University of California, Davis

BIS 2A

Britt final exam key V2

1)The energy for photosynthesis is provided by

1. 1. The splitting of water
   2. The absorption of photons by chlorophyll
   3. The citric acid cycle
   4. Glycolysis
   5. The reduction of NADH

2) There was no O₂ in the early earth’s atmosphere because

1. O₂ was poisonous to life
2. The early Earth was very hot, and O₂ is very reactive, so it reacted with other chemicals.
3. The element O was very rare on Earth, until there was photosynthesis.
4. It was devoured by early forms of life.

 3) Avery, MacLoed, and McCarty were able to genetically transform “rough” pneumococcus to “smooth”, using an extract from killed pneumococci, They concluded that DNA rather than protein was the genetic material because:

1. The extract, when treated with protease, still transformed cells
2. The extract, when treated with protease, no longer transformed cells
3. The extract, when treated with DNAse, still transformed cells
4. The extract, when treated with DNase, no longer transformed cells
5. Both a and d

4) The net movement of uncharged molecules across a membrane barrier from a lower concentration to a higher concentration can be best explained by which of the following? a.diffusion

1. 2. active transport
   3. osmosis
   4. facilitated diffusion e.exocytosis

5) In respiration, most of the ATP is generated as a result of…

1. Fermentation
2. Substrate-level phosphorylation, during glycolysis
3. Substrate-level phosphorylation, during the citric acid cycle
4. Oxidation of NADH via the ETC and PMF
5. Reduction of pyruvate

Refer to the diagram below to answer the next two questions. The Y axis represents the standard potential energy (G^0’) of the molecules A,B.C,D and their intermediates. A,B,C, and D are present at steady state at 1 Molar concentration in the cell.

6. Which of the following statements is false?
   1. The reaction C+D -> A+B releases energy, and so might be used to power another reaction
   2. The reaction A+B -> C+D releases energy, and so might be used to power another reaction
   3. In the cell, some molecules of C + D will react to form A + B.
   4. In the cell, the net (overall) direction of this reaction is A + B -> C + D
   5. In the cell, some molecules of A + B will react to form C + D

7. Addition of a catalyst to the above reaction will:
   1. Increase the ?G of the reaction
   2. Decrease the ?G of the reaction
   3. Decrease the activation energy
   4. Increase the activation energy
   5. Change the equilibrium constant

8. In what intracellular compartment is ATP synthesized via oxidative phosphorylation?
   1. The nucleus
   2. The cytoplasm
   3. The golgi
   4. The mitochondrial intermembrane space
   5. The mitochondrial matrix

9. What is the external source of electrons used to reduce NADP+ during oxygenic photosynthesis?
   1. They are produced when sugars are oxidized
   2. They are transported across the thylakoid membrane by ATP synthase
   3. They are extracted from water by photosystem II
   4. They are derived from metals in the environment
   5. b and c

10. The light reactions of photosynthesis are similar to those of respiration in that:
    1. Both processes consume O₂
    2. Both processes consume NADH
    3. Both processes produce ATP
    4. Both processes involve an electron transport chain
    5. c and d

11. In the purine nucleotide biosynthetic pathway drawn below, the numbers represent enzymes required for the synthesis of GMP and AMP. The biosynthetic pathway branches at IMP. This pathway is subject to allosteric feedback regulation. If the concentration of AMP is too high, and the concentration of GMP is too low, a good way to adjust the concentration would be to have:

1. 1. AMP act as a negative regulator of enzyme 1
   2. GMP act as a positive regulator of enzyme 2
   3. AMP act as a negative regulator of enzyme 3
   4. GMP act as a negative regulator of enzyme 5
   5. AMP act as a negative regulator of enzyme 5

6

12) A protein’s tertiary structure is stabilized by…

1. Hydrophobic R groups
2. Ionic bonds
3. Ribosomes
4. Hydrogen bonds
5. a, b, and d

13. Which of the following mutations will reduce the ability of the cell to synthesize proteins correctly?
    1. A mutation that reduces the accuracy of DNA polymerase
    2. A mutation that reduces the accuracy of RNA polymerase
    3. A mutation that affects the accuracy of an aminoacyl-tRNA synthetase
    4. A mutation that alters the anticodon of a tRNA
    5. All of the above mutations will result in the synthesis of defective proteins

14. How can deleterious (“bad”, nonfunctional) mutations be eliminated?

1. Through DNA replication
2. Through DNA repair
3. Through natural selection
4. Through polymerization
5. Through Watson-Crick base pairing

15. A helicase is required for DNA replication. Helicases…
    1. excise mismatched bases
    2. charge tRNAs with amino acids
    3. degrade RNA primers
    4. seal up nicks in the backbone of DNA
    5. separate the two strands of the double helix

16. A DNA sequence reads: 5' ATCGATCG 3'

The sequence of the strand annealing to it (= the complementary strand) would read

1. 1. 5' CGATCGAT 3'
   2. 5' TAGCTAGC 3'
   3. 5' GCATGCAT 3'
   4. 5' ATCGATCG 3'
   5. 5' TACGTACG 3'

17. The synthesis of a polymer usually requires energy (i.e., is an endergonic reaction). What provides the energy for the polymerization of DNA?
    1. hydrolysis of ATP to ADP plus phosphate
    2. hydrolysis of ATP to AMP + pyrophosphate (PPi), and subsequent cleavage of PPi to phosphate
    3. release of phosphate from the incoming dNTP (deoxynucleotide)
    4. release of pyrophosphate (PPi) from the incoming dNTP (deoxynucleotide) and subsequent cleavage of PPi to phosphate
    5. hydrolysis of the complementary strand

Use this information to answer questions 18 and 19:

The maternity plant produces plantlets (“babies”) along the edge of the leaf. Individual leaf cells divide and form somatic embryos, i.e. embryos that are not derived from zygotes. Each embryo develops into a plantlet that will fall off the leaf and grow next to the mother plant. The plant also forms seed from normal flowers, which can cross- or self-pollinate.

18. The genotype of the "mother" plant at the "z" locus is Zz. The most likely set of genotypes of 12 plants produced from the leaf edge of this plant is:

a. all Zz                                                                                  b. 6 ZZ, 6 zz

c. 3 zz, 6 Zz, 3 ZZ                                                                d. 3 zz, 9 ZZ

e. 1 zz, 2 Zz, 9 ZZ

19. The genotype of the "mother" plant at the "z" locus is Zz. The most likely genotypes of 12 plants grown from the seeds of this self-pollinated plant are:

a. all Zz                                                                                  b. 6 ZZ, 6 zz

c. 3 zz, 6 Zz, 3 ZZ                                                                d. 3 zz, 9 ZZ

e. 1 zz, 2 Zz, 9 ZZ

20. Bacteria, Archaea, and Eukarya represent the three major subdivisions of living organisms. Which of the answers below is correct?

21. A eukaryotic chromosome is typically                                                and it is replicated                                     

                   .:

21. 1. linear, from a single origin
    2. linear, from multiple origins
    3. linear, from two terminal origins
    4. circular, from a single origin
    5. circular, from multiple origins
22. The mitochondrion evolved from:
    1. A bacterium that could use O₂ as an external electron acceptor for respiration.
    2. The endoplasmic reticulum
    3. The hydrogenosome
    4. The plastid
    5. An aberrant cell division that maintained one cell inside the other

23. Male pattern baldness is an X-linked recessive trait. It can also be observed in women, if they are homozygous for the recessive allele. If 40% of the men in a population display “male” pattern baldness, what is the expected frequency of women with “male” pattern baldness?

a. 36%

b. 40%

c. 48%

d. 16%

e. 8%

24) Which of the following molecules would be expected to most rapidly diffuse across a cell membrane?

1. glucose
2. a sodium ion
3. water
4. CH₄
5. an amino acid

25. The object marked as F in the diagram above is/are:

1. two paired chromosomes
2. two chromatids engaged in crossing over
3. one chromatid
4. one chromosome, before replication
5. one chromosome, after replication

26. The eukaryotes represent a major subdivision of living organisms. Consider the processes and structures below and choose the answer that best summarizes their status in eukaryotes:

27. Enhancers                 are       frequently          found        in      eukaryotic           genes,         but      not      in prokaryotic genes. One possible explanation for this difference is:
    1. The       prokaryotic            genome         is      too      compact  to             include         regulatory sequences
    2. The prokaryotic genome is too large for enhancers to function
    3. The density of genes (#genes  per kilobase) is  too high for enhancers to specifically affect a single operon
    4. An entire operon cannot be regulated by an enhancer

28. Bacteria tend to locate their genes in functional clusters  (operons)- where  the genes involved in a pathway involving several enzymes, and their regulatory sequences, will be co-located. In eukaryotes, however, the genes involved a biosynthetic pathway are randomly distributed throughout the genome. Why does this discrepancy exist?
    1. The eukaryotic genome is older, and the genes have moved apart due to random mutations
    2. DNA is often inherited via vertical, as well as horizontal, transmission in eukaryotes
    3. DNA is often inherited via horizontal, as well as vertical, transmission in prokaryotes
    4. Meiotic recombination breaks up gene clusters in eukaryotes, but not prokaryotes
    5. Meiotic recombination breaks up gene clusters in prokaryotes, but not eukaryotes

29. A mutant is isolated with a nonsense mutation in the 5’ end of the RNA encoding the lac repressor. Such a mutant would be expected to:
    1. Strongly and constitutively express the lac operon
    2. Strongly express the lac operon only in the presence of trp
    3. Strongly express the lac operon only in the presence of lactose
    4. Strongly express the lac operon only when lactose is absent
    5. Strongly express the lac operon only the presence of high levels of cAMP.

30. A mutant is isolated with a mutation in the trp operator that prevents binding of the trp repressor. Such a mutant would be expected to:
    1. Strongly and constitutively express the trp biosynthesis genes
    2. Strongly express the trp biosynthetic genes only in the presence of trp
    3. Strongly express the trp biosynthetic genes only in the presence of lactose
    4. Strongly express the trp biosynthetic genes only when trp is absent
    5. Strongly express the trp biosynthetic genes only the presence of high concentration of cAMP.

The plant Coleus sometimes turns its leaves red. A hormone (here called “signal”) is involved in the regulation of flowering and leaf color. The indicated pathway (below) is hypothesized to explain its action.

31. The T-bar symbol depicted above is consistent with the action of a protein that:
    1. modifies nucleosomes and causes chromatin condensation at its target gene
    2. modifies nucleosomes and causes chromatin decondensation at its target gene
    3. binds the fully translated RNA causing its release from the ribosome
    4. binds the promoter of a gene, facilitating RNA polymerase II binding
    5. removes introns from RNA
32. The arrow symbol depicted above is consistent with the action of a protein that:
    1. modifies nucleosomes and causes chromatin condensation at its target gene
    2. modifies nucleosomes and causes chromatin decondensation at its target gene
    3. binds partially translated RNAs, causing their release from the ribosome
    4. binds the promoter of a gene, inhibiting RNA pol II binding
    5. removes introns from RNAs.
33. "Signal" is a small molecule. Predict the correct outcome when “signal” is applied to the plant, according to the pathway in the figure above.

34. Generally speaking, the genes for regulatory proteins that directly detect environmental signals (for example, the Lac Repressor protein) are transcribed…
    1. All the time (constitutively)
    2. Only when the genes they regulate are transcribed
    3. Only when the genes they regulate are not transcribed
    4. Only when the signal they detect is present

35. The fastest way to negatively regulate an enzymatic activity would be:
    1. Mutate the gene
    2. Turn off transcription of its gene
    3. Prevent translation of its mRNA
    4. Prevent degradation of the protein
    5. Allosteric inhibition of the protein

36. The least costly (least ATPs!) way to negatively regulate an enzymatic activity would be:
    1. Mutate the gene
    2. Turn off transcription of its gene
    3. Prevent translation of its mRNA
    4. Prevent degradation of the protein
    5. Allosteric inhibition of the protein

37. Chromatin silencing is often observed in eukaryotes, but not prokaryotes. Chromatin silencing may have evolved to cope with:
    1. The lack of operons in the eukaryotic genome
    2. The high mutational load found in eukaryotic genomes
    3. The high number of transposons and retrotransposons in the eukaryotic genome
    4. The very compact (many genes per kilobase pair) genome of most eukaryotes
    5. The presence of centrosomes in eukaryotes

38. Methylation of C at CG sequences is often found in silenced genes. This type of methylation contributes to epigenetic gene silencing because:
    1. Methyl-C cannot be read correctly by DNA polymerase
    2. Methyl-C cannot be read correctly by RNA polymerase
    3. Methylation patterns at CG are heritable from one cell generation to the next
    4. Methylation patterns cannot be transmitted from one cell generation to the next
    5. b and c

39. The eukaryotic RNA polymerase II, required for mRNA production, cannot recognize promoters without assistance. What additional proteins/events are required for RNA pol II to bind to the promoter?
    1. The DNA must be heavily methylated
    2. The DNA must have recently undergone replication
    3. Transcription factor II D must be bound to the TATA box
    4. The nucleosomes must be removed from the gene

1. 5. The statement above is not true- RNA polymerase II can recognize promoters on its own.

40. Transposons are often silenced in eukaryotic genomes. Although their transposase genes carry functional promoter DNA sequences and encode functional proteins, the protein is not expressed. Which of the following is NOT a mechanism for gene silencing?
    1. Methylation of DNA by micro RNAs
    2. Acetylation of histones at transposon sequences
    3. Degradation of transposon mRNA via micro RNAs
    4. Compaction of chromatin structure at transposons
    5. Deacetylation of histones at transposon sequences.

41. Head to tail polarity in drosophila is established in part by gradient in the bicoid messenger RNA in the egg. This gradient can be observed in the egg below, where bicoid mRNA is darkly stained. The progeny of female drosophila homozygous for a knockout in the bicoid gene (bb) are 100% headless (but have 2 tails), regardless of the male’s genotype. In contrast, the progeny of two flies that are heterozygous for the bicoid knockout (Bb) all have heads. How can you explain this pattern of inheritance?
    1. The bicoid knockout is lethal when homozygous.
    2. The bicoid knockout cannot be inherited from the father.
    3. The mRNA in the egg comes only from the mother.
    4. The gene is always silenced if it is inherited from the father.

42. Leucine is an amino acid required for protein synthesis. How might the leucine biosynthetic operon be regulated?
    1. Induced by leucine, by having leucine inhibit a repressor
    2. Induced by leucine, by having leucine activate an activator
    3. Repressed by leucine, by having leucine activate a repressor
    4. Repressed by leucine, by having leucine inactivate an activator
    5. Both c and d would work.

43. Methylation, phosphorylation, and acetylation are all ways of modifying…
    1. DNA
    2. RNA
    3. RNA polymerase
    4. Nucleosomes
    5. TFIID

44. A promoter includes
    1. a site that binds tRNA
    2. a site that binds RNA polymerase
    3. a site that binds regulatory factors
    4. a site that binds ribosomes
    5. b and c

45. What is the catalytic activity of the lac repressor?
    1. It polymerizes RNA
    2. It degrades DNA
    3. It transports lactose across membranes
    4. It digests lactose
    5. It has no catalytic activity

46. For complete expression of the Lac operon two criteria must be met, the absence of glucose and the presence of lactose. The reason for this is:
    1. All bacteria can use glucose, so it’s best to use this up first, before another bacterium does.
    2. Glucose catabolism genes are expressed all the time; they’re part of “core metabolism”
    3. It is a waste of energy to produce proteins that are not needed, therefore, they do not turn on the Lac operon until lactose is available.
    4. a and b
    5. a, b, and c are all true

47. In E. coli, cAMP is a signal for

1. a drop in the intracellular ATP levels
2. a drop in the extracellular lactose levels
3. a rise in the extracellular lactose levels
4. a rise in the extracellular lactose ATP levels
5. a rise in the extracellular glucose levels

48. Our cells respond to growth factors (mitogens) through the signal transduction cascade depicted below. The majority of the activating events shown require…
    1. a protein kinase
    2. a protein phosphatase
    3. a protease
    4. a transcription factor
    5. methylation

49. A mutation occurs in the MEK gene in one of your cells, changing the amino acid that is targeted for phosphorylation by Raf into an amino acid that is a “phosphomimic”­ it makes the protein behave like the phosphorylated version of that protein.  As a result:
    1. Nothing happens­ that amino acid isn’t essential for the protein’s function
    2. Cell division is constantly stimulated in that cell, causing it to become a tumor
    3. Cell division is shut down permanently in that cell
    4. The growth factor receptor migrates to the nucleus

50.   The signal transduction pathway depicted above requires many intermediate steps before the signal reaches the final “effectors” of the response (including transcription factors). What is one advantage of having multiple steps, instead of a direct connection between the growth factor receptor and the target genes?
    1. Having a greater variety of signal transduction genes involved means its less likely that a single mutation could mess up the pathway.
    2. Having multiple steps allows the cell to amplify the initial signal.
    3. It’s not possible for a membrane­bound receptor to also be a transcription factor.
    4. a and b are true
    5. Having multiple activating steps makes it less likely that the response will be induced by accident.