under a microscope. Identifying local and nonlocal parasite lineages We sampled song sparrows throughout Ontario, and sequenced Plasmodium lineages infecting them, to identify parasite lineages that were present at the western site and absent from the eastern site and vice versa. We identified 11 unique Plasmodium lineages based on mitochondrial DNA sequence (Figure 1). Lineage P-SOSP10 was detected only at the western site and not at the eastern site. Thus, P-SOSP10 was used as the ‘western’ lineage in the cross-infection experiment. Conversely, lineage P-SOSP9 was detected only at eastern site and not at the western site, so we used this as the ‘eastern’ lineage in the cross-infection experiment. Cross-infection experiment Although Plasmodium parasites are normally transmitted to birds by t h e b i t e s o f i n f e c t e d mosquitoes, they can also be transferred between individual birds through cross-infection experiments, that is, by inoculating a small volume of blood from an infected bird into another bird. We collected blood from each of two already-infected ‘parasite donors’, i.e. an eastern bird infected with the eastern parasite (P-SOSP9) and a western bird infected with the western parasite (P-SOSP10). Each sample of infected blood was inoculated into ‘experimental’ birds from both the eastern and western site. We inoculated 20 eastern birds and 20 western birds with the eastern parasite (P-SOSP9). Similarly, we inoculated another 20 eastern birds and 20 western birds with the western parasite (P-SOSP10). As controls, we inoculated 20 eastern and 20 western birds with blood from an uninfected bird. Results Figure 2 shows infection success for eastern and western birds, exposed to eastern vs. western parasites.
