Ch 1 - 6

Question 1

If a stretch of DNA on the parental strand of a replicating chromosome has the sequence 5′–AGCTCGATCGGCTA–3′, what will the sequence of the newly synthesized strand made from this stretch of template be?

Answer 1

3′–TCGAGCTAGCCGAT–5′

The parental strand will be used as a template for replication using complementary base-pairing rules. So, the new strand will be complementary and antiparallel to the other.

Question 2

Meselson and Stahl performed a classic experiment to explore three models for the mechanism of DNA replication. Which of the models held that the two parental strands would remain associated after replication?

a) dispersive
b) liberal
c) conservative
d) semiconservative

Answer 2

c) conservative

The conservative model of replication posited that after replication of the parental strands, the original parental molecule remained intact. In this model, the two newly replicated strands would associate, forming their own new molecule.

Question 3

Using the technique shown below, Meselson and Stahl did an experiment where they grew cells in a heavy medium for many generations, then after a single generation in a light-medium, they observed a single band of intermediate weight after centrifugation. This experiment ruled out which model of DNA replication?

a) dispersive
b) semiconservative
c) conservative
d) liberal

Answer 3

c) conservative

After one round of DNA replication in light-medium, the DNA molecules were of an intermediate weight, meaning that they had a mixture of light and heavy nitrogen isotopes. This could arise from either dispersive or semiconservative mechanisms but ruled out the conservative mechanism since it would produce molecules that are completely made up of light isotope as well as the parental molecules of all heavy isotope.

Question 4

What is the name of the DNA sequence where replication begins?

a) replication origin
b) replication fork
c) DNA template
d) initiator

Answer 4

a) replication origin

DNA synthesis begins at replication origins, where local strand separation is promoted by initiator proteins.

Question 5

The sequence at which DNA replication begins tends to have which characteristic?

a) AT-rich
b) GC-rich
c) A-rich
d) G-rich

Answer 5

a) AT-rich

Replication origins tend to be very rich in A-T base pairs to facilitate local strand separation by initiator proteins. A-T base pair is held together by fewer hydrogen bonds than is a G-C base pair. Therefore, A-T base pairs are easier to pull apart.

Question 6

In which direction, and on which strands does DNA replication proceed from a replication origin on a chromosome?

a) only on one strand in the 5’ → 3’ direction
b) only on one strand in the 3’ → 5’ direction
c) in both directions, on both strands
d) from the left to the right, on both strands

Answer 6

c) in both directions, on both strands

The DNA replication machinery can only add nucleotides in the 5′ to 3′ direction, but both strands are replicated using a clever workaround by the cell. In addition, replication forks move in both directions from a replication origin.

Question 7

The bonds that link two DNA strands together are

a) electrostatic interactions
b) phosphodiester bonds
c) phosphoanhydride bonds
d) hydrogen bonds

Answer 7

d) hydrogen bonds

Two DNA strands are held together by many weak noncovalent hydrogen bonds between complementary base pairs.

Question 8

The energy for the polymerization reaction in DNA synthesis is powered by

a) the formation of phosphodiester bonds between nucleotides
b) the breaking of the hydrogen bonds between complementary base pairs
c) the breaking of high-energy phosphate bonds in the deoxynucleotides
d) the association/disassociation cycles of the DNA polymerase enzyme

Answer 8

c) the breaking of high-energy phosphate bonds in the deoxynucleotides

The formation of the phosphodiester bonds requires an input of energy. This is provided by the hydrolysis of the high-energy phosphoanhydride bonds in the incoming nucleotide triphosphates that are the building blocks for the DNA polymer. Pyrophosphate is further hydrolyzed to inorganic phosphate, which makes the polymerization reaction effectively irreversible.

Question 9

Shown below is a replication bubble. At which location is the DNA polymerase adding nucleotides in a continuous manner?

a) C and D
b) A and B
c) B and C

D) A and D

Answer 9

d) A and D

DNA polymerase can only add nucleotides continuously in the 5′ to 3′ direction - when it is moving along the template strand in a 3′ to 5′ direction. This would be the locations marked A and D.

Question 10

How does Polymerase help maintain the accuracy of DNA replication?

a) polymerase fixes its mistakes by adding the right base during the next round of replication
b) after DNA replication is complete, the polymerase re-scans the entire chromosome to check for errors
c) polymerase never makes mistakes; bases are added based on strict complementarity
d) DNA polymerase can cut out improperly base-paired nucleotides and add the correct one during synthesis

Answer 10

d) DNA polymerase can cut out improperly base-paired nucleotides and add the correct one during synthesis

Concurrent with the polymerization reaction, polymerase double-checks the base pair that it just formed and, if it is incorrect, it will snip it out. Then the right base will be added to form a correct base pair.

This proofreading mechanism is possible only for DNA polymerases that synthesize DNA exclusively in the 5’-to-3’ direction. This is because if this “backward” polymerase were to remove an incorrectly paired nucleotide from the 5’ end, it would create a chemical dead end – a strand that could no longer be elongated. Polymerization cannot proceed, as no high-energy bond is available to drive the reaction

Question 11

How are the primers from which DNA synthesis starts different from the DNA itself?

a) The primers have three phosphates on each nucleotide
b) The primers are not properly based paired so they can be removed
c) The primers are made up of RNA not DNA
d) The primers do not have a 3’ –OH

Answer 11

c) the primers are made up of RNA not DNA

RNA primers are made by an enzyme called primase to provide a “seed” from which the DNA polymer can grow. The primer provides a base-paired 3′−OH for starting the reaction. DNA polymerase cannot start a new polynucleotide chain by joining together two triphosphates without the need for a base-paired 3’ end as a starting point.

Question 12

What is the name of the enzyme that fills the sequence gaps after primers are removed from a newly synthesized DNA strand?

a) DNA ligase
b) primase
c) nuclease
d) repair polymerase

Answer 12

d) repair polymerase

After a nuclease removes the primers, there are gaps in the DNA molecule. A repair polymerase can come in and fill the gaps to produce a continuous strand.

Question 13

What is the name of the protein that binds the two separated DNA strands to keep them from base pairing again before they can be replicated?

a) single-stranded binding protein
b) primase
c) sliding clamp
d) helicase

Answer 13

a) single-stranded binding protein

As a replication fork travels along the chromosome, the helicase pries apart the two strands of the double-helix. Single-strand binding protein holds these two strands apart and keeps them from reforming base pairs by binding to the single-stranded DNA.

Question 14

What is the function of a topoisomerase in DNA replication?

a) it separates the DNA strands
b) it binds the template DNA to hold in place
c) it relieves the tension in DNA strands
d) it coils the DNA fro tighter packaging

Answer 14

c) it relieves the tension in DNA strands

The unwinding of the DNA by helicase produces tension in the DNA molecule because the DNA cannot freely rotate. Topoisomerases cut the DNA to allow the DNA to freely rotate and relieve this tension, avoiding supercoils and tangles, and then reseals the DNA.

Question 15

Shown below is the end of a newly replicated chromosome. Which strand will telomerase elongate?

a) both strands
b) the template of the lagging strand
c) the incomplete newly synthesized lagging strand
d) neither of the strands, this is done by polymerase

Answer 15

b) the template of the lagging strand

Telomerase extends telomere repeat sequences at the end of chromosomes by extending the template of the lagging strand. The incomplete lagging strand will be extended by a polymerase, using the extended template.

The strand is incomplete because when the final RNA primer on the lagging strand is removed, there is no enzyme that can replace it with DNA. The template is extended beyond the DNA that is to be copied. The telomerase adds the telomere repeat sequences at the 3’ end of the template strand (carries its own RNA template). After the lagging strand replication is complete, a short stretch of single-stranded DNA remains at the ends of the chromosome, but the newly synthesized lagging strand contains all the information present in the original DNA

Question 16

Which of the following is NOT a common source of DNA damage for cells in our bodies?

a) replication fork problems
b) UV light
c) soap
d) spontaneous loss of amino groups on cytosine

Answer 16

c) soap

DNA is continuously undergoing thermal collisions with other molecules.

Depurination (spontaneous reaction) removes a purine base from a nucleotide (does not break the backbone), giving rise to lesions that resemble missing teeth.

Deamination (spontaneous reaction) is the loss of an amino group from cytosine in DNA to produce the base uracil.

UV light promotes thymine dimers.

Mishaps at the replication fork can cause a double-strand break.

Question 17

What is the name of the type of damage caused by an improperly paired base in the DNA?

a) thymine dimer
b) double-strand break
c) depurination
d) mismatch

Answer 17

d) mismatch

Sometimes mispaired bases escape the proofreading of DNA polymerase during replication. These are called mismatches and can be repaired after replication is complete.

Question 18

What is the first step that must occur to repair damage on one strand of the double helix?

a) The DNA backbone on both strands must becut to make a double-strand break.
b) The damaged region must be removed
c) The gap caused by the damage must be filled
d) The DNA backbone must be sealed

Answer 18

b) the damaged region must be removed.

Repair of damaged bases occurs through three steps. The first step is to remove the damaged base and the neighboring bases. Then, the gap can be filled using the undamaged strand as a template and the backbone sealed by ligase.

Question 19

In bacteria, how does the cell recognize which strand is the newly synthesized strand and thus contains the mismatch?

a) the newly synthesized strand is methylated
b) the newly synthesized strand is unmethylated
c) the sequences on both strands are cut
d) the sequence on the newly synthesized strand is made of RNA

Answer 19

b) the newly synthesized strand is unmethylated

In bacterial DNA, adenine bases are methylated. This occurs slowly, so newly synthesized strands remain unmethylated for a time after replication and give mismatch repair enzymes time to recognize the damaged strand.

Question 20

Which double-strand break repair mechanism is a simple ligation mechanism?

a) DNA mismatch repair
b) nonhomologous end joining
c) DNA ligase
d) homologous recombination

Answer 20

b) nonhomologous end joining

Nonhomologous end-joining is a mechanism for sealing DNA double-strand breaks using DNA ligase. This “quick and dirty” mechanism rapidly seals the break, but it comes with a price: nucleotides are often lost at the site of repair. A nuclease chews back the broken ends to produce flush ends. The flush ends are then stitched together by a ligase.

Homologous recombination is more complicated, with end processing and copying of the damaged region from an intact copy of that sequence.

Question 21

Homologous recombination

Answer 21

Flawlessly repair DNA double-strand breaks

Most often occurs shortly after a cell’s DNA has been replicated before cell division when the duplicated helices are sill physically close to each other.

Question 22

Single nucleotide changes from unrepaired DNA damage that lead to uncontrolled cell division can lead to which of the following conditions?

a) sickle cell anemia
b) aging
c) cancer
d) cell death

Answer 22

c) cancer

DNA damage that is unrepaired can lead to mutations that alter the regularity of cell division. Uncontrolled cell division that occurs at the expense of the organ system or organism is called cancer.

Question 23

Which of the following are critical components of the helicase mechanism of action necessary to unwind DNA?

a) dissociation of the helicase subunits
b) conformational changes of subunits
c) oscillating loops pulling the single-stranded DNA through a central hole
d) binding of four helicase subunits to the double-stranded DNA
e) ATP binding and hydrolysis

Answer 23

b) conformational changes of subunits
c) oscillating loops pulling the single-stranded DNA through a central hole
e) ATP binding and hydrolysis

Six helicase subunits go through cycles of ATP binding and hydrolysis to change conformation. The conformational changes lead to a pulling of the single-stranded DNA through the central hole.

Question 24

Which of the following is a function of the protein component of chromosomes?

a) It helps to translate the genetic message
b) It contains the enzymes that replicate the DNA
c) It carries a portion of the genetic information
d) It packages the DNA strands

Answer 24

d) It packages the DNA strands

Chromosomes are made up of protein and DNA. The DNA carries the genetic information, and the protein component helps store the long DNA molecules in the cell, in addition to controlling access to the DNA molecules.

Question 25

When Griffith injected heat-killed infectious bacteria mixed with live harmless bacteria, he found that the mice died because

a) the live harmless bacteria were transformed into infectious bacteria
b) the heat-killed infectious bacteria were ablet o kill the mice on their own
c) the live harmless bacteria were able to kill the mice on their own
d) the heat-killed infectious bacteria were not actually dead

Answer 25

a) the live harmless bacteria were transformed into infectious bacteria

Griffith found that heat-killed infectious bacteria alone or live harmless bacteria on their own were NOT able to infect and kill the mice. However, when he mixed the two together, the mice died and an infectious strain of bacteria was found in the blood of the mice. This experiment provided the first strong evidence that genes are made of DNA.

Question 26

The type of bond that holds together neighboring subunits in a single strand of DNA is a

a) phosphoanhydride bond
b) hydrogen bond
c) peptide bond
d) phosphodiester bond

Answer 26

d) phosphodiester bond

DNA is a long polymer of nucleotides held together by phosphodiester bonds. This forms a single strand of DNA. The two strands of DNA are held together by hydrogen bonds between the nucleotide bases.

Question 27

The consistent diameter of the DNA double helix arises because of which property?

a) phosphodiester linkages have a consistent bond length
b) base pairing of pyrimidines with purines
c) the use of deoxyribose sugar with a 3’ –OH
d) the antiparallel nature of the two strands of DNA

Answer 27

b) base pairing of pyrimidines with purines

DNA is made of up two strands where purines bases (2 carbon rings) pair with pyrimidines (1 carbon ring). If purines and purines base-paired, or pyrimidines and pyrimidines base-paired, the helix would be wider or narrower (respectively) along its length due to the different sizes of the base combinations.

Question 28

If one end of a DNA strand has a phosphate group on it, the chemical group on the other end must be

a) phosphate
b) methyl
c) carboxyl
d) hydroxyl

Answer 28

d) hydroxyl

The hydroxyl group and the phosphate group of neighboring nucleotides bond together during DNA polymerization, which leads to a polarity of the DNA strand. One end has a phosphate group, the other end has an unbound hydroxyl group.

Question 29

The part of the DNA molecule that carries the information for producing proteins is

a) the 3’ hydroxyl terminus
b) the sequence of the deoxyribose sugar groups
c) the phosphate groups at the 5’ end
d) the order of the nucleotide bases

Answer 29

d) the order of nucleotide bases

The linear sequence of the nucleotide bases in the DNA molecule is what contains the information for making proteins. This is decoded by other molecules that translate this information to link specific amino acids into proteins.

Question 30

The structural feature of DNA that hints at the mechanism for its replication is the

a) double helix
b) constant diameter
c) complementary base pairing
d) antiparallel strands

Answer 30

c) complementary base pairing

The complementary base pairing of A with T and G with C suggests that one strand could be used to make a new copy of the other strand. Although the other answers are features of DNA, they do not directly produce a way for the molecule to be copied.

Question 31

Which of the following is found only in eukaryotic genomes, and NOT in prokaryotic genomes?

a) packaging proteins
b) chromosomes
c) telomeres
d) deoxyribose sugar

Answer 31

c) telomeres

Prokaryotic DNA has the same chemical structure as eukaryotic DNA and is packaged by proteins to fit in the cell. Prokaryotes, however, usually have a single circular chromosome, and therefore no telomeres, which are the specialized sequences at the ends of linear chromosomes.

Question 32

The technique whereby human chromosomes are stained and identified is called a

a) karyotype
b) chromosome display
c) genotype
d) phenotype

Answer 32

a) karyotype

Scientists can stain chromosomes from a human tissue sample in order to identify all the chromosomes. Identification is done from the staining pattern or by using chromosome sequence-specific fluorescent dyes and can be helpful in identifying abnormalities in the size, number, or sequence rearrangement of chromosomes.

Question 33

Which of the following describes the chromosomal makeup of a somatic cell of a human biological male?

a) 23 pairs of autosomes + 1 Y chromosome
b) 23 pairs of autosomes + 1 X chromosome + 1 Y chromosome
c) 22 pairs of autosomes + 1 X chromosome + 1 Y chromosome
d) 23 pairs of autosomes

Answer 33

c) 22 pairs of autosomes + 1 X chromsome + 1 Y chromosome

Humans typically have 46 chromosomes, 22 pairs of autosomes, and two sex chromosomes. Biological females have a pair of X chromosomes, and biological males have an X chromosome and a Y chromosome.

Question 34

___________ is a sequence of DNA that contains the information required for making a particular functional RNA or protein.

a) genetics
b) a genome
c) a gene
d) a chromosome

Answer 34

c) a gene

A gene is a segment of DNA that has the instructions for making an RNA or protein. The genome is the total amount of genetic information carried in the complete set of chromosomes in an organism.

Question 35

What is the relationship between genome size and organismal complexity?

a) the more complex the organism, the larger the genome, without exception
b) the more complex the organism, the smaller the genome, without exception
c) more complex organisms generally have larger genomes but there are many notable exceptions
d) there is no relationship between genome size and organismal complexity, it is random

Answer 35

c) more complex organisms generally have larger genomes but there are many notable exceptions.

More complex organisms have more functions requiring specialized proteins and RNA and thus generally have more genes and larger genomes. However, there are many exceptions to this rule, as many plants and a particular amoeba have much larger genomes than humans.

Question 36

Which of the following is true about “junk DNA”?

a) it is found at the ends of chromosomes from over replication during each cell cycle
b) it serves no biological function and is all just a remnant from evolution
c) it usually codes for proteins but the proteins are nonfunctional in the cell
d) portions of junk sequences are conserved between species and thus may be functional

Answer 36

d) portions of junk sequences are conserved between species and thus may be functional

Junk DNA is just a name given to portions of DNA for which the function is currently unknown. Some of these sequences have sequence similarity between species, suggesting they are conserved through evolution and thus may have unique functions.

Question 37

Which of the following specialized DNA sequences provides an attachment point for the segregation of duplicated chromosomes?

a) nucleosomes
b) replication origins
c) telomeres
d) centromeres

Answer 37

d) centromeres

Centromeres are a sequence of DNA upon which specialized structures are formed to allow the mitotic spindle machinery to attach to each sister chromatid of a duplicated chromosome.

Question 38

What is a function of the nucleolus?

a) to form the nuclear lamina
b) to enclose sex chromosomes into a membrane-bound organelle
c) to assemble ribosomal RNA and proteins into ribosomes
d) to assist in the segregation of chromosomes during mitosis

Answer 38

c) to assemble ribosomal RNA and proteins into ribosomes

The nucleolus is an organization of chromosomal regions that contain genes for ribosomal RNAs. It is a structure large enough to be seen in the light microscope in the interphase nucleus. These RNAs are transcribed in the nucleolus and are assembled with ribosomal proteins to make ribosomes.

Question 39

When are chromosomes in their most compact form?

a) during mitosis
b) during replication
c) they are always equally compacted
d) during interphase

Answer 39

a) during mitosis

During mitosis, the chromatin structure that packages chromosomes gets even more compact than usual. This is to compress the chromosomes into discrete bodies that are easier to separate and organize.

Question 40

Which of the labels in the following figure shows a nucleosome core particle?

Answer 40

B

Approximately 147 bp DNA is wrapped around a core of eight histone proteins to form a “nucleosome core particle.” This is the basic unit of chromatin. The nucleosome is the core particle plus the ~200 bp of linker DNA between core particles.

Question 41

What types of bonds are formed between histone proteins and DNA to form nucleosome core particles?

a) nonpolar covalent bonds
b) phosphodiester bonds
c) electrostatic interactions
d) polar covalent bonds

Answer 41

c) electrostatic interactions

Histone proteins are highly positively charged due to their composition from a large number of basic amino acids. This allows them to interact with the negatively charged DNA backbone.

Question 42

How does methylation of histone tails affect the accessibility of DNA?

a) it relaxes the chromatin to make DNA more accessible
b) histone methylation has no effect on chromatin
c) it can have different effects depending on the location
d) it compacts the chromatin to make DNA less accessible

Answer 42

c) it can have different effects depending on the location

The addition of methyl groups to modify histone tails can have various effects on the chromatin structure, depending on the histone protein modified and location in the protein. Most importantly, these histone modifications serve as docking sites to recruit different regulatory proteins and do not have a direct effect on the chromatin themselves.

Histone-modifying enzymes work in concert with the chromatin-remodeling complexes to condense and relax stretches of chromatin, allowing local chromatin structure to change rapidly according to the needs of the cell.

Question 43

Chromatin-remodeling complexes

Answer 43

protein machines that use the energy of ATP hydrolysis to change the position of the DNA wrapped around nucleosomes, rendering the DNA more accessible (or less accessible) to other proteins in the cell.

During mitosis, many of the complexes are inactivated, which may help mitotic chromosomes maintain their tightly packed structure.

Question 44

Shown below is a schematic of an interphase chromosome. Which region is most likely to contain the highest density of genes?

Answer 44

B) euchromatin

Heterochromatin tends to be relatively gene-poor, whereas euchromatin is relatively gene-rich. This is because heterochromatin is so compact that the gene expression machinery cannot access the DNA in chromatin for transcription.

The most striking example of the use of heterochromatin to keep genes silences is found in the interphase X chromosomes of female mammals.

Question 45

Heterochromatin can spread along a chromosome until it encounters a

a) barrier DNA sequence
b) heterochromatin-specific protein
c) unmodified histone tail
d) histone-modifying enzyme

Answer 45

a) barrier DNA sequence

Heterochromatin spreading can occur when histone-modifying complexes bind to the mark they catalyze and establish the modification in neighboring histones. The modifications that direct the formation of the most common type of heterochromatin include the methylation of lysine 9 in the tail of histone H3. A wave of condensed chromatin propagate along the chromosome until it encounters a barrier sequence. Barrier sequences function to block neighboring histones from the modification activity and stop the spread of heterochromatin.

Question 46

What can happen if heterochromatin spreads inappropriately into an area with active genes?

a) the genes will resist and remain active
b) the active genes can become silenced
c) the region will convert into a telomere
d) heterochromatin cannot spread into areas with active genes

Answer 46

b) the active genes can become silenced

Due to its compact nature, genes in heterochromatin are often inactive because they are not accessible. Heterochromatin can sometimes accidentally spread into regions where it is not meant to and cause silencing of genes in that region. Such inappropriate packaging of genes in heterochromatin can cause diseases like anemia.

Question 47

Which type of noncovalent interaction can involve either the polypeptide backbone or the amino acid side chains?

a) van der Waals attractions
b) hydrophobic forces
c) hydrogen bonds
d) electrostatic interactions

Answer 47

c) hydrogen bonds

The backbone of the polypeptide consists of uncharged polar covalent bonds. Because the bonds in the backbone are neither charged nor hydrophobic (i.e. nonpolar), hydrophobic forces and electrostatic interactions do not involve the backbone. Hydrogen bonds, however, are formed between atoms in the polar covalent bonds found in both amino acid side chains and in the carboxyl group and amino groups of the polypeptide backbone.

Question 48

A folded protein structure with which free-energy (G) value would likely have the most stable conformation?

a) 10
b) 5
c) 15
d) 1

Answer 48

d) 1

Free energy is designated by the letter G. If the free-energy value decreases during protein folding, the reaction is energetically favorable. Proteins generally fold along the most energetically favorable pathway, so the conformation with the lowest free-energy (G) value, 1, is the correct answer.

Question 49

If protein folding is determined by the sequence of amino acids in the polypeptide chain, why are chaperone proteins needed to assist folding in the cell?

a) some proteins cannot fold on their own
b) certain proteins easily aggregate with other proteins
c) proteins constantly unfold and refold
d) protein folding is energetically unfavorable

Answer 49

b) certain proteins easily aggregate with other proteins

All proteins can fold on their own without assistance due to the energetically favorable interactions between amino acid side chains and/or the polypeptide backbone. However, the cytoplasm is crowded and some proteins can form favorable interactions with other proteins before they have a chance to fold. These aggregates could impede proper folding.

Question 50

Hydrogen bonding between N–H and C=O groups of every fourth amino acid within a polypeptide chain results in which type of folding pattern?

a) antiparallel b-sheet
b) parallel b-sheet
c) amyloid structure
d) a-helix

Answer 50

d) a-helix

The hydrogen bonds that form a β-pleated sheet structure occur between the N–H and C=O groups of amino acids in different segments of a single polypeptide chain lying side by side. Amyloid structures are β sheets that interlock with each other through their side chains. The α helices are formed by hydrogen bonds between every fourth amino acid in the primary structure.

Question 51

Which of the following is true about amyloid protein structures?

a) they always cause neurogenerative diseases
b) they are weak and brittle
c) they consist of stacked b-sheets
d) they are made up of helical protein fibers

Answer 51

c) they consist of stacked b-sheets

Amyloid protein structures are β sheets that interlock with each other through their side chains and form stacks. Since they are made from many β sheets, they form strong structures that have many roles in the cell, including the formation of important cellular compartments. However, some amyloid structures, but not all, are abnormal and cause disease.

one good role is that amyloid structures make some of the interiors of intracellular condensates. The stacked beta-sheets come together to produce a “hydrogel” that pulls other molecules into the condensate.

Question 52

A stretch of amino acids in a polypeptide chain that is capable of independently folding into a defined structure is called a

a) domain
b) backbone
c) subunit
d) ligand

Answer 52

a) domain

A domain is a sequence of amino acids in a polypeptide chain that adopts a defined folding pattern based on the interactions of the side chains, as well as contributions from the polypeptide backbone molecules. This is distinct from a subunit, which is a term used for a single, complete polypeptide chain that can interact with other subunits to form a larger complex.

Question 53

Which of the following levels of protein structure involves the interaction of more than one polypeptide chain into a three-dimensional structure?

a) primary
b) secondary
c) tertiary
d) quaternary

Answer 53

d) quaternary

The primary structure is the linear order of amino acids in a polypeptide chain. The secondary structure is the formation of organized arrangements to form segments like α helices and β sheets. The tertiary structure is the overall three-dimensional shape of a protein. The quaternary structure is the assembly of multiple folded polypeptide chains into a larger complex.

Question 54

Which of the following shows protein organizational units in the correct order from smallest to largest?

a) subunit < domain < complex
b) domain < subunit < complex
c) complex < domain < subunit
d) complex < subunit < domain

Answer 54

b) domain < subunit < complex

A domain is a portion of an amino acid sequence that folds into a three-dimensional structure and is the smallest of the organizational units. A subunit is a single, complete polypeptide chain and may be made up of one or more domains. Two or more subunits can assemble into a complex.

Question 55

A binding site on the surface of a protein interacts specifically with another protein through

a) many weak noncovalent interactions
b) a few strong noncovalent interactions
c) many weak covalent interactions
d) a few strong covalent interactions

Answer 55

a) many weak noncovalent interactions

Covalent interactions are rarely used between protein molecules because they are difficult to break, often requiring an enzyme. Interactions between proteins and their partners need to be reversible but very specific. A specific interaction, but one that is able to be altered, can be achieved through the formation of many weak noncovalent interactions between proteins and their binding partners.

Question 56

Disulfide bonds stabilize protein shape outside the cell by

a) covalent bonds between cysteines
b) noncovalent bonds between charged side chains
c) hydrophobic interactions within the lipid bilayer
d) hydrogen bonds with water molecules

Answer 56

a) covalent bonds between cysteines

Using mechanisms such as noncovalent bonds between charged side chains, proteins fold into their final conformation based on their amino acid sequence inside the cell. However, in the harsh environment outside the cell, this structure needs to be stabilized to keep its final form and function. Disulfide bonds are covalent cross-linkages between cysteine groups juxtaposed in the three-dimensional structure, and they act to hold the shape of the protein.

Question 57

The figure below shows a depiction of an antibody. Which label correctly identifies the region(s) of the antibody that contains variable amino acids for the binding of a specific antigen?

Answer 57

A

Unique antigen-binding sites in antibodies are formed by varying the terminal amino acid sequences in the heavy chain and the light chain, which come together into a unique three-dimensional conformation for specific side-chain interactions with the antigen. Since each antibody is made up of two heavy chains and two light chains, there are two antigen-binding sites per antibody.

Question 58

The Michaelis constant (K_M) of an enzyme is a measure of

a) the rate at which the enzyme converts the substrate to product
b) the binding strength of enzyme to substrate
c) the energetic favorability of the reaction
d) the activation energy

Answer 58

b) the binding strength of enzyme to substrate

Enzyme activity is measured using two values. The first is the rate at which the enzyme converts the substrate to the product, which is called Vmax. This value is determined by measuring the rate of product formation in conditions where all enzyme binding sites are occupied by a substrate. The Michaelis constant measures the relative binding strength of the enzyme to a substrate, determined by the concentration of substrate at which the enzyme operates at half of its Vmax.

Question 59

Which of the following is INCORRECT regarding how enzymes lower the activation energy of a reaction?

a) enzymes encourage the substrates to change shape toward a transition state that favors the reaction
b) enzymes align substrates to promote a reaction between them
c) enzymes rearrange electrons in the substrates in a way that favors the reaction
d) enzymes reduce the free energy of the products of the reaction

Answer 59

d) enzymes reduce the free energy of the products of the reaction

Enzymes lower the activation energy of a reaction by promoting changes in the substrate that will encourage the reaction to occur. This can include changes in shape, alignment of two substrates, or changes in the electron density to encourage bond breakage and formation. Enzymes do not change the overall free energy of the reaction.

Question 60

The function of feedback inhibition of an enzymatic pathway is to

a) turn off the synthesis of a product when it is in abundance
b) accumulate large amounts of important biological molecules
c) irreversibly shut down a biosynthetic pathway
d) increase concentrations of intermediates to drive the reaction forward

Answer 60

a) turn off the synthesis of a product when it is in abundance

Feedback inhibition keeps the cell from accumulating unnecessarily large amounts of biological compounds by reversibly inhibiting an enzyme in the biosynthetic pathway for that compound. These mechanisms shut down the enzymes early in the pathway so that wasteful intermediates do not accumulate.

Question 61

When a ligand binds to an allosteric enzyme’s regulatory site, it changes the activity of that enzyme by

a) directly blocking the active site
b) inducing a conformational change
c) acting as a chaperone
d) denaturing the enzyme

Answer 61

b) inducing a conformational change

Once a regulatory molecule or ligand binds to a regulatory site, an allosteric protein undergoes a conformational change that is transmitted to the active site. Inducing the conformational change can change the activity of the enzyme, but it does NOT directly block the active site.

Question 62

How does phosphorylation of a protein affect its activity?

a) always increases the activity
b) always decreases activity
c) never affects the activity
d) could increase or decrease activity

Answer 62

d) could increase or decrease activity

Phosphorylation of amino acid side chains in a protein changes their charge to a negative charge. It could lead to changes in the conformation of the protein, differences in binding to partners, and either increased or decreased activity of an enzyme. Thus, the effects of protein phosphorylation are particular to the protein itself.

Question 63

How does the binding of GTP to a GTP-binding protein affect its activity?

a) always activates the protein
b) always inactivates the protein
c) does not affect the protein
d) could activate or inactivate the protein

Answer 63

a) always activates the protein

Proteins in the GTP-binding protein family are always in their active conformation when GTP is bound, in contrast to phosphorylation of a protein, which can activate or inactivate a protein. The hydrolysis of GTP to GDP returns the protein to the inactive state.

Question 64

Chemical modifications like phosphorylation and acetylation of proteins occur on ___________ of amino acids and can affect the interaction of proteins with other cell components or structures.

a) amino groups
b) carboxyl groups
c) side chains
d) peptide bonds

Answer 64

c) side chains

The chemical modification of phosphorylation and acetylation occurs on the side chain of the amino acid and can affect how proteins interact with other proteins or components of the cell. This effect can occur through the direct alteration of binding sites if the phosphorylation and acetylation occur on an amino acid in a binding site, or through a conformational change that indirectly alters the binding site.

Question 65

Shown below is the ATP hydrolysis cycle of a motor protein. What sentence BEST describes the state of the motor protein in “C”?

a) the hydrolysis of ATP to ADP caused a conformational change in the protein
b) the binding of ADP and inorganic phosphate displaced ATP
c) The release of ADP led to a forward shift of the protein along the filament
d) The binding of ATP activated the protein for movement

Answer 65

a) the hydrolysis of ATP to ADP caused a conformational change in the protein

(A) is the state where no ATP or ADP is bound. The protein is bound to the filament with one of its two filament-binding domains, while the other binding domain is unbound. (B) shows that upon ATP binding, a conformational change moves the unbound domain forward one step to interact with the filament. In (C), ATP is hydrolyzed to ADP, creating a conformational change to release the rear filament-binding domain and bring it forward.

Question 66

Biochemical subcompartments that form inside the nucleus are distinct from their immediate surroundings because of the

a) formation of a lipid membrane
b) high concentrations of interacting proteins and RNA
c) fusion of oil droplets
d) disassembly of scaffold proteins

Answer 66

b) high concentrations of interacting proteins and RNA

Biochemical subcompartments can form inside of organelles by the aggregation of interacting proteins, RNA, and protein complexes that perform a concerted function. These “intracellular condensates” are simply concentrations of molecules; they are not bound by a membrane, nor are they hydrophobic droplets.

Question 67

Which method is used for separating proteins based on specific interactions with other molecules?

a) mass spectrometry
b) gel electrophoresis
c) affinity chromatography
d) x-ray chromatography

Answer 67

c) affinity chromatography

Affinity chromatography separates proteins from one another by using a known molecular interaction with the protein of interest. Gel electrophoresis and mass spectrometry separate protein fragments by size and charge. X-ray crystallography is a method for determining the three-dimensional structure of a protein; it is not used for separation.

Question 68

Which method is most suitable for determining the three-dimensional structure of an extremely large integral membrane protein complex?

a) NMR spectroscopy
b) mass spectrometry
c) x-ray crystallography
d) cryoelectron microscopy

Answer 68

d) cryoelectron microscopy

Large proteins produce confounding signals on NMR spectra, and large integral membrane complexes are hard to crystallize for x-ray crystallography. Structural studies of large integral membrane protein complex could be performed with cryoelectron microscopy. In this method, the purified protein complex is embedded in ice and a beam of electrons is used to collect projected images in multiple orientations.

Question 69

What is a protein family?

a) a set of proteins that have the same number of domains
b) a group of proteins from unrelated species
c) a structurally related group of proteins
d) a collection of proteins that have all been crystallized

Answer 69

c) a structurally related group of proteins

A protein family is a set of proteins that have similar primary structures (amino acid sequences) and thus have related three-dimensional structures. If the structure of one member of a protein family is determined, it can be used to make predictions about the structure of other members of the protein family.

Question 70

Match the following secondary structure characteristics with the correct categories: alpha-helix only, beta-sheet only, both alpha-helix and beta-sheet

a) one full turn every 3.6 amino acids
b) can be formed by many sequences
c) cylindrical structure
d) side chains alternating above and below the structure
e) consists of antiparallel or parallel strands
f) formed by hydrogen-bonding between backbone atoms

Answer 70

a) alpha helix
b) both alpha-helix and beta-sheet
c) alpha helix
d) beta-sheet
e) beta-sheet
f) alpha-helix and beta-sheet

Question 71

Predict what would happen to the secondary structure of a protein if an alcohol molecule that disrupts hydrogen-bonding was added.
Select ALL the correct answers.

a) The alpha helices would unfold, disrupting protein structure.
b) Nothing would happen to the protein; hydrogen-bonding is not important for secondary structure.
c) Individual amino acids would be hydrolyzed from the protein, disrupting protein structure.
d) The beta-sheets would unfold, disrupting protein structure.

Answer 71

A; B

Question 72

Reactions that build larger molecules in the cell are called ___________; reactions that break down molecules into smaller ones are called ___________.

a) anabolic; metabolic
b) catabolic; anabolic
c) metabolic; anabolic
d) anabolic; catabolic

Answer 72

d) anabolic; catabolic

Metabolism is the sum total reactions in a cell, which are of two types. Catabolic reactions are the reactions where cells break down the large molecules in foodstuffs into their constituent building blocks. Anabolic reactions are biosynthetic—they build larger molecules from smaller building blocks.

Question 73

Living systems can generate and maintain order without violating the second law of thermodynamics because they generate

a) macromolecules
b) heat
c) order
d) decreased entropy

Answer 73

b) heat

Even though living systems produce ordered structures from smaller building blocks, this does not violate the second law of thermodynamics because they still create disorder by releasing energy in the form of heat. Heat is energy in its most disordered form.

Question 74

Why is sunlight the ultimate source of energy for nearly all living things on Earth?

a) because all animals must consume plants for obtaining energy
b) because animals obtain heat energy from the sun, which keeps them alive
c) because the Eart makes a full rotation on its axis which provides light once a day
d) because photosynthetic organisms produce food molecules using light energy

Answer 74

d) because photosynthetic organisms produce food molecules using light energy

Organic molecules for energy are obtained by animals through their diet, by consuming either plants or animals that eat plants. Plants produce organic molecules using energy from the sun, thus the source of energy from food molecules in plants is ultimately from the sun.

Question 75

Cellular respiration ___________ energy and produces ___________, whereas photosynthesis ___________ energy and produces ___________.

a) consumes; water + carbon dioxide; produces; oxygen + sugars
b) produces; oxygen + sugars; consumes; water + carbon dioxide
c) consumes; oxygen + sugars; produces; water + carbon dioxide
d) produces; water + carbon dioxide; consumes; oxygen + sugars

Answer 75

d) produces; water + carbon dioxide; consumes; oxygen + sugars

Cellular respiration is the process by which cells break down food molecules for energy, producing water and carbon dioxide as waste products. Photosynthesis uses energy from the sun to make food molecules (sugars) and oxygen.

Question 76

In an enzymatic reaction, a molecule gains an electron. This is known as a(n) ___________ reaction.

a) electronegative
b) oxidation
c) reduction
e) hydrogenation

Answer 76

c) reduction

The gain of electrons by a molecule is called reduction; the loss of electrons is called oxidation.

Question 77

Why is CO2 an end product of cellular respiration?

a) because it captures light energy for photosynthesis
b) because it is the most stable form of carbon in our atmosphere
c) because plants can use it for respiration
d) because it can accept electrons and produce a reduced form of carbon

Answer 77

b) because it is the most stable form of carbon in our atmosphere

Cellular respiration is the breakdown of carbon-containing molecules through oxidation to release energy. Because the most energetically stable form of carbon in the presence of oxygen is carbon dioxide, this is the final end product of respiration.

Question 78

If a reaction is energetically favorable (exergonic), then it must produce a(n)

a) increase in activation energy
b) decrease in heat
c) increase in entropy
d) decrease in reaction time

Answer 78

c) increase in entropy

Reactions that are energetically favorable increase the entropy (or disorder) of the universe. This can be through either the release of free energy in different forms, or the release of energy as heat (energy in its most disordered form).

Question 79

Enzymes increase the speed of a chemical reaction because they

a) lower the activation energy needed to start the reaction
b) make the reaction spontaneous
c) increase the temperature to provide the necessary boost of energy
d) make the reaction more energetically favorable

Answer 79

a) lower the activation energy needed to start the reaction

Enzymes work by lowering the threshold energy input, called the activation energy, needed to start a reaction.

Question 80

If the products of a reaction have more free energy than the reactants, then that reaction is

a) energetically favorable
b) not energetically favorable
c) spontaneous
d) physically impossible

Answer 80

b) not energetically favorable

Reactions that have products with LESS free energy than the reactants will release free energy and create more disorder in the system. These will proceed spontaneously and are energetically favorable. However, when products of a reaction have MORE free energy than the reactants, there is no release of energy, and the reaction is not energetically favorable.

Question 81

What is the value of ΔG at equilibrium?

a) it depends on the concentration of substrates and products
b) positive
c) zero
d) negative

Answer 81

c) zero

When equilibrium is reached, the forward and reverse reactions are equal, so there is no net change in the number of reactants and products. The reaction does not proceed forward or backward at this point and ΔG is equal to zero.

Question 82

What is the relationship between ΔG and ΔG⁰?

a) ΔG⁰ is the free-energy change at standard temperature
b) ΔG⁰ is the free-energy change at standard pressure
c) ΔG⁰ is the free-energy change at standard temperature and concentrations
d) ΔG⁰ is the free-energy change at standard concentrations

Answer 82

c) ΔG⁰ is the free-energy change at standard temperature and concentrations

The value of the free-energy change of a reaction, ΔG, is dependent upon the concentration of reactants and products, as well as the temperature of the system. ΔG⁰ is the standard free-energy change at standard temperature and concentrations to allow comparison of different reactions.

Question 83

For a reaction Y → X with a very high equilibrium constant, K, which of the following is true? You can refer to the following two equations in formulating your answer.

a) The standard free-energy change is large and positive.
b) The forward reaction rate greatly exceeds the reverse reaction rate at equilibrium
c) The amount of product and substrate are equal at equilibrium
d) The amount of product will be higher than the amount of substrate at equilibrium

Answer 83

d) The amount of product will be higher than the amount of substrate at equilibrium

If the equilibrium constant, K, is very high, it means that there is far more of the product X than substrate Y at equilibrium (i.e., when forward and reverse reaction rates are balanced). This corresponds to a free-energy change that is large and negative.

Question 84

For the following uncatalyzed reaction at equilibrium, how would the reaction change if it was now an enzyme-catalyzed reaction at equilibrium?

a) Levels of X would increase
b) Levels of Y would increase
c) Reverse reaction B would decrease
d) Forward reaction A would increase

Answer 84

d) Forward reaction A would increase

Enzymes lower the activation energy for reactions and thus increase the probability that the energy barrier can be surpassed for the reaction to proceed. They do not change the equilibrium point (the relative levels of reactants and product at equilibrium).

Question 85

What is the role of activated carriers in cells?

a) they capture energy from energy-releasing reactions and transfer it to other reactions.
b) they carry energy from anabolic reactions for use in catabolic reactions
c) they are enzymes that catalyze the reactions that break down foodstuffs for energy generation in the cell.
d) they are enzymes that catalyze biosynthetic reactions and make them feasible at the temperature of a cell.

Answer 85

a) they capture energy from energy-releasing reactions and transfer it to other reactions.

Activated carriers collect electrons from the oxidation of molecules in catabolic reactions and transfer them to anabolic reactions that require electrons for the reduction of molecules.

Question 86

Which of the following is a reason why ATP hydrolysis has a negative ΔG⁰?

a) ATP has a high concentration in cells
b) The removal of the phosphate is energetically favorable
c) the products have more free energy than the reactants
d) the reverse reaction is extremely fast

Answer 86

b) The removal of the phosphate is energetically favorable

The products of the ATP hydrolysis reaction—ADP and free phosphate—are more stable and have a lower free energy. Release of free phosphate is energetically favorable because it relieves the repulsion of the negative charges of the neighboring phosphate groups and the aqueous environment to make hydrogen bonds with the phosphate.

Question 87

In the following condensation reaction, how does ATP power the formation of product A–B?

a) ADP binds to reactant A–OH and releases inorganic phosphate
b) A phosphate is first transferred to reactant B–H to form a high energy intermediate
c) a phosphate is first transferred to reactant A–OH to form a high energy intermediate.
d) ATP is split into free phosphate and the energy is released as heat

Answer 87

c) a phosphate is first transferred to reactant A–OH to form a high energy intermediate.

ATP powers the formation of energetically unfavorable bonds between two molecules by first breaking its own high-energy phosphoanhydride bond and transferring the phosphate to another molecule. This new phosphate linkage can then be broken to power the bond formation between molecules.

Question 88

What purpose does NADPH serve in biosynthetic reactions like the one pictured below?

a) accepting electrons from an oxidation reaction
b) releasing energy from high-energy phosphate groups
c) storing energy in high-energy phosphate groups
d) donating electrons for a reduction reaction

Answer 88

d) donating electrons for a reduction reaction

NADPH is an activated carrier of electrons that participates in oxidation-reduction reactions. NADPH is produced in reactions in which 2 electrons are removed from a substrate. The oxidized form of the carrier molecule (NADP+) receives these two electrons as a hydrogen atom plus an electron (a hydride ion). NADPH holds its hydride ion in a high-energy linkage; this ion can easily be transferred to other molecules.

Question 89

Why is the oxidation of NADPH energetically favorable?

a) The biosynthetic reactions that are coupled to NADPH oxidation are energetically favorable.
b) oxidation of NADPH breaks a high-energy phosphoanhydride bond
c) NADPH is the form of the molecule that can gain two high-energy electrons
d) the oxidized form of NADPH is more stable than the reduced form

Answer 89

d) the oxidized form of NADPH is more stable than the reduced form

NADPH carries two high-energy electrons (holds the hydride ion in a high-energy linkage). Losing these electrons in an oxidation reaction is energetically favorable because the oxidized form, NADP+, is more stable than the reduced form (NADPH).

Question 90

Which of the following has a higher concentration in the cell to allow it to be available to accept electrons from the oxidation of food molecules?

a) NADPH
b) NADP+
c) NADH
d) NAD+

Answer 90

d) NAD+

Cells maintain high concentrations of NAD+ so that it is readily available to accept electrons from the oxidation of food molecules in catabolic reactions.

NADPH operates chiefly with enzymes that catalyze anabolic reactions.

NADH has a special role as an intermediate in the catabolic reactions that generate ATP through the oxidation of food molecules.

Inside the cell, the ratio of NAD+ and NADH is kept high, whereas the ratio of NADP+ and NADPH is kept low. This allows plenty of NAD+ to act as an oxidizing agent and plenty of NADPH to act as a reducing agent.

Question 91

Condensation reactions are energetically ___________ and hydrolysis reactions are energetically ___________.

a) unfavorable; favorable
b) favorable; unfavorable
c) favorable; favorable
d) unfavorable; unfavorable

Answer 91

a) unfavorable; favorable

Condensation reactions reduce the disorder in the universe and are energetically unfavorable. Hydrolysis reactions increase the disorder in the universe and are energetically favorable.

Question 92

The chemical reactions that take place in a cell, as compared to the chemistry in the nonliving matter, primarily occur

a) between carbon-based molecules
b) between simple molecules
c) in extreme temperatures
d) in a hydrophobic environment

Answer 92

a) between carbon-based molecules

The chemistry of living things follows all the same chemical and physical laws of the chemistry that occurs in the nonliving world. The difference is that the chemical compounds of life are mostly carbon-based polymers that undergo complex reactions in an aqueous environment.

Question 93

Two atoms held together solely by ionic bonds are referred to as a(n)

a) salt
b) molecule
c) element
d) isotope

Answer 93

a) salt

Atoms are the simplest form of an element that still retain the properties of that element. When multiple atoms come together, it is called a molecule, but when those bonds are solely ionic bonds, it is often called a salt.

Question 94

Given the periodic table below, how many electrons does an oxygen atom have?

Answer 94

8

The atomic number is the number of protons in the nucleus of an atom. For an atom to remain electrically neutral, the number of electrons is equal to the number of protons. Thus oxygen, with an atomic number of 8, has 8 electrons.

Question 95

Glucose has a molecular weight of 180 grams per mole. To prepare 1 liter of a 100 mM solution, how many grams of glucose would you dissolve in water for a total volume of 1 liter?

Answer 95

18

Glucose has a molecular weight of 180 g/mole, which means that dissolving this mass of glucose in a liter would give a 1 M solution. Because 100 mM is 0.1 M, it is 1/10 of the concentration. Using 1/10 of the molecular weight (18 g) in the same volume would give a 1/10 concentration, thus a 0.1 M or 100 mM solution.

Question 96

How many bonds are made by a carbon atom? Why?

Answer 96

4

Carbon has four electrons in its outer shell and it needs eight to fill it. To fill its outer shell, carbon can share electrons with four different atoms, forming four bonds.

Question 97

A molecule with a lot of polar covalent bonds is likely to

a) form ionic bonds
b) be highly soluble in water
c) be hydrophobic
d) contain only atoms with low electronegativity

Answer 97

b) be highly soluble in water

Polar covalent bonds are formed when two atoms unequally share an electron, forming partial positive and partial negative charges at either end of the bond. Water also has partial positive and negative charges, and thus will easily interact with other molecules’ partial charges, thus making it soluble.

Question 98

A base is a molecule that will ___________ a proton in water.

Answer 98

accept

Bases accept protons in water, removing a proton from a water molecule. If the hydrogen atoms in water are more attracted to the electronegative atom of a base, they can be stripped off as a proton.

Question 99

Ionic bonds are ___________ than covalent bonds and are ___________ common in cells.

a) stronger; less
b) weaker; more
c) stronger; more
d) weaker; less

Answer 99

d) weaker; less

Ionic bonds are weak bonds commonly found in salts that form ions in aqueous solution. Because the cell environment is primarily aqueous, ionic bonds are dissociated by interaction with water.

Question 100

Which of the following chemical groups could confer nonpolar/hydrophobic characteristics on the region of a molecule in which it is found?

a) amino group (–NH₂)
b) carboxyl group (–COOH)
c) phosphate groups (–PO₃^2–)
d) methyl groups (–CH₃)

Answer 100

d) methyl groups (–CH₃​)

Bonds between carbon and hydrogen are nonpolar and molecules with mostly nonpolar bonds are hydrophobic. Bonds between H and electronegative atoms like O and N are polar and tend to impart hydrophilic characteristics.

Question 101

Which way do the fatty acid tails of a phospholipid face in a cell membrane?

a) both directions
b) they lay parallel to the direction of the membrane
c) toward the outside of the cell
d) toward the inside of the cell

Answer 101

a) both directions

To form cell membranes, phospholipids form a bilayer, with the fatty acid tails facing each other. Thus, one layer has the fatty acid tails facing inward and the other layer has the fatty acid tails facing outward.

Question 102

Why do unsaturated fatty acid tails form relatively more fluid membranes?

Answer 102

Unsaturated fatty acids have double bonds in their hydrocarbon chains, and so have kinks in them; these keep them further apart. Thus, membranes made from unsaturated fatty acids are more fluid because they cannot pack as closely together as saturated fatty acids, which have no double bonds and are straighter.

Question 103

A nucleotide is different from a nucleoside because a nucleotide has a

a) nitrogen-containing base
b) phosphate group(s)
c) deoxyribose sugar
d) ribose sugar

Answer 103

b) phosphate group(s)

A nucleotide is a sugar (either ribose or deoxyribose), attached to a nitrogen-containing base, and one or more phosphate groups. A nucleoside simply lacks any phosphate groups.

Question 104

What is one of the main differences between DNA and RNA?

Answer 104

RNA uses the sugar ribose, which has a 2′ and a 3′ −OH, whereas DNA uses the sugar deoxyribose, which only has a 2′ −OH. This leads to fundamental differences in the self-replication and stability of these molecules.

Question 105

The technique that scientists used to determine that hemoglobin was a single large macromolecule rather than a loose conglomeration of small organic molecules was

a) electron microscopy
b) DNA sequencing
c) hydrolysis
d) ultracentrifugation

Answer 105

d) ultracentrifugation

Scientists at first did not believe that molecules could be as large as hemoglobin, and thought they were a lot of smaller molecules held together in close association. Separating molecules by size in ultracentrifugation allowed scientists to see if macromolecules were just one molecule of a large size, or many small molecules.

Question 106

All living cells

a) are nearly the same size
b) share the same basic chemistry
c) need to replicate within another cell
d) have the same overall shape

Answer 106

b) share the same basic chemistry

Question 107

Self-replication of living cells occurs through the catalytic action of

a) DNA
b) RNA
c) nucleotides
d) proteins

Answer 107

d) proteins

Question 108

Which of the following microscopy techniques would allow the most detailed image showing the 3-D structure of a flagellum?

a) scanning electron microscopy
b) high-resolution fluorescence light microscopy
c) conventional light microscopy
d) transmission electron microscopy

Answer 108

a) scanning electron microscopy

Of all the microscopy techniques listed, electron microscopy provides the most high-resolution and detailed images. Scanning electron microscopy examines the outer surface of cell structures and gives a 3-D image.

Question 109

Which of the following microscopy techniques allows the observation of living cells (i.e., NOT fixed cells)?

a) there are no microscopy techniques suitable for the observation of living cells
b) fluorescence microscopy
c) scanning electron microscopy
d) transmission electron microscopy

Answer 109

b) fluorescence microscopy

Light microscopy, which also includes fluorescence microscopy, can be used to view living cells. Electron microscopy requires cell fixation and extensive preparation of the samples, which kills the cells.

Question 110

Archaea and bacteria are two separate domains of prokaryotes that were distinguished by examining

a) their cell structures
b) the presence vs absence of a nucleus
c) the presence vs absence of chloroplasts
d) their DNA sequences

Answer 110

d) their DNA sequences

Archaea and bacteria are similar in size and cell appearance under a microscope, but they differ significantly in their genome sequences. Because they are both prokaryotes, they lack a nucleus and any organelles.

Question 111

The main function of the nucleus is to

a) house the DNA in a separate cell compartment
b) make proteins using the genetic material
c) allow cells to segregate their chromosomes
d) provide energy for the cell

Answer 111

a) house the DNA in a separate cell compartment

DNA, in the form of chromosomes, is stored in the nucleus. Keeping the DNA separate from other cell components helps eukaryotic cells regulate access to genetic information.

Question 112

The inner membrane of the mitochondrion appears ___________ because it provides___________.

a) folded; a large surface area for staging energy production processes
b) smooth; a large surface area for staging energy production processes
c) smooth; a smaller surface for the organization of proteins in respiration
d) folded; a smaller surface for the organization of proteins in respiration

Answer 112

a) folded; a large surface area for staging energy production processes

In mitochondria, the outer membrane is smooth, while the inner membrane is highly folded and convoluted. The proteins that are responsible for energy production via respiration are found in the inner membrane.

Question 113

Lysosomes and peroxisomes both perform a series of reactions to break down molecules, but one difference between them is that

a) lysosomes break down toxic molecules
b) peroxisomes digest molecules for reuse
c) lysosomes are large and form an interconnected network
d) peroxisomes contain hydrogen peroxide

Answer 113

d) peroxisomes contain hydrogen peroxide

Both peroxisomes and lysosomes are small organelles that break down materials, but lysosomes generally digest biomolecules like proteins so that the building blocks can be reused. Peroxisomes deactivate toxic materials, often using hydrogen peroxide.

Question 114

What are the different types of filaments in the cytoskeleton and their cellular function?

Answer 114

microtubules – chromosome segregation

intermediate filaments – strength support

actin filaments – cell movement

Microtubules form the mitotic spindle that functions in the accurate chromosome segregation, intermediate filaments have a primary function of providing strength and support to the cell, and actin filaments are involved in dynamic processes such as cell motility.

Question 115

Which of the following is a role of the cytoskeleton in plant cells?

a) actin filaments guide cell movement
b) microtubules form tracks for movement of cell components
c) intermediate filaments form strong links to the plant cell wall
d) intermediate filaments support chloroplast structure

Answer 115

b) microtubules form tracks for movement of cell components

While plant cells are immobile, actin filaments and microtubules still play a major role in the movement of organelles and cell components within the cell. Intermediate filaments are not found in the cytoplasm of plant cells.

Question 116

The rules of which fundamental process were elucidated in the bacterium Escherichia coli (E. coli)?

Answer 116

DNA Replication

The rules behind the process of DNA replication (as well as transcription and translation) were determined by experiments with the bacterial model organism E. coli. These simple prokaryotes are single-celled and do not have a cell cycle or mitosis, thus are best for studies of the most evolutionarily conserved cell processes.

Question 117

The fruit fly, Drosophila melanogaster (D. melanogaster) is an excellent model for studying fundamentals of development because

a) they reproduce every 30 minutes on a plate.
b) many of the genes involved in the development of the fruit fly are also found in humans
c) they are transparent and allow visualization of cell division and movement
d) the precise formation of each of their 959 body cells has been mapped

Answer 117

b) many of the genes involved in the development of the fruit fly are also found in humans

Fruit flies have been studied for over a century, and their complex body plan can be manipulated genetically to uncover which genes govern the development of which parts. Many of these genes have direct counterparts in the human genome.

Question 118

Genome size does not precisely determine gene number. What is a plausible explanation as to why A. thaliana, a plant, has a similarly sized genome as the fruit fly, but twice the number of protein-coding genes?

a) flies are more complex organisms with neural behaviors
b) plants have more genes that were duplicated during evolution
c) plants have more repeated noncoding DNA sequences
d) flies are motile and must need more genes for movement

Answer 118

b) plants have more genes that were duplicated during evolution

Plants have genes that were duplicated during evolution, and thus their protein-coding gene number is higher. Repeated noncoding sequences cannot explain this observation, since these do not get counted in the number of protein-coding genes.

Question 119

What do homologous genes strongly suggest?

Answer 119

They they evolved from a common ancestral gene

Question 120

A segment of DNA in the genome that is not a protein-coding gene

a) can be used to regulate gene activity
b) must encode a functional RNA
c) is junk DNA left over from mistakes in evolution
d) is completely unimportant to the cell

Answer 120

a) can be used to regulate gene activity

The genome of a complex organism is made up of protein-coding genes, sequences that encode functional RNAs, and regulatory sequences used to finely tune gene activity. Other DNA noncoding sequences seem to be unimportant to the cell, but we might just not understand their function yet.

Question 121

protozoan

Answer 121

A free-living, nonphotosynthetic, single-celled, motile eukaryote

Question 122

acetyl CoA

Answer 122

Activated carrier that donates the carbon atoms in its readily transferable acetyl group to many metabolic reactions, including the citric acid cycle and fatty acid biosynthesis; the acetyl group is linked to coenzyme A (CoA) by a thioester bond that releases a large amount of energy when hydrolyzed.

Question 123

Vmax

Answer 123

The maximum rate of an enzymatic reaction, reached when the active sites of all of the enzyme molecules in a sample are fully occupied by substrates

Question 124

turnover number

Answer 124

The maximum number of substrate molecules that an enzyme can convert into product per second.

Question 125

Michaelis constant (K_M)

Answer 125

The concentration of substrate at which an enzyme works at half its maximum velocity; serves as a measure of how tightly the substrate is bound.

Question 126

scaffold protein

Answer 126

Protein with multiple binding sites for other macromolecules, holding them in a way that speeds up their functional interactions.