BIOL 230 Molecular and Cell Biology Exam 3 Smartwork5 Questions - ULL

What generally is the fate of mutations to the genome that have harmful consequences to an organism? Choose one: A. They are usually eliminated from the population by natural selection. B. They give rise to new species. C. They do not occur. D. They are always fixed by DNA repair. E. They tend to persist and spread.

A. They are usually eliminated from the population by natural selection. Harmful mutations do sometimes occur and can cause significant problems to the individual harboring that mutation, which usually leads to reduced reproductive fitness. The reduced ability to reproduce means that most harmful mutations will be eliminated from a population. In contrast, some mutations produce favorable changes and will tend to persist and spread throughout a population. This dichotomy between the reproductive outcomes of individuals that harbor different forms of mutation is the basis for evolution by natural selection.

What type of bond connects base pairs? Choose one: A. hydrogen B. ionic C. van der Waals D. covalent

A. hydrogen A DNA molecule is composed of two polynucleotide chains (DNA strands) held together by hydrogen bonds between the paired bases. Each strand of a DNA double helix contains a sequence of nucleotides that is exactly complementary (due to hydrogen-bonding) to the nucleotide sequence of its partner strand. The sugar-phosphate backbone of each strand is composed of a phosphodiester bond between individual nucleotides.

To begin transcription, eukaryotic RNA polymerase recognizes nucleotide sequences in what region of the DNA? Choose one: A. promoter region B. terminator region C. initiator region D. replication origin E. G-C-rich region

A. promoter region To begin transcription, RNA polymerase recognizes nucleotide sequences in the promoter region of the DNA. The promoter region is within the approximately 50 nucleotides that are before, or upstream of, the transcription start site. RNA polymerase binds tightly to the promoter and opens the double helix, revealing a small region of single-stranded DNA immediately in front of the promoter. One of these strands will serve as the template for RNA synthesis, which will continue along the DNA strand until a separate terminator site is reached at the end of the gene.

At a replication fork, how is the lagging strand synthesized? Choose one: A. without the use of a template B. in the incorrect 3'-to-5' direction C. discontinuously D. continuously E. in the correct 3'-to-5' direction

C. discontinuously At a replication fork, the lagging strand is synthesized discontinuously. The Okazaki fragments are then joined together to form a continuous new DNA strand. In contrast, the leading strand is synthesized continuously and does not need the same extensive ligation of DNA fragments. Like all nucleic acid, polymerization of the lagging strand occurs in the 5'-to-3' direction and never occurs in the opposite direction.

When compared to each other, the two replication forks that form at an origin of replication move in which direction? Choose one: A. in the 5'-to-3' direction B. in the 3'-to-5' direction C. in opposite directions D. toward the template strand E. toward the origin

C. in opposite directions DNA replication begins at an origin of replication where two Y-shaped replication forks form. The replication forks represent the location where the original DNA double helix is being pried apart into two single strands, each of which will function as a template for the replication of a new daughter strand. As replication proceeds, the two forks move away from each other, creating two new DNA molecules, each consisting of a template strand and a new daughter strand. In eukaryotic chromosomes, there are multiple origins of replication, and hence multiple replication forks, as shown in the figure below. Eventually, replication forks meet, and the separate newly synthesized segments are linked together by the enzyme DNA ligase, resulting in a single daughter DNA strand.

In addition to its role in DNA repair, homologous recombination is also responsible for generating genetic diversity during what process? Choose one: A. mitosis B. fertilization C. meiosis D. DNA replication E. independent assortment of chromosomes

C. meiosis During meiosis, homologous recombination helps generate genetic diversity. This recombination event, which is also called crossing-over when it occurs during meiosis, allows for the exchange of genetic information between pairs of replicated chromosomes during the formation of sperm and eggs. The result is gametes that are not identical, in part because any one chromosome is a mosaic of its original diploid pair.

Which biochemical reaction is catalyzed by a ribozyme? Choose one: A. DNA polymerization during DNA replication B. RNA polymerization during transcription in eukaryotes C. peptide bond formation in protein synthesis D. RNA polymerization during transcription in prokaryotes E. peptide bond hydrolysis by proteases

C. peptide bond formation in protein synthesis Peptide bond formation in protein synthesis is catalyzed by a ribozyme. When the tRNA molecules associate with the mRNA, the rRNA of the ribosomal subunits orients the newly incoming amino acid with the growing peptide chain in a way that encourages the addition reaction between the two. Even though the ribosome is composed of both RNA and protein, the peptidyl transferase reaction is accomplished by the RNA, not protein, component of the ribosome.

Proteasomes act primarily on proteins that have been marked for destruction by the attachment of which small protein? Choose one: A. histone B. prion C. ubiquitin D. termination factor E. protease

C. ubiquitin The proteasome is a large, multi-subunit complex that breaks down proteins in the cell. The determination of which proteins get targeted for destruction by the proteasome is carefully regulated using the small protein tag called ubiquitin to identify proteins that should be degraded. Specialized enzymes tag the proteins that are destined for rapid degradation with a short chain of ubiquitin molecules. Damaged proteins or amino acid sequences on short-lived proteins are recognized as ones to be ubiquitylated and degraded in proteasomes.

What is the "central dogma"? Choose one: A. Much like all viruses, all cells use DNA to encode genetic information. B. The ability of a cell to survive depends on the accurate duplication of genetic information carried in its DNA. C. The earliest cells on Earth most likely used RNA to store and copy genetic information. D. Within the cell, genetic information flows from DNA to RNA to protein. E. For a given individual, the genetic information contained in the DNA of every cell is identical.

D. Within the cell, genetic information flows from DNA to RNA to protein. The central dogma states that genetic information flows from DNA to RNA to protein. At each step in this process, one biomolecule serves as a template, or a molecular instruction, for the synthesis of another biomolecule. The genetic code (DNA) is transcribed into RNA by the activity of RNA polymerase. Messenger RNA is translated at the ribosome and serves as a template that directs the placement of amino acids during protein synthesis.

At which site on the DNA of a gene does RNA polymerase release its newly made RNA? Choose one: A. promoter B. poly-A tail C. TATA box D. terminator E. stop codon

D. terminator Signals in the nucleotide sequence of a gene tell RNA polymerase where to start and stop transcription. Transcription is initiated by the promoter region, which is not transcribed. Transcription is stopped at the terminator, which, unlike the promoter, is actually transcribed. The terminator sequence functions to stop transcription by causing RNA polymerase to release both the newly formed transcript and the DNA template.

What does depurination refer to? Choose one: A. the loss of thymine due to damage from UV radiation B. the breaking of the DNA backbone C. the accumulation of mutations and subsequent loss of purity of a nucleotide sequence D. the loss of A or G bases from DNA E. the loss of G or C bases from DNA

D. the loss of A or G bases from DNA Depurination is random, spontaneously occuring event involving the loss of adenine (A) or guanine (G) bases from DNA. Depurination removes a purine base while keeping the phosphodiester bond intact, leaving a gap in the DNA's complementary base pairing. When depurination happens, the bond between the nucleotide sugar and base is the one that breaks; there is no breakage of covalent bonding within the sugar-phosphate backbone of DNA.

What is the main function of the TATA-binding protein? Choose one: A. to inhibit transcription until transcriptional activators bind B. to inhibit DNA replication until S phase C. to help initiate DNA replication D. to promote initiation of transcription

D. to promote initiation of transcription The TATA-binding protein is a subunit of RNA polymerase II that helps initiate transcription. When the TATA-binding protein binds to the TATA sequence in DNA, the protein causes the DNA helix to bend. The kink that occurs is thought to signal assembly of the transcriptional complex and initiation of transcription.

Identify which statement(s) is/are true regarding telomere characteristics. Choose one: A. Telomeres allow duplicated chromosomes to become separated into daughter cells during M phase. B. Telomeres are found in all living cells. C. Telomeres cap the ends of linear chromosomes and prevent them from being recognized by the cell as broken DNA in need of repair. D. Telomeres contain repeated nucleotide sequences that are required to replicate the ends of linear chromosomes. E. All of the above are true regarding telomere characteristics. F. Only statements B and D are true regarding telomere characteristics. G. Only statements C and D are true regarding telomere characteristics.

G. Only statements C and D are true regarding telomere characteristics. (C. Telomeres cap the ends of linear chromosomes and prevent them from being recognized by the cell as broken DNA in need of repair.) (D. Telomeres contain repeated nucleotide sequences that are required to replicate the ends of linear chromosomes. Telomeres contain repeated nucleotide sequences that are required to replicate the ends of linear chromosomes. These telomeric repeated nucleotide sequences cap the ends of linear chromosomes and prevent them from being recognized by the cell as broken DNA in need of repair. Bacteria typically have circular chromosomes, and thus have no telomeres on their chromosome ends.

Match each of the following protein functions during DNA replication with its correct protein on the figure. A. Unwinds the DNA double helix B. Prevents the separated DNA strand from reannealing C. Helps DNA polymerase to be processive (to continue on the DNA) D. Creates the RNA primer E. Synthesizes a new DNA strand F. Leading-strand of DNA G. Lagging-strand of DNA

see image DNA replication begins with helicase (yellow protein) unwinding the two strands of DNA. Single-stranded binding proteins (brown proteins) bind to the single-stranded DNA to prevent the two strands from re-forming before DNA synthesis can occur. This is particularly important for the lagging strand, which is looped out before another primer is added and the next Okazaki fragment is synthesized. Primase (blue protein) adds an RNA primer to each strand. This provides the starting point for DNA polymerase (green proteins) to begin synthesizing the new DNA strand. Primase works repeatedly on the lagging strand (bottom strand in figure) as the DNA continues to be unwound. The sliding clamp (red proteins) binds and holds the DNA polymerase on the DNA strand, allowing it to continue for a longer distance along the DNA before releasing. The leading strand (top in figure) is synthesized continuously. The lagging strand (bottom in figure) is synthesized discontinuously in a series of Okazaki fragments. The RNA primers are removed from the Okazaki fragments and replaced with DNA nucleotides. The ligase closes the final gap between Okazaki fragments so that the two final DNA copies are finished.

To determine if DNA replication is semiconservative, dispersive, or conservative (see image below), Matt Meselson and Frank Stahl performed a classic experiment in which they grew E. coli for many generations in a heavy isotope of nitrogen (15N) and then transferred the bacteria to media containing 14N for a single round of DNA replication. Density of the DNA was determined via centrifugation and the result of intermediate density DNA (50% 14N, 50% 15N within a double helix) was consistent with both semiconservative and dispersive DNA replication. Another round of replication can discriminate between semiconservative and dispersive DNA replication. What kind of DNA is expected if these E. coli are grown for a second generation in14N-containing media? Choose one: A. 50% intermediate density, 50% light density B. 25% intermediate density, 75% light density C. same as result after one round of replication; all intermediate-density DNA D. 50% light density, 50% heavy density

A. 50% intermediate density, 50% light density After a single round of replication following the transfer of the 15N E. coli to 14N media, all the resulting daughter DNA will be intermediate density. After a second round of replication, one new DNA duplex will be light (the one that used the 14N strand as template) and the other will remain intermediate (the one that used 15N strand as template). The intermediate density DNA will never disappear entirely during successive divisions; instead, it will become a smaller and smaller proportion of the DNA strands.

What evidence suggests that the large amount of excess "junk" DNA in a genome may serve an important function? Choose one: A. A portion of "junk" DNA is highly conserved in its DNA sequence among many different eukaryotic species. B. Deletions or mutations that fall within "junk" DNA tend to always be harmful to the organism. C. The nucleotide sequence of "junk" DNA is very similar to that of genes involved in important cell functions. D. The sheer quantity of "junk" DNA suggests it must have some function in the life of the organism. E. All organisms have excess "junk" DNA.

A. A portion of "junk" DNA is highly conserved in its DNA sequence among many different eukarotic species. A portion of "junk" DNA being highly conserved in its DNA sequence among many different eukaryotic species is clear evidence suggesting that the large amount of excess "junk" DNA in a genome may serve an important function. Genetic conservation indicates that a selective pressure is acting on that stretch of DNA, meaning that survival is enhanced in organisms that harbor it. If there were no selective pressure, one would expect mutations to randomly accumulate without consequence over time in the "junk" DNA, resulting in low levels of sequence conservation among disparate organisms, but this is not the case.

The translation of an mRNA begins at the start codon. What is the sequence of this codon? Choose one: A. AUG B. UAG C. UGG D. AGU

A. AUG The translation of all mRNAs begins with a start codon, which has the sequence AUG and also codes for the amino acid methionine. While many amino acids can be coded for by multiple codons, there is only one codon that specifies the placement of methionine, and hence the start codon. This simple fact serves a very important purpose, as it helps ensure that the proper reading frame is established at the beginning of translation.

Which of the following events occur when TATA-binding protein binds to the DNA? Choose one or more: A. Binding leads to assembly of the rest of the transcription complex at the initiation site. B. Four α helices separate the two strands of DNA. C. The DNA backbone is kinked nearly 90 degrees. D. An eight-stranded β-sheet domain of the TATA-binding protein lies on the DNA helix.

A. Binding leads to assembly of the rest of the transcription complex at the initiation site. C. The DNA backbone is kinked nearly 90 degrees. D. An 8-stranded β-sheet domain of the TATA-binding protein lies on the DNA helix. The TATA-binding protein contains an eight-stranded β-sheet domain that lies on the DNA helix. Binding of the protein causes the DNA to bend nearly 90 degrees. This kink in the DNA is thought to lead to assembly of the transcription complex at the initiation site, leading to transcription at the gene. Other subunits separate the two strands of DNA.

In their 1953 paper on the double-helical structure of DNA, Watson and Crick famously wrote: "It has not escaped our notice that the specific pairing we have postulated immediately suggests a possible copying mechanism for the genetic material." What did they mean? Choose one: A. Each strand in a DNA double helix contains all the information needed to produce a complementary partner strand. B. The sugar-phosphate backbone of DNA holds the helix together in a way that allows the genetic information to be copied. C. Sexually reproducing organisms swap their DNA. D. When a cell divides, each DNA helix is split between the daughter cells.

A. Each strand in a DNA double helix contains all the information needed to produce a complementary partner strand. Because each DNA strand in a double helix contains a sequence of nucleotides that is complementary to the sequence of its partner strand, each strand can serve as a template to direct the synthesis of a new strand identical in sequence to its former partner. Watson and Crick were hinting at the "semiconservative" mechanism of DNA replication that we now know occurs.

AMP-PNP is a non-hydrolyzable analog of ATP that can bind to proteins in a similar manner as ATP but is no longer hydrolyzed. Predict what would happen to helicase activity if AMP-PNP were added to a DNA replication reaction. Choose one: A. Helicase would no longer function, since the AMP-PNP is not hydrolyzed. ATP binding and hydrolysis induce the conformational changes that facilitate DNA unwinding by helicase. B. Helicase would function normally, since AMP-PNP is not the same as ATP and would not bind to helicase. C. Helicase would function normally, since AMP-PNP is an analog of ATP and would function in the place of ATP. D. Helicase would function faster than normal, since the AMP-PNP is not hydrolyzed. ATP binding induces conformational changes, but hydrolysis slows DNA unwinding by helicase.

A. Helicase would no longer function, since the AMP-PNP is not hydrolyzed. ATP binding and hydrolysis induce the conformational changes that facilitate DNA unwinding by helicase. ATP hydrolysis is a critical component of the helicase mechanism of DNA unwinding. ATP binding and hydrolysis lead to conformational changes in the six helicase subunits, which facilitates DNA unwinding. Since AMP-PNP is an analog of ATP, it will bind to helicase similar to ATP, but the AMP-PNP will no longer be hydrolyzed to ADP and Pi. The helicase subunits will be locked in their ATP-bound conformation and unable to change conformation. Helicase will no longer function and DNA unwinding will stop.

Which statement is true about the association of histone proteins and DNA? Choose one: A. Histone proteins have a high proportion of positively charged amino acids, which bind tightly to the negatively charged DNA backbone. B. Histone proteins have a high proportion of negatively charged amino acids, which bind tightly to the positively charged DNA backbone. C. Histone proteins insert themselves into the major groove of DNA. D. Each histone protein has a deep groove into which a DNA double helix tightly fits. E. Histone proteins form hydrogen bonds with the nucleotide bases of DNA.

A. Histone proteins have a high proportion of positively charged amino acids, which bind tightly to the negatively charged DNA backbone. Histone proteins have a high proportion of positively charged amino acids, which bind tightly to the negatively charged DNA backbone. Because the entire DNA backbone is negatively charged, the electrostatic interaction between histones and DNA can occur throughout the length of the DNA molecule, regardless of nucleotide sequence. It is important to note that the lysine and arginine residues that interact with the DNA backbone are distant from those in the histone tails that undergo modification to regulate chromatin form.

What is the role of the sliding clamp during replication? Choose one: A. It keeps DNA polymerase attached to the template while the polymerase synthesizes a new strand of DNA. B. It keeps the RNA primer attached to the lagging strand. C. It hydrolyzes ATP to push the DNA polymerase along the DNA template. D. It loads DNA helicase onto the replication fork. E. It unwinds the double helix at the replication fork to allow DNA polymerase to progress along the DNA.

A. It keeps DNA polymerase attached to the template while the polymerase synthesizes a new strand of DNA. Choose one:A. It keeps DNA polymerase attached to the template while the polymerase synthesizes a new strand of DNA.B. It keeps the RNA primer attached to the lagging strand.C. It hydrolyzes ATP to push the DNA polymerase along the DNA template.D. It loads DNA helicase onto the replication fork.E. It unwinds the double helix at the replication fork to allow DNA polymerase to progress along the DNA.

Everyone is exposed regularly to ionizing radiation found in the soil, water, and air and from cosmic rays. In fact, 80% of the ionizing radiation people are exposed to comes from naturally occurring sources. Ionizing radiation can cause double-strand breaks in the DNA. Often, the DNA breaks have missing nucleotides at the broken ends. What type of repair would likely be used, and what would be the result of repairing this type of damage? Choose one: A. Nonhomologous end joining would be used to join the DNA, but errors would still remain. B. Direct repair would be used and would lead to properly repaired DNA. C. Nonhomologous end joining would be used and would lead to properly repaired DNA. D. Direct repair would be used to join the DNA, but errors would still remain.

A. Nonhomologous end joining would be used to join the DNA, but errors would still remain. Nonhomologous end joining is one mechanism for repairing double-strand DNA breaks. The newly joined DNA would still contain errors because nucleotides are often lost and missing around the region of the break. Ionizing radiation is often the cause of missing nucleotides. These missing nucleotides are not replaced in nonhomologous end joining, so the resulting repaired DNA has been altered.

Which of the following statements about nucleosomes is false? Choose one: A. Nucleosomes are found only in mitotic chromosomes. B. A nucleosome consists of DNA wrapped around eight histone proteins, plus a short segment of linker DNA. C. Nucleosomes convert a DNA molecule into a chromatin thread about one-third the length of the initial DNA. D. Nucleosomes can be seen in the electron microscope. E. Nucleosomes represent the first and most fundamental level of chromatin packing.

A. Nucleosomes are found only in mitotic chromosomes. Nucleosomes represent the first and most fundamental level of chromatin packing. A nucleosome consists of DNA wrapped around eight histone proteins, plus a short segment of linker DNA. Nucleosomes package DNA into a chromatin fiber, which greatly condenses the overall length of the DNA molecule. At interphase, DNA typically is found in the chromatin fiber arrangement. However, during cell division, the chromatin fiber condenses even more to form the mitotic chromosome, which is easy to visualize under the light microscope.

Which type of molecule has the potential to perform the catalytic act of reproducing itself? Choose one: A. RNA B. DNA C. proteins D. polysaccharides

A. RNA RNA is a unique molecule in the living world because it can carry out a number of critical functions. Like DNA, it can function as a molecule of heredity. Indeed, this can still be observed in many viruses that are present on Earth today. Because of its single-stranded nature, which allows extensive complementary hydrogen-bonding with itself, RNA can take on a variety of three-dimensional conformations that allow it to have catalytic activity. Therefore, it is widely thought that RNA preceded DNA and proteins in the evolutionary history of life. This idea is shown in broad terms in the figure.

Choose all of the following that correctly describe a characteristic of mismatch repair. Choose one or more: A. Regions of improper base-pairing between parent and daughter strand are detected and repaired. B. DNA polymerase and ligase fill in the gap. C. Helicase unwinds the DNA in the mismatched area. D. Exonuclease removes the newly synthesized DNA in the mismatched area.

A. Regions of improper base-pairing between parent and daughter strand are detected and repaired. B. DNA polymerase and ligase fill in the gap. C. Helicase unwinds the DNA in the mismatched area. D. Exonuclease removes the newly synthesized DNA in the mismatched area. Mismatch repair starts with the detection of a region of improper base-pairing on newly synthesized DNA. Prokaryotes and eukaryotes each have mechanisms for recognizing the daughter strand of DNA, and helicase unwinds the DNA in the mismatched area. Once unwound, an exonuclease removes a region of DNA from the daughter strand around the mismatched area. DNA polymerase then replaces the nucleotides with the correct base-pairing nucleotides, and ligase seals the gap, resulting in a complete double-stranded DNA with proper base-pairing.

Predict what might happen to telomeres in Tetrahymena cells if the Tetrahymena template RNA were mutated from 3'-ACCCCAAC-5' to 3'-AGCCCAAC-5'. Choose one: A. Telomeres would become shorter in every generation compared to normal cells. B. Telomere sequence would be altered to 5'-GGGTTC-3'. C. There would be no changes in telomeres compared to normal cells. D. Telomere sequence would be altered to 5'-CTTGGG-3'.

A. Telomeres would become shorter in every generation compared to normal cells. The AC at the beginning of the telomere RNA template base-pairs with the 3' end of the chromosome to bring the telomerase RNA into correct alignment to begin synthesis of the next telomere repeat. The telomerase 3'-AC-5' base-pairs with the 5'-TG-3' of the chromosome. If the telomere RNA sequence in this region changes to 3'-AG-5', the telomerase RNA will no longer base-pair with the 5'-TG-3' at the end of the chromosome. The telomerase RNA is the required template for addition of new nucleotides to the 3' end of the chromosome. Telomere repeats will not be added if telomerase RNA does not base-pair to the chromosome. The chromosome telomeres will therefore shorten with each generation.

What potential outcomes are possible after replication in a DNA molecule with a depurination modification that is left unrepaired? Choose one or more: A. The DNA molecule contains the normal sequence. B. The DNA molecule is converted into RNA. C. The DNA molecule is missing one nucleotide pair. D. The DNA molecule contains an extra three nucleotide pairs.

A. The DNA molecule contains the normal sequence. C. The DNA molecule is missing one nucleotide pair. Replication of double-stranded DNA that has undergone depurination on an A or G nucleotide on one of the strands can yield a daughter DNA molecule that is either missing one nucleotide pair or that contains the normal sequence. On the strand that harbors the damaged nucleotide, the replication machinery can simply skip over the depurinated base and move to the next nucleotide, thereby producing a daughter DNA molecule that is missing that entire nucleotide pair. On the other DNA strand, the sequence is not modified, so during replication a normal, complementary daughter strand is synthesized, yielding a completely normal double-stranded DNA. It's important to remember that the cell's normal DNA repair machinery usually prevents these mutations from being transmitted to daughter DNA molecules. The two potential outcomes are shown in the figure below.

Some applications in biology, such as polymerase chain reaction (PCR), require melting the DNA double helix into single strands of DNA. This can be accomplished by heating the DNA. As DNA is heated, why does the double helix structure denature into single strands of DNA but not into individual nucleotides? In other words, why do the single strands remain intact even though the double helix does not? Choose one: A. The double helix is held together with hydrogen bonds, while the single strands are linked by phosphodiester bonds. B. The double helix is held together with phosphodiester bonds, while the single strands are linked by hydrogen bonds. C. The double helix structure is important for DNA replication, while the single strands carry the information. D. The single strands are wrapped around histones, which protects them from denaturing.

A. The double helix is held together with hydrogen bonds, while the single strands are linked by phosphodiester bonds. Within each strand of DNA, adjacent nucleotides are connected via covalent phosphodiester bonds. In contrast, relatively weaker hydrogen bonds between complementary base pairs hold the two strands of the double helix together. These structural differences support the function of DNA. DNA must be able to unwind into single strands during DNA replication and transcription, and it is important that each strand remains intact to serve as a template for new DNA strand synthesis or RNA production. The structure of DNA, with covalent phosphodiester bonds linking adjacent nucleotides within a single strand and hydrogen bonds (three for G-C base pairs and two for A-T base pair) linking complementary base pairs and thereby holding the two single strands into a double helix, explains both parts of this question.

Once heterochromatin has been established, it will often spread until it encounters which of the following? Choose one: A. a barrier DNA sequence B. a gene C. a nucleosome D. histone H3 methylated on lysine 9 E. euchromatin

A. a barrier DNA sequence The form of chromatin in which a stretch of an interphase chromosome finds itself largely depends on the modification of histone tails. When heterochromatin-specific proteins bind to a chromosome, they tend to also modify the tails of neighboring nucleosomes, causing the spread of heterochromatin. This spread continues until the remodeling proteins reach a DNA sequence that acts as a barrier. Barrier sequences can function by having a different histone modification that favors the euchromatin conformation. For example, a common barrier DNA sequence modification is the addition of an acetyl group to lysine 9 on histone H3, which blocks the heterochromatin-inducing modification of methylation on that same lysine residue of histone H3.

The loss of purine bases from a strand of DNA is typically caused by which of the following? Choose one: A. a spontaneous chemical reaction B. ultraviolet radiation C. replication error D. deamination E. double-strand DNA break

A. a spontaneous chemical reaction The loss of purine bases from a strand of DNA is typically caused by a spontaneous chemical reaction as a result of the random thermal motion that all molecules inside of cells experience. Depurination is a hydrolysis reaction where the adenine or guanine base is hydrolyzed from the 1' carbon of the nucleotide. The phosphodiester bonds holding together the sugar-phosphate backbone are not affected, leading this form of DNA damage to sometimes be referred to as resembling "missing teeth."

Which of the following does not cause a mutation? Choose one: A. evolution B. failure of DNA repair systems C. metabolic activity D. UV radiation E. replication errors

A. evolution Mutations can arise spontaneously as a result of exposure to harmful conditions or as a result of defects in DNA replication and repair machinery. Any mutations that are not repaired will fall into one of three categories: beneficial, harmful, or neutral (meaning they pose no harm or benefit to the individual). A mutation that is harmful will be selected against and be removed from the population. In contrast, those that are beneficial will confer a survival advantage and not only persist, but become more prevalent in successive generations of the population. This process of preferential survival based on the particular variant of a gene that an individual harbors is the basis of evolution. Therefore, evolution acts upon mutations that arise in a population, but is not the cause of the mutation itself.

What kinds of bonds link the two strands of a double helix to each other? Choose the most specific answer. Choose one: A. hydrogen B. hydrophobic C. ionic D. covalent

A. hydrogen Electrostatic attractions occur between any charged groups on polar molecules. The bonds that hold together two strands of DNA are a form of electrostatic attraction involving a hydrogen atom. Hydrogen bonds form between the nucleotide bases in each complementary DNA strand. On the other hand, a type of covalent bond called the phosphodiester bond forms between the 5' phosphate group of one nucleotide and the 3' hydroxyl group of another nucleotide, linking nucleotides into a polymer, or strand.

What type of enzyme removes damaged DNA from the rest of the DNA molecule? Choose one: A. nuclease B. polymerase C. primase D. ligase E. helicase

A. nuclease During DNA repair, nuclease breaks the phosphodiester bonds that hold the damaged or incorrect nucleotide in the DNA strand (Step 1 in the figure below). Once removed, a repair polymerase replaces the damaged nucleotide, using the other strand as the template to insert the proper nucleotide (Step 2), and then ligase seals the remaining nick by forming the remaining phosphodiester bond (Step 3).

The DNA helicase animation shows the bacteriophage T7 helicase unwinding DNA. Which of the following are critical components of the helicase mechanism of action necessary to unwind DNA? Choose one or more: A. oscillating loops pulling the single-stranded DNA through a central hole B. ATP binding and hydrolysis C. dissociation of the helicase subunits D. conformational changes of subunits E. binding of four helicase subunits to the double-stranded DNA

A. osciliating loops pulling the single-stranded DNA through a central hole B. ATP binding and hydrolysis D. conformational changes of subunits Six identical helicase subunits bind to the double-stranded DNA by surrounding one single strand of the DNA. Each of the six subunits alternates between binding to ATP, ATP hydrolysis leading to binding of ADP and Pi, and empty. The conformation or shape of subunits changes depending on whether the subunit is bound to ATP, ADP and Pi, or empty. It is thought that these conformational changes cause loops in the central hole of the helicase enzyme to bind DNA and may pull the single-stranded DNA through the hole, further unwinding the DNA double strand. The helicase subunits stay associated during the process of unwinding DNA.

Which is not specifically targeted for destruction by the proteasome? Choose one: A. phosphorylated proteins B. damaged proteins C. misfolded proteins D. short-lived, regulatory proteins E. oxidized proteins

A. phosphorylated proteins Phosphorylation alone does not specifically target a protein for destruction by the proteasome. Phosphorylation controls the activity of many proteins in the cell. Although phosphorylation can trigger ubiquitin modification of a protein and its subsequent degradation, the phosphorylation itself is not directly recognized by the proteasome as a signal for degradation. Many other changes to proteins, whether by damage or by the activity of other regulatory proteins, are targeted by the proteasome. Examples include misfolded or damaged proteins as well as regulatory proteins that contain sequences that make them especially prone to tagging by ubiquitin.

What type of enzyme fills in the gap after damaged DNA has been removed? Choose one: A. polymerase B. ligase C. primase D. helicase E. nuclease

A. polymerase A "repair polymerase" fills in the gap after damaged DNA has been removed. This enzyme is very similar to the DNA polymerase that catalyzes the reaction of 5'-to-3' addition of new DNA nucleotides to a growing DNA strand. Additionally, it has the same type of proofreading activity to ensure that the nucleotide that has been inserted to replace the error is, indeed, the correct one.

What is the best term for an RNA molecule that possesses catalytic activity? Choose one: A. ribozyme B. enzyme C. RNase D. ribosomal RNA

A. ribozyme Similar to catalytic proteins that are called enzymes, RNA molecules that possess catalytic activity are called ribozymes. The ribosome, with its catalytic RNA core, is an excellent example of a ribozyme. Ribosomes are complexes of RNA and proteins, and when the entire three-dimensional structure of the ribosome's subunits was determined, it became apparent that the catalytic site for peptide bond formation is far away from the closest protein subunit, indicating that the catalytic activity must reside in the rRNA. Thus, the rRNAs—not the proteins—are responsible for the ribosome's overall structure and its ability to choreograph and catalyze protein synthesis. In addition to ribosomes, many other types of ribozymes performing a variety of functions have been found, as noted in the table shown here.

The assembly of general transcription factors at a eukaryotic promoter typically begins at what site? Choose one: A. the TATA box B. the sigma site C. the TFIID sequence D. the start codon

A. the TATA box Within the promoters of eukaryotic genes is often a region called the TATA box, located about 30 bases upstream of the transcription start site. Its name reflects that it is a short sequence composed largely of T and A nucleotides. To initiate transcription, TFIID first binds to the TATA box, causing a profound bend in the DNA double helix (see figure below) that allows other transcription factors to recognize this site. The assembly of other transcription factors along with RNA polymerase at this site allows for transcription to begin.

Synthetic biologists are trying to create cells from raw material. One step in the process is encapsulating genetic material into a compartment. Researchers of the origin of life think that the earliest cells on Earth may have used RNA as their genetic material instead of DNA. As biologists consider which genetic material to use in creating their synthetic cells, which of the following characteristics of a single-stranded RNA genome should they keep in mind? Choose one: A. RNA is highly chemically stable relative to DNA. B. A newly synthesized RNA strand is not identical to the template strand. C. No enzymes can produce RNA. D. RNA cannot serve as a template.

B. A newly synthesized RNA strand is not identical to the template strand. In contrast to the double helical structure of DNA—in which the result of replication is two new double helices, each identical to the original parent DNA molecule (barring replication errors)—replication of a single-stranded RNA molecule produces an RNA molecule complementary rather than identical to the original strand. This complementary strand can then serve as a template to create an RNA molecule identical to the original RNA strand. Single-stranded RNA thus requires two rounds of replication instead of one.

The Encyclopedia of DNA Elements (ENCODE) project aims to discover functional elements in DNA. One technique employed is DNase-seq. This technique employs a DNase enzyme that digests accessible regions of chromatin, while inaccessible regions remain undigested. The undigested DNA is then sequenced. Which of the following is likely true of the DNase-seq results? Choose one or more: A. Unsequenced DNA regions were likely bound to nucleosomes. B. Centromeric DNA is likely to be sequenced in all samples. C. The sequenced DNA will be the same from cell type to cell type. D. Unsequenced DNA is likely part of euchromatin.

B. Centromeric DNA is likely to be sequenced in all samples. D. Unsequenced DNA is likely part of euchromatin. DNA bound by nucleosomes or tightly packed into heterochromatin (such as centromeres) will not be digested and hence will be sequenced. The digested sequences reflect euchromatin regions of DNA that may be actively transcribed. Since different cell types transcribe different genes, the regions of the genome that are euchromatin vs. heterochromatin may differ between cell types. Along with using DNAse-seq, the ENCODE project is also mapping histone modifications and transcription factor binding along the genome and analyzing spatial relationships among regions from different chromosomes.

How does the tRNA synthetase enzyme charge a tRNA with the correct amino acid? Choose one: A. A single tRNA synthetase in the cell is responsible for recognizing specific anticodon sequences and adding the correct amino acid to each tRNA. B. Each different tRNA synthetase has one region that recognizes the tRNA anticodon and a second region that attaches the matching amino acid to the CCA at the 3' end of the tRNA. C. Different tRNA synthetases recognize specific sequences in the final three nucleotides on the 3' end of the tRNA and attach the matching amino acid. D. Different tRNA synthetases recognize specific sequences in the D and T loops of the tRNA and attach the matching amino acid.

B. Each different tRNA synthetase has one region that recognizes the tRNA anticodon and a second region that attaches the matching amino acid to the CCA at the 3' end of the tRNA. There are many aminoacyl-tRNA synthetases in the cell. Each one makes multiple contacts with the specific tRNA. One part of the enzyme recognizes the anticodon base triplet. A second region carries the matching amino acid that is added to the CCA at the 3' end of the tRNA.

Another step in PCR requires small single-stranded DNA primers to anneal to a target sequence on denatured DNA. Two primers are used. Ideally these primers will have similar melting temperatures. The melting temperature is defined as the temperature at which 50% of the DNA is in the single-stranded form. Which of the following primers will have the highest melting temperature? Choose one: A. G G G A A A T T T C C C B. G G G G A A A T T T C C C C C. A A A A G G G C C C T T T T D. A A A A G G C C T T T T

B. G G G G A A A T T T C C C C Since G-C base pairs contain an extra hydrogen bond, it takes more heat energy to separate them. Thus primers with greater G-C content have a higher melting temperature. There are software programs, such as Primer-BLAST, that aid in designing PCR primers to amplify a given DNA sequence. The structure of DNA, with covalent phosphodiester bonds linking adjacent nucleotides within a single strand and hydrogen bonds (three for G-C base pairs and two for A-T base pair) linking complementary base pairs and thereby holding the two single strands into a double helix, explains both parts of this question.

What is a polyribosome? Choose one: A. a ribosome translating a polycistronic mRNA molecule B. a cluster of ribosomes simultaneously translating the same mRNA, but positioned at different sites along the mRNA C. a ribosome that is in the process of translating an mRNA into a polypeptide D. a mutant ribosome that contains additional ribosomal subunits E. a ribosome translating an mRNA molecule that encodes a membrane-embedded protein

B. a cluster of ribosomes simultaneously translating the same mRNA, but positioned at different sites along the mRNA Even though translation by ribosomes moves quickly, the efficiency of protein synthesis is greatly increased by the ability of multiple ribosomes to bind the same transcript simultaneously. This structure is called a polyribosome, for the many ribosomes bound to a single mRNA molecule. Polyribosomes are found in both bacteria and eukaryotes and are sometimes also called polysomes.

You have joined a lab studying DNA replication in E. coli. The graduate student you are working with has identified a mutation in primase that makes primase very inefficient. Your project is to characterize the cells with this mutation. Predict the defects you would most likely see in the mutant E. coli cells. Choose one or more: A. a delay in the unwinding of DNA B. a longer total time to replicate DNA C. a delay in DNA polymerase beginning synthesis D. inefficient Okazaki fragment joining E. rapid lagging-strand synthesis but slow leading-strand synthesis

B. a longer total tume to replicate DNA C. a delay in DNA polymerase beginning synthesis Primase synthesizes the RNA primer for DNA polymerase to use in starting synthesis. If primase is slow to synthesize the RNA primer, there will be a delay in the start of DNA polymerase synthesizing DNA and it will take longer for DNA replication to finish. This slowdown in replication will be more severe on the lagging strand than the leading strand since primase creates new primers for every Okazaki fragment on the lagging strand. Unwinding of the DNA can still occur because helicase functions before primase. Okazaki fragments can still be joined normally by ligase since that occurs once Okazaki fragments have already formed.

Which option correctly describes the two strands of DNA in a double helix? Choose one: A. held together by covalent bonds between bases B. antiparallel in orientation C. identical D. parallel in orientation

B. antiparallel in orientation The two strands of DNA are considered to be antiparallel in orientation. This means that one strand runs 5' to 3' against the 3'-to-5' orientation of the other strand, as shown in the figure below. In double-stranded DNA, base-pairing is such that A always pairs with T, and G always pairs with C, hence the term "complementary base-pairing." Base pairs are held together via hydrogen-bonding, which allows for both a stable double-stranded structure and the ability to separate the strands to allow for DNA replication and transcription.

Which of the following represents the specialized DNA sequence that attaches to microtubules and allows duplicated eukaryotic chromosomes to be separated during M phase? Choose one: A. nucleosome B. centromere C. histone D. telomere E. centrosome F. mitotic spindle

B. centromere The specialized DNA sequence that allows duplicated eukaryotic chromosomes to be separated during M phase is a centromere. During this stage of the cell cycle, the DNA coils up, adopting a more compact structure and ultimately forming highly compacted, or condensed, mitotic chromosomes. Once the chromosomes have condensed by histones, the centromere allows the mitotic spindle to attach to each duplicated chromosome in a way that directs one copy of each chromosome to be segregated to each of the two daughter cells. The telomeres mark the ends (tips) of each chromosome, and these structures contain repeated nucleotide sequences that are required for the ends of chromosomes to be fully replicated. They also serve as a protective cap that keeps the chromosome tips from being mistaken by the cell as broken DNA in need of repair. Nucleosomes are the basic units of eukaryotic chromosome structure and are composed of histone proteins. Nucleosomes convert the DNA molecules in an interphase nucleus into a chromatin fiber that is approximately one-third the length of the initial DNA. As this histone association with DNA increases, the chromosome becomes much more visible in the microscope.

Researchers often want to isolate a certain type of RNA. For some RNA species, this can be accomplished via affinity chromatography, using beads coated with chains of poly-deoxythymidine (poly-dT). The desired RNA will stick to the beads while unwanted RNAs will flow through the column. The retained RNA can then be eluted. What RNA species can be purified using this method? Choose one: A. eukaryotic rRNA B. eukaryotic mRNA C. bacterial rRNA D. bacterial mRNA

B. eukaryotic mRNA In eukaryotes (but not prokaryotes), mRNAs receive several modifications, such as a 5' methylguanosine cap and a 3' poly-A tail. These modifications increase mRNA stability and aid nuclear export. The poly-A tail allows researchers to isolate eukaryotic mRNAs away from noncoding RNAs, which lack this feature.

Ionizing radiation can also cause damage to the nitrogenous bases of DNA. Repair of the damaged bases takes several steps. Which of the following repair mechanisms can be used to repair the nitrogenous bases of DNA damaged by ionizing radiation? Choose one: A. mismatch repair B. excision repair C. direct repair D. nonhomologous end joining

B. excision repair Excision repair is most often used to repair nitrogenous bases damaged by ionizing radiation. Direct repair is a simple one-step process. In excision repair, the damaged base is flipped out of the helix and removed by a DNA glycosylase, leaving a baseless site called an AP site. An AP endonuclease, and in some cases phosphodiesterase, cut on either side of the gap, and DNA polymerase and DNA ligase repair the gap. The damage from ionizing radiation is severe enough to require the multiple steps in excision repair instead of the simpler direct repair.

Determine whether the following statement is true or false: When a cell divides, its chromatin structure is completely reset. A. true B. false

B. false The ability for a parent cell to faithfully transmit gene expression patterns to its daughter cells is an important feature that allows differentiated tissues in multicellular organisms to exist. There are multiple mechanisms that allow for this "cell memory" to persist from generation to generation, including the maintenance of chromatin structures. An extreme example of this is the X-inactivation that is found in female mammals and is responsible for the coat color phenotype of the calico cat mentioned in the hint. The figure below outlines the inheritance patterns at work in X-inactivation.

Determine whether the following statement is true or false: Homologous recombination occurs only in eukaryotes. This statement is: A. true B. false

B. false The proteins that carry out homologous recombination have been conserved in virtually all cells on Earth. Both bacterial cells and eukaryotic cells can undergo homologous recombination. When bacteria exchange genes by way of horizontal gene transfer, and when some of that sequence integrates into the bacterial chromosome, then homologous recombination has indeed taken place in the bacterial cell. When DNA sequence becomes integrated into the bacterial chromosome, then those genes contained therein become part of the genome of that cell.

What is the general name given to the most highly condensed form of chromatin? Choose one: A. 30-nm chromatin fiber B. heterochromatin C. euchromatin D. nucleosome E. X chromatin

B. heterochromatin Heterochromatin is the name given to the most highly condensed form of chromatin, and it typically makes up about 10% of an interphase chromosome. In mammalian chromosomes, heterochromatin is concentrated around the centromere region and in the telomeres at the chromosome ends. The rest of the interphase chromosome exists in a more relaxed form, called euchromatin. Although we use the term euchromatin to refer to chromatin that exists in a less condensed state than heterochromatin, it is now clear that both euchromatin and heterochromatin are composed of mixtures of different chromatin structures.

In the absence of repair, what would the replication of a double helix containing a mismatch yield? Choose one: A. two DNA molecules containing different mismatches at the site of the original error B. one DNA molecule with the normal sequence and one DNA molecule with a mutated sequence C. one DNA molecule with the normal sequence and one DNA molecule with a mismatch D. two DNA molecules containing the mismatchE. two DNA molecules with a mutated sequence

B. one DNA molecule with the normal sequence and one DNA molecule with a mutated sequence. In the absence of repair, following replication, neither daughter DNA molecule will contain the original mismatch, but one will have a mutated sequence that does not match the original parent sequence. Thus, replication of a double helix containing a mismatch would yield one DNA molecule with the normal sequence and one DNA molecule with a mutated sequence. The mutation in one of the daughter DNA molecules will be permanent and hence replicated each time that DNA molecule is copied.

Nucleotides in each strand of DNA are held together by what type of bonds? Choose one: A. peptide B. phosphodiester C. ionic D. hydrogen

B. phosphodiester The DNA double helix is held together by hydrogen bonds between the nucleotide bases of each DNA strand, but phosphodiester bonds, which are covalent, link the sugars and phosphate groups that form the backbone of each strand of the double helical DNA.

Prokaryotes have chromosomes that are circular in structure. Which of the following would such chromosomes lack? Choose one: A. complementary base pairs B. telomeres C. replication origin D. sugar-phosphate backbone E. DNA double helix

B. telomeres Prokaryotes have chromosomes that are circular in structure, so these chromosomes would lack telomeres, which are a feature of linear chromosomes. While prokaryotes and eukaryotes exhibit differences in the structure of their chromosomes, the structure of the DNA itself is largely the same between these two groups of organisms.

What are the specialized DNA sequences that are at the ends of most eukaryotic chromosomes called? Choose one: A. histones B. telomeres C. centromeres D. nucleosomes E. centrosomes

B. telomeres The specialized DNA sequences that are at the ends of most eukaryotic chromosomes are called telomeres. Telomeres contain repeated nucleotide sequences that are required for the ends of chromosomes to be fully replicated and also serve as a protective "cap" that keeps the chromosome tips from being mistaken by the cell as broken DNA in need of repair. This region of condensed heterochromatin is shown in the figure below. Also visible in this diagram is the centromere, or waist, of the chromosome, which is the attachment point for the spindle fibers that will separate chromosomes during cell division.

Which part of a protein is synthesized by a ribosome first? Choose one: A. It depends on whether the cell is eukaryotic or prokaryotic. B. the N-terminus C. It depends on the protein. D. the C-terminus E. It depends on where the ribosome binds to the mRNA.

B. the N-terminus Remember that the covalent peptide bond that links amino acids into polymers involves the amino group of one amino acid and the carboxyl group of the other. Therefore, in any polypeptide, there will always be a free amino group at one end of the molecule and a free carboxyl group at the other. The nature of the peptidyl transferase activity of the ribosome is such that the amino group of the new amino acid being added reacts with the carboxyl group of the growing polypeptide. Therefore, the first amino acid in the polypeptide (that is, the one specified for by the first codon) will have its amino group remain free. This is referred to as the N-terminus. Additionally, because translation proceeds 5' to 3' along the mRNA molecule, the amino acids coded for at the 5' end of the transcript will be located at the N-terminus of the polypeptide.

Consider the image of a double-stranded polynucleotide in the upper panel and a single-stranded polynucleotide in the lower panel. This image represents the process of _________ , which is an event that occurs __________ in gene expression. Blank #1: A. transcription B. translation C. replication Blank #2: A. early B. late

Blank #1: A. transcription Blank #2: A. early The process in the figure, which represents using DNA as a template to construct RNA, is called transcription. Transcription is when DNA is used as a template for RNA synthesis. The enzyme RNA polymerase catalyzes transcription. Transcription of a gene produces an RNA complementary to one strand of DNA. In the upper panel, the bottom strand of DNA is called the template strand because it is used to guide the synthesis of the RNA molecule. The nontemplate strand of the gene is sometimes called the coding strand because its sequence is equivalent to the RNA product, as shown. Transcription happens much earlier in genetic expression because to translate an mRNA to make protein, the mRNA must be created first.

Consider this image that represents DNA replication along with the "end-replication problem," and then answer the question. Using your knowledge of telomerase enzyme and chromosomal size, choose the statement that correctly describes the information in the figure. Choose one: A. Telomerase activity would solve the end-replication problem, as this enzyme is involved in telomere shortening. B. Telomerase activity would complicate the end-replication problem, as this enzyme is involved in telomere shortening. C. Telomerase activity would solve the end-replication problem, as this enzyme is involved in telomere lengthening. D. Telomerase activity would complicate the end-replication problem, as this enzyme is involved in telomere lengthening.

C. Telomerase activity would solve the end-replication problem, as this enzyme is involved in telomere lengthening. Telomerase activity would solve the end-replication problem because this enzyme is involved in telomere lengthening. Telomerase prevents linear eukaryotic chromosomes from shortening with each cell division. To complete the replication of the lagging strand at the ends of a chromosome, the template strand is first extended beyond the DNA that is to be copied. To achieve this, the enzyme telomerase adds to the telomere repeat sequences at the 3ʹ end of the template strand, which then allows the newly synthesized lagging strand to be lengthened by DNA polymerase. The telomerase enzyme itself carries a short piece of RNA with a sequence that is complementary to the DNA repeat sequence; this RNA acts as the template for telomere DNA synthesis.

Which of the following occurs when a cell repairs a double-strand DNA break by the process of nonhomologous end joining? Choose one: A. The DNA sequence at the site of repair matches that of a homologous chromosome. B. The DNA sequence at the site of repair contains a short segment of telomere DNA. C. The DNA sequence at the site of repair is altered by a short deletion. D. The DNA sequence at the site of repair is altered by a short addition. E. The original DNA sequence at the site of repair is reconstituted with 100% accuracy.

C. The DNA sequence at the site of repair is altered by a short deletion. Cells can repair double-strand breaks in one of two ways: nonhomologous end joining or homologous recombination. When a cell repairs a double-strand DNA break by the process of nonhomologous end joining, the DNA sequence at the site of repair is altered by a short deletion. The process begins by having nuclease enzymes "clean" the broken DNA ends before joining them back together with DNA ligase. The nuclease activity results in a loss of nucleotides at the site of repair. Unlike homologous recombination, which requires an undamaged DNA molecule as a template to correct the damaged DNA, there is no need for DNA polymerase activity in nonhomologous end joining, and hence no nucleotides are added at the site of damage on the DNA molecule.

What problem with replication of linear chromosomes does telomerase address? Choose one: A. The lagging strand continues farther during replication, so chromosomes would lengthen in each replication cycle without telomerase. B. The leading strand stops short of the 5' end during replication, so chromosomes would shorten in each replication cycle without telomerase. C. The lagging strand stops short of the 3' end during replication, so chromosomes would shorten in each replication cycle without telomerase. D. The leading strand continues farther during replication, so chromosomes would lengthen in each replication cycle without telomerase.

C. The lagging strand stops short of the 3' end during replication, so chromosomes would shorten in each replication cycle without telomerase. The lagging-strand synthesis does not reach the 3' end of the chromosome. DNA polymerase requires a 3'-OH for adding additional nucleotides. The RNA primer provides this 3'-OH. The RNA primers are then removed from between Okazaki fragments on the lagging strand, and the 3'-OH on the Okazaki fragment is used to fill in the bases for the removed RNA primer. At the 3' end of the chromosome, there is no additional Okazaki fragment and there is no 3'-OH to add to. The 3' end of the chromosome will not be fully replicated. During each round of DNA synthesis, the lagging strands will shorten. Telomerase solves this problem by adding additional nucleotides to the 3' end of the original DNA on the lagging strand. An additional RNA primer can be added to the now longer strand and used to create additional double-stranded DNA. In this way, the overall chromosome length can be maintained. Synthesis on the leading strand is complete as the DNA polymerase can add nucleotides up to the end of the existing strand.

How do chromatin-remodeling complexes work? Choose one: A. They bind to nucleosomes in the 30-nm fiber and induce another level of packing, obscuring DNA from binding by other proteins. B. They use the energy from GTP hydrolysis to alter the arrangement of nucleosomes, rendering certain regions of the DNA more accessible to other proteins. C. They use the energy from ATP hydrolysis to alter the arrangement of nucleosomes, rendering certain regions of the DNA more accessible to other proteins. D. They add methyl groups to the tails of histones in order to attract other proteins. E. They remove acetyl groups from the tails of histones, rendering the DNA more accessible to other proteins.

C. They use the energy from ATP hydrolysis to alter the arrangement of nucleosomes, rendering certain regions of the DNA more accessible to other proteins. Chromatin-remodeling complexes are multi-subunit protein machines that utilize ATP to reposition DNA along the histone. Using the figure below, you can visualize the action of the chromatin-remodeling complex as moving a string (the DNA) over a spool (the histone). In this figure, the blue bands on the DNA have been included to help show the movement of the DNA over the histone "spool" and in the process, certain regions of the DNA have now become more accessible while others are more sequestered away.

Which chemical group is at the 3' end of a DNA strand? Choose one: A. a carboxyl group B. a phosphate group C. a hydroxyl group D. a nitrogenous base

C. a hydroxyl group A hydroxyl group is located at the 3' end of a DNA strand. The phosphate group on a nucleotide is attached to the 5' carbon of the pentose sugar and reacts with the 3' hydroxyl group to form the phosphodiester bond that links individual nucleotides into a DNA polymer, as shown in the figure below. Because of the nature of the phosphodiester bond, DNA strands have a distinct polarity, with one end harboring a free hydroxyl group (the 3' end) and the other harboring a free phosphate group (the 5' end).

Amino acids are attached to their tRNA molecules by which of the following? Choose one: A. hydrogen bonds B. peptide bonds C. aminoacyl-tRNA synthetases D. RNA ligase

C. aminoacyl-tRNA synthetases For translation to accurately transmit the protein-coding information contained within a gene, there needs to be a direct, linear relationship between the mRNA codon, a tRNA anticodon, and the amino acid attached to the tRNA. The recognition of a codon by the anticodon is accomplished through complementary base-pairing. However, the attachment of the proper amino acid to a tRNA harboring the correct anticodon specificity is accomplished by a group of enzymes known as aminoacyl-tRNA synthetases. Cells have enough different versions of this enzyme so that there is a separate enzyme for recognizing each anticodon and then covalently attaching the appropriate amino acid. Once the amino acid has been linked to the tRNA, the tRNA is said to be "charged" and is ready to bind its specific codon on the ribosome-bound mRNA, resulting in the addition of the proper amino acid to the growing polypeptide chain.

Determine whether the following statement is true or false and why: Homologous recombination occurs only between DNA molecules that are identical in nucleotide sequence without any variation. Choose one: A. true, for all eukaryotic diploid cells B. true, but only for cells that have more than 10 chromosomes C. false, because only similar sequence is needed D. false, because crossing-over does not happen

C. false, because only similar sequence is needed Homologous recombination occurs between DNA molecules that are similar in nucleotide sequence. Homologous recombination requires extensive regions of matching sequence, but the DNA molecules need not be entirely identical. For example, homologous recombination by crossing-over is important during prophase I of meiosis. This contributes to the genetic diversity of gametes. Diploid chromosome content is not a determining factor in whether or not homologous recombination can take place.

The karyotype below was obtained from a human cell. Based on the chromosome spread, what cell type could have been used? Choose one: A. male human liver cell B. egg cell C. female human liver cell D. sperm cell

C. female human liver cell With the exception of the gametes (sperm and eggs) and highly specialized cells that lack DNA entirely (such as mature red blood cells), human cells each contain two copies of every chromosome, one inherited from the mother and one from the father. The maternal and paternal versions of each chromosome are called homologous chromosomes (homologs). The only nonhomologous chromosome pairs in humans are the sex chromosomes in males, who inherit an X chromosome from the mother and a Y chromosome from the father. Females contain two X chromosomes, one inherited from each parent.

In the light microscope, DNA molecules are most visible in which of the following? Choose one: A. during interphase B. when they are in their most extended form C. in a cell that is dividing D. in the form of euchromatin E. when they are stripped of histones

C. in a cell that is dividing In the light microscope, DNA molecules are most visible in a cell that's dividing. During cell division, the mitotic chromosomes are so highly condensed that individual chromosomes can be seen in the light microscope. Each of these chromosomes contains a single, very long DNA molecule. Interphase chromosomes, even though they have regions that are condensed into heterochromatin in addition to less condensed euchromatin, are rarely visible in the light microscope. Indeed, any visibility of interphase chromatin with the light microscope will be a result of the relatively little regions of heterochromatin.

Which of the following best defines a mutation? Choose one: A. change in DNA sequence that causes a change in an amino acid in a protein B. harmful change in a DNA sequence C. permanent change in a DNA sequence D. mistake created by faulty mismatch repair E. by-product of natural selection

C. permanent change in a DNA sequence A mutation is defined as a permanent change in DNA sequence, arising from ineffective DNA repair and/or errors during DNA replication, which could alter an organism's ability to survive and reproduce. Mutations that are not repaired become fixed in the genome of that cell line and are then inherited by cell descendants. Because genetic mutations can impact phenotype, they are propagated or eliminated by natural selection. Changes in DNA sequence can produce small variations that underlie the differences between individuals of the same species; when such changes accumulate over hundreds of millions of years, they provide the variety in genetic material that makes one species distinct from another.

Which model of DNA replication is represented in this image? Choose one: A. conservative model B. dispersive model C. semiconservative model

C. semiconservative model DNA replication is semiconservative because the two parental strands separate and each serves as a template for synthesis of a new strand. Thus, each daughter helix is composed of one old and one new strand of DNA. The first round of replication produces two hybrid molecules, each containing one strand from the original parent and one newly synthesized strand. Dispersive replication predicts that each generation of replicated DNA molecules will be a mosaic of DNA from the parent strands and the newly synthesized DNA. Conservative replication predicts that the parent molecule remains intact after being copied. In this case, the first round of replication would yield the original parent double helix and an entirely new double helix. Neither conservative nor dispersive replication is responsible for the replication of DNA.

The mismatch repair system recognizes mismatched base pairs, removes a portion of the DNA strand containing the error, and then resynthesizes the missing DNA using the correct sequence as a template. But what if the mismatch repair system instead removed a piece of the DNA strand that contained the correct sequence? What would replication of this improperly repaired sequence produce? Choose one: A. two DNA molecules with different mutations B. two DNA molecules that are missing one nucleotide pair C. two DNA molecules bearing the same mutation D. two DNA molecules with a gap where the correct sequence was excised E. one DNA molecule with a mutation and one DNA molecule with a mismatch

C. two DNA molecules bearing the same mutation Mismatch repair fills in the gaps left when nucleotides are removed, so the removal of the correct nucleotide would produce a DNA molecule with an altered sequence, because the new mutation would now be used as a template for "DNA repair." Further replication would produce daughter DNAs with this same mutation. Therefore, replication of this improperly repaired sequence would produce two DNA molecules bearing the same mutation.

Which of the following statements is not true? Choose one: A. A cell can permanently condense and silence an entire chromosome during development. B. A cell will temporarily decondense its chromatin to allow access to specific DNA sequences for replication, repair, or gene expression. C. When a cell divides, its chromatin structures will typically be inherited by its daughter cells. D. A cell will temporarily decondense its chromatin to silence genes during differentiation. E. A cell will temporarily decondense its chromatin to give proteins rapid, localized access to specific DNA sequences.

D. A cell will temporarily decondense its chromatin to silence genes during differentiation. DNA carries a vast amount of information that cells utilize to perform their various and specific functions. Regulation of the genetic information on chromosomes can occur at many steps, with regulation of chromatin structure being fundamental among them. Interphase chromosomes contain chromatin in its most relaxed state, yet much of it remains tightly condensed in a form termed heterochromatin. Genes found within heterochromatin are inaccessible and not expressed, which is most notably observed in X chromosome inactivation in mammalian females. On the other hand, relaxed chromatin, or euchromatin, is accessible to transcription, replication, and repair machinery and genes in this type of chromatin are able to be expressed.

Which of the following is true for most genes? Choose one: A. A gene is a segment of DNA that contains the instructions for making a particular protein. B. A gene is a segment of DNA that contains the instructions for making a particular RNA. C. A gene is a unit of heredity that contains instructions that dictate the characteristics of an organism. D. All of the above are true regarding genes. E. None of the above are true regarding genes.

D. All of the above are true regarding genes. All of these options are correct regarding genes. A gene is a segment of DNA that (1) contains the instructions for making a particular protein, (2) contains the instructions for making a particular RNA, and (3) contains instructions that dictate the characteristics of an organism. These are all interrelated by the central dogma of molecular biology in terms of information flow from gene to RNA to protein. The DNA that makes up a gene is considered the genotype, and if genetic expression takes place, then the protein that is translated from the RNA transcript will dictate phenotype.

Which of the statements below is supported by the Avery, McCarty, and MacLeod experiment and the data in the figure above?Which of the statements below is supported by the Avery, McCarty, and MacLeod experiment and the data in the figure above? Choose one: A. Lipids from one strain of bacteria can alter the phenotype of another strain if those lipids are taken up by the other strain. B. Protein from one strain of bacteria can alter the phenotype of another strain if that protein is taken up by the other strain. C. Carbohydrates from one strain of bacteria can alter the phenotype of another strain if those carbohydrates are taken up by the other strain. D. DNA from one strain of bacteria can alter the phenotype of another strain if that DNA is taken up by the other strain. E. RNA from one strain of bacteria can alter the phenotype of another strain if that RNA is taken up by the other strain.

D. DNA from one strain of bacteria can alter the phenotype of another strain if that DNA is taken up by the other strain. The Avery, McCarty, and MacLeod experiment and the data in the figure demonstrate that DNA from one strain of bacteria can alter the phenotype of another strain if that DNA is taken up by the other strain. The researchers prepared an extract from the disease-causing S strain of pneumococci and showed that the "transforming principle" that would permanently change the harmless R strain pneumococci into the pathogenic S strain is DNA. This was the first conclusive evidence that DNA could serve as the genetic material.

An RNA molecule (for example, rRNA or tRNA) within a cell can fold into complex three-dimensional shapes for which reason? Choose one: A. It is double-stranded. B. It contains the base uracil rather than thymine. C. It contains the sugar ribose rather than deoxyribose. D. It is single-stranded. E. It is more primitive than DNA.

D. It is single-stranded. The product of transcription is a single-stranded RNA molecule that can form hydrogen bonds with itself. Because the nucleotides are not already base-paired, as they would be if the nucleic acid were double-stranded, these intramolecular hydrogen bonds are possible. There are other chemical differences between DNA and RNA, but it is the single-stranded nature of the RNA molecule that dictates this behavior. An example of the hydrogen-bonding that gives rise to three-dimensional shapes is shown in the following figure.

One of the first steps in obtaining a karyotype (such as that shown below of a cancer cell) is treating cells with a drug that stalls cells in mitosis. Why must cells arrest in mitosis for karyotype analysis? Choose one: A. Only in mitosis are homologous chromosomes paired up. B. Only mitotic cells contain homologous chromosomes, because mitosis happens after DNA replication. C. Only mitotic chromosome DNA is separated into single strands, allowing for staining by dyes. D. Only mitotic chromosomes are highly condensed and visible with a light microscope.

D. Only mitotic chromosomes are highly condensed and visible with a light microscope. Karyotype analysis can detect chromosomal abnormalities such as extra or missing chromosomes. Some of these abnormalities are inherited and others arise during carcinogenesis. Mitotic chromosomes are much more highly condensed than interphase chromosomes, so cells are treated with a compound to arrest cells in mitosis and then painted with tags that recognize specific chromosomes. The resulting image is manipulated to place homologous chromosomes together in descending size order (i.e., chromosome number 1 is the longest). The sex chromosomes are shown at the end. This is an image of a normal human karyotype.

Telomerase was first described in the ciliate Tetrahymena thermophila by Elizabeth Blackburn and her student Carol Greider. They, along with Jack Szostak, subsequently won a Nobel Prize for this discovery. As the animation shows, the template RNA sequence in Tetrahymena is 3'-ACCCCAAC-5'. The telomerase protein and RNA template together extend the 3' end of the chromosome by adding 5'-GGGTTG-3' repeats to the chromosome. The complementary strand is then synthesized by DNA polymerase α. Blackburn's lab altered the sequence of the telomerase RNA. Predict what might happen to telomeres in Tetrahymena cells if the Tetrahymena template RNA were mutated to 3'-ACCCCGAC-5'. Choose one: A. Telomeres would become shorter every generation compared to normal cells. B. Telomere sequence would be altered to 5'-GTCGGG-3'. C. There would be no changes in telomeres compared to normal cells. D. Telomere sequence would be altered to 5'-GGGCTG-3'.

D. Telomere sequence would be altered to 5'-GGGCTG-3'. As the Telomere Replication animation shows, the AC at the beginning of the telomerase RNA base-pairs with the existing 3' end of the chromosome. The next bases in the telomerase RNA act as a template for synthesis of a new DNA repeat on the 3' end of the chromosome. The telomerase reverse transcriptase enzyme will add complementary nucleotides to the DNA to match the RNA. If the RNA sequence in the template region changes, the DNA sequence will also change to remain complementary to the template RNA sequence. A cytosine (C) will be added to the DNA to base-pair with the G in the telomerase template RNA. Experiments that mutated the telomerase RNA sequence were completed in the laboratory of Elizabeth Blackburn, who demonstrated that the DNA telomere sequence was changed to be complementary to the altered telomerase RNA template (Yu, G. et al, 1990 Nature 344:126-132).

In the 1920s, bacteriologist Fred Griffith demonstrated that a heat-killed, infectious pneumococcus produced a substance that could convert a harmless form of the bacterium into a lethal one. Fifteen years later, researchers prepared an extract from the disease-causing S strain of pneumococci and showed that this material could transform the harmless R-strain pneumococci cells into the infectious S-strain form. This change to the bacteria was both permanent and heritable, suggesting that this "transforming principle" represents the elusive genetic material of the cells. The researchers subjected their extract to a variety of tests to determine the chemical identity of the "transforming principle." In one experiment, they treated the material with enzymes that destroy all proteins. This treatment did not affect the ability of the extract to transform harmless bacteria into an infectious form. From this result, what could the researchers conclude? Choose one: A. The transforming principle is not DNA. B. The transforming principle is not genetic material. C. DNA acts as the genetic material. D. The genetic material is not protein. E. Proteins act as the genetic material.

D. The genetic material is not protein. Because the transforming principle caused a permanent and heritable change in the bacteria, it was considered to represent the genetic material, which is why the researchers were intent on determining its chemical identity. The results ruled out protein as the genetic material because experimentally destroying the proteins had no effect on the ability of the extract to transform harmless bacteria into an infectious form. However, these results did not conclusively prove that the genetic material is DNA.

The genetic code was originally deciphered, in part, by experiments in which synthetic polynucleotides with repeating sequences were used as mRNAs to direct protein synthesis in cell-free extracts. Under these conditions, ribosomes could be made to start translation anywhere within the RNA molecules, with no start codon necessary. What peptide would be made by translation from a synthetic mRNA made entirely of adenine (poly-A)? Choose one: A. a polymer of alanine: Ala-Ala-Ala... B. the peptide Met-Lys-Lys... C. a polymer of methionine: Met-Met-Met... D. a polymer of lysine: Lys-Lys-Lys... E. a polymer of phenylalanine: Phe-Phe-Phe...

D. a polymer of lysine: Lys-Lys-Lys... The genetic code was originally deciphered, in part, by experiments in which synthetic polynucleotides with repeating sequences were used as mRNAs to direct protein synthesis in cell-free extracts. Under these conditions, ribosomes could be made to start translation anywhere within the RNA molecules, meaning that no start codon was necessary. A polymer of lysine (Lys-Lys-Lys...) would be made by translation from a synthetic mRNA made entirely of adenine (poly-A). Because no start codon was necessary, the ribosome could recognize any of the three reading frames of this synthetic polynucleotide (...AAAAAAAAAAAAAAA...) and begin translation. Because the synthetic polynucleotide is just repeating adenines, all reading frames are "AAA," which codes for lysine. Similarly, a synthetic poly-U mRNA as shown below would yield a polymer composed entirely of phenylalanine.

When does homologous recombination most likely occur in order to flawlessly repair double-stranded DNA breaks? Choose one: A. after the broken ends have been "cleaned" by a nuclease B. when the damage is overlooked by the fast-acting mismatch repair system C. whenever a double-strand break is detected D. after the cell's DNA has been replicated E. before DNA is replicated, to avoid propagating the error

D. after the cell's DNA has been replicated To make an accurate repair in a DNA molecule, a template with the correct sequence is necessary. This is relatively simple to do with sequence errors on a single strand, because the complementary strand can dictate the proper nucleotide to insert. However, when there is a double-stranded break, the repair is more serious. Nonhomologous end joining can repair these types of breaks, but because there is no template to dictate the correct sequence, this type of repair results in a loss of some nucleotides at the repair site. If a double-stranded break happens shortly after DNA replication but before the cell has divided, homologous recombination can repair the error while maintaining fidelity to the original DNA sequence. This type of repair is complex and is outlined in the figure below. Briefly, a short stretch of the broken DNA molecule is digested by nuclease enzymes to create free 3' ends, which can "invade" the undamaged homologous chromosome and bind to its complementary sequence. Upon doing so, the undamaged DNA serves as a template for DNA polymerase to resynthesize the broken DNA strand, which then releases and repairs with its original partner strand. Remaining gaps in the broken strand are now filled and then ligated to complete the repair.

The tails of the core histone proteins can be chemically modified by the covalent addition of what type of chemical group? Choose one: A. acetyl B. methyl C. phosphate D. all of these E. none of these

D. all of these The tails of the core histone proteins can be chemically modified by the covalent addition of methyl, acetyl, or phosphate groups. Each modification alters the physiology of the histone by modulating the packing of the chromatin fiber. Acetylation of histone tails, for instance, can reduce the affinity of the tails for adjacent nucleosomes, thereby promoting the euchromatin form. Alternatively, methylation of histone tails has the opposite effect and promotes condensation into heterochromatin.

What is the term that describes the complex of DNA and proteins that makes up a eukaryotic chromosome? Choose one: A. centromere B. centrosome C. centriole D. chromatin

D. chromatin The complex of DNA and proteins that makes up a eukaryotic chromosome is called chromatin. Chromatin functions to condense the long, linear DNA into a more compact form. The fundamental unit of chromatin is the nucleosome, which is composed of the DNA molecule wrapped tightly around histone proteins. Nucleosomes appear like "beads on a string" and can be further condensed into more compact arrangements.

At a replication fork, how is the leading strand synthesized? Choose one: A. in the incorrect 5'-to-3' direction B. in the correct 3'-to-5' direction C. without the use of a template D. continuously E. discontinuously

D. continuously At a replication fork, the leading strand is synthesized continuously. A sliding clamp protein allows the DNA polymerase to move along the leading-strand template without falling off. In constrast, the lagging strand is synthesized discontinuously with multiple Okazaki fragments that need to be ligated together. All DNA synthesis involves a template strand and occurs in the correct 5'-to-3' direction as nucleic acids cannot be synthesized in the 3'-to-5' direction.

Which sugar is not readily made from formaldehyde in experiments simulating conditions on primitive Earth? Choose one: A. ribose B. glucose C. fructose D. deoxyribose

D. deoxyribose Many carbohydrates are readily made from formaldehyde in experiments that simulate predicted conditions on primitive Earth. However, this is not true for deoxyribose. Ribose, like the simple carbohydrates glucose and fructose, is easily formed from formaldehyde (HCHO), which is believed to be an early organic molecule found on Earth before life arose. However, even though deoxyribose differs from ribose only by its lack of a single hydroxyl group, it is harder to make. In living cells found on Earth today, deoxyribose is only formed from an enzyme-catalyzed reaction that uses ribose as its substrate. This leads to the prediction that ribose is a more ancient sugar than deoxyribose, and by extension, it also means that RNA likely appeared in living organisms before DNA.

What type of enzyme seals the newly added (repaired) DNA to the rest of the DNA molecule? Choose one: A. helicase B. DNase C. polymerase D. ligase E. primase

D. ligase Ligases join DNA fragments by catalyzing the formation of phosphodiester bonds between them, during both DNA replication and mismatch repair. During replication, ligase joins the individual Okazaki fragments together on the lagging strand. During mismatch repair, ligase links the newly inserted nucleotide to the rest of the strand. While similar to DNA polymerase in its ability to form phosphodiester bonds, ligase is different in that it doesn't add free nucleotides to the end of a growing strand of DNA; rather, it links together nicks in the backbone of existing DNA molecules.

What structure in an interphase eukaryotic cell is the site of ribosomal RNA transcription? Choose one: A. nucleosome B. nuclear lamina C. ribosome D. nucleolus

D. nucleolus The nucleolus is the obvious, large, dark-staining region found in most eukaryotic nuclei and is the site of ribosomal RNA transcription. In humans, the genes that code for ribosomal RNA are found on the tips of five different chromosome pairs, which are brought together in proximity to one another, as shown in the figure below. Proteins involved in ribosome formation are also found in this area. This dense collection of protein, DNA, and RNA results in the characteristic dark appearance of the nucleolus.

In the living cell, histone proteins pack DNA into a repeating array of DNA-protein particles called what? Choose one: A. heterochromatin B. beads on a string C. euchromatin D. nucleosomes E. octamers

D. nucleosomes Histone proteins pack DNA into a repeating array of DNA-protein particles called nucleosomes, which are the fundamental level of chromatin organization. Nucleosomes convert the DNA molecules in an interphase nucleus into a chromatin fiber that is approximately one-third the length of the initial DNA. These chromatin fibers, when examined with an electron microscope, contain clusters of closely packed nucleosomes. Although long strings of nucleosomes form on most chromosomal DNA, chromatin in the living cell rarely adopts the extended beads-on-a-string form; however, chromatin fibers that have been experimentally unpacked (or decondensed) after isolation do show the "beads-on-a-string" appearance of the nucleosomes. During cell division, chromatin fibers condense even more, resulting in the distinct mitotic chromosome that is readily observed under the light microscope.

Which structure is normally on the 5' end of a DNA strand? Choose one: A. sulfur group B. hydroxyl group C. nitrogenous base D. phosphate group

D. phosphate group On the 5' carbon of the pentose sugar (deoxyribose), a phosphate group is attached. The hydroxyl group of the sugar molecule is positioned on the 3' carbon, while the nitrogenous base is positioned on the 1' carbon. The differing chemical groups on the sugar molecules that compose DNA, together with the nature of the phosphodiester bond that links nucleotide monomers into a DNA polymer, result in the 5'-to-3' polarity that is used to describe DNA.

Researchers sought to grow blood vessels in a lab for implantation into patients with clogged coronary arteries. To avoid immune rejection, they wanted to grow vessels by starting with patients' own vascular smooth muscle cells (SMCs). However, most patients requiring coronary artery replacement are elderly, and their vascular SMCs do not divide often enough to grow a new vessel. To get around this problem, the researchers introduced into the cells the gene coding for a particular protein in a format through which the gene would be transcribed and the protein produced. What gene was introduced into the elderly patients' SMCs to allow continued cell divisions? Choose one: A. primase B. DNA ligase C. DNA polymerase D. telomerase

D. telomerase In linear eukaryotic chromosomes, the lagging strand shortens during each cell division. The chromosomes end in telomeres, repetitive DNA sequences that do not code for proteins. Once chromosomes reach a critically short state, division ceases in order to prevent loss of coding DNA sequences and recognition of the chromosome ends as "broken." The enzyme telomerase, expressed in cells that need to keep dividing throughout the lifetime of an organism, extends telomere sequences. The article "Relevance and safety of telomerase for human tissue engineering" showed that telomerase expression allowed for continued cell proliferation without leading to a cancer phenotype.

In 1952, Alfred Hershey and Martha Chase, working with a virus called T2, conducted what is now considered a landmark experiment to determine whether genes are made of DNA or protein. When this virus, which is made entirely of DNA and protein, infects E. coli, it injects its genetic material into the bacterial cell, leaving the empty virus head stuck to the cell surface. To determine whether it was DNA or protein that enters the infected bacterial cell, the researchers radioactively labeled one batch of T2 with the isotope 35S, which resulted in radiolabeled viral proteins, and a second batch of T2 with 32P, which resulted in radiolabeled DNA. They then incubated the radioactive viruses with E. coli. After allowing a few minutes for the viruses to transfer their genetic material to the bacterial cells, the researchers used a blender to shear the empty virus heads from the bacterial cell surface. They then used a centrifuge to separate the infected bacteria from the empty virus heads: spinning the sample at high speed caused the heavier, infected bacteria to pellet at the bottom of the centrifuge tube, while the lighter, empty virus heads remained in solution. Using this protocol, what should the researchers have seen in terms of the distribution of radioactivity in the centrifuged sample? Choose one: A. 35S in the pellet, 32P in the solution B. no radioactivity in the pellet C. an equal amount of 35S and 32P in the pellet and the solution D. no radioactivity in the solution E. 32P in the pellet, 35S in the solution

E. 32P in the pellet, 35S in the solution For the Hershey and Chase experiment that was performed in 1952, radioactivity was expected in the solution and also in the pellet. The type of radioactivity in each is what differed and conclusively elucidated that DNA was indeed the genetic and hereditary material. They found the distribution of radioactivity in the centrifuged sample to be 32P in the pellet and 35S in the solution. This demonstrated that DNA is the genetic material since the viruses inject their genetic material into the bacterial cells. The head of the virus is composed mainly of protein; thus the 35S remained with the empty virus heads in the solution. The success of the experiment depended on the shearing off of the empty virus heads from the surface of bacterial cells. Results would not have been as clear had these viral particles remained attached to the bacterial cell surface.

The production of a continuous new strand of DNA using the many separate Okazaki fragments (in other words, the joining of the already made fragments) found on the lagging strand requires all of the following except which one? Choose one: A. nuclease B. ATP C. repair polymerase D. DNA ligase E. DNA primase

E. DNA primase The production of a continuous new strand of DNA from the many separate Okazaki fragments found on the lagging strand does not require DNA primase. Primase is instead needed to place the RNA primer used to initiate DNA synthesis, whereas the joining of Okazaki fragments occurs after their synthesis is completed. All of the other enzymes and molecules listed play a role in linking the separate Okazaki fragments into a single, continuous strand.

If DNA replication were conservative (although we know it is not), what would Meselson and Stahl have seen following the first round of replication in E. coli that had been switched from a heavy (15N-containing) nutrient medium to a light (14N-containing) nutrient medium? Choose one: A. All the DNA would be light. B. All the DNA would be heavy. C. Half the DNA would be heavy, and the other half would be intermediate in density. D. Half the DNA would be intermediate in density, and one-quarter would be heavy while one-quarter would be light. E. Half the DNA would be heavy, and the other half would be light.

E. Half the DNA would be heavy, and the other half would be light. If DNA replication were conservative, Meselson and Stahl would have observed half the DNA to be heavy and the other half to be light following the first round of replication in E. coli that had been switched from a heavy (15N-containing) nutrient medium to a light (14N-containing) one. In the conservative model, the heavy, parent molecule would remain intact after being used as a template to produce the light, newly synthesized DNA. DNA intermediate in density must contain equal amounts of heavy and light isotopes—a feature not predicted by the conservative model of replication.

How does ultraviolet radiation in sunlight typically damage DNA? Choose one: A. It removes bases from nucleotides in DNA. B. It converts cytosine into uracil. C. It breaks the sugar-phosphate backbone of DNA. D. It breaks hydrogen bonds between the two strands of DNA. E. It causes two adjacent pyrimidine bases to become covalently linked.

E. It causes 2 adjacent pyrimidine bases to become covalently linked. Ultraviolet radiation in sunlight typically damages DNA by causing two adjacent pyrimidine bases to become covalently linked. These covalently linked thymine bases are called thymine dimers, and their structure is shown in the figure below. Not surprisingly, this type of mutation typically appears in the skin, but under normal circumstances it is repaired by mismatch repair mechanisms. Individuals with the disorder xeroderma pigmentosum cannot repair thymine dimers, leaving them highly susceptible to skin cancer.

In the late 1920s, bacteriologist Fred Griffith was studying Streptococcus pneumoniae. This bacterium comes in two forms: one that is highly infectious (called the "S strain" because it forms colonies that appear smooth when grown on a nutrient plate in the lab) and one that is relatively harmless (called the "R strain" because its colonies appear rough). When injected into mice, the S strain is lethal, whereas the R strain causes no ill effect. Griffith confirmed that when the S strain is killed by heating, it is no longer infectious. But he then discovered that if he injected mice with both the heat-killed S strain pneumococci and the live, harmless R strain bacteria, the animals died of pneumonia. Furthermore, their blood was swarming with live, S strain bacteria that, when grown in culture, remained infectious and lethal. Based on these results, what could Griffith conclude? Choose one: A. The infectious S strain of bacteria cannot be killed by heating. B. The R strain of bacteria is more deadly than previously thought. C. DNA is conclusively confirmed as the genetic material of cells. D. S. pneumoniae is a poor choice for investigating the molecular basis of heredity, as most bacteria behave unpredictably in the laboratory. E. Some substance in the infectious S strain can change the harmless R strain into the more lethal form.

E. Some substance in the infectious S strain can change the harmless R strain into the more lethal form. Griffith's results opened the door to experiments that would demonstrate that DNA is the genetic material. But his work did not, on its own, provide information about the identity of the substance that carries hereditary information. Based on his results, Griffith could conclude that some substance in the infectious S strain could change the harmless R strain into the more lethal form. The Griffith experiments were performed in the early 1900s, well before the Watson and Crick publication, along with a series of other publications including those by Avery, McCarty, and MacLeod and by Hershey and Chase, all of which led to the acceptance that DNA was the hereditary biomolecule.

Where is heterochromatin not commonly located? Choose one: A. telomeres B. centromeres C. silenced X chromosomes D. gene-poor regions of chromosomes E. chromosomal regions carrying genes that encode ribosomal proteins

E. chromosomal regions carrying genes that encode ribosomal proteins Chromosomal regions carrying genes that encode ribosomal proteins are active in gene transcription; therefore, you would not expect to see heterochromatin in these areas. The genes that encode ribosomal proteins are highly expressed because cells require large numbers of ribosomes to carry out protein synthesis. Thus, these chromosomal regions, which cluster together in the nucleolus, tend to be less tightly condensed. Telomeric DNA and centromeric DNA do not encode any genes, and as such, these regions are commonly heterochromatic like an inactivated X chromosome.

When DNA replication proceeds along a template, which of the following best describes the directionality of synthesis? Choose one: A. from the centromere to the telomeres B. in the 3'-to-5' direction C. in both the 3'-to-5' and the 5'-to-3' directions D. from telomere to telomere E. in the 5'-to-3' direction

E. in the 5'-to-3' direction DNA synthesis always covalently links nucleotides in the 5'-to-3' direction, meaning that nucleotides are added to the 3' end of a growing DNA strand. DNA replication does proceed in both directions from an origin of replication, but because of the antiparallel arrangement of the individual strands in double-stranded DNA, the newly synthesized strand always grows by adding new nucleotides to the free 3' end of the molecule.

What structure is responsible for selecting and transporting only properly processed eukaryotic mRNAs into the cytoplasm? Choose one: A. snRNPs B. spliceosome C. RNA polymerase D. ribosomes E. nuclear pore complex

E. nuclear pore complex The nuclear pore complex is the gateway for molecules to move between the nucleus and the cytosol, and is therefore responsible for allowing only properly processed mRNA molecules to enter the cytosol. A properly spliced and modified mRNA molecule is able to bind a cohort of proteins, including cap-binding protein, poly-A-binding protein, and others that direct it to and permit its passage through the nuclear pore complex, as shown in the figure below. Transcripts with improper splicing or other defects will be unable to pass through the pore complex and will remain in the nucleus, where they will be degraded and the nucleotides recycled for use in other transcripts.

What does each eukaryotic chromosome contain? Choose one: A. one long ribonucleotide sequence B. one single long gene C. one long DNA strand D. one long sequence of amino acids E. one long double-stranded DNA molecule

E. one long double-stranded DNA molecule Genomic information is encoded as DNA, not RNA or amino acids, so there are no ribonucleotides or amino acids that make up the nucleotide sequence of the eukaryotic chromosome. Each eukaryotic chromosome, even though it may contain thousands of individual genes, is composed of just one long double-stranded DNA molecule.

Identify the iteration of the Griffith experiment that conclusively demonstrated that transformation must have taken place. Drag the label to the target next to this iteration. A. transformation

see image Griffith showed that heat-killed infectious bacteria can transform harmless live bacteria into pathogens. This was demonstrated in the lower panel of the image. Streptococcus pneumoniae comes in two forms that differ both in their microscopic appearance and in their ability to cause disease. Cells of the pathogenic strain, which are lethal when injected into mice, are encased in a slimy, glistening polysaccharide capsule. When grown on a plate of nutrients in the laboratory, this disease-causing bacterium forms colonies that look dome-shaped and smooth; hence, it is designated the S form. The harmless strain of the pneumococcus, on the other hand, lacks this protective coat and instead forms colonies that appear flat and rough; hence, it is referred to as the R form. As illustrated in this diagram, Griffith found that a substance present in the pathogenic S strain could permanently change, or transform, the nonlethal R strain into the deadly S strain.

Label the following image of a tRNA with the appropriate descriptor that matches each region of the tRNA. A. CCA sequence where the amino acid is covantly attached to the tRNA B. D loop and T loop C. Region that forms complementary base pairs with the mRNA D. Amino acid that will be added to the growing polypeptide in the ribosome

see image The 3' end of the tRNA is found at one end and contains a CCA sequence to which the amino acid is attached. This amino acid will be added to the growing polypeptide in the ribosome. The anticodon is found in the anticodon loop at the other end of the tRNA. The anticodon forms complementary base pairs with the mRNA, resulting in the proper amino acid being brought into the ribosome. The D and T loops are found in the middle of the tRNA.

The image below depicts the Hershey and Chase experiment. Identify where viral protein and viral DNA would be found by dragging the labels to the correct targets. A. DNA B. protein

see image The results of the Hershey and Chase experiment in this image are consistent with the results from the Avery, McCarty, and MacLeod experiments, which determined that DNA was the "transforming principle" that changed nonpathogenic bacteria into pathogenic bacteria. Hershey and Chase showed definitively that genes are made of DNA. Radioactively labeled viruses were allowed to infect E. coli, and the mixture was then disrupted by brief pulsing in a Waring blender, followed by centrifugation to separate the infected bacteria from the empty viral heads. When the researchers measured the radioactivity, they found that much of the 32P-labeled DNA had entered the bacterial cells, while the vast majority of the 35S-labeled proteins remained in solution with the spent viral particles. Furthermore, the radioactively labeled DNA also made its way into subsequent generations of virus particles, confirming that DNA is the heritable, genetic material.