Molecular Biology Of The Cell 6th Edition by Bruce Alberts -Test Bank

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Molecular Biology Of The Cell 6th Edition by Bruce Alberts -Test Bank

 

MOLECULAR BIOLOGY OF THE CELL, SIXTH EDITION

CHAPTER 2: CELL CHEMISTRY AND BIOENERGETICS

© Garland Science 2015

 

 

 

 

1.1.         Which of the following elements is not normally found in cells?

  1. Copper
  2. Iron
  3. Silver
  4. Cobalt
  5. Zinc

 

1.2.         A hydrophobic molecule is typically …

  1. able to form hydrogen bonds with itself but not with water.
  2. able to form hydrogen bonds with water.
  3. charged.
  4. hard to dissolve in a solvent.
  5. incapable of interacting favorably with water.

 

1.3.         For each of the following pairs, indicate whether they interact via hydrogen bonds (H) or ionic bonds (I), or do not favorably interact (N). Your answer would be a four-letter string composed of letters H, I, and N only, e.g. HNNI.

(  )  ATP and Mg2+

(  )  Urea and water

(  )  Glucose and the enzyme hexokinase (which uses glucose as a substrate)

(  )  A phospholipid tail and inorganic phosphate

 

1.4.         Which of the following chemicals do you NOT expect to be readily dissolved in water?

  1. Uric acid
  2. Hexane
  3. Glycerol
  4. Ethanol
  5. Potassium chloride

 

1.5.         Weak noncovalent attractions in the cell can be very strong in a nonaqueous environment. Some of these attractions are as strong as covalent interactions in a vacuum (their bond energy is approximately 340 kJ/mole), but become more than twenty-five times weaker (their bond energy becomes approximately 13 kJ/mole) in water. What type of attraction shows this phenomenon?

  1. Electrostatic attractions
  2. Hydrogen bonds
  3. van der Waals attractions
  4. Hydrophobic force
  5. All of the above

 

1.6.         The bond energies associated with noncovalent attractions in the cell are too weak to resist disruption by thermal motion. However, cellular macromolecules can interact specifically AND strongly with each other (or fold by themselves) merely via such interactions. How is this possible?

  1. The bond energies increase radically when two interacting molecules approach each other.
  2. The interacting molecules also fortify their binding via covalent bonds to keep them from dissociation.
  3. Many weak bonds together in a complementary geometry can afford a strong binding.
  4. The cell lowers its internal temperature to reduce thermal motion of molecules and enhance the weak attractions.

 

1.7.         What is the pH of a 10–8 M solution of hydrochloric acid? Round the pH value to the nearest integer, e.g. 10.

  1. 8
  2. 7
  3. 6
  4. 5
  5. 4

 

1.8.         The cell can change the pH of its internal compartments using membrane transport proteins that pump protons into or out of a compartment. How many protons should be pumped into an endocytic vesicle that is 10–15 liters in volume and has a neutral pH in order to change the pH to 5? Avogadro’s number is 6 × 1023. Omit complications such as the membrane potential, buffers, and other cellular components.

  1. 6000
  2. 60,000
  3. 120,000
  4. 600,000
  5. 6,000,000

 

1.9.         Which of the following is true regarding a fatty acid molecule in water?

  1. It is positively charged at physiological pH, but can become neutral when the pH is high enough.
  2. It is positively charged at physiological pH, but can become neutral when the pH is low enough.
  3. It is negatively charged at physiological pH, but can become neutral when the pH is high enough.
  4. It is negatively charged at physiological pH, but can become neutral when the pH is low enough.
  5. It is not charged at physiological pH.

 

1.10.       The amino acid serine has an amino group, a carboxyl group, and a hydroxyl group. Which of the following better represents the structure of this amino acid at neutral pH?

 

 

1.11.       The three families of cellular macromolecules are polymerized and depolymerized by a general mechanism involving water. Each of them has a set of monomers whose polymerization changes the total free energy of the system. Which of the following statements is true regarding these macromolecules?

  1. Each polymerization step requires free-energy input and proceeds by the consumption of one water molecule.
  2. Each depolymerization step requires free-energy input and proceeds by the consumption of one water molecule.
  3. Each polymerization step requires free-energy input and proceeds by the release of one water molecule.
  4. Each depolymerization step requires free-energy input and proceeds by the release of one water molecule.

 

1.12.       Sort the following from a low to a high contribution to the total mass of an E. coli bacterium. Your answer would be a four-letter string composed of letters A to D only, e.g. DCBA.

(A) Water

(B) Sugars

(C) Proteins

(D) Nucleic acids

 

1.13.       Which of the following statements is true regarding cellular metabolism?

  1. A living organism decreases the entropy in its surroundings.
  2. During catabolism, heat is generated, and the cell uses this heat to perform work during anabolism.
  3. The heat released by an animal cell as part of its metabolic processes comes from the bond energies in the foodstuffs that are consumed by the animal.
  4. Living organisms defy the second law of thermodynamics, but still obey the first law.

 

1.14.       The folding of proteins can be considered a simple conversion from the unfolded to the natively folded state. At about 27°C (or 300 K), the free-energy change of folding for a particular protein is measured to be –40 kJ/mole. If the enthalpy change (ΔH) of folding is –640 kJ/mole, what is the entropy change (ΔS) of folding for this protein? Write down your answer with the appropriate sign (+ or –) and in kJ/mole/K, e.g. –1000 kJ/mole/K.

 

1.15.       Which of the following correctly summarizes the overall process of photosynthesis?

  1. CO2 + O2 → H2O + sugars
  2. CH2O + CO2 + O2 → H2O + sugars
  3. CO2 + H2O → H2 + CO2
  4. CO2 + H2O → O2 + sugars

 

1.16.       Which of the following statements is true regarding reactions involving oxidation and reduction?

  1. The carbon atom is more oxidized in formaldehyde (CH2O) than in methanol (CH3OH).
  2. Oxidation of food in all organisms requires oxygen.
  3. A molecule is oxidized if it gains an electron (plus a proton) in a reaction.
  4. A dehydrogenation reaction is a reduction.
  5. In an organic molecule, the number of C–H bonds increases as a result of oxidation.

 

1.17.       Enzymes are the cell’s catalyst crew. They make the life of the cell possible by carrying out various reactions with astounding performance. Which of the following is NOT true regarding cellular enzymes?

  1. Enzymes lower the activation energy of the reactions that they catalyze.
  2. Enzymes can specifically drive substrate along certain reaction pathways.
  3. Enzymes can push energetically unfavorable reactions forward by coupling them to energetically favorable reactions.
  4. Enzymes are proteins, but RNA catalysts, called ribozymes, also exist.
  5. Enzymes can change the equilibrium point for reactions that they catalyze.

 

1.18.       In the following diagram showing the reaction pathway for a simple single-substrate enzymatic reaction, which of the quantities corresponds to the activation energy of the forward reaction?

 

 

  1. (ab)
  2. (a + b)
  3. (ac)
  4. (a + c)
  5. (bc)

 

1.19.       In the following diagram showing the distribution of thermal energy in a population of substrate molecules, the energy thresholds indicated by numbers represent …

 

  1. the activation energy at high and low temperature.
  2. the reaction rate at high and low pH.
  3. the activation energy with and without an enzyme.
  4. the reaction rate at high and low substrate concentrations.
  5. the activation energy at high and low substrate concentrations.

 

1.20.       A cellular enzyme catalyzes the catabolic reaction shown below. Its coenzyme is shown in the box. Which of the following is correct regarding this reaction?

 

 

  1. The substrate is reduced in this reaction and the coenzyme is converted from state 1 to state 2.
  2. The substrate is oxidized in this reaction and the coenzyme is converted from state 1 to state 2.
  3. The substrate is reduced in this reaction and the coenzyme is converted from state 2 to state 1.
  4. The substrate is oxidized in this reaction and the coenzyme is converted from state 2 to state 1.

 

1.21.       The molecules inside the cell constantly collide with other molecules and diffuse through the cytoplasm in a random walk. The average net distance traveled by such a molecule after a certain time period t is proportional to the square root of t, i.e. (t)0.5, as well as to its diffusion coefficient. If, on average, it takes a molecule 100 milliseconds to travel a net distance of 0.5 µm from its starting point, how long would it normally take for the same molecule to travel a net distance of 5 µm from the same starting point?

  1. 0.2 second
  2. 0.3 second
  3. 1 second
  4. 10 seconds
  5. 0.32 seconds

 

1.22.       Sort the following molecules from a low to high rate of diffusion inside the cytosol. Your answer would be a four-letter string composed of letters A to D only, e.g. ADCB.

(A) Myoglobin (a protein)

(B) Glycine (an amino acid)

(C) Ribosome (a protein–RNA complex)

(D) CO2

 

1.23.       The equilibrium constant for the reaction that breaks down each molecule of substrate A to one molecule of B and one molecule of C is equal to 0.5. Starting with a mixture containing only molecules A at 1 M concentration, what will be the concentration of molecule A after reaching equilibrium under these conditions?

  1. 0.5 M
  2. 0.25 M
  3. 0.125 M
  4. 0.333 M
  5. 0.667 M

 

1.24.       The free-energy change (ΔG) for a simple reaction, A → B, is 0 kJ/mole at 37°C when the concentrations of A and B are 10 M and 0.1 M, respectively. What is the free-energy change for the reaction when the concentrations of A and B are instead 0.01 M and 1 M, respectively? Recall that Δ = –5.9 × log(Keq). Write down your answer as a number with the appropriate sign (+ or –) and in kJ/mole, e.g. +11.8 kJ/mole.

 

 

1.25.       Imagine the reaction A → B with a negative ΔG value under experimental conditions. Which of the following statements is true about this reaction?

  1. The reaction is energetically unfavorable.
  2. The reaction proceeds spontaneously and rapidly under these conditions.
  3. Increasing the concentration of B molecules would increase the ΔG value (toward more positive values).
  4. The reaction would result in a net decrease in the entropy (disorder) of the universe.
  5. The reaction cannot proceed unless it is coupled to another reaction with a positive value of ΔG.

 

1.26.       In the first reaction of the glycolytic pathway, the enzyme hexokinase uses ATP to catalyze the phosphorylation of glucose, yielding glucose 6-phosphate and ADP. The ΔG° value for this reaction is –17 kJ/mole. The enzyme glucose 6-phosphatase catalyzes a “reverse” reaction, in which glucose 6-phosphate is converted back to glucose, and a phosphate is released. The ΔG° value for this reaction is –14 kJ/mole. What is the ΔG° value for the following reaction?

ATP + H2O → ADP + Pi

 

  1. –3 kJ/mole
  2. +3 kJ/mole
  3. –31 kJ/mole
  4. +31 kJ/mole
  5. –15.5 kJ/mole

 

1.27.       The enzyme phosphoglucose isomerase converts glucose 6-phosphate to its isomer fructose 6-phosphate in the second step of glycolysis. The equilibrium constant for the reaction is 0.36. Evaluating the ΔG° of the reaction (ΔG° = –5.9 × log Keq), decide which of the following conclusions is true.

  1. The ΔG° is negative, therefore the reaction proceeds in the forward direction.
  2. The ΔG° is negative, but whether or not the reaction proceeds would depend on ΔG, not ΔG°.
  3. The ΔG° is positive, but in a cell that is active in glycolysis, the reaction can still proceed in the forward direction.
  4. The ΔG° is positive, therefore the reaction proceeds in the reverse direction.

 

1.28.       Which of the following represents an “activated” carrier molecule?

  1. AMP
  2. NADH
  3. NAD+
  4. NADP+
  5. CoA

 

1.29.       ATP is the main energy currency in cells, and it can especially be used to drive condensation reactions that produce macromolecular polymers. How does ATP normally catalyze the condensation reaction, which by itself is energetically unfavorable?

  1. It transfers its terminal phosphate to an enzyme and is released as ADP.
  2. It transfers its two terminal phosphates to an enzyme, and is released as AMP.
  3. It covalently attaches to both of the substrates.
  4. It transfers either one or two terminal phosphate(s) to one of the substrates and is released as either ADP or AMP.
  5. It covalently attaches to the enzyme, forming an enzyme–AMP adduct.

 

1.30.       Despite their overall similarity, NADH and NADPH are not used indiscriminately by the cell. What are the distinctive features of these two carrier molecules?

  1. NADPH has an extra phosphate near its nicotinamide ring, giving it distinct electron-transfer properties.
  2. In the cell, NADH is usually in excess over NAD+, but NADP+ is usually in excess over NADPH.
  3. NADH is normally involved in anabolic reactions, whereas NADPH is normally involved in catabolism.
  4. Both NADPH and NADH are recognized by the same enzymes with similar affinities, since the extra phosphate group in NADPH is not involved in such recognition.
  5. In the cell, NADH is found mostly in the form that acts as an oxidizing agent, whereas NADPH is found mostly in the form that acts as a reducing agent.

 

1.31.       In an enzymatic reaction involving NADH or NADPH, reduction of a substrate accompanies the oxidation of these carrier molecules to NAD+ or NADP+, respectively. What else typically happens in such a reaction?

  1. A molecule of water is released to the solution upon completion of the reaction.
  2. A proton is released during the oxidation of the carriers.
  3. A proton is taken up by the substrate that is being reduced.
  4. A proton is taken up by the carrier molecule that is being oxidized.
  5. A phosphate group is transferred to the substrate.

 

1.32.       What is the reaction performed on the molecule labeled as substrate in the following diagram? What is the name of the activated carrier?

 

  1. This is a methylation reaction and the activated carrier is ATP.
  2. This is a methylation reaction and the activated carrier is S-adenosylmethionine.
  3. This is a carboxylation reaction and the activated carrier is ATP.
  4. This is a carboxylation reaction and the activated carrier is carboxylated biotin.
  5. This is an acetylation reaction and the activated carrier is acetyl CoA.

 

1.33.       Under anaerobic conditions, glycolysis provides most of the ATP that the cell needs. In animal cells, pyruvate, the end product of glycolysis, is converted to lactic acid by lactate dehydrogenase, as shown below:

CH3(CO)COO + X →CH3(CHOH)COO + Y

What is the correct carrier pair (in place of X and Y) in this reaction?

  1. X is (ADP + Pi), and Y is (ATP)
  2. X is (NADP+), and Y is (NADPH + H+)
  3. X is (NAD+), and Y is (NADH + H+)
  4. X is (NADH + H+), and Y is (NAD+)
  5. X is (NADP++ H+), and Y is (NADPH)

 

1.34.       Macromolecules in the cell can be made from their monomers using one of two polymerization schemes. One is called head polymerization, in which the reactive bond required for polymerization is carried on the end of the growing polymer. In contrast, in tail polymerization, the reactive bond is carried by each monomer for its own incorporation. In the figure, indicate the polymerization scheme and the type of macromolecule.

  1. Head polymerization of a protein
  2. Tail polymerization of a protein
  3. Head polymerization of a polysaccharide
  4. Head polymerization of a nucleic acid
  5. Tail polymerization of a nucleic acid

 

1.35.       What is the end product of glycolysis in the cytoplasm of eukaryotic cells? How many carbon atoms does the molecule have?

  1. Acetyl CoA; it has two carbon atoms attached to coenzyme A
  2. Phosphoenolpyruvate; it has three carbon atoms
  3. Glucose 6-phosphate; it has six carbon atoms
  4. Pyruvate; it has three carbon atoms
  5. Glyceraldehyde 3-phosphate; it has three carbon atoms

 

1.36.       The substrate for the glycolytic enzyme glyceraldehyde 3-phosphate dehydrogenase is glyceraldehyde 3-phosphate (with one phosphate group) while its product is 1,3-bisphosphoglycerate (with two phosphate groups). Where does the extra phosphate group come from?

  1. From combining two molecules of the substrate
  2. ATP
  3. Fructose 1,6-bisphosphate
  4. Pi
  5. NADH

 

1.37.       Steps 6 and 7 of glycolysis are catalyzed by the enzymes glyceraldehyde 3-phosphate dehydrogenase and phosphoglycerate kinase, respectively. Together, they …

  1. result in the oxidation of an aldehyde to a carboxylic acid.
  2. produce both ATP and NADH.
  3. couple the oxidation of a C–H bond to the activation of carrier molecules.
  4. catalyze the only glycolytic reactions that create a high-energy phosphate linkage directly from inorganic phosphate.
  5. All of the above.

 

1.38.       Arsenate is a toxic ion that can interfere with both glycolysis and oxidative phosphorylation. Arsenate resembles Pi (inorganic phosphate) and can replace it in many enzymatic reactions. One such reaction is catalyzed by glyceraldehyde 3-phosphate dehydrogenase in step 6 of glycolysis. Upon completion of the reaction, instead of the normal product, 1,3-bisphosphoglycerate, the mixed anhydride 1-arsenato-3-phosphoglycerate is formed; this undergoes rapid spontaneous hydrolysis into arsenate plus 3-phosphoglycerate, the latter being a normal product of step 7 in glycolysis. What would be the effect of arsenate poisoning in glycolysis?

  1. It results in more ATP and NADH molecules generated for every glucose molecule.
  2. It results in fewer ATP molecules generated per glucose molecule, but NADH generation is not directly affected.
  3. It brings glycolysis to an abrupt stop.
  4. It results in fewer ATP and NADH molecules generated per glucose molecule.
  5. It does not affect the number of ATP or NADH molecules generated per glucose molecule.

 

1.39.       Which of the following is true regarding energy production and storage in plants and animals?

  1. Plant and animal cells make starch for long-term energy storage.
  2. Most of the ATP in a plant cell has been generated in the chloroplast and transported to other parts of the cell.
  3. Oxidation of one gram of starch releases more energy than oxidation of fat, but since starch absorbs a lot of water, it is not as efficient as fat in energy storage.
  4. Animals, but not plants, can store fats in the form of triacylglycerol (triglyceride).
  5. Plant seeds often contain large amounts of fats and starch.

 

1.40.       What are the molecules that normally supply carbon and oxygen atoms, respectively, for the citric acid cycle?

  1. Oxaloacetate, oxaloacetate
  2. Acetyl CoA, O2
  3. Oxaloacetate, O2
  4. Acetyl CoA, H2O
  5. Pyruvate, pyruvate

 

3.41.       Indicate if each of the following descriptions matches lipids (1), nucleic acids (2), polysaccharides (3), or proteins (4). Your answer would be a four-digit number composed of digits 1 to 4 only, e.g. 1332.

(  ) Their monomers contain phosphorus and nitrogen.

(  ) They constitute almost half of the cell’s dry mass.

(  ) They are the main constituent of all cellular membranes.

(  ) They are largely hydrophobic and can store energy.

 

1.42.       Sort the following molecules (A to E) based on the oxidation of the carbon atom, from higher to lower oxidation states. Your answer would be a five-letter string composed of letters A to E only, e.g. ADCBE. Put the letter corresponding to the highest oxidation level on the left.

 

3.43.       Indicate true (T) and false (F) statements below regarding glycolysis. Your answer would be a four-letter string composed of letters T and F only, e.g. TTTT.

(  )  Molecular oxygen is used in glycolysis to oxidize glucose.

(  )  Along the glycolytic pathway, ATP is both consumed and generated.

(  )  In the course of glycolysis, one molecule of NADH is formed per molecule of glucose.

(  )  Following the production of one molecule of fructose 1,6-bisphosphate, the rest of the glycolytic pathway generates four molecules of ATP.

 

3.44.       Fill in the blank in the following paragraph.

“During intense ‘anaerobic’ physical exercise, the high energy demand in the muscle cells leads to an accumulation of lactic acid in these cells and their surrounding tissues. Similarly, the yeast Saccharomyces cerevisiae can produce ethanol when grown anaerobically. The lactate or ethanol production takes place in a process called …”

 

3.45.       Sort the following molecules based on the amount of energy that is released when their phosphate bond is hydrolyzed as indicated. Your answer would be a four-letter string composed of letters A to D only, e.g. ADCB. Put the molecule with the highest amount of hydrolysis energy on the left.

(A) ATP when hydrolyzed to ADP

(B) Glucose 6-phosphate when hydrolyzed to glucose

(C) 1,3-bisphosphoglycerate when hydrolyzed to 3-phosphoglycerate

(D) Phosphoenolpyruvate when hydrolyzed to pyruvate

 

3.46.       Indicate true (T) and false (F) statements below regarding fatty acid metabolism. Your answer would be a four-letter string composed of letters T and F only, e.g. TTTT.

(  )  Most animals derive their energy from fatty acids between meals.

(  )  Fatty acids are converted to acetyl CoA in the cytosol, which is then transported into mitochondria for further oxidation.

(  )  Fatty acids are stored in fat droplets in the form of triacylglycerols.

(  )  The breakdown of fatty acids into each acetyl CoA unit requires the hydrolysis of two ATP molecules.

 

3.47.       Indicate whether each of the following descriptions matches glycolysis (G) or the Krebs cycle (K). Your answer would be a four-letter string composed of letters G and K only, e.g. GGGK.

(  )  It oxidizes acetyl CoA to CO2.

(  )  In eukaryotic cells, it is carried out in the cytosol.

(  )  It produces FADH2.

(  )  α-Ketoglutarate, one of its intermediates, is used to synthesize the amino acid glutamic acid.

 

3.48.       Indicate whether each of the following molecules is an intermediate in glycolysis (G) or in the tricarboxylic acid cycle (T). Your answer would be a four-letter string composed of letters G and T only, e.g. GGTT.

(  )  Fumarate

(  )  Malate

(  )  Phosphoenolpyruvate

(  )  Succinate

 

 

The Citric Acid Cycle: Questions 49-52

The citric acid cycle is summarized in the following figure. Answer the following question(s) about this cycle.

 

3.49.       In step 1 of the citric acid cycle drawn above, what is the molecule indicated with a question mark?

  1. O2
  2. ATP
  3. H2O
  4. H+
  5. Pyruvate

 

3.50.       In the citric acid cycle shown above, which steps produce CO2 as a by-product? List all such steps by their number, from the smallest number to the largest. Your answer would be a number composed of digits 1 to 8 only, e.g. 258.

 

3.51.       In the citric acid cycle shown above, which steps produce either NADH or FADH2? List all such steps by their number, from the smallest number to the largest. Your answer would be a number composed of digits 1 to 8 only, e.g. 258.

 

3.52.       Aconitase catalyzes an isomerization reaction in the citric acid cycle shown above, in which H2O is first removed and then added back to the substrate. Which step is catalyzed by this enzyme? Write down the step number as your answer, e.g. 5.

 

 

 

3.53.       The electron carriers NADH and FADH2 donate their electrons to the electron-transport chain in the inner mitochondrial membrane, leading to ATP synthesis powered by an H+ gradient across the membrane. If, on average, the oxidation of each NADH or FADH2 molecule in this pathway results in the production of 2.5 and 1.5 molecules of ATP, respectively, how many ATP (and GTP) molecules are produced on average as a result of the complete oxidation of one molecule of acetyl CoA in the mitochondrion? Consider only the citric acid cycle and oxidative phosphorylation.

  1. 10
  2. 12
  3. 13.5
  4. 14.5
  5. 15

 

 

3.54.       Indicate true (T) and false (F) statements below regarding the cellular metabolism of nucleotides and amino acids. Your answer would be a four-letter string composed of letters T and F only, e.g. TTTT.

(  )  Nitrogen fixation occurs in the mitochondria in most animal cells to generate amino acids.

(  )  All 20 natural amino acids must be provided in our diet and are therefore “essential.”

(  )  There are NO essential nucleotides that must be provided in the diet.

(  )  Catabolism of amino acids in our body leads to the production of urea which is excreted.

 

 

 

Answers

  1. Answer: C

Difficulty: 1

Section: The Chemical Components of a Cell

Feedback: Metal ions such as copper, iron, cobalt, and zinc are used as cofactors that are necessary for the function of some enzymes.

  1. Answer: E

Difficulty: 1

Section: The Chemical Components of a Cell

Feedback: Hydrophobic molecules usually have no charge and form no or few hydrogen bonds, and are therefore not favored by the network of hydrogen bonds in liquid water. They do dissolve in nonpolar organic solvents.

  1. Answer: IHHN

Difficulty: 2

Section: The Chemical Components of a Cell

Feedback: ATP is negatively charged and can form ionic bonds with magnesium ions. Urea is highly soluble in water due to its hydrogen-bonding capacity. Similarly, interaction of a polar molecule like glucose with the active site of an enzyme can be mediated by hydrogen bonds and other noncovalent (or even covalent) bonds. In contrast, the fatty acid tails in phospholipids are hydrophobic and do not favorably interact with negatively charged phosphate molecules.

  1. Answer: B

Difficulty: 3

Section: The Chemical Components of a Cell

Feedback: Hexane is an alkane hydrocarbon incapable of hydrogen-bonding with water molecules, which results in an entropically unfavorable state when the two interact. All the other mentioned chemicals can be readily dissolved in water because they have polar bonds or can dissociate into ions.

  1. Answer: A

Difficulty: 2

Section: The Chemical Components of a Cell

Feedback: The probing of the charged ions by water molecules greatly reduces the bond energy of ionic bonds (electrostatic interactions) in aqueous solutions. Hydrogen bonds are also weakened in water, but they are not that strong in a vacuum to begin with.

  1. Answer: C

Difficulty: 1

Section: The Chemical Components of a Cell

Feedback: Although each noncovalent bond is weak, when many of them are formed simultaneously (in a complementary interface), their energies can sum to produce a tight binding.

  1. Answer: B

Difficulty: 3

Section: The Chemical Components of a Cell

Feedback: The concentration of hydronium ions would be the sum of those obtained from the dissociation of water and the acid: [H3O+] = 10–7 + 10–8 = 1.1 × 10–7. The pH value will then be calculated as: pH = –log [H3O+] = –log [1.1 × 10–7] = 7 – log (1.1) = 6.96. This is very close to neutral pH.

  1. Answer: A

Difficulty: 3

Section: The Chemical Components of a CellFeedback: The initial number of hydronium ions would be: [H3O+]1 = 10–15 L × 10–7 mole/L = 10–22 mole. The final number at pH 5 would be: [H3O+]2 = 10–15 L × 10–5 mole/L = 10–20 mole. The difference is: [H3O+]2 – [H3O+]1 = 9.9 × 10–21 mole. This is equivalent to approximately 6000 protons that need to be pumped in.

  1. Answer: D

Difficulty: 3

Section: The Chemical Components of a Cell

Feedback: Due to the presence of the carboxyl group, a fatty acid molecule carries a negative charge at neutral pH. However, lowering the pH can reverse the ionization of this group to the neutral (protonated) state.

  1. Answer: C

Difficulty: 2

Section: The Chemical Components of a Cell

Feedback: The amino and carboxyl groups are common to all amino acids. The serine side chain contains a hydroxyl group.

  1. Answer: C

Difficulty: 2

Section: The Chemical Components of a Cell

Feedback: The polymerization reaction generally requires a free-energy input. Also, the addition of each monomer to the growing polymer is a condensation reaction that is accompanied by the release of one water molecule. The opposite reaction (depolymerization) involves hydrolysis and consumes one water molecule.

  1. Answer: BDCA

Difficulty: 2

Section: The Chemical Components of a Cell

Feedback: Water accounts for about 70% of the total mass in a typical cell. In the remaining “dry mass,” proteins constitute about half, the nucleic acids RNA and DNA are next, and polysaccharides (and their sugar monomers) are still less abundant.

  1. Answer: C

Difficulty: 2

Section: Catalysis and the Use of Energy by Cells

Feedback: For a living animal cell, heat-generating reactions from burning of foodstuffs are “coupled” to other reactions that increase order inside the cell. Concomitantly, there is an increase in the overall entropy of the universe (cell plus its environment), with no violation of the laws of thermodynamics for a spontaneous process.

  1. Answer: –2 kJ/mole/K

Difficulty: 3

Section: Catalysis and the Use of Energy by Cells

Feedback: For the folding reaction, the free-energy change can be written as:

ΔG = ΔHTΔS

Therefore:

ΔS = (ΔH – ΔG)/T = (–640 kJ/mole + 40 kJ/mole) / (300 K) = –2 kJ/mole/K

The negative value of ΔS means a decrease in entropy. This is not unexpected since folding results in the formation of a single conformation (or a limited set of conformations) out of an enormous number of possible coils.

  1. Answer: D

Difficulty: 2

Section: Catalysis and the Use of Energy by Cells

Feedback: Photosynthesis consumes water and atmospheric CO2 to make simple sugars and the by-product oxygen.

  1. Answer: A

Difficulty: 3

Section: Catalysis and the Use of Energy by Cells

Feedback: Oxidation involves the full or partial removal of electrons from an atom, and does not necessarily involve oxygen. In the cell, organic molecules usually release a proton to their surrounding when oxidized in a dehydrogenation reaction, decreasing the number of C–H bonds in the molecule.

  1. Answer: E

Difficulty: 2

Section: Catalysis and the Use of Energy by Cells

Feedback: Enzymes catalyze most cellular reactions by lowering the activation energy, but they cannot change the equilibrium constant of the reactions that they catalyze; that is, both forward and reverse reactions are sped up by the same factor. However, they can selectively drive substrates along one of various cellular metabolic pathways, and can also couple unfavorable reactions to spontaneous heat-generating reactions.

  1. Answer: A

Difficulty: 2

Section: Catalysis and the Use of Energy by Cells

Feedback: The activation energy corresponds to the height of the energy barrier between the reactant and the product, and is the minimum amount of energy that should be provided in order for the reaction to proceed.

  1. Answer: C

Difficulty: 2

Section: Catalysis and the Use of Energy by Cells

Feedback: In the presence of an enzyme (line 1), the fraction of substrate molecules that have enough thermal energy to proceed through the reaction is increased compared to that in the uncatalyzed reaction (line 2).

  1. Answer: B

Difficulty: 3

Section: Catalysis and the Use of Energy by Cells

Feedback: This is the reaction catalyzed by the enzyme succinate dehydrogenase in the citric acid cycle. Succinate is oxidized to fumarate, and the FAD carrier is reduced to FADH2. By subsequently donating its two electrons to the electron-transport chain, FADH2 will be converted back to FAD for another round of the reaction.

  1. Answer: D

Difficulty: 3

Section: Catalysis and the Use of Energy by Cells

Feedback: The net distance of 5 µm is 10 times higher than 0.5 µm, and would on average take 10 seconds (i.e. 102 × 100 milliseconds) to reach.

  1. Answer: CABD

Difficulty: 2

Section: Catalysis and the Use of Energy by Cells

Feedback: In general, larger molecules diffuse more slowly compared to smaller molecules. Interaction with other molecules (including the solvent) and the shape of the molecule will also affect the diffusion coefficient.

  1. Answer: A

Difficulty: 3

Section: Catalysis and the Use of Energy by Cells

Feedback: The equilibrium constant for this reaction is calculated as:

Keq = 0.5 = [B]eq [C]eq / [A]eq

Since the initial mixture contains only molecule A, it follows that:

[B]eq = [C]eq = 1 M – [A]eq

Combining these equations and solving for [A]eq, we will have:

[A]eq = 0.5 M

which means the molecules B and C will also be present at 0.5 M at equilibrium.

  1. Answer: +23.6 kJ/mole

Difficulty: 4

Section: Catalysis and the Use of Energy by Cells

Feedback: The free-energy change can be written as:

ΔG = ΔG° + RT ln([B]/[A])

When ΔG is equal to zero, the system is at chemical equilibrium, and

ΔG° = – RT ln([B]eq/[A]eq) = – RT ln(10–2) = –5.9 × log (10–2) = +11.8 kJ/mole

When the concentrations are changed, we have:

ΔG = ΔG° + RT ln([B]/[A]) = ΔG° + RT ln(102) = ΔG° – RT ln(10–2) = 2 ΔG° = +23.6 kJ/mole

  1. Answer: C

Difficulty: 3

Section: Catalysis and the Use of Energy by Cells

Feedback: The negative ΔG value indicates that the reaction is favorable under these conditions and would increase the entropy of the universe. However, unless we know the steps of the reaction, the ΔG value cannot predict the reaction rate, because the latter depends on the activation-energy barrier. Finally, the ΔG value changes as the concentrations of reactants and products change. As the products accumulate, the reaction will eventually reach an equilibrium, where ΔG is equal to zero.

  1. Answer: C

Difficulty: 3

Section: Catalysis and the Use of Energy by Cells

Feedback: The two reactions described in the question can be written as:

ATP + glucose            → ADP + glucose 6-phosphate                      ΔG°= –17 kJ/mole

glucose 6-phosphate    + H2O  → glucose + Pi                                    ΔG°= –14 kJ/mole

Combining these reactions yields the ATP hydrolysis reaction presented in the question. Since the free-energy changes are additive, the ΔG° value for this combined reaction is the sum of the ΔG° values for the two reactions above:

(–17 kJ/mol) + (–14 kJ/mole) = –31 kJ/mole.

  1. Answer: C

Difficulty: 3

Section: Catalysis and the Use of Energy by Cells

Feedback: Since the equilibrium constant is less than 1, the log Keq term is negative, making the ΔG° positive, which means the reaction is unfavorable under standard conditions. But inside a cell performing glycolysis, a lower concentration of fructose 6-phosphate than of glucose 6-phosphate can drop ΔG to a negative value. The reaction thus proceeds in the forward direction, providing a continuous supply of substrate for the next step in the pathway. Note that due to “coupling” of the reactions in the glycolytic pathway, and even though some steps can have positive ΔG° values, the overall negative ΔG° of the pathway can drive the entire chain of reactions in the forward direction, even under the standard conditions.

  1. Answer: B

Difficulty: 1

Section: Catalysis and the Use of Energy by Cells

Feedback: The activated carrier molecules carry chemical groups in high-energy linkages and can deliver the group or the energy (or both) to metabolic reactions when necessary. They then need to be activated again. NADH is an activated carrier, while its oxidized form NAD+ is not.

  1. Answer: D

Difficulty: 1

Section: Catalysis and the Use of Energy by Cells

Feedback: By transferring either a phosphate group or a pyrophosphate group to a hydroxyl group on one of the monomers involved in the polymerization, ATP “activates” the monomer, making the overall reaction favorable.

  1. Answer: E

Difficulty: 3

Section: Catalysis and the Use of Energy by Cells

Feedback: The extra phosphate in NADPH does not affect its electron-transfer properties, but makes it different enough to be recognized by a different set of enzymes. NADPH operates chiefly with enzymes that catalyze anabolic reactions, which normally need reducing power. Accordingly, NADPH is found mostly in its reduced form (i.e. in excess over NADP+) in the cell. The opposite is true for NADH, which is normally involved in catabolic reactions.

  1. Answer: C

Difficulty: 2

Section: Catalysis and the Use of Energy by Cells

Feedback: In a reduction reaction, NADPH (or NADH) is oxidized, donating a hydride ion to the substrate. A substrate can also capture a proton from the surroundings, creating two C–H bonds. The carrier is converted to its oxidized form (NADP+ or NAD+).

  1. Answer: D

Difficulty: 2

Section: Catalysis and the Use of Energy by Cells

Feedback: The reaction shown is catalyzed by the enzyme pyruvate carboxylase, which uses a covalently bound carboxylated biotin to carboxylate pyruvate and produce oxaloacetate.

  1. Answer: D

Difficulty: 3

Section: Catalysis and the Use of Energy by Cells

Feedback: Under anaerobic conditions, NAD+ can be recycled in this reaction (in which pyruvate is reduced), so that glycolysis can continue in the absence of oxidative phosphorylation.

  1. Answer: E

Difficulty: 2

Section: Catalysis and the Use of Energy by Cells

Feedback: Each nucleotide monomer is activated—at the expense of hydrolysis of two ATP molecules—into an intermediate carrying a reactive phosphoanhydride bond.

  1. Answer: D Difficulty: 1    Section: How Cells Obtain Energy from Food

Feedback: In glycolysis, two pyruvate molecules (each with three carbon atoms) are produced from each molecule of glucose (with six carbon atoms).

  1. Answer: D

Difficulty: 2

Section: How Cells Obtain Energy from Food

Feedback: In step 6 of glycolysis, this enzyme couples the oxidation of the substrate with the production of NADH, as well as incorporation of inorganic phosphate. The Pi is then transferred to ADP to generate ATP in step 7.

  1. Answer: E

Difficulty: 2

Section: How Cells Obtain Energy from Food

Feedback: Please refer to Figure 2–48.

  1. Answer: B

Difficulty: 3

Section: How Cells Obtain Energy from Food

Feedback: Since step 7 is bypassed, the ATP molecules that are naturally generated in that step are no longer produced; however, NADH is still made as before in the first half of step 6. Arsenate also has other effects on cell metabolism that collectively make it a toxic compound. Please refer to Figure 2–48.

  1. Answer: E

Difficulty: 1

Section: How Cells Obtain Energy from Food

Feedback: Compared to the polysaccharides glycogen (in animals) and starch (in plants), fat is more efficient as a long-term energy storage both per gram and per volume. It can be stored as triglycerides in both plants and animals, although the types of fatty acids vary. In plant cells, chloroplasts generate sugars that can be oxidized by the mitochondria to generate ATP for the cell. The ATP produced in the chloroplast by photosynthesis cannot be transported out of that organelle.

  1. Answer: D

Difficulty: 2

Section: How Cells Obtain Energy from Food

Feedback: One molecule of acetyl CoA enters the cycle by combining with oxaloacetate.

  1. Answer: 2411

Difficulty: 2

Section: The Chemical Components of a Cell

Feedback: The nucleotides contain one to three phosphate groups and a nitrogen-containing base, and are polymerized to form long nucleic acid molecules such as DNA. Proteins are made of amino acids and make up half of the dry mass of the cell, i.e. approximately 15% of the total cell weight. Lipids have large hydrophobic fatty acid chains and, in addition to forming bilayer membranes, can store food energy and release it when necessary.

  1. Answer: DAECB

Difficulty: 3

Section: Catalysis and the Use of Energy by Cells

Feedback: As a general rule, in organic molecules, a lower number of C–H bonds corresponds to more oxidized carbon atoms.

  1. Answer: FTFT

Difficulty: 2

Section: How Cells Obtain Energy from Food

Feedback: During the stepwise oxidation of glucose in the course of glycolysis, two molecules of ATP are used to make fructose 1,6-bisphosphate, which is then cleaved and eventually converted to two molecules of pyruvate, generating four molecules of ATP and two molecules of NADH.

  1. Answer: fermentation

Difficulty: 1

Section: How Cells Obtain Energy from Food

Feedback: Fermentation is an energy-yielding pathway and is often anaerobic.

  1. Answer: DCAB

Difficulty: 2

Section: How Cells Obtain Energy from Food

Feedback: Hydrolysis of phosphoenolpyruvate to pyruvate is the most exergonic (releases the highest amount of energy). ATP can be generated from ADP upon the hydrolysis of 1,3-bisphosphoglycerate to 3-phosphoglycerate. ATP hydrolysis can be used to drive the phosphorylation of glucose.

  1. Answer: TFTF

Difficulty: 2

Section: How Cells Obtain Energy from Food

Feedback: Between meals, fatty acids stored in the fat droplets in adipocytes in the form of triacylglycerol are released by hydrolysis and enter the bloodstream. Upon entry into other cells, they are transported to the mitochondria where they are mostly converted to acetyl CoA in a step-by-step manner, each step producing one FADH2 and one NADH molecule.

  1. Answer: KGKK

Difficulty: 2

Section: How Cells Obtain Energy from Food

Feedback: In the citric acid cycle (Krebs cycle), which takes place in the mitochondria, the carbon atoms of acetyl CoA are oxidized and released as CO2, while NADH, FADH2, and GTP are generated in the process. Many intermediates of the citric acid cycle and glycolysis are precursors for the biosynthesis of important small molecules in the cell.

  1. Answer: TTGT

Difficulty: 2

Section: How Cells Obtain Energy from Food

Feedback: Phosphoenolpyruvate is converted to pyruvate in the last step of glycolysis. Succinate, fumarate, and malate are three consecutive citric acid cycle intermediates leading to the regeneration of oxaloacetate.

  1. Answer: C

Difficulty: 3

Feedback: In the first step of the citric acid cycle, CoA is hydrolyzed by water after the formation of a citryl CoA intermediate.

  1. Answer: 34

Difficulty: 3

Feedback: Steps 3 and 4 are catalyzed by isocitrate dehydrogenase and the α-ketoglutarate dehydrogenase complex, respectively, and involve decarboxylation of the substrates and the release of carbon dioxide.

  1. Answer: 3468

Difficulty: 3

Refer to: The Citric Acid Cycle          Section: How Cells Obtain Energy from Food

Feedback: Steps 3, 4, and 8 produce NADH, while step 6 produces FADH2.

  1. Answer: 2

Difficulty: 2

Feedback: Aconitase converts citrate to isocitrate through an aconitate intermediate created by dehydration of the substrate.

  1. Answer: A

Difficulty: 3

Section: How Cells Obtain Energy from Food

Feedback: The complete oxidation of a molecule of acetyl CoA results in the production of three NADH molecules plus one FADH2 and one GTP (or ATP) molecule. Therefore, as a result of oxidative phosphorylation, a total of 10 molecules are generated:

(3 × 2.5) + (1 × 1.5) + 1 = 10

  1. Answer: FFTT

Difficulty: 2

Section: How Cells Obtain Energy from Food

Feedback: All known nitrogen-fixing cells are prokaryotic microorganisms. Animals rely on their dietary intake of protein and nucleic acids as sources of useful nitrogen. However, only 9 of the 20 amino acids that make up proteins and none of the nucleotides that make up nucleic acids are essential; the remainder can be synthesized from other ingredients in the diet. When amino acids in our body are degraded, their nitrogen atoms eventually appear in urea molecules which are excreted.

 

 

 

 

 

MOLECULAR BIOLOGY OF THE CELL, SIXTH EDITION

CHAPTER 4: DNA, CHROMOSOMES, AND GENOMES

© Garland Science 2015

 

 

1.55.       In a double-stranded DNA molecule, one of the chains has the sequence CCCATTCTA when read from the 5′ to the 3′ end. Indicate true (T) and false (F) statements below regarding this chain. Your answer would be a four-letter string composed of letters T and F only, e.g. TTFT.

(  )  The other chain is heavier, i.e. it has a greater mass.

(  )  There are no C residues in the other chain.

(  )  The 5′-terminal residue of the other chain is G.

(  )  The other chain is pyrimidine-rich.

 

1.56.       Indicate which numbered feature (1 to 5) in the schematic drawing below of the DNA double helix corresponds to each of the following. Your answer would be a five-digit number composed of digits 1 to 5 only, e.g. 52431.

 

 

(  )  Hydrogen-bonding

(  )  Covalent linkage

(  )  Phosphate group

(  )  Nitrogen-containing base

(  )  Deoxyribose sugar

 

1.57.       Complete the DNA sequence below such that the final sequence is identical to that of the complementary strand. Your answer would be a seven-letter string composed of letters A, C, T, and G only, e.g. TTCTCAG.

 

5′- C T T T A G A               -3′

 

1.58.       A DNA nucleotide pair has an average mass of approximately 660 daltons. Knowing the number of nucleotides in the human genome, how many picograms of DNA are there in a diploid human nucleus? Avogadro’s number is 6 × 1023. Write down the picogram amount without decimals (round the number to the closest integer), e.g. 23 pg.

 

1.59.       Which of the following features of DNA underlies its simple replication procedure?

  1. The fact that it is composed of only four different types of bases
  2. The antiparallel arrangement of the double helix
  3. The complementary relationship in the double helix
  4. The fact that there is a major groove and a minor groove in the double helix

 

1.60.       Which of the following correlates the best with biological complexity in eukaryotes?

  1. Number of genes per chromosome
  2. Number of chromosomes
  3. Number of genes
  4. Genome size (number of nucleotide pairs)

 

1.61.       Indicate true (T) and false (F) statements below about the human genome. Your answer would be a six-letter string composed of letters T and F only, e.g. FTFFFT.

(  )  Only about 1.5% of the human genome is highly conserved.

(  )  Almost half of our genome is composed of repetitive sequences.

(  )  Genes occupy almost a quarter of the genome.

(  )  There are roughly as many pseudogenes in the human genome as functional genes.

(  )  Transposable elements occupy almost 10% of our genome.

(  )  On average, exons comprise 1.5% of our genes.

 

1.62.       Chromosome 3 contains nearly 200 million nucleotide pairs of our genome. If this DNA molecule could be laid end to end, how long would it be? The distance between neighboring base pairs in DNA is typically around 0.34 nm.

  1. About 7 mm
  2. About 7 cm
  3. About 70 cm
  4. About 7 m
  5. None of the above

 

1.63.       For the Human Genome Project, cloning of large segments of our genome was first made possible by the development of yeast artificial chromosomes, which are capable of propagating in the yeast Saccharomyces cerevisiae just like any of the organism’s 16 natural chromosomes. In addition to the cloned human DNA, these artificial vectors were made to contain three elements that are necessary for them to function as a chromosome. What are these elements? Write down the names of the elements in alphabetical order, and separate them with commas, e.g. gene, histone, nucleosome.

 

1.64.       Indicate whether each of the following descriptions better applies to a centromere (C), a telomere (T), or an origin of replication (O). Your answer would be a seven-letter string composed of letters C, T, and O only, e.g. TTTCCTO.

(  )  It contains repeated sequences at the ends of the chromosomes.

(  )  It is NOT generally longer in higher organisms compared to yeast.

(  )  Each eukaryotic chromosome has many such sequences.

(  )  There are normally two such sequences in each eukaryotic chromosomal DNA molecule.

(  )  There is normally one such sequence per eukaryotic chromosomal DNA molecule.

(  )  It is where DNA duplication starts in S phase.

(  )  It attaches the chromosome to the mitotic spindle via the kinetochore structure.

 

1.65.       The eukaryotic chromosomes are organized inside the nucleus with a huge compaction ratio of several-thousand-fold. What is responsible for such a tight packaging?

  1. The various chromatin proteins that wrap and fold the DNA
  2. The nuclear envelope which encapsulates the chromosomes
  3. The nuclear matrix that provides a firm scaffold
  4. All of the above

 

1.66.       The two chromosomes in each of the 22 homologous pairs in our cells …

  1. have the exact same DNA sequence.
  2. are derived from one of our parents.
  3. show identical banding patterns after Giemsa staining.
  4. usually bear different sets of genes.
  5. All of the above.

 

1.67.       Compared to the human genome, the genome of yeast typically has …

  1. more repetitive DNA.
  2. longer genes.
  3. more introns.
  4. longer chromosomes.
  5. a higher fraction of coding DNA.

 

1.68.       Indicate true (T) and false (F) statements below regarding histones. Your answer would be a six-letter string composed of letters T and F only, e.g. TTFFFF.

(  )  The histones are highly acidic proteins.

(  )  The histone fold consists of three α helices.

(  )  The core histones are much more conserved than the H1 histone.

(  )  The N-terminal tails of the core histones undergo a variety of reversible post-translational modifications.

(  )  Every nucleosome core is made up of three polypeptide chains.

(  )  The H1 histone is absent in the 30-nm fibers.

 

1.69.       Indicate which feature (1 to 4) in the schematic drawing below of a chromatin fiber corresponds to each of the following. Your answer would be a four-digit number composed of digits 1 to 4 only, e.g. 2431.

 

(  )  Nucleosome core particle

(  )  Linker DNA

(  )  Histone octamer

(  )  Non-histone protein

 

1.70.       In assembling a nucleosome, normally the …(1) histone dimers first combine to form a tetramer, which then further combines with two … (2) histone dimers to form the octamer.

  1. 1: H1–H3; 2: H2A–H2B
  2. 1: H3–H4; 2: H2A–H2B
  3. 1: H2A–H2B; 2: H1–H3
  4. 1: H2A–H2B; 2: H3–H4
  5. 1: H1–H2; 2: H3–H4

 

1.71.       The chromatin remodeling complexes play an important role in chromatin regulation in the nucleus. They …

  1. can slide nucleosomes on DNA.
  2. have ATPase activity.
  3. interact with histone chaperones.
  4. can remove or exchange core histone subunits.
  5. All of the above.

 

1.72.       Which of the following is true regarding heterochromatin in a typical mammalian cell?

  1. About 1% of the nuclear genome is packaged in heterochromatin.
  2. The DNA in heterochromatin contains all of the inactive genes in a cell.
  3. Genes that are packaged in heterochromatin are permanently turned off.
  4. The different types of heterochromatin share an especially high degree of compaction.
  5. Heterochromatin is highly concentrated in the centromeres but not the telomeres.

 

1.73.       The position effect variegation (PEV) phenotype described in this chapter can be used to identify new genes that regulate heterochromatin formation. For instance, strains of Drosophila melanogaster with the White variegation phenotype have been subjected to mutagenesis to screen for dominant mutations (in other genes) that either enhance or suppress PEV, meaning the mutations result in either lower or higher red pigment production, respectively. Which of the following mutations is expected to be an enhancer of variegation?

  1. A mutation that results in the loss of function of the fly’s HP1 (heterochromatin protein 1) gene.
  2. A loss-of-function mutation in a gene encoding a histone deacetylase that deacetylates lysine 9 on histone H3.
  3. A gain-of-function mutation in a gene encoding a histone methyl transferase that trimethylates lysine 9 on histone H3, resulting in a hyperactive form of the enzyme.
  4. A gain-of-function mutation in a gene encoding a histone acetyl transferase that normally acetylates lysine 9 on histone H3, resulting in higher expression of the protein.

 

1.74.       The acetylation of lysines on the histone tails …

  1. loosens the chromatin structure because it adds positive charges to the histone.
  2. recruits the heterochromatin protein HP1, resulting in the establishment of heterochromatin.
  3. can be performed on methylated lysines only after they are first demethylated.
  4. is sufficient for the formation of an open chromatin structure.
  5. is a covalent modification and is thus irreversible.

 

1.75.       Nucleosomes that are positioned like beads on a string over a region of DNA can interact to form higher orders of chromatin structure. Which of the following factors can contribute to the formation of the 30-nm chromatin fiber from these nucleosomes?

  1. Interactions that involve the histone tails of neighboring nucleosomes
  2. Interaction of the linker histone H1 with each nucleosome
  3. Binding of proteins to DNA or the histones
  4. ATP-dependent function of chromatin remodeling complexes
  5. All of the above

1.76.       Indicate whether each of the following histone modifications is generally associated with active genes (A) or silenced genes (S). Your answer would be a four-letter string composed of letters A and S only, e.g. SSAS.

(  )  H3 lysine 9 acetylation

(  )  H3 serine 10 phosphorylation

(  )  H3 lysine 4 trimethylation

(  )  H3 lysine 9 trimethylation

 

1.77.       Indicate whether each of the following histone modifications adds a negative charge to the histone (A), removes a positive charge from the histone (B), or does neither of these (C). Your answer would be a four-letter string composed of letters A, B, and C only, e.g. CABA.

(  )  H3 lysine 9 acetylation

(  )  H3 serine 10 phosphorylation

(  )  H3 lysine 4 trimethylation

(  )  H3 lysine 9 trimethylation

 

1.78.       To study the chromatin remodeling complex SWR1, a researcher has prepared arrays of nucleosomes on long DNA strands that have been immobilized on magnetic beads. These nucleosomes are then incubated with an excess of the H2AZ–H2B dimer (which contains the histone variant H2AZ) in the presence or absence of SWR1 with or without ATP. She then separates the bead-bound nucleosomes (bound fraction) from the rest of the mix (unbound fraction) using a magnet, elutes the bound fraction from the beads, and performs SDS-PAGE on the samples. This is followed by a Western blot using an antibody specific to the H2AZ protein used in this experiment. The results are shown below, with the presence (+) or absence (–) of ATP, SWR1, or the H2AZ–H2B dimer in each incubation reaction indicated at the top of the corresponding lane.

 

Which of the following statements is confirmed by the Western blot shown?

  1. SWR1 deposits H2AZ histones into the nucleosome arrays.
  2. SWR1 function is not ATP-dependent.
  3. The antibody used in this experiment binds to the SWR1 complex.
  4. All of the above.

 

1.79.       Indicate whether each of the following descriptions better matches the major histones (M) or the histone variants (V). Your answer would be a six-letter string composed of letters M and V only, e.g. VVMVMV.

(  )  They are more highly conserved over long evolutionary time scales.

(  )  They are present in much smaller amounts in the cell.

(  )  They are synthesized primarily during the S phase of the cell cycle.

(  )  Their incorporation often requires histone exchange.

(  )  They are often inserted into already-formed chromatin.

(  )  They are assembled into nucleosomes just behind the replication fork.

 

1.80.       A chromatin “reader complex” …

  1. is always coupled to a “writer complex” and spreads specific chromatin modifications.
  2. can recognize any histone code.
  3. binds tightly to the chromatin only when a specific set of histone marks is present.
  4. can only bind to a single specific histone mark.
  5. has at least five protein subunits.

 

1.81.       The centromeric regions in the fission yeast Schizosaccharomyces pombe are wrapped by nucleosomes containing the CENP-A histone H3 variant, and are flanked by clusters of tRNA genes that separate them from the surrounding pericentric heterochromatin. If the tRNA clusters are removed from this region, the HP1-bound heterochromatin spreads further to cover the centromeric regions. The tRNA genes are transcribed by strong RNA polymerase III promoters, which can associate with transcription factors and recruit chromatin-modifying enzymes.

Based on these observations, indicate which blanks (A to E) in the paragraph below correspond to each of the following phrases. Your answer would be a five-letter string composed of letters A to E only, e.g. BCDEA.

 

“The …(A) are not sufficient to prevent heterochromatin expansion to the centromeric regions. Instead, the …(B) are acting as …(C) in S. pombe, similar to the role of the …(D) in the β-globin locus in chickens and humans. Likely candidates for the histone-modifying enzymes recruited by the RNA polymerase III complexes are …(E).”

 

(  )  HS4 element

(  )  chromatin boundaries

(  )  histone acetyl transferases

(  )  tRNA genes

(  )  CENP-A-containing histones

 

1.82.       In human cells, the alpha satellite DNA repeats …

  1. have a specific sequence indispensable for the seeding event that leads to chromatin formation.
  2. can be seen to be packaged into alternating blocks of chromatin, one of which contains the histone H3 variant CENP-A.
  3. are sufficient to direct centromere formation.
  4. are necessary for centromere formation.
  5. All of the above.

 

1.83.       It has been shown that inhibition of a key chromatin remodeling complex known as NuRD, by deleting one of its subunits, can result in a significant increase in the efficiency of reprogramming of somatic cells into pluripotent stem cells. The reprogramming is normally done by the induced expression of a battery of transcription factors in the somatic cells, but is typically not very efficient. Such an observation suggests that the NuRD complex is normally involved in …

  1. erasing the epigenetic memory in somatic cells.
  2. maintaining the epigenetic memory in somatic cells.
  3. preventing DNA replication.
  4. formation of extended loops from chromosome territories.

 

1.84.       Imagine a chromosome translocation event that brings a gene encoding a histone acetyl transferase enzyme from its original chromosomal location to a new one near heterochromatin. Which of the following scenarios is definitely NOT going to happen?

  1. The gene gets silenced due to heterochromatin expansion, leading to the misregulation of gene expression for a number of critical genes.
  2. The translocation event also brings along a chromatin barrier that can prevent heterochromatin expansion into the gene, and there is no phenotypic anomaly.
  3. Since the gene encodes a histone acetyl transferase, it resists heterochromatin expansion by acetylating its own histones.
  4. The level of the gene product decreases due to a position effect, leading to an imbalance in the chromatin state of the cell that results in the activation of programmed cell death.

 

1.85.       Lampbrush chromosomes …

  1. are transcriptionally inactive.
  2. are readily observed in the oocytes of humans and insects.
  3. have thousands of duplicated DNA molecules arranged side by side.
  4. are mitotic chromosomes with two sister chromatids attached together only at the centromere.
  5. are thought to have a structure that is relevant to mammalian chromosomes in interphase.

 

1.86.       Findings from a number of experiments on human chromatin have suggested that the DNA in our chromosomes is organized into loops of various lengths. Approximately how long is a typical loop (in nucleotide pairs of DNA)?

  1. 50
  2. 2000
  3. 100,000
  4. 10 million
  5. 50 million

 

1.87.       You have performed a chromosome conformation capture (3C) experiment to study chromatin looping at a mouse gene cluster that contains genes A, B, and C, as well as a regulatory region R. In this experiment, you performed in situ chemical cross-linking of chromatin, followed by cleavage of DNA in the nuclear extract with a restriction enzyme, intramolecular ligation, and cross-link removal. Finally, a polymerase chain reaction (PCR) was carried out using a forward primer that hybridizes to a region in the active B gene, and one of several reverse primers, each of which hybridizes to a different location in the locus. The amounts of the PCR products were quantified and normalized to represent the relative cross-linking efficiency in each analyzed sample. You have plotted the results in the graph below. The same experiment has been done on two tissue samples: fetal liver (represented by red lines) and fetal muscle (blue lines).

Interaction of the A, B, and C genes with the regulatory R region is known to enhance expression of these genes. Indicate true (T) and false (F) statements below based on the results. Your answer would be a four-letter string composed of letters T and F only, e.g. TTTF.

 

(  )  The B gene would be predicted to have higher expression in the fetal liver compared to the fetal muscle tissue.

(  )  Interaction between the R region and the B gene involves the A gene looping out.

(  )  Interaction between the R region and the B gene involves the C gene looping out.

(  )  In fetal muscle, the B gene definitely does not engage in looping interactions with any other elements in the cluster.

 

1.88.       Polytene chromosomes are useful for studying chromatin because they …

  1. are smaller than regular chromosomes and easier to manipulate.
  2. lack heterochromatin.
  3. have distinct visible banding patterns.
  4. can make polyploid cells.
  5. All of the above.

 

1.89.       Five major types of chromatin were identified in studies performed on Drosophila melanogaster cells, although much more remains to be learned about chromatin diversity and dynamics. Which of the following is correct regarding these findings in Drosophila?

  1. These results were obtained using the 3C technique, which determines the positions of loops in the chromatin.
  2. According to these results, there are four types of heterochromatin and only one type of euchromatin.
  3. The Polycomb form of chromatin belongs to the euchromatin type.
  4. In addition to these five major types of chromatin, there seem to be additional minor types as well.
  5. The pattern of chromatin types in the chromosomes is constant across the different cell types in a multicellular organism.

 

1.90.       For each of the following classifications, indicate whether you would expect to find an actively transcribed gene in the first category (1) or the second (2). Your answer would be a six-digit number composed of digits 1 and 2 only, e.g. 222121.

(  )  1: Heterochromatin, or 2: euchromatin

(  )  1: Chromosome puffs, or 2: condensed chromosome bands

(  )  1: Nuclear periphery, or 2: the center of the nucleus

(  )  1: Within the chromosome territory, or 2: extended out of the territory

(  )  1: Apart from, or 2: close to actively transcribed genes within the nucleus

(  )  1: 11-nm “beads-on-a-string” fibers, or 2: 30-nm fibers

 

1.91.       A gene that had been turned off in a liver cell has just been induced to be highly expressed as the cell responds to a new metabolic load. What observations do you expect to accompany this change?

  1. More than 100 proteins would become associated with the gene for its transcription.
  2. The nuclear position of the gene would change to place it in a “transcription factory.”
  3. Chromatin modifications associated with the gene would change in favor of higher expression.
  4. All of the above.

 

1.92.       Fill in the blank in the following paragraph regarding chromatin organization. Do not use abbreviations.

“Our ~6.4-giganucleotide nuclear DNA is organized into 46 chromosomes, each occupying a territory inside the interphase nucleus. In each chromosome, the chromatin is thought to be composed of ‘open’ and ‘closed’ chromatin compartments. At the meganucleotide scale, each compartment is organized into knot-free arrangements called … that allow tight packing and simultaneously avoid entanglement.”

 

1.93.       What are the consequences for the cell of the “fractal globule” arrangement of chromosome segments?

  1. The chromosomes cannot be condensed maximally this way.
  2. The neighboring regions of DNA are furthest from each other in the three-dimensional space.
  3. The ability of the chromatin fiber to fold and unfold efficiently is maintained.
  4. Dense packing is permitted, but the ability to easily fold and unfold the chromatin is prohibited.
  5. None of the above.

 

1.94.       Which of the following is NOT true about the nuclear subcompartments?

  1. Nucleoli, Cajal bodies, and speckles are examples of such subcompartments.
  2. Each specialized subcompartment has a distinct biochemical environment and a selected set of proteins and/or RNA molecules.
  3. The subcompartments are constitutively present in a cell except during nuclear divisions.
  4. They are likely to be organized by a tethered network of macromolecules in gel-like structures.

 

1.95.       What features do Cajal bodies, interchromatin granule clusters, and nucleoli have in common?

  1. High permeability to the surrounding nucleoplasm
  2. A network of macromolecules bound together by covalent linkages
  3. A lipid bilayer membrane
  4. The same set of RNA and protein molecules
  5. The same size

 

1.96.       As each cell in our body prepares for mitosis, its chromosomes start to look different. What are the changes in chromosome appearance that accompany the entry into M phase? Indicate true (T) and false (F) descriptions below. Your answer would be a five-letter string composed of letters T and F only, e.g. TTFFT.

(  )  The chromosomes become readily visible by the naked eye.

(  )  The chromosomes coil up further to become about 10 times shorter.

(  )  Each chromosome is condensed and then replicated to form two sister chromatids.

(  )  The typical diameter of a mitotic chromosome arm is about 70 nm.

(  )  The two sister chromatids are disentangled from each other by the time chromosome condensation is complete.

 

1.97.       The genetic information carried by a cell is passed on, generation after generation, with astonishing fidelity. However, genomes are still altered over evolutionary time scales, and even their overall size can change significantly. Which of the following genome-altering events has increased the size of the mammalian genome the most?

  1. Transposition
  2. Point mutation
  3. Chromosomal deletion
  4. Chromosomal inversion
  5. Chromosomal translocation

 

1.98.       Fill in the blanks in the following paragraph regarding genome evolution. In your answer, separate the two missing phrases with a comma, e.g. protein, plasma membrane. Do not use abbreviations.

“Conservation of genomic sequences between humans and chickens is mainly due to … selection, whereas the conservation observed between humans and chimpanzees is mostly due to the short time available for mutations to accumulate. Even the DNA sequences at the … position of synonymous codons are nearly identical between humans and chimpanzees.”

 

 

Reference: Phylogenetic Tree Questions 45 and 46

Phylogenetic trees based on nucleotide or amino acid sequences can be constructed using various algorithms. One simple algorithm is based on a matrix of pairwise genetic distances (divergences) calculated after multiple alignment of the sequences. Imagine you have aligned a particular gene from different hominids (humans and the great apes), and have estimated the normalized number of nucleotide substitutions that have occurred in this gene in each pair of organisms since their divergence from their last common ancestor. You have obtained the following distance matrix.

 

  A B C
B 0.40    
C 3.61 3.79  
D 1.36 1.44 3.94

 

Answer the following question(s) based on this matrix.

 

 

 

1.99.       If species A in the distance matrix represents human, indicate which of the other species (B to D) represents chimpanzee, gorilla, and orangutan, respectively. Your answer would be a three-letter string composed of letters B, C, and D only, e.g. DCB.

 

 

1.100.     The following tree can be constructed from these distances assuming a constant molecular clock, meaning that the length of each horizontal branch corresponds to evolutionary time as well as to the relative genetic distance from the common ancestor that gave rise to that branch. Indicate which one of the species in the matrix (B to D) corresponds to branches 1 to 3, respectively. Your answer would be a three-letter string composed of letters B, C, and D only, e.g. DCB.

 

 

1.101.     The regions of synteny between the chromosomes of two species can be visualized in dot plots. In the example shown in the following graph, a chromosome of a hypothetical species A has been aligned with the related chromosome in species B. Each dot in the plot represents the observation of high sequence identity between the two aligned chromosomes in a window located at the two corresponding chromosome positions. A series of close dots can make a continuous line. Choosing a sufficiently large window size allows a “clean” dot plot with solid lines that show only the long stretches of identity, allowing ancient large-scale rearrangements to be identified. Several chromosomal events can be detected in such dot plots. Indicate which feature (a to g) in the dot plot is best explained by each of the following events. Your answer would be a seven-letter string composed of letters a to g only, e.g. cdbagef. Each letter should be used only once.

 

 

(  )  A duplication that exists in both species

(  )  A duplication in species A only

(  )  A triplication in species B only

(  )  An inversion without relocation

(  )  An inversion combined with relocation

(  )  A deletion in species A

(  )  A translocation in species A from a different chromosome

 

 

1.102.     In each of the following comparisons, indicate whether the molecular clock is expected to tick faster on average in the first (1) or the second (2) case. Your answer would be a four-digit number composed of digits 1 and 2 only, e.g. 2222.

(  )  1: The exons, or 2: the introns of a gene

(  )  1: The mitochondrial, or 2: the nuclear DNA of vertebrates

(  )  1: The first, or 2: the third position in synonymous codons

(  )  1: A gene, or 2: its pseudogene counterpart

 

1.103.     Most fish genomes are at least 1 billion nucleotide pairs long. However, the genome of the puffer fish Fugu rubripes is quite small at only about 0.4 billion nucleotide pairs, even though the number of Fugu genes is estimated to be comparable to that of its relatives which have larger genomes. What do you think mainly accounts for the Fugu genome being this small?

  1. Evolutionary advantage of extremely small exon sizes in the Fugu lineage
  2. Unusual disappearance of all intronic sequences from the Fugu genome
  3. Increased abundance of transposable elements in the Fugu genome
  4. Increased occurrence of mitotic whole-chromosome loss in the Fugu lineage
  5. Low relative rate of DNA addition compared to DNA loss in the Fugu lineage

 

1.104.     The copy number of some human genes, such as the salivary amylase gene AMY1, varies greatly between different individuals. The salivary amylase breaks down some of the dietary starch into smaller sugars. In the case of AMY1, a positive correlation has been observed between the copy number and the amount of amylase in the saliva. Gene copy number per diploid genome can be estimated by performing a quantitative polymerase chain reaction (PCR) using primers specific to the gene of interest. You have performed such PCR experiments on samples from two human populations that have traditional diets with low and high starch levels, respectively, and have plotted the data in the histogram below. Which population (A or B) in the histogram is likely to be the one with traditionally higher dietary starch? Write down A or B as your answer.

 

 

 

1.105.     Indicate true (T) and false (F) statements below regarding a genome and its evolution. Your answer would be a four-letter string composed of letters T and F only, e.g. FFTF.

(  )  The genome of the last common ancestor of mammals can be investigated only if a DNA sample of the ancestor can be obtained.

(  )  All of the “ultraconserved” elements found in the human genome have been shown to encode long noncoding RNAs.

(  )  If a mouse carrying a homozygous deletion of a highly conserved genomic sequence survives and shows no noticeable defect, the highly conserved sequence has to be functionally unimportant.

(  ) The “human accelerated regions” are genomic regions that are found in humans with no homologs in chimpanzees or other animals.

 

1.106.     The globin gene family in mammals, birds, and reptiles is organized into α- and β-globin gene clusters that are located on two different chromosomes. In most fish and amphibians, however, the globin genes are close to each other on one chromosome. At which point (A to E) in the following simplified phylogenetic tree is a chromosomal translocation likely to have happened that placed the α- and β-globin genes on two separate chromosomes?

 

 

 

1.107.     To discover genes that have undergone accelerated evolution in the human lineage, you compared the amino acid sequences of dozens of proteins from orthologous protein-coding genes in humans, chimpanzees, and mice. For each gene, you build an unrooted phylogenetic tree in which the branch lengths (a, b, or c) correspond to the number of amino acid substitutions in that branch, as depicted below. Primates and rodents diverged ~90 million years ago, and humans and chimpanzees diverged ~5.5 million years ago. For each individual gene shared by the three species, you therefore define the “normalized substitution rate” parameters h and k as h = (a/5.5)/[c/([2 × 90] – 5.5)], and k = (b/5.5)/[c/([2 × 90] – 5.5)]. Based on these definitions, which genes are more likely to be responsible for “uniquely human” traits?

 

 

  1. Genes with very high h and k values
  2. Genes with very low h and k values
  3. Genes with very high h values but not very high k values
  4. Genes with very high k values but not very high h values

 

1.108.     Which of the following would most reliably suggest that a genomic sequence is functionally important?

  1. The presence of a long open reading frame in the sequence
  2. Multispecies conservation of the sequence
  3. Low copy number variation of the sequence
  4. The presence of active chromatin marks over the sequence

 

1.109.     Imagine a human protein containing 33 repeats of a simple domain arranged in tandem. In contrast, a homolog found in bacteria contains only one domain. What is the minimum number of duplication events that can account for the evolution of this protein since our divergence from bacteria? Write down the number as your answer, e.g. 200.

 

1.110.     Indicate true (T) and false (F) statements below regarding human genetic variations. Your answer would be a four-letter string composed of letters T and F only, e.g. TFTF.

(  )  The genomes of two randomly chosen humans are expected to be identical with respect to at least 99.99% of the nucleotides.

(  ) Copy number variations can contain genes.

(  )  If the frequency of a point mutation in a population is only 0.1%, with no mutation at this site in the rest of the population, then the variation is NOT considered to constitute a single-nucleotide polymorphism.

(  )  Most of the common genetic variants in the current human population could have been present in a human ancestral population of only about 10,000 individuals.

 

1.111.     Assume two isolated human communities with 500 individuals in each. If the same neutral mutation happens at the same time in two individuals, one from each community, what is the probability that it will be eventually fixed in both of the populations? How would the result change if the two communities fully interbreed? Write down the numbers in scientific notation and separate the two answers with a comma, e.g. 10–5, 3 × 10–2.

 

Answers

  1. Answer: TTFF

Difficulty: 2

Section: The Structure and Function of DNA

Feedback: The sequence of the other chain is 5′-TAGAATGGG-3′, which makes it slightly heavier because it is mainly composed of the bulkier purine bases.

  1. Answer: 43215

Difficulty: 1

Section: The Structure and Function of DNA

Feedback: In the double-stranded DNA, the sugar-phosphate backbones form two covalently continuous chains, while the nitrogenous bases from one chain are hydrogen-bonded to those of the other chain according to the Watson–Crick model.

  1. Answer: TCTAAAG

Difficulty: 2

Section: The Structure and Function of DNA

Feedback: The final sequence should be 5′-CTTTAGATCTAAAG-3′. The complementary strand would then have the exact same sequence. This is an example of a palindromic sequence.

  1. Answer: 7 pg

Difficulty: 3

Section: The Structure and Function of DNA

Feedback: There are about 6.4 × 109 nucleotide pairs (np) in a diploid nucleus. The average mass of 660 daltons is equivalent to 660 grams per mole (6 × 1023) of the nucleotide pairs. Thus, the total mass of nuclear DNA is:

(6.4 × 109 np) × (660 g/mole) / (6 × 1023 np/mole) = ~7 × 10–12 g = ~7 pg

  1. Answer: C

Difficulty: 1

Section: The Structure and Function of DNA

Feedback: In principle, replication would have been conceptually as simple without any of the other features.

  1. Answer: C

Difficulty: 1

Section: Chromosomal DNA and its Packaging in the Chromatin Fiber

Feedback: Biological complexity correlates better with the number of genes than it does with genome size, number of chromosomes, or number of genes per chromosome.

  1. Answer: FTTTFF

Difficulty: 2

Section: Chromosomal DNA and its Packaging in the Chromatin Fiber

Feedback: Only about 5% of our genome is highly conserved. Our relatively long genes (including their exons, introns, and some regulatory sequences) cover almost a quarter of the genome, but only about 1.5% of our genome is composed of exonic sequences. These exons constitute roughly 6% of our genes. In contrast, a whopping 50% of our genome is made of various repeated sequences, most notably the transposable DNA elements. Please refer to Table 4–1 for the data.

  1. Answer: B

Difficulty: 3

Section: Chromosomal DNA and its Packaging in the Chromatin Fiber

Feedback: Multiplying the size of the chromosome by the distance between consecutive bases gives the end-to-end distance as follows: (200 × 106 nucleotide pairs [np]) × (0.34 × 10–9 m/np) = ~0.068 m = ~7 cm.

  1. Answer: centromere, origin of replication, telomere

Difficulty: 2

Section: Chromosomal DNA and its Packaging in the Chromatin Fiber

Feedback: These three elements are required in a functional chromosome.

  1. Answer: TTOTCOC

Difficulty: 1

Section: Chromosomal DNA and its Packaging in the Chromatin Fiber

Feedback: There is normally one centromere per chromosomal DNA molecule, and two per mitotic chromosome. The mitotic kinetochore structure forms at the centromere. In contrast, the telomeres are at the two ends of each linear chromosome. There are usually many replication origins per eukaryotic chromosome. Replication origins and centromeres are both generally much longer in higher eukaryotes compared to yeast.

  1. Answer: A

Difficulty: 1

Section: Chromosomal DNA and its Packaging in the Chromatin Fiber

Feedback: The various chromatin proteins, including histone and non-histone proteins, wrap and fold the DNA to achieve an astonishing compaction ratio. The nuclear envelope and the nuclear matrix are dispensable for this effect, as evident by the high compaction seen in mitotic chromosomes when the nucleus is disassembled.

  1. Answer: C

Difficulty: 1

Section: Chromosomal DNA and its Packaging in the Chromatin Fiber

Feedback: The homologs are derived from both parents, normally have the same set of loci, and show the same banding pattern. However, the sequences are not expected to be identical.

  1. Answer: E

Difficulty: 1

Section: Chromosomal DNA and its Packaging in the Chromatin Fiber

Feedback: The yeast has a concise genome with a much higher ratio of coding to noncoding DNA.

  1. Answer: FTTTFF

Difficulty: 1

Section: Chromosomal DNA and its Packaging in the Chromatin Fiber

Feedback: The highly basic histone proteins are made of the histone fold that consists of three α helices connected by two loops. Additionally, each core histone has an N-terminal tail which, along with the rest of the protein, can be modified post-translationally. Each nucleosome core particle contains eight histone proteins, two copies of each type. The less well conserved histone H1 is not part of the nucleosome core, but is present in the 30-nm fibers.

  1. Answer: 2431

Difficulty: 1

Section: Chromosomal DNA and its Packaging in the Chromatin Fiber

Feedback: The nucleosome core particle is composed of a histone octamer wrapped by 147 nucleotide pairs of DNA, and is connected to its neighbors via a linker DNA of variable length. Non-histone proteins are also abundant in the chromatin.

  1. Answer: B

Difficulty: 1

Section: Chromosomal DNA and its Packaging in the Chromatin Fiber

Feedback: The H3–H4 and the H2A–H2B dimers appear to be stable intermediates in histone assembly and exchange. The H3–H4 tetramers are also stable, and are thought to be assembled (and inherited) mostly as a single unit.

  1. Answer: E

Difficulty: 1

Section: Chromosomal DNA and its Packaging in the Chromatin Fiber

Feedback: The ATP-dependent chromatin remodeling complexes can move the nucleosomes on DNA, interact with histone chaperones, and exchange histones or remove them from DNA.

  1. Answer: D

Difficulty: 2

Section: Chromatin Structure and Function

Feedback: Heterochromatin covers a significant fraction of the mammalian genome including the centromere and telomere neighborhoods, but contains very few, mostly inactive, genes. Even though heterochromatin has a very high degree of compaction, its genes can still become active under appropriate conditions through the remodeling of the chromatin into more open arrangements. There are many genes that are not packaged in heterochromatin but are nevertheless transcriptionally inactive.

  1. Answer: C

Difficulty: 3

Section: Chromatin Structure and Function

Feedback: An enhancer of variegation would facilitate the spread of heterochromatin into the nearby White gene, and consequently decrease red pigmentation. For example, the production of a hyperactive methyl transferase enzyme would result in elevated levels of H3 lysine 9 methylation, leading to enhanced heterochromatin expansion. The other described mutations bring about suppression of variegation.

  1. Answer: C

Difficulty: 2

Section: Chromatin Structure and Function

Feedback: Histone acetylation on lysine residues is generally an activating mark favoring more open chromatin structures (in contrast to what HP1 does), but this effect is mediated mostly through the recruited non-histone proteins that recognize this mark, not the chemical modification itself. The acetylation eliminates the positive charge on lysine, which helps—but is not sufficient alone—to loosen the nucleosome packaging. All known natural covalent modifications on the histone tails are reversible.

  1. Answer: E

Difficulty: 2

Section: Chromatin Structure and Function

Feedback: Chromatin structure is highly dynamic, undergoing tightly regulated conformational rearrangements. In the formation of the 30-nm fiber, in addition to the interactions of histone tails (such as the H4 tail) with neighboring nucleosomes and the 1-to-1 deposition of the linker histone H1, non-histone proteins that bind to the nucleosomes also play important roles. Among these are the chromatin remodeling complexes that can slide the nucleosomes in order to attain optimal positioning on the DNA.

  1. Answer: AAAS

Difficulty: 1

Section: Chromatin Structure and Function

Feedback: Acetylation of lysine 9, phosphorylation of serine 10, and trimethylation of lysine 4 in histone H3 are associated with active genes. Trimethylated lysine 9 in this histone, however, is a mark associated with silenced genes.

  1. Answer: BACC

Difficulty: 1

Section: Chromatin Structure and Function

Feedback: Histone H3 lysine 9 acetylation removes a positive charge from the histone and is associated with actively transcribed genes and open chromatin conformations; similar effects arise from H3 serine 10 phosphorylation, which adds a negative charge to the histone. Trimethylation of H3 lysine 4 is also associated with active genes. It keeps the positive charge on the lysine, as does the H3 lysine 9 trimethylation that is associated with silenced genes and heterochromatin formation.

  1. Answer: A

Difficulty: 3

Section: Chromatin Structure and Function

Feedback: Remodeling complexes can catalyze the exchange of histone subunits (including the histone variants) in nucleosomes with the help of histone chaperones. In this experiment, addition of the SWR1 complex and ATP in the presence of excess H2AZ–H2B dimers results in the appearance of H2AZ-containing nucleosomes in the arrays, i.e. a band appears for the bound fraction. This indicates that the SWR1 complex transfers the H2AZ histones to the immobilized nucleosomes. There is no reason to believe that the antibody binds to the SWR1 complex.

  1. Answer: MVMVVM

Difficulty: 2

Section: Chromatin Structure and Function

Feedback: Compared to the histone variants, the major histones are more highly conserved, are much more abundant, and become available in a burst of synthesis at S phase in order to associate with the newly replicated, histone-deficient, DNA molecules.

  1. Answer: C

Difficulty: 2

Section: Chromatin Structure and Function

Feedback: There are various chromatin reader complexes in the nucleus, each recognizing a limited set of histone mark combinations, not merely a single specific histone mark. A reader complex is highly specific thanks to its modular design: it binds tightly only if the several histone marks that it recognizes are present. A reader complex may bear a number of recognition modules all linked on one single protein, and does not have to be associated with a writer complex.

  1. Answer: DCEBA

Difficulty: 2

Section: Chromatin Structure and Function

Feedback: In this example, the tRNA genes serve as barrier sequences to block the spread of heterochromatin by recruiting histone-modifying enzymes such as histone acetyl transferase enzymes. This is similar to the role of the HS4 sequence in protecting the β-globin locus in our red blood cells. Heterochromatin can spread over the centromeric regions in the absence of the tRNA genes in the fission yeast, suggesting that the presence of CENP-A-containing histones does not provide a sufficient protection.

  1. Answer: B

Difficulty: 2

Section: Chromatin Structure and Function

Feedback: Human alpha satellite DNA has been observed to be organized in an alternating pattern, implying a folding scheme that positions the CENP-A-containing nucleosomes in proximity to the proteins of the outer kinetochore. Even though it enhances the seeding step in de novo centromere formation, the alpha satellite DNA is neither necessary nor sufficient for centromere formation: centromeres can form on a DNA molecule that lacks these repeats altogether; furthermore, the same repeat sequences are also found at non-centromeric chromosomal regions.

  1. Answer: B

Difficulty: 2

Section: Chromatin Structure and Function

Feedback: The NuRD chromatin remodeling complex is thought to be normally involved in maintaining the repressive epigenetic state of several developmentally important genes in the somatic cells, thus resisting the pluripotency induction and preventing efficient reprogramming of these cells into stem cells. Inhibition of this complex relieves this repression, allowing a more efficient reprogramming.

  1. Answer: C

Difficulty: 2

Section: Chromatin Structure and Function

Feedback: All the other scenarios can happen with some degree of likelihood. In contrast, the reasoning in (C) is flawed. The fact that a gene encodes a histone acetyl transferase does not mean that it can automatically resist heterochromatin expansion. That would require a mechanism for targeting of the protein to its own genetic locus.

  1. Answer: E

Difficulty: 2

Section: The Global Structure of Chromosomes

Feedback: The gigantic lampbrush chromosomes are meiotically paired chromosomes first found in amphibian eggs and are readily seen by light microscopy. Even though they have not been observed in mammals, studying their structure has provided important insights into the ways the chromosomes (such as those in a mammalian interphase nucleus) fold into loops of different lengths.

  1. Answer: C

Difficulty: 1

Section: The Global Structure of Chromosomes

Feedback: The typical loop size is between 50,000 and 200,000 nucleotide pairs, although significantly longer loops may also exist. A DNA stretch of 2000 nucleotide pairs would have only about 10 nucleosomes. On the other hand, 10 million nucleotide pairs would be comparable to the size of a small chromosome arm.

  1. Answer: TTFF

Difficulty: 3

Section: The Global Structure of Chromosomes

Feedback: The 3C data show high relative cross-linking between the B gene and the R region in the liver (red) but not the muscle (blue) sample. This suggests that, in the liver cells, these two elements spatially interact by looping out the region containing the A gene, and supports the prediction that the B gene is expressed at higher levels in the liver compared to muscle. Even though no significant looping interaction is detected between the B gene and the R region in the muscle sample, the gene might still interact with other elements in the locus, as suggested by the elevated cross-linking efficiency with the neighboring C gene in this sample.

  1. Answer: C

Difficulty: 2

Section: The Global Structure of Chromosomes

Feedback: The enormous polytene chromosomes have proven very useful in visualizing chromosome organization; they can be readily studied by light microscopy and show distinct banding patterns corresponding to certain genomic positions, providing clues as to how the chromatin is organized into domains at a large scale.

  1. Answer: D

Difficulty: 2

Section: The Global Structure of Chromosomes

Feedback: The five major chromatin types were found by analyzing localization data for tens of proteins and histone marks. Such data can be obtained through techniques such as chromatin immunoprecipitation. Three heterochromatin (compacted) and two euchromatin (open) chromatin types were identified in this way, including the known classical HP1 heterochromatin and the developmentally important Polycomb heterochromatin. Note that the chromatin types are not fixed during development and across different cell types, and that in addition to the five major types, other minor chromatin forms appear to be present.

  1. Answer: 212221

Difficulty: 3

Section: The Global Structure of Chromosomes

Feedback: More open chromatin is usually associated with higher transcriptional activity. Thus, an actively transcribed gene is likely to be found in euchromatin (which is generally not associated with the nuclear lamina), and more so in the extended 11-nm fibers. In a polytene chromosome, “puffs” represent the sites where active transcription occurs. An active gene can even leave the territory of its parent chromosome and extend out into “factories” of transcription along with other active genes.

  1. Answer: D

Difficulty: 2

Section: The Global Structure of Chromosomes

Feedback: High transcriptional activity is accompanied by the recruitment of a large number of proteins including the RNA polymerase transcription machinery. This may coincide with the migration of the gene from its previous environment in the chromosome territory out toward the foci of active transcription. The chromatin profile of the gene is also expected to change toward a more open state.

  1. Answer: fractal globules

Difficulty: 1

Section: The Global Structure of Chromosomes

Feedback: Most regions of our chromosomes are folded into a conformation referred to as a fractal globule. This knot-free arrangement facilitates maximally dense packing, but also preserves the ability of the chromatin fiber to unfold and fold.

  1. Answer: C

Difficulty: 2

Section: The Global Structure of Chromosomes

Feedback: The fractal globule model is permissive to maximal packing and, at the same time, allows the chromatin to fold and unfold easily. In this model, compared to a “random coil” arrangement, the spatial distance between two neighboring regions of the DNA is shorter on average.

  1. Answer: C

Difficulty: 1

Section: The Global Structure of Chromosomes

Feedback: Nuclear subcompartments such as nucleoli, interchromatin granule clusters (or speckles), and Cajal bodies carry out specialized functions using distinct sets of macromolecules organized in gel-like structures. These structures are highly permeable and highly selective at the same time. They are formed only when needed and are not constitutively present.

  1. Answer: A

Difficulty: 1

Section: The Global Structure of Chromosomes

Feedback: Each nuclear subcompartment is a non-membrane-bound network of a specific set of RNA and protein molecules linked via noncovalent interactions to produce a distinct biochemical environment, despite their high permeability. They come in various sizes.

  1. Answer: FTFFT

Difficulty: 2

Section: The Global Structure of Chromosomes

Feedback: The cells achieve an extra 10-fold compaction ratio of their DNA by condensing the chromatin into mitotic chromosomes that can be seen using a light microscope. This is done after the DNA replication in S phase, and ends in the complete disentanglement of the sister chromatids. A mammalian mitotic chromatid is normally between 0.5 and 1 µm in diameter. A 70-nm-thick chromosome would be only about twice as wide as a 30-nm fiber.

  1. Answer: A

Difficulty: 2

Section: How Genomes Evolve

Feedback: Transposition by various “parasitic” mobile DNA elements has changed the mammalian genome so profoundly that almost half of our genome is composed of recognizable products of transposition. In addition, chromosomal duplications can be facilitated by transposable elements.

  1. Answer: purifying, third

Difficulty: 1

Section: How Genomes Evolve

Feedback: Human and chimpanzee genomes are nearly identical, even in the third position of synonymous codons, where there is no functional constraint on the nucleotide sequence. This reflects the short time available for mutations to accumulate in the two closely related lineages. In contrast, the sequence conservation found between the genes of humans and chickens is almost entirely due to purifying selection.

  1. Answer: BDC

Difficulty: 2

Section: How Genomes Evolve

Feedback: The lowest pairwise distance in the matrix (0.40) is between species A and B. Species A and B are each separated from their common ancestor by a distance of 0.20 (= 0.40 / 2). Next, the average distance of species A and B from species D is only 1.40 (= [1.36 + 1.44] / 2), which is lower than the average distance from species C—that is, 3.7 (= [3.61 + 3.79] / 2)—and so on. Based on these distances and the known phylogeny of the hominid family, one would expect species B to be chimpanzee, species C to be orangutan, and species D to be gorilla.

  1. Answer: BDC

Difficulty: 2

Section: How Genomes Evolve

Feedback: Based on the distances and the known phylogeny of the hominid family, one would expect species B (1) to be chimpanzee, species C (3) to be orangutan, and species D (2) to be gorilla.

  1. Answer: bcfedag

Difficulty: 4

Section: How Genomes Evolve

Feedback: A diagonal line in this plot indicates a continuous region of synteny between the two chromosomes. Because of a genomic region that is absent (e.g. deleted) from one end of the chromosome of species A, the long diagonal line is shifted up (represented by a in the plot). A symmetrical set of parallel lines (as at b) on the two sides of a diagonal reflects the existence of a repeat in both chromosomes, whereas the parallel line on one side of the diagonal at c indicates duplication only in species A. A chromosomal inversion shows up as a diagonal with a negative slope, which may or may not be accompanied by translocation of the inverted region (as at d and e, respectively). Two tandem duplications in species B may appear as a triplication (as at f) represented by three parallel lines. Finally, g represents a region at the end of the chromosome in species A that is absent from the chromosome of species B.

  1. Answer: 2122

Difficulty: 3

Section: How Genomes Evolve

Feedback: The molecular clock runs faster for sequences that are subject to less purifying selection.

  1. Answer: E

Difficulty: 2

Section: How Genomes Evolve

Feedback: The balance between the rates of DNA addition and DNA loss has been biased toward a net DNA loss in the Fugu lineage, resulting in a “cleansing” of functionally unnecessary sequences including most transposable elements and large parts of the introns. The exons, which are under purifying selection, were not affected as significantly.

  1. Answer: B

Difficulty: 3

Section: How Genomes Evolve

Feedback: Since the correlation of gene copy number and expression level has been established for this gene, it is reasonable to expect the histogram for the population with higher dietary starch to be shifted toward higher gene copy numbers.

  1. Answer: FFFF

Difficulty: 2

Section: How Genomes Evolve

Feedback: By comparing genomic DNA sequences from various existing mammals, as well as non-mammals, much can be inferred about the genome of the last common ancestor that lived about 100 million years ago. Although some of the ultraconserved elements may encode long noncoding RNAs, the function of the majority of these elements is still a mystery. Since even a tiny selective advantage can support the conservation of a particular DNA sequence, lack of an “obvious” defect in a laboratory strain of mouse does not preclude functional importance. The “human accelerated regions” show accelerated genetic changes during recent evolution in the human lineage, but are not entirely novel genetic elements; in fact, they had been highly conserved before the sudden spurt of nucleotide changes.

  1. Answer: C

Difficulty: 2

Section: How Genomes Evolve

Feedback: It appears that a chromosome translocation event happened about 300 million years ago in the lineage leading to mammals and birds after it diverged from the amphibian lineage.

  1. Answer: C

Difficulty: 3

Section: How Genomes Evolve

Feedback: Genes with very high h and k values have shown an accelerated rate of change in both human and chimpanzee lineages compared to the mouse lineage. Even though such genes are interesting candidates to study general hominid evolution, the genes that have evolved uniquely in humans are probably those with high h but low k values. Note that in the normalized substitution rates, a and b are divided by 5.5 because each of them represents amino acid changes during 5.5 million years since the divergence of human and chimpanzee lineages. The parameter c represents the changes that occurred in the mouse lineage since its divergence from the primate lineage (90 million years ago), as well as changes that occurred in the primate lineage before human and chimpanzee lineages diverged (between 90 million and 5.5 million years ago). Therefore, c is normalized by dividing by the combined period (90 + 90 – 5.5).

  1. Answer: B

Difficulty: 1

Section: How Genomes Evolve

Feedback: Multispecies conservation of a genomic sequence is a strong indication of its functional importance.

  1. Answer: 6

Difficulty: 3

Section: How Genomes Evolve

Feedback: Duplication of DNA segments has been a widespread evolutionary event that can happen over long or short blocks of the genome. It can exponentially expand the block to create tandem repeated arrays. For example, in an ideal case, only five duplications are needed to create 32 (that is, 25) repeats of an original DNA segment that encodes a protein domain. One more is needed to make the total 33 repeats. In reality, however, more than six duplication events might have been involved.

  1. Answer: FTTT

Difficulty: 2

Section: How Genomes Evolve

Feedback: A number of variations—such as scattered rare mutations and abnormalities, single-nucleotide polymorphisms, simple tandem repeat expansions [such as those of the (CA)n repeats], and copy number variations—makes each two human individuals of the same sex different in, at least, about 1 in every 1000 nucleotides of their genomes; therefore, 99.99% identity is an overestimation. Nearly half of the copy number variations contain genes. For a variation at a nucleotide position to be considered a single-nucleotide polymorphism, it must be common enough such that the probability that two individuals have a different nucleotide at this position is at least 1%. It is estimated that the current pattern of human genetic variation was mostly in place when our ancestral population size was about 10,000.

  1. Answer: 10–6, 10–3

Difficulty: 3

Section: How Genomes Evolve

Feedback: The probability that the mutation becomes fixed in each isolated population of 500 individuals is approximately 1/(2 × 500) = 10–3. The combined probability for fixation in two independent populations is thus equal to 10–3 × 10–3 = 10–6. If the two populations fully interbreed, they can be considered as a single population of 1000 people with two mutations. The fixation probability is therefore calculated to be 2/(2 × 1000) = 10–3. Hence, it is more likely for the same rare neutral variations to become fixed in the entire population in these communities if they highly interbreed.

 

 

 

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