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Exam TWO GENE-BIOL 3000 Fall 2013
810-______-____________
This exam has 27 questions on 7 pages. Pay attention to point totals in allocating your
time. No calculators. In the case of math, either show your work as intended or explain
the concept verbally. Answer as succinctly and legibly as possible in the space provided.
The multiple choice questions will be answered on a Scantron form.
Putting your name ON BACK OF EXAM and 810# above indicates you agree to abide
by UGA’s academic honesty policy and procedures
(this step worth 100 points, we cannot grade otherwise!)
1. You are doing a lab project on allozyme marker diversity in beetles and find 2 alleles at the Gpi locus, called 100 and 120 for how far they migrate on a gel in one hour. After collecting and genotyping 100 individuals, you have found the 100/100 genotype in 81 individuals, and the 100/120 genotype in 19 individuals. State a specific hypothesis about evolution in this population (4), describe how you would test your hypothesis (4; 8 total points). My specific hypothesis is that these data are not in Hardy-­‐Weinberg equilibrium because it seems strange that there are no 120/120 homozygotes, and the mechanism I suspect is selection against these homozygotes (why doesn’t it go to fixation for the 100 allele? Perhaps overdominance?). I will test this hypothesis by estimating the allele frequencies from these data (There are 162+19 copies of the 100 allele, and 19 copies of the 120 allele) and from those identifying what the HWE predicted genotype frequencies would be. The difference between my observations and expectation can be tested with a chi-­‐squared statistical test. It would be good to have a larger sample size, since there is a marginal probability of zero 120 homozygotes at this allele frequency; I also wish I had data from a separate independent locus to ensure that there is random mating with respect to the Gpi locus. Other answers may apply. TAs provide rubric. Full potential credit if HWE was discussed. 4 points for valid hypothesis (ie: pop not in HWE, selection against 120/120) 4 points for expected vs observed allele frequencies (HWE equation) and Chi-­‐squared test Half credit if no HWE is discussed. 2 points for valid hypothesis 2 points for valid method of testing that hypothesis 2. Climate change is happening; the oceans are warmer and more acidic, glaciers and icecaps are receding. Assume you know π (nucleotide diversity) for 10,000 species and you have been asked to report on the probability that each species becomes endangered or goes extinct in the next 100 years because it cannot adapt to this change. Draw a graph that relates π to “potential for adaptation”. Why are these two things related? (5) Because π is a measure of effective population size, and effective population size tells us about standing variation (and additive variation), which is what is necessary for adaptation to happen. A graph that shows a positive relationship between “potential for adaptation”/additive genetic variation and π would be one way to illustrate this. TAs provide rubric. 2.5 for positive association between “potential to adapt” and π. No credit for graphs showing association between anything other than π and potential to adapt. _______ out of ______ points on this page 1 2.5 for showing they understand that selection acts of standing variation and the more variation in a population, the greater for the potential to adapt. 3. Define and give an example of a genetic bottleneck OR founder effect? What is the evolutionary importance of such events for natural populations? (4) Either event is an example where when it happens, genetic diversity is reduced to that contained in the individuals that make it through. It reduces allelic richness, reduces homozygosity (less so), reduces effective population size, has the potential to increase inbreeding. 1 point for definition, 1 point for example, 2 points for understanding of evolutionary importance (those described above) 4. Explain the distinction between broad sense and narrow sense heritability, why is this distinction important? (2) Explain how you would estimate narrow sense heritability (use drawings if appropriate) (2; 4 total points). Broad-­‐sense is all genetic variation divided by all phenotypic variation, in other words proportion of phenotypic variance caused by genetic variance. However, this includes dominance and interaction terms, among other things. Because narrow is additive variance only and that is what selection operates most directly on. These two questions go together. TAs provide rubric. 1 point for distinction or definition of H and h, 1 point for why the distinction is important, 2 points for how to estimate h (I accepted both the slope of a parent-­‐offspring regression and estimating h by selecting on a trait and looking at the shift in the mean trait value in the next generation (heritability curves) This would be estimated by the slope of a parent-­‐offspring regression, where the mean value of the trait in the parents is plotted against the mean value in the offspring (full sibs) and standard regression to get the slope of those points. If it is zero, heritability is zero. To the extent it is non-­‐zero, that is the extent to which there can be a response to selection. Room for a drawing. TAs provide rubric. 5. Matching. Below are two lists of concepts we discussed, and what those mechanisms can do in a population. In the eight blanks below, write a letter-­‐number combination that describes a way these are connected. For example, if choice Z on the left was “selection” and choice 13 on the right was “relative fitness” you could put Z13 in a blank. Some items will be used multiple times, some not at all. We will only grade the first eight! (2 points each) A1, A3, B5, B6, C2, C6, D1, D3, E1, E3, E4, F2, F7, G1, G3 A. Heritability 1. Response to selection B. Linkage Disequilibrium 2. Maintenance of variation C. Nonrandom Mating 3. Parent-­‐Offspring Regression D. Additive Variation 4. Environmental Variability E. Phenotypic Variation 5. Haplotype F. Major Histocompatibility Complex 6. Increased Homozygosity _______ out of ______ points on this page 2 G. Breeders Equation ___ex: B5____
____a1____
___e4____
____g1____
7. Overdominance ___f7____
____c6___
___a3_____
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6. Myzus persicae is an aphid that damages peach crops, but also lays nymphs on vegetable crops. The graph at right is called _reaction norm_ and is used to identify _plasticity or GxE effects_. (2 pts) 1 point for “reaction norm” 1 point for “plasticity”, “phenotypic plasticity”, or GxE 7a. Describe an ideal population that is unlikely to be evolving. (5) Basically the 5 HWE assumptions. 7b. That ideal population above has the following genotypes: A A (found in 36%),
A A (found in 9%), A A (found in 1%), A A (found in 36%), A A (found in 12%),
A A (found in 6%). What will be the allele frequencies in the next generation? (2
pts)
1
2
2
2
3
3
3
1
2
1
1
3
A1 0.6, A2 0.3, A3 0.1 (2 points) 8. Describe what a quantitative trait locus is (2), what information you need to find
a QTL, how is that information used to identify the QTL (2).
A QTL is a locus that is responsible for some of the phenotypic variation observed in a
(polygenic) trait. In order to identify it we need marker loci, which can be any sort of
genetic marker, and hopefully enough of them to densely cover the chromosomes of the
species. We also need an F2 generation that is displaying portions of the overall
phenotypic variation of the two parental types that are of interest. Then we can see which
markers are statistically more likely to show up in F2 individuals that display a particular
component of the trait. TAs provide rubric. _______ out of ______ points on this page 3 9. In A and B to the right, body height of twins are
plotted against each other. Describe all possible reasons
for the phenotypic variance illustrated in both cases;
what is different between the two? (4)
TAs provide rubric. Involves all components of variance, difference is that there is only environmental variance with monozygotic twins (and perhaps epigenetic but we won’t worry about that!) 10. When very strong positive selection acts on an allele that was previously rare in a population, of course the allele goes up in frequency. Aside from this fact, illustrate and fully explain a way that evolutionary geneticists would identify this selection event from genomic data (4). What they should get to are extended haplotypes (which were illustrated in lecture)
or regions of very low diversity (also illustrated), both in accordance with selective
sweeps. In each case, they should mention linkage disequilibrium among nearby
sites, this is the reason for both phenomena.
TAs provide rubric. 11. In a population of barnacles, the average number of reproductive offspring
produced each generation by each hermaphroditic individual is 1. studying a
population of these barnacles carefully, you find there is a rare polymorphism with an A
allele at 10% frequency and a G at 90% frequency. AA individuals on average
produce 1.10 reproductive offspring, with other genotypes producing (on average)
0.99 offspring. predict the frequency of the A allele in the next generation. you may
do this with numbers or with algebra.
_______ out of ______ points on this page 4 a. what is the AA genotype frequency?
b. what is the AA fitness ratio?
c. what is the AG genotype frequency?
d. what is its fitness ratio?
e. now what should be the frequency of A in the next generation given this information?
(2 points each) They should be able to get A and C no problem, that is super basic. B and D they just need to understand that the ratio of w11 to mean w is 1.1, and is just less than 1 for the heterozygote, and that should get them to those answers. The crux is (e). Here they need to put together the genotype frequency of homozygotes (0.01) and the fitness ratio (1.1) plus half of the genotype frequency of the heterozygotes (here it should be 0.18/2 = 0.09) times that fitness ratio (so, 0.09X0.99 = 0.09-­‐0.009 = 0.0891). An astute student may notice that the allele frequency has barely gone up in one generation (p(t+1)=0.1001) and that is the peril of a rare allele. And yet if the initial allele frequencies had been 0.5/0.5 in one generation it could have changed by 0.27… TAs provide rubric. Grading Rubric for Exam 2, Page 2 7a. This question was looking for the five assumptions of Hardy-­‐Weinberg Equilibrium. One point was awarded for listing each of the following assumptions (5 points total): no migration (gene flow), infinite population size (no drift), no mutation, no selection, and random mating. 7b. This question was looking for allele frequencies and not genotype frequencies, which were given in the question. We were looking for actual numeric values as well. Thus, no points were awarded if the student simply stated “The allele frequencies will not change from one generation to the next because it is in Hardy-­‐
Weinberg Equilibrium”. The answers were A1=.6, A2=.3, A3=.1. Full credit was given if these numbers were provided. Full credit was also awarded if the students provided the correct equations to arrive at that those values but did not complete the math, since calculators were not allowed on this exam. Half credit was given if the student was able to provide some answers correctly, but the others were incorrect. 8. Two points were awarded if the student provided the correct definition of a QTL, which is a locus that is responsible for some of the phenotypic variation observed in a polygenic trait. A variant on that answer that was similar to that definition was awarded full credit. Only one out of the two possible points was awarded if the student omitted the fact that the trait had to be polygenic but still said that it was a gene that encoded for a trait. For the second half of the question, one point was given if the student mentioned that genetic markers are required for QTL analysis. One point was also awarded if the student mentioned that a F2 generation was required for QTL analysis. _______ out of ______ points on this page 5 9. Two points were awarded for this question if the student mentioned that monozygotic twins were more genetically similar than dizygotic twins, thus arriving at the differences observed in the figure. Full credit was awarded if the students mentioned that the only reason we observe any differences in the monozygotic twin comparisons arise from environmental variation alone, while the differences observed in the dizygotic twins involved both environmental and genetic variation. Two points were removed if the student stated that monozygotic twins encountered the exact same environment since they developed in the same womb. Points were deducted from this because the environment was NOT EXACTLY the same for these individuals. 10. No points were awarded for this question if the student said something along the lines of “sequences a population at generation 1, come back 50 generations later and sequence the population again and see if we observe an changes in allele frequency.” Two points were awarded if the student just mentioned selective sweeps but failed to define what it was. Full credit was awarded if the student provided a picture or explanation of what a selective sweep was, including mentioning linkage of genes to positively selected alleles. 11. Two points were given for each of the five “sub-­‐questions” for this question. Again, like in question 7b, full credit was awarded if the correct numerical values were provided or if the student provided the correct equations that would produce the correct numerical values if worked out (again, because the students were not allowed to use a calculator in this exam. Full credit was given for a sub-­‐question if the equations were correct but the numerical values were incorrect due to the fact that they had gotten a previous question wrong (i.e. propagated error). a. (.1)2 = .01 b. 1.1 c. (.1) x (.9) x 2 = .18 d. .99 e. (.01 x 1.1) + (.1 x .9 x .99) = .1001 Multiple Choice, Exam 2. 16 questions worth 2 points each 1. Which of these can cause allele frequencies to change in a population? a. Lineage sorting b. Epistasis c. Low population size d. Physical linkage e. Fractional meiosis 2. “Random mating” doesn’t mean lack of mate choice, it means a chosen mate is randomly selected with respect to a. marker loci b. gender c. phenology d. fitness _______ out of ______ points on this page 6 3.
4.
5.
6.
e. all of the above If the Luria-­‐Delbruck study shows that mutations happen constantly during DNA replication, which factors will have the greatest effect on the number of polymorphic sites in a population? a. Population size b. Recombination c. Body size d. Environmental change e. Epistasis Which of these conditions must be met for two loci to be in linkage equilibrium? a. Frequency of an allele at 2nd locus is dominant to all alleles at 1st locus b. Point mutations must have happened recently c. Frequency of any haplotype is obtained by multiplying frequency of alleles d. None of the above Not a condition for Hardy-­‐Weinberg to be met: a. No mutation b. Large population size c. Allelic variation d. Random mating e. Variation in fitness The selection coefficient a. Ranges from 0 to 1 b. Represents proportional difference in fitness between genotypes c. Depends on environment d. Only meaningful for heritable traits e. All of the above 7. Narrow-­‐sense heritability of a trait excludes variation caused by a. Genotype of parents b. Genotype of offspring c. Gene interactions d. A and B e. A and C 8. Selection is least effective when a. Population size is very small b. Heritability is very large c. Population size is very large d. Environment changes rapidly e. It is artificial 9. If a single trait exhibits only two clear phenotypic states, the underlying genetics: _______ out of ______ points on this page 7 a. May involve allelic dominance b. May involve plasticity c. Is unlikely to be a quantitative trait d. All of the above e. Not enough information 10. Humans prefer to catch big fishes, and yet have selected for marine fishes to grow more slowly a. True b. Not True 11. You have a population in which you can observe 6 phenotypes for a given trait. If you cross any 2 individuals, the offspring will be in one or more of these 6 classes. This trait is most likely: a. multilocus b. epistatic c. at mutation-­‐selection equilibrium d. involves dominant alleles e. Mendelian 12. The mean tail length of Anolis sagrei is 45.5mm, with standard deviation 10mm. You collect 10 adults from southern GA and allow them to mate, and measure tail length in offspring. You find it is now 41.7mm, with standard deviation 10mm. Heritability of tail length must be: a. Influenced by a bottleneck b. Not enough information c. Between 0 and 0.5 d. Between 0.5 and 1 e. Variable 13. If you sequence DNA at a single locus from 48 individuals, and you find 4 distinct sequences – allele A is at frequency 50%, allele B at frequency 25%, allele C at frequency 12.5%, and allele D is found in a. 6.25% of the population b. 12.5% of the population c. 25% of the population d. not enough information 14. At a single Mendelian locus, with alleles D and d, you know that D is dominant to d. You have a male and a female, each with dominant phenotype. If all their offspring also have the dominant phenotype, at least one of the parents must have been a. homoplasious b. homologous c. homozygous d. homophonic _______ out of ______ points on this page 8 e. homogeneous? 15. In a standard deck of cards, there are 4 equally represented “suits”: spades, clubs, hearts, diamonds. You pick a card out of the deck. Your friend is asked to do the same. What is the probability that both of you have picked a diamond card? a. 0.5000 = 1/2 b. 0.2500 = 1/4 c. 0.0625 = 1/16 d. 0.0500 = 1/20 e. 0.0385 = 2/52 16. The nuclear genome of most eukaryotes is composed <5% of regions that encode proteins. How can a mutation that happens in the remaining 95% affect the phenotype? a. It happens in a promoter region b. It happens in the plastids (mitochondria, chloroplasts) c. it involves major duplication or deletion of portion of the chromosome d. all of the above e. none of the above _______ out of ______ points on this page 9