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BIOLOGY 3B Natural Selection Week 3 Darlak WHICH BEAR IS THE “MOST FIT”? EXPLAIN YOUR ANSWER NAME George Ben Spot Sam SIZE 10 ft 1,200 lb 8.5 ft 1,000 lb 9 ft 1,100 lb 9 ft 1,000 lb # OF CUBS FATHERED 19 25 20 20 # OF CUBS SURVIVING TO ADULTHOOD 2 0 1 5 COMMENTS George is very large, very healthy. The strongest bear. Ben mated with the greatest number of females. When the area that Spot lived in was destroyed by fire, Spot was able to move to a new area and change his feeding habits. Sandy was killed by an infection resulting from a cut in his foot. AGE AT DEATH 13 yr 16 yr 12 yr 9 yr AGENDA 04/20 Modes of Selection Check Selection and Survival Wrap 5 Drivers of Evolution Quiz Tomorrow! Types of Adaptations Steps of Selection Modes of Selection 2B Essay Due Wednesday Standard 2B Friday 04/24 RECORD YOUR GROUP DATA HERE P3 Group Children Short Long Neck Neck Allele (N) Allele (n) 1 2 3 4 5 6 7 8 9 10 Grand Children Short Neck Allele (N) Long Neck Allele (n) Great-Grand Children Short Neck Allele (N) Long Neck Allele (n) Using the class generated data below: 1. Explain in your own words what happened to the Tortoise population. 2. Did evolution occur? If so, which of the 5 evolutionary forces was at work? 3. How do you know if evolution occurred? Use data below to support your answer. 4. What type of adaptation was modified? Biotic or abiotic Pressure? Class Average Allele Frequencies Children Grand Children Great-Grand Children Short Neck (N) .53 .30 .12 Long Neck (n) .47 .70 .88 SELECTION & SURVIVAL OF THE GALAPAGOS Selection & Survival of the Galapagos P1 0.900 0.800 0.700 Children Short Neck Allele (N) 0.600 Children Long Neck Allele (n) 0.500 Grand Children Short Neck Allele (N) 0.400 Grand Children Long Neck Allele (n) 0.300 0.200 Great-Grand Children Short Neck Allele (N) 0.100 Great-Grand Children Long Neck Allele (n) 0.000 Allele (N) Short Neck Allele (n) Allele (N) Long Neck Short Neck Children Allele (n) Allele (N) Long Neck Short Neck Grand Children Allele (n) Long Neck Great-Grand Children SELECTION & SURVIVAL OF GALAPAGOS TORTOISE Selection & Survival of Galapagos Tortoise p4 1.000 0.900 0.800 Children Short Neck Allele (N) 0.700 Children Long Neck Allele (n) 0.600 Grand Children Short Neck Allele (N) 0.500 Grand Children Long Neck Allele (n) 0.400 Great-Grand Children Short Neck Allele (N) 0.300 Great-Grand Children Long Neck Allele (n) 0.200 0.100 0.000 Allele (N) Allele (n) Allele (N) Allele (n) Allele (N) Allele (n) Short Neck Long Neck Short Neck Long Neck Short Neck Long Neck Children Grand Children Great-Grand Children 1. WHAT TYPE OF SELECTION? Example: Birds with varying sizes of beaks, after a drought, the large beak birds have selective advantage 2. WHAT TYPE OF SELECTION? Example: Tourists in the desert like to pick up a souvenir from their travels, they pick up the medium spine cactus. Leaving the “homelier” low spine and avoiding the “prickly” high spine cactus. 3. WHAT TYPE OF SELECTION? Example: Birth weight in humans is highly variable, however, a child that is of average weight has a higher chance of being born and of good health, as to mature and reproduce. Nature selects against the extreme phenotypes. 1. WHAT TYPE OF SELECTION? Directional Selection Selection Favors One Extreme Variation of a Trait Example: Birds with varying sizes of beaks, after a drought, the large beak birds have selective advantage 2. WHAT TYPE OF SELECTION? Disruptive Selection Individuals with either extreme trait are selected for. Example: Tourists in the desert like to pick up a souvenir from their travels, they pick up the medium spine cactus. Leaving the “homelier” low spine and avoiding the “prickly” high spine cactus. 3. WHAT TYPE OF SELECTION? Stabilizing Selection Selection favors average individuals of the population. Example: Birth weight in humans is highly variable, however, a child that is of average weight has a higher chance of being born and of good health, as to mature and reproduce. Nature selects against the extreme phenotypes. AGENDA 04/21 Quiz 5 Drivers of Evolution Hardy-Weinberg 2B Essay Due Wednesday Standard 2B Friday 04/24 WHAT DRIVES EVOLUTION? P. 16 Changes in Allelic Frequency or the Gene Pool size Natural Selection Genetic Drift (small Pop) Gene Flow Sexual Selection Mutations 5 DRIVERS NATURAL SELECTION OF FAVORABLE TRAITS 5 drivers of Evolution GENETIC DRIFT SMALL POPULATIONS Founder Effect New small population Bottle neck effect Small # of survivors of a large population GENETIC DRIFT Impact on small Populations Decreasing Variation Decreasing Variation small populations GENETIC DRIFT EXAMPLES 1. BLOOD TYPE DISTRIBUTION Race United States Blackfoot Navajo Mexico Chinese India O 45% 24% 76% 84% 34% 33% A 42% 76% 24% 11% 31% 24% B 10% --------4% 28% 35% AB 3% --------1% 7% 8% GENETIC DRIFT - SMALL POPULATIONS 2. FOUNDER EFFECT COMMON COCKLEBUR (XANTHIUM STRUMARIUM) Spread by “hitchhiking” as a bur Few plants colonizes new area, low genetic variability One trait (allele) may show up more in new population than larger ancestral population GENETIC DRIFT OF SMALL POPULATION 3. BOTTLE NECK EFFECT SEAL HUNTING Large Population reduced to very small in 1800’s Surviving population had different allele frequency and little genetic diversity 5 Drivers of Evolution GENE FLOW Immigration- to come into a population Emigration- to leave a population 5 drivers of Evolution SEXUAL SELECTION Strong force of evolution Females/Males only choose mates based on a certain trait. Males/Females with the trait are “fit” SEXUAL SELECTION = NON-RANDOM Gentleman, which woman would you select as a perspective mate? SEXUAL SELECTION = NON-RANDOM Ladies, which man would you select as a perspective mate? 5 Drivers of Evolution MUTATIONS Mutations can introduce new phenotypes May be beneficial or detrimental New genetic material for selection to work with WHAT DRIVES EVOLUTION? P. 16 Changes in Allelic Frequency or the Gene Pool size Natural Selection Genetic Drift (small Pop) Gene Flow Sexual Selection Mutations Mechanisms of Evolution HARDY-WEINBERG http://youtu.be/xPkOAnK20kw THE HARDY-WEINBERG EQUATION IS USED TO TEST WHETHER A POPULATION IS EVOLVING For a population to remain in Hardy-Weinberg equilibrium for a specific trait, it must satisfy five conditions: 1. 2. 3. 4. 5. No Genetic Drift (Very large population) No gene flow (immigration/ emigration) No mutations Random mating (no Sexual selection) No natural selection EQUILIBRIUM EQUATION p2 + 2pq + q2 = 1 and p + q = 1 p = frequency of the dominant allele in the population q = frequency of the recessive allele in the population p2 = percentage of homozygous dominant individuals q2 = percentage of homozygous recessive individuals 2pq = percentage of heterozygous individuals HOW TO REMEMBER? p = frequency of the dominant allele in the population q = frequency of the recessive allele in the population PP = individuals have 2 alleles so p2 q2 = percentage of homozygous recessive individuals q= 1 allele p2 = percentage of homozygous dominant individuals p= 1 allele QQ= individuals have 2 alleles so q2 2pq = percentage of heterozygous individuals PQ = individuals have 1 of each allele so 2pq The Hardy-Weinberg equation can be used to test whether a population is evolving Imagine that there are two alleles in a blue-footed booby population: W and w – W is a dominant allele for a nonwebbed booby foot – w is a recessive allele for a webbed booby foot Copyright © 2009 Pearson Education, Inc. W Webbing w No webbing The Hardy-Weinberg equation is used to test whether a population is evolving Consider the gene pool of a population of 500 boobies – – – 320 (64%) are homozygous dominant (WW) 160 (32%) are heterozygous (Ww) 20 (4%) are homozygous recessive (ww) Copyright © 2009 Pearson Education, Inc. p2 2pq q2 Phenotypes Genotypes WW Ww ww Number of animals (total = 500) 320 160 20 Genotype frequencies 320 ––– 500 Number of alleles in gene pool (total = 1,000) Allele frequencies (p, q) 160 ––– 500 = 0.64 160 W + 160 w 640 W 800 1,000 20 ––– 500 = 0.04 = 0.32 = 0.8 W p 200 1,000 40 w = 0.2 w q The Hardy-Weinberg equation can be used to test whether a population is evolving Frequency of dominant allele (W) = 0.80 = p – 80% of alleles in booby population are W Frequency of recessive allele (w) = 0.20 = q – 20% of alleles in booby population are w Copyright © 2009 Pearson Education, Inc. THE HARDY-WEINBERG EQUATION CAN BE USED TO TEST WHETHER A POPULATION IS EVOLVING What about the next generation of boobies? – Probability that a booby sperm or egg carries W = 0.8 or 80% – Probability that a sperm or egg carries w = 0.2 or 20% Copyright © 2009 Pearson Education, Inc. Gametes reflect allele frequencies of parental gene pool W egg p = 0.8 Eggs w egg q = 0.2 Next generation: Genotype frequencies Allele frequencies Sperm W sperm p = 0.8 w sperm q = 0.2 WW p2 = 0.64 Ww pq = 0.16 wW qp = 0.16 ww q2 = 0.04 p2 + 2pq + q2 0.64 WW 0.32 Ww 0.8 W 0.04 ww 0.2 w THE HARDY-WEINBERG EQUATION CAN BE USED TO TEST WHETHER A POPULATION IS EVOLVING Frequency of all three genotypes is 100% or 1.0 – – p2 + 2pq + q2 = 100% = 1.0 homozygous dominant + heterozygous + homozygous recessive = 100% Copyright © 2009 Pearson Education, Inc. WARM-UP 04/21 1. 2. 3. Based on the condition given, is the population Evolving or in Equilibrium (Hardy-Weinberg). Random Mating EQUILIBRIUM no sexual selection Large Population EQUILIBRIUM no small population No Natural Selection EQUILIBRIUM All traits are beneficial 6. Many mutations EVOLVING Adding new alleles No immigration EQUILIBRIUM No mixing Peahens choosing Peacocks for healthy tails EVOLVING 7. Propeller Seeds traveling 100’s of miles EVOLVING 4. 5. 8. Sexual Selection Cactus plant in stable desert conditions EQUILIBRIUM No Natural Selection Mixing AGENDA 04/21 Quiz Back/Take 2B Essay Due Hardy-Weinberg Problems p. 17 Homework: pp. 17-18 Standard 2B FRIDAY! Be ready to create NS scenario for standard 2B ESSAY PEER-REVIEW & GRADE 4 Steps (10 pts each, if explained correctly) 40 pts Overpopulation Variation Selection Adaptation Type of Selection 20 pts Do they explain which one? Is it correct? 10 pts Correct Graph provided. 10 pts Type of Adaptation Explanation of adaptation and typed of pressure Conventions Intro & Conclusion 20 pts 10 pts 10 pts 0.488 or 48.8% WINGED TRAIT IS DOMINANT 7/21= 0.333 √ 0.33 = 0.577 1- 0.57 = 0.423 0.4232 = 0.179 2 (.577)(0.423) = 0.488 PRACTICE A population of birds contains 16 animals with red tail feathers and 34 animals with blue tail feathers. Blue tail feathers are the dominant trait. What is the frequency of the red allele? q2 = 16/50 = 0.32 so q = What is the frequency of the blue allele? p = 1 - 0.566 = √.32 = 0.566 0.434 What is the frequency of heterozygotes? 2 (0.566)(0.434) = 0.491 What is the frequency of birds homozygous for the blue allele? P2 = (0.434)2 = 0.188 2. You have sampled a population in which you know that the percentage of the homozygous recessive genotype (aa) is 36%. Using that 36%, calculate the following: A. The frequency of the "aa" genotype. Given as 36% = 0.36 = q2 B. The frequency of the "a" allele. q = √ 0.36 = 0.6 C. The frequency of the "A" allele. p + q = 1, so p = 1- 0.6 = 0.4 D. The frequencies of the genotypes "AA" and "Aa." AA = p2 = (0.4)2 = 0.16 Aa = 2pq = 2(.6)(.4) = 0.48 E. The frequencies of the two possible phenotypes if "A" is completely dominant over "a." p2 + 2pq = (0.4)2 + 2(.6)(.4) = 0.64 Or 0.16 + 0.48 = all Dominant = 64% HOMEWORK #2-9 WARM-UP 04/23 1. 2. The yellow bellied three-toed skink (Saiphos equalis) is a lizard of New South Wales, in Australia, that appears to be undergoing the change from laying eggs to live birth as the result of a series of mutations. Skinks living on the coast tend to lay eggs, probably because the warm weather is predictable and sufficient for embryonic development. Those skinks living in the cooler mountains tend to give birth to live young, the mother’s body providing a more stable temperature. It is to be predicted that these two populations will at some point separate into different species. Physiological What type of Adaptation? Biotic or Abiotic pressure? Abiotic - temperature AGENDA 04/23 Homework Check pp.16-17 2B Review Retake Pre-test on Clickers I can Statements 2B Essay 3. There are 100 students in a class. Ninety-six did well in the course whereas four blew it totally and received a grade of F. Sorry. In the highly unlikely event that these traits are genetic rather than environmental, if these traits involve dominant and recessive alleles, and if the four (4%) represent the frequency of the homozygous recessive condition, please calculate the following: A. The frequency of the recessive allele. q2 is given as 4% = 0.04 so q = √ 0.04 = 0.2 B. The frequency of the dominant allele. q = 0.2 so p = 1- 0.2 = 0.8 C. The frequency of heterozygous individuals. 2pq = 2(0.2)(0.8) = 0.32 4. Within a population of butterflies, the color brown (B) is dominant over the color white (b). And, 40% of all butterflies are white. Given this simple information, which is something that is very likely to be on an exam, calculate the following: A. The percentage of butterflies in the population that are heterozygous. if q2 = 0.40 so q = √0.40 = 0.63 and p = 1 - 0.63 = 0.37 So, 2pq = 2(0.63)(0.37)= 0.46 or 46% B. The frequency of homozygous dominant individuals. P2 = (0.37)2 = 0.14 or 40% (bb) + 46% (Bb)= 86% so 100% - 86% = 14% (BB) 5. After graduation, you and 19 of your closest friends (lets say 10 males and 10 females) charter a plane to go on a round-the-world tour. Unfortunately, you all crash land (safely) on a deserted island. No one finds you and you start a new population totally isolated from the rest of the world. Two of your friends carry (i.e. are heterozygous for) the recessive cystic fibrosis allele (c). Assuming that the frequency of this allele does not change as the population grows, what will be the incidence of cystic fibrosis on your island? ______ 2pq = 2/20 = 0.1 q = 2 (c alleles)/40 total alleles = 0.05 or 5% p = 1- 0.05 = 0.95 Let’s just check our answer 2pq = 2(.05)(.95) = 0.095 (~ 0.1) 6. Cystic fibrosis is a recessive condition that affects about 1 in 2,500 babies in the Caucasian population of the United States. Please calculate the following. The frequency of the recessive allele in the population. q2 is given as 1/2500 = .0004 so q = √ 0.0004 = 0.02 The frequency of the dominant allele in the population. q = 0.02 so p = 1 - 0.02 = 0.98 The percentage of heterozygous individuals (carriers) in the population. ____ q = 0.02, p = 0.98, so 2pq = 0.039 or 3.9% 7. This is a classic data set on wing coloration in the scarlet tiger moth (Panaxia dominula). Coloration in this species had been previously shown to behave as a single-locus, two-allele system with incomplete dominance. Data for 1612 individuals are given below: White-spotted (AA) = 1469 Intermediate (Aa) = 138 Little spotting (aa) = 5 Total = 1612 Calculate the allele frequencies ( p and q ) aa = q2 = 5/1612 = 0.0031 q2 = 0.0031, q = √0.0031 = 0.055 so p = 1 – 0.055 = 0.945 p2 = (0.945)2 = 0.893 2pq = 2(.945)(.055) =.103 Check It….. AA = 89.3% Aa = 10.3% aa = 0.31% 8. The allele for a widow's peak (hairline) is dominant over the allele for a straight hairline. In a population of 500 individuals, 25% show the recessive phenotype. How many individuals would you expect to be homozygous dominant and heterozygous for the trait? Given q2 = 0.25 q = 0.5, so 1 – 0.5 = 0.5 = p So p2 = 0.25, or 25% of 500 = 125 homozygous dominant So 2pq= 2 (0.5)(0.5) = .5 or 50% of 500 = 250 heterozygous 9. The allele for a hitchhiker's thumb is recessive compared to straight thumbs, which are dominant. In a population of 1000 individuals, 510 show the dominant phenotype. How many individuals would you expect for each of the three possible genotypes for this trait. 1000 – 510 = 490 recessive individuals Given q2 = 490 recessive/1000 = 0.49 q = 0.7 so 1 – 0.7 = 0.3 = p So p2 = 0.09, or 9% of 1000 = 90 homozygous dominant So 2pq= 2 (0.7)(0.3) = .42 or 42% of 1000 = 420 420 heterozygous individuals EXTRA CREDIT 10. REAL WORLD APPLICATION PROBLEM Choose a human trait to study and survey a population at your school. (Aim for at least a sample size of 50 to get meaningful results). Use your sample to determine the allele frequencies in the human population. Traits (dominant listed first) Hitchhiker's Thumb vs Straight Thumbs Widow's peak vs straight hairline PTC taster vs non-taster Short Big toe vs long big toe Free earlobes vs attached earlobes Tongue rolling vs non-rolling Bent little fingers vs straight little fingers Arm crossing (left over right) vs right over left Ear points vs no ear points Warm-up 04/24 FIND FOUR INCORRECT WORDS: 1. 2. 3. 4. Organisms produce less offspring than can survive. In any population, mutations exist. Individuals with certain “useful” or beneficial variations perish and pass on their variations to the next generation. Overtime, offspring with “beneficial” variations will make up half of the population. 1. 2. 3. 4. LESS offspring Organisms produce MORE than can survive. MUTATIONS exist. In any population, VARIATIONS Individuals with certain “useful” or beneficial variations SURVIVE PERISH and pass on their variations to the next generation. Overtime, offspring with “beneficial” variations will make up HALF MOST of the population. AGENDA 04/24 Standard 2B Intro to Engineering Steps of Natural Selection 3 Types of Adatations 3 Modes of Selection 5 Drivers of Evolution Hardy-Weinberg I Spy …….. Midterm Next Week (no passes or improve grade)