Biology 1114: Exam 2 Study Guide 6. SELECTION AND FITNESS A. Evolution By Natural Selection a. Selection- the nonrandom differential survival and reproduction of certain phenotypes based on selective pressures that results in adaptive evolution b. Natural SeleDon't forget about the age old question of dna/rna study guide
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ction Requires: i. Variation in a population for a given phenotype ii. Selective pressure iii. Differential survival/reproduction of phenotypes iv. Competition v. Heritability of the phenotype vi. Alleles for adaptive phenotype more common in next generation B. Example of Identifying Characteristics of Selection a. Rosemary and Peter Grant in the Galapogos i. Took a long term survey of finches ii. Allowed for a real-time observation of evolutionary processes iii. Endemic- species found in one location and no where else on Earth 1. Penguins, many bird species, giant tortoise b. Variation i. Preexisting variation was observed c. Selective pressure- some change in the environment that is potentially limiting i. Drought ii. A ton of plants died d. Differential survival/reproduction i. Important to note changing population sizes and phenotype distribution ii. Selection differential (S)- quantifies shift in mean phenotype of population before and after selection event 1. S = P* - P 2. S =the mean population trait after the event - the mean population trait before the event 3. Indicates how strong the selective pressure is 4. Does not depend on population size if the mean percentages stay the same e. Competition i. How might a phenotypic change relate to survival differences via competition? f. Heritability of phenotypic characteristici. the proportion of phenotypic variation attributable to genotype 1. h2 = (R/S) = (O* - O-)/(P* - P-) ii. use given data to calculate the population response to selection (R) 1. compare mean phenotypic values of post-selection offspring to hypothetical offspring of pre-existing parental generation 2. R = O* - O 3. assume hypothetical pre-selection offspring mirrors the parent pre-selection mean C. Categories of Selection a. Describe selection based on how the next generation differs from the original generation b. Directional selection – shift from one extreme to another i. Curve shifts in one direction c. Disruptive selection – shift from middle ground to either extreme i. Curve forms 2 high peaks d. Stabilizing selection – shift from an extreme to middle ground i. Curve shifts to peak in the middle of where the 2 extremes are D. Components of Differential Reproductive Success a. Viability – probability that an individual bearing genotype will survive b. Fecundity – number of gametes per individual c. Gamete viability – alleles that impact longevity or quality of sperm/eggs d. Mating success- number and quality of mates e. Fertilization success- alleles impact the probability that fertilization will occur E. Fitness- the average reproductive output (R average) of all organisms bearing a given genotype; integrates differences in survival and reproduction a. Individuals with a higher fitness pass their alleles to the next generation at a greater rate than individuals with lower fitness b. If individual fitness differences are due to allele differences, alleles associated with higher fitness more frequent in next generation c. Fitness calculated for genotypes i. Why do we need to sample many individuals of the same genotype to calculate R? Why can’t we use only one individual to represent the genotype? 1. Environmental effects: individuals of the same genotype differ in reproductive output because of their environment a. EX: mate availability, # of reproductive females, food availability, nest site availability, weather, disease d. Calculating fitness i. To find R average for a specific genotype: 1. Count the number of offspring that have that specific genotype and take the average ii. Relative fitness (W)- relative measure of the survival and reproduction among known genotypes 1. Divide R average of each genotype by the R average of the fittest genotype 2. The fittest genotype’s relative fitness always equals 1 (W = 1) 3. If there were NO differences in relative fitness between genotypes, what is your prediction for changes in allele frequency in the next generation? a. Allele frequencies would NOT change from the parental to the offspring generation b. Each genotype is equally likely to survive and reproduce 7. SOCIETAL ISSUES AND EVOLUTION A. Evolution and Social Issues a. Understanding a human health crisis through evolution B. Gram Positive and Negative Bacteria a. Key difference: Cell wall thickness and membrane organization b. Gram negative i. 5 layers from inside to out ii. Cell membrane periplasmic space cell wall another periplamic space outer membrane iii. Doesn’t retain purple stain 1. Cell wall is thinner than in gram positive, so it can’t hold a stain as well iv. Much more resistant to antibiotics than gram positive bacteria c. Gram positive i. Just 3 layers from inside to out ii. Cell membrane periplasmic space cell wall (peptidoglycan) iii. Stains purple C. Antibiotics- many act by binding with targeted surface receptors on plasma membrane of bacteria a. Disrupt the cell wall or cell membrane; disrupt bacterial ribosome b. Kills bacteria c. Mechanisms of Antibiotic Resistance: i. Low outer membrane permeability to antibiotics1. Antibiotics can’t reach receptors on plasma membrane to bind ii. Surface receptor mutations 1. Antibiotic can’t bind to receptor iii. Antibiotic efflux pumps 1. Antibiotic is actively pumped out by bacterium D. Explaining Addie’s Infections a. For a given mechanism of antibiotic resistance, a bacterial population has variation b. The heritability of the antibiotic resistance was high, so the new generations of bacteria were increasingly resistant to the antibiotics E. NDMI-1 gene a. How does it spread through bacteria populations? i. Traditionally- we think about genes spreading through populations from parents to offspring ii. Horizontal gene transfer- Movement of genes between organisms without reproduction iii. Plasmids- small circular pieces of DNA independent of chromosomal DNA 1. Genes for antibiotic resistance usually on bacterial plasmids b. Horizontal Gene Transfer via Plasmids i. Donor with resistance gene on plasmid ii. Donor forms connection with recipient bacterium iii. 1 strand of the double-stranded plasmid DNA moved to recipient bacterium iv. both donor and recipient bacteria make complementary DNA strand v. both donor and recipient now antibiotic resistant and can pass resistance genes to other bacteria 8. COEVOLUTION AND SPECIAL CASES OF SELECTION A. Special Cases of Selection a. Frequency dependent selection- the fitness of a phenotype depends on how common it is in the population i. EX: scale eating fish 1. Scale eaters are left or right mouthed; determines attack direction 2. Right mouthed is completely dominant over left 3. Prey species guard against attack from most common phenotype 4. So, the frequency of the most common phenotype will go down in coming generations because its prey knows to look out for it b. Coevolution- the joint evolution of two or more interacting species in which pressure from each species acts as a selective agent on the other i. What type of selection is the cheetah and gazelle an example of? 1. Speed is under strong directional selection pressure in the cheetah and the gazelle populations 2. Slow gazelles are more likely to get eaten; have lower relative fitness 3. Slow cheetahs are less likely to eat; have lower relative fitness 4. Cheetahs and gazelles exert selective pressure on each other ii. For 2 or more populations to coevolve, what conditions are necessary? 1. Geographical overlap of the species 2. Reciprocal effects on phenotype traits as a result of differential survival/reproduction a. What does reciprocal mean? b. Each species acts as a selective pressure on the phenotype of the others i. Fast gazelles = selective pressure on cheetahs to be fast and vise versa c. Effects on fitness: i. A given phenotype of species A is more fit in the presence of species B 9. SEX AND SEXUAL SELECTION A. Modes of reproduction a. Asexual Reproduction i. Formation of offspring without the fusion of egg and sperm ii. Offspring are genetically identical to parent iii. Modes of asexual reproduction: 1. Budding- offspring develop as outgrowth of parent 2. Fission- parent splits into 2 organisms of roughly equal size 3. Root Suckering- many clonal individuals sprout from the roots of an individual a. EX: Pando the quaking aspen (“I spread” in Latin) i. Started roughly 80,000 years ago ii. Started with one single aspen iii. Covers more than 100 acres iv. Is Pando one individual or many close individuals? 4. Parthenogenosis- unfertilized egg develops into offspringa. Whiptail lizards i. Entire species is female, reproduce parthenogenically ii. Still display courtship/mating behavior 1. Associated with hormone levels during breeding season iii. Females mounted by “male-like” females more likely to ovulate and lay more eggs b. Sexual Reproduction i. Formation of offspring by fusion of a female gamete (egg) and male gamete (sperm) ii. Offspring are genetically distinct and inherit traits form both parents iii. Offspring are not identical to either parent B. Does Sex Make Sense? a. From evolutionary perspective, the reproductive mode that produces the most offspring should be the most frequent b. Sexual reproduction: i. Leads to a stable population number 1. Only half of individuals can bear offspring (limited to females) ii. Cannot mate in isolation; a mate that wants to mate with you is required c. Asexual reproduction: i. Leads to a much bigger population 1. Not limited by the number of females 2. All individuals can bear offspring ii. Can reproduce in isolation d. So why sex? i. Ubiquity- seen in eukaryotes 1. Prokaryotes also have modes of recombination ii. Historic extinction rates 1. Asexual lineages have higher extinction rates than sexual lineages do C. Advantages of Sex a. Variation i. Sexual 1. Beneficial alleles can be rapidly combined in offspring a. Can rapidly spread through a population 2. Produces new combinations of alleles ii. Asexual 1. Offspring are clones of parents a. Beneficial alleles are not combined in offspring 2. Each lineage would have to independently gain beneficial alleles a. Since mutation is random, takes much longeriii. Why is variation beneficial? 1. Allows population response to changing environment via selection 2. Makes a population more likely to survive future environmental changes b. Fewer Deleterious Alleles i. Sexual 1. Elimination of deleterious phenotypes 2. Results in a much more vigorous population ii. Asexual 1. Since offspring are clones, all parental mutations are passed to offspring 2. Over time, due to random chance, all subsets of the population acquire deleterious alleles a. Now 1 deleterious allele is the norm 3. Over time more and more deleterious alleles will become the norm in a population 4. Mutation meltdown- over time in asexual populations, proportions of low fitness individuals with many deleterious alleles increases a. Eventual consequence = extinction b. Mutation meltdown + lower variation explains greater extinction rates of asexual lineages D. Sexual Selection- individuals with certain inherited characters are more likely than others to attract mates a. Form of natural selection b. Why is it usually males competing for females? i. Reproductive differences E. Reproductive Differences Between Males and Females a. Females i. Gametes: eggs are limiting 1. Produce limited number per breeding opportunity ii. Parental care 1. When it occurs, often falls on female a. Gestation, lactation, guarding, feeding, etc… 2. Greater energetic cost to raising young iii. Gametes limited, high PI b. Males i. Gametes: excess of sperm 1. Not limited by sperm available during breeding season ii. Parental care 1. Males often provide no parental care- mate and leave 2. Lower energetic cost to raise young iii. Gametes NOT limited, low PIc. Operational sex ratio (OSR) – ratio of males in a population that are ready to mate with females in that population that are ready to mate i. Skewed male OSR results from differences in number of ready gametes produced between males and females 1. Lots of male, few of female d. Differential parental investment (PI) – measured as each parent’s contribution to rearing offspring e. What mating strategy would optimize fitness for individuals of each sex? i. Female: 1. Find a mate with beneficial alleles or good resources so offspring have a better chance of survival 2. Be very choosy ii. Male: 1. Mate with as many partners as possible and leave rearing to them if possible iii. These strategies are in direct conflict 1. Average reproductive output = the same between sexes F. Variation within Selection a. Female reproductive success: Relatively uniform i. Not limited by mating opportunities because males are trying to mate with as many as possible ii. Limited egg resources iii. Limited by resources required to raise young b. Male reproductive success: Variable i. Limited by mating opportunities 1. Females chose only high quality males 2. Males that can demonstrate quality are preferred c. Average reproductive success between sexes it equal, but males experience the most variation within themselves on who is having the most offspring G. Variable reproductive success a. Selection acts on variation i. Stronger in males because of greater variation in reproductive success b. Selection for characteristics that allow greater access to mates i. Characteristics that allow defense of territories/resources and thus greater access/opportunity to mate ii. Characteristics that result in greater female choice 10. INTRA AND INTERSEXUAL SELECTION A. Intra-sexual Selection – results from competition between members of one sex for access/opportunity to mate with members of the other sex a. Usually male-male competitionb. Competition for access to females OR a resource females need c. EX: Elephant seal mating season i. Males fight to defend stretches of beach where females gather ii. Social hierarchy of males: alpha 1 access to males iii. Variation in reproductive success: 1. 95% of females mated 2. 5 most active males of over 100 account for 48-83% of reproduction 3. only about 10% of males mated at all 4. strong selective pressure for size iv. Why such large annual differences in reproductive dominance of 5 most active males? 1. As number of females increases, percent copulations by 5 most active males decreases 2. “Sneaker” copulations by peripheral males a. intra-sexual selection sometimes indirect B. Indirect Male-Male Competition: Intrasexual selection a. Sneaking and Cuckoldry i. EX: Sunfish 1. Dominant males build nests to attract females, guard eggs against other males post fertilization 2. Reproduction via spawning: a. Female releases eggs, male releases cloud of sperm over eggs 3. Evolution encourages cheating 4. Size class of sunfish: multiple cheating strategies a. Sneaker males – tiny; make “spawning rushes” i. Wait for spawning, rush in and spew out sperm as fast as possible b. Satellites – too big to be sneakers, but roughly the size of females i. Dominants tolerate them at the nest because they appear female, then they mate with females c. Cuckoldry – both strategies i. Males unwittingly invest parental care into unrelated offspring b. Sperm competition c. Male guarding d. Removing other males’ sperm e. Barbed and curved penis morphology i. Damselfly and dragonflies C. Intersexual Selection – results from preference by members of one sex for members of the other sex that bear a certain phenotype a. Usually females choosing malesb. What kinds of characteristics would you predict act as cues for female choice? i. Features suggesting health/feeding capability ii. Features that emphasize distinction between males and females – minimize wasted effort 1. Often results in pronounced sexual dimorphism a. Difference in appearance of females and males of the same species c. Zahavi’s Handicap Hypothesis i. Individuals with well developed sexually selected characteristics have survived a test ii. Extreme morphologies entail costs: 1. Energy to produce characteristic 2. Greater risk of being eaten a. More conspicuous to predators b. less agile iii. individuals that survive and thrive DESPITE this handicap demonstrate superior genotype iv. survival with characteristic therefore an “honest indicator” of fitness 1. a positive correlation indicates that a trait is an honest indicator 2. a negative correlation indicates that a trait is NOT an honest indicator d. Relating signals to fitness and honest signaling i. Measures of male fitness 1. Number of active nests in a male’s territory 2. Clutch initiation date a. Desirable males chosen to mate with first b. Less desirable males mate later ii. Objective measure of mate quality 1. Body condition: scored based on size and weight 2. Higher body condition = better health iii. Statistics: 1. Correlation coefficient (r) – value indicating the direction and strength of a linear relationship between 2 continuous variables (-1 to a. Direction – positive, negative, or none i. Positive correlation (>0) – as one variable increases, so does the other variable ii. Negative correlation (<0) – as one variable increases, the other variable decreases b. Strength – closer to 1 or -1 i. Close to 0 = weak/no relationshipii. -1 = very strong negative correlation iii. 1 = very strong positive correlation 2. R2 – proportion of variation in the dependent variable explained by variation in the independent variable a. EX: about 51% of variation in the number of active nests is explained by variation in male collar area i. For the positive and negative correlation graphs b. Why are R2 values for both positive and negative correlation graphs the same? i. No info on direction of relationship iv. Data interpretation: 1. Slope – rate of change in dependent variable per unit increase in independent variable 2. p – probability that the relationship is due to random chance D. Sexual Selection a. Are females always choosy? i. Polygamous 1. Greater PI for females 2. Similar average reproductive success between sexes a. but much more variation in the males 3. Females choosy 4. Most common ii. Monogamous 1. Similar PI between sexes 2. Similar average reproductive success between sexes a. And similar variation iii. Polyandrous 1. Greater PI for males 2. Similar average reproductive success between sexes a. but much more variation in females 3. Males choosy