Intro Biology BIS 001B
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This 10 page Class Notes was uploaded by Charity Daniel on Tuesday September 8, 2015. The Class Notes belongs to BIS 001B at University of California - Davis taught by Robert Kimsey in Fall. Since its upload, it has received 30 views. For similar materials see /class/187625/bis-001b-university-of-california-davis in Biological Sciences at University of California - Davis.
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Date Created: 09/08/15
l N E 4 V39 0 BIS 1B Winter 2007 Answers to study questions Population genetics lectures 3 5 How does the concept of allele frequencies in a de ned Mendelian cross between two individuals differ from the concept of allele frequencies in an interbreeding population of organisms Alleles in aMendelian cross have discrete frequency classes of 0 05 or 1 in each individual because an individual has only two alleles per locus In a population allele frequencies can vary continuously from 00 to 10 and can change in response to various population level forces Define evolution in terms of population genetics Evolution can be defined as a change in allele or genotype frequencies Changes in phenotype frequencies could be due to changes in environmental conditions only What does it mean to say that adaptation can only be defined relative to a specific environment An adaptive trait in one environment may possibly be non adaptive in another environment For example long thin ears may be adaptive in hot tropical climates where they aid in cooling the body but not adaptive in an arctic climate where they cause heat loss Can the process of natural selection prepare a population of organisms for unanticipated future challenges In general no Natural selection acts on existing variation and in the context of current environmental conditions Any preparation for the future preadaptation is fortuitous What is genetic drift What are some conditions that increase the importance of genetic drift Does genetic drift mean that evolution has occurred Why or why not Genetic drift is change in allele frequencies due to random sampling in a finite population Random sampling can occur during gamete formation or during survival and reproduction of individuals Any process that decreases population size such as a founder event or a population bottleneck or that maintains a population at a small size will tend to increase the intensity of genetic drift relative to that in continuously large populations Because genetic drift causes changes in allele frequencies acting in the long run to decrease the amount of genetic variability present in a population it does cause evolution What does it mean to say that the HardyWeinberg equilibrium is the null hypothesis against which we can test for natural selection or nonrandom mating gt1 9 gt0 For any given population we can start with the null hypothesis that the population is not subject to any of the forces including selection and non random mating that might in uence and change allele frequencies andor genotype frequencies If the observed genotype J 39 in thisr r 39 differ 39 39 from those expected under Hardy Weinberg equilibrium we can reject the null hypothesis and explore alternative hypotheses Among these alternatives are the hypotheses that natural selection or non random mating are responsible for the differences 1 In qualitative terms what should happen if each of the assumptions underlying the Hardy Weinberg equilibrium is relaxed Assumptions 1 infinite or at least very large population size if this assumption is violated allele and genotype frequencies can change due to chance genetic drift 2 random mating if there is non random mating this can cause changes in genotype frequencies but not in allele frequencies 3 no selection if selection occurs this can bring about changes in both allele and genotype frequencies 4 no mutation if mutation occurs then new alleles appear and allele and genotype frequencies therefore change but only slightly 5 no gene flow if this assumption is violated both allele and genotype frequencies may change if the pool of immigrants is genetically distinct from the population receiving the immigrants The ultimate s0urces of all genetic variation is are polypeptide sequences mutations ribosomal DNA genetic recombination gene ow rap96x W mutations Which of the following does not change allele frequencies b mutation d gene ow f natural selection a assortative nonrandom mating genetic drift e recombination O a and e 10 In addition to the ABO blood group system in humans there is also the MN system where M and N are codominant alleles at a single locus The following numbers of the different types were found by Stern 1973 in a sample ofUS residents typeM 1787 type MN 3039 typeN 1303 a What are the observed genotype frequencies b What are the allele gene frequencies c Given the allele frequencies what are the expected genotype frequencies assuming Hardy Weinberg equilibrium conditions hold Do these differ from the observed frequencies Explain Total population size is 1 787 3039 1303 6129 So Q genotype frequencies freq l lI 1 7876129 0292 freq lHV 30396129 0496 freqNN 13036129 0212 6 allele frequencies freq lU 0292 05496 054 we can call this freqN 0212 05496 046 we can call this q 9 expected frequencies under H W equilibrium frequ p2 0502 0292 eq lIN 2pq 2054046 0496 freqNN cf 0212 These expected frequencies match those that were observed Thus there is no evidence that the H W assumptions were violated in the natural population 11 The frequencies of 3 genotypes in a population are freqAA 06 freqAa 04 and freqaa 00 What will be the frequencies in the next generation assuming a Hardy Weinberg equilibrium LetfreqA p 06 0504 08 Let freqa q 00 0504 02 Then under H W freqAA in the next generation p2 082 064 freqAa 2pq 20802 032 freqaa q2 022 004 12 In a sample human population the following numbers of individuals with different ABO blood group genotypes were found genotype AA 340 genotype A0 341 genotype BB 279 genotype BO 279 genotype AB 79 genotype 00 701 What are the allele frequencies in this sample Are the genotype frequencies those expected under HW equilibrium conditions Total sample size 2019 freqA freqAA 05freqAO 05freqAB 3402019 053412019 05792019 0272 we can call this freqB 2792019 052792019 05792019 0227 we can call this q freqO 7012019 053412019 052792019 0501 we can call this r Under H ardy Weinberg the expected frequency of each homozygote is simply the square of the corresponding allele frequency and the expected frequency of each heterozygote is two times the product of the component allele frequencies To see this build a Punnett square with each of the three alleles along the top and the left hand side freqAA p2 02732 0074 freqAO 2pr 202720501 0273 freqBB 5f 02232 0052 freqBO 2qr 20 2270 501 0227 freqAB 2pq 202720227 0123 freqOO r2 05002 0251 Comparing these predicted genotype frequencies with those observed in the initial population reveals large di erences for example the frequency of AA initially observed was 3402019 0168 the predicted frequency is 0074 Thus the initial population was not in HW equilibrium 13 In a population of snails that live on a rocky shore a marine biologist finds that snails are striped SS spotted Ss or plain colored ss The heterozygote Ss is thus intermediate A sample of the population taken during the spring reveals 422 striped 845 spotted and 491 plain individuals A sample of the same population taken during the fall of the same year reveals 389 striped 796 spotted and 533 plain colored snails a What are the frequencies of the genotypes in the original spring sample What are the frequencies of the alleles b What are the frequencies of the genotypes in the subsequent fall sample What are the frequencies of the alleles c How might you explain the change in allele frequencies that occurred during the season d Does the fall population show HardyWeinberg equilibrium frequencies If not how might you explain the deviation from HW frequencies a 6 9 The spring genotype frequencies are freqSS 422422 845 491 4221758 0240 freqSs 8451758 0481 freqss 4911758 0279 The spring allele frequencies are freqS freqSS 05freqSs 0240 050 481 048 freqs freqss 05freqSs 0279 050481 0519 In the fall sample freqSS 389389 796 533 3891 718 0226 freqSs 7961718 0463 freqss 5331718 0310 The fall allele frequencies are freqS freqSS 05freqSs 0226 050 463 0458 freqs freqss 05freqSs 0310 050463 542 The allele frequency changes are relatively small They could be due to a number of di erent factors including selection in favor of plain colored snails immigration of plain colored snails that is plain colored snails entering the population we are studying emigration of striped or spotted snails that is striped or spotted snails leaving the population we are studying or random sampling e quotects genetic drift d Based upon allele frequencies observed during the fall we can calculate expected H W genotype frequencies freqSS 04582 0210 freqSs 20458 0 542 0496 freqss 0542 2 0294 So we have slight deviations from these H ardy Weinberg expectations these deviations could be due to the same processes that caused the changes in allele frequencies across the season 14 Draw a graph that shows how for a locus with two alleles the frequency of heterozygotes changes as the frequencies of the alleles vary from 0 to 10 under the assumption of Hardy Weinberg equilibrium The data to be plotted can be generated in a simple table freer1 ew q freqAB 2pc freqAA p2 echB 92 0 1 0 0 1 01 0 9 01 8 0 01 0 81 0 2 0 8 0 32 0 04 0 64 0 5 0 5 0 5 0 25 0 25 06 04 048 036 016 etc The resulting graph looks like this fre q A A fre q B B fre q A B l 09 08 07 06 05 04 03 02 01 0 Genotype frequencies I I I I l I I I I 0 01 02 03 04 05 06 07 08 09 l freqA p 15 Many insects are wing polymorphic with winged and wingless morphs Assume that winged W is dominant to wingless w An ethologist a biologist studying behavior is interested in mate choice in a species of wing polymorphic bugs and examines mating in a laboratory setting using different combinations of bugs E The ethologist places 50 Wingless bugs and 50 homozygous winged bugs together in a chamber In this parental generation what are the frequencies of the W and w alleles U The ethologist notices that there is 100 positive assortative mating Assuming no selection or drift what will be the genotype and allele frequencies in the offspring generation generation 1 O The ethologist now forces random matings among these individuals by placing randomly chosen males and females together in small boxes Assuming HardyWeinberg conditions what will be the allele genotype and phenotype frequencies in the next generation generation 2 Q freqWW 50100 050 freqww 50100 050 freqW 050 andfreqw 050 b The same as previous generation Q Allelefrequencies will be the same iefreqW freqw 050 Expected genotype frequencies freqWW p2 052 025 freqWw 2pq 20505 050 freqww 052 025 Explain the idea that evolution proceeds according to a principle of accumulation of design New adaptations typically represent additions or modi cations of pre existing structures Life has been around on the planet for 35 billion years sufficient time for the gradual accumulation of di erent features in different evolutionary lineages If mutation is a rare process how can it provide sufficient variation for evolution to proceed The estimated mutation rates per locus appear to be rather low about 10396 per gamete but 1 many populations are quite large so that for example among the individuals in a large population of size 10 one expects about 10 individuals to carry a new mutation at a given locus and 2 there are many gene loci eg 1 in humans so that over all loci one expects each individual to carry about 01 new mutations It is often found that a bacterial population that has recently evolved resistance to an antibiotic is at a competitive disadvantage ie has lower fitness compared to a non resistant bacterium IF there are no antibiotics present Why This suggests that there is a cost of resistance so that although the R resistant individuals have higher fitness than the S susceptible individuals when both are exposed to antibiotics the R individuals achieve this resistance through some biochemical physiological or behavioral mechanisms that isare costly in energy time or nutrients Hence in an environment without antibiotics S individuals have higher relative fitness 19 We showed in class that when selection acts against a recessive trait we can calculate the change in frequency of the favored allele A using the equation the change in the frequency ofp deltap p p psqzlsqz where p is the frequency of the favored allele in the next generation p is the frequency of the favored allele in the current generation q is the frequency of the disfavored allele a in the current generation and s is the selection coef cient of the homozygous recessives If we begin with a population where we have experimentally halted selection so that Hardy Weinberg equilibrium genotype frequencies are attained we find that 99 of the population has the favored phenotype and 1 of the population expresses the disfavored allele What are p and q What are the initial genotype frequencies If we allow selection to act again and the disfavored phenotype has only 50 of the lifetime reproductive success of the favored phenotype what are s p q and deltap Why is deltap so small We are told that freqaa 1 0 01 But freqaa also q2 from the H W model Therefore a freqa 0 01 12 010 Hence p freqA 1 7 q 090 The initial genotype frequencies are freqAA 09 081 freqAa 2090010 018 freqaa 0 01 as given s 050 delta p 09005001021 0500102 000450995 0004522613 0005 p39 090 0005 0905 q39 1 p39 0095 The change in frequency of p delta p is small because most of the disfavored alleles are found in heterozygotes where they are sheltered from selection because of the dominance relationship betweenA and a Selection only acts against the individuals that are homozygous for the recessive allele and these individuals are very rare 20 Recall that under heterozygote advantage the equilibrium frequency of an allele A1 is ts t where s is the selection coef cient of the A1A1 homozygote and t is the selection coef cient of the other AzAz homozygote Now consider a locus with two alleles X and Y in which the relative tnesses ofthe three genotypes XX XY YY are 08 10 and 09 respectively What is the equilibrium frequency of the X allele N N The relative fitness of the XX genotype is 08 so its selection coef cient s is 1 08 02 The relative fitness of the Y Y genotype is 09 so its selection coe icient t is 1 09 01 Thus the equilibriumfrequency oftheXallele is39 ts t 0102 01 0333 Why does the frequency distribution of phenotypes for polygenic traits resemble a bell curve As more and more genes contribute to a phenotype a larger and larger number of possible gene combinations is created Phenotypes near the middle of the distribution can be produced by many different combinations of genes whereas phenotypes near the ends extremes of the distribution can be produced by only a few gene combinations Furthermore not all gene combinations are equally likely and the commonest combinations are the ones that produce average phenotypic values These e ects combine to produce the bell shaped curve Environmental effects also tend to smooth the distribution Discuss qualitatively the expected outcome of different modes of selection on a continuous trait Stabilizing selection is expected to decrease the variance of the trait while the mean stays the same Directional selection is expected to shift the mean of the trait in the direction being selected the variance may or may not change Disruptive selection is expected to increase the variance of a trait without changing the mean If disruptive selection is su iciently strong it might also produce a bimodal distribution of the trait Draw graphs showing how directional truncation selection on a polygenic trait causes evolutionary change in the frequency distribution of trait values Show the frequency distribution for the parental generation the selected portion of the parental population and the offspring population Label the parental population s mean trait value the mean trait value of the selected group and the mean trait value of the offspring generation If S the selection differential 10 and R the response to selection 5 what is the heritability of the trait Parental generation Frequency Truncation selection Xp S Xs H Offspring generation Frequency R Xo Trait value The heritability of the trait can be calculated using R h2S h2 ms 510 05
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