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    B I O L O G Y 1002BMarch 17, 2013 Time: 2 Hours

    Student number Room Row

    Print Name _____________________________________________________

    I N S T R U C T I O N S - F O L L O W T H E C H E C K L I S T !

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    Instructor Haffie

    Course Bio 1002B

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    Exam Code 111 Very important

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    2) There are 11 pages in this test. Check your paper carefully.3) There is only one fully correct answer for each question. Part

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    Clarifications: Please defend your arguments on this pageonly. Comments will not be accepted after the test.

    Q # Ans Clarification Alt Ans

    40 D If you mean Denis has a dog named Watson, then my answer changes B

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    March 2013 Biology 1002B Page 1 of 11

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    Circle the best single letter choice for each of the following questions before transferring

    your answers to your computer sheet. Note: Questions may have 3, 4 or 5 choices.

    1. One of the key transitions in the development of life occurred after the rise in O2 concentration

    in the atmosphere.

    What was the main source of this O2?

    A. Respiration in archaea and bacteria.

    B. Respiration in early eukaryotes.

    C. Photosynthesis in bacteria.

    D. Photosynthesis in chloroplasts.

    2. Membrane surface area and cell volume are two physical variables that are relevant in

    explaining the difference in size and complexity between prokaryotic and eukaryotic cells.

    Which of the following statements about these variables is correct?

    1. The ratio of plasma membrane area to total cell volume increased in early eukaryotes

    which allowed for greater morphological complexity.

    2. The total membrane area dedicated to oxidative phosphorylation is greater in eukaryotes.

    3. Although physically smaller, prokaryotic cells have more total cell volume than a typical

    eukaryotic cell.

    4. In prokaryotic cells, the membrane location of ATP synthase is the same as that for

    lactate transport.

    A. 1,2 and 3

    B. 1 and 3C. 2 and 4

    D. 4 only

    E. All of 1,2,3 and 4 are correct

    3. Genomes sizes among related eukaryotes vary much more widely than genome sizes among

    related bacteria.

    How is this difference in variability best explained?

    A. Evolutionary processes (e.g. natural selection) exert less constraint on genome size in eukaryotes

    than in bacteria.

    B. Genome size is proportional to energy production - cells that make more ATP have more DNA.

    C. Eukaryotes have a much wider range of chromosome number than bacteria.

    D. Organisms with greater morphological complexity have larger genomes.

    4. Is it true that all known eukaryotes have mitochondria?

    A. Yes, but some eukaryotes have mitochondria that do not contain DNA.

    B. Yes, but some have only one mitochondrion per cell.

    C. No, but all eukaryotes contain mitochondria-derived genes.

    D. No, but those that don't contain mitochondria contain chloroplasts instead.

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    March 2013 Biology 1002B Page 2 of 11

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    5. Once endosymbiosis was first established, it appears that selection favoured organisms with

    smaller organelle genomes over evolutionary time. As a result, modern organelle genomes are

    greatly diminished in size relative to those of free-living bacteria.

    Which of the following is a likely explanation for why diminished genome size would have been

    favoured?

    1. Organelle proteins are smaller, requiring fewer amino acids. Therefore selection has

    favoured organisms with fewer tRNA genes in organelles.

    2. Proteins that are rapidly damaged by reactive oxygen species (ROS) can be replaced more

    efficiently if their genes are located in the nucleus. Therefore, selection has favoured

    organisms where these genes have transferred out of the organelles into the nucleus.

    3. Organelles have typical prokaryotic gene structure. Therefore, selection has favoured

    organisms in which organelle genes have lost their introns.

    4. Following endosymbiosis, several genes would have been present in both the organelle and

    nuclear genomes. Therefore, selection has favoured organisms that lost the organelle copy

    of such redundant genes.

    A. 1,2 and 3

    B. 1 and 3

    C. 2 and 4

    D. 4 only

    E. All of 1,2,3 and 4 are correct

    6. Human mitochondrial genomes code for 37 genes but mitochondria produce only 13 proteins.

    What would most likely account for mitochondrial genomes having more genes than needed to

    code for the proteins produced?

    A. Several genes code for products that are not translated.

    B. Several genes have been inactivated over evolutionary time and no longer code for proteins.

    C. Several genes, and the proteins they code for, are of viral origin. They are not human

    mitochondrial genes.

    D. Several genes are coded on both the top and bottom strands of DNA. Therefore they code for

    the same proteins.

    7. Which of the following sequences in template DNA would likely code for a hairpin transcription

    terminator loop?

    A. 3 GCCCAAAGCTAGCTATTTGGGC

    B. 3 GCGCGCGCGCATATATATATAT

    C. 3 ATGCATGCTACGTACGTACGTA

    D. 3 AUGCAUGCUAAUGCAUGCUAA

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    March 2013 Biology 1002B Page 3 of 11

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    8. The rps10 gene codes for one of several protein components of the chloroplast ribosome.

    This gene is found in the nuclear genome in some flowering plant species but remains in the

    chloroplast genome of other species.

    If the rps10 gene arrived in the nucleus in its DNA form, it contained several associated signals

    related to its expression. Which of the following signals would have been present on rps10 and

    likely functional in the nuclear/cytoplasmic environment?

    1. enhancer

    2. promoter

    3. SD box

    4. start codon

    A. 1,2 and 3B. 1 and 3

    C. 2 and 4D. 4 onlyE. All of 1,2,3 and 4 are correct

    9. Using the section of the genetic code below, identify the sequence of DNA template that would

    code for the polypeptide sequence of Pro-His-Arg.

    A. 3CCACAACGG5

    B. 3GGTGTGGCT5

    C. 3GGUGUGGCU5

    D. 3CCTCATCGT5

    10. Some viruses that infect E. coliproduce an anti-terminator protein that causes RNA

    polymerase to ignore the termination signal of one virus gene and continue transcribing right

    into the next gene.

    What might this anti-terminator protein do?

    A. Bind to DNA, causing a loop that prevents the RNA polymerase from reaching the termination

    signal in DNA.

    B. Bind to mRNA to prevent it from complementary base pairing with itself at the terminator signal.

    C. Bind to the termination factor, keeping it away from ribosomes that reach the stop codon.

    D. Bind to the polyadenylation signal, preventing the action of the RNAase clipping enzyme.

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    March 2013 Biology 1002B Page 4 of 11

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    11. Which of the following features is a characteristic of tRNA anticodons?

    1. Are catalytic, making the tRNA a ribozyme.

    2. For the codon of 5UAA, there is a complementary anticodon 3AUU.

    3. Bind to specific amino acids.

    4. Can be coded in DNA of mitochondrial genomes.

    A. 1,2 and 3

    B. 1 and 3

    C. 2 and 4D. 4 only

    E. All of 1,2,3 and 4 are correct

    12. Although most genes have been lost from mitochondria since the original endosymbiosis was

    established, some remain in the mitochondrial genome.

    Which of the following genes would you most likely find remaining in modern plant

    mitochondria?

    A. components of Rubisco

    B. cytochrome C

    C. snRNA

    D. polyA polymerase

    E. None, modern plants dont have mitochondria.

    13. Ribosomes are ribonucloprotein particles in that they are composed mostly of rRNA with some

    associated ribosomal proteins.

    How would the genes coding for ribosomal RNAs be different than the genes coding for

    ribosomal proteins?

    A. Only the protein coding genes would have transcription termination signals.

    B. Only the RNA coding genes would have anticodons.

    C. Only the protein-coding genes would have polyT sequences giving rise to polyA tails.

    D. Only the protein-coding genes could suffer a nonsense mutation.

    14. Modern Chlamydomonas cells contain one large chloroplast that requires thousands

    of proteins.

    Where are these chloroplast proteins translated?

    1. On the rough endoplasmic reticulum.

    2. Inside the chloroplast.

    3. In the nucleus.

    4. In the cytoplasm.

    A. 1,2 and 3

    B. 1 and 3C. 2 and 4

    D. 4 only

    E. All of 1,2,3 and 4 are correct

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    15. This picture of the structure and expression of the lac operon, showing the coding regions of

    lacI, Z, Y and A genes, in a different style than youve seen before.

    If you were to label the location of the

    transcription terminator sequence in thisimage, how many would there be in DNA?

    A. 1

    B. 2

    C. 3

    D. 4

    16. This Figure shows the activity of the galactosidease

    enzyme (coded by LacZgene) in a mutant strain of

    E. colirelative to a normal strain.

    Mutant (dotted line); Normal (solid line);

    Arrow indicates the addition of lactose to

    the growth medium.

    Which sequence is likely affected by this mutation?

    1. SD box of operon mRNA, such that translation efficiency is decreased.

    2. lacY, such that lactose is not transported into the cell as well.

    3. lacZ gene, such that the enzyme has a lower affinity for substrate.

    4. lacI gene, such that the repressor has a lower affinity for DNA.

    A. 1,2 and 3

    B. 1 and 3C. 2 and 4D. 4 only

    E. All of 1,2,3 and 4 are correct

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    17. We discussed three types of boxes that are present as signals in DNA.

    Which of the following boxes is transcribed?

    1. homeobox

    2. TATA box

    3. SD box

    A. 1 and 3

    B. 2 and 3

    C. 3 onlyD. All of 1,2 and 3

    18. If you imagine a messenger RNA molecule in the cytoplasm of a zebra cell, which of the

    following will likely affect how much protein is made by translation of this message?

    A. The presence of appropriate snRNPs.

    B. The length of the polyA tail.

    C. The strength of hydrogen bonds holding the mRNA to ribosomal RNA.D. The ability of the mRNA to pair with itself to form a helix-turn-helix structure.

    19. Promoters, promoter-proximal regions and enhancers are important regulatory sequences for

    many eukaryotic protein-coding genes.

    Which of the following characteristics do these three different sequences have in common?

    1. Understood by the cell as double-stranded DNA.

    2. Mutations in their sequence may result in increased gene expression.

    3. Contain binding sites for proteins.

    4. Must be upstream of the gene they influence.

    A. 1,2 and 3

    B. 1 and 3

    C. 2 and 4D. 4 only

    E. All of 1,2,3 and 4 are correct

    20. There are a variety of mechanisms of genetic regulation that all fall under the broad category of

    RNA interference (RNAi).

    When does this category of regulation occur during gene expression?

    A. before/during transcription

    B. after transcription; before/during translation

    C. after translation

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    March 2013 Biology 1002B Page 7 of 11

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    21. Imagine that a methylation survey reveals that the region of DNA

    shown in this picture is frequently methylated in the brain cells of

    biology students who do not attend lectures but is not methylated in

    brain cells of students who regularly attend lectures.

    What type of sequence do you see in this DNA?

    A. promoter-proximal region

    B. start codon

    C. zinc-finger domain of gene coding a transcription factor

    D. intron splice signal

    22. Deletion of a single AT base pair from codon number 4 can cause a frameshift mutation in a

    protein-coding gene.

    Which of the following additional mutations will restore the reading frame back to normal such

    that the original stop codon will still function?

    (Note that the amino acid sequence will not necessarily be restored back to normal).

    1. Adding a base pair into each of the next two codons.

    2. Adding a GC base pair back in where the AT pair was deleted.

    3. Adding one base to the next codon and deleting one base from the one after that.

    4. Deleting a base pair from each of the next two codons.

    A. 1,2 and 3

    B. 1 and 3

    C. 2 and 4

    D. 4 only

    E. All of 1,2,3 and 4 are correct

    23. Complementary DNA (cDNA) is frequently created during genetic engineering projects by

    allowing reverse transcriptase to make double-stranded DNA from mature mRNA.

    Imagine collecting mature mRNA from the cytoplasm of adult Drosophila wing cells and reverse

    transcribing it all into a collection of double-stranded cDNAs.

    Which of the following signals would you likely find (as DNA) among this collection of cDNAs?

    1. promoter

    2. Reaper gene start codon

    3. intron4. 3UTR

    A. 1,2 and 3

    B. 1 and 3

    C. 2 and 4

    D. 4 only

    E. All of 1,2,3 and 4 are correct

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    24. The figure below shows four genes in the genome of a goldfish. Shaded areas represent the

    binding site at each genes enhancer region for 6 different activator proteins.

    Assuming that an enhancer is only functional if all activator proteins are bound, which pair of

    genes can be selectively enhanced in a specific tissue without enhancing the remaining two

    genes?

    1. Genes L and W

    2. Genes L and P

    3. Genes T and P4. Genes T and W

    A. 1,2 and 3B. 1 and 3

    C. 2 and 4

    D. 4 only

    E. All of 1,2,3 and 4 are correct

    25. Which of the following cellular components is created on one side of the nuclear membrane but

    functions on the other side?

    1. snRNA

    2. microRNA

    3. peptide tag that targets proteins to mitochondria

    4. histone deacetylase

    A. 1,2 and 3

    B. 1 and 3C. 2 and 4

    D. 4 only

    E. All of 1,2,3 and 4 are correct

    26. Which of the following statements is consistent with the claim multi-cellularity is polyphyletic?

    A. Multicellular plants are closely related to multicellular animals; unicellular plants are

    closely related to unicellular animals.

    B. Unicellular organisms have a common ancestor but multicellular organisms do not.

    C. The common ancestor of all modern multicellular organisms was unicellular.

    D. All unicellular organisms have the necessary genes to be multi-cellular.

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    27. In what way are identical twins not likely identical with respect to DNA and its expression?

    1. One twin might be homozygous for a missense mutation; the other twin might be

    heterozygous for this mutation.

    2. One twin might show higher heritability for disease than the other twin.

    3. One twin might inactivate one X chromosome; the other twin might inactivate both

    X chromosomes.

    4. One twin might make different gametes during meiosis than the other twin.

    A. 1,2 and 3

    B. 1 and 3

    C. 2 and 4

    D. 4 only

    E. All of 1,2,3 and 4 are correct

    28. In class we made predictions about what information comparative genomic studies would reveal

    about the genomes ofChlamydomonas vs. Volvox.

    Which of the following predictions for a relatively dramatic increase in the Volvoxgenome wassupported by the data?

    A. number of total base pairs

    B. average number of genes in gene families

    C. average number of base pairs per gene

    D. number of chromosomes

    29. This image shows the main gene products involved in programmed

    cell death in the nematode worm model system, C. elegans.

    If you discovered a mutant worm that did not show programmed cell

    death during development, where might the mutation be?

    1. Loss of function of death signal receptor gene

    (no binding of death signal).

    2. Loss of function of CED 4 (no interaction with CED 3)

    3. Loss of function of CED 3 (no enzyme activity).

    4. Loss of function of CED 9 gene (insensitive to signals from receptor).

    A. 1,2 and 3

    B. 1 and 3

    C. 2 and 4

    D. 4 only

    E. All of 1,2,3 and 4 are correct

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    30. Recall that a concentration gradient of BICOID protein in fertilized Drosophila eggs establishes

    the anterior-posterior axis of developing embryos. Failure of this mechanism is lethal because

    embryos do not develop a head.

    If you imagine a male and female fly, both heterozygous for the same severe nonsense mutation

    in the bicoidgene, what would happen to their homozygous recessive bicoid/bicoidmutant

    offspring?

    A. The homozygous recessive offspring would all die as headless embryos.

    B. The homozygous recessive male and female offspring would all survive to adulthood but all of

    theiroffspring would die as headless embryos.

    C. The homozygous recessive female offspring would survive to adulthood but only theiroffspring

    would die as headless embryos.

    D. The homozygous recessive male offspring would survive to adulthood but only theiroffspring

    would die as headless embryos.

    31. This figure speculates on the possible

    arrangement of homeobox-containing

    genes in a hypothetical commonancestor ofDrosophila, Amphioxus

    (amphibian) and Mouse. (Note that,

    in mouse, the genes are present on

    four different chromosomes.)

    Which pair of genes is orthologous?

    A. S and T

    B. S and V

    C. S and W (onAmphioxus)

    D. T and V

    32. In a chromatography investigation, which of the following data would you need if you wanted to

    compare the Rf of a particular chlorophyll vs a particular carotenoid?

    1. Distance travelled by the chlorophyll.

    2. Distance travelled by the solvent.

    3. Distance travelled by the carotenoid.

    4. Average distance travelled by all chlorophylls and carotenoids.

    A. 1,2 and 3

    B. 1 and 3

    C. 2 and 4D. 4 only

    E. All of 1,2,3 and 4 are correct

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    33. In response to shortened day length and cool temperatures, many trees begin a period of

    senescence when the breakdown of chlorophyll exceeds chlorophyll production. The leaves of

    these trees change colour from green to yellow and orange.

    Based on your knowledge from the photosynthesis lab, which of the following explanations for

    this change is most likely?

    A. Total pigment production decreases at this time such that only yellowish to orange

    wavelengths of light are reflected.

    B. Chlorophyll breakdown at this time allows additional yellowish to orange pigments to

    become visible.

    C. When chlorophyll breaks down during this time, it appears yellowish to orange.

    D. Chlorophylls that are yellowish to orange are produced during this time.

    34. After the fermentation lab, Jamal and Laura decided to make some alcohol at home by mixing

    table sugar with water and adding yeast. Although the fermentation got off to a good start, it

    stopped before all of the sugar was consumed.

    What should Jamal and Laura do differently in their second batch to prevent wasting sugar?

    A. Add more yeast.

    B. Reduce the sugar in the starting mix.

    C. Add some sodium dodecyl sulfate.

    D. Keep stirring the mix during fermentation to get more oxygen in.

    35. The relationship between Elysia and Vaucheria makes an interesting story for

    educational outreach.

    Which of the following characteristics is found in this system?

    A. An animal that makes its own reduced carbon.

    B. An algae that grows photosynthetically, even after its chloroplasts have been stolen.

    C. An underwater plant that makes oxygen by photosynthesis and then uses the oxygen as

    a terminal electron acceptor in respiration.

    D. A chloroplast that functions without any genes.