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Biochemistry 403 > Exam 3 > Flashcards

Exam 3 Flashcards

1
Q

2009

In the structure below the orientation at which C atom would determine if this is a D or L sugar

  • 1
  • 2
  • 3
  • 4
  • 5
A

4

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2
Q

2009

When two carbohydrates are epimers

  • They differ in length by one carbon
  • They rotate plane-polarized light in the same direction
  • One is an aldose, the other a ketose
  • One is a pyranose, the other a furanose
  • They differ only in the configuration around one carbon atom
A

They differ only in the configuration around one carbon atom

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3
Q

2009

From the abbreviated name of hte compound Gal(B1→4)Glc, we know that

  • the galactose residue is at the reducing end
  • the glucose residue is the B anomer
  • the glucose is in its pyranose form
  • the compound is dextrorotatory
  • the compound is lactose
A

the compound is lactose

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4
Q

2009

In glycoproteins, the carbohydrate moiety is always attached through the amino acid residue

  • glutamine or arginine
  • aspartate or glutamate
  • asparagine, serine, or threonine
  • glycine, alanine, or aspartate
  • tryptophan, aspartate or cysteine
A

asparagine, serine, or threonine

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5
Q

2009

Which is found in glocogen?

  • A
  • B
  • C
  • D
  • E
A

E

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6
Q

2009

Which is a ketose?

  • A
  • B
  • C
  • D
  • E
A

D

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7
Q

2009

Which is an L-sugar?

  • A
  • B
  • C
  • D
  • E
A

C

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8
Q

2009

Which is a pair of epimers?

  • A & B
  • B & C
  • C & D
  • D & E
  • C & A
A

B

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9
Q

2009

Which of the following statements about starch and glycogen is false?

  • Both serve primarily as structural elements in cell walls
  • Both starch and glycogen are stored intracellularly as insoluble granules
  • Amylose is unbranched; amylopectin and glycogen contain many (α1→6) branches
  • Both are homopolymers of glucose
  • Starch is found in plants, glycogen is found in animals
A

Both serve primarily as structural elements in cell walls

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10
Q

2009

If glycogen is partially hydrolysed and all possible disaccharides are isolated, how many would be found?

  • 1
  • 2
  • 3
  • 4
  • 5
A

2

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11
Q

2009

Neuraminic acid is a 9 carbon atom keose (i.e. nonose). The total number D diastereomers of neuraminic acid (ignore anomers is)

  • 2
  • 4
  • i
  • 16
  • 32
A

16

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12
Q

2009

Neuraminic acid is formed from 2-acetamido-D and the metabolite phosphoennol pyruvate. How many 9C atom sugars can be formed?

  • 1
  • 2
  • 4
  • 8
  • 16
A

2

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13
Q

2009

An unknown oligosaccharide is reduced at its reducing end with an appropriate reducing agent. It was found that 504 mg of the oligosaccharide consumed the equivalent of 1 millimole of hydrogen. The molecular weight of oligosaccharide is

  • 126
  • 252
  • 378
  • 504
  • 1008
A

504

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14
Q

2009

An unknown oligosaccharide is reduced at its reducing end with an appropriate reducing agent. It was found that 504 mg of the oligosaccharide consumed the equivalent of 1 millimole of hydrogen. This oligosaccaride most like is.

  • an aldo- or ketohexose
  • an aldo- or ketodecanose (deca=10)
  • a reducing disaccharide consisting of 2 aldohexoses
  • a non-reducing disaccharide consisting of 2 aldohexoses
  • a reducing trisaccharide consisting of 3 hexoses
A

a reducing trisaccharide consisting of 3 hexoses

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15
Q

2009

Ther term amphipathic means

  • branched, with at least two branch points
  • having one region that is positively charged and one region that is negatively charged
  • different on the inside and outside of the lipid bilayer
  • having one region that is polar and one that is nonpolar
  • having two different types of bonds
A

having one region that is polar and one that is nonpolar

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16
Q

2009

The polar head group of cholesterol is

  • the alkyl side chain
  • glycerol
  • the steroid nucleus
  • the hydroxyl group
  • choline
A

the hydroxyl group

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17
Q

2009

Required biological membranes are “self-sealing” due to all of the following except:

  • The amphipathic character of the lipids
  • hydrophobic interactions between lipids
  • hydrogen bonding between the head groups of the lipids and H2O
  • an increase in entropy of the system upon sealing
  • covalent interactions among lipids
A

covalent interactions among lipids

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18
Q

2009

A cell has available deoxyribose molecules, inorganic phosphates and the standard purine and pyrimidine molecules with nothing attached to them. The cell makes a deoxytetranucleotide with a hydroxyl at one end and a phosphate at the other. How many water molecules does it remove to make the deoxytetranucleotide?

  • none
  • four
  • eight
  • nine
  • eleven
A

eleven

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19
Q

2009

A cell has available deoxyribose molecules, inorganic phosphates and the standard purine and pyrimidine molecules with nothing attached to them. The cell makes a deoxytetranucleotide with a hydroxyl at one end and a phosphate at the other.

If the cell had ribose molecules instead of deoxyribose available, how many water molecules would it have had to remove to make the tetranucleotide with a hydroxyl at one end and a phosphate at the other?

  • none
  • four
  • eight
  • nine
  • eleven
A

eleven

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20
Q

2009

Frederick Griffith injected mice with various strains of pneumococcus. Which of the following formulations would have caused the mice to die?

  • live nonencapsulated nonvirulant bacteria
  • live encapsulated virulent bacteria which are heat killed prior to injection
  • a mixture of heat-killed nonencapsulated nonvirulent bacteria and heat-killed encapsulated virulent bacteria
  • a mixture of heat-killed encapsulated virulent bacteria and live nonencapsulated nonvirulent bacteria
  • none of the above
A

a mixture of heat-killed encapsulated virulent bacteria and live nonencapsulated nonvirulent bacteria

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21
Q

2009

Which of the following would be a violation of Chargaff’s rules with respect to the base composition of double stranded DNA?

  • A = T
  • C = G
  • T + C = A + G
  • A + T = G + C
  • all of the above follow Chargaff’s rules
A

A + T = G + C

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22
Q

2009

Which of the following is not true of the following DNA sequence?

5’ … TGCGATACTCATCGCA … 3’

  • it could form a hairpin
  • it matched with its complementary strand, it could form a cruciform structure
  • if matched with its complementary strand, it could be palindromic except for four bases
  • if matched with its complementary strand, it would be a mirror repeat
  • all of the above are true
A

if matched with its complementary strand, it would be a mirror repeat

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23
Q

2009

Hoogsteen base pairing can lead to formation of

  • duplex DNA
  • triplex DNA
  • tetraplex DNA
  • hairpins
  • hatpins
A

triplex DNA

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24
Q

2009

The melting temperature, Tm of DNA is found to follow the expression

Tm = 81.5° + 0.41°(%GC) - 675°/(length of DNA in base pairs)

An oligonucleotide 15 basepairs in length is found to have a melting temperature of 65.2°. What is its GC content?

  • 80%
  • 70%
  • 60%
  • 50%
  • 40%
A

70%

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25
Q

2009

Deamination of cystone occurs spontaneously and is uusally detected by the DNA repair system, which restores the cytosine. Occassionally the lesion in not detected. What would be the most likely effect in this case?

  • A GC base pair would be replaced by an AT base pair at the next cell division
  • An AT base pair would be replaced by a GC base pair at the next cell division
  • The resulting base would be deleted form the DNA
  • The cell would be unable to replicate its DNA
  • none of the above
A

A GC base pair would be replaced by an AT base pair at the next cell division

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26
Q

2009

Nicotinamide adenine dinucleotide and flavin adenine dinucleotide are adenosine-containing co-factors used in many enzymatic reactions. Which of the following best describes their role?

  • They assist the cell in proeolytic hydrolysis of damaged proteins
  • They inhibit enzymatic attack on the cells lining the intestinal tract
  • They serve as oxidizing or reducing agents in redox reactions
  • They assis in the folding of certain proteins
  • None of the above
A

They serve as oxidizing or reducing agents in redox reactions

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27
Q

2009

In the Hersey-Chase experiment the finding of 32P in the cells infected with 32P labeled bacteriophage but no 35S in cells infected with 35S labeled bacteriophage demonstrated that

  • the protein coat of the phage damaged the cells, letting the 32P through the membrane
  • the vital genome was carried by the viral DNA and not by the coat
  • the viral genome was single stranded
  • the cells can extract 32P from the medium but not 35S
  • none of the above
A

the vital genome was carried by the viral DNA and not by the coat

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28
Q

2009

What is the correct order of the following steps in the solid state synthesis of oligonucleotides?

  1. reaction of a nucleotide which is activated at the 3’ position and blocked at 5’ and at the base
  2. deblocking of the protected 5’ position
  3. attachment of a nucleotide blocked at the 5’ position and at the base to the solid support
  4. oxidation at the phosphorous atom by iodine
  5. cleavage of the chain from the support
  6. removal of blocking groups from the bases
  7. removal of methyl groups from the phosphates
  • 3, (2, 1, 4) repeatedly 6, 7, 5
  • 1, 2, (3, 4, 5) repeatedly 6, 7
  • 3, (4, 1, 2) repeatedly 6, 7, 5
  • 1, 3, (2, 4, 7) repeatedly 6, 5
  • none of the above
A

3, (2, 1, 4) repeatedly 6, 7, 5

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29
Q

Which of the following contributes to the packing of DNA into cells?

  • supercoiling
  • topoisomerase
  • DNA binding proteins
  • palindromic sequences
  • all of these
A

all of these

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30
Q

2011

The metabolic role of an allosteric stimulator is to

  • prevent an enzyme from being active when the stimulator is bound to the protein.
  • shift the substrate binding curve to the right on a Vo versus [S] plot.
  • help ensure that the enzyme is active throughout the normal range of substrate concentration.
  • prevent the cell from accessing abnormally high or low substrate concentrations.
  • none of the above
A

help ensure that the enzyme is active throughout the normal range of substrate concentration.

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31
Q

2011

Which of the following events in the binding of a substrate to an enzyme active site constitutes a decrease in entropy if all other factors are ignored?

  • the formation of a single ES from an E and an S
  • an increase in enzyme and substrate rigidity
  • the expulsion of water of hydration from the molecular surfaces
  • more than one of the above
  • none of the above
A

more than one of the above

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32
Q

2011

The half-time of a first order reaction is the time required for

  • 0.693 (69.3%) of the reactant to be consumed.
  • half the amount of reactant present to be consumed.
  • football teams to rest between halves of the game.
  • the time for the amount of product to decrease by half.
  • more than one of the above.
A

half the amount of reactant present to be consumed

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33
Q

2011

Which of the following DNA sequences, if hydrogen bonded to its complementary sequence in WatsonCrick base pairs, would have the highest melting temperature Tm?

  • ACGCACGC
  • ATACATAC
  • ATGCATGC
  • ATTAATTA
  • All these will melt at the same Tm
A

ACGCACGC

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34
Q

2011

The metabolic role of an allosteric inhibitor is to

  • A. ensure that the enzyme is unable to bind stimulators.
  • B. shift the substrate binding curve to the right on a Vo versus [S] plot.
  • C. help ensure that the enzyme is active throughout the normal range of substrate concentration.
  • D. prevent the cell from accessing abnormally high or low substrate concentrations.
  • E. none of the above
A

B. shift the substrate binding curve to the right on a Vo versus [S] plot.

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35
Q

2011

Thymine and uracil differ in that

  • A. at the 5 position of the pyrimidine ring thymine has a methyl group which uracil lacks.
  • B. at the 5 position of the pyrimidine ring uracil has a methyl group which thymine lacks.
  • C. uracil has two keto groups, whereas thymine only has one.
  • D. in the 6 position thymine has an amino group which uracil lacks.
  • E. none of the above
A

A. at the 5 position of the pyrimidine ring thymine has a methyl group which uracil lacks.

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36
Q

2011

Frederick Griffith showed that

  • A. heat-killed virulent pneumococcus bacteria, if incubated with heat killed non-virulent pneumococci and injected into mice, caused the mice to die of pneumonia.
  • B. heat-killed non-virulent pneumocci, if incubated with live virulent pneumococci and injected into mice, had no effect on the mice.
  • C. heat-killed virulent pneumococci, if incubated with live non-virulent pneumococci and injected into mice, caused the mice to die of pneumonia.
  • D. non-virulent pneumococci, if injected into mice, mutated to the virulent form and caused the mice to die of pneumonia.
  • E. more than one of the above
A

C. heat-killed virulent pneumococci, if incubated with live non-virulent pneumococci and injected into mice, caused the mice to die of pneumonia.

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37
Q

2011

Avery, MacLeod and McCarthy incubated DNA from virulent pneumococci with live non-virulent pneumococci and injected the culture into mice, which died of pneumonia. Their conclusion that the DNA carried the genetic information for virulence was criticized because their DNA preparation might have contained trace amounts of protein, which could have carried the information. How did they show this was not the case?

  • A. Their DNA showed only one band on a gel.
  • B. Treatment of the preparation with proteases destroyed the transforming activity of the preparation but treatment with nucleases had no effect.
  • C. Injection of pneumococcal proteins into the mice had no effect on them.
  • D. Nuclease treatment of the preparation destroyed the transforming activity, but protease treatment did not destroy it.
  • E. none of the above
A

D. Nuclease treatment of the preparation destroyed the transforming activity, but protease treatment did not destroy it.

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38
Q

2011

In the Hershey-Chase experiment bacteriophage was grown in a medium containing radioactive 32P, while another phage sample was grown in a medium containing radioactive 35S. Each of these phage preparations was then used to infect a bacterial culture, and soon after infection the cultures were processed in a blender to remove the viral heads, which were separated from the bacteria by centrifugation. Which of the following possible results were consistent with DNA acting as the viral genetic material rather than protein?

  • A. The cells infected with 35S-containing phage were radioactive, and the viral progeny were also radioactive.
  • B. The viral heads from the 35S-containing phage were radioactive, and the viral progeny were also radioactive.
  • C. The cells from the 32P-containing phage were radioactive, as were the viral progeny, but the heads from this culture were not radioactive.
  • D. Both the cells and the viral heads from the 32P-containing phage were radioactive.
  • E. None of these
A

C. The cells from the 32P-containing phage were radioactive, as were the viral progeny, but the heads from this culture were not radioactive.

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39
Q

2011

  1. Some viruses appear not to follow Chargaff’s rule that A = T and G = C. The most likely explanation of this finding is
  • A. the data are wrong owing to poor measurements.
  • B. the viruses do not cause disease.
  • C. the viruses infect bacteria but not higher organisms.
  • D. the viral genetic material is single stranded.
  • E. more then one of the above.
A

D. the viral genetic material is single stranded.

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40
Q

2011

Which of the following DNA sequences, if paired with its complemtary sequence is NOT a palindrome?

  • A. TTAGCACGTGCTAA
  • B. TTAGCACCACGATT
  • C. AATCGTGCACGATT
  • D. All these are palindronic palindromic
  • E. None of these is
A

B. TTAGCACCACGATT

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41
Q

2011

DNA concentrations are usually determined from the ultraviolet absorbance at 260 nm. Why would the presence of protein in the preparation lead to an erroneous result for the DNA concentration?

  • A. Although proteins absorb maximally near 280 nm, they have significant absorbance at 260 nm, which would add to the total absorbance of the sample..
  • B. The presence of proteins reduces the ability of the DNA to absorb UV light.
  • C. The proteins cause the DNA to absorb more light.
  • D. The proteins block the UV light.
  • E. Proteins have no effect on the absorbance of the sample and thus introduce no error at all.
A

A. Although proteins absorb maximally near 280 nm, they have significant absorbance at 260 nm, which would add to the total absorbance of the sample..

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42
Q

2011

Deamination of cytosine yields uracil. If the resulting double stranded DNA is replicated with standard Watson-Crick base pairing, the original cytosine will be replaced by

  • A. thymine
  • B. guanine
  • C. uracil
  • D. adenine at this location.
  • E. none of these; there will be no change
A

thymine

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43
Q

2011

The compaction of DNA into chromosomes in eucaryotes requires

  • A. histone proteins
  • B. scaffolding proteins
  • C. cleavage of the DNA into small fragments
  • D. RNA which is complementary to the sequence of one of the DNA strands
  • E. more than one of the above
A

E. more than one of the above

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44
Q

2011

Which of the following is NOT involved in compaction of DNA and/or maintenance of chromosome structure?

  • A. formation of DNA-containing loops
  • B. SMC proteins
  • C. condensins
  • D. cohesions
  • E. al the above are involved.
A

E. al the above are involved.

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45
Q

2011

Competitive inhibition of a Michaelis-Menten enzyme would lead to

  • A. Lineweaver-Burk plots which intersect on the 1/[S] axis.
  • B. Lineweaver-Burk plots which intersect on the 1/vo axis.
  • C. Lineweaver-Burk plots which intersect show curvature like vo versus [S] plots.
  • D. Lineweaver-Burk plots which don’t intersect at all.
  • E. none of the above
A

B. Lineweaver-Burk plots which intersect on the 1/vo axis.

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46
Q

2011

DNA from two species is cut into small fragments by nucleases. The samples are then run out on a gel, which is probed (treated) with short radioactive DNA molecules. The banding patterns are seen to be similar. One may conclude that

  • A. the species are closely related in evolution.
  • B. the species are distantly related in evolution.
  • C. the species are evolutionarily unrelated.
  • D. one can draw no conclusion at all, as the experiment cannot detect evolutionary relationships.
  • E. none of these
A

A. the species are closely related in evolution

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47
Q

2011

Cellulose is virtually insoluble in water, whereas amylose is relatively soluble. Explain this disparity in solubility.

  • A. Water cannot gain access to the cellulose.
  • B. Water cannot gain access to the amylose as it is too curled up.
  • C. Amylose consists of α linked glucoses permitting extensive hydrogen bonding between the surrounding water and -OH groups of the glucose resides
  • D. Cellulose consists of β linked glucose units forming rod-like molecules which H bonds formed between -OH of glucose units in the different chains
  • E. C and D
A

E. C and D

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48
Q

2011

From the abbreviated name of the compound Gal(β1→4)Glc, we know that:

  • A. the glucose residue is the β anomer
  • B. the galactose residue is at the reducing end
  • C. C-4 of glucose is joined to C-1 of galactose by a glycosidic bond.
  • D. the compound Is dextrorotatory.
  • E. the glucose in is its furanose form.
A

C. C-4 of glucose is joined to C-1 of galactose by a glycosidic bond.

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49
Q

2011

In glycoproteins the carbohydrate portion is always attached through the amino acid residues:

  • A. Tryptophan, aspartate, or cysteine.
  • B. asparagine, serine, or threonine.
  • C. glycine, alanine, or aspartate.
  • D. aspartate or glutamate
  • E. glutamine or arginine.
A

B. asparagine, serine, or threonine.

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50
Q

2011

When the linear form of glucose cyclizes, the product is a(n):

  • A. glycoside.
  • B. hemiacetal.
  • C. anhydride.
  • D. lactone.
  • E. oligosaccharide.
A

B. hemiacetal

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51
Q

2011

The biochemical property of lectins that is the basis for most of their biological effects is their ability to bind to:

  • A. specific oligosaccharides.
  • B. specific peptides
  • C. specific lipids
  • D. amphipathic molecules
  • E. hydrophobic molecules.
A

A. specific oligosaccharides

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52
Q

2011

Assume that a hypothetical oxidising agent oxidises a reducing sugar in a 1:1stoichiometry. An unknown sugar (342 mg) consumes 12.5ml of 0.08 M solution of this oxidising agent. The molecular weight of this sugar is

  • A. 180
  • B. 150
  • C. 120
  • D. 342
  • E. 171
A

D. 342

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53
Q

2011

Oxidation of glucose by glucose oxidase is specific and proceeds by the below reaction. The H2O2 concentration can be measured photometrically. If 100 µL of blood on analysis gave 5.00 µmoles of H2O2, the concentration of glucose in blood is

  • A. 0.5 molar
  • B. 0.5 grams/100ml
  • C. 0.05 molar
  • D 0.005 molar
  • E. 0.05 millimolar
A

C. 0.05 molar

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54
Q

2011

What is the maximum number of D-stereoisomers for hexo-uronic acids (ignore anomers)?

  • A. 2
  • B. 4
  • C. 8
  • D. 16
  • E. 1
A

C. 8

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55
Q

2011

A newly isolated D-aldohexose is reduced at C-1 to give the corresponding glycitol(s), which has (have) an internal plane of symmetry. What is (are) the structure(s) of the original hexose?

  • A. 1
  • B. 2
  • C. 3
  • D. 4
  • E. All 1-5
A

C. 3

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56
Q

2011

Treatment of a mutarotated mixture of D-fructose by methylation gave methyl β- 1, 3, 4, 6-tetra-O-methyl D-fructose as the main product. Therefore this D-fructose contains a high proportion of

  • A. α-pyranose
  • B. β-pyranose
  • C. α-furanose
  • D. β-furanose
  • E. Both B and C
A

D. β-furanose

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57
Q

2011

The 5-epimer of D-glucuronate (GlcUA) is

  • A. L-GlcUA
  • B. D-GalUA
  • C. L-IdoUA
  • D. D-ManUA
  • E. more than one of the above
A

C. L-IdoUA

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58
Q

2011

β galactosidase hydrolyzes lactose. An unknown trisaccharide is converted by β galactosidase to maltose and galactose. The structure of the trisaccharide could be:

  • A. Gal 1 4 glc 1 4 glc
  • B. Gal 1 6 glc 1 4 glc
  • C. Gal 1 3 glc 1 4 glc
  • D. Gal 1 2 glc 1 4 glc
  • E. All of the above.
A

E. All of the above

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59
Q

2014

Consider the DNA denaturation curves in the figure below. Which of the two samples has the higher pertcentage of AT base pairs? (From the posted exam)

  • A. the one to the right
  • B. the one to the left
  • C. Their AT content is the same.
  • D. Neither sample has a significant number of AT base pairs, so one cannot distinguish them
  • E. Experiments of this kind have nothing to do with AT content.
A

B. the one to the left

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60
Q

2014

Which of the following is an anomeric pair?

  • A. α-D-glucose and β-L-glucose
  • B. α-D-glucose and β-D-glucose
  • C. D-glucose and D-fructose
  • D. D-glucose and L-glucose
  • E. D-glucose and L-fructose
A

B. α-D-glucose and β-D-glucose

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61
Q

2014

Glycogen is the storage form of glucose in mammalian species. It is a branched polymer containing only glucose residues. If glycogen is partially hydrolysed, how many different types of disaccharide and trisaccharide can be formed?

  • A. 2, 2
  • B. 3, 3
  • C. 2, 4
  • D. 3, 3
  • E. 2, 3

How many different methylated glucoses are obtained when glycogen is completely methylated followed by hydrolysis?

  • A. 1
  • B. 2
  • C. 3
  • D. 4
  • E. more than 4
A
  • C. 2, 4
  • C. 3
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62
Q

2014

When a bacterium such as E. coli is shifted from a warmer growth temperature to a cooler growth temperature, it compensates by:

  • A. increasing its metabolic rate to generate more heat.
  • B. putting longer-chain fatty acids into its membranes.
  • C. putting more unsaturated fatty acids into its membranes.
  • D. shifting from aerobic to anaerobic metabolism.
  • E. synthesizing thicker membranes to insulate the cell.
A

C. putting more unsaturated fatty acids into its membranes.

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63
Q

2014

The type of membrane transport that uses ion gradients as the energy source is:

  • A. facilitated diffusion.
  • B. passive transport.
  • C. secondary active transport.
  • D. primary active transport.
  • E. simple diffusion.
A

C. secondary active transport

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64
Q

2014

In the study of the rate of membrane lipid diffusion (using fluorescent recovery after laser photobleaching) at 37°C was compared with the rate at 10°C. At 10°C you would expect the rate of:

  • A. Lateral diffusion to be faster.
  • B. Lateral diffusion to be slower.
  • C. Lateral diffusion to be the same.
  • D. Translayer movement (flip-flop) to increase.
  • E. None of the above.
A

B. Lateral diffusion to be slower.

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65
Q

2014

Consider the lower fatty acid in the phosphatidyl choline shown below. Which of the following is a correct description of its structure?

  • A. 16:2Δ9,12.
  • B. 18:2 ω6,ω9
  • C. 18:2 Δ9,12
  • D. 18:2 Δ6,9
  • E. B and C
A

B and C

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66
Q

2014

How many epimeric forms of an aldohexose can exist?

  • A. two
  • B. four
  • C. eight
  • D. sixteen
  • E. none of these.
A

D. sixteen

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67
Q

2014

Which of the following contains an ether-linked alkyl group?

  • A. Cerebrosides
  • B. Gangliosides
  • C. Phosphatidyl serine
  • D. sphingomyelin
  • E. Archae membrane lipids
A

E. Archae membrane lipids

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68
Q

2014

Which of the following is not a reducing sugar?

  • A. Ribose
  • B. glucose
  • C. fructose
  • D. glyceraldehyde
  • E. sucrose
A

E. sucrose

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69
Q

2014

β-D-galactose is likely to be less stable than β-D-glucose because

  • A. β-D-galactose cannot assume the chair form.
  • B. it has an axial hydroxyl in the chair form, whereas β-D-glucose does not.
  • C. all its hydroxyls as well as the sixth carbon are equatorial.
  • D. none of its non-hydrogen substituents on the pyranose ring are equatorial.
  • E. none of the above
A

B. it has an axial hydroxyl in the chair form, whereas β-D-glucose does not.

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70
Q

2014

Why are fish oils recommended as dietary supplements for people with cardiovascular disease?

  • A. They are rich in eicosapentaenoic acid (EPA), which is 20:4(Δ5,8,11,14,17)
  • B. They are rich in stearic acid.
  • C. They are rich in docosahexaenoicacid (DHA), which is 22:6(Δ4,7,10,13,16,19).
  • D. A and C
  • E. None of the above. There is something fishy about this question
A

D. A and C

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71
Q

2014

If all the single bonds of an unsaturated fatty acyl chain in a phospholipid, a fatty acid such as palmitoleic acid, 16:1(Δ9 ), or linoleic acid, 18:2(Δ9,12), were free to rotate within a membrane, which would sweep out a larger volume of space in the course of the free rotations?

  • A. the forms in which all the double bonds are cis
  • B. the forms in which some double bonds are cis and others trans
  • C. the forms in which all the double bonds are trans
  • D. all would sweep out equal volumes of space
  • E. cannot tell from data given.
A

A. the forms in which all the double bonds are cis

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72
Q

2014

Arachidonic acid has the formal representation of 20:4(Δ5,8,11,14 ). If the carboxyl group has no proton on it, how many atoms of hydrogen does the molecule contain?

  • A. twenty
  • B. four
  • C. eight
  • D. thirty-one
  • E. none of these
A

D. thirty-one

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73
Q

2014

How many polar atoms are there in a diacyl glycerol phosphate which lacks a head group attached to the phosphate. Include the phosphorous but not any hydrogens in your count.

  • A. three
  • B. six
  • C. nine
  • D. more than nine
  • E. none of these
A

C. nine

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74
Q

2014

In the stabilization of chromosome structure in the course of intedacting the DNA with histone cores, the role of topoisomerase is to

  • A. wind the DNA around each core histone particle.
  • B. introduce new negative supercoils as each nucleosome is formed.
  • C. attach the DNA to the chromosomal scaffolding proteins.
  • D. relax positive supercoils in the part of the DNA which has not yet bound histones.
  • E. none of the above
A

D. relax positive supercoils in the part of the DNA which has not yet bound histones.

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75
Q

2015

Which of the following is maltose?

  • A. Fru(2β→α1)Glc
  • B. NAcGal(β1→4)NAcGlu
  • C. Glc(α1→1α)Glc
  • D. Glc(α1→4)Glc
  • E. none one of these
A

D. Glc(α1→4)Glc

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76
Q

2015

The type of membrane transport that uses ion gradients as the energy source is:

  • A. facilitated diffusion.
  • B. passive transport.
  • C. secondary active transport.
  • D. primary active transport.
  • E. simple diffusion.
A
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77
Q

2015

You are studying the uptake of L-leucine by epithelial cells of the mouse intestine. Measurements of the rate of uptake of L-leucine and several of its analogs, with and without Na+ in the assay buffer, yield the results given in the table. Kt (analogous to Km) for D-Leu is ~20x higher than for the L-Leu but similar for L-Leu and L-Val. This is because:

  • A. The binding site recognizes the difference in configuration at the α carbon of the L and D Leu.
  • B. Vmax is similar for L and D-Leu therefore the transporter does not discriminate.
  • C. The structure of L-Leu and L-Val are quite similar.
  • D. B and C
  • E. A and C
A

E. A and C

78
Q

2015

Which of the following is a reducing sugar?

  • A. Fru(2β→α1)Glc
  • B. NAcGal(β1→4)NAcGlu
  • C. Glc(α1→1α)Glc
  • D. Glc(α1→4)Glc
  • E. more than one of these
A

E. more than one of these

79
Q

2015

Primary active transport of solutes across a membrane

  • A. is active because the concentration gradient favors transport.
  • B. requires concurrent flow of another solute down its concentration gradient.
  • C. in mediated by ionophores.
  • D. is driven by ATP hydrolysis.
  • E. more than lone of these.
A

D. is driven by ATP hydrolysis.

80
Q

2015

Consider the DNA denaturation curves in the figure below. Which of the two samples has the lower pertcentage of AT base pairs?

  • B. the one to the left
  • C. Their AT content is the same.
  • D. Neither sample has a significant number of AT base pairs, so one cannot distinguish them
  • E. Experiments of this kind have nothing to do with AT content.
A

A. the one to the right

81
Q

2015

Why do some membrane proteins not diffuse laterally within the bilayer membrane?

  • A. They are too massive to move.
  • B. They are attached to cytoskeletal proteins such as spectrin
  • C. The ATP cost of moving them is too great to allow much movement
  • D. They are part of membrane rafts.
  • E. none of the above.
A

B. They are attached to cytoskeletal proteins such as spectrin

82
Q

2015

The phi/psi plot of carbohydrates

  • A. illustrates polypeptide conformations not prevented by steric hindrance.
  • B. illustrates the energies of various polysaccharide conformations.
  • C. shows the energies of axial versus equatorial conformations.
  • D. explains why some carbohydrates form helical structures
  • E. none of the above
A

B. illustrates the energies of various polysaccharide conformations.

83
Q

2015

SNAPs and SNAREs can be involved in

  • A. catching small animals.
  • B. aiding vesicles within cells to fuse with the plasma membrane, thus enabling secretion of compounds from the cell.
  • C. anchoring proteins within membranes to limit their lateral diffusion.
  • D. contributing to primary active transport.
  • E. none of these.
A

B. aiding vesicles within cells to fuse with the plasma membrane, thus enabling secretion of compounds from the cell.

84
Q

2015

What is the difference between an enantiomer and an epimer?

  • A. Epimers are complete mirror images of one another, while enantiomers differ in configuration at only one carbon.
  • B. Enantiomers are complete mirror images of one another, while epimers differ in configuration at only one carbon.
  • C. They do not differ; the words are alternate names for the same thing.
  • D. Enantiomers have only three carbons, while epimers may have three or more.
  • E. One of them is an aldose, while the other is a ketose. (from a posted exam)
A

B. Enantiomers are complete mirror images of one another, while epimers differ in configuration at only one carbon.

85
Q

2015

Secondary active transport

  • A. uses ATP hydrolysis to drive transport of one solute, which then flows back down its concentration gradient, helping a second solute to cross the membrane.
  • B. occurs after primary active transport capacity has been exhausted.
  • C. requires the use of an ionophore.
  • D. requires a hydrogen ion gradient.
  • E. more than one of the above.
A

A. uses ATP hydrolysis to drive transport of one solute, which then flows back down its concentration gradient, helping a second solute to cross the membrane.

86
Q

2015

Why does treatment of bacterial cells with lysozyme cause them to die provided the medium outside the cells has a lower ionic strength than the cellular interior?

  • A. The lysozyme hydrolyzes peptide parts of the cell wall, weakening it and allowing water to flow in through the membrane until the cells burst.
  • B. The lysozyme catalyzes transport of essential cellular components out of the cell.
  • C. The lysozyme catalyzes hydrolysis of β1→4 linkages in cell wall peptidoglycans, weakening the wall and allowing water to flow in through the membrane until the cells burst.
  • D. The lysozyme catalyzes transport of water out of the cells until dehydration leads to their death.
  • E. none of the above.
A

C. The lysozyme catalyzes hydrolysis of β1→4 linkages in cell wall peptidoglycans, weakening the wall and allowing water to flow in through the membrane until the cells burst.

87
Q

2015

Consider the following three processes:

  • # 1: ∆H = -1367 kJ/mol, -T∆S = +41 kJ/mol
  • # 2: ∆Η = −82 kJ/mol, -T∆S = -136 kJ/mol
  • # 3: ∆Η = +110 kJ/mol, -T∆S = -140 kJ/mol

Which of them is entropy driven?

  • A. #1
  • B. #2
  • C. #3
  • D. More than one process
  • E. None of the processes

Considering the same three processes as in the question above, which will have the largest equilibrium constant?

  • A. #1
  • B. #2
  • C. #3
  • D. All the reactions are favorable, so the equilibrium constants will be about the same.
  • E. cannot tell from data given.
A
  • C. #3
  • A. #1
88
Q

2015

The free energy change for ATP hydrolysis to ADP and Pi is -30.5 kJ/mol, while the free energy of hydrolyzing glucose-6-phosphate (G6P) to glucose and Pi is -13.8 kJ/mol. What is the free energy of phosphorylating glucose with ATP to make G6P?

  • A. -44.3 kJ/mol
  • B. -30.5 kJ/mol
  • C. +13.8 kJ/mol
  • D. -16.7 kJ/mol
  • E. none of these
A

D. -16.7 kJ/mol

89
Q

2015

The structure below was copied from the notes and the text. It is a galactosyl lipid found in chloroplasts. There is something wrong with the structure as drawn. What is wrong?

  • A. It has only cis double bonds, when it should have some trans.
  • B. One of the fatty acids should be saturated.
  • C. There is no phosphate group.
  • D. The galactose structure is incorrect.
  • E. None of these; the structure is correct.
A

D. The galactose structure is incorrect.

90
Q

2015

UDP-glucose has a single glucose molecule in an ester link between C1 of the glucose and the terminal phosphate of the UDP. Hydrolysis of UDP-glucose yields UMP and glucose-1-phosphate. This reaction is involved in mobilization of glucose from glycogen. How many high energy bonds are involved in the hydrolysis?

  • A. one
  • B. two
  • C. three
  • D. more than three
  • E. cannot tell from data given.
A

A. one

91
Q

2015

Hydrolysis of phosphoenolpyruvate (PEP) occurs with a standard free energy change of -61.9 kJ/mol. The reaction can drive phosphorylation of ADP to make ATP, whose free energy of hydrolysis in -30.5 kJ/mol. What percentage of the ∆Go’ of PEP hydrolysis is used to phosphorylate ADP?

  1. A. 100%
  2. B. 30.5%
  3. C. 49.3%
  4. D. 61.9%
  5. E. none of these
A

C. 49.3%

92
Q

2015

An unknown oligosaccharide is reduced at its reducing end with an appropriate reducing agent. It was found that 504 mg of the oligosaccharide consumed the equivalent of 1 millimole of hydrogen. The molecular weight of oligosaccharide is 504. Atomic weights: C = 12, O = 16, H = 1. This oligosaccharide most likely is

  • A. an aldo- or ketohexose.
  • B. an aldo- or ketodecanose (deca ≡ 10).
  • C. a reducing disaccharide consisting of 2 aldohexoses.
  • D. a non-reducing disaccharide consisting of 2 aldohexoses.
  • E. a reducing trisaccharide consisting of 3 hexoses.
A

a reducing trisaccharide consisting of 3 hexoses

93
Q

2015

Membrane proteins:

  • A. are sometimes covalently attached to lipid moieties.
  • B. are sometimes covalently attached to carbohydrate moieties.
  • C. are composed of the same 20 amino acids found in soluble proteins.
  • D. diffuse laterally in the membrane unless they are anchored
  • E. have all of the properties listed above
A

E. have all of the properties listed above

94
Q

2015

PEP hydrolysis as in the question above yields more free energy than required for ATP formation. What happens to the additional PEP free energy?

  • A. It is all dissipated as heat
  • B. It drives the reaction to completion in the face of unfavorable cellular concentrations.
  • C. Nothing happens to it; it is conserved for future use.
  • D. It raises the entropy of the surrounding water
  • E. none of the above
A

B. It drives the reaction to completion in the face of unfavorable cellular concentrations.

95
Q

2015

The fluidity of a lipid bilayer will be increased by:

  • A. decreasing the number of unsaturated fatty acids.
  • B. decreasing the temperature.
  • C. increasing the length of the alkyl chains.
  • D. increasing the temperature.
  • E. substituting 18:0 (stearic acid) in place of 18:2 (linoleic acid).
A

D. increasing the temperature

96
Q

2015

In biochemical redox reactions

  • A. ATP is always hydrolyzed.
  • B. Electrons are transferred from one compound to another.
  • C. The overall free energy depends on how the standard electrode potentials of the half reactions combine and on what the actual concentrations of reactants and products are.
  • D. The hydrogen half-cell must always be present.
  • E. more than one of the above
A

E. more than one of the above

97
Q

2015

Facilitated diffusion through a biological membrane is:

  • A. driven by a difference of solute concentration.
  • B. driven by ATP.
  • C. endergonic.
  • D. generally irreversible.
  • E. not specific with respect to the substrate
A

A. driven by a difference of solute concentration.

98
Q

2015

If three different amino acids or nucleotides or monosaccharides are used to construct tripeptides or trinucleotides or trisaccharides, which results in the greater number of trimers

  • A. amino acids
  • B. nucleotides
  • C. monosaccharides
  • D. A=B but C>A or B
  • E. C and D
A

E. C and D

99
Q

2018

What is the difference between a pyranose and a furanose?

  • A. Pyranose rings have six carbons in them, while furanoses have only five carbons.
  • B. Pyranose rings, if connected by glycosidic bonds, may have either α or β configuration at the anomeric carbon, while furanoses have only β.
  • C. Pyranose rings may have either axial or equatorial substituents on them, while furanoses have only axial substituents.
  • D. Only ketoses can form pyranoses, while aldoses cannot.
  • E. None of the above is correct.
A

E. None of the above is correct.

100
Q

2018

What is the difference between N-acetylgalactosamine and N-acetylglucosamine?

  • A. One has the N-acetyl group on C-2, while the other has it on C-4.
  • B. In N-acetylglucosamine all the substituents larger than a hydrogen atom must axial, while in N-acetylgalactosamine they must be equatorial.
  • C. One is always a pyranose, while the other is always a furanose.
  • D. They differ in the chiral configuration at C-4
  • E. more than one of the above
A

D. They differ in the chiral configuration at C-4

101
Q

2018

How many of the following are NOT epimeric pairs?

  • D-glucose and D-mannose
  • L-mannose and L-fructose
  • D-lactose and D-sucrose
  • D-glucose and L-glucose
  • D-glucose and D-glucosamine
  • A. One
  • B. Two
  • C. Three
  • D. Four
  • E. none of the five are epimeric pairs
A

D. Four

102
Q

2018

Which of the following is maltose?

  • A. Fru(2β→α1)Glc
  • B. NAcGal(β1→4)NAcGlu
  • C. Glc(α1→1α)Glc
  • D. Glc(α1→4)Glc
  • E. none one of these
A

D. Glc(α1→4)Glc

103
Q

2018

Which of the following is a reducing sugar?

  • A. Fru(2β→α1)Glc
  • B. NAcGal(β1→4)NAcGlu
  • C. Glc(α1→1α)Glc
  • D. Glc(α1→4)Glc
  • E. more than one of these
A

E. more than one of these

104
Q

2018

Which of the following amino acid residues is not a point of oligosaccharide attachment in glycoproteins?

  • A. Thr
  • B. Asn
  • C. Ser
  • D. Glu
  • E. more than one of these
A

D. Glu

105
Q

2018

Which of the following normally has branches within its polymer structure?

  • A. cellulose
  • B. glycogen
  • C. DNA
  • D. proteins
  • E. none of these
A

B. glycogen

106
Q

2018

In the table below which of the materials is most likely to contain large amounts of long chain saturated fatty acids?

  • A. beef fat
  • B. fish oil
  • C. chicken fat
  • D. olive oil
  • E. All would be expected to contain about the same amounts of long chain saturated fatty acids
A

A. beef fat

107
Q

2018

As temperature rises, membranes become more flexible and permeable to diffusion of molecules into and out of the cell. Bacterial cells, when grown initially at room temperature and then shifted to higher temperature compensate for the increased fluidity. How might they do this?

  • A. by synthesizing and adding to their membrane lipids fatty acids with longer chains.
  • B. by slowing their metabolism to the rate characteristic at room temperature.
  • C. by synthesizing and adding to their membrane lipids fatty acids with fewer double bonds
  • D. by changing the head group added to the phosphate of fatty acid glycerophosphates
  • E. A and C
A

E. A and C

108
Q

2018

2,3,7,8-tetrachlorodibenzo-p-dioxin (TCDD) has the structure shown below. Its half-life for excretion from humans is approximately 7.1 years, which is very slow. The reason it is so slow is that although it moves to the upper part of the large intestine, most of it is reabsorbed back into the blood stream in the lower part of the large intestine. Rats fed 2,3,7,8-TCDD and then fed squalene excrete the toxin in their feces. Why does feeding squalene have this effect?

  • A. Squalene causes breakdown of the dioxin, and the breakdown products are readily eliminated.
  • B. Squalene prevents deposition of the dioxin into the upper part of the large intestine.
  • C. Being strongly hydrophobic, as is dioxin, and conformationally flexible, which dioxin is not, squalene binds to the dioxin, preventing reabsorption.
  • D. Squalene blocks the intestinal cell sites at which dioxin is reabsorbed.
  • E. none of the above
A

C. Being strongly hydrophobic, as is dioxin, and conformationally flexible, which dioxin is not, squalene binds to the dioxin, preventing reabsorption.

109
Q

2018

Which of the following best describes the MAP kinase family of enzymes?

  • A. MAPK phosphorylates MAPKK, which phosphorylates MAPKKK, which then phosphorylates other proteins.
  • B. They can randomly phosphorylate one another as well as other proteins.
  • C. They are growth factors.
  • D. MAPKKK phosphorylates MAPKK, which phosphorylates MAPK, which then phosphorylates other proteins.
  • E. none of the above
A

D. MAPKKK phosphorylates MAPKK, which phosphorylates MAPK, which then phosphorylates other proteins.

110
Q

2018

Adenyl cyclase makes cAMP from ATP and phosphodiesterase hydrolyzes cAMP, yielding AMP. Both are stimulated by Ca++. Why does this not simply waste ATP?

  • A. Cyclization and hydrolysis can occur at different parts of the cell, each for a different purpose.
  • B. Cyclization and hydrolysis can occur at different times depending on the needs at the time.
  • C. The use of ATP in this situation serves to generate heat.
  • D. The energy released in the process drives other processes.
  • E. A and B
A

E. A and B

111
Q

2018

What are Scatchard plots used for?

  • A. to convert Michaelis-Menten curves to a linear form
  • B. to determine the number of ligand binding sites on a protein
  • C. to determine Vmax of an enzyme
  • D. to determine the dissociation constant of a ligand from a protein
  • E. B and D
A

E. B and D

112
Q

2018

In fluorescence resonance energy transfer (FRET) why is the wavelength of the emitted light longer than the donor’s emission wavelength?

  • A. The emitter has gained energy directly from the incoming light.
  • B. Protein to protein interaction changes the protein conformations.
  • C. Some of the excited state’s energy is lost on transfer from donor to emitter.
  • D. The energy difference between the emitting protein’s excited and ground states is less than for the donor’s excited and ground states.
  • E. None of these is correct
A

D. The energy difference between the emitting protein’s excited and ground states is less than for the donor’s excited and ground states.

113
Q

2018

Fluorescence resonance energy transfer (FRET) can be used to measure the cAMP concentration inside cells. In the assay green fluorescent protein (GFP) is covalently attached to the catalytic subunit of cAMP dependent protein kinase A (PKA), while blue fluorescent protein (BFP) is covalently attached to the regulatory subunit of the enzyme. In the absence of cAMP the catalytic and regulatory subunits are associated through a non-covalent interface. They separate when cAMP binds. GFP when by itself absorbs light at 395 nm and emits at 545 nm. BFP by itself absorbs at 380 nm and emits at 460 nm. If light is shined on the sample at 380 nm, at what wavelength would one see a decrease in fluorescence emission upon cAMP binding inside a cell?

  • A. 380 nm
  • B. 395 nm
  • C. 460 nm
  • D. 545 nm
  • E. more than one of these.
A

D. 545 nm

114
Q

2018

Why do the female offspring of mating male mice whose X chromosomes carry red fluorescent protein with females whose X chromosomes carry green fluorescent protein yield some cells that glow red alone while others glow green when illuminated with light of the appropriate wavelengths?

  • A. The cells blend the proteins from their parental X chromosomes.
  • B. Some cells shut off one of their X chromosomes, while other cells shut off the other X chromosome.
  • C. Some mice grow red hair, while others grow green hair.
  • D. Each X chromosome operates in a different organ, the other X being suppressed in that organ.
  • E. This is a hairy question

The mosaic patters may allow a study of the differences between left brain and right brain functions. How might this be so?

  • A. Left and right brain cells all show both red and green fluorescence depending on which part of the cell one illuminates.
  • B. The silent X chromosome’s gene products dominate the fluorescence.
  • C. The distribution of red and green fluorescent cells is the same on the two sides of the brain.
  • D. The pattern of red and green fluorescent cells varies from one side of the brain to the other.
  • E. None of the above would enable a study of brain function.
A
  • B. Some cells shut off one of their X chromosomes, while other cells shut off the other X chromosome
  • D. The pattern of red and green fluorescent cells varies from one side of the brain to the other.
115
Q

2018

What is the sequence of events that allows an action potential to move down a nerve cell axon?

  • A. sodium entry followed by potassium exit
  • B. potassium entry followed by sodium exit
  • C. calcium entry followed by ATP hydrolysis
  • D. the simultaneous entry and exit of both sodium and potassium.
  • E. none of the above
A

A. sodium entry followed by potassium exit

116
Q

2019

Injection with which of the following would lead mice used as test animals to die of pneumonia?

  • A. a mixture of DNA from heat-killed virulent bacteria and live non-virulent bacteria to which DNAase had been added prior to injection.
  • B. live non-virulent bacteria
  • C. a mixture of live non-virulent bacteria and heat-killed virulent bacteria
  • D. virulent bacteria which had been killed by heat
  • E. more than one of the above
A

C. a mixture of live non-virulent bacteria and heat-killed virulent bacteria

117
Q

2019

Suppose one had a supply of ribose, phosphate and the bases adenine and thymine. How many water molecules would have to be removed (dehydration) to make the dinucleotide p-AT-OH?

  • A. three
  • B. five
  • C. two
  • D. none
  • E. None of these is correct.
A

B. five

118
Q

2019

Why does the genetic material of some viruses not follow Chargaff’s rules?

  • A. Those viruses have RNA as their genetic material.
  • B. Their genetic material is not double stranded.
  • C. Those viruses have unconventional bases such as pseudouridine in their genetic material.
  • D. Those viruses have significant amounts of thymine dimers in their genetic material.
  • E. none of these
A

B. Their genetic material is not double stranded.

119
Q

2019

Could a palindromic DNA base sequence form a hairpin conformation?

  • A. yes
  • B. no
  • C. It would depend on the exact sequence.
  • D. It could do so only if rich in GC base pairs
  • E. none of the above
A

A. yes

120
Q

2019

What are Hoogstein structures?

  • A. DNA that has unusual base pairs
  • B. DNA that forms triple helices
  • C. tetraplex DNA
  • D. long palindromic sequences
  • E. more than one of the above
A

E. more than one of the above

121
Q

2019

Could absorbance measurements of a DNA sample at 260 nm yield a concentration in units of picograms per microliter?

  • A. yes
  • B. no
  • C. only with DNA of high GC content
  • D. only with DNA with high AT content
  • E. cannot tell from data given.
A

A. yes

122
Q

2019

Lyle, Lyle the Nile crocodile gives the appearance of a somnolent beast. However, Lyle is capable of very fast movement for a brief period (to catch the little fishes). Glucose residues (for energy production) can be rapidly removed (at 250 µmoles/minute in 1.00g muscle tissue) from the glycogen stored in crocodile muscle. Crocodile muscle contains 0.81% w/w of glycogen. For how long could Lyle could sustain rapid movement? (Note: a glucose residue in glycogen has molecular weight of 162)

  • A. 250 minutes 0.81
  • B. 0.81 minutes 250
  • C. 5 minutes
  • D. 12 seconds
  • E. 12 minutes
A

D. 12 seconds

FYI: 2017 says 5 minutes

123
Q

2019

If glycogen is partially hydrolysed, how many disaccharides and trisaccharides can be formed?

  • A. 2, 2
  • B. 3, 3
  • C. 2, 4
  • D. 3, 3
  • E. 2, 3
A

C. 2, 4

124
Q

2019

In Figure X. which exoglycosidase(s) will completely hydrolyse the trisaccharide?

  • A. β mannosidase
  • B. α glucosidase
  • C. α galactosidase
  • D. β fructosidase
  • E. C & D

The monosaccharides in Figure X are:

  • A. Glc, Fru, Gal
  • B. Glc, Fru, Man
  • C. Gal, Man, Fru
  • D. Gal, Man, Rib
  • E. Glc, Gal, Fru
A
  • E. C & D
  • C. Gal, Man, Fru
125
Q

2019

Which of the following is a characteristic of sucrose?

  • A. It is a disaccharide of glucose.
  • B. It is Glc(α1→4)Fru
  • C. It is Fru(β2→4)Glc
  • D. It has no free anomeric carbon
  • E. none of these
A

D. It has no free anomeric carbon

126
Q

2019

Consider the proteoglycan to the left. How many non-reducing ends does the oligosaccharide at the top right of the diagram have?

  • A. none
  • B. one
  • C. three
  • D. five
  • E. None of these

If the proteoglycan to the left is NOT subjected to any hydrolysis, how many reducing ends does the oligosaccharide at the top right of the diagram have?

  • A. none
  • B. one
  • C. three
  • D. five
  • E. None of these
A
  • D. five
  • A. none
127
Q

2019

In the table below which of the materials is most likely to contain large amounts of long chain saturated fatty acids?

  • A. beef fat
  • B. fish oil
  • C. chicken fat
  • D. olive oil
  • E. All would be expected to contain about the same amounts of long chain saturated fatty acids
A

D. olive oil

128
Q

2019

In one catalytic cycle, the Na+/K+ ATPase transporter transports:

  • A. 3 Na+ out, 2 K+ in, and converts 1 ATP to ADP + Pi.
  • B. 2 Na+ out, 3 K+ in, and converts 1 ATP to ADP + Pi.
  • C. 3 Na+ in, 2 K+ out, and converts 1 ATP to ADP + Pi.
  • D. 1 Na+ out, 1 K+ in, and converts 1 ATP to ADP + Pi.
  • E. 2 Na+ out, 3 K+ in, and converts 1 ADP + Pi to ATP.
A

A. 3 Na+ out, 2 K+ in, and converts 1 ATP to ADP + Pi.

129
Q

2019

A fatty acid has the formal representation of 22:5(Δ4,7,10,13,16). If the carboxyl group has no proton on it, how many atoms of hydrogen does the molecule contain?

  • A. thirty-three
  • B. ten
  • C. five
  • D. twenty-two
  • E. none of these.
A

A. thirty-three

130
Q

2019

How do sphingolipids differ from glycerophospholipids?

  • A. Sphingolipids have three fatty acids in ester linkage, whereas glycerophospholipids have only two.
  • B. Glycerophospholipids have three fatty acids in ester linkage, whereas sphingolipids have only two
  • C. Glycerophospholipids have one fatty acid in ester linkage, whereas sphingolipids have two.
  • D. Sphingolipids have one fatty acid in ester linkage, whereas glycerophospholipids have two.
  • E. None of the above is correct.
A

E. None of the above is correct.

131
Q

2019

How do most bacterial lipids differ from the lipids of some archaea, particularly those that live at elevated temperatures?

  • A. Most bacterial lipids have two or three fatty acids in ester linkage to glycerol, while the archaeal lipids in question have no fatty acids in them.
  • B. The archaeal lipids have hydrocarbon chains anchored at both ends by ether linkages to glycerol.
  • C. The hydrocarbon chains of the archaeal lipids are made up of isoprene units.
  • D. The hydrocarbon chains in the archaeal lipids are about twice as long as typical fatty acids.
  • E. All the above are correct.
A

E. All the above are correct

132
Q

2019

What is the net charge at pH 7 on a glycerophospholipid which has choline linked to the phosphate?

  • A. zero
  • B. +1
  • C. -1
  • D. -2
  • E. none of these
A

A. zero

133
Q

2019

What is removed from a glycerophospholipid by phospholipase A2?

  • A. The fatty acid at C1.
  • B. The fatty acid at C2
  • C. the phosphate with whatever else is attached to it
  • D. the head group attached to the phosphate
  • E. none of the above
A

B. The fatty acid at C2

134
Q

2019

What is removed by the sphingomyelinase associated with Nieman-Pick disease?

  • A. phosphocholine
  • B. N-acetylgalactoseamine
  • C. glucose
  • D. galactose
  • E. more than one of the above
A

B. N-acetylgalactoseamine

135
Q

2019

Movement of positively charged ions across a membrane from a compartment with an excess of negative charge to one with excess positive charge

  • A. requires that it move down its concentration gradient.
  • B. generates energy in the form of ATP.
  • C. requires that its concentration in the region of excess positive charge exceeds the concentration in the region with excess negative charge.
  • D. requires that oppositely charged ions move with it E. none of the above
A

A. requires that it move down its concentration gradient.

(E3-2020 states more than one of the above)

136
Q

2016

The sequence of one strand of DNA is 5’ CTAGCT 3’. The sequence of the complementary strand would be

  • A. 5’ AGCTAG 3’
  • B. 5’ TCGATC 3’
  • C. 5’ CTAGCT 3’
  • D. 5’ GCTAGC 3’
  • E. 5’ GATCGA 3’
A

A. 5’ AGCTAG 3’

137
Q

2016

Guanine, whose predominant structure is shown at the top of the figure, tautomerizes to a limited extent, yielding the form at the lower left. If DNA is treated with excess dimethyl sulfate, the tautomerization equilibrium is pulled toward the tautomer by LeChatlier’s principle, and all of it is converted to the O6 -methyl form as shown. What is the likely effect on base pairing in the DNA?

  • A. There will be no effect at all.
  • B. GC base pairing will become stronger.
  • C. O6 -methylguanine cannot form normal GC base pairs at all.
  • D. O6 -methylguanine will form base pairs with thymine.
  • E. cannot tell without more data.
A

C. O6 -methylguanine cannot form normal GC base pairs at all.

138
Q

2016

Who is Karst Hoogstein?

  • A. the discoverer of A-form double stranded DNA
  • B. the discoverer of the unusual base pairing in triplex DNA
  • C. the discoverer of guanine tetraplex DNA
  • D. the discoverer of Z-form DNA E. none of the above
A

B. the discoverer of the unusual base pairing in triplex DNA

139
Q

2016

In a palindromic region of DNA

  • A. the two strands have the same sequence if both are read in their 5’ to 3’ directions except for short sequences that connect the base paired arms.
  • B. the two strands have the same sequence if one is read 5’ to 3’, while the other is read in the 3’ to 5’ direction except for short sequences that connect the base paired parts.
  • C. both strands are needed to form a hairpin loop.
  • D. each strand is symmetric within itself, its left half read 5’ to 3’ being the same as its right half read 3’ to 5’.
  • E. None of the above is correct
A

A. the two strands have the same sequence if both are read in their 5’ to 3’ directions except for short sequences that connect the base paired arms.

140
Q

2016

A bacterium is grown in two separate cultures which differ only in the nitrogen isotope supplied to the culture. In one all the nitrogen is N14, while in the other it is N15 . The cell count in the two cultures is the same. The DNA is extracted from the cultures, heated to separate the strands, and the two hot solutions are mixed and allowed to cool slowly. What fraction of the resulting double stranded DNA will have one N14 strand and one N15 strand?

  • A. one sixth
  • B. one third
  • C. one fourth
  • D. two thirds
  • E. none of these
A

D. two thirds

141
Q

2016

If a cytosine base in a double stranded DNA becomes deaminated and the DNA is then replicated twice, the consequence will be

  • A. normal base pairing will lead to replacement of the deaminated base by a new cytosine.
  • B. the GC base pair will be replaced by an AT base pair at the second replication.
  • C. the deamination will be fatal at the second replication.
  • D. G will pair abnormally with T.
  • E. none of the above.
A

B. the GC base pair will be replaced by an AT base pair at the second replication.

142
Q

2016

From the abbreviated name of the compound Glc(β1→4)Gal, we know that: (modified from the posted exam)

  • A. the galactose residue is the β anomer.
  • B. the galactose residue is at the reducing end.
  • C. C-4 of glucose is joined to C-1 of galactose by a glycosidic bond.
  • D. C-4 of galactose is joined to C-1 of glucose by a glycosidic bond.
  • E. the glucose in is its furanose form.
A

D. C-4 of galactose is joined to C-1 of glucose by a glycosidic bond

FYI: as per exam 2017, B or D is correct

143
Q

2016

In which of the following pairs is one member of the pair NOT the C-2 keto of the other?

  • A. D-glucose and D-mannose
  • B. D-ribose and D-ribulose
  • C. D-xylose and D-xylulose
  • D. D-glucose and D-fructose
  • E. more than one of the above
A

A. D-glucose and D-mannose

144
Q

2016

Which of the following is NOT a reducing sugar?

  • A. Trehalose (Glc(α1→1α)Glc)
  • B. Lactose (Gal(β1→4)Glc)
  • C. Sucrose (Fru(2β→α1)Glc)
  • D. the riboses in a nucleic acid
  • E. more than one of the above
A

E. more than one of the above

145
Q

2016

Which of the following statements about the glycoprotein shown below is correct?

  • A. The oligosaccharide is O-linked to the protein.
  • B. There are three reducing chain ends in the oligosaccharide.
  • C. There are probably two α1→6 linkages to manose residues, but one cannot be sure without further data.
  • D. There are three non-reducing chain ends in the oligosaccharide.
  • E. C and D are both correct.
A

E. C and D are both correct.

146
Q

2016

What are the glycodsidic linkages in the two disaccharide units shown below?

  • A. (a) is α1→4 while (b) is β1→4
  • B. (b) is α1→4 while (a) is β1→4
  • C. both are α1→4
  • D. both are β1→4
  • E. α and β cannot be distinguished in these representations.
A

A. (a) is α1→4 while (b) is β1→4

147
Q

2016

Shrimp skins are crunchy because they contain

  • A. heparin
  • B. chitin
  • C. cellulose
  • D. amylose
  • E. none of these are found in shrimp skins
A

B. chitin

148
Q

2016

The NS domains of heparin sulfate can bind the protein antithrombin (AT) as discussed in cless. When AT binds, its conformation changes leading to binding of the blood factor X (X = 10 here) or to binding thrombin. The result is

  • A. the breakdown of blood clots.
  • B. inhibition of blood clot formation.
  • C. promotion of blood clot formation in response to wounds.
  • D. hydrolysis of thrombin
  • E. none of these
A

B. inhibition of blood clot formation.

149
Q

2016

One molecule of vasopressin activates a single molecule of phospholipase-C (PL-C). The PL-C molecule catalyzes hydrolysis of ten molecules of phosphatidylinositol-4,5-bisphosphate, releasing ten molecules of inositol-1,4,5-trisphosphate (IP3) and leaving behind in the membrane ten diacyl glycerol (DAC) molecules. Each IP3 triggers the endoplasmic reticulum to release 1000 calcium ions. Each DAC binds one calcium ion. Once the calcium is bound, the DAC-calcium complex activates one molecule of protein kinase-C (PK-C). Each PK-C catalyzes phosphorylation of 100 other proteins, using ATP to do so. How many of these other protein molecules are phosphorylated as a result?

  • A. 10
  • B. 100
  • C. 1000
  • D. 10000
  • E. none of these is correct
A

D. 10000

150
Q

2016

What is the effect of incorporating cholesterol into lipid bilayer membranes?

  • A. The membranes become more permeable to small molecular weight solutes.
  • B. The membranes become less flexible.
  • C. The membranes become better able to hold membrane-spanning proteins.
  • D. The cholesterol serves as a binding site for various proteins
  • E. The membranes become more fluid.
A

B. The membranes become less flexible

151
Q

2016

In the table below which of the materials is least likely to contain large amounts of long chain saturated fatty acids?

  • A. beef fat
  • B. fish oil
  • C. chicken fat
  • D. olive oil
  • E. All would be expected to contain about the same amounts of long chain saturated fatty acids
A

D. olive oil

152
Q

2013

Considering the structure of double stranded DNA, what kinds of bonds hold one complementary strand to the other?

  • A. ionic
  • B. covalent
  • C. Van der Waals
  • D. hydrogen
  • E. hydrophobic and hydrophilic
A

D. hydrogen

153
Q

2013

Which of the following is NOT a reducing sugar?

  • A. Trehalose (Glc(α1⇔1α)Glc)
  • B. Lactose (Gal(β1→4)Glc)
  • C. Sucrose (Fru(2β⇔α1)Glc)
  • D. the riboses in a nucleic acid
  • E. more than one of the above
A

E. more than one of the above

154
Q

2013

The sequence of one strand of DNA is 5’ TCGATC 3’. The sequence of the complementary strand would be

  • A. 5’ AGCTAG 3’
  • B. 5’ TCGATC 3’
  • C. 5’ CTAGCT 3’
  • D. 5’ GCTAGC 3’
  • E. 5’ GATCGA 3’
A

E. 5’ GATCGA 3’

155
Q

2013

From the abbreviated name of the compound Gal(β1→4)Glc, we know that: (from the posted exam)

  • A. the glucose residue is the β anomer.
  • B. the galactose residue is at the reducing end.
  • C. C-4 of glucose is joined to C-1 of galactose by a glycosidic bond.
  • D. the compound is dextrorotatory.
  • E. the glucose in is its furanose form.
A

C-4 of glucose is joined to C-1 of galactose by a glycosidic bond

156
Q

2013

In which of the following ways do the NS domains of heparin sulfate participate in preventing blood clotting?

  • A. by inducing a conformational change in antithrombin (AT) when the AT binds, enabling binding of Factor Xa to the AT. This immobilizes the Factor Xa, thereby preventing it from taking part in clotting.
  • B. by catalyzing liberation of oligosaccharides which block the active sites necessary for clotting.
  • C. by inducing a conformational change in AT when the AT binds, enabling binding of thrombin to the AT and to an adjacent NS domain. This immobilizes the thrombin, thereby preventing it from taking part in clotting.
  • D. by binding lipoprotein lipase, which is thereby activated and releases fatty acids that inhibit clotting.
  • E. A and C
A

E. A and C

157
Q

2013

Which of the following is a characteristic of membrane transporters?

  • A. All require an energy input to move polar compounds across membranes.
  • B. They transport fully formed membranes from one place to another.
  • C. They reduce the activation energy barrier of transport, thereby increasing the flow of material across the membrane.
  • D. They only act to move molecules from outside the cell to the inside.
  • E. None of these is a characteristic of transporters.
A

C. They reduce the activation energy barrier of transport, thereby increasing the flow of material across the membrane.

158
Q

2013

What are v-SNAREs and t-SNARES?

  • A. They are involved in membrane fusion in nerve cells.
  • B. They aid fusion by coiling up like screws, pushing the membrane leaflets together.
  • C. Their action leads to release of neurotransmitter molecules into the extracellular space.
  • D. They are traps specially designed to catch certain types of small animals.
  • E. more than one of the above
A

E. more than one of the above

159
Q

2013

Which of the following is NOT one of Chargaff’s rules?

  • A. The base composition of DNA generally varies from one species to another.
  • B. DNA specimens isolated from different tissues of the same species have the same base composition.
  • C. The base composition of DNA in a given species does not change with the organism’s age, nutritional state or changing environment.
  • D. In all DNAs, regardless of species, the number of adenine residues equals the number of thymine residues, and the number of guanine residues equals the number of cytosine residues.
  • E. All of the above are Chargaff’s rules.
A

E. All of the above are Chargaff’s rules.

160
Q

2013

Tay-Sachs disease, Gaucher’s disease and Nieman-Pick disease are genetic dieeases

  • A. in which enzymes of phospholipid breakdown are missing or not fully functional.
  • B. in which enzymes of sphingolipid breakdown are missing or not fully functional.
  • C. in which enzymes needed to form unsaturated fatty acids are missing or not fully functional.
  • D. of carbohydrate metabolism.
  • E. none of the above
A

B. in which enzymes of sphingolipid breakdown are missing or not fully functional.

161
Q

2013

GU base pairs occur in

  • A. DNA
  • B. RNA
  • C. both RNA and DNA
  • D. neither RNA nor DNA
  • E. only between a DNA strand and an RNA strand
A

B. RNA

162
Q

2013

The fatty acid shown below has been esterified to picolinic acid and subjected to mass spectrometry yielding the fragmentation pattern indicated by the vertical dashed lines. What is the fragment molecular weight indicated by the arrow in the figure?

  • A. 92
  • B. 108
  • C. 164
  • D. 371
  • E. none of these
A

C. 164

163
Q

2013

The rise per residue in the direction of an Į-helix axis is 1.5 Å. If it requires 19 residues for a transmembrane helix of glycophorin to span the red cell membrane, the thickness of the membrane is _____ Å, and one phospholipid molecule in the membrane has an average length of approximately _____ Å.

  • A. 28.5 Å, 14.3 Å
  • B. 19 Å, 1.5 Å
  • C. 14.3 Å, 28.5 Å
  • D. 19 Å, 9.5 Å
  • E. none of these
A

A. 28.5 Å, 14.3 Å

164
Q

2013

GPI-anchored proteins are held to the membrane by

  • A. membrane-spanning protein Į-helices
  • B. isoprene groups
  • C. the fatty acids of a glycerol phosphate
  • D. cholesterol
  • E. none of these
A

C. the fatty acids of a glycerol phosphate

165
Q

2013

If a flippase and a floppase were each able to operate on any phosphatidyl serine or phosphatidyl ethanolamine, and if both were active to the same extent at the same time, the result would be

  • A. no net effect on membrane leaflet composition.
  • B. equivalent to that of a scramblase.
  • C. the expenditure of much ATP.
  • D. equilibration of membrane leaflet composition
  • E. B, C, and D
A

E. B, C, and D

166
Q

2017

Which of the following pairs is interconverted in the process of mutarotation?

  • Α. α-D-glucose and β-D-glucose
  • B. D-glucose and L-glucose
  • C. D-glucose and D-fructose
  • D. D-glucose and D-glucosamine
  • E. D-glucose and D-galactose
A

Α. α-D-glucose and β-D-glucose

167
Q

2017

Which of the following is an epimeric pair?

  • A. D-glucose and D-mannose
  • B. D-lactose and D-sucrose
  • C. D-glucose and D-glucosamine
  • D. L-mannose and L-fructose
  • E. D-glucose and L-glucose
A

A. D-glucose and D-mannose

168
Q

2017

Which of the following amino acid residues is not a point of oligosaccharide attachment in glycoproteins?

  • A. Thr
  • B. Gly
  • C. Ser
  • D. Asn
  • E. more than one of these
A

B. Gly

169
Q

2017

G proteins hydrolyze bound GTP spontaneously, thereby becoming inactive in converting ATP to cAMP. This is part of the normal shut-off mechanism. The A1 subunit of cholera toxin inactivates the GTPase activity, so the G protein continues making cAMP, which makes protein kinase A (PKA) overactive. How does the PKA lead to the symptoms of cholera?

  • A. It closes a channel which would leak water from the cell.
  • B. It phosphorylates both a chloride exchanger and a Na + -H+ exchanger. These cause sodium to exit, and water exits the cell with it to maintain cytosol concentrations.
  • C. It collapses the epinephrine cascade.
  • D. It reduces NAD+ .
  • E. more than one of the above.
A
  • B. It phosphorylates both a chloride exchanger and a Na + -H+ exchanger. These cause sodium to exit, and water exits the cell with it to maintain cytosol concentrations.
170
Q

2017

What is Scatchard analysis used for?

  • A. to determine the number of ligand binding sites on a protein
  • B. to determine the rate at which ligands are bound
  • C. to separate receptor proteins from their ligands
  • D. to determine the dissociation constant for a ligand-protein complex
  • E. more than one of the above
A

E. more than one of the above

171
Q

2017

Which of the following best describes fluorescence?

  • A. A photon is absorbed at low energy (long wavelength), gains some energy from the surroundings and is emitted at higher energy (shorter wavelength).
  • B. Absorption of a photon promotes an electron from a ground state to an excited state.
  • C. Light absorption causes cis-trans isomerization of the absorbing chromophore.
  • D. A photon is absorbed at high energy (short wavelength), loses some energy to the surroundings and is emitted at lower energy (longer wavelength).
  • E. none of these
A

D. A photon is absorbed at high energy (short wavelength), loses some energy to the surroundings and is emitted at lower energy (longer wavelength).

172
Q

2017

. In fluorescence resonance energy transfer (FRET)

  • A. Interaction of two different fluorescent proteins causes light emission at the longer emission wavelength of the two.
  • B. The energy of the photons absorbed by one protein is transferred to the second and emitted as heat.
  • C. Interaction of two different fluorescent proteins causes light emission at the shorter emission wavelength of the two.
  • D. Light emission requires the hydrolysis of cAMP
  • E. The absence of light in the student’s mind causes the student to fret.
A

A. Interaction of two different fluorescent proteins causes light emission at the longer emission wavelength of the two.

173
Q

2017

When mice carrying GFP in their X chromosomes are bred with mice having RFP on their X chromosomes, it is found in female offspring that

  • A. each of their cells expresses both fluorescent proteins.
  • B. some cells fluoresce red, while others fluoresce green.
  • C. the fluorescence pattern shows that in some cells the X chromosome from the male parent is shut down, while in other cells the female parent’s X is shut down.
  • D. the fluorescence pattern may indicate differences in brain function depending on which parent’s X is active.
  • E. more than one of the above.
A

E. more than one of the above.

174
Q

2017

In the insulin receptor, which is a tyrosine kinase, insulin binding to the extracellular part of the receptor

  • A. causes the intracellular kinase domains to come together to start the cascade.
  • B. causes each intracellular kinase domain to phosphorylate tyrosines on the other.
  • C. the charges introduced by phosphorylation of the kinase domains force a conformational change which exposes the kinase active site.
  • D. A and B
  • E. B and C
A

E. B and C

175
Q

2020

Why do the phi-psi pairs of much of the Ramachandran plot represent “forbidden” conformations?

  • The pi clouds of forbidden conformations are repelled by partial negative charges of nearby peptides
  • The pi clouds of allowed conformations are stabilized by partial positive charges of nearby peptides
  • Forbidden conformations are due to steric hindrance of atoms.
  • None of the above.
A

Forbidden conformations are due to steric hindrance of atoms.

176
Q

2020

Consider the thermal denaturation curves of two species’ DNA. The DNA in sample A melts at 83C, while the the DNA of sample B melts at 87C. Which sample has the higher percentage of AT base pairs?

  • Sample A
  • Sample B
  • Their AT content is the same
  • Neither sample has a significant number of AT base pairs, so one cannot distinguish them.
  • Experiments of this kind have nothing to do with AT content.
A

Sample A

177
Q

2020

Consider the proteoglycan structure. How many non-reducing ends are present in the oligosaccharides attached to the protein?

  • none
  • three
  • five
  • none of these
A

none of thest

178
Q
A
179
Q

2020

Consider the proteoglycan structure. How many reducing ends are present in the oligosaccharides attached to the protein?

  • none
  • seven
  • five
  • none of those
A

none

180
Q

2020

A fatty acid has the formal representation of 20:5(Δ4,7,10,13,16). If the carboxyl group has no proton on it, how many atoms of hydrogen does the molecule contain?

  • 33
  • 10
  • five
  • 29
  • none of these
A

29

181
Q

2020

How many of the following are epimeric pairs?

D-glucose and D-mannose

L-mannose and L-fructose

D-lactose and D-sucrose

D-glucose and L-glucose

D-glucose and D-glucosamine

  • one
  • two
  • three
  • four
  • None of the five are epimeric pairs.
A

one

182
Q

2020

In the conversion of lactate to pyruvate as happens after muscle exercise, which of the following reactions would be coupled to the conversion?

  • NAD+ + H2 –> NADH + H+
  • NADH + H+ ⇌ NAD+ + H2
  • ATP + H2O à ADP + Pi
  • ADP + Pi ⇌ ATP + H2O
  • None of these would serve to drive the reaction.
A

NAD+ + H2 –> NADH + H+

183
Q

2020

Consider the oxidation of ethanol to acetaldehyde by NAD+. The reduction potentials are

Acetaldehyde + 2H+ + 2e- –> ethanol Eo’ = -0.197 V

NAD+ + 2H+ + 2e- –> NADH + H+ Eo’ = -0.320 V

The standard change in electrode potentials at pH 7, DeltaEo’, is given by (Eo’ of the electron acceptor) – (Eo’ of the donor), and the standard free energy change is given by DeltaGo’ = -n F DeltaEo’. Would the reaction proceed in the direction of ethanol production under standard conditions? The number of electrons involved in the conversion = n. F, the Faraday is 96,500 J/(V-mol).

  • no
  • yes
  • it would depend on the []
  • It would be at equilibrium
  • cannot tell from data given
A

yes

184
Q

2020

In which of the following cases would the reaction described be enthalpy driven against an unfavorable entropy? (all units are kJ/mol)

  • DeltaG = -218, DeltaH = -82
  • DeltaG = -1326, DeltaH = -1367
  • DeltaG = -30, DH = +110
  • more than one of these
  • none of these
A

DeltaG = -1326, DeltaH = -1367

185
Q

2020

Consider diffusion of a solute, A, across a membrane from zone 1 to zone 2. In which of the following cases would the diffusion be favorable?

  • ln ([A2] / [A1]) > 0
  • ln ([A2] / [A1]) = 0
  • ln ([A2] / [A1]) < 0
  • [A2] / [A1] > 1
  • more than one of these
A

ln ([A2] / [A1]) < 0

186
Q

2020

The equilibrium constant for the isomerization of glucose 6-phosphate (G6P) to fructose 6-phosphate (F6P) is 0.504. In which of the following situations would the free energy be best buffered against a change of the fraction of G6P that is isomerized?

  • The fraction of the material in the G6P form is approximately one third
  • The fraction of the material in the G6P form is approximately one half
  • The fraction of the material in the G6P form is approximately two thirds
  • The fraction of the material in the F6P form is approximately two thirds
  • More than one of these is correct
A

The fraction of the material in the G6P form is approximately two thirds

187
Q

2020

If all materials were to be in their standard states, would hydrolysis of ATP be sufficient to reverse the hydrolysis of UDP-glucose to UMP and glucose 1-phosphate

  • yes
  • no
  • Yes, but only if sufficient magnesium were present
  • Insufficient data are available in the text or the class materials to answer this question
  • no, but only because UDP-glucose cannot be hydrolyzed
A

yes

188
Q

2020

The phosphate trianion, PO4-3, has more resonance stabilization than does the dianion, HPO4-2. The third pKa for phosphoric acid is 12.2. Why does it require such strongly alkaline conditions to remove the proton from HPO4-2?

  • Having lost two protons, the dianion’s environment contains a high concentration of them, which disfavors dissociation of the third proton
  • The Kd is unfavorable
  • The concentration of the dianions is too large
  • The negative charge on the trianion attracts the proton very strongly.
  • The statement is incorrect, as the conditions at pH = pKa are not alkaline.
A

The negative charge on the trianion attracts the proton very strongly.

189
Q

2020

The glyceraldehyde 3-phosphate dehydrogenase reaction is endergonic (requires energy input), while the subsequent reaction catalyzed by phosphoglycerate kinase is exergonic. How does the kinase reaction make the endergonic reaction run from left to right?

  • It couples the reaction to ATP hydrolysis
  • It recycles 3-phosphoglycerate back to glyceraldehyde 3-phosphate to drive the endergonic reaction to the right.
  • The NADH produced in the endergonic reaction is reoxidized to NAD+ to drive the endergonic reaction to the right
  • The endergonic reaction product is drained away by the kinase reaction, pulling the endergonic reaction to the right.
  • None of the above is correct
A

The endergonic reaction product is drained away by the kinase reaction, pulling the endergonic reaction to the right.

190
Q

2020

Glycogen phosphorylase and alpha-amylase both remove a glucose residue from glycogen. How do the reactions differ?

  • The phosphorylase acts at branch points, whereas amylase removes unbranched residues
  • The phosphorylase acts in liver and muscle cells, whereas amylase acts in saliva
  • The phosphorylase attached a phosphate to the glucose it removes, whereas the amylase does not
  • The phosphorylase requires ATP, whereas the amylase does not.
  • More than one of the above is correct
A

More than one of the above is correct

191
Q

2020

In the first bypass of glycolysis in gluconeogenesis reactions are catalyzed by a carboxykinase and by a carboxylase. How do the two reactions differ?

  • The carboxylase adds a CO2, whereas the kinase removes one
  • The carboxylase removes a CO2, whereas the kinase adds one
  • There is no difference. The reactions are equivalent.
  • The kinase requires ATP, whereas the carboxylase does not.
  • None of these is correct.
A

The carboxylase adds a CO2, whereas the kinase removes one

192
Q

2020

In the pentose phosphate pathway, which of the following reactions occur more than once

  • isomerase
  • epimerase
  • transketolase
  • all except epimerase
  • aldolase
A

all except epimerase

Biochemistry 403 (12 decks)

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