Bio 1 - Exam 3: Mastering Bio Questions

A ________________ mutation causes an early Stop codon to occur.

nonsense

Which of these is a G-protein-linked receptor?

A

Which of these is a signal molecule?

A

In which of the cultures was the concentration of toxins the highest? A. Control B. Peptide 1 C. Peptide 2 D. Peptides 1 + 2

A. Control

A protein kinase activating many other protein kinases is an example of _____. A. mutualism B. a cellular response C. amplification D. sensitization E. deactivation

C. amplification By activating many other molecules the initial signal is amplified.

A signal molecule is also known as a(n) _____. A. receptor B. initiator C. protein D. key E. ligand

E. ligand A ligand is a signal molecule.

miRNAs can control gene expression by what action? A. binding to mRNAs and degrading them or blocking their translation B. seeking out viral DNA and destroying it C. degrading proteins as soon as they are formed D. binding to DNA and preventing transcription of certain genes E. inhibiting the catalytic activity of rRNA

A. binding to mRNAs and degrading them or blocking their translation miRNAs can effectively "silence" genes by binding to mRNA transcripts. The mRNAs are either broken down by enzymes or are unable to physically interact with the ribosomes to complete translation.

You can tell that this is an image of a DNA nucleotide and not an RNA nucleotide because you see a _____. A. double-stranded molecule, not a single-stranded molecule B. phosphate group, not a uracil C. sugar with two, and not three, oxygen atoms D. uracil nitrogenous base, not a thymine nitrogenous base E. thymine nitrogenous base, not a uracil nitrogenous base

Answer: C. sugar with two, and not three, oxygen atoms Explanation: DNA nucleotides are composed of deoxyribose sugars, whereas RNA nucleotides are composed of ribose sugars.

In the accompanying image, a nucleotide is indicated by the letter _____.

B

Which of these is(are) pyrimidines? A and B B, C, and D B and C A, B, and C C, D, and E

C, D, and E Pyrimidines are single-ring structures.

Thyroid hormones bind to _____ receptors. A. plasma membrane B. ion-channel C. tyrosine-kinase D. steroid E. G-protein-linked F. intracellular

F. intracellular Thyroid hormones are able to pass through the plasma membrane.

The bonds or interactions between stacked nucleotide units that help hold the DNA molecule together are _____________

van der Waals interactions.

Which of these is responsible for initiating a signal transduction pathway?

A This is a signal molecule. The attachment of a signal molecule to a plasma membrane receptor initiates a signal transduction pathway.

Which of the following regulatory DNA sequences might be located thousands of nucleotides away from the transcription start site of a gene? A. Enhancer B. Promoter C. TATA box D. Promoter-proximal element

A. Enhancer Enhancers can function thousands of nucleotides away from the promoter and transcription start site.

Which of the following diagrams most clearly shows the details of the bonds between nitrogenous bases of complementary nucleotide pairs? A. Space filing model B. Structural diagram C. Ribbon Diagram D. Ladder Diagram

Answer: B. Structural diagram Explanation: While the complementary nature of the interactions between the base pairs is demonstrated in all four diagrams, only the structural diagram shows the molecular details--the specific number and location of hydrogen bonds that form between complementary base pairs.

A signal transduction pathway is initiated when a _____ binds to a receptor. A. G protein B. tyrosine kinase C. signal molecule D. calmodulin E. cyclic AMP

C. signal molecule The binding of a signal molecule to a receptor initiates a signal transduction pathway.

Which of these acts as a second messenger?

D

Which of these is an ion-channel receptor?

D

Which of these receptor molecules would allow Na+ to flow into the cell?

D This is an ion-channel receptor.

Which of these receptors is NOT a membrane receptor?

E This receptor is not associated with the plasma membrane.

The _______________________ is/are arranged sequentially after the promoter.

genes of an operon

The bonds or interactions that hold together complementary bases from opposite strands of DNA are _______________

hydrogen bonds.

A(n) _______________is a specific nucleotide sequence in DNA that binds RNA polymerase, positioning it to start transcribing RNA at the appropriate place.

promoter

Rank the three cultures grown with peptides from lowest toxin production to highest toxin production.

see graph from previous card

A hydroxyl is present at the 3' end of the growing DNA strand. What is at the 5' end? A. a ribose B. a nitrogenous base C. a deoxyribose D. a phosphate group

Answer: D. a phosphate group Explanation: The 5' phosphate is an important player in the reaction that joins the next deoxyribonucleotide onto the growing strand.

Which of these acts as a second messenger? A. adenylyl kinase B. cyclic AMP C. protein kinase D. G protein E. G-protein-linked receptor

B. cyclic AMP Cyclic AMP can act as second messengers.

The antiparallel arrangement of double-stranded DNA is due to the phosphate group being bonded to the 3' carbon on one strand and the 5' carbon on the complementary strand. True or False

False

True or false? Regulatory and basal transcription factors regulate transcription by binding to the promoter.

False Basal transcription factors do indeed bind to the promoter, but regulatory transcription factors bind to promoter-proximal elements and enhancers.

A(n) _____________ codes for a protein, such as a repressor, that controls the transcription of another gene or group of genes.

regulatory gene

What materials does DNA polymerase require in order to synthesize a complete strand of DNA? Select all that apply. All four deoxyribonucleotides triphosphates (containing A, C, T, or G) ATP Inorganic phosphate 3'-OH end of the new DNA strand Single-stranded DNA template

Answers: All four deoxyribonucleotides triphosphates (containing A, C, T, or G) 3'-OH end of the new DNA strand Single-stranded DNA template Explanation: In order for DNA polymerase to synthesize a complete new strand of DNA, it requires a template to determine the order of bases on the new strand, a 3'-OH end to add more nucleotides onto, and the full set of four kinds of nucleotides (A,C,T,G) if they are needed to complement the template strand.

In a DNA double helix an adenine of one strand always pairs with a(n) _____ of the complementary strand, and a guanine of one strand always pairs with a(n) _____ of the complementary strand. A. uracil ... cytosine B. thymine ... cytosine C. cytosine ... thymine D. guanine ... adenine E. cytosine ... uracil

B. thymine ... cytosine

Which of the three cultures grown with peptides produced a toxin concentration similar to the control culture? A. Peptide 1 B. Peptide 2 C. Peptides 1 + 2 D. All of the above E. None of the above

E. None of the above The concentration of toxins in the control culture was substantially higher than in all cultures grown with peptides.

In eukaryotes, binding of RNA polymerase II to DNA involves several other proteins known as transcription factors. Many of these transcription factors bind to the DNA in the promoter region (shown below in green), located at the 3' end of the sequence on the template strand. Although some transcription factors bind to both strands of the DNA, others bind specifically to only one of the strands.

Transcription factors do not bind randomly to the DNA. Information about where each transcription factor binds originates in the base sequence to which each transcription factor binds. The positioning of the transcription factors in the promoter region determines how the RNA polymerase II binds to the DNA and in which direction transcription will occur.

The phosphate attached to the 5' carbon of a given nucleotide links to the 3' -OH of the adjacent nucleotide. True or False

True

A(n) _________________ is a protein that inhibits gene transcription. In prokaryotes, this protein binds to the DNA in or near the promoter.

repressor

A ___________ mutation does not change the wild-type amino acid sequence.

silent

A(n) ____________is a stretch of DNA consisting of an operator, a promoter, and genes for a related set of proteins, usually making up an entire metabolic pathway.

operon

The bonds or interactions that hold together adjacent nucleotides in the sugar-phosphate backbone of DNA are ________________.

covalent bonds

Regulatory proteins bind to the ________________to control expression of the operon.

operator

_____ catalyzes the production of _____, which then opens an ion channel that releases _____ into the cell's cytoplasm. A. Phospholipase C ... IP3 .... Ca2+ B. Phospholipase C ... cyclic AMP ... Ca2+ C. Adenylyl cyclase ... IP3 .... Ca2+ D. Protein kinase ... PIP2 ... Na+ E. Adenylyl cyclase ... cyclic AMP ... Ca2+

A. Phospholipase C ... IP3 .... Ca2+ Phospholipase C cleaves IP3 from a membrane protein, and IP3 then binds to a calcium channel on the ER.

Calcium ions that act as second messengers are stored in _____. A. endoplasmic reticula B. chloroplasts C. peroxisomes D. lysosomes E. mitochondria

A. endoplasmic reticula The ER stores calcium ions.

Which of the following regulatory elements is not composed of DNA sequences? A. Enhancers B. Activators C. Silencers D. Promoter-proximal elements

B. Activators Activators are proteins that are involved in transcription initiation.

_____ is a signal molecule that binds to an intracellular receptor

D Steroids are nonpolar and can diffuse through the plasma membrane.

Which of these is a receptor tyrosine kinase?

C

Use the table to sort the following ten codons into one of the three bins, according to whether they code for a start codon, an in-sequence amino acid, or a stop codon. Drag each item to the appropriate bin.

Nearly every mRNA gene that codes for a protein begins with the start codon, AUG, and thus begins with a methionine. Nearly every protein-coding sequence ends with one of the three stop codons (UAA, UAG, and UGA), which do not code for amino acids but signal the end of translation.

In DNA replication in bacteria, the enzyme DNA polymerase III (abbreviated DNA pol III) adds nucleotides to a template strand of DNA. But DNA pol III cannot start a new strand from scratch. Instead, a primer must pair with the template strand, and DNA pol III then adds nucleotides to the primer, complementary to the template strand. Each of the four images below shows a strand of template DNA (dark blue) with an RNA primer (red) to which DNA pol III will add nucleotides. In which image will adenine (A) be the next nucleotide to be added to the primer?

In the example above, DNA pol III would add an adenine nucleotide to the 3' end of the primer, where the template strand has thymine as the next available base. You can tell which end is the 3' end by the presence of a hydroxyl (-OH) group. The structure of DNA polymerase III is such that it can only add new nucleotides to the 3' end of a primer or growing DNA strand (as shown here). This is because the phosphate group at the 5' end of the new strand and the 3' -OH group on the nucleoside triphosphate will not both fit in the active site of the polymerase.

Did deletion of any of the possible control elements cause an increase in reporter gene expression? How can you tell? A. Deletion of control element #1 caused an increase in reporter gene expression; that construct resulted in the highest level of mRNA. B. Deletion of control element #1 or #2 caused an increase in reporter gene expression; both constructs resulted in over 100% of the control level of mRNA. C. All of the deletions caused an increase in reporter gene expression; all of them still resulted in reporter mRNA being made. D. Deletion of control element #3 caused an increase in reporter gene expression; that construct resulted in less reporter mRNA than the control.

Answer: B. Deletion of control element #1 or #2 caused an increase in reporter gene expression; both constructs resulted in over 100% of the control level of mRNA. Explanation: For cells incubated with the DNA constructs in which element #1 or #2 was deleted, the amount of reporter mRNA made was over 100% of the amount of reporter mRNA made by the cells in the control group. This result indicates that the deletion of control element #1 or #2 causes an increase in reporter gene expression.

Let's review the working model for the signaling pathway that leads to the formation of shmoo projections in yeast cells. Fus3 kinase and formin proteins are generally distributed evenly throughout a yeast cell. Based on the model in the diagram, why does the shmoo projection emerge on the same side of the cell that bound the mating factor? A. All of the Fus3 kinase and formin molecules in the cell become activated by binding of mating factor, but the actin strands form only on the side toward the potential mate. B. The G protein-coupled mating factor receptors are located on only one side of the cell, and that is where the signal transduction occurs to initiate shmoo formation. C. The only formin molecules that get phosphorylated and thus activated are those near the G protein-coupled receptor that binds mating factor.

Answer: C. The only formin molecules that get phosphorylated and thus activated are those near the G protein-coupled receptor that binds mating factor. Explanation: Because the mating factor is diffusing from the direction of the other mating-type yeast, it is the G protein-coupled receptors on that side of the cell that bind mating factor, become activated, and start the phosphorylation cascade that activates Fus3, which activates formin, which leads to localized shmoo formation.

All your cells contain proto-oncogenes, which can change into cancer-causing genes. Why do cells possess such potential time bombs? A. Proto-oncogenes protect cells from infection by cancer-causing viruses. B. Proto-oncogenes are necessary for the normal control of cell growth and division. C. Proto-oncogenes are genetic junk that has not yet been eliminated by natural selection. D. Proto-oncogenes are unavoidable environmental carcinogens. E. Cells produce proto-oncogenes as a by-product of mitosis.

B. Proto-oncogenes are necessary for the normal control of cell growth and division. Proto-oncogenes can become oncogenes when a mutation or other genetic change increases the activity of the encoded protein.

e diagram below shows a segment of DNA containing an imaginary gene (Z) and the primary RNA transcript that results from the transcription of gene Z. Exons are represented in green and introns are represented in blue. Which of the following choices represent mRNA molecules that could be produced from the primary RNA transcript by alternative RNA splicing? (In each choice, the yellow part on the left represents the 5' cap, and the yellow part on the right represents the poly-A tail.) A. ACEGI B. ACEI C. BFHI D. BDFH E. ACGI F. BCDFH G. ACDI H. BDEH

Answers: A. ACEGI B. ACEI E. ACGI Explanation: Alternative RNA splicing produces different mRNA molecules from the same primary RNA transcript. During alternative RNA splicing, all introns are removed, and some exons may also be removed. The removal of different exons produces different mRNA molecules, which are then translated into different proteins. Alternative RNA splicing can greatly expand the number of proteins produced from the same gene.

Do the data suggest that any of these possible control elements are actual control elements? A. Only control elements 1 and 2 appear to be control elements. B. Only control element 3 appears to be a control element. C. All three appear to be control elements. D. None of the possible control elements appear to be actual control elements.

C. All three appear to be control elements. All three elements appear to be control elements because when they were deleted, the levels of reporter mRNA differed from the level produced by the intact enhancer construct.

Fluorescent dyes were used in the experiment to distinguish the old and new cell walls of the yeast cells. First, the existing cell walls of each strain were stained with a green fluorescent dye. These green-stained cells were then exposed to mating factor and then stained with a red fluorescent dye that only labels new cell wall growth. Growth of the cell on all sides (symmetric growth) is indicated by a uniform yellow color, resulting from merged green and red stains. This occurs normally in wild-type cells that have not been exposed to mating factor. The image below shows the fluorescence pattern in wild-type yeast cells that have been exposed to mating factor. Note the asymmetric growth. When designing an experiment, scientists make predictions about what results will occur if their hypothesis is correct. One of their hypotheses was that Fus3 kinase is required for the signal transduction pathway leading to shmoo formation. If this hypothesis is correct, what result should be observed in the ΔFus3 strain? A. The ΔFus3 strain should form multiple red-stained shmoos. B. The ΔFus3 strain should look the same as the wild type and form shmoos that stain red. C. The ΔFus3 strain should not form shmoos, and the cells should not have a red zone in their walls. D. The ΔFus3 strain should form green-stained shmoos.

C. The ΔFus3 strain should not form shmoos, and the cells should not have a red zone in their walls. If Fus3 kinase is critical to shmoo formation, then cells that don't produce Fus3 kinase should not be able to form shmoos.

What do you predict would happen if the yeast had a mutation that prevented the G protein from binding GTP? A. Phosphorylated Fus3 kinase would accumulate in the cell in the presence of mating factor. B. Shmoos would form even in the absence of mating factor. C. Mating factor would be unable to bind to the G protein-coupled receptor. D. No shmoo would form in response to mating factor.

D. No shmoo would form in response to mating factor. If the G protein could not bind GTP, it would not become activated, and so it would not start the phosphorylation cascade. Return to Assignment

If deletion of a control element causes an increase in gene expression, what must be the normal role of that control element? A. To repress gene expression; without the control element, activators are not able to bind to the enhancer, and the level of gene expression decreases. B. To activate gene expression; without the control element, activators are not able to bind to the enhancer, and the level of gene expression decreases. C. To activate gene expression; without the control element, repressors are not able to bind to the enhancer, and the level of gene expression increases. D. To repress gene expression; without the control element, repressors are not able to bind to the enhancer, and the level of gene expression increases.

D. To repress gene expression; without the control element, repressors are not able to bind to the enhancer, and the level of gene expression increases.

True or false? One possible way to alter chromatin structure such that genes could be transcribed would be to make histone proteins more positively charged.

False The positive charge on histone proteins allows them to interact tightly with negatively charged DNA, thus inhibiting transcription. To disrupt this interaction, the histone proteins would have to be made more negatively charged.

This is an image of a(n) _____. A. amino acid B. nucleotide C. nucleic acid D. thiol E. none of the above

B. nucleotide

In a single nucleotide, the phosphate group is attached to the 5' carbon of the sugar unit. True or False

True

Use the table to sort the following ten codons into one of the three bins, according to whether they code for a start codon, an in-sequence amino acid, or a stop codon. Drag each item to the appropriate bin.

start/ methionine AUG stop codon UAA UAG UGA amino acid CAC ACU GCA UGC AUC AAA

What was the control treatment in this experiment? A. the construct that had no DNA deleted from the enhancer B. the temperature, pH and salt concentration of the incubation medium C. the construct that resulted in the lowest amount of reporter mRNA D. the reporter gene

A. the construct that had no DNA deleted from the enhancer The control treatment was the construct that had no DNA deleted from the enhancer. All the other treatment groups were compared to that control group.

Did deletion of any of the possible control elements cause a reduction in reporter gene expression? How can you tell? A. Deletion of element #3 caused a reduction in reporter gene expression; that construct resulted in less than 50% of the control level of mRNA. B. Deletion of elements #2 and #3 caused a reduction in reporter gene expression; those constructs resulted in less than the highest level of mRNA. C. None of the deletions caused a reduction in reporter gene expression; all of them still resulted in reporter mRNA being made.

A. Deletion of element #3 caused a reduction in reporter gene expression; that construct resulted in less than 50% of the control level of mRNA. For cells incubated with the DNA construct in which element #3 was deleted, the amount of reporter mRNA made was less than 50% of the amount of reporter mRNA made by the cells in the control group. This result indicates that the deletion of control element #3 causes a reduction in reporter gene expression.

Which of the following events in transcription initiation likely occurs last? A. RNA polymerase binds to the promoter of the gene. B. Basal transcription factors form a basal transcription complex. C. TBP is recruited to the promoter. D. Regulatory transcription factors bind to enhancers.

A. RNA polymerase binds to the promoter of the gene. RNA polymerase is recruited only when other transcription factors, including TBP, are assembled at the promoter.

Which statements about the modification of chromatin structure in eukaryotes are true? Select all that apply. A. Some forms of chromatin modification can be passed on to future generations of cells. B. Methylation of histone tails in chromatin can promote condensation of the chromatin. C. Acetylation of histone tails is a reversible process. D. Acetylation of histone tails in chromatin allows access to DNA for transcription. E. DNA is not transcribed when chromatin is packaged tightly in a condensed form. F. Deacetylation of histone tails in chromatin loosens the association between nucleosomes and DNA.

A. Some forms of chromatin modification can be passed on to future generations of cells. B. Methylation of histone tails in chromatin can promote condensation of the chromatin. C. Acetylation of histone tails is a reversible process. D. Acetylation of histone tails in chromatin allows access to DNA for transcription. E. DNA is not transcribed when chromatin is packaged tightly in a condensed form. see next card for explanation

One of their hypotheses was that formin is required for the signal transduction pathway leading to shmoo formation. If this hypothesis is correct, what result should be observed in the Δformin strain? A. The Δformin strain should not form shmoos, and the cells should not have a red zone in their walls. B. The Δformin strain should look the same as the wild type and form shmoos that stain red. C. The Δformin strain should form green-stained shmoos. D. The Δformin strain should form multiple red-stained shmoos.

A. The Δformin strain should not form shmoos, and the cells should not have a red zone in their walls. If formin is critical to shmoo formation, then cells that don't produce formin should not be able to form shmoos.

If deletion of a control element causes a reduction in gene expression, what must be the normal role of that control element? A. To activate gene expression; without the control element, activators are not able to bind to the enhancer, and the level of gene expression decreases. B. To repress gene expression; without the control element, repressors are not able to bind to the enhancer, and the level of gene expression decreases. C. To activate gene expression; without the control element, repressors are not able to bind to the enhancer, and the level of gene expression increases. D. To repress gene expression; without the control element, repressors are not able to bind to the enhancer, and the level of gene expression increases.

A. To activate gene expression; without the control element, activators are not able to bind to the enhancer, and the level of gene expression decreases.

A(n) _____ is an example of a signal molecule that can bind to an intracellular receptor and thereby cause a gene to be turned on or off. A. steroid B. ion C. nucleic acid D. protein E. carbohydrate

A. steroid Steroids bind to intracellular receptors, which can then bind to, and regulate, the expression of genes.

The diagrams show the intact DNA sequence (top) and the three experimental DNA sequences. A red X indicates the possible control element (1, 2, or 3) that was deleted in each experimental DNA sequence. The area between the slashes represents the approximately 8 kilobases of DNA located between the promoter and the enhancer region. The horizontal bar graph shows the amount of reporter gene mRNA that was present in each cell culture after 48 hours relative to the amount that was in the culture containing the intact enhancer region (top bar = 100%). What was the independent variable in this experiment? A. the possible control element that was deleted B. the distance between the promoter and the enhancer C. the relative level of reporter gene mRNA D. the length of time that the cells were incubated

A. the possible control element that was deleted The independent variable was the possible control element that was deleted from the experimental sequences because the researchers manipulated that variable in each treatment.

What was the dependent variable in this experiment? A. the relative level of reporter gene mRNA B. how many of the artificial DNA molecules were taken up by the cells C. the length of time that the cells were incubated D. the distance between the promoter and the enhancer

A. the relative level of reporter gene mRNA The dependent variable was the relative level of reporter gene mRNA because that was the response that was measured in the experiment.

In addition to testing shmoo formation in the two mutant strains of yeast, the scientists also tested shmoo formation in wild-type yeast. What is the purpose of including wild-type yeast cells in the experiment? A. to show normal shmoo formation under the experimental conditions B. to show the difference between symmetrical and asymmetrical growth C. to show what happens to shmoo formation when both Fus3 kinase and formin are absent D. to allow the scientists to see the interaction between Fus3 kinase and formin

A. to show normal shmoo formation under the experimental conditions Wild-type yeast cells act as a control for observation of shmoo formation because wild-type cells have both Fus3 and formin present under the experimental conditions.

Life as we know it depends on the genetic code: a set of codons, each made up of three bases in a DNA sequence and corresponding mRNA sequence, that specifies which of the 20 amino acids will be added to the protein during translation. Imagine that a prokaryote-like organism has been discovered in the polar ice on Mars. Interestingly, these Martian organisms use the same DNA → RNA → protein system as life on Earth, except that -there are only 2 bases (A and T) in the Martian DNA, and -there are only 17 amino acids found in Martian proteins. Based on this information, what is the minimum size of a codon for these hypothetical Martian life-forms? 2 bases 3 bases 4 bases 5 bases 6 bases The answer cannot be determined from the information provided. Submit

Answer: 5 bases Explanation: In the most general case of x bases and y bases per codon, the total number of possible codons is equal to xy . In the case of the hypothetical Martian life-forms, is the minimum codon length needed to specify 17 amino acids is 5 (25 = 32), with some redundancy (meaning that more than one codon could code for the same amino acid). For life on Earth, x = 4 and y = 3; thus the number of codons is 43, or 64. Because there are only 20 amino acids, there is a lot of redundancy in the code (there are several codons for each amino acid).

In a nucleotide, the nitrogenous base is attached to the sugar's _____ carbon and the phosphate group is attached to the sugar's _____ carbon. A. 1' ... 5' B. 1' ... 2' C. 2' ... 3' D. 1' ... 3' E. 2' ... 1'

Answer: A. 1' ... 5' Explanation: The nitrogenous base is attached to the sugar's 1' carbon and the phosphate group is attached to the sugar's 5' carbon.

Given a template strand of 3'-ATGCTTGGACA-5' and a partially-made complementary strand containing only 5'-TAC-3', what would be the sequence of the new strand of DNA (including the 5'-TAC-3') if the only additional nucleotides available to DNA polymerase were those containing the bases G, A, and C? A. 5'-TACGAACC-3' B. 5'-TAC-3'; All four nucleotides are required for DNA polymerase to function. C. 5'-GAACC-3' D. 3'-TACGAACCTGT-5'

Answer: A. 5'-TACGAACC-3' Explanation: DNA polymerase will continue to add nucleotides onto the growing strand as long as it has nucleotides with the bases required to complement the template strand. If it is missing one kind of base, it will stop at that point on the strand.

Various types of chemical bonds or interactions maintain the three-dimensional (3D) structure of large biological molecules like DNA. Not all types of bonds or interactions are shown in all diagrams. The types of bonds or interactions shown depend on the emphasis of the particular diagram. Which of the following diagrams most clearly shows the overall 3D shape and atomic composition of DNA? A. Space filing model B. Structural diagram C. Ribbon Diagram D. Ladder Diagram

Answer: A. Space filing model Explanation: The space-filling model shows each atom making up the two strands, and reveals the helical shape and the double-stranded structure of the DNA molecule.

After transcription begins, several steps must be completed before the fully processed mRNA is ready to be used as a template for protein synthesis on the ribosomes. Which three statements correctly describe the processing that takes place before a mature mRNA exits the nucleus? Noncoding sequences called introns are spliced out by molecular complexes called spliceosomes. A translation stop codon is added at the 3' end of the pre-mRNA. A poly-A tail (50-250 adenine nucleotides) is added to the 3' end of the pre-mRNA. Coding sequences called exons are spliced out by ribosomes. A cap consisting of a modified guanine nucleotide is added to the 5' end of the pre-mRNA.

Answer: Noncoding sequences called introns are spliced out by molecular complexes called spliceosomes. A poly-A tail (50-250 adenine nucleotides) is added to the 3' end of the pre-mRNA. A cap consisting of a modified guanine nucleotide is added to the 5' end of the pre-mRNA. Explanation: Once RNA polymerase II is bound to the promoter region of a gene, transcription of the template strand begins. As transcription proceeds, three key steps occur on the RNA transcript: Early in transcription, when the growing transcript is about 20 to 40 nucleotides long, a modified guanine nucleotide is added to the 5' end of the transcript, creating a 5' cap. Introns are spliced out of the RNA transcript by spliceosomes, and the exons are joined together, producing a continuous coding region. A poly-A tail (between 50 and 250 adenine nucleotides) is added to the 3' end of the RNA transcript. Only after all these steps have taken place is the mRNA complete and capable of exiting the nucleus. Once in the cytoplasm, the mRNA can participate in translation.

Which of the following terms associated with transcription describe regions of nucleic acid? Select all that apply. promoter terminator gene RNA polymerase

Answer: promoter gene terminator Explanation: The gene itself, the promoter, and the terminator are all DNA deoxyribonucleic acid) sequences. The RNA (ribonucleic acid) that is produced is another. The RNA polymerase enzyme that performs transcription is a protein and therefore not a nucleic acid.

The diagram below shows two stretches of DNA in the genome of an imaginary eukaryotic cell. The top stretch of DNA includes the fantasin gene, along with its promoter and one of its enhancers. The bottom stretch of DNA includes the imaginin gene, its promoter, and one of its enhancers. The slash marks (//) indicate that more than 1,000 nucleotides separate the promoter and enhancer of each gene. Which statements about the regulation of transcription initiation in these genes are true? Select all that apply. A. The imaginin gene will be transcribed at a high level when repressors specific for the imaginin gene are present in the cell. B. Control elements C, D, and E are distal control elements for the imaginin gene. C. Both the fantasin gene and the imaginin gene will be transcribed at high levels whenever general transcription factors are present in the cell. D. Control elements A, B, and C are proximal control elements for the fantasin gene. E. Both the fantasin gene and the imaginin gene will be transcribed at high levels when activators specific for control elements A, B, C, D, and E are present in the cell. F. The fantasin gene and the imaginin gene have identical enhancers. G. The fantasin gene will be transcribed at a high level when activators specific for control elements A, B, and C are present in the cell.

Answers: B. Control elements C, D, and E are distal control elements for the imaginin gene. E. Both the fantasin gene and the imaginin gene will be transcribed at high levels when activators specific for control elements A, B, C, D, and E are present in the cell. G. The fantasin gene will be transcribed at a high level when activators specific for control elements A, B, and C are present in the cell. Explanation: Only certain genes are transcribed in a eukaryotic cell at any particular time. The regulation of transcription initiation depends on the interaction of specific transcription factors with specific control elements in enhancers. In the imaginary eukaryotic cell used as an example here, the enhancers for the fantasin gene and imaginin gene are unique because they contain different sets of control elements (A, B, and C for the fantasin gene; C, D, and E for the imaginin gene). Each gene will be transcribed at a high level when activators specific for all of the control elements in its enhancer are present in the cell.

Which of the following terms describes the DNA-protein complexes that look like beads on a string? A. Histones B. Nucleosome C. 30-nanometer fiber D. Chromatin

B. Nucleosome The "beads on a string" appearance of nucleosomes comes from the wrapping of DNA around a core of eight histone proteins.

A model helps scientists form testable hypotheses. What hypothesis was being tested with the ΔFus3 strain? A. Fus3 causes shmoo formation in mating yeast. B. Fus3 is required for formin phosphorylation. C. Fus3 is required for the signal transduction pathway leading to shmoo formation. D. Mating factor is required to start the signal transduction pathway leading to shmoo formation.

C. Fus3 is required for the signal transduction pathway leading to shmoo formation.

Which of these extracellular signal molecules could diffuse through a plasma membrane and bind to an intracellular receptor? A. cellulose B. glycerol C. estrogen D. glucose E. starch

C. estrogen Nonpolar molecules can diffuse through the plasma membrane and bind to intracellular receptors.

Why is a frameshift missense mutation more likely to have a severe effect on phenotype than a nucleotide-pair substitution missense mutation in the same protein? A. A substitution missense mutation causes the protein to be shorter and thus non-functional. B. A frameshift missense mutation will cause an early Stop codon, but a substitution missense might be silent. C. A frameshift missense will cause the codons to be out of order, but a substitution missense does not change the order of the codons. D. A substitution missense affects only one codon, but a frameshift missense affects all codons downstream of the frameshift.

D. A substitution missense affects only one codon, but a frameshift missense affects all codons downstream of the frameshift. Many genetic diseases are caused by missense mutations.

You are studying a bacterium that utilizes a sugar called athelose. This sugar can be used as an energy source when necessary. Metabolism of athelose is controlled by the ath operon. The genes of the ath operon code for the enzymes necessary to use athelose as an energy source. You have found the following: The genes of the ath operon are expressed only when the concentration of athelose in the bacterium is high. When glucose is absent, the bacterium needs to metabolize athelose as an energy source as much as possible. The same catabolite activator protein (CAP) involved with the lac operon interacts with the ath operon. Based on this information, how is the ath operon most likely controlled? Drag the labels onto the diagram to identify the small molecules and the states of the regulatory proteins. Not all labels will be used.

Metabolism of the sugar athelose in this hypothetical system is controlled by an operon that exhibits both positive control and negative control. Transcription of the ath operon is turned on when athelose is present (negative control), and sped up when the bacterium runs out of glucose and must rely on athelose for energy (positive control).

Sort the statements into the appropriate bins depending on whether or not each operon would be transcribed under the stated conditions.

The trp operon is regulated through negative control only. When tryptophan is present, the operon genes are not transcribed. The lac operon is regulated through both negative control and positive control. Negative control: -When lactose is absent, the repressor protein is active, and transcription is turned off. -When lactose is present, the repressor protein is inactivated, and transcription is turned on. Positive control: -When glucose is absent, another regulatory protein (CAP) binds to the promoter of the lac operon, increasing the rate of transcription if lactose is present.

Use the codon table to determine which mRNA triplets code for the amino acid cysteine, Cys. Select all that apply. UGU ACA UGC UGG AGU

UGU UGC There are only two mRNA triplets that code for the amino acid Cys: UGU and UGC.

A(n) __________________is a specific small molecule that binds to a bacterial regulatory protein and changes its shape so that it cannot bind to an operator, thus switching an operon on.

inducer

Which of these is a difference between a DNA and an RNA molecule? A. DNA contains nitrogenous bases, whereas RNA contains phosphate groups. B. DNA is usually double-stranded, whereas RNA is usually single-stranded. C. DNA is a polymer composed of nucleotides, whereas RNA is a polymer composed of nucleic acids. D. DNA contains five-carbon sugars, whereas RNA contains six-carbon sugars. E. DNA contains uracil, whereas RNA contains thymine.

Answer: B. DNA is usually double-stranded, whereas RNA is usually single-stranded. Explanation: With some exceptions, DNA is a double-stranded molecule and RNA is a single-stranded molecule.

What is the role of DNA polymerase during DNA synthesis? A. DNA polymerase provides the free energy to catalyze the endergonic addition of a nucleotide onto the 3' end of a growing DNA strand. B. DNA polymerase is the enzyme that catalyzes the addition of a nucleotide onto the 3' end of a growing DNA strand. C. DNA polymerase removes inorganic phosphate from the template strand of DNA to catalyze the polymerization reaction. D. DNA polymerase catalyzes the synthesis of the template strand of DNA.

Answer: B. DNA polymerase is the enzyme that catalyzes the addition of a nucleotide onto the 3' end of a growing DNA strand. Explanation: DNA polymerase is the enzyme complex responsible for synthesizing a new strand of DNA, using an existing strand as a template.

Suppose that the triplet of nucleotides indicated in bold (AGC) spans two codons, that is, CTA and GCC. If the triplet AGC were deleted from this DNA coding sequence, what effect would it have on the resulting protein? 5'-ATGCTAGCCTATCGTAAC-3' A. All of the amino acids up to the deletion would be altered due to the frameshift. B. The two flanking codons would be altered, but the rest of the amino acid sequence would be the same because there would be no frameshift. C. The entire amino acid sequence would be altered due to the frameshift. D. All of the amino acids after the deletion would be altered due to the frameshift.

Answer: B. The two flanking codons would be altered, but the rest of the amino acid sequence would be the same because there would be no frameshift. Explanation: When a deletion of a set of three nucleotides that is out of frame to the reading frame of codons occurs, it only affects the codons flanking it. Once the ribosome reads past that point, the rest of the codons are in frame because a full triplet was deleted at one time. Return to Assignment

Do these data alone suggest that humans infected with antibiotic-resistant S. aureus can be effectively treated with peptides 1 + 2? A. No. The results do not demonstrate that toxin production by S. aureus is reduced when treated with peptides 1 + 2, so humans should not be treated with peptides 1 + 2. B. Yes. The results demonstrate that toxin production in vitro is reduced when treated with peptides 1 + 2, so humans should also be treated with peptides 1 + 2. C. No. Not enough information is provided about the integrity of the study design (for example, number of samples, replicates) and the statistical difference between treatment groups. More data is needed to confirm the results. D. Yes. With in vitro results this clear, human trials are unnecessary.

Answer: C. No. Not enough information is provided about the integrity of the study design (for example, number of samples, replicates) and the statistical difference between treatment groups. More data is needed to confirm the results. Explanation: Before a treatment approach for S. aureus infections in humans is prescribed, human trials are needed. Further information should also be provided about the integrity of the study design and the statistical strength of the experimental results stemming from the in vitro experiments.

Duplication of chromosomes occurs during S phase of the cell cycle. Duplication requires the separation of complementary DNA strands to allow for DNA replication. Which of the following statements best explains how weak hydrogen bonds function better in this situation than stronger bonds would? Duplicated DNA molecules must be extremely flexible in order to fit sister chromatid pairs into the nucleus of a eukaryotic cell. Weak hydrogen bonds between complementary strands are easily bent and modified, allowing for this type of flexibility. Complementary DNA strands are separated or "unzipped" for the replication process. Weak hydrogen bonds between complementary strands are easily disrupted during DNA replication because they are not high-energy chemical bonds. Complementary DNA strands are easily mutated by environmental factors. These changes lead to evolution. Therefore, weak hydrogen bonds support continued evolution by allowing mutations during DNA replication. Hydrogen is used in many metabolic functions; therefore its function connecting complementary DNA strands stores hydrogen atoms for use in the G2 phase of the cell cycle.

Answer: Complementary DNA strands are separated or "unzipped" for the replication process. Weak hydrogen bonds between complementary strands are easily disrupted during DNA replication because they are not high-energy chemical bonds. Explanation: Hydrogen bonds connect complementary bases from opposite strands of the DNA molecule. Hydrogen bonds are relatively weak, low-energy interactions that allow complementary DNA strands to "unzip" during S phase replication.

Addition of a nucleotide onto a DNA strand is an endergonic reaction. What provides the energy to drive the reaction? A. Binding of the pre-existing new strand, the template strand, and the incoming nucleotide to the active site of the DNA polymerase B. The dehydration reaction between the 5'-phosphate of the incoming nucleotide and the 3'-OH of the growing strand of DNA C. Complementary bases on the template and the incoming nucleotide are attracted to each other, releasing free energy. D. Release of pyrophosphate from the incoming nucleotide, and then hydrolysis of the pyrophosphate to inorganic phosphate

Answer: D. Release of pyrophosphate from the incoming nucleotide, and then hydrolysis of the pyrophosphate to inorganic phosphate Explanation: Each deoxyribonucleotide enters the reaction as a triphosphate, and hydrolysis of the phosphates releases the free energy needed for the nucleotide to bind to the growing strand.

During translation, nucleotide base triplets (codons) in mRNA are read in sequence in the 5' → 3' direction along the mRNA. Amino acids are specified by the string of codons. What amino acid sequence does the following mRNA nucleotide sequence specify? 5′−AUGGCAAGAAAA−3′ Express the sequence of amino acids using the three-letter abbreviations, separated by hyphens (e.g., Met-Ser-Thr-Lys-Gly).

Answer: Met-Ala-Arg-Lys Explanation: An amino acid sequence is determined by strings of three-letter codons on the mRNA, each of which codes for a specific amino acid or a stop signal. The mRNA is translated in a 5' → 3' direction.

Before a molecule of mRNA can be translated into a protein on the ribosome, the mRNA must first be transcribed from a sequence of DNA. What amino acid sequence does the following DNA nucleotide sequence specify? 3′−TACAGAACGGTA−5′ Express the sequence of amino acids using the three-letter abbreviations, separated by hyphens (e.g., Met-Ser-His-Lys-Gly). View Available Hint(s)

Answer: Met-Ser-Cys-His Explanation: Before mRNA can be translated into an amino acid sequence, the mRNA must first be synthesized from DNA through transcription. Base pairing in mRNA synthesis follows slightly different rules than in DNA synthesis: uracil (U) replaces thymine (T) in pairing with adenine (A). The codons specified by the mRNA are then translated into a string of amino acids.

Suppose that a portion of double-stranded DNA in the middle of a large gene is being transcribed by an RNA polymerase. As the polymerase moves through the sequence of six bases shown in the diagram below, what is the corresponding sequence of bases in the RNA that is produced? Enter the sequence of bases as capital letters with no spaces and no punctuation. Begin with the first base added to the growing RNA strand, and end with the last base added.

Answer: UGAGCC Explanation: There are three principles to keep in mind when predicting the sequence of the mRNA produced by transcription of a particular DNA sequence. The RNA polymerase reads the sequence of DNA bases from only one of the two strands of DNA: the template strand. The RNA polymerase reads the code from the template strand in the 3' to 5' direction and thus produces the mRNA strand in the 5' to 3' direction. In RNA, the base uracil (U) replaces the DNA base thymine (T). Thus the base-pairing rules in transcription are A→U, T→A, C→G, and G→C, where the first base is the coding base in the template strand of the DNA and the second base is the base that is added to the growing mRNA strand.

What role does a transcription factor play in a signal transduction pathway? A. By binding to a plasma membrane receptor it initiates a cascade. B. It relays a signal from the cytoplasm to the plasma membrane. C. It activates relay proteins. D. By binding to DNA it triggers the transcription of a specific gene. E. It is a plasma membrane protein that binds signal molecules.

D. By binding to DNA it triggers the transcription of a specific gene. This is the function of a transcription factor.

What hypothesis was being tested with the Δformin strain? A. Mating factor is required to start the signal transduction pathway leading to shmoo formation. B. Phosphorylation of formin is required for the signal transduction pathway leading to shmoo formation. C. Formin causes shmoo formation in mating yeast. D. Formin is required for the signal transduction pathway leading to shmoo formation.

D. Formin is required for the signal transduction pathway leading to shmoo formation.

The diagram below shows an mRNA molecule that encodes a protein with 202 amino acids. The start and stop codons are highlighted, and a portion of the nucleotide sequence in the early part of the molecule is shown in detail. At position 35, a single base-pair substitution in the DNA has changed what would have been a uracil (U) in the mRNA to an adenine (A). Based on the genetic code chart above, which of the following would be the result of this single base-pair substitution? a silent mutation (no change in the amino acid sequence of the protein) a nonsense mutation resulting in early termination of translation a frameshift mutation causing a single amino acid change in the protein a missense mutation causing a single amino acid change in the protein a frameshift mutation causing extensive change in the amino acid sequence of the protein

Answer: a nonsense mutation resulting in early termination of translation Explanation: The effect of a single base substitution depends on the new codon formed by the substitution. To identify the new codon, it is first necessary to determine the reading frame for the amino acid sequence. The first codon starts with base 1, the second codon with base 4, the third with base 7, and so on. In this problem, the codon that contains the single base substitution begins with base 34. The original codon (UUA, which encodes the amino acid leucine) is converted by the single base substitution to UAA, which is a stop codon. This will cause premature termination of translation, also called a nonsense mutation.

Which of these is a membrane receptor?

B

Which of these is a receptor molecule?

B

A nitrogenous base is indicated by the letter _____.

C

The binding of signal molecules to _____ results in the phosphorylation of tyrosines.

C The binding of signal molecules to tyrosine-kinase receptors activates tyrosine-kinase enzymes, which phosphorylate tyrosines.

Based on these data alone, identify the most appropriate hypothesis that explains the effect of peptides 1 and 2 on the quorum-sensing pathways of S. aureus. A. Peptides 1 and 2 acted non-additively, suggesting that they inhibit the same quorum-sensing pathway. B. Peptides 1 and 2 acted additively, suggesting that they inhibit the same quorum-sensing pathway. C. Peptides 1 and 2 acted additively, suggesting that they inhibit different quorum-sensing pathways. D. Peptides 1 and 2 acted non-additively, suggesting that they inhibit different quorum-sensing pathways.

C. Peptides 1 and 2 acted additively, suggesting that they inhibit different quorum-sensing pathways. Explanation: Peptides 1 and 2 act additively because when peptides 1 and 2 are used in combination, they reduce toxin secretion more than either peptide does on it own. Therefore, you can conclude that peptides 1 and 2 inhibit different quorum-sensing pathways.

Think about the DNA coding sequence of a gene. If an A were swapped for a T, what kind of mutation could it cause and why? Think about the DNA coding sequence of a gene. If an A were swapped for a T, what kind of mutation could it cause and why? A. It could cause a silent mutation because A and T are complementary to each other so it is not really a substitution mutation. B. It could cause a frameshift nonsense or frameshift missense mutation because it would change the reading frame of the codon triplet. C. It could cause a nonsense mutation because the sequence would no longer be the same, so the protein would be shorter and non-functional. D. It could cause a silent, missense, or nonsense mutation because those are the types that can be caused by a nucleotide-pair substitution like this one.

D. It could cause a silent, missense, or nonsense mutation because those are the types that can be caused by a nucleotide-pair substitution like this one. When the sequence of bases in a gene is changed by a single nucleotide-pair substitution, it can have different effects on the product depending on where it is and whether or not the mutated codon codes for the same amino acid as the original codon.

The letter A indicates a _____. A. nitrogenous base B. sugar C. nucleotide D. phosphate group E. none of the above

D. phosphate group

A signal transduction pathway is initiated when a _____ binds to a receptor. A. G protein B. tyrosine kinase C. calmodulin D. signal molecule E. cyclic AMP

D. signal molecule The binding of a signal molecule to a receptor initiates a signal transduction pathway.

Which of these is the second of the three stages of cell signaling? A. gene activation B. reception C. binding of a neurotransmitter to a plasma membrane receptor D. transduction E. cell response

D. transduction Transduction is the second of the three stages of cell signaling.

Which of the following sequences shows a frameshift mutation compared to the wild-type mRNA sequence? Which of the following sequences shows a frameshift mutation compared to the wild-type mRNA sequence? A. wild-type 5'-AUGCAUACAUUGGAGUGA-3' mutant 5'-AUGCAUGUGACAUUGGAGUGA-3' B. wild-type 5'-AUGCAUACAUUGGAGUGA-3' mutant 5'-AUGCAUACGUUGGAGUGA-3' C. wild-type 5'-AUGCAUACAUUGGAGUGA-3' mutant 5'-AUGCAUACAGAGUGA-3' D. wild-type 5'-AUGCAUACAUUGGAGUGA-3' mutant 5'-AUGCAUACAUCUGGAGUGA-3'

D. wild-type 5'-AUGCAUACAUUGGAGUGA-3' mutant 5'-AUGCAUACAUCUGGAGUGA-3' In order for an insertion or deletion to cause a frameshift mutation, it must cause the reading frame of each triplet of bases after it to be shifted by one or two places.

Nucleic acids are assembled in the _____ direction. A. 5' to 1' B. 4' to 5' C. 2' to 3' D. 1' to 5' E. 5' to 3'

E. 5' to 3' New nucleotides are added to the 3' end of a growing polynucleotide.

RNA plays important roles in many cellular processes, particularly those associated with protein synthesis: transcription, RNA processing, and translation. Drag the labels to the appropriate bins to identify the step in protein synthesis where each type of RNA first plays a role. If an RNA does not play a role in protein synthesis, drag it to the "not used in protein synthesis" bin.

Explanation: In eukaryotes, pre-mRNA is produced by the direct transcription of the DNA sequence of a gene into a sequence of RNA nucleotides. Before this RNA transcript can be used as a template for protein synthesis, it is processed by modification of both the 5' and 3' ends. In addition, introns are removed from the pre-mRNA by a splicing process that is catalyzed by snRNAs (small nuclear RNAs) complexed with proteins. The product of RNA processing, mRNA (messenger RNA), exits the nucleus. Outside the nucleus, the mRNA serves as a template for protein synthesis on the ribosomes, which consist of catalytic rRNA (ribosomal RNA) molecules bound to ribosomal proteins. During translation, tRNA (transfer RNA) molecules match a sequence of three nucleotides in the mRNA to a specific amino acid, which is added to the growing polypeptide chain. RNA primers are not used in protein synthesis. RNA primers are only needed to initiate a new strand of DNA during DNA replication.

In this activity, you will demonstrate your understanding of antiparallel elongation at the replication forks. Keep in mind that the two strands in a double helix are oriented in opposite directions, that is, they are antiparallel. Drag the arrows onto the diagram below to indicate the direction that DNA polymerase III moves along the parental (template) DNA strands at each of the two replication forks. Arrows can be used once, more than once, or not at all.

Explanation: DNA polymerase III can only add nucleotides to the 3' end of a new DNA strand. Because the two parental DNA strands of a double helix are antiparallel (go from 3' to 5' in opposite directions), the direction that DNA pol III moves on each strand emerging from a single replication fork must also be opposite. For example, in the replication fork on the left, the new strand on top is being synthesized from 5' to 3', and therefore DNA pol III moves away from the replication fork. Similarly, the new strand on the bottom of that same replication fork is being synthesized from 5' to 3'. But because the bottom parental strand is running in the opposite direction of the top parental strand, DNA pol III moves toward the replication fork. In summary, at a single replication fork, one strand is synthesized away from the replication fork, and one strand is synthesized toward the replication fork. When you look at both replication forks, note that a single new strand is built in the same direction on both sides of the replication bubble.

The DNA double helix is composed of two strands of DNA; each strand is a polymer of DNA nucleotides. Each nucleotide consists of a sugar, a phosphate group, and one of four nitrogenous bases. The structure and orientation of the two strands are important to understanding DNA replication. Drag the labels to their appropriate locations on the diagram below. Pink labels can be used more than once.

Explanation: The DNA double helix is constructed from two strands of DNA, each with a sugar-phosphate backbone and nitrogenous bases that form hydrogen bonds, holding the two strands together. Each DNA strand has two unique ends. The 3' end has a hydroxyl (-OH) group on the deoxyribose sugar, whereas the 5' end has a phosphate group. In the double helix, the two strands are antiparallel, that is, they run in opposite directions such that the 3' end of one strand is adjacent to the 5' end of the other strand.

DNA is a double-stranded molecule made up of complementary, antiparallel strands. Based on what you know about complementary base pairing, fill in the rest of the details in the figure.

Explanation: The nucleotide pairs in double-stranded DNA follow the base-pairing rules: A with T, and G with C. The complementary strands are antiparallel, with one strand running 5' to 3', and its complement running 3' to 5'. The 3' end of a DNA strand has an exposed -OH group, and the 5' end has a phosphate group.

Place the events in the transcription of a gene in their proper order from left (first event) to right (last event). Rank from first event to last event.

Explanation: Transcription begins when a molecule of RNA polymerase binds to a promoter. Transcription continues through the gene, producing the RNA. Once RNA polymerase reaches the terminator, the RNA is released, and RNA polymerase falls off the DNA.

The -OH group on the 3' carbon of the sugar unit is the attachment site for the nitrogenous base. True or False

False

A _______________ mutation causes a wild-type amino acid to be replaced by a different amino acid.

missense

DNA is composed of two strands that are bound together, resembling a rope ladder with rigid rungs. This DNA ladder is twisted, forming what is called the double helix. The structure of the DNA double helix depends on the complementary pairing of bases between the two strands.

Replication of DNA requires that the two strands of the helix separate, as shown in the image below. New daughter molecules are constructed by the sequential addition of nucleotides and the formation of base pairs between the new strand and the parent (template) strand. The replication of the double helix results in two daughter molecules, each composed of one parent strand and one new strand. The enzymes that accomplish the replication of DNA are called DNA polymerases.

In contrast to the leading strand, the lagging strand is synthesized as a series of segments called Okazaki fragments. The diagram below illustrates a lagging strand with the replication fork off-screen to the right. Fragment A is the most recently synthesized Okazaki fragment. Fragment B will be synthesized next in the space between primers A and B. Drag the labels to their appropriate locations in the flowchart below, indicating the sequence of events in the production of fragment B. (Note that pol I stands for DNA polymerase I, and pol III stands for DNA polymerase III.)

Synthesis of the lagging strand is accomplished through the repetition of the following steps. Step 1: A new fragment begins with DNA polymerase III binding to the 3' end of the most recently produced RNA primer, primer B in this case, which is closest to the replication fork. DNA pol III then adds DNA nucleotides in the 5' to 3' direction until it encounters the previous RNA primer, primer A. Step 2: DNA pol III falls off and is replaced by DNA pol I. Starting at the 5' end of primer A, DNA pol I removes each RNA nucleotide and replaces it with the corresponding DNA nucleotide. (DNA pol I adds the nucleotides to the 3' end of fragment B.) When it encounters the 5' end of fragment A, DNA pol I falls off, leaving a gap in the sugar-phosphate backbone between fragments A and B. Step 3: DNA ligase closes the gap between fragments A and B. These steps will be repeated as the replication fork opens up. Try to visualize primer C being produced to the right (closest to the replication fork). Fragment C would be synthesized and joined to fragment B following the steps described here.

Complementary base pairing relies on the number of hydrogen bonds that each base can make. True or False

True

The diagram below shows a replication bubble with synthesis of the leading and lagging strands on both sides of the bubble. The parental DNA is shown in dark blue, the newly synthesized DNA is light blue, and the RNA primers associated with each strand are red. The origin of replication is indicated by the black dots on the parental strands.

earliest h,a b,g c,f d,e latest As soon as the replication bubble opens and the replication machinery is assembled at the two replication forks, the two primers for the leading strands (primers a and h) are produced. The production of the first primers on the lagging strands (those closest to the origin of replication, b and g) is delayed slightly because the replication forks must open up further to expose the template DNA for the lagging strands. After completion of the first segments of the lagging strands, additional template DNA must be exposed before the second primers (c and f) can be produced. And after completion of the second segments, additional template DNA must be exposed before the third primers (d and e) can be produced. In summary, because of the way the replication bubble expands, the lagging strand primers near the origin of replication were produced before the primers near the replication forks.

The DNA in a cell's nucleus encodes proteins that are eventually targeted to every membrane and compartment in the cell, as well as proteins that are targeted for secretion from the cell. For example, consider these two proteins: Phosphofructokinase (PFK) is an enzyme that functions in the cytoplasm during glycolysis. Insulin, a protein that regulates blood sugar levels, is secreted from specialized pancreatic cells. Assume that you can track the cellular locations of these two proteins from the time that translation is complete until the proteins reach their final destinations. For each protein, identify its targeting pathway: the sequence of cellular locations in which the protein is found from when translation is complete until it reaches its final (functional) destination. (Note that if an organelle is listed in a pathway, the location implied is inside the organelle, not in the membrane that surrounds the organelle.)

explanation: There are two general targeting pathways for nuclear-encoded proteins in eukaryotic cells. Proteins that will ultimately function in the cytoplasm (PFK, for example) are translated on free cytoplasmic ribosomes and released directly into the cytoplasm. Proteins that are destined for the membranes or compartments of the endomembrane system, as well as proteins that will be secreted from the cell (insulin, for example), are translated on ribosomes that are bound to the rough ER. For proteins translated on rough ER, the proteins are found in one of two places at the end of translation. If a protein is targeted to a membrane of the endomembrane system, it will be in the ER membrane. If a protein is targeted to the interior of an organelle in the endomembrane system or to the exterior of the cell, it will be in the lumen of the rough ER. From the rough ER (membrane or lumen), these non-cytoplasmic proteins move to the Golgi apparatus for processing and sorting before being sent to their final destinations.

Place the blue labels in their proper locations on this diagram showing the process of transcription. Then, use the pink labels to identify the corresponding RNA nucleotide that belongs in each pink target. Pink labels can be used once, more than once, or not at all. Drag the appropriate labels to their respective targets.

not pictured: RNA being produced Explanation: RNA polymerase reads the template strand of the DNA to produce the final RNA. In doing so, the enzyme follows the base pairing rules for RNA, with U complementary to A.

During transcription in eukaryotes, a type of RNA polymerase called RNA polymerase II moves along the template strand of the DNA in the 3'→5' direction. However, for any given gene, either strand of the double-stranded DNA may function as the template strand. For any given gene, what ultimately determines which DNA strand serves as the template strand? which of the two strands of DNA carries the RNA primer the location of specific proteins (transcription factors) that bind to the DNA the location along the chromosome where the double-stranded DNA unwinds the base sequence of the gene's promoter

the base sequence of the gene's promoter

Which of these nitrogenous bases is found in DNA but not in RNA? thymine guanine adenine uracil cytosine

thymine DNA contains thymine; RNA does not.