Molecular biology of the cell 6th edition bruce alberts test bank

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MOLECULAR BIOLOGY OF THE CELL, SIXTH EDITION Chapter 8: ANALYZING CELLS, MOLECULES, AND SYSTEMS © Garland Science 2015

1.

You have used fluorescence-activated cell sorting to separate blood T lymphocytes from a healthy person and a patient infected with a virus. To sort the cells, you used antibodies against two cell-surface receptors, CD4 and CD8, each conjugated to a different fluorescent dye. Based on the results below, which cell type do you think is depleted by the viral infection: CD4+ helper T cells (H) or CD8+ cytotoxic T cells (C)? In these dot plots, each cell is represented by a dot whose coordinates are the intensities of CD4 and CD8 signals. Write down H or C as your answer.

Virus-infected patient

CD8 signal

CD8 signal

Healthy control

CD4 signal

CD4 signal

2.

Fill in the blank in the following paragraph regarding plant cell cultures. Do not use abbreviations.

“Plant cells can be easily rendered totipotent to give rise to all plant tissues on their own. This can be done by culturing a plant tissue in sterile nutrient media to produce an amorphous cell mass called a …”

3.

Indicate whether each of the following descriptions applies to transformed cell lines (T), nontransformed cell lines (N), or both (B). Your answer would be a four-letter string composed of letters T, N, and B only, e.g. BBNN. ( ) They can be thawed and recultured after long-term storage in liquid nitrogen. ( ) They often grow only when attached to a surface, such as that of a culture dish. ( ) They proliferate to very high densities. ( ) They usually cause tumors when injected into immunocompromised animals.

4.

What are the advantages of monoclonal antibodies over antisera?

A. They can be produced in higher quantities.

B. They can be produced with higher purity.

C. They have a more uniform specificity.

D. They can be made against molecules that constitute only a minor component of a complex mixture.

E. All of the above.

5.

How are antibody-producing hybridoma cell lines immortalized to provide an unlimited source of monoclonal antibodies?

A. By ectopic expression of telomerase B. By transformation with a retrovirus C. By cell fusion D. By overproduction of introduced oncogenes E. By irradiation 6.

To create cellular factories for monoclonal antibody production, cells derived from an immortalized B lymphocyte cell line (M) are cultured together with antibody-secreting B lymphocytes (B) on a so-called HAT medium. The medium contains an inhibitor of DHFR, an enzyme essential for de novo nucleotide biosynthesis. Additionally, it contains substrates for the only alternative nucleotide biosynthetic pathway called the salvage pathway. The enzyme HGPRT is essential for the salvage pathway. One of the cells mentioned above lacks the gene encoding one of the enzymes mentioned. Considering the role of the HAT medium in selecting for hybrid cells that are both immortalized and secrete antibodies, which cell (B or M) do you think is missing an enzyme? Which enzyme is missing in this cell?

A. M – DHFR B. M – HGPRT C. B – DHFR

D. B – HGPRT 7.

Sort the following cellular components to reflect the centrifugal force required to sediment them by preparative ultracentrifugation, from low to high force required. Your answer would be a four-letter string composed of letters R, M, L, and N, e.g. LRNM. (L) Lysosomes (M) Microsomes (N) Nuclei (R) Ribosomes (free cytosolic) Answer: NLMR

Difficulty: 2

Section: Purifying Proteins

Feedback: During cell fractionation by ultracentrifugation, progressively higher speeds are used to fractionate different components. Nuclei sediment in the early steps whereas macromolecular complexes such as ribosomes come out in later steps.

8.

Due to friction with molecules of the surrounding solution, sedimenting particles in a mixture reach a terminal velocity when subjected to ultracentrifugation. The sedimentation coefficient (S) for a particle is defined as the ratio of terminal velocity (vt) to the centrifugal acceleration applied to the particle (ac).

S = vt / ac If each Svedberg unit (S value) is defined as 10–13 seconds, approximately how long would it take for a free 50S ribosomal subunit to travel the length of a 9 cm ultracentrifuge tube subjected to a centrifugal force of 50,000 times gravity (g)? Recall that g is approximately 10 m/s2.

A. 1 min B. 10 min C. 50 min D. 5 h E. 10 h 9.

Indicate whether each of the following descriptions better applies to equilibrium sedimentation (E) or velocity sedimentation (V). Your answer would be a four-letter string composed of letters E and V only, e.g. EEEE. ( ) It separates subcellular components based on their buoyant density. ( ) It can conveniently separate 15N-containing and 14N-containing DNA.

( ) It uses a shallow sucrose gradient to stabilize the sedimentation bands against convection. ( ) In this method, subcellular components may either move up or down to form discrete bands.

10.

For the two following graphs, the same purified protein has been subjected to velocity sedimentation and gel filtration, as indicated. Two absorption peaks are visible in each graph, corresponding to the monomeric and dimeric forms of the protein. Which peaks correspond to the protein dimer?

Velocity sedimentation

A280 (relative protein amount)

1

Top

Bottom

Fractions

Gel filtration 3

1

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

2

4

Fractions

100

11.

In purifying proteins by column chromatography, elution can be performed in different ways, depending on the type of matrix. For example, bound proteins can be eluted by gradually increasing or decreasing salt concentration in the solution applied to the column. For which of the following matrices do you think increasing and decreasing salt concentration, respectively, are used for elution?

A. Ion-exchange; hydrophobic B. Affinity; ion-exchange C. Hydrophobic; affinity D. Hydrophobic; ion-exchange E. Ion-exchange; gel-filtration 12.

Which of the following chromatography methods resembles immunoprecipitation in the purification of proteins from mixtures?

A. Ion-exchange chromatography B. Gel-filtration chromatography C. Hydrophobic interaction chromatography D. Affinity chromatography 13.

How is tandem affinity purification distinct from other purification schemes such as

simple affinity or ion-exchange chromatography?

A. In its use of antibodies B. In its use of proteases C. In its use of glutathione S-transferase D. In its use of viral peptides E. In its use of limited washing steps 14.

In SDS-PAGE of proteins, …

A. the proteins are unfolded while being separated.

B. the proteins are separated by size, mostly independent of their native charge.

C. large proteins move more slowly.

D. an ionic detergent is used.

E. All of the above.

15.

Migration of proteins during separation by SDS polyacrylamide-gel electrophoresis (SDS-PAGE) is generally related to the logarithm of their molecular weight, as depicted in the qualitative graph below. If a protein mixture containing proteins of evenly distributed molecular

Relative migration rate

weights (e.g. 10, 20, 30, 40, etc.) is separated by SDS-PAGE, which of the following staining patterns would you expect to observe?

Log (MW)

A

B

C

D

16.

In isoelectric focusing, is the low-pH end close to the anode (A; positive electrode) or cathode (C)? Would you use octylglucoside (O; a nonionic detergent) or SDS (S) for separation of proteins by this technique? Would a highly basic protein be near the high-pH (H) or low-pH (L) end of the strip at the end of the experiment? Write down your answer as a three-letter string using the letters in the parentheses above, e.g. ASH. 17.

Three subunits of a protein complex have been resolved in a two-dimensional polyacrylamide gel, as shown in the schematic drawing below. Indicate whether each of the

following better describes the a, b, or c subunit. Your answer would be a three-letter string composed of letters a, b, and c only, e.g. cca.

SDS migration

Low pH

High pH c

pH gradient

a

b

( ) It has the smallest apparent molecular weight. ( ) It has the highest negative charge. ( ) It has the lowest isoelectric point.

18.

You have discovered a mutant protein that forms promiscuous disulfide bonds with many other cellular proteins in the endoplasmic reticulum. Using an antibody against the protein, you immunoprecipitate proteins that interact with it. You then separate these proteins using a special type of two-dimensional polyacrylamide-gel electrophoresis (PAGE): in this method, separation in the first dimension is carried out by SDS-PAGE in the absence of β-mercaptoethanol, while separation in the second dimension is carried out by SDS-PAGE in the presence of βmercaptoethanol. The result is shown in the following schematic drawing. Apparent molecular weights are indicated on the left. What do you think is the molecular weight of the mutant protein? What is the molecular weight represented by the question mark?

First dimension

?

15K

Second dimension

150K

A. 15K, 45K B. 30K, 45K C. 15K, 30K D. 30K, 30K E. 30K, 60K 19.

Where do individual peptides get fragmented in a tandem mass spectrometry (MS/MS)

set-up?

A. Electrospray ion source B. Mass filter C. Inert gas chamber D. Mass analyzer E. Detector 20.

Which mass spectrometry instrument is most suitable for determining all the proteins present in an organelle?

A. MALDI-TOF B. MS/MS C. LC-MS/MS D. GC-MS 21.

You performed a co-immunoprecipitation experiment to identify cellular proteins associated with protein X. For the experiment, you used antibodies against protein X that were

immobilized on beads. You incubated these beads with cell lysates from either wild-type cells or cells lacking the gene encoding protein Y. After washing and eluting the bound protein off the beads, you separated each bound and unbound fraction by SDS-PAGE. Finally, you performed immunoblotting using antibodies against either protein Y or protein Z. The results of the Western blots are presented in the following schematic drawings. According to these results, do you think protein Y enhances (E) or inhibits (I) binding of protein X to protein Z? Write down E or I as your answer.

Unbound

Unbound

Bound

Y-null

Bound

Wild type

Bound

Y-null

Unbound

Fraction:

Protein Z

Wild type

Bound

Lysate:

Protein Y

Unbound

Western antibody:

22.

The binding of protein A to two other proteins, B and C, has been studied with fluorescence anisotropy, and the results of the measurements have been plotted in the following schematic diagram. Given equal kon rate constants, which complex do you think has a longer half-life? What is the value of Ka (the association constant) for this complex?

Anisotropy (mP)

200 A-B A-C 150 -

100 0

10

15

40

Concentration of A (µM)

A. A-B; 105 L/mol B. A-B; 10–5 L/mol C. A-C; 1.5 × 105 L/mol D. A-C; 6.7 × 104 L/mol E. A-C; 6.7 × 10–4 L/mol 23.

Indicate whether each of the following descriptions better applies to the determination of atomic structure of proteins by x-ray diffraction (X) or by nuclear magnetic resonance (NMR) spectroscopy (N). Your answer would be a five-letter string composed of letters X and N only, e.g. NNXNN. ( ) It is more suitable for complexes larger than about 100 kD. ( ) It requires samples to crystallize. ( ) It can better reveal dynamics of molecular motions. ( ) It involves the calculation of electron-density maps. ( ) It is used to determine the solution structure of proteins.

24.

In an unfolded (random coil) protein, amino acid residues are exposed to the solvent and share more or less the same environment, whereas each residue in a folded protein has its own unique neighborhood. This fact can be exploited in using NMR to study protein folding. The two schematic diagrams below represent two-dimensional NMR spectra for the same protein in either its folded (native) or unfolded state. The chemical shifts, which depend on the local neighborhood of each atom, are plotted in these diagrams. Which diagram (A or B) do you think corresponds to the folded protein? Write down A or B as your answer.

A

B

25.

You have carried out a genetic screen in a species of fish, and have identified a new gene in this organism. You have cloned and sequenced the gene, but do not know anything about the biochemical function of the gene product. Which of the following is the most sensible next step in order to get clues about the function?

A. Purifying the protein product and determining its structure by x-ray diffraction B. Screening for small molecules that perturb gene function C. Using the BLAST algorithm to scan sequence databases for similar sequences D. Using MS/MS to sequence the gene product without the need for purification E. Mutating the gene and performing reverse genetics 26.

In sequence alignments such as those generated by a BLAST search, the significance of the alignment can be presented as an E-value, which specifies how likely it is for a match this good to be found in a database of a given size by chance. You have scanned a large sequence database with a sequence query, hoping to find a significant match. Which one would you look at: (A) a match with an E-value of 1, or (B) one with an E-value of 0.001? Write down A or B as your answer.

27.

Indicate whether each of the following manipulations or procedures used in current recombinant DNA technology commonly rely on an enzyme (E) or are done nonenzymatically (N). Your answer would be a six-letter string composed of letters E and N only, e.g. EEEEEE. ( ) Cleavage of DNA at specific sites

( ) DNA ligation ( ) DNA cloning ( ) Nucleic acid hybridization ( ) Synthesis of DNA of any desired specific sequence ( ) DNA sequencing 28.

The EcoRI restriction enzyme recognizes and cuts at the sequence GAATTC. Such an enzyme is expected to cut a typical double-stranded DNA at 1 in every … sites.

A. about 400 B. about 4000 C. about 40,000 D. about 400,000 E. about 4 million 29.

The AgeI restriction enzyme recognizes the sequence ACCGGT and cuts each strand between the first (A) and the second (C) nucleotide. This can be represented as A|CCGGT, with the vertical bar indicating the cleavage position. If a DNA fragment is cut with AgeI at both ends, cutting a plasmid with which of the following enzymes would then allow insertion of the fragment into the plasmid by simple ligation of sticky ends?

A. FseI, with the recognition sequence GGCCGG|CC B. AatI, with the recognition sequence AGG|CCT C. BstZI, with the recognition sequence C|GGCCG D. MroI, with the recognition sequence T|CCGGA E. None of the above 30.

Indicate whether ordinary agarose-gel electrophoresis (A), polyacrylamide-gel electrophoresis (P), or pulsed-field gel electrophoresis (F) is used to better separate each of the following DNA molecules. Your answer would be a three-letter string composed of letters A, P, and F only, e.g. APF. ( ) Small DNA oligonucleotides such as hybridization probes and PCR primers ( ) Plasmid DNA ( ) Bacterial chromosomes 31.

Your friend has cut a 3000-nucleotide-pair circular plasmid with either or both restriction enzymes A and B, and has separated the digestion products by electrophoresis. The results are shown in the schematic drawing below, with the approximate size of the fragments indicated

3000 3000

Cut with A and B

Cut with B

Cut with A

Uncut

above each band. Based on these results, what is the closest distance (in nucleotide pairs) between an A restriction site and a B restriction site on this plasmid?

2600 1600 1000 400 400

A. 400 B. 600 C. 1000 D. 1200 E. 1600 32.

A researcher has labeled the 5′ end of a DNA primer with 32P radioisotopes. The primer can hybridize to a complementary sequence near the 3′ end of a long, spliced mRNA molecule composed of three exons. The mRNA molecule can interact with two RNA-binding proteins: one of them (protein 1) binds to the first exon and the other (protein 2) binds to the second exon. The researcher uses reverse transcriptase to extend the primer in the presence of the mRNA and either of these proteins, then separates the cDNA products by gel electrophoresis, and finally visualizes the labeled DNA bands by autoradiography. The results are presented schematically in the following drawing. Assuming that the reverse transcriptase falls off the template when confronted by bound proteins, which lane (A or B) in the gel corresponds to the reaction in the presence of protein 1? Write down A or B as your answer.

A

B

33.

You have devised genetic screens to identify gain- and loss-of-function mutations in the Escherichia coli DNA ligase gene. One of the screens (1) seeks mutant cells that are infected more readily with a ligase-deficient strain of T4 phage. The other screen (2) seeks mutant cells that have increased radiation sensitivity at high temperatures. Indicate whether each of the following descriptions better applies to screen 1 (1) or screen 2 (2). Your answer would be a three-digit number composed of digits 1 and 2 only, e.g. 122. ( ) It would identify gain-of-function DNA ligase mutants. ( ) It would identify loss-of-function DNA ligase mutants. ( ) It would identify conditional mutants.

34.

Bacterial artificial chromosomes (BACs) …

A. are derived from the E. coli chromosome.

B. are present in high copy numbers per cell.

C. can maintain DNA sequences of hundreds of thousands of nucleotide pairs.

D. are generally not stable.

E. All of the above.

35.

Indicate whether each of the following descriptions better applies to cDNA libraries (C) or genomic DNA libraries (G). Your answer would be a five-letter string composed of letters C and G only, e.g. CGCGC. ( ) Their production requires use of reverse transcriptase. ( ) They are enriched in protein-coding genes. ( ) Different sequences are represented in them based on their transcription level. ( ) They contain intronic sequences.

( ) They are cell-type specific.

36.

You have used a variation of fluorescence in situ hybridization (FISH) called flow-FISH. In this technique, DNA probes are labeled with fluorescence and hybridized to their targets inside the cells in a tissue. The cells are then sorted using a fluorescence-activated cell sorter based on their fluorescent signal intensity. The probes in your experiment are designed to hybridize to human telomeric sequences at the ends of the chromosomes. You perform flowFISH using these probes on two human cell types, an early primary fibroblast culture and a late secondary culture that is showing the signs of replicative cell senescence. The sorting results are presented in the following histograms. Which histogram (1 or 2) would you expect to correspond to the primary culture?

Percentage of cells

2 1

Fluorescent signal intensity per cell

37.

The specificity of nucleic acid hybridization is remarkable. Even a single mismatch between a probe and a target sequence can be detected depending on hybridization conditions.

This sensitivity can be employed in applications such as SNP haplotyping. Consider a probe that is complementary to a 20-nucleotide-long genomic region including a single polymorphic nucleotide. The probe is perfectly complementary to the more common allele, but has one mismatch in the case of the minor allele. The temperature-dependent hybridization of the probe to the single-strand genomic fragment is schematically presented in the following graph. To detect hybridization, a fluorescent dye is added to the reaction. The dye fluoresces strongly only when intercalated between the bases of double-stranded DNA, and is therefore used to quantify double-strand formation or dissociation. Which “melting curve” (1 or 2) in the graph do you think corresponds to the more common SNP allele? Which curve shows a higher melting temperature?

Fluorescence

2 1

Temperature

A. 1; 1 B. 1; 2 C. 2; 1 D. 2; 2 38.

The following schematic graph shows temperature change over time in a cycle of a PCR assay. Which steps (A to C) in the cycle correspond to hybridization (annealing), strand separation (denaturation), and DNA synthesis (extension), respectively? Your answer would be a three-letter string composed of letters A to C, e.g. ACB.

Temperature

A C B Time

39.

Each cycle of PCR doubles the amount of DNA synthesized in the previous cycle. A pair of primers is used to direct the specific synthesis of a desired DNA segment. The following diagram shows the products of PCR amplification after three cycles. The DNA strands of the desired segment are indicated in yellow boxes in the diagram. As shown, at the end of the third cycle, 8 out of 16 strands have the desired length and are boxed. After four more cycles (i.e. at the end of the seventh cycle), how many strands of DNA are expected to be synthesized? How many of them have the desired length? Write down the two numbers and separate them with a comma, e.g. 16,8.

Desired segment

BEFORE PCR

END OF FIRST CYCLE

END OF SECOND CYCLE

END OF THIRD CYCLE

40.

In a forensic lab, DNA samples from a mother (M) and her child (C) have been analyzed at three short tandem repeat (STR) loci using PCR and gel electrophoresis. The same analysis has been performed for five men (1 to 5). According to the results below, which men can be eliminated as possible candidates for the biological father of the child? Your answer would be a number composed of digits 1 to 5 only, e.g. 145.

M

C

1

2

3

4

5

41.

You would like to clone a gene by PCR. The sequence of the two ends of the coding strand for the gene is provided below. Which following pair of primers would direct the amplification of this gene? All sequences are written from 5′ to 3′.

ATGAAATCTACGTTTCAC……CCCCCAGTACCCCCCTTA A. ATGAAATCTACGTTTCAC ; CCCCCAGTACCCCCCTTA B. ATGAAATCTACGTTTCAC ; TAAGGGGGGTACTGGGGG C. TACTTTAGATGCAAAGTG; ATTCCCCCCATGACCCCC D. TACTTTAGATGCAAAGTG ; CCCCCAGTACCCCCCTTA E. GTGAAACGTAGATTTCAT; ATTCCCCCCATGACCCCC 42.

A piece of DNA has been sequenced by automated dideoxy sequencing, in which each dideoxyribonucleoside triphosphate was labeled with a fluorescent tag of a different color. The data corresponding to a small segment with the sequence 5′-TGCCACA-3′ is shown in the following diagram. Knowing that ddGTP has been labeled with green fluorescence, what color do you think was the fluorescent tag used to label ddTTP in the sequencing experiment?

Direction of electrophoresis

A. Blue B. Green C. Red D. Yellow 43.

In Sanger sequencing, does each labeled DNA molecule have a fluorescently labeled dideoxyribonucleotide at its 5′ end or its 3′ end? Do the labeled nucleotides lack the 2′ or 3′ hydroxyl group?

A. 3′ end; 2′ hydroxyl group only B. 3′ end; 3′ hydroxyl group only C. 5′ end; 2′ hydroxyl group only D. 5′ end; 3′ hydroxyl group only E. 3′ end; both hydroxyl groups 44.

Indicate whether each of the following descriptions better applies to Illumina® sequencing (I), Ion Torrent™ sequencing (T), or both sequencing technologies (B). Your answer would be a four-letter string composed of letters B, I, and T only, e.g. IBBB. ( ) It uses fluorescently labeled nucleotides. ( ) It uses PCR-generated copies of DNA. ( ) It is a second-generation sequencing technology and employs a cyclic wash-andmeasure paradigm. ( ) It relies on the fidelity of DNA polymerase for its accuracy.

45.

A piece of DNA has been sequenced by Ion Torrent™ sequencing. The result corresponding to a six-nucleotide segment of this DNA is presented in the graph below, in which pH change is recorded after the addition of each of the four nucleotide triphosphates in cycles.

What is the sequence of this segment? All sequences are written from 5′ to 3′.

Number of protons released

2 1 T A C G T A C G T A C G T A C G Sequence of nucleotide added

A. TACGTA B. CGTACG C. CCGTAA D. CGGTTA E. ATTGGC

Reading frames

46.

There are six possible reading frames for any double-stranded DNA. In the following schematic drawing of a genomic region in Escherichia coli, stop codons in each of these frames are indicated by black bars. In which reading frame do you think the open reading frame (ORF) is translated in this region? Write down the frame number as your answer, e.g. 2.

3 2 1 –1 –2 –3 Position on chromosome 47.

A portion of RNA-seq data obtained from two tissue samples is plotted in the following schematic diagram. Which region (A to E) corresponds to an alternative exon?

Number of reads

Liver

Muscle

A

B

C

D

E

Position on chromosome 48.

A protein made from an expression vector can be so plentiful as to account for about 10% of total cell protein. You would like to produce a protein of interest that can be expressed at such a level in Escherichia coli. You plan to induce the expression and collect the cells when cell density in the culture medium reaches about 109 cells/mL. Proteins make up about 15% of total cell weight, and each E. coli cell typically has a mass of about 1 picogram (i.e. 10–12 g).

Assuming that almost 50% of the protein is lost during purification, how many liters of culture should you prepare in order to obtain 15 milligrams of pure protein? Write down your answer in liters, with two decimal places and without units, e.g. 15.00.

49.

Two loci on a human chromosome are about 1,500,000 nucleotide pairs apart and are found to recombine in about 2% of gametes during gametogenesis. Consistently, each genetic map unit (which corresponds to 1% recombination) in humans is equivalent to a physical distance of about … nucleotide pairs.

A. 30 million B. 7.5 million C. 3 million D. 0.75 million E. 0.3 50.

The p53 gene is an important tumor suppressor gene in our genome. It codes for a transcription regulatory protein that can assemble into tetramers. A number of mutations in p53 render the protein inactive. These mutant protein molecules form mixed tetramers with wild-type p53 proteins and, in this case, the tetramer is inactive. The mutations can therefore be categorized as …

A. null.

B. gain-of-function.

C. recessive lethal.

D. dominant-negative.

E. suppressor.

51.

For a complementation test to work, the mutations under study must be…

A. recessive.

B. dominant.

C. gain-of-function.

D. null.

E. conditional.

52.

You have isolated five mutations (1 to 5) in the yeast Saccharomyces cerevisiae that make the haploid cells unable to grow in the absence of histidine. Each haploid mutant can be mated with any of the other ones, forming diploid cells that either can (+) or cannot (–) grow in the absence of histidine, as indicated in the following complementation table. How many complementation groups do these mutations represent? Each complementation group typically corresponds to a separate gene.

1

2

3

4

5

–

+

–

+

+

1

–

+

–

+

2

–

+

+

3

–

+

4

–

5

A. 1 B. 2 C. 3 D. 4 E. 5

53.

The wild-type nematode Caenorhabditis elegans has a bent body with the shape of a wave. In the presence of increasing concentrations of ethanol, however, the body bend is

progressively diminished. Loss-of-function mutations in SLO-1, a gene that codes for an ion channel, confer ethanol resistance and allow the worm to maintain its body bend up to higher ethanol concentrations. The channel function is defined based on the ability of the channel to conduct ions across the membrane. According to these observations, would you expect the channel to open or close in the presence of ethanol in wild-type worms? Would a gain-offunction mutation in SLO-1 tend to increase or decrease body bend?

A. Open; increase B. Open; decrease C. Close; increase D. Close; decrease 54.

You have studied mutations in a transcription regulatory protein from Bacillus subtilis by an in vivo reporter gene assay carried out in Escherichia coli. The cis-regulatory sequence that the protein normally recognizes was inserted upstream of a gene encoding an enzyme that can produce a visible blue product. E. coli cells were then transformed with the reporter construct along with the gene encoding one of the various mutant forms of the transcription regulatory protein. Since the transcription regulatory protein is an inhibitor, the colonies turn blue only when the mutant protein is nonfunctional and unable to inhibit gene expression. Interestingly, you have identified a number of “cold-sensitive” loss-of-function mutations that show the mutant phenotype at 22°C but not at 37°C. Which of the following correctly describes the phenotype of these cold-sensitive mutants?

A. Their colonies turn blue at permissive (low) temperatures.

B. Their colonies turn blue at permissive (high) temperatures.

C. Their colonies turn blue at nonpermissive (low) temperatures.

D. Their colonies turn blue at nonpermissive (high) temperatures.

55.

In a bacterium that is normally cylindrical in shape, loss of either of two genes A and B results in round cells. Additionally, gain-of-function mutations in gene A can rescue the phenotype of gene B loss-of-function mutants. What phenomenon best describes this observation? The product of which gene is more likely to function upstream of the other one in determining cell shape in this bacterium?

A. Suppression; gene A B. Suppression; gene B C. Complementation; gene A D. Complementation; gene B

56.

Consider a biochemical pathway in petal cells of a plant, in which a white precursor is turned into a final red pigment through a yellow and an orange intermediate by the activity of three enzymes encoded by genes A, B, and C. The petal color in plants harboring functional (capital letter) or nonfunctional (lowercase letter) alleles of these genes is presented in the following table. What do you think is the order in which the enzymes act in the pathway? Your answer would be a three-letter string composed of letters A to C, e.g. BCA. Genotype

Petal color

ABC ABc

Red Orange

AbC Abc aBC aBc abC Abc

White White Yellow Yellow White White

57.

Loss-of-function mutations in either imp or yfgL genes in Escherichia coli result in an increased permeability of the outer membrane to small molecules and therefore confer antibiotic sensitivity, as the antibiotics can enter the cell more readily. Interestingly, double mutants, in which both genes are defective, are resistant to antibiotics. Similar to the wild-type strain, but unlike the single mutants, they can grow on media containing antibiotics. This is an example of …

A. synthetic lethality.

B. complementation.

C. synergy.

D. a recessive mutation.

E. None of the above.

58.

Susceptibility to drug and alcohol addiction is partially genetic. For example, the human dopamine receptor DRD2 is associated with cocaine and alcohol abuse; individuals carrying an A nucleotide, instead of the more common C nucleotide, at a certain position within the sixth intron of the DRD2 gene are at a significantly increased risk of abuse. This variation in sequence is an example of …

A. a SNP.

B. a CNV. C. an indel.

D. epistasis.

E. None of the above.

59.

If two genomic polymorphic sites were randomly and independently associated with each other, one would expect the frequency of observing a combination of their allelic forms in an individual to be equal to the multiplication product of the probability of the alleles occurring separately. In many cases, however, this is not true. The degree of deviation from random association can be quantified and summarized in diagrams such as the one below. In this diagram, eight SNPs in a genomic region are shown and their pairwise deviation from random association is color-coded by different shades of red, such that the darkest red shade indicates the furthest deviation from random association. In contrast, those pairs that associate randomly are colored white. For example, SNPs A and F seem to associate randomly as indicated by the corresponding box marked with black borders. From this diagram, two haplotype blocks can be readily detected. What subset of the shown SNPs is not part of any of the two blocks? Write down the letter(s) as your answer, in alphabetical order, e.g. CEH. Position on chromosome

A

B

C

D

E

F

G

H

60.

Indicate whether each of the following descriptions better applies to forward (F) or reverse (R) genetics. Your answer would be a four-letter string composed of letters F and R only, e.g. RFFF. ( ) It involves new gene discovery by genetic screens.

( ) It may also be called genome editing. ( ) In this approach, mutations are introduced deterministically in a known gene. ( ) It is involved in creating knockout mice lacking a particular gene.

61.

You have created transgenic mice that carry the gene encoding green fluorescent protein (GFP) under the control of a ubiquitous promoter. The GFP-encoding gene, however, is inactive unless recombination by Cre removes a blocking sequence between the promoter and the gene.

These mice have been mated with those harboring the Cre recombinase under the control of the insulin promoter. The Cre recombinase is fused to the estrogen receptor, which keeps the fusion in the cytosol unless it binds to tamoxifen, in which case it allows nuclear entry. After mating, you select adult mice harboring both of these transgenes and inject them with either tamoxifen or water. Which of the following would you expect to observe later as a result when you examine brain and pancreas tissue samples under a fluorescence microscope?

A. GFP signal is observed in the brain in the presence of tamoxifen only.

B. GFP signal is observed in the brain in the absence of tamoxifen only.

C. GFP signal is observed in the pancreas in the presence of tamoxifen only.

D. GFP signal is observed in the pancreas in the absence of tamoxifen only.

E. GFP signal is observed in the brain in either the presence or the absence of tamoxifen.

62.

You mated male and female heterozygous Oct4 knockout mice to study the function of this important gene. Among the progeny, one-third were homozygous wild type and the rest were heterozygous. You then repeated the cross, but this time you genotyped several embryos at different stages of early development and grouped them as either homozygous wild type (+/+), heterozygous (+/–), or homozygous null (–/–). You have summarized your results in the following table, in which the number of embryos in different groups and at different developmental stages are presented. According to these results, complete loss of Oct4 is …

Embryonic stage

Genotype

8-cell embryos

64-cell embryos

Implanted embryos

+/+

34

27

40

+/–

71

50

88

–/–

29

35

0

A. NOT lethal.

B. synthetic lethal.

C. zygotic lethal.

D. embryonic lethal.

63.

Which of the following is NOT correct regarding both Cas9 and EcoRI? A. They are both endonucleases.

B. They both create double-strand breaks in DNA. C. They both recognize their target sequences with the help of guide RNAs. D. They are both part of bacterial defense mechanisms against foreign DNA. E. They are both greatly useful in manipulating DNA and studying gene expression and function.

64.

A pool of five barcoded yeast mutants, each deleted for a different gene (genes A to E), has been grown under three different conditions (1 to 3). After the growth phase was completed, genomic DNA was isolated from the entire pool and the relative amount of each mutant was determined by quantifying the amount of DNA containing each barcode. The results are presented in the following graph. Which gene (A to E) is essential under condition 1 but not under conditions 2 or 3?

Relative DNA abundance

Conditions: 1 2 3

A

B

C

D

E

Deletion mutant 65.

In which of the following organisms is systemic RNAi possible through feeding the animal with the RNA? A. Caenorhabditis elegans B. Drosophila melanogaster C. Mus musculus

D. Homo sapiens E. All of the above 66.

Which of the following can limit the use of RNA interference in studying gene expression and function?

A. RNAi does not efficiently inactivate all genes.

B. Certain tissues are resistant to the action of RNAi. C. Off-target effects are common in RNAi treatments.

D. All of the above.

67.

You have used a reporter gene system to investigate the contribution to gene expression

of three cis-regulatory DNA sequences (A to C) from a key developmental gene. Reporter gene expression from constructs containing all three or only a subset of the regulatory sequences is summarized in the diagram below. According to these findings, indicate whether each of the following conclusions is (Y) or is not (N) acceptable. Your answer would be a four-letter string composed of letters Y and N only, e.g. NNNN. cis-regulatory DNA sequences A

B

Relative expression level

C 100% 99% 0% 85% 0% 190%

( ) Region B is likely to normally bind an activator and turn the gene on. ( ) Region C is sufficient for activation of gene expression. ( ) Region A is necessary for activation of gene expression. ( ) Region C is necessary for activation of gene expression.

Fluorescence

68.

The following schematic graph shows the result of two quantitative RT-PCR experiments. Total mRNA from two tissue samples (1 and 2) was isolated and subjected to RTPCR using primers designed to amplify a tissue-specific gene. Added to the reaction was a fluorescent dye that fluoresces only when bound to double-stranded DNA. According to the graph, which tissue has a higher level of this mRNA? By how much?

1

10

2

20

30

Number of PCR cycles A. About 2-fold higher in tissue 1 B. About 1000-fold higher in tissue 1 C. About 2-fold higher in tissue 2 D. About 1000-fold higher in tissue 2 69.

Which of the following methods provides the most sensitive and accurate way of measuring and comparing mRNA levels of a set of genes in two different tissues. The genes are expressed at relatively low levels in these tissues, and do not necessarily code for protein.

A. RNA-seq B. RT-PCR C. DNA microarray analysis D. ChIP-seq E. Ribosome profiling 70.

You have grown cultures of the yeast Saccharomyces cerevisiae in a complex medium alone or supplemented with either glucose or ethanol. In the presence of glucose, available energy is abundant and the cells can therefore grow by fermenting glucose. In the presence of ethanol, however, the cells switch to a mostly aerobic respiration to meet their energy demands.

They also shut down many of their anabolic pathways and activate catabolic pathways to produce more energy to better survive under these conditions. After the growth phase, you extract total mRNA from the cells, convert them to cDNA, and label them with a fluorescent

molecule. You then hybridize the cDNAs to a yeast DNA microarray. Part of the results obtained by cluster analysis after fluorescent measurements is summarized in the following simplified diagram. Here, red and green indicate increase and decrease, respectively, in expression of a gene compared to that in the nonsupplemented medium. Which row (1 or 2) do you think corresponds to the glucose-supplemented medium?

Ribosome biogenesis

Electrontransport chain

1 2

71.

Consider a protein composed of only 21 amino acids encoded by a short ORF that is located between two known genes in a mammalian genome. Due to its length, the ORF had not been annotated as a gene. Which of the following methods can demonstrate that it is indeed a gene that is actively translated to protein?

A. RNA-seq B. RT-PCR C. DNA microarray analysis D. ChIP-seq E. Ribosome profiling 72.

The results of two sets of chromatin immunoprecipitation experiments on two cell types (1 and 2) are shown in the following simplified graphs for the same genomic region containing the first two exons of a gene. Antibodies against a transcription regulatory protein (X) or against a subunit of RNA polymerase II were used in the ChIP-seq experiments, as indicated. Do you think the protein is a transcriptional activator or repressor? The results from a parallel RNA-seq experiment are provided in the third graph. Which profile (a or b) in the graph would you expect to correspond to cell type 1?

Number of sequence reads

ChIP-seq Protein X

1 2

RNA Pol II 2

1 Exon 1 Genomic position

Exon 2

Number of sequence reads

Total RNA-seq

a b Genomic position

A. Activator; profile a B. Repressor; profile a C. Activator; profile b D. Repressor; profile b 73.

RNA-seq and ribosome profiling experiments have been carried out on the same cells.

The following simplified graphs show the results for the same genomic region containing a gene with three exons. Which graph (1 or 2) do you think corresponds to the RNA-seq results? What feature is represented by the arrow?

Number of sequence reads

1

2

Genomic position A. Graph 1; stop codon B. Graph 1; polyadenylation site C. Graph 2; stop codon D. Graph 2; polyadenylation site 74.

Sort the following steps in the common procedure to create transgenic plants. Your answer would be a four-letter string composed of letters A to D, e.g. DABC. (A) Callus growth (B) Agrobacterium infection (C) Shoot/root induction (D) Removal of leaf tissue from a plant 75.

Consider the binding of a transcription regulatory protein (A) to a promoter DNA (p) according to the following scheme.

A + p → A:p At steady state, which of the following formulae correctly yields the bound promoter fraction, i.e. [A:p]/([A:p] + [p])? The association constant is designated as K. A. [A]/(K + [A]) B. K[A]/(1 + K[A]) C. K[A]/(K + [A]) D. [A]/(1 + K[A]) 76.

When a promoter p is not saturated with bound regulatory protein A, doubling the total concentration of A would …

A. increase steady-state [A:p] exactly twofold.

B. increase steady-state [A:p] over twofold.

C. increase steady-state [A:p] less than twofold.

D. decrease steady-state [A:p] approximately twofold.

E. decrease steady-state [A:p] over twofold.

77.

In the following schematic graph, the fraction of promoters (px) bound to the regulatory protein A is plotted at various concentrations of A. Which of the following quantities represents X in the graph?

Fraction bound

100%

50%

0% X

[A]

A. [Total px] B. [px] C. kon / koff D. koff / kon E. kon[px] / koff 78.

Consider a transcription regulatory protein (A) that can bind to a promoter (p). After reaching an initial steady state, the concentration of A is suddenly increased tenfold. The following graph shows the formation of A:p over time after this change for two systems, one of which is described by higher koff and kon values compared to the other. Which curve (1 or 2) corresponds to this system? Write down 1 or 2 as your answer.

[A:p] (relative value)

5 1 2

1 Time

[A:p] (relative value)

79.

Consider a transcription regulatory protein (A) that can bind to the promoter (p) of a target gene. After reaching an initial steady state, the concentration of A is suddenly increased tenfold. The following graph shows the formation of A:p over time after this change. What is the slope of the curve at time zero?

5

1 0

Time

A. 2 koff B. 2 kon C. 5 koff D. 5 kon E. 9 koff 80.

Everything else being equal, if the mean lifetime of a protein is doubled, its concentration at steady state would …

A. increase approximately twofold.

B. increase less than twofold.

C. increase over twofold.

D. decrease approximately twofold.

E. decrease less than twofold.

81.

Consider two proteins A and B with equal transcription and translation rates. The mean lifetime of A, however, is twice as long as that of B. Starting with no protein, the time it takes for a protein to reach 99% of its steady-state concentration is …

A. the same for proteins A and B. B. two times longer for A compared to B. C. two times longer for B compared to A. D. less than twofold longer for A compared to B. E. less than twofold longer for B compared to A. 82.

You have set up a system of differential equations to describe the concentration of a

protein as a function of time based on transcription, translation, and degradation rates. You simulate two conditions (1 and 2) that are identical except that the transcription rate for the protein is set tenfold higher in one of them (1) compared to the other (2). Starting from a concentration of zero, the time required to reach 50% of the final steady-state concentration is …

A. the same for conditions 1 and 2.

B. tenfold longer for 1 compared to 2.

C. tenfold longer for 2 compared to 1.

D. less than tenfold longer for 1 compared to 2.

E. less than tenfold longer for 2 compared to 1.

83.

Which equation better describes the steady-state concentration of a protein (X) that is regulated by the simple binding of a transcription repressor (R) to its cis-regulatory DNA sequences? β is the transcription rate, m is the translation rate, K is the association constant in the binding of R to the regulatory sequences, and τX is the mean lifetime of the protein X. A. βmτX / (1 + K[R]) B. βmτX / (1 + [R]) C. βmτX K / (1 + K[R]) D. βmτX [R] / (1 + K[R]) E. βmτX [R] / (K + [R]) 84.

The following graph shows the change in the concentration of a protein (A) over time as a fraction of the final steady-state level. The two curves (1 and 2) in the graph are simulations with similar parameters except that one of them involves a negative feedback loop in which A activates the expression of a repressor protein which in turn represses the expression of A. Which

[A] (relative value)

curve (1 or 2) corresponds to the simulation with the negative feedback loop? Write down 1 or 2 as your answer.

1 1 2

0 Time

85.

Consider a genetic network consisting of gene A whose product activates the transcription of gene R. Imagine that the concentration of protein A is suddenly changed by a significant amount. The following graphs show how the concentration of protein R would then change over time to reach steady state again. The two curves (1 and 2) in each graph correspond to systems that have different transcription rates for A. For one of the two graphs (A or B) a negative feedback loop is present, in which protein R represses the transcription of gene A. Which conclusion is correct according to these results?

A

1 2

B 1 2

A. The negative feedback loop makes the system slower in reaching the steady state.

B. With negative feedback, the steady-state concentration is less sensitive to fluctuations in kinetic rate constants.

C. Higher transcription rates for gene A result in lower steady-state levels of protein R both with and without negative feedback.

D. All of the above.

86.

Consider the transcriptional circuits depicted in the following schematic drawings. These circuits share identical kinetic parameters but have a different number of stages. Which circuit (A or B) creates more robust oscillations? Write down A or B as your answer.

A

B

87.

Consider a promoter sequence with n binding sites for an activator protein. Which Hill coefficient (h) describes noncooperative binding, i.e. independent, noninteracting binding sites?

A. h = n B. h = 1 C. h = 0 D. h = n – 1 E. h < 0 88.

Bistability in a biological system can be accompanied by hysteresis. Consider switchlike cellular responses to a signal molecule, as shown in the following graphs. In each graph, the solid curve corresponds to the change in cellular response when the signal concentration is being increased over time, whereas the dashed curve corresponds to the change when the signal concentration is being decreased. Which graph do you think corresponds to a system with a greater degree of hysteresis? What mechanism is responsible for this behavior?

B

Cellular response

A

Signal strength

Signal strength

A. Graph A; negative feedback B. Graph A; positive feedback C. Graph B; negative feedback D. Graph B; positive feedback

Concentration of Y

89.

The schematic graph below represents the nullclines (solid curves) and a few selected trajectories (dotted curves) for a system composed of two proteins (X and Y) with a complex regulatory behavior. According to the graph, indicate whether each of the following statements is (Y) or is not (N) acceptable. Your answer would be a four-letter string composed of letters Y and N only, e.g. YYYY.

Concentration of X

( ) The system is bistable.

( ) There are three steady states in the system. ( ) The system contains positive feedback. ( ) The system contains delayed negative feedback.

90.

The quantitative output of a gene depends in part on the relative affinities of transcription regulators. In the following graphs, the red color indicates high gene expression and blue color indicates low gene expression. The expression level is plotted as a function of the concentrations of two transcription regulatory proteins, an activator (A) and a repressor (R). Every other parameter being identical, which graph corresponds to a system where the affinity of protein A for the cis-regulatory sequences is higher than that of protein R? Which graph corresponds to a system that shows a “leakier” expression?

1

2

A. Graph 1; Graph 1 B. Graph 1; Graph 2 C. Graph 2; Graph 1 D. Graph 2; Graph 2 91.

In the schematic graphs below, the red color indicates high gene expression and blue color indicates low gene expression. The expression level is plotted as a function of the concentrations of two transcription regulatory proteins, A1 and A2. Which graph (1 or 2) do you think corresponds to an “AND NOT” combinatorial logic? Write down 1 or 2 as your answer.

1

2

92.

Some feed-forward motifs are capable of generating a brief pulse of gene activation in response to a sustained input signal: in such motifs, the signal activates an activator (A) and a repressor (R), both of which control the expression of the same gene. The quantitative output of the gene depends in part on the relative affinities of A and R for binding to the cis-regulatory sequences of the target gene. Under which of the following conditions is a brief pulse of gene activation observed in response to a sustained input signal? KA and KR are the association constants for A and R, respectively.

A. KA = KR B. KA << KR C. KA >> KR D. KAKR > 1 93.

Indicate which of the following network motifs represent negative feedback, positive feedback, incoherent feed-forward, and coherent feed-forward, respectively. Your answer would be a four-letter string composed of letters A to D only, e.g. BACD. A

B

C

D

94.

Indicate whether each of the following better applies to a coherent (C) or incoherent (I) feed-forward motif. Your answer would be a three-letter string composed of letters C and I only, e.g. CCC. ( ) It can provide cellular responses with robustness despite perturbations in the input signal. ( ) A cellular pathway with such a motif generally requires a sustained (long) input signal to generate a full response. ( ) A cellular pathway with such a motif can generate a brief pulse of response in the presence of a sustained input signal.

95.

Imagine a bistable system composed of two repressor proteins X and Y, each of which

binds to the cis-regulatory sequences of the gene encoding the other protein, i.e. X represses the expression of Y, and Y represses the expression of X. A researcher engineers the genes such that they encode protein X fused to green fluorescent protein (GFP) and protein Y fused to red fluorescent protein (RFP). Cells containing the fusion constructs are then grown and sorted by fluorescence-activated cell sorting (FACS) based on their GFP and RFP signals. The results are presented in the following dot plot. Each cell is represented by a dot whose coordinates correspond to GFP and RFP fluorescence in the cell. According to these results, which protein (X or Y) do you think binds to its target DNA sequence with a higher degree of cooperativity?

GFP

Write down X or Y as your answer.

RFP

Answers: 1. Answer: H Difficulty: 3 Section: Isolating Cells and Growing them in Culture Feedback: The cells with a high CD4 signal (that is, CD4+) are depleted in the patient.

2. Answer: callus Difficulty: 1 Section: Isolating Cells and Growing them in Culture Feedback: A single totipotent cell from such a callus can be grown into a small clump of cells from which a whole plant can be regenerated.

3. Answer: BNTT Difficulty: 2 Section: Isolating Cells and Growing them in Culture Feedback: Cell lines can be stored in liquid nitrogen at –196°C for an indefinite period and retain their viability when thawed. Unlike nontransformed cell lines, transformed cell lines often grow without attaching to a surface and can proliferate to a much higher density in a culture dish. They can usually cause tumors if injected into a susceptible animal.

4. Answer: E Difficulty: 2 Section: Isolating Cells and Growing them in Culture Feedback: Monoclonal antibodies have several advantages over the polyclonal antibodies present in antisera.

5. Answer: C Difficulty: 2 Section: Isolating Cells and Growing them in Culture Feedback: Fusion of B cell clones with immortalized myeloma cells (a transformed B cell line) creates hybrid cells that are also capable of indefinite proliferation.

6. Answer: B Difficulty: 3 Section: Isolating Cells and Growing them in Culture Feedback: The transformed myeloma (M) cells are immortalized but are not selected for in HAT media. This is because they lack HGPRT and their DHFR-dependent pathway is also inhibited. Non-fused B cells can still make their nucleotides by the salvage pathway but can only proliferate for a few generations. The hybrid cells, on the other hand, are both immortalized and capable of carrying out the salvage pathway.

7. Answer: NLMR Difficulty: 2 Section: Purifying Proteins

Feedback: During cell fractionation by ultracentrifugation, progressively higher speeds are used to fractionate different components. Nuclei sediment in the early steps whereas macromolecular complexes such as ribosomes come out in later steps.

8. Answer: E Difficulty: 3 Section: Purifying Proteins Feedback: The terminal velocity for the 50S particle is calculated as: vt = S × ac = (50S × 10–13 s/S) × (50,000 × 10 m/s2) = 2.5 × 10–6 m/s = 0.9 cm/h It would therefore take about 10 hours for the 50S particle to travel about 9 cm.

9. Answer: EEVE Difficulty: 2 Section: Purifying Proteins Feedback: In equilibrium sedimentation, each subcellular component moves up or down the steep sucrose or cesium chloride gradient to reach a position corresponding to its density. It is sensitive enough to separate DNA molecules containing different nitrogen isotopes. Velocity sedimentation also uses gradients, but these are shallow and are used to limit convective mixing.

10. Answer: C Difficulty: 2 Section: Purifying Proteins Feedback: The larger dimer would sediment faster and would appear near the bottom of the ultracentrifuge tube (peak 2). In gel filtration, the dimer would come off the column in earlier fractions (peak 3).

11. Answer: A Difficulty: 3 Section: Purifying Proteins Feedback: Increasing salt concentration would weaken the charge–charge interactions between bound proteins and an ion-exchange matrix. Decreasing salt concentration, on the other hand, weakens the hydrophobic interactions between bound proteins and a hydrophobic matrix.

12. Answer: D Difficulty: 1 Section: Purifying Proteins Feedback: Immunoprecipitation is a variation on the theme of affinity chromatography.

13. Answer: B Difficulty: 2 Section: Purifying Proteins Feedback: Since tandem affinity purification (TAP)-tagging relies on proteolytic cleavage for elution of the protein from the first column, the protein can be washed

extensively at this step to remove most contaminants. Additionally, it can be purified further due to the presence of another, non-cleaved, tag.

14. Answer: E Difficulty: 1 Section: Analyzing Proteins Feedback: SDS (sodium dodecyl sulfate) is a negatively charged detergent that can release proteins from their binding partners and unfold them into extended polypeptide chains. It also masks each protein’s intrinsic charge. Consequently, during SDS polyacrylamide-gel electrophoresis (SDS-PAGE), proteins are separated mainly based on their size, and not their shape or intrinsic charge; larger proteins move more slowly because the gel matrix hinders their migration to a higher extent compared to smaller proteins.

15. Answer: C Difficulty: 2 Section: Analyzing Proteins Feedback: Within the linear range shown in the graph, smaller proteins are better resolved (separated) in SDS-PAGE compared to larger proteins.

16. Answer: AOH Difficulty: 2 Section: Analyzing Proteins Feedback: If the pH is lower than the isoelectric point of a protein, the protein is protonated (positively charged) and should therefore be repulsed by the nearby anode toward regions with higher pH. Ionic detergents are not suitable for this charge-sensitive technique. A basic protein has a high (basic) isoelectric point.

17. Answer: baa Difficulty: 2 Section: Analyzing Proteins Feedback: The b subunit seems to be the smallest. The a subunit has the lowest isoelectric point and is thus the most acidic (has the highest negative charge).

18. Answer: C Difficulty: 3 Section: Analyzing Proteins Feedback: All protein complexes separated on this gel presumably contain the mutant protein. Those on the diagonal are associated noncovalently with the mutant protein, and therefore run identically in the two dimensions and end up on the diagonal. The proteins that had formed disulfide bonds with the mutant protein, however, are dissociated from it in the second dimension, resulting in two spots for each such complex, one of which is the common 15K mutant protein. The protein also appears to have dimerized with itself (large spot at the intersection), corresponding to a molecular weight of 30K, indicated by the question mark.