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Biology 1951 – Fall 2021 – Exam 3 (100 points)

1. Gene Regulation (25 points)

A prokaryotic cell can use three sugar sources, Sugar A, Sugar B, Sugar C. Sugar B and Sugar C are chemically related, but Sugar A is the preferred energy source. Three proteins, Gopherase 1, Gopherase 2 and Gopherase 3 code for enzymes that are necessary to break down Sugar B and Sugar C, but are not involved in the breakdown of Sugar A. The genes for these three enzymes are located in the Gopher operon. The table provides information about the regulation of the Gopher operon.

Sugar A

Sugar B

Sugar C

Level of Transcription

Present

Present

Present

0

Present

Absent

Present

0

Present

Present

Absent

0

Present

Absent

Absent

0

Absent

Present

Present

2

Absent

Absent

Present

1

Absent

Present

Absent

1

Absent

Absent

Absent

0

Below are two of the equations we generated in class to represent gene regulation.

Equation 1: (1-R)(1+A)*S*P = T

Equation 2: (1-R)(1+A)*TF1*TF2*TF3*TF4*TF5*P*C=T

Where R=Repressor, A=Activator, TF=Transcription Factor, S=Sigma, P=RNA polymerase, C=chromatin

1A. (2 points)Which of the two equations would you choose to modify to represent this situation? Explain why specifically referring to variables in the equation.

I would choose Equation _____________

Because

1B. (7 points) How would you modify the equation you chose to describe the regulation of the Gopher operon? Rewrite the equation and provide a key for any new variables you introduce. Be sure to use subscripts to indicate which sugar is controlling the regulatory proteins. Use A for Activator and R for Repressor. (e.g., AF would indicate that the Activator is under control of Sugar F.)

1C. (8 points) Explain your reasoning for why you made the modifications in 1B. If you did not change the equation or part of the equation, explain that as well. Make sure to cover the number of regulatory terms included (terms with R or A in them), and why the regulatory terms are structured the way they are (both why activator or repressor and why that mathematical representation (for example, if you were doing this for our class equation, you would be explaining why the term for the activator is A+1 and not 2A).

1D. (8 points) In a situation where Sugar B and Sugar C are present and Sugar A is absent, show what proteins are bound at the Gopher operon on the picture below and indicate the level and direction of transcription including approximation of transcription start and ending points.

1. Draw the proteins that are bound to the DNA and show where they are bound on the picture below.

2. Draw the proteins that are not bound to the DNA below the picture.

3. Indicate the level and direction of transcription with approximation of starting and ending points.

Label the regulatory proteins so it is clear which sugar is controlling their presence or absence and the role of the protein (e.g., AR would indicate that the Activator is under control of Sugar R). Proteins can be represented by circles or other shapes.

2. Meiosis and Cell Reproduction (25 points)

Plants in the genus Arabidopsis are model organisms in biological research. The species Arabidopsis arenosa is a model organism for meiosis and autopolyploidy because it naturally occurs in both diploid (2n) and tetraploid forms (4n). Both types of plants undergo the same pattern of meiosis, producing eventual gametes with a ploidy half that of the parent cell. The species karyotype has 8 types of chromosome (n=8). For this question, assume that tetraploid plants have four homologs of each chromosome type.

2A. (2 points) Fill is the table below for diploid plants. If pairs are not present, enter a 0 or -.

G1

G2

Beginning of mitosis or meiosis

End of mitosis

End of meiosis I

End of meiosis II

# chromosomes

# homologous chromosome pairs

Ploidy (e.g. 2n)

# sister chromatid pairs

2B. (4 points) Fill is the table below for tetraploid plants. If pairs are not present, enter a 0 or -.

G1

G2

Beginning of mitosis or meiosis

End of mitosis

End of meiosis I

End of meiosis II

# chromosomes

# homologous chromosome pairs

Ploidy (e.g. 2n)

# sister chromatid pairs

2C. (3 points) For your table in 2B for tetraploid plants, explain your reasoning for the numbers in the row on # sister chromatid pairs. You can use pictures in your explanation as well. (1-3 sentences each).

G1:

G2:

Beginning of mitosis or meiosis:

End of mitosis:

End of meiosis I:

End of meiosis II:

2D. (5 points) See the Exam 3 Assignment on Canvas to download a copy of the Morgan et al. 2020 article on meiotic evolution in autotetraplods. This research refers to the “numerous challenges” that new tetraploid A. arenosa plants might face and the strong selection likely present for mechanisms that promote stable meiosis. What is meant by stable meiosis and what challenges might tetraploid plants face in cell reproduction? In other words, what is the result of successful meiosis and what are the consequences if it doesn’t go right? (4-8 sentences. Your answer can refer to meiosis and tetraploids in general and doesn’t need to refer specifically to the article.)

2E. (3 points) In the research article (Morgan et al. 2020), the research focuses on prophase I and metaphase I during meiosis I. Why are these critical phases in meiosis? Why would they be the focus of meiosis evolution? Your answers to 2A and 2B may help you here. (2-6 sentences)

2F. (4 points) According to the research article (Morgan et al. 2020), describe ONE structural change that seems to have evolved that promotes stable meiosis. What potential problem in meiosis does the change address? Refer to the structure(s) involved in meiosis and what problem they might solve, not the genes that may be associated with the structure(s). Write in your own words, using minimal quotes from the article. No need to understand all parts of the article – focus on overall results. (3-6 sentences)

2G. (4 points) Some species of Arabidopsis can reproduce via self-pollination (akin to asexual reproduction) and cross-pollination (sexual reproduction). Why might self-pollination be advantageous in some circumstances? Consider that Arabidopsis is often considered a weed. Why might cross-pollination be advantageous in other circumstances? Think about the costs and benefits of asexual vs. sexual reproduction. (4-8 sentences)

3. Inheritance (25 points)

In a different land and a different time, you are a dragon breeder. Giant green dragons with horns that breathe red fire and have wings fetch the most money. The inheritance pattern for these traits is shown in Table 1.

Table 1. Genotypes and phenotypes of different dragon traits

Trait

Homozygous Dominant

Heterozygous

Homozygous Recessive

Size

BB – big

Bb – big

bb – small

Color

GG – green

Gg – green

gg – blue

Horns

HH – horns

Hh – horns

hh – no horns

Fire

RR – yellow fire

Rr – yellow fire

rr – red fire

Wings

NN – wings

Nn – wings

nn – no wings

You have a male and female dragon with the valuable phenotype that you have recently bought to breed to produce more dragons with the valuable phenotype to sell.

Figure 1: The genotypes and phenotypes of the two dragons

The allele combinations and expected proportions produced by the male and female are shown in Table 2.

Table 2. Proportion of Sperm and Egg Types

Sperm Types

Proportion of Sperm Types

Egg Types

Proportion of Egg Types

rNgBH

.12

rNgBH

.02

rNgbh

.12

rNgbh

.02

rNgBh

.08

rNgBh

.03

rNgbH

.08

rNgbH

.03

rnGBH

.12

rnGBH

.02

rnGbh

.12

rnGbh

.02

rnGBh

.08

rnGBh

.03

rnGbH

.08

rnGbH

.03

rNGBH

.03

rNGBH

.08

rNGbh

.03

rNGbh

.08

rNGBh

.02

rNGBh

.12

rNGbH

.02

rNGbH

.12

rngBH

.03

rngBH

.08

rngbh

.03

rngbh

.08

rngBh

.02

rngBh

.12

rngbH

.02

rngbH

.12

3A. (3 points) Provide a mathematical model(s) or diagram(s) for calculating the proportion of the sperm or eggs. Use variables (not numbers, unless they would not change) and provide a key. Make sure the model meets our criteria for mathematical models of biological phenomenon.

3B. (11 points) The mathematical model/diagram should apply to all calculations. To demonstrate that it works, explain how the model/diagram would be used to calculate the probability/proportion of rrNGBh sperm. Don’t just plug numbers in. Explain where those numbers came from referring to the following biological processes:

1) Segregation of alleles during meiosis

2) Independent assortment of chromosomes during meiosis

3) Genetic recombination

Remember to describe these processes in your explanation, not just repeat these phrases. Explain why you multiply or add.

3C. (11 points) Dragon buyers are fickle. The trendy dragon is now one that is small and blue has no horns, has wings and can breathe red fire as shown here:

Write the possible genotype(s) of this dragon:

Write a mathematical model for calculating the probability of producing any offspring genotype (include a key). Explain how this mathematical model represents the biological process of fertilization.

Use the mathematical model to make a determination of whether the dragon breeder will be able to make a living by breeding the male and female to produce the small blue dragon with wings and no horns that breathes red fire. Make sure you show how you would calculate the probability of producing this dragon. Show your work/explain your reasoning so that we can see how you derived your answer and where your numbers came from. (A pair of dragons produce 4 offspring every year.)

Work:

Can the dragon breeder make a living by producing the two dragons he already has to get the new trendy dragon? Explain your answer.

4. Animal Behavior (25 points)

Image result for lioness roaring

Suppose you are a graduate student working on field research in the Serengeti National Park, Tanzania, studying a newly discovered apparently altruistic behavior seen in female African lions. It has been observed that female lions sometimes leave their pride group, on their own, to patrol their pride’s territory. If a neighbor lion from another pride is found, they will alert their pride-mates by roaring. Given what is currently known about lions and the basis of their social behavior, answer the following questions:

4A. (4 pts) What is the definition of altruism and why does this behavior fit this definition? Be specific, including information on what you learned about lions in class. (2-4 sentences)

 

4B. (3 pts) Kin selection is one hypothesis for the evolution of altruistic behavior. Applying Hamilton’s Rule, determine the level of relatedness required between lions to make this patrolling behavior advantageous, if the benefit (b) = 1.3 and the cost (c) = 0.39. Show a simple calculation.

4C. (4 pts) Does the value of relatedness from 4B suggest that kin selection could explain this behavior in lions? Explain. Consider what you know of the biology of a lion pride from class material, especially expected patterns of relatedness between females within a pride. (3-4 sentences)

 

4D. (7 points) Reciprocal altruism is another explanation for the evolution of altruism. What is reciprocal altruism and could this apply in this case for lions? Apply what you’ve learned in class about reciprocal altruism and lion behavior, and again use the values benefit (b) = 1.3 and cost (c) = 0.39. Consider the general context needed for reciprocal altruism to function. The GameBug simulator (http://ess.nbb.cornell.edu/) we used in class may be helpful here but is not required. (4-8 sentences)

4E. (7 points) Group selection is yet another explanation for the evolution of altruism. What is group selection and could this apply in this case for lions? Apply what you’ve learned in class about group selection and lion behavior, and again use the values benefit (b) = 1.3 and cost (c) = 0.39. Consider the general context needed for group selection to function. The GameBug simulator (http://ess.nbb.cornell.edu/) we used in class may be helpful here but is not required. (4-8 sentences)

1

Promoter Protein 1 Protein 3Operator Protein 2Upstream DNA sequences

Downstream DNA sequences

Promoter

Protein 1 Protein 3Operator Protein 2

Upstream DNA

sequences

Downstream DNA

sequences

H h

r N n

Ggr

Male

bB

• N and g kept together 80% of the time

• B and H kept together 60% of the time

H h

r N n

gGr

Female

Bb

• N and G kept together 80% of the time

• b and H kept together 60% of the time

H

h

r

Nn

G

g

r

Male

b

B

•N and g kept together

80% of the time

•B and H kept together

60% of the time

H

h

r

Nn

g

G

r

Female

B

b

•N and G kept together

80% of the time

•band H kept together

60% of the time

,

Derived alleles of two axis proteins affect meiotic traits in autotetraploid Arabidopsis arenosa Chris Morgana,1, Huakun Zhanga,b,1, Clare E. Henrya, F. Chris H. Franklinc, and Kirsten Bombliesa,d,2

aDepartment of Cell and Developmental Biology, John Innes Centre, NR4 7UH Norwich, United Kingdom; bKey Laboratory of Molecular Epigenetics of Ministry of Education, Northeast Normal University, 130024 Changchun, China; cSchool of Biosciences, The University of Birmingham, B15 2TT Edgbaston, Birmingham, United Kingdom; and dInstitute of Molecular Plant Biology, Department of Biology, Swiss Federal Institute of Technology (ETH) Zürich, 8092 Zürich, Switzerland

Edited by Anne M. Villeneuve, Stanford University, Stanford, CA, and approved March 11, 2020 (received for review November 6, 2019)

Polyploidy, which results from whole genome duplication (WGD), has shaped the long-term evolution of eukaryotic genomes in all kingdoms. Polyploidy is also implicated in adaptation, domestica- tion, and speciation. Yet when WGD newly occurs, the resulting neopolyploids face numerous challenges. A particularly pernicious problem is the segregation of multiple chromosome copies in mei- osis. Evolution can overcome this challenge, likely through modi- fication of chromosome pairing and recombination to prevent deleterious multivalent chromosome associations, but the molec- ular basis of this remains mysterious. We study mechanisms un- derlying evolutionary stabilization of polyploid meiosis using Arabidopsis arenosa, a relative of A. thaliana with natural diploid and meiotically stable autotetraploid populations. Here we inves- tigate the effects of ancestral (diploid) versus derived (tetraploid) alleles of two genes, ASY1 and ASY3, that were among several meiosis genes under selection in the tetraploid lineage. These genes encode interacting proteins critical for formation of meiotic chromosome axes, long linear multiprotein structures that form along sister chromatids in meiosis and are essential for recombi- nation, chromosome segregation, and fertility. We show that de- rived alleles of both genes are associated with changes in meiosis, including reduced formation of multichromosome associations, re- duced axis length, and a tendency to more rod-shaped bivalents in metaphase I. Thus, we conclude that ASY1 and ASY3 are compo- nents of a larger multigenic solution to polyploid meiosis in which individual genes have subtle effects. Our results are relevant for understanding polyploid evolution and more generally for under- standing how meiotic traits can evolve when faced with challenges.

meiosis | polyploid | genome duplication | adaptation | evolution

Whole genome duplication, which results in polyploidy, in-creases genome complexity, and plays roles in speciation, adaptation, and domestication (1–5). Yet when polyploids are newly formed they face numerous challenges (1, 4, 6, 7); one of the biggest is the reliable segregation of the additional copies of each chro- mosome in meiosis (1, 7, 8). In diploids, each chromosome has just one homologous partner it can pair and recombine with in meiosis (9), but in polyploids more homologous partners are available, and this can result in multivalent associations, as well as unpaired uni- valents that indicate failures in pairing or recombination (1, 7, 8). Evolved polyploids rarely form multivalents or univalents, suggest- ing that preventing them is an important aspect of meiotic stability in polyploids (8, 10). The molecular basis of multivalent prevention and meiotic stabilization in polyploids remains almost entirely mysterious. A major factor in the evolution of meiotic stability in poly-

ploids seems to involve modulation of crossing over among ho- mologous chromosomes (8). Solutions to polyploid meiosis fall into two major phenotypic groups that follow the distinction between allo- and autopolyploids. Allopolyploids have a hybrid origin and thus carry two or more at least partially distinct “subgenomes” (4, 11). Stable bivalent formation in allopolyploids

involves strengthening pairing partner choice such that chromo- somes recombine preferentially with partners from the same subgenome (6, 12). In contrast, autopolyploids arise from within- species genome duplications (4, 11), do not contain distinguish- able subgenomes, and lack consistent pairing preferences (6–8, 13). The ability to primarily form bivalents in autopolyploids has been proposed to rely in large part on a reduction in crossover (CO) rates, ideally to one per chromosome, which at least in theory can suffice to prevent multivalent formation (8). Indeed, meiotically stable autopolyploids generally have low CO rates (10, 14, 15), and neopolyploid fertility negatively correlates with the diploid CO rate (16, 17). Evolved autopolyploids also usually have distal COs (8, 18–20); why this is important is less clear. We use Arabidopsis arenosa as a model to understand the

molecular basis of autopolyploid meiotic stabilization. This species is a close relative of A. thaliana with naturally occurring diploid and autotetraploid populations (21, 22). The tetraploid lineage arose just once, albeit with subsequent gene flow from diploids (23, 24). Meiosis in autotetraploid A. arenosa is stable, while that of neopolyploids is not, the latter being characterized by abundant multivalent associations, univalents, chromosome mis-segregation, and low fertility (14, 25). Meiosis in evolved

Significance

Genome duplication is an important factor in the evolution of eukaryotic lineages, but it poses challenges for the regular segregation of chromosomes in meiosis and thus fertility. To survive, polyploid lineages must evolve to overcome initial challenges that accompany doubling the chromosome com- plement. Understanding how evolution can solve the challenge of segregating multiple homologous chromosomes promises fundamental insights into the mechanisms of genome main- tenance and could open polyploidy as a crop improvement tool. We previously identified candidate genes for meiotic stabilization of Arabidopsis arenosa, which has natural diploid and tetraploid variants. Here we test the role that derived al- leles of two genes under selection in tetraploid A. arenosa might have in meiotic stabilization in tetraploids.

Author contributions: C.M., F.C.H.F., and K.B. designed research; C.M. and H.Z. performed research; C.E.H. contributed new reagents/analytic tools; C.M. and K.B. analyzed data; and C.M., F.C.H.F., and K.B. wrote the paper.

The authors declare no competing interest.

This article is a PNAS Direct Submission.

This open access article is distributed under Creative Commons Attribution License 4.0 (CC BY).

Data deposition: All image files used in this study are deposited in the ETH Zürich Re- search Collection and are freely available at https://www.research-collection.ethz.ch/ handle/20.500.11850/386103 (DOI: 10.3929/ethz-b-000386103). 1C.M. and H.Z. contributed equally to this work. 2To whom correspondence may be addressed. Email: [email protected]

This article contains supporting information online at https://www.pnas.org/lookup/suppl/ doi:10.1073/pnas.1919459117/-/DCSupplemental.

First published April 9, 2020.

8980–8988 | PNAS | April 21, 2020 | vol. 117 | no. 16 www.pnas.org/cgi/doi/10.1073/pnas.1919459117

D o w

n lo

a d e d a

t U

n iv

e rs

ity o

f M

in n e so

ta L

ib ra

ri e s

o n D

e ce

m b e r

1 4 , 2 0 2 1

tetraploid A. arenosa has several key features, including that COs are few in number, often just over one per bivalent, and that neotetraploids have considerably more multivalents (correlated with low fertility) than evolved tetraploids (14). To understand which genetic changes might be responsible for the evolution of these traits, we previously used genome scans of A. arenosa to identify loci that show strong evidence of having been targets of natural selection in tetraploids (14, 26). Among these are mul- tiple genes encoding proteins important for meiotic processes such as cohesion, axis formation, synapsis, and homologous recombination. In this study, we compare effects on tetraploid meiosis of

derived (tetraploid) and ancestral (diploid) alleles of two of the genes that we previously found to be under selection in tetra- ploid A. arenosa, ASYNAPSIS1 (ASY1) and ASYNAPSIS3 (ASY3). ASY1 and ASY3 are homologs of the yeast axis proteins Hop1 and Red1, respectively (27, 28). The axes are protein structures that form along the lengths of replicated chromosomes in mei- otic prophase I and are essential for chromosome pairing, syn- apsis, and homologous recombination (9). Mutants for ASY1 or ASY3 are defective in synapsis, have low CO rates, high univalent rates, and are nearly sterile (28, 29). Both proteins are also im- portant in yeast for directing repair partner choice to the ho- molog rather than the sister chromatid, an important feature of meiotic recombination (30–33). Hop1 and Red1 interact directly in yeast (34–36), and this is functionally important for re- combination and synapsis (35). Like their yeast counterparts, plant homologs also interact (28, 37). Thus, these proteins are good candidates for collaboratively causing the changes in the recombination rate and/or pattern that we see in tetraploids, which we test here using genetic and cytological approaches.

Results Effects of Alternate ASY1 Alleles on Metaphase I Phenotypes. To begin testing the function of the derived alleles of the axis proteins in tetraploid A. arenosa, we took advantage of naturally segregating variation. While most A. arenosa populations carry only the tet- raploid (T) allele of ASY1, we previously identified some tetra- ploid populations that segregate diploid (D) alleles of ASY1 as rare variants (26). Thus, we generated a PCR marker to detect a ploidy-differentiated polymorphism (Materials and Methods) and used this to identify plants grown from seeds collected from Tri- berg, Germany (TBG), with the genotype A