Notes for Final - Bio 1001 2023

d) 4 only e) 1, 2, 3, and 4 Increasing climate trends - Global surface temperature (land and ocean) - Ocean heat context - Carbon dioxide - Sea level - Methane - Nitrous oxide - Specific humidity - Middle tropospheric temperature Decreasing climate trends - Arctic and Antarctic sea ice extent - Ocean pH - Middle stratospheric temperature - Global glacier mass balance - Greenland ice sheet mass balance Lecture 25 - How will plants and other animals cope? The industrial revolution - Start of the Anthropocene - Coal fire → lots of air pollution, affected peppered moths - There was a directional change in the moth population → frequency of dark-coloured moths increased at that time (example of the industrial revolution) - because lichens and blackened urban tree trunks and walls were killed off, the pale form of the moth was more visible to predators - Hardly any white moths were found in areas of industrial growth. Conversely, forests with no pollution still had white moths (demonstrates adaptation) Sex determination - Certain species (like turtles), can depend on the temperature - <27.7°C → Male - >31°C → Female - Current temperature fluctuations can affect the distribution of males and females in the population Example: The Univoltine

- However, climate change is gradual - Birds can fly very long distances - Some animals use other animals to help them move Tropicalization - Movement of organisms in response to warmer temperatures - Generally causes increase in tropical specials and decrease in temperate species - Mangroves replacing salt marshes - Think about food chains! - Movement of organisms will disrupt the ecosystem - if there are predators better suited to the environment or who can outcompete for food - Beneficial / decremental? - Possible consequences: hybridization - Immigrant may gain genes allowing them to survive in new environment - Gain of diversity Asyncrony - Species following different environmental cues and migrate at different times - Herbivore and its’ host plant - Host plant comes out at different time than herbivore migrates - Herbivore arrives → no food source Example: Monarch + Red Admiral Butterflies - Arrived in large numbers due to high temperatures and unusual wind patterns in April, 2019 - Red admiral: semi-tropical, migrates in summer - Food source was above ground, successful oviposition - Temperature dropped → larvae died - Monarch - Arrived, couldn’t lay eggs - Milkweed had not come out yet → died Migrant species - Migrate north in spring when there are short days and hot temperatures - Migrate south in fall when there are short days and cold temperatures - Allows species to stay in optimal conditions year-round - Ex: Mythmina Unipuncta - Ideal temperature below 27°C - Conditions have changed - More days above 29°C in both summer and winter breeding grounds - Pest → rare species

b) 1 and 3 c) 2 and 4 d) 4 only e) 1, 2, 3, 4 Question: which of the following is not an effect of tropicalization? a) Hybridization between immigrant and native species b) Replacement of salt marshes by mangroves c) The evolution of the new traits in the native species in response to novel predators d) Reduced genetic diversity in all cases Cycle 11: Main takeaways - The Anthropocene is the current geological age where humans have been the dominant factor in environmental changes - We are currently in a climate crisis - need to stay below the tipping point of 1.5°C global increase - The industrial revolution started the anthropocene - Population density can influence natural selection - The ability of species to migrate can influence their mating and survival in these changing environmental conditions and extreme weather events - The effects of migration should be considered Cycle 11: Vocabulary - Anthropocene - Tipping point - Virtual water

Simple (Direct) Life cycle - 1 parasite, 1 host - When a parasite species uses ONE host species to complete their life cycle and survive - Eg. mistletoe - Ectoparasite: parasitic plant - Roots penetrate into tree - Can exploit OR provide nutrients (i.e. water) - E.g. Tapeworms - Endoparatide (Endo = inside) - Lives in the intestine - Shed segments full of eggs that, once ingested, can infect a new host Complex (inDirect) life cycle - 1 parasite, MULTIPLE hosts - When a parasite species uses AT LEAST 2 different host species to complete their life cycle and survive

Summer conditions (Toward light) Fall Conditions (Away from light) - Manipulates the aphids to move from the undersurface to the upper surface of the leaf just before death - Reduced developmental time through passive thermoregulation, thus decreasing the probability of mortality - Manipulates the aphids to move front he undersurface of leaf and to hide in the sheltered sited - This reduces the probability of mortality during the winter from snow, ice, etc. Food webs - Interaction of species within an environment → complex interactions - Also have interactions within an organism at a given trophic level - Interact for a given food source - Guild → share at least one prey and occupy the same trophic level - Intraguild predators Host Manipulation - Host manipulation is a parasite-induced alteration of a host’s phenotype that increases parasite fitness - Improves the probability of transmission - With host manipulation, more parasites are transmitted to the host - cockles are easier to access

Question: The node represents the common ancestor of which taxa? a) A, B, C, D b) B, C, D, E c) D, E, F, G, H d) All of the above Question: of the organisms in the evolutionary tree below, dandelion, spruce tree, tulip, and cactus are all land plants. Based on this information and the evolutionary tree below, which of the following statements is false? a) Spruce trees are equally related to dandelions and tulips b) Amoeba are equally related to tulips and green algae c) Tulips are more closely related to cacti than to red algae d) Red algae are more closely related to land plants than to green algae - Spruce trees are more commonly related to dandelions Question: According to figure 1, which of the following groups of species are monophyletic? Assume each of the groups named below also includes the group’s most recent common ancestor. 1. F, G, H, and I 2. F, G, H, I, J, and K 3. H and I 4. F, G, and H a) 1, 2, and 3 b) 1 and 3 c) 2 and 4 d) 4 only e) 1, 2, 3, and 4 are all monophyletic groupings Traditional groupings - Many groupings we know today are non-monophyletic - This is because we started grouping organisms before we understood their history!

Outgroup - An outgroup consists of species related to a clade, but not included in it - Outgroup comparison is used to - Identify ancestral and derived traits - Identify the root (common ancestor of the ingroup and outgroup) Is this trait Ancestral or Derived? Question: What is the major goal of outgroup analysis? a) To determine whether a trait is shared or unique b) To determine whether or not a group is monophyletic c) To determine whether a trait is probably ancestral or derived d) To determine whether similarities between two or more taxa are due to homology or homoplasy Question: According to Figure 1, which of the following is most likely to be a symplesiomorphy? a) A trait that is present only in species H and I b) A trait that is present only in species J and K c) A trait that is present only in species K d) A trait that is present in species F, G, H, I, and J, but not species K - Shared, ancestral - Look at the clade of interest as a whole!! (our clade of interest contains species F, G, H, I, J, K)

Question: Which of the following statements is correct? 1. Trait B is a derived trait of the ingroup 2. Trait C is a symplesiomorphy 3. Trait A is an ancestral trait 4. Trait B is probably not useful for determining evolutionary relationships a) 1, 2, and 3 b) 1 and 3 c) 2 and 4 d) 4 only e) All of 1, 2, 3, and 4 Important terminology: - Homology - A similarity that reflects common ancestry - Pterodactyl limbs in bats, dinosaurs - Convergence - Not closely related organisms develop similar features through evolution - Evolution can be seen in molecular state (DNA mutation cause change) - Homoplasy - Misleading similarities or dissimilarities - Train missing/gained for separate lineages Most parsimonious phylogeny - Principle or parsimony - A particular train is unlikely to evolve independently in separate evolutionary lineages - The simplest possible approach is the best. Minimize the number of homoplasies - Most likely evolutionary history - Count the number of evolutionary changes → The evolutionary tree with fewer evolutionary changes will be the most parsimonious one Cladistics - Goal - Make a phylogeny - Method - Compared shared and different traits to guess which organisms are closely vs. more related - Outgroup comparison - Shortcoming - Homoplasy (creating cladistics using physical traits can sometimes be misleading, hence why we have molecular sequences)

Building a phylogenic tree 1. Use parsimony 2. Group organisms based on relevant traits 3. Determine the outgroup 4. Determine the organism with the fewest differences from the outgroup 5. Repeat Question: You have been transported to another universe where there are creatures with many excess body parts. The Keeper to let you back home will only let you pass if you can create the phylogenetic tree that is the most likely to be correct based on cladistics. Try to draw out your phylogenetic trees. Using molecular sequences to create phylogenetic trees:

Question: An evolutionary tree was constructed for six hypothetical organisms based on an 18-nucleotide DNA sequence for a common gene. Based on this tree, the 18 nucleotide DNA sequence of the common ancestor at node X is inferred to be: GTC CTC AAT GTT GTT AGA (the first G is position 1, and the last A is position 18.) The nucleotide changes that have occurred during the evolutionary history of these organisms are shown on the tree. For example, 13C means that the nucleotide at position 13 changed to a C. Which of the following two organisms have the most similar DNA sequences for this region? a) Gamma and Epsilon b) Gamma and Delta c) Alpha and Beta d) Epsilon and Kappa Solution: DNA sequence of common ancestor at the gene → GTC CTC AAT GTT GTT AGA Alpha: G A C CTC AAT GTT C T A AGA Beta: GTC CTC AAT GTT C T A AGA Gamma: GTC CTC T AT GTT C TT AGA Delta: GTC CTC T AT GTT C T A AGA Epsilon: GTC CTC T AT GTT C TT AGA Kappa: GTC A TC AA C GTT GTT AGA Study Tips for the Cycle: → Make sure you know the vocabulary - Derived - Ancestral - Phylogenetic tree - Outgroup - Ingroup - Parsimony - Homology - Homoplasy - Monophyletic - Cladistics - Evolutionary systematics - Taxa - Node - Root - Clade - Sister clade

Cycle 8 - Life on earth, species and speciation Definition of Species - Ecological concept - Populations that are adapted to specific niches in the environment - Eg. mosquitoes feeding on humans vs birds may evolve differently - Morphological concept - Individuals of species share measurable traits that distinguish them from other species - Biological concept - Interbreeding populations that don't reproduce with other species - All individuals in a species can successfully mate and produce viable, fertile offspring - Genetic cohesiveness → Populations of same species experience a gene flow that mixes their genetic material - Genetic distinctiveness → Different species cant exchange genetic info - Phylogenetic concept - Populations that share recent evolutionary history - Independent branch on phylogenetic tree = species - Common morphological ancestry for sexually reproducing organisms - Implies reproductive isolation, but does not prove Species concept Ecological Morphological Phylogenetic Biological Advantages Explains the role of the environment in speciation Easy to classify physical traits to recognize species Applies to all groups of organisms (asexual, extinct) Testable and describes gene flow of species Disadvantages Unable to explain the existence of the same species in different environments Does not reveal much about evolutionary history Cannot distinguish different species with similar physical characteristics Does not describe gene flow This does not apply to asexual and extinct organisms

Clinical Variation - Gradient of traits in a geographic range - Varying environment conditions favour different traits → evolution within a species Ring species - Population migrates around geographic barrier → two adjacent populations that cannot interbreed - Considered the same species due to shared alleles and gene pools through intermediates Prezygotic Isolating Mechanisms - Temporal isolation → mating at different times - Ecological isolation → different habitats - Mechanical isolation → different reproductive structures - Behavioural isolation → different mating signals (courtship displays) - Gamete mortality (gamete isolation) → incompatibility between sperm and egg Postzygotic Isolating Mechanisms - Hybrid inviability → conflict geens prevents development - Fertilization occurs but the hybrid is frail of has early death - Hybrid sterility → survives but is unable to produce functional gametes - Hybrid breakdown → hybrid develops and can mate with other hybrids and parent species - Second generation will have higher fatality, lower fertility - Long-term reproductive isolation rather than immediate Question: There are over 20 species of bushbabies (which are small nocturnal primates) living in Africa, each with distinctly shaped genitalia. Bushbabies sometimes try to mate with an individual from a different species, but their genitalia do not fit together. Which type of reproductive isolating mechanism is being described? a) Hybrid inviability b) Gametic isolation c) Hybrid breakdown d) Behavioural isolation e) Mechanical isolation

Question: Baltimore and Bullock’s orioles look different, and they are not each other’s closest relatives. They interbreed where their ranges overlap, and produce hybrid offspring of normal fitness. Which of the species concepts below would consider these birds as belonging to the same species? a) Morphological Species Concept b) Biological Species Concept c) Phylogenetic Species Concept d) None of the above e) All of the above Allopatric vs Speciation - Modes of speciation based on species isolated by location - Allopatric → two populations that are geographically separated potentially causing the evolution of reproductive isolating mechanisms - Sympatric → reproductive isolation that evolves between subgroups that arise within one population Allopatric Speciation - Two populations that are geographically separated potentially causing the evolution of reproductive isolating mechanisms - Examples: - A physical barrier that subdivides a large population - A small population becomes separated from a species' main geographical location - Natural disasters ——-----------------------------

Secondary contact - Two previously isolated populations reunite - Not required for speciation to occur Secondary contact → potential outcomes - Reinforcement - Increases the rate of speciation - If reproductive isolation occurs, there is speciation - When the two species are back in contact, they can't interbreed - Fusion - Slows down the rate of speciation - Population successfully interbreed, they can “fuse” back together - Gene flow can continue → remains one species Autopolypoid Mutations on Speciation - Autoploidy mechanisms - Chromosomes fail to separate in cell division = with no reduction of the number of chromosomes = and the gamete is a different ploidy than the parent - Offspring species can’t interbreed with the original parent species

Question: During allopatric speciation, which of the following reproductive isolating mechanisms could evolve? a) Gametic isolation b) Ecological isolation c) Hybrid sterility d) Temporal isolation e) All of the above mechanisms could evolve Question: 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. Question: Which of the following statements are incorrect about modes of speciation? a) Hybrid species that form after secondary contact may or may not be the same species as the original population before the geographic barrier appeared b) Gametes that underwent autopolyploidy mutation receive the number of chromosomes that would be found in a somatic cell c) Secondary contact is not essential for speciation to occur d) Autopolyploid offspring can interbreed with their original parent species because they are still the same species e) All of the above are correct statements Question: A large population of wolves becomes geographically subdivided into two smaller, allopatric populations. Several generations later, the geographic barrier disappears and the two populations come back into contact with each other. Individuals cannot freely mate with individuals from either population and hybrid offspring have reduced fitness. Which of the following best describes this situation? a) Secondary contact and species fusion b) Temporal isolation c) Secondary contact and reinforcement d) Sympatric speciation e) Ecological isolation

Question: Which of the following statements regarding secondary contact is correct? 1. During reinforcement, hybrids have a decreased fitness because individuals that mate within their populations have a selective advantage 2. Components of the biological species concept can explain why reinforcement causes speciation and fusion continues gene flow 3. After secondary contact, the two reuniting species can produce a sterile offspring 4. Postzygotic isolation is the mechanism for continued interbreeding or fusion a) 1, 2 & 3 b) 1 & 3 c) 2 & 4 d) All of 1, 2, 3, & 4 Evolutionary Myths and Quirks - Why the idea that “humans are descended from chimps” is incorrect - We just share a common ancestor, that doesn’t mean chimps are our ancestors - Why some traditional groupings of organisms do not reflect evolutionary relationships - Just because some organisms are grouped, does not mean they are as closely evolutionary-related as they seem - Also, shared similarities might not always reflect shared ancestry, it could be convergent evolution - “Pedigree Collapse” - The idea that some of your ancestors are not ALL unique individuals - Eg. great x13 aunt on mom’s side and great x9 grandmother on dad’s side are the same person MRCA and LUCA - MRCA (blue circle) - The most recent common ancestor - The organism from which a set of other organisms are descended - Eg. in the diagram, the MRCA of all archaea is pictured - LUCA (green circle) - The last universal common ancestor - The most recent common ancestor of all living organisms - NOT the first thing to have ever lived, it’s just what survived

Phylogenetic Trees → Key takeaways 1. Reading from nodes to tips is going forward in time (and vice versa) 2. The tips represent organisms living today 3. The branches represent evolutionary relationships between ancestors and descendants 4. The nodes represent the ancestors of organisms *(MRCA and LUCA are the nodes) Phylogenetic Species Concept Revisited - Another way to think about the phylogenetic species concept is in terms of phylogenetic trees - Use shared, genetic history to classify species - When a “new” trait arises, it becomes a branching point, it would be a speciation event Question: Genomic sequencing has provided evidence that Neanderthals and Denisovans likely interbred with ancestral human populations. Which of the following statements could be used to successfully argue that Humans, Neanderthals, and Denisovans are the same species based on the phylogenetic species concept? a) Humans, Neanderthals, and Denisovans overlapped in their geographic distributions and could have interacted frequently with one another. b) The presence of Neanderthal sequences in the genomes of existing humans indicates that hybrids were likely viable. c) The physical features of Denisovans were more similar to Humans and Neanderthals than to any other hominin. d) Humans, Neanderthals, and Denisovans share a more recent common ancestor than any other hominin. Question: Based on the information provided for each organism, which organism would most likely be considered a species based on the biological species concept? a) Broccoli. Broccoli is an edible green plant in the cabbage family, along with cauliflower, Brussels sprouts, and kale. Broccoli, cauliflower, brussel sprouts and kale are closely related cultivars and can easily interbreed producing new foods such as kalettes and broccoflower.

b) Pseudomonas aeruginosa. Pseudomonas Aeruginosa is known as a common species of bacteria and many strains can cause disease in plants and animals, including humans. The genes shared by all strains of Pseudomonas Aeruginosa are on average 95 % similar in sequence, but many genes in a genome are not found in all strains and are exchanged with other bacteria. c) Bottom-dwelling three-spined stickleback (a type of fish). Bottom-dwelling populations eat different foods and are bigger than populations that swim in the water. When mixed, sperm from bottom-dwelling populations can fertilize eggs from populations that swim but do not mate in the wild due to assortative mating based on size and different breeding behaviours. d) All of these organisms would be considered a species based on the biological species concept. Cycle 7 (Summary) - All About Sex Reproduction - Asexual Reproduction - Asexual reproduction in plants → new individuals are genetically identical (bulbs, corms, tubers, rhizomes, adventitious roots) - Asexual reproduction in animals → fragmentation, budding, parthenogenesis: females produce offsprings without fertilization from sperm - Thelytoky → female offspring are produced - Arrhenotoky → male offspring are produced - Sexual Reproduction - Sexes in plants - Monoecious → flower of both sexes in plant - Dioecious → male and female flowers are on different plants - Hermaphrodites → the flower has male and female parts (but not much selfing occurs) - Sexes in animals - Monoecious hermaphrodite → species with male and female sex organs - Sequential hermaphrodite - Dioecious → individuals of separate sexes - Some will not exhibit sexual dimorphism (when sexes look different) - Sequential hermaphrodites - Change from male to female or female to male based on a series of cues (olfactory, visual, auditory, tactile)

- Size advantage model - Protrandry: male to female - Protogyny: female to male - Why sex? - Costs → time spent, male-male competition, courtship, rejection, natural enemies, STDs - Benefits → diversity of offspring (chances for gaining beneficial alleles, and losing negative ones) through sexual reproduction Sexual Competition - Mating Systems - Monogamy → one partner - Polygamy - Polygyny → male mating with many females - Polyandry → female mating with many males - Courtship behaviour and mating - Visual, auditory, chemical, and tactile - Habitat influences which cues are used during mating - Sexual selection - Females are generally the choosier sex (but there are exceptions) - The sex that invests more in offspring (higher parental investment) is the choosier sex - Can also be dependent on ecological situations (role reversal) When females choose - Male strategy → mate as often as possible (best strategy to leave his genes) - Male-male competition - A cost of sexual reproduction - Post-mating sperm competition - Second mating increases reproductive success of female - Second male may not sire any progeny (competition can dependent on sperm quality) - Could be that female chooses (has not been ruled out) - Evolution of adaptations to reduce sperm competition - Increased ejaculate size - Remove sperm from previous mating - Mate guarding How do females choose? - Visual signals (eg. size) - Nuptial gifts - Good gene hypothesis and symmetry (more symmetrical implies better genes)

- MHC (group of genes on cell surfaces) - Different MHC genes (fitness against parasites varies based on MHC genes) - Goal is to obtain a diversity of MHC genes for protection of progeny against disease Question: You are excited to discover a new species of fish. For males in this species, body size does not influence fitness. 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 Cycle 6 - Families to Populations On multiple choice exams, should be able to identify (lecture 11) 1. How the dominance status of an allele that IS NOT related to fitness influences its frequency 2. How the dominance status of an allele that IS related to fitness influence its frequency 3. Whether selection always results in evolution 4. How heterozygote advantage and heterozygote disadvantage affect allele frequencies and genetic variation 5. Expected genotype frequencies in a population that is in Hardy-Weinberg equilibrium (HWE, also called genetic equilibrium) 6. The conditions that must be met for HWE to occur 7. Whether or not a population appears to be in HWE at a particular locus, given a set of observed genotype or allele frequencies 8. What it means for a population to be in HWE at a particular locus, and what it means if the population is not in HWE at a particular locus Fundamental definitions - Allele - Slight variations of the same gene → different genotype/phenotype - Locus - Location of a gene on a chromosome - Gene pool - Collection of all possible alleles for a particular gene

Mendelian inheritance - Key points - Use punnet squares - Predicts offspring geno/phenotype - Assumes - Population “in equilibrium” - Random crosses - No selection for/against anything! - Therefore… not accurate in real populations Equilibrium versus Selection Genotypic and Allele Frequency - Genotype frequency → how common is that genotype in the entire population - Allele frequency → how common is that allele in the entire population Formula: number of allele or genotype / total number

Genotype and Allele Frequency (drawing a genotype and allele frequency table) Hardy-Weinberg Equilibrium - HWE → allele frequencies stable, no microevolution How to know: - Cross the calculated allele frequencies - If output genotype Frequency = original frequency, → HWE What Disrupts HWE? So what’s the “formula”...? - The Hardy Weinberg principle → → → Flower population – HWE?

Mathematical Relations of HWE - P= frequency of one allele (usually dominant) - Q= frequency of the other allele (usually recessive) If you square “p” or “q” - Genotype frequency of p- or q- homo zygous If you calculate 2*p*q ( or 2pq) - Genotype frequency of hetero zygous If you square-root a homozygous genotype frequency… - → frequency of that allele → (p+q) must equal 1 On multiple choice exams, you should be able to identity… - lecture 12 1. Effect of heterozygote advantage or disadvantage on genetic variation 2. Relative fitness for each genotype, given a set of absolute fitnesses 3. What type of selection is operating, given a set of relative fitnesses 4. The effect of different types of selection in maintaining or decreasing genetic variation 5. How genetic drift can affect allele frequencies even in the absence of bottlenecks or founder events 6. How non-random mating affects the distribution of genotypes, and whether it affects allele frequencies 7. Which assumptions of Hardy-Weinberg equilibrium have likely been violated, given an observed set of genotype or phenotype frequencies (i.e. which evolutionary agents could be acting on a population based on deviation from HWE) Quantifying fitness Fitness: determined by number of surviving offsprings that survives long enough to reproduce - Absolute fitness (W): number of surviving offspring (that reproduces) for each genotype - It is a measurable quantity - Ex. number of eggs - Relative fitness (w): absolute fitness divided by absolute fitness of the most fit genotype (W/Wmax) - Most fit genotype has w=1 - All other have w<1 (W/Wmax) - Example from lecture: 3 different pigs ith different genotypes - Which one is the most fit? - What is the relative fitness of BB?

Types of selection - Compare values of relative fitness to determine: ● w YY < w Yy > w yy → heterozygote advantage ● w YY = w yy > w Yy → heterozygote disadvantage ● w YY = w Yy > w yy → dominant advantage ● w YY = w Yy < w yy → recessive advantage ● w YY = w yy = w Yy → no selection - Why is heterozygous equal to dominant in the bolded types of selections? - Selection acts on the phenotypes of individuals - how they look - Similar to Mendelian genetics - Note: Homozygous Dominant YY and Heterozygous Yy yellow peas look the same - Selection acts on phenotypes, not genotypes - They are different genotypes, but the same phenotypes. - Example: if I were an animal and I only liked eating yellow peas, it doesn’t matter whether it’s heterozygous or homozygous dominant, the yellow pea is subject to the same selection forces. - Recessive disadvantage will not completely eradicate the allele since it can hide in heterozygotes, once again. - Heterozygote fitness advantage - chosen over both homozygous phenotypes, whether dominant or recessive Dominance status, fitness, and frequencies - Dominance status of an allele that is not related to fitness does not affect its frequency over time - Dominance status of an allele that is related to fitness affects its frequency over time Selection against a dominant allele: w AA = w Aa < w aa - Dominant alleles will be removed from the population after some time - No presence of homozygous/heterozygous dominant phenotype, only the presence of homozygous recessive phenotype Selection against a recessive allele: w AA = w Aa > w aa - The frequency of a recessive allele decreases but never disappears completely; it can hide in heterozygotes - Frequency of recessive phenotype will occur occasionally but will be removed quickly due to selection Heterozygote Advantage vs Disadvantage - Heterozygote Advantage: w AA < w Aa > w aa - Allele frequencies will stabilize near 0.5 - Once frequencies stabilize, selection is still occurring but evolution is not

- Genetic variation maintained (balancing selection) - Heterozygote Disadvantage: wAA = waa > wAa - More common allele frequency increases to 1, less common allele vanishes - Rare allele disappears not because its harmful, but they are most likely to be found in heterozygotes - Genetic variation decreases Quantitative selection - Remember: most phenotypes are quantitative phenotypes - Physically expressed with a distribution in population - Example: weight, height, skin tone etc. - Qualitative phenotypes: can be described as a category - Example blood type - Types of selection include: - Directional selection - Stabilizing selection - Disruptive selection - Balancing selection Does selection always = evolution? - No! - Selection is differential survival or reproductive success - It is one of many causes of evolution - Evolution means a change in heritable characteristics over multiple generations - How can I tell if evolution is occurring?? - Allele frequencies are changing - Can select for certain alleles without changing the overall frequencies - Ex. Heterozygote advantage Non-random Mating Mating type Increases homozygosity Increases heterozygosity Assortative mating (inbreeding) ✓ Dissociative mating (inbreeding avoidance) ✓ - Non-random mating does not change allele frequencies, just genotype frequencies - Based on HWE calculations, you should be able to list what mating types may be occurring

Non-Random mating - Consequences of assortative mating - Increase in homozygosity - Increases probability of harmful recessive alleles to be expressed - Inbreeding depression - Consequences of disassortative mating - Increase in heterozygosity - Can work with selection to select advantageous phenotypes Genetic Drift - Random sampling error - an error that occurs in small populations - Allele frequencies change due to chance - Population bottlenecks - Founder effect - Over many generations, one allele may completely disappear or completely take over - Unpredictable - Eg. if one allele increases in frequency and one decreases in frequency, you cant predict what will happen in the next generation - Reduced variability - Some alleles become more common or less common Phenotype advantages Type of Advantage Equation form Effect on A Effect on a Type of Selection… Dom. advantage w AA = w Aa >> w aa Nears freq of 1 Nears freq of 0 Directional —one extreme favoured Rec. advantage w AA = w Aa << w aa Extinction (goes to 0) Fixation (goes to 1) Directional —one extreme favoured Heterozygous advantage w AA << w Aa >> w aa Near freq of 0.5 Near freq of 0.5 Balancing + stabilizing middle ground favoured Homozygote disadvantage w AA >> w Aa << w aa Commoner allele fixates. Rarer allele disappears. Disruptive —two extremes favoured. No middle ground

HWE Question: In corn, yellow kernel colour is governed by a dominant allele; white, by its recessive allele. A random sample of 1000 kernels from a population that is in Hardy-Weinberg Equilibrium reveals that 910 are yellow and 90 are white. What is the frequency of the yellow allele in this population? - If you square-root the genotype frequency of any homozygous genotype, you get the frequency of the allele that makes up that homozygous genotype. - Let’s call the dominant yellow gene, the p, and the recessive white, the q in this scenario. So, what is q, or the frequency of the recessive white allele? - We’re not given any regular genotype information here, and we cannot construct a genotype/allele frequency table, since we don’t have the exact number of heterozygous individuals, who would also be yellow. We know that the homozygous dominant and heterozygous add up to 910, but that’s all we know. When we’re missing the number of heterozygotes, we cannot construct a frequency table - Instead, we know that 90 are homozygous recessive, and are thus white. We know if we can just find the homozygous recessive genotype frequency, we can simply square root it to obtain the recessive ALLELE frequency. Notice how since we square-rooted the recessive genotype, we get the recessive allele frequency, not the dominant. - There are 90 recessives in 1000 individuals, so our frequency is 90/1000, or 0.09. This is our recessive genotype frequency. - If we square root this number (0.09 serves as our q-squared), we would just get the q. sqrt(0.09) gives us 0.3, so that is our allele frequency for q. - Since we are interested in the frequency of the yellow allele (p), we can use the formula p + q = 1; where p = 1 - 0.3 - Therefore, our final answer for the frequency of the yellow allele (p) in this population is 0.7 HWE Question: Which of the following hypothetical populations is in Hardy-Weinberg equilibrium at the A locus? In each case, population size is 100, and numbers refer to observed numbers of each genotype. 1. 36AA, 48Aa, 16aa 2. 25AA, 50Aa, 25aa 3. 1AA, 18Aa, 81aa 4. 40AA, 20Aa, 40aa a) 1, 2, & 3 only b) 1 and 3 only c) 2 and 4 only d) 4 only e) All 1, 2, 3, and 4 are correct - The question asks if each situation is in HWE. For each population, try to do this: - HWE is attained if the genotypes you get using these frequencies match the ones given already. So, calculate the genotype frequencies for each population. - Calculate the allele frequencies for A and a, by adding up each respective allele and dividing by the total number of alleles present in a population. - For the next step, let’s assign p to dominant allele A, and q to recessive allele a. - Either do a monohybrid cross with each allele frequency, or use the relation… - Square p → genotype freq of homozygous dominant (AA)

- 2 * p * q → genotype freq of heterozygous (Aa) - Square q → genotype freq of homozygous recessive (aa) - Compare these genotypes, with the original ones calculated in Step 1. - Any discrepancy hints that the population is NOT in HWE. Population 4 falls under this, therefore A is true. HWE Question: In snails, shell texture is determined by a single gene. An allele for rough shells is dominant over an allele that results in smooth shells. For a large population of 500 snails, 20 have smooth shells. If this population is in hardy-weinberg equilibrium, what would be the frequency of rough heterozygote snails? a) 0.98 b) 0.80 c) 0.64 d) 0.48 e) 0.32 - This one is super tough! This population is in HWE, and we want to find the heterozygote frequency, and mathematically, (because the population is in HWE, we can use the relations we talked about), that is equal to (2pq). But what is p and q? - The rough shell is dominant, so let’s call that our p-allele. The soft shell is recessive, so it gets the q, but you can flip them if you want. - We have 20 smooth snails out of 500, and we know that if we can find the genotype frequency of the homozygous recessive, we can square-root it to get the allele frequency of the recessive allele, q. - The genotype frequency of the homozygous recessive is 20/500, or 0.04. Square-rooting this gives us 0.2 which is our q. But how do we get p? - The other relation is that p and q add up to 1. Naturally, p = 0.8 as q = 0.2. Finally, to get the heterozygote frequency, we do 2pq… We get 0.32. Question: Consider two alleles, L1 and L2, at the L locus controlling whisker length in a population of mice. There are millions of mice in this population, so we can ignore the effects of genetic drift. The starting frequencies of alleles L1 and L2 are 0.8 and 0.2, respectively. Under which of the following scenarios will L1 eventually reach a frequency of 1 and L2 eventually, disappear completely? 1. wL1L1 > wL1L2 > wL2L2 2. wL1L1 > wL1L2 < wL2L2 3. wL1L1 > wL1L2 = wL2L2 4. wL1L1 < wL1L2 > wL2L2 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. - 1 and 3 are due to dominant advantage, causing allele 2 to disappear completely - 2 also removes L2 to be removed completely because L2 is a rare allele (small frequency value)

Question: Consider two alleles, A1 and A2, at the A locus in a population of ducks. This population is so large that we can ignore the effects of genetic drift. Which of the following scenarios will eventually result in allele A1 reaching a frequency of 1 and allele A2 disappearing completely? 1. wA1A1 = wA1A2 = wA2A2 2. wA1A1 = wA1A2 > wA2A2 3. wA1A1 < wA1A2 > wA2A2 4. wA1A1 > wA1A2 > wA2A2 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. - Recall that if A1 is at a relative frequency of 1, it is the most fit of the genotypes. - 4. Is an example of the disadvantage of A2 - the only one. See equal signs and greater than. notice that A1 is more selected for, even within the heterozygous individual. - The question also does not state which allele is dominant, but it doesn’t matter in these questions. What is important is that A1 is a preferred allele in this question, so any allele that isn’t A1 will be selected against. - FOR 2. Allele A2 in heterozygous species is masked as a recessive trait, and doesn’t completely disappear - FOR 4. Allele A2 will disappear cuz the homozygous dominant always overpowers heterozygous Question: In a hypothetical population of badgers, claw length is determined by the C locus. Badgers with genotype CC have long claws; badgers with genotype Cc have medium claws; and badgers with genotype cc have short claws. In a colony of 50 badgers, 32 individuals have long claws (genotype CC), 16 have medium claws (Cc), and 2 have short claws (cc). Which of the following processes is most likely to be occurring at this locus? a) gene flow (the two cc individuals have probably immigrated from another colony) b) heterozygote disadvantage c) selection favouring the c allele d) random mating Cycle 5 - Genetics and Inheritance Homologous Alleles - Alleles → variation in genome sequence - Homologous allele → same locus - Know where alleles are located on a single or replicated chromosome - Would cross over at this location - Homologous chromosomes can carry the same or different alleles → heterozygosity or homozygosity - Dominant allele: only need one copy of allele to express phenotype - Recessive allele: need both copies of allele to express phenotype

Blending theory - Offpsrings are a blend of parental phenotypes (NOT TRUE) Mendelian genetics - People thought inheritance was through blending - Mendel carefully studied pea plants - Height, flower colours, seed characteristics - Crossed plants and analyzed them - quantitative Mendel’s model - Variation in traits due to different alleles - Allele for blue eyes vs brown eyes - Law of segregation: alleles segregate randomly into gametes - 50% chance that either alleles end up in a gamete - Organisms inherit two alleles for each trait - One paternal and one maternal - Appearance of heterozygots determined by dominant alleles - Noticed despite having one of each allele, consistently only one showed - Proved blending theory false Question: One of Mendel’s most powerful tools was the dihybrid cross involving independently assorting genes. The cell at right is from such a dihybrid individual in meiosis 1 with the genotype Tt Yy. the location of the T allele is shown. Which of the following statements about the labels on this diagram are correct? 1. B shows the likely location of a Y allele 2. D shows the likely location of a t allele 3. C shows the likely location of a T allele 4. A shows the likely location of a t allele a) 1, 2, & 3 only b) 1 and 3 c) 2 and 4 d) 4 only e) 1, 2, 3, & 4 Question: Mendel crossed homozygous tall plants having smooth seeds (TTRR) with homozygous short plants having wrinkled seeds (ttrr) to create F1 hybrids (TtRr). Crossing these hybrids together revealed that the alleles of these two genes show independent assortment If we pretend that chromosomes remain condensed throughout the cell cycle, which of the following diagrams best illustrates the location of alleles in the TtRr hybrid in G2 of the cell cycle? a) A b) B c) C d) D

Question: You and your partner are attempting to figure out the possible phenotypes for your child. You have sequenced both of your genomes and have received the following results: Hair colour: B is blonde, b is black Eye colour: G for green, g for gray Height: T for tall, t for short Your partner wants a baby with blonde hair, green eyes, and is tall. What is the chance of this happening? a) ⅛ b) 3/16 c) 1 d) 1/3 Question: You and your partner are attempting to figure out the possible phenotypes for your child. You have sequenced both of your genomes and have received the following results: Hair Colour: B is blonde, b is black Eye Colour: G for green, g for gray Height: T for tall, t for short Your partner wants a baby with blonde hair, green eyes, and is tall. What is the chance of this happening? a) ⅛ b) 3/16 c) ¼ d) 27/64 Dihybrid Punnet Squares

Incomplete Dominance - Heterozygotes have intermediate expression Codominance - Both alleles are equally expressed in heterozygotes Sex linked - May be carried on either the X or Y chromosome - Most commonly on the X chromosome - Pay attention to how mutation may affect male offspring Epistasis - Non-mendelian inheritance - One gene interfered with expression of another gene expression - 9:3:4 phenotypic ratio Epistasis Question: The figure below shows a hypothetical pathway controlling fur colour in wild bears. The active allele of Gene B codes for an enzyme that converts brown pigments to black pigments. The active allele of Gene C codes for an enzyme that degrades all pigment. Inactive alleles code for non-functional enzymes with no effect on phenotype. What will be the phenotypic distribution of offspring from a cross of BbCc x BbCc? a) 12 Colourless: 3 Brown: 1 Black b) 9 Colourless: 4 Brown: 3 Black c) 12 Colourless: 3 Black: 1 Brown d) 9 Colourless: 7 Black

Question: There are many genes involved in determining whether a pumpkin will be shaped like a disk, or a ball, or elongated. Two such genes are of particular interest to Halloween pumpkin growers: we will call them “disk, D” and “zucchini, Z” If a grower makes a dihybrid cross and observes a ratio of 9 disk: 6 ball: 1 elongated among the offspring, what woud be the predicted shape of a squash with the genotype DDzz? a) There is no way to predict b) Disk c) Elongated d) Ball Question: 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 the 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. Assuming that all the genes assort independently, what is the likelihood that your first bord child will develop as a boy and have blood type MN and have wet ear wax? a) 1/8 b) 1/16 c) 3/32 d) 1/64 e) ⅜ - You could equally do an 8x8 punnet square but making 4 diff ones and multiplying them is faster Question: The colour of your hair is controlled by epistasis. The production of black pigment is controlled by gene B. The production of black pigment is blocked by the gene product of gene C and when this occurs, the hair will be white. When homozygous recessive for both B and C, there is no hair growth. If you and your partner have BbCc genotypes, what is the proportion of children with black hair, white hair and no hair? a) 9:3:3:1 b) 12:3:1 c) 10:4:2 d) 9:6:1

Cycle 4 - DNA Mutations and Repair DNA repair mechanisms → how does DNA get damaged? → how does DNA get repaired? Sources of DNA damage - Exogenous Sources → coming from OUTSIDE the cell - UV sun - Chemicals - IR - Endogenous Source → coming from inside the cell - Cell metabolism (ROS) - Replication Errors Mutations vs. DNA Damage - Mutation → change in the double stranded DNA sequence - Genetic variation refers to mutations, not DNA damage - Effect of mutation is dependent on its location - Substitution, indel, inversions - DNA damage → any change to DNA that is not double stranded - Can lead to mutation Reactive Oxygen Species (ROS) - Very electronegative and unstable - ROS will take electrons away from any sort of molecule nearby to regain stability (targets DNA) Oxygen paradox - While oxygen is crucial to many cell functions, too much of it can ultimately lead to cell death - Causes oxidative stress - Thankfully, antioxidants (either naturally occurring in cells or gained through diet) help our cells combat the effects of ROS DNA proofreading during replication 1. DNA polymerase recognizes its made a mistake 2. Recognition of damage can occur when DNA polymerase detects distortion in the DNA backbone 3. DNA polymerase moves backwards and removes the incorrect base through its 3’-5’ exonuclease activity (can remove bases one at a time from the end of the chain) 4. DNA polymerase moves forward again and puts the correct base

Mismatch (DNA Pol.)/Excision (exogenous damage) Repair 1. Repair enzymes detect the distortion 2. Mismatch/Repair enzymes bind to the damages region and excise (cleave) the backbone through endonuclease activity 3. DNA polymerase fills in the gap 4. Ligase seals the nick Thymine Dimers and Photolyase - Thymine dimers form under UV light (damage), causing two T bases to covalently bond together (forming a dimer) - Repaired by photolyase + white light - Mammals lack photolyase → excision repair Non-homologonous end joining (NHEJ) - Double-stranded breaks often caused by ROS, or ionization - Non-homologous end-joining (NHEJ) is the repair mechanisms for double-stranded breaks - Cell does not use a template so it is more likely (but not guaranteed) to make error - Sloppy, can result in mutations Most of your DNA is junk! - 90% → junk - 10% → essential (2% → coding) Different forms of DNA damage and their repair mechanisms & Summary

Question: Imagine that a malfunctioning X-ray machine accidentally exposes living cells on your hand to excessive ionizing irradiation. Which of the following consequences is most likely? a) Mutations will likely result from repair of double-strand breaks in chromosomes b) The cell will arrest at the metaphase cell cycle checkpoint c) Photolyase will go to work repairing thymine dimers d) Transposable elements will randomly start inserting into various coding regions - We learned that ionizing radiation can lead to double-stranded breakages in DNA. These are repaired by NHEJ, which often leads to mutation since the cell sloppily tries to reassemble the broken DNA strands without a template Question: which of the following mechanisms of DNA repair does not involve creating new covalent sugar-phosphate bonds in the backbone of DNA in order to restore proper base pairing? a) Non-homologous end joining repair (NHEJ) of double-strand breaks b) Excision repair of mismatched bases c) Proof-reading of mismatched bases d) Photolyase repair of thymine dimers - photolyase is able to remove the double bond that makes up the dimer, unlike with excision repair (since excision repair only occurs in organisms where photolyase is absent) Genetic Variation due to mutation - Types of mutation Point mutations - Silent → the base change will result in a change in the mRNA, however, it will code for the same amino acid (recall that there are multiple codon variations for the same amino acid) - Missense → the base change will result in a change in the mRNA and will change the amino acid it codes for. This could or could not cause an issue. It depends on where it occurs and the type of amino acid that replaces the original amino acid. - Nonsense → the base change will result in a change in the mRNA; it will code for a premature stop codon. This could potentially lead to a non-functional protein.

Single nucleotide polymorphisms (SNPs) - Single nucleotide changes “in pairs” - Most common type of genetic variation between individuals - Occur mostly between genes - Analyzing SNPs lets us predict an individual’s - Response to drugs - Susceptibility to environmental factors - Risk of developing particular disease InDel Mutations - Insertion due to backward slippage in the synthesized strand - Deletion due to forward slippage in the template strand

Question: Imagine you’re a chemist and you have synthesized a new compound for colouring fabric. However, during testing you realize that your new compound stabilizes the loop in the structure shown Which of the following types of genomic variation will your new compound likely promote? a) Aneuploidy b) Indel c) SNP d) Translocation - A key indicator that this would be an indel mutation is that loop. A forward or backward slippage can cause a loop to form, resulting in either an insertion or deletion of base pairs. Question: one of the mechanisms that leads to DNA mutations is a process known as DNA polymerase slippage which occurs during DNA replication. Which of the following statements about DNA polymerase slippage is correct? a) Backward slippage will lead to an insertion mutation b) DNA polymerase slippage often leads to a point mutation c) Forward slippage will lead to DNA polymerase reading more base pairs than it should d) None of the above - When Polymerase slips backwards, this leads to an insertion mutation on the synthesized strand as it doesn’t realize that the bases have already been copied. Question: Single nucleotide polymorphisms (SNPs) are the most common type of genetic variation among people. Which of the following mechanisms can result in the formation of SNPs? 1. Mismatch during DNA replication. 2. Slippage during DNA replication. 3. Formation of tautomers. 4. 3’ to 5’ exonuclease activity of Polymerase 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. - Mismatch during DNA replication occurs when DNA Polymerase makes a mistake and adds a non-complementary base, creating a distortion in the backbone. For example, it might copy a G where a T should be. This can lead to a single base nucleotide difference in the DNA sequence. Role of tautomeric shifts in mutagenesis - Key concepts to remember 1. Tautomeric shifts are spontaneous events 2. Tautomeric shifts DO NOT results in a mismatch … it is not recognized by the cell as a mismatch … there is just a shift in preferential base pairing - Thymine and guanine are mainly in KETO form … but it can shift - KETO → ENOL (Enol T-G) - Adenine and Cytosine are mainly in AMINO form … BUT it can shift - AMINO → IMINO (Imino A-C)

Tautomeric Shift - DNA polymerase reads one strand at a time, with no way of knowing what’s on the complementary strand or what was originally in the parent strand! Mutagens as Tautomerically Unstable Base Analogues - DNA polymerase 3 can insert unstable analogues in the place of correct nucleotides - “Unstable” because they are more prone to tautomerization - The dominant form of 5BU is its KETO form, where it pairs with an A - The rarer form of 5BU is its ENOL form, where it pairs with a G Base Analogue Incorporation + Tautomeric Shift

Transition vs Tranversion Mutations - Transition mutation - Purine to Purine - Pyrimidine to pyrimidine - Transversion mutation - Purine to pyrimidine - Pyrimidine to purine Transposable Elements as Biological Mutagens - What are they? - Short regions of DNA that are able to move (“jump”) around the genome - How were they discovered? - The first TEs were discovered in Maize corn. The gene that codes for purple pigment is inactivated by the movement of TE so we get no purple - Where can we find them? - Typically, TEs are found in “safe havens” between genes - How many TEs can a genome have? - Many. depends on the organism, but TEs are one reason why our genomes are so large! - Implication? - Genome size does not necessarily reflect the number of henes an organism has - Alive or dead? - Most are silenced/deactivated by most and no longer undergo transposition. A few are still active. Fewer still interfere with genes and can cause disease or gene shuffling Multifactorial Diseases - What is a multifactorial disease? - A disease that can be caused by a variety of factors, both environmental (exercise, diet, lifestyle) and genetics (SNPs) - Example: - SNPs rs10757274 and rs 2383206 double the risk of heart disease - However, lifestyle, diet, and exercise play a role in disease prevention and onset

Question: Recall that the “normal” tautomeric form of 5 Bromo-uracil (5BU keto) behaves as a base analogie of thymine which is incorporated into growing DNA chains by DNA polymerase. Imagine a normal cell that now has a 5BU keto present in its DNA structure. If, during the next round of replication, the 5BU keto switches to its abnormal tautomeric form 5BU enol, which preferentially base pairs with G, which of the following statement is correct? a) In the third round of replication, we will see a parent cell with a transversion mutation producing 2 daughter cells each having a transversion mutation b) In the third round of replication, we will see a parent cell with a transition mutation producing 2 daughter cells each having a transition mutation c) In the second round of replication, we will see a parent cell with a transition mutation producing 2 daughter cells each having a transition mutation d) In the second round of replication, we will see a parent cell with a transversion mutation producing 2 daughter cells each having a transversion mutation Question: the most common transposable element in humans is the Alu sequence. It is approximately 300 base pairs long and it is estimates to make up about 15% of the human genome. Based on your knowledge of transposable elements in humans, which of the following conclusions can you make? a) If Alu is often inserted into exons of coding regions, deleterious effects can occur in the cell b) The Alu transposable element is very mobile in humans c) The Alu transposable element is most likely a retrotransposon d) Most humans have about 100 copies of the Alu transposable element Cycle 3 - Inheritance of Sameness - How is DNA organized in cell nuclei over the course of the cell cycle?

- How is DNA organized in the cell nuclei over the course of meiosis? Taking a look at n vs C - “N” identifies how many nuclear chromosomes are unique - Coefficient of n identifies how many sets of n there are Notes: 1. Total number of chromosomes = number of centromeres 2. N does NOT change during mitosis 3. Coefficient of n is also known as ploidy - “C” = amount of DNA (pg or base pairs) for 1n - Coefficient of c = number of copies of the entire genome Notes: 1. The word “genome” is synonyms with 1C! 2. Tip: the coefficient of C can be found by continue the number of chromatids per homologous grouping 3. The value of C cannot be determined just by looking at a karyotype Chromosome close up

Question: Consider a tetraploid organism with 6 chromosomes per set. During prophase, what is: a) The coefficient of n? 4n b) The coefficient of C? 8c G1 S G2 P M A T C N 4 4 4 4 4 4 4 4 C 4 8 8 8 8 8 8 4 Question: the image at the right shows all of the chromosomes in a dividing wing cell of an insect. What is the coefficient of C for this cell? a) 12C b) 6C c) 4C d) 2C - May be initially difficult to tell if this insect is diploid or triploid. However, the two chromosomes on the far right are distinct in shape from the others. Since there are 2 chromosomes with the unique “y” shape, we can reasonably conclude that the insect is diploid with 2n=6. - To determine the coefficient of C, count the number of chromatids per homologous pair. There are 4 chromatids per homologous pair, therefore this cell is 4C. Question: bread wheat is one of the “big three” globally important crops, accounting for 20% of the calories consumed by people. Leaf cells of a wheat plant are hexaploid (6n). That is, somatic cells contain 6 copies of each of the species’ 7 chromosomes. One set of wheat chromosomes contains 17 billion base pairs; over five times more than one set of human chromosomes! Which of the following statements about a wheat karyotype is true? 1. The cells used for karyotype analysis would contain 84 chromatids 2. The cells used for karyotype analysis would be 12C 3. The cells used for karyotype analysis would have passed the G2/Mitosis cell cycle check-point 4. The cells used for karyotype analysis would be undergoing apoptosis a) 1, 2, & 3 only b) 1 & 3 only c) 2 & 4 only d) 4 only e) 1, 2, 3, & 4 are correct - Plant is 6n → meaning 6 sets of unique chromosomes and each set has 7 unique chromosomes - Therefore, the plant before the S phase has 6 sets x 7 chromosomes = 42 chromosomes - Karyotype is taken during metaphase, therefore, the 42 chromosomes are replicated and now we have 42 chromosomes x 2 chromatids/chromosome = 84 chromatids → thus 1 is correct - The plant has 6 sets of unique single-stranded chromosomes (6C). During S phase, each chromosome is double-stranded → the coefficient of C likewise doubles (12 C) → 2 is correct - Karyotype made in metaphase. For cell to reach that point it passed G2/M checkpoint → 3 correct - The cell is not undergoing apoptosis since it has passed all the checkpoints → thus 4 is incorrect

n and C have no correlational relationship - The n-values of 2 organisms could be the same and their C-valyes vastly different - n 1 = n 2 - C 1 >>> C 2 - The n-values of 2 organisms could be very different and their C-values the same - n 1 >>> n 2 - C 1 = C 2 - A larger n-value does not mean a charger C-value - n 1 > n 2 - C 1 < C 2 C-value paradox - Despite observations that the C-value of an organisms tends to increase with its “complexity”, this is not always true - Less complicated organisms CAN have larger genomes than more complicated organisms - There is debate over what “complexity” means Question: the “C-value enigma” has perplexed genome scientists for decades. Which one of the following questions best summarizes this enigma? a) Why do most prokaryotes have higher C values than some eukaryotes? b) Why do some cells of a given organism have 1C of DNA while other cells have 2C or even 4C? c) Why do some species have dramatically different C values than related species? d) Why is the C value of a species not related to the number of chromosomes in its karyotype? Graphs to know!! Question: imagine a cell entering the cell cycle to undergo mitosis. Which of the following graphs illustrates the correct amount of DNA during each phase of the cell cycle? - Correct answer is A - identify that the cell is undergoing MITOSIS, meaning that the cell will start and end as diploid, not haploid as it would it during meiosis - The amount of DNA doubles during S phase as replication occurs - Stays 4C until the end of mitosis, then goes back to 2C

Question: Which set of axis labels would be best suited to this graph? a) Y-axis “n-value → X-axis “phases of mitotic cell cycle” b) Y-axis “ploidy” → A-axis “phases of meiotic cell cycle” c) Y-axis “Coefficient of C” → X-axis “phases of meiotic cell cycle” d) Y-axis “coefficient of C” → X-axis “phases of mitotic cell cycle” - Mitotic cell cycle (ploidy does not change, and only one reduction occurs) DNA: The basics - Made up of 4 nucleotides - Purines = Afenine (A) and Guanine (G) - Pyrimidines = Thymine (T) and Cytosine © - Base parings - A-T; 2 H-bonds - C-G; 3 H-bonds - Phosphodiester bonds link nucleotides together - DNA has a direction - 5’end → phosphate group - 3’ end → OH group - DNA runs antiparallel A replication is Semi-conservative - Helicase unwinds the double helix to separate the 2 parent strands - Parent strands becomes a template for synthesis of a new strand - Happens during S phase before mitosis Question: the figure at the right is a cartoon illustration of the replication of one pair of homologous chromosomes (4) during S-phase in a chicken cell. Look carefully to notice that each vertical line represents one backbone of a double helix. Aminopurine is a base analogue that is incorporated into elongating DNA strands (instead of Adenine) by DNA polymerase. Imagine that our chicken cell is fed aminopurine during S-phase. Which of the following illustration most accurately represents the expected distribution of aminopurine (*) in replicated chromosomes following S-phase? a) A b) B c) C d) D

DNA polymerase - DNA polymerase 3 extends primers by adding nucleotides to form new DNA complementary to template strand - Synthesizes new strand 5’-3’ - Reads template strand 3’-5’ - Polymerase 1 cleaves off RNA primer and replaces it with DNA nucleotides - 5’-3’ exonuclease activity - U → T Question: Recall that, during DNA replication, RNA Primers are removed and replaced with DNA through the actino of DNA polymerase 1. Consider an RNA primer with the sequence 5’ AUGCCAUC 3’. Which of the following sequences of DNA would replace this RNA primer? a) 5’ ATGCCATC 3’ b) 5’ CTACCGTA 3’ c) 5’ TACGGTAG 3’ d) 5’ GCTGGCAT 3’ - When the RNA primer is replaced with DNA, uracil is replaced with thymine. So the DNA strand would be the same as the RNA, but with T instead of U. Continuous vs. Discontinuous - Leading strand = continuous replication towards the replication fork - Lagging strand = discountious replication away from replication form (Okazaki fragments) ** remember that both leading and lagging strand begin with an RNA primer! However, only the lagging strands have okazaki fragments Replication bubbles - Multiple origins (bubbles) increase replication efficiency - Each replication origin has 2 replication forks travelling in oppositive directions - Any one particular strand of DNA is replicated continuously at one fork but discontinuously at the other fork Filling in the gaps!! - DNA polymerase 1 → removes and replaces the RNA primers - DNA ligase: fills in the nicks left after removal of the primers Replication bubbles

Question: the image shows a partially completed stretch of a replication bubble at the Ori. Dark solid lines indicate primers and dashed line indicated newly synthesized DNA. What os the following numbers represents a 3’ end of DNA or RNA? a) 1, 2, 3 b) 1 and 3 c) 2 and 4 d) 1, 2, 3, and 4 Question: as their name suggest, both DNA polymerase 1 (Pol1) and DNA polymerase 3 (Pol3) create polymers of DNA (by adding new nucleotides to the growing chain). In which of the following ways does the activity of Pol1 differ from Pol3? 1. Pol1 reads template DNA 3’ to 5’ while Pol3 reads template DNA 5’ to 3’ 2. Pol1 is only active in lagging strand synthesis; Pol3 is active in both lagging and leading strand synthesis 3. Pol1 is active in creating Okazaki fragments, Pol3 is not. 4. Pol1 can add new bases to DNA only; Pol3 can add new bases to RNA and/or DNA a. 1, 2, &3 b. 1 & 3 only c. 2 & 4 only d. 4 only e. 1, 2, 3, & 4 Cell Senescence - Telomeres are repeating TTAGGG sequences at the end of your chromosomes - They act as buffer regions - Hayflick limit: the number of times a somatic cell can divide before reaching the end of the TTAGGG region - Cell Senescence → irreversible cell cycle arrest Telomerase - DNA replication leaves the 5’ end of nerly-synthesized strands shorter than their complementary strand - Telomerase restores the length of chromosomes (but does not prevent DNA shortening!) - Telomerase binds to the 3’ end of the template strand and extends the length of the template strand using a short RNA template built into the enzyme - RNA primer synthesied on newly extended template strand - DNA polymerase 3 extends the primer using complementary base pairing - Remember: in which cells would telomerase be active? stem cells, germ cells, cancer cells, etc. (not active in somatic cells)

Question: The image at the right shows the action of telomerase. Which arrow indicated the 5’ end of DNA? a) A b) B c) C d) D e) Both B and C are correct - Telomerase extends the 3’ end of the template strand (position B) - DNA is antiparallel, so position A must be the 5’ end of the DNA - Telomerase uses an RNA template → therefore C and D are not correct Cycle 2 - Cell Cycle & Cell Division Prokaryotes vs. Eukaryotes - Prokaryote - DNA is in nucleoid - Replication occurs through binary fission - Eukaryote - DNA is in nucleus - Replication though mitosis and meiosis Prokaryotic Cell Cycle - Binary Fission Eukaryotic Cell Cycle

Mitosis → Cell division into 2 identical daughter cells Mitosis Micrograph Example Human Cell Types

Why is cell division necessary? 1. Tissue repair 2. Multi-Cellular Growth 3. Regeneration - Maintain hight Surface area:volume - Larger cell → more volume → larger diffusion distance - At a large enough volume, SA cannot keep up with the demands of the cell - Must divide to survive Cell Cycle Checkpoints - G1/S assesses for - Cell size and DNA damage - Growth factor prpesence required to enter synthesis (S stage) - G2/M assesses for - Proper chromosomal replication - DNA damage - Molecules required to enter mitosis - Mitotic spindle assesses for - Attachment of spindles at kinetochores - Chromosomes are lined up evenly on metaphase plate - Mitigates aneuploidy (abnormal number of chromosomes) Positive Regulation - Positive regulators: Includes cell cycle - Cyclins: bind to CDK to form Cyclin-CDK complex - CDK: phosphorylation cascade intermediate Negative Regulation - Negative regulators: halts cell cycle - p53 → detects damage & upregulates p21 transcription - p21 → binds to CDK to inhibit Cyclin-CDK complex P53: Guardian of the genome

Question: Sources of genetic variation in mitosis include…? a) Recombination b) Random alignment of chromosome in metaphase c) Independent assortment d) A & B e) None of the above Question: choose the statement that is correct about cell cycles: a) Negative regulators stopping the cell cycl will not result in apostasis b) Binary fission and mitosis both produce two daughter cells that are identical to their respective parent cell c) Binary fission produces two daughter cells that are unique to the parent cell d) DNA in the nucleus of a prokaryote is replicated before binary fission e) Eukaryotes and prokaryotes have the same process relating to cell division Question: imagine you look into a microscope and see a cell that is in the S phase of the cell cycle. Which of the following statements could be correct about the cell? 1. The cell is undergoing anaphase 2. The cell have passed the G2/M checkpoint 3. The cell is receiving signals to differentiate 4. The cell is replicating its DNA a) 1, 2, & 3 b) 1 & 3 c) 2 & 4 d) 4 only e) 1, 2, 3, & 4 Question: one key idea from the lecture on cell cycling is that not all cells divide all the time. Which of the following statements correctly describes the conditions when a cell might be in G0? a) Cells thar enter G0 have different probabilities of re-entering the cell cycle, depending on what type of cell they are b) Cells enter G0 only if they have DNA damage c) Cells in G0 cannot function d) Cells in G0 most likely have high turn over rates (replicate often) Question: which of the following incorrectly describes the function of a cell cycle checkpoint? a) G1/S: checks for cell size, DNA damage, and growth factor needed for S phase b) G2/M: confirms tha nuclear membrane has dissolved and that spindle poles have formed c) M ensures spindles have correctly attached and chromosomes have lined up evenly on metaphase plate d) All of the above are correct descriptions Question: which stage in the cell cycle diagram would end at a “checkpoint” mediated by p53? a) 1, 2, & 3 b) 1 & 3 c) 2 & 4 d) 4 only e) 1, 2, 3, & 4

Question: which of the following can lead to increased cell division? 1. Loss of p53 2. Loss of CDK 3. Loss of p21 4. Loss of cyclins a) 1, 2, and 3 b) 1 and 3 c) 2 and 4 d) 4 only e) 1, 2, 3, & 4 Question: what role to cell cycle checkpoints play in the “inheritance of sameness”? 1. Checkpoints ensure that DNA polymerase does not read the wrong template strand 2. Checkpoints ensure that all kinetochores are attached to spindles before proceeding to anaphase 3. Checkpoints ensure that organelles are properly replicated 4. Checkpoints ensure that damage is repaired before DNA is replicated a) 1, 2, & 3 b) 1 and 3 c) 2 and 4 d) 4 only e) 1, 2, 3, & 4 Counting Chromosomes - One chromosome = one centromere - Sister chromatids are identical (before recombination) - Homologous chromosomes are related, but not identical Meiosis - Cell division into 4 unique daughter cells Meiosis Micrograph Example

Meiosis 1 is Reductional - Homologous chromosomes separate - The number of chromosomes is halved (46 → 23) - The cells produced are now haploid Meiosis 2 is Equational - Sister chromatids separate - The number of chromosomes is unchanged (23 → 23) - Similar to mitosis Recombination → source of genetic variation - Swapping chromosome segments in prophase 1 of meiosis Linked genes - Allele on the same chromosome are more likely to end up together - The closer they are the lower the chance recombination will separate them Genetic Variation → what are some sources of genetic variation? - Variation in meiosis - Homologous recombination - Crossing over between sister chromatids (prophase 1) - Independent assortment - Chromosomes segregate independent of one another (anaphase) - Random alignment - Chromosomes line up at metaphase plate randomly (metaphase) - Random fertilization Mitosis vs. Meiosis

Nondisjunction Causes Aneuploidy Down Syndrome Risk Increases with Mother’s Age - Incidence increases with age of oocyte - Why might this be? Life cycles of different organisms

Question: which of the following statements are correct regarding meiosis and mitosis? a) Meiosis results in four haploid daughter cells, while mitosis results in two diploid daughter cells b) Meiosis occurs in somatic cells, while mitosis occurs in gametes c) Meiosis involves only one round of cell division, while mitosis involves two rounds of cell division d) Meiosis is responsible for growth and tissue repair Question: what phase of meiosis does this micrograph show? a) Anaphase b) Telophase 2 c) Metaphase 1 d) Anaphase 2 e) Prophase Question: which one of the following statement about metaphase in mitosis or meiosis is correct? a) Homologous chromosomes are pulled to opposite poles in metaphase 1 of meiosis b) Single chromosomes are lined up at the metaphase plate in metaphase of mitosis c) Metaphase in mitosis is the same as metaphase 1 in meiosis d) There are two stages of metaphase in mitosis e) None of the above statements are correct Question: Which one of the following accurately describes the number of chromosomes in a human cell at various stages? a) Germ cell -> 23, Gamete -> 23, Zygote -> 46 b) Germ cell -> 46, Gamete -> 23, Zygote -> 23 c) Germ cell -> 23, Gamete -> 23, Zygote -> 23 d) Germ cell -> 46, Gamete -> 23, Zygote -> 46 e) None of the above are correct Question: Alleles A and B are separated by a very short distance on one homolog of Chromosome 1, while allele C is found on the other homolog. Assuming Meiosis occurs without nondisjunction, and no crossing over occurs, what is the likelihood a gamete will inherit all three alleles? a) 100% b) Less than 100% but greater than 75%. c) Less than 75% but greater than 50% d) Less than 50% but greater than 0% e) 0% Question: XYY syndrome is a genetic condition in which a male has an extra Y chromosome. Assuming the XYY karyotype resulted from a single error in chromosome partitioning, in which of the following stages of meiosis might the error have occurred? 1. Meiosis I in the mother 2. Meiosis I in the father 3. Meiosis II in the mother 4. Meiosis II in the father a) 1, 2, and 3 b) 1 and 3 c) 2 and 4 d) 4 only e) All of 1, 2, 3, and 4

Question: Which of the following statements are CORRECT regarding non-disjunction events? 1. A non-disjunction event in the 1st meiotic division will result in all gametes exhibiting aneuploidy. 2. Trisomy 21 occurs from a meiotic non-disjunction event. 3. A non-disjunction event in the 2nd meiotic division will result in 50% of the produced gametes having a normal amount of chromosomes. 4. Non-disjunction events occur due to a malfunction in prophase I or II. a) 1, 2, and 3 b) 1 and 3 c) 2 and 4 d) 4 only e) All of 1, 2, 3, and 4 Cycle 1 - Evolution is Happening Let’s get viral! Key characteristics of viruses - Viruses are not on the tree of life - Viruses are obligate parasites - They cannot reproduce on their own - They must infect a cell (host) in order to replicated - Require host machinery for metabolism - HIV is a virus and it is a retrovirus (viruses made of RNA that use RT to convert their RNA to DNA) Origins of HIV - HIV is a zoonotic disease (SIV) - Diseases that were originally present in non-human animals and transferred to humans (spillover event) - More harmful in the new host (bc it has resided in the old host longer) & more likely between closely related species Why is it so hard to eradicate viruses? - Viruses have very little of their own machinery - As obligate parasites, they rely on the host cell for their reproduction… - This means targeting a part of the virus’ cycle could also harm the host cell it hijacked, and cause even more damage to the host organism

AZT HIV treatment - HIV has a high mutation rate - due to reverse transcriptase - Lots of errors! - AZT - base analog of thymidine - no 3’ OH (N3 instead); - Blocks DNA elongation and stop action of reverse transcriptase AZT … is it a long term antiviral treatment? - The reality was that once we exposed HIV to this drug, the virus managed to become resistant… how?? - ONLY 2 nucleotide changes in reverse transcriptase… now there is a proofreading function (knows the difference between AZT and thymidine) - Antiviral drug effectiveness decreases over time How did HIV evolve resistance to AZT? - RT makes mistakes in copying HIV genome (from RNA to DNA), resulting in a mutation - Remember … reverse transcriptase is not so good at its job and DOES make mistakes - The mutation in the next generation of RT created might confer resistance to AZT or it might not - Where these mutations occur in the genome is completely RANDOM - If there are AZT resistant mutants, they would accumulate in the population as more rounds of replication and AZT treatment occur - Overtime, the genetic makeup of the virus population changes… - Resistant viruses are favoured by natural selection and make up a greater portion of the total population than before Some key concepts/learning outcomes! - Mutations are RANDOM - Mutations are always occurring, regardless of the environment - Organisms do not “mutate” in response to the environment … they are always just occurring - These mutations allow for variation in the HIV population - Some viruses become resistant, some stay susceptible, etc. - However, once the AZT treatment was applied to the HIV population, this treatment provided the right environment for non-random selection - Now in this environment, the drug resistant virions are able to survive and reproduce - They have an advantage - “Mutation proposes, selection disposes” - RANDOM mutations, Non-random selection (based on the environment) Drug cocktail therapies - Drug cocktails → using multiple drugs as therapy - Dilutes possibility of total resistance

The vaccine effort VS evolution - Viruses divide very fast - Mutations accumulate very fast - Viruses change very fast - Faster than we can catch up - Evolution - Happens over generations - Driven by mutation Question: Three flasks each contain equal numbers of bacteria, reproducing at equal rates. - Flask X → treated with the antibiotic ampicillin; - Flask Y → treated with two antibiotic drugs (ampicillin and bacticillin); - Flask Z → not treated with any antibiotic drugs. 1. Bacteria in which flask is most likely to develop antibiotic resistance through mutation? A 2. After a long time, which flask will see bacteria resistant to ampicillin only dominate? B a) All flasks are equally likely for this. b) Flask X. c) Flask Y. d) Flask Z. e) Flasks X and Y. Question: Which of the following traits is shared by all life, if we assume viruses are considered alive in this question? a) Genetic information is contained within a protein shell. b) Genes are transcribed from a RNA code to a DNA message. c) DNA or RNA genomes are copied without error or mutations. d) Copies of DNA or RNA genomes are inherited by offspring. Scientific Theories - Scientific theory must be - Testable: can be revised, and experiments can be run on it - Falsifiable: if contradicting evidence appears, can the “theory” be claimed “false”? - Examples: - “All organisms are related, and change over time” (falsifiable) - “Every organisms of species X only reproduces once per lifetime” (falsifiable) - “If i try hard enough, i can find a taylor swift eras tour ticket for <$1000” (not falsifiable/borderline “belief”) The theory of evolution, 1. - “All organisms are related, and change over time” - The “last universal common ancestor” diverged/evolved into all species of life today - New species constantly forming from previous ones - Thus → all organisms are related

The theory of evolution, 2 - “ all organisms are related and change over time” Question: Which statement agrees the most with the Lamarckian (not Darwinian) view of “evolution?” a) Nectar-sucking birds, on average, have longer beaks than 100 years ago. b) All life on Earth originated from one single species billions of years ago. c) Species change is mostly driven by external factors, not necessarily because the organisms want to change. d) Organisms can simply decide to change over their lifespan, and these purposeful changes drive species-level change. The theory of evolution, 3 - Natural selection is not evolution! - Natural selection is a process OF evolution Summary of the Theory of Evolution - Core principles of the theory: 1. Organisms change/diverge over long periods of time 2. All organisms are related, however distant 3. Evolution changes populations, not individuals 4. Natural selection is one agent driving evolution Question: Which statement describes possible evidence that could falsify the theory of evolution? a) A new type of microorganism is discovered whose cells are identical to one another and never mutate. b) There are species that undergo mutation, but their reproductive speeds are so slow that they do not appreciably change. c) A new species of frogs is discovered in one specific place on Earth, nowhere else. d) The theory of evolution can never shown to be false. Misconceptions about evolution (true or false) 1. In our modern world, evolution is occurring at a significantly lower rate than it did millions of years ago (false) 2. Individuals with a greater fitness are almost always stronger, faster, and larger, seeing that these traits are conductive to reproductive success (false) 3. Chimpanzees evolved into humans (false) 4. An individual organisms cannot evolve over the course of its lifetime (true) 5. A bacterium plated in an antibiotic medium is equally as likely to develop a mutation that confers

antibiotic resistance than bacter plated on nutrient agar (true) 6. Evolution does not result in organisms that are perfectly suited to their environment (true) Evidence for evolution (“Macro” scale) - Biogeography → similar species in different places, but a few changes have occurred - Comparative morphology → different species have similar morphological characteristics - Geology → earth has been around for a very long time! - Fossils → show that life on earth today is different than it was in the past; allows comparisons

Question: Which of the following statements is NOT part of the theory of evolution? a) Over their lifetimes, individuals evolve to become better suited to their environment. b) Over generations, populations evolve to become better suited to their environment. c) In an environment with tall trees, mutations in giraffe genomes that result in longer necks will occur just as frequently as those that result in shorter necks. d) All life on earth is related through descent from a common ancestor. Question: A school of fish becomes trapped in a dark underwater cave. Many generations later, the descendants of these fish still live in the cave, but the eyes of the descendants are much smaller than those of their ancestors. Which of the following explanations most likely accounts for the scenario above? a) The cave contained a chemical pollutant that increased the mutation rate at a gene influencing eye size. Over time, individual fish accumulated mutations and everyone’s eyes became smaller. b) Fish in the cave no longer needed to use their eyes, because the cave was so dark. Over time, due to lack of use, eyes became smaller. c) In the original school of fish, individuals differed in their eye size. In the cave, having large eyes was a waste of energy and small-eyed individuals produced more offspring. Mutations that reduced eye size were favoured by natural selection, and the frequency of small-eyed individuals increased. d) In the cave, having large eyes was a waste of energy, so mutations occurred that reduced eye size were more likely to occur.