Unit 6 Review Page: Objectives, Flashcards, and Quizzes

Table of Contents

  1. Learning Objectives
  2. Flashcards
  3. Multiple Choice Quiz 1
  4. Multiple Choice Quiz 2
  5. Practice FRQs
  6. Click on Challenges
    1. DNA and DNA Replication
    2. Transcription and Translation
    3. Operons
    4. Eukaryotic Gene Regulation

Unit 6 Review Video

1. Unit 6 Learning Objectives

Topic 6.1: DNA and RNA Structure and Function

[note: some of this is also covered in  Topic 1.6]

  1. Compare and contrast the functions of DNA and RNA
    • DNA is the molecule of heredity in organisms. RNA can play that role in viruses.
    • In organisms, RNA’s function is information transfer related to protein synthesis (mRNA, tRNA, and rRNA)
    • RNA also has a catalytic role in certain processes (splicing out introns, influencing gene expression).
  2. Compare and contrast DNA in eukaryotic and prokaryotic organisms.
    • Prokaryotes have circular chromosomes. Eukaryotes have multiple linear chromosomes.
    • Plasmids are small circles of double-stranded DNA found in prokaryotes (and in some eukaryotes)
  3. Describe the features of DNA that make it suited to be the molecule of heredity.
    • The stability of the double helix allows for information storage.
    •  Base pairing rules allow for accurate replication.
      • Complementary nitrogenous bases pair through hydrogen bonding.
      • In DNA, A (adenine) bonds with T (thymine); C (cytosine) bonds with G (guanine)
      • Purines (G and A) have two rings and bind with pyrimidines (C, T, U), which have a single ring.

Topic 6.2: DNA Replication

  1. Describe the overall process of DNA replication
    • DNA polymerase adds new nucleotides at the 3’ end of a growing strand (5′ to 3′ synthesis)
    • Because DNA polymerase can only add to an existing strand, an RNA primer is required.
  2. Explain how DNA replication is semiconservative
    • One strand acts as the template for the synthesis of a new complementary strand.
  3. List the key enzymes involved in DNA Replication (and describe the function of each)
    • Helicase breaks hydrogen bonds to unwind the strands.
    • Topoisomerase prevents supercoiling
    • Polymerase (see above)
    • Ligase creates sugar-phosphate bonds between adjacent DNA fragments.
  4. Explain the concept of leading and lagging strands, and explain how replication is different on each.
    • Leading strand: continuous replication.
    • Lagging strand: discontinuous replication (Okazaki fragments).

Topic 6.3: Transcription

  1. Describe the flow of genetic information within cells (AKA the central dogma)
    • DNA makes RNA makes protein
  2. Describe what happens during transcription
    • RNA polymerase uses the sequence of DNA nucleotides in the template strand (AKA the noncoding strand, minus strand, or antisense strand) to synthesize complementary RNA.
    • RNA polymerase binds at a promoter, a DNA sequence just upstream of the transcription start site.
    • The binding of RNA polymerase can be regulated by transcription factors (see topics 6.5 and 6.6 below)
    • The template strand varies depending on the gene that’s being transcribed.
    • RNA polymerase synthesizes in the 5’ to 3’ direction as it reads DNA in the 3’ to 5’ direction
  3. Describe the roles and key features of the 3 types of RNA in protein synthesis
    • mRNA carries information from DNA to the ribosome. Information is encoded in codons: 3 RNA nucleotides that specify a particular amino acid.
    • tRNAs have an amino acid binding site and an anticodon (a sequence complementary to a codon). The specific binding of anticodon to codon ensures that the amino acid sequence in a polypeptide matches the sequence specified in mRNA
    •  rRNA is the catalytic part of the ribosome, connecting amino acids in a polypeptide chain.
  4. Describe the additional RNA processing that occurs in eukaryotic cells
    • Addition of a poly-A tail.
    • Addition of a GTP cap.
    • Excision of introns and splicing together and retention of exons.
  5. Explain how the organization of eukaryotic genetic material into introns and exons can increase phenotypic variation.
    • Through alternative splicing, exons can be spliced together in alternative ways allowing for the production of multiple protein versions from the same mRNA transcript.

Topic 6.4. Translation

  1. Compare translation in prokaryotes and eukaryotes
    • In all cells, translation occurs at ribosomes. In eukaryotes, some ribosomes are embedded into the rough E.R. (“bound ribosomes”) while other ribosomes float freely in the cytoplasm (“free ribosomes”)
    • In prokaryotes, translation and transcription can occur simultaneously. In eukaryotes, transcription is in the nucleus, and translation in is the cytoplasm.
  2. Define the genetic code, describe its key features, and use a genetic code chart to predict the amino acid sequence that will be generated by a given sequence of mRNA.
    • The genetic code is a set of 3-letter sequences of nucleotides called codons that code for specific amino acids.
    • The code is redundant, with many codons coding for the same amino acid.
    • The code is nearly universal, with only a few variants throughout the living world.
    • The code includes punctuation: codons that signal for translation to start and stop.
  3. Describe the process of translation.
    • Initiation: the small subunit of a ribosome binds with the start codon. A tRNA with an anticodon matching the start codon binds, bringing the first amino acid. The large subunit binds the small subunit.
    • Elongation: As specified by the codon on the mRNA, tRNAs with the designated amino acid bind with the mRNA at the ribosome. The ribosome catalyzes a peptide bond between the newly arrived amino acid and the growing polypeptide chain.
    • Termination: when a stop codon is reached, a release factor causes the polypeptide to be released, and the ribosome dissociates from the mRNA.
  4. Explain how retroviruses violate the central dogma (note: this is covered in the 6.7 extension topic below).
    • Retroviruses use RNA as their genetic code.
    • Reverse transcriptase uses RNA as a template for creating DNA, which is then incorporated into the host’s chromosome. The incorporated virus (a provirus) then exploits the host cell’s replication, transcription, and translation machinery for the replication of new retroviruses.

Topics 6.5 and 6.6. Regulation of Gene Expression and Cell Specialization

  1. Define regulatory sequences.
    • Regulatory sequences are segments of DNA that control the expression of genes, usually by increasing or decreasing the rate of transcription.
  2. Describe how prokaryotic cells use operons to control gene expression.
    • Operons are clusters of genes under the control of a single promoter.
    • The expression of operons is under the control of a regulatory protein, which binds to an operator region that is just downstream from the promoter.
  3. Explain the role of transcription factors in regulating eukaryotic gene expression.
    • Transcription factors are proteins that bind near or at the promoter to regulate the binding of RNA polymerase.
    • Transcription factors can promote, block, or inhibit transcription.
    • Negative regulatory molecules (including small RNAs) inhibit gene expression by binding to DNA and blocking transcription.
  4. Explain how the phenotype of a multicellular organism is determined by gene expression.
    • All cells in a eukaryotic organism have the same DNA.
    • Cells differentiate into specific tissues because they express genes for tissue-specific proteins.
    • These tissue-specific genes are activated through the induction of transcription factors during embryonic development.
    • Induction and gene activation unfold in a hierarchical sequence.
    • Small RNAs also play a role in the regulation of transcription and translation.
  5. Explain how gene expression can be coordinated in eukaryotes
    • In eukaryotes, genes in different tissues can share regulatory sequences that can be activated or repressed by transcription factors to coordinate gene expression.
  6. Define epigenetics.
    • Epigenetics involves reversible modifications to DNA that influence gene expression without changing the DNA sequence.

Topic 6.7 Learn-Biology Extension: Viruses

Note from Mr. W: With two exceptions, the objectives below are not in the College Board’s Course and Exam description. In the context of the COVID-19 Pandemic (not to mention AIDS and other viral diseases), teaching AP Bio without teaching about viruses is unimaginable. Here’s a set of objectives for teaching and learning about viruses. Also, because of the objectives about horizontal gene transfer that are in topic 6.7, it makes sense to teach or learn this material about viruses first.

  1. Explain what a virus is, and how viruses straddle the boundary between living and non-living.
  2. Describe the basic structure of viruses, starting with phage, and including animal viruses.
  3. Describe, compare and contrast, the lytic and lysogenic cycles.
  4. As an illustrative example of a retrovirus, describe the life cycle of HIV. This covers the learning objective about retroviruses from Topic 6.4, above.
  5. Explain how viruses develop their own genetic diversity through recombination and mutation, and how they introduce variation into other species through transduction. This covers part of the learning objective from Topic 6.7 below.
  6. Describe the life cycle and overall biology of SARS-CoV-2.
  7. Explain how mRNA vaccines work to provide protection against virtual infection.
  8. Explain how Rapid Antigen Tests work.

Topic 6.7. Mutation and Horizontal Gene Transfer

  1. Define mutation
  2. Compare and contrast somatic and germline mutations: where they occur, and what their consequences are.
  3. Describe various types of point mutations
    • silent (no change in protein coded for by the mutated gene)
    • missense (one amino acid is substituted for another one)
    • nonsense (a stop codon is substituted for an amino acid)
    • insertions and deletion errors leading to a frameshift (change in reading frame), causing extensive missense (or nonsense).
  4. Explain how mutations come about.
    • Causes can include radiation, reactive chemicals, or errors in DNA replication or DNA repair.
  5. Explain how mutations can be harmful, beneficial, or neutral.
    • Harmful: result in a protein that is nonfunctional or harmful (illustrative examples include mutations in the CFTR gene leading to cystic fibrosis)
    • Beneficial: improves the function of a protein (illustrative examples include the MC1R mutation that leads to adaptive melanism, or the mutation leading to lactase persistence in certain human populations).
    • Neutral: the resulting protein is the same (silent mutation) or similar (because of the type of amino acid substitution).
  6. Explain the overall importance of mutation to evolution.
    • Mutation provides the raw material upon which natural selection acts.
  7. Connect the events of mitosis or meiosis to mutation (see related topics in Unit 5)
    • Changes in chromosome number (polyploidy) resulting from mitosis or meiosis can create new species.
    • Changes in chromosome number caused by nondisjunction during meiosis results in a variety of human developmental disorders (Down syndrome), or chromosomal differences (Turner Syndrome)
  8. Define horizontal gene transfer, and describe various types of mechanisms of horizontal gene transfer
    • Horizontal gene transfer is the uptake of genetic information (as opposed to the inheritance of genetic information from a parent.
    • It occurs primarily in prokaryotes and viruses via transformation, conjugation, and transduction.
    • When two viruses infect the same cell, their genetic information can be combined, leading to viral progeny with novel gene sequences

Topic 6.8. Biotechnology

  1. Explain the basic goals of genetic engineering.
    • Analyzing or manipulating DNA.
  2. Describe the basic method and purpose of electrophoresis
    • Separating molecules of DNA, RNA, or protein according to size and charge, usually for analytical purposes.
  3. Describe the polymerase chain reaction (PCR)
    • During PCR, DNA or RNA fragments are amplified (small amounts are made into larger samples that can be analyzed).
  4. Describe the purpose of bacterial transformation
    • Introduces DNA into bacterial cells, usually to get these cells to express desired proteins (such as with genetically engineered insulin or clotting factor)
    • Transformation is usually preceded by inserting novel genes into plasmids, which are then used as a vector to introduce these genes into bacterial cells where these genes can be replicated and expressed.
  5. Describe the purpose of nucleic acid sequencing
    • Determining the order of nucleotides in DNA or RNA.

2. Unit 6 Cumulative Flashcards

Note: if you’re logged in on the Biomania phone app, you can get the same credit for completing these flashcards. Your progress on the app will be remembered on the website, and vice versa.

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[h] Unit 6 Cumulative Flashcards

[i]

[start]

[q json=”true” yy=”4″ unit=”6.Gene_Expression_and_Regulation” dataset_id=”Unit 6 Cumulative Flashcards|4381144a7f78″ question_number=”1″ topic=”6.1-2.DNA_and_RNA,_DNA_Replication”] Compare and contrast how genetic information is stored in bacteria and eukaryotes.

[a] Bacteria store their DNA in looped, circular chromosomes. Their genome size ranges from 100,000 to 10,000,000 base pairs. Bacterial DNA is “naked” (not wrapped around a protein scaffold).

Eukaryotic DNA is organized into multiple linear chromosomes in which the DNA is wrapped up around histone proteins. Eukaryotic genomes are much larger than bacterial genomes. The human genome, for example, consists of 3.2 billion base pairs. Some plant genomes consist of 150 billion base pairs.

[q json=”true” yy=”4″ unit=”6.Gene_Expression_and_Regulation” topic=”6.1-2.DNA_and_RNA,_DNA_Replication” dataset_id=”Unit 6 Cumulative Flashcards|436e73eb5f78″ question_number=”2″] Compare and contrast the functions of DNA and RNA

[a]

  • DNA is the molecule of heredity in organisms. RNA can play that role in viruses.
  • In organisms, RNA’s function is information transfer related to protein synthesis. The RNAs involved are mRNA, tRNA, and rRNA.
  • In eukaryotes, RNA is also involved in the regulation of gene expression, including splicing out introns from pre-mRNA to create mRNA, and in regulating protein synthesis.

[q json=”true” yy=”4″ unit=”6.Gene_Expression_and_Regulation” dataset_id=”Unit 6 Cumulative Flashcards|43597f805b78″ question_number=”3″ topic=”6.1-2.DNA_and_RNA,_DNA_Replication”] What are plasmids? What’s their function, and how are they used in genetic engineering?

[a] Plasmids (2) are small, extrachromosomal loops of DNA commonly found in bacteria, less commonly in archaea, and rarely in eukaryotes. Plasmids are involved in horizontal gene transfer between bacterial cells through a process called bacterial conjugation: the transferred genes are often for antibiotic resistance. Plasmids are commonly used in genetic engineering as a vector for replicating DNA and for expressing engineered genes within bacterial cells.

[q json=”true” yy=”4″ unit=”6.Gene_Expression_and_Regulation” dataset_id=”Unit 6 Cumulative Flashcards|43448b155778″ question_number=”4″ topic=”6.1-2.DNA_and_RNA,_DNA_Replication”] Describe the structure of DNA, and explain how its structure allows it to serve as the molecule of heredity. Your response should include DNA’s monomers, backbone, base pairing, and overall anti-parallel structure.

[a] DNA is a double-stranded helical molecule composed of nucleotide monomers. Each monomer consists of a 5-carbon sugar (deoxyribose), a phosphate group, and one of four nitrogenous bases. Each strand of the double helix consists of covalently bonded deoxyribose sugars and phosphate groups, which comprise DNA’s sugar-phosphate backbone. Within the helix, the bases can bind with one another through hydrogen bonds. The bonding follows base pairing rules: adenine binds only with thymine, and guanine only with cytosine. For the nucleotides to bind, they have to be oriented upside down relative to one another, making the two strands antiparallel.

[q json=”true” yy=”4″ unit=”6.Gene_Expression_and_Regulation” topic=”6.1-2.DNA_and_RNA,_DNA_Replication” dataset_id=”Unit 6 Cumulative Flashcards|432f96aa5378″ question_number=”5″] Explain how DNA’s structure allows it to serve as the molecule of heredity.

[a] Two features of DNA’s structure allow it to serve as the molecule of heredity.

  1. The four nitrogenous bases in DNA can occur in any order. That allows the sequence of bases to serve as an informational code that specifies sequences of RNA and protein, both of which work to carry out the activities that make life possible.
  2. The specificity of the base pairing rules (A:T and C:G) allows each strand to serve as a template for the synthesis of a complementary strand during DNA replication, ensuring the high-fidelity transmission of genetic information from parent cells to daughter cells.

[q json=”true” yy=”4″ unit=”6.Gene_Expression_and_Regulation” dataset_id=”Unit 6 Cumulative Flashcards|431aa23f4f78″ question_number=”6″ topic=”6.1-2.DNA_and_RNA,_DNA_Replication”] DNA replication is semiconservative. On a big picture level, describe how semiconservative replication occurs (and what that term means).

[a] Durin DNA replication,  a team of enzymes, using each strand of the double helix as a template, synthesizes new daughter strands. As a result, each daughter DNA double helix consists of one conserved strand from the parent molecule, and another strand that was synthesized anew, making the process semi-conservative.

[q json=”true” yy=”4″ unit=”6.Gene_Expression_and_Regulation” dataset_id=”Unit 6 Cumulative Flashcards|4305add44b78″ question_number=”7″ topic=”6.1-2.DNA_and_RNA,_DNA_Replication”] Describe how DNA replication starts. In your response, include the roles of the following:  origin of replication, replication bubble, RNA primase, and primer.

[a] Replication begins when an enzyme called helicase (1) finds a sequence called the origin of replication and separates the double-stranded DNA, exposing two single strands in a structure called a replication fork (7). An enzyme called RNA primase (5) lays down a short stretch of complementary RNA called a primer (4). The primer enables DNA polymerase (2) to bind and to use the template strand (dark blue) as a guide for synthesizing a complementary strand (light blue), always adding new nucleotides at the 3’ end of a growing strand.

[q json=”true” yy=”4″ unit=”6.Gene_Expression_and_Regulation” dataset_id=”Unit 6 Cumulative Flashcards|42f0b9694778″ question_number=”8″ topic=”6.1-2.DNA_and_RNA,_DNA_Replication”] How is DNA replication at the leading strand different from replication at the lagging strand? 

[a] In the leading strand (4), DNA replication is relatively continuous, as DNA polymerase (3) follows the opening replication fork. In the lagging strand, DNA polymerase synthesizes in the opposite direction from the opening replication fork. This results in short sequences that are called Okazaki fragments (5).

[q json=”true” yy=”4″ unit=”6.Gene_Expression_and_Regulation” dataset_id=”Unit 6 Cumulative Flashcards|42d970f25f78″ question_number=”9″ topic=”6.1-2.DNA_and_RNA,_DNA_Replication”] Describe the roles of ligase and topoisomerase in DNA replication.

[a] After DNA polymerase has laid down all of the complementary nucleotides that it can, an enzyme called ligase (N) creates a sugar-phosphate bond between DNA fragments. Throughout the replication process, enzymes called topoisomerases (not shown) nick DNA’s sugar-phosphate backbone, preventing the DNA from overwinding, and then resealing the break.

[!]6.3.Transcription and RNA Processing[/!]

[q json=”true” yy=”4″ unit=”6.Gene_Expression_and_Regulation” dataset_id=”Unit 6 Cumulative Flashcards|42bfd46f9378″ question_number=”10″ topic=”6.3.Transcription_and_RNA_Processing”] List the three principal forms of RNA and describe the function of each one.

[a]

  • mRNA (messenger RNA) brings instructions from DNA to ribosomes, which translate mRNA into protein. mRNA’s information is encoded in 3 base sequences called codons, each of which codes for an amino acid, a start codon, or a stop codon.
  • tRNA (transfer RNA) brings specific amino acids to ribosomes, allowing the ribosomes to bind amino acids together into polypeptides, following the sequence laid down by mRNA. tRNAs have a region called an anticodon: three RNA nucleotides that complement (and bind with) complementary codons on mRNA (which is the basis of the information transfer between RNA and protein).
  • rRNA (ribosomal RNA) makes up the catalytic part of ribosomes and binds amino acids together during protein synthesis.

[q json=”true” yy=”4″ unit=”6.Gene_Expression_and_Regulation” dataset_id=”Unit 6 Cumulative Flashcards|42ad34107378″ question_number=”11″ topic=”6.3.Transcription_and_RNA_Processing”] Explain the overall flow of genetic information of cells.

[a] The basic flow of information in cells is captured in the central dogma of molecular genetics: DNA makes RNA makes protein. More specifically, information flows from a sequence of DNA triplets to mRNA codons to amino acids.

[q json=”true” yy=”4″ unit=”6.Gene_Expression_and_Regulation” dataset_id=”Unit 6 Cumulative Flashcards|42983fa56f78″ question_number=”12″ topic=”6.3.Transcription_and_RNA_Processing”] What happens during transcription?

 

[a]

During transcription, an enzyme called RNA polymerase binds with a promoter on DNA. The RNA polymerase then transcribes the sequence of DNA bases on DNA’s template strand into a sequence of RNA. RNA polymerase reads the DNA in the 3’ to 5’ direction, and syntheses RNA in the 5’ to 3’ direction. When the RNA polymerase reaches a terminator region, it dissociates from the DNA, ending transcription.

[q json=”true” yy=”4″ unit=”6.Gene_Expression_and_Regulation” topic=”6.3.Transcription_and_RNA_Processing” dataset_id=”Unit 6 Cumulative Flashcards|4280f72e8778″ question_number=”13″] How is transcription different in prokaryotes and eukaryotes?

[a] In prokaryotes (1), transcribed RNA (d) can immediately be translated by ribosomes (f) into protein (e). In eukaryotes, RNA (b) is processed within the nucleus before it enters the cytoplasm for translation (the details of which are covered in another card within this topic).

[q json=”true” yy=”4″ unit=”6.Gene_Expression_and_Regulation” dataset_id=”Unit 6 Cumulative Flashcards|42526640b778″ question_number=”14″ topic=”6.3.Transcription_and_RNA_Processing”] Define and describe “template strand,” “minus strand,” “noncoding strand,” or “antisense strand” in relationship to RNA transcription

[a]

When DNA is transcribed into RNA, the template strand (also called the non-coding strand, the antisense strand, and the minus strand is what gets transcribed. The complementary strand is the coding strand. That’s because it has the same sequence of nucleotides as the mRNA will (once the thymines are replaced by uracils and the introns are removed). It’s also called the sense strand or positive strand.

[q json=”true” yy=”4″ unit=”6.Gene_Expression_and_Regulation” dataset_id=”Unit 6 Cumulative Flashcards|423fc5e19778″ question_number=”15″ topic=”6.3.Transcription_and_RNA_Processing”] What is pre-mRNA? Describe some of the post-transcriptional modification that has to happen to pre-mRNA in eukaryotes before it can be translated into protein.

[a] In eukaryotic cells, pre-mRNA (2) is what’s transcribed from a protein-coding gene. Before it can be translated into protein, pre-mRNA is processed with the addition of a 5’ GTP cap (“g”), a 3’ poly-A tail (j), excision of introns (“f”) and splicing together of exons.

[q json=”true” yy=”4″ unit=”6.Gene_Expression_and_Regulation” topic=”6.3.Transcription_and_RNA_Processing” dataset_id=”Unit 6 Cumulative Flashcards|42287d6aaf78″ question_number=”16″] What are the functions of the 5′ GTP cap and the 3′ poly-A-tail that’s added to mRNA during eukaryotic RNA processing?

[a]

The 5’ GTP cap (“g”) protects the mRNA from breakdown by enzymes in the cytoplasm, and also assists the mRNA in leaving the nucleus and binding with a ribosome. The 3’ poly-A tail (j) makes the mRNA more stable and delays its enzymatic breakdown in the cytoplasm.

[q json=”true” yy=”4″ unit=”6.Gene_Expression_and_Regulation” dataset_id=”Unit 6 Cumulative Flashcards|421388ffab78″ question_number=”17″ topic=”6.3.Transcription_and_RNA_Processing”] Describe, on a big-picture level, how transcription is regulated in eukaryotes.

[a] Eukaryotes possess regulatory DNA sequences that interact with regulatory proteins to control transcription. These sequences include promoters (e) and enhancers (a). Interactions between activator proteins (b), DNA bending proteins (f), mediator proteins (g), and general transcription factors (h) enable RNA polymerase (i) to bind, making transcription possible.

[q json=”true” yy=”4″ unit=”6.Gene_Expression_and_Regulation” topic=”6.3.Transcription_and_RNA_Processing” dataset_id=”Unit 6 Cumulative Flashcards|41fe9494a778″ question_number=”18″] Explain how the organization of eukaryotic genetic material into introns and exons can increase phenotypic variation.

[a] Through alternative splicing, exons can be spliced together in alternative ways allowing for the production of multiple protein versions from the same mRNA transcript.

[!]6.4.Translation[/!]

[q json=”true” yy=”4″ unit=”6.Gene_Expression_and_Regulation” topic=”6.4.Translation” dataset_id=”Unit 6 Cumulative Flashcards|41e9a029a378″ question_number=”19″] Explain how free ribosomes (at 3)  become bound (at 4).

[a]

All ribosomes are initially freely floating in the cytoplasm (1). During protein synthesis, some polypeptides are made with a signal sequence (B): a group of amino acid residues that are not part of the final protein, but serve as a “shipping tag.” This signal sequence is bound by a signal recognition particle (A), which, in turn, binds with receptors on the ER membrane. This binding connects the ribosome that was translating this polypeptide to the ER membrane (D). As this bound ribosome synthesizes its protein (E), it delivers it into the ER lumen (the fluid-filled space within the E.R.). Bound ribosomes remain bound until they finish synthesizing their proteins. At the end of the process, these bound ribosomes dissociate, and their subunits return to the cytoplasm (5).

[q json=”true” yy=”4″ unit=”6.Gene_Expression_and_Regulation” dataset_id=”Unit 6 Cumulative Flashcards|41d257b2bb78″ question_number=”20″ topic=”6.4.Translation”] How is translation/protein synthesis different in prokaryotes and eukaryotes?

[a] In prokaryotes (1), transcription and translation are coupled: protein synthesis begins while transcription of mRNA is still taking place. In eukaryotes (2), the processes of transcription and translation are spatially and temporally separate: transcription occurs in the nucleus, and translation is in the cytoplasm.

In prokaryotes, a single mRNA can code for multiple proteins within a specific metabolic pathway; in eukaryotes, the mRNA codes for only a single polypeptide. In eukaryotes, pre-mRNA is modified before translation occurs (the details of which are covered in another card).

[q json=”true” yy=”4″ unit=”6.Gene_Expression_and_Regulation” dataset_id=”Unit 6 Cumulative Flashcards|41bd6347b778″ question_number=”21″ topic=”6.4.Translation”] What is the first phase of translation called? Describe how the process begins.

[a] Translation begins with initiation. The small subunit of the ribosome binds with mRNA and then slides along the mRNA until it reaches the start codon, AUG. The small subunit pauses, waiting until a tRNA with a matching anticodon and carrying the amino acid methionine binds with the start codon. This sets the stage for the large subunit to bind with the small subunit, ending the initiation phase.

[q json=”true” yy=”4″ unit=”6.Gene_Expression_and_Regulation” dataset_id=”Unit 6 Cumulative Flashcards|41a86edcb378″ question_number=”22″ topic=”6.4.Translation”] Describe the elongation phase of protein synthesis.

[a] After initiation, elongation follows. The ribosome has 3 binding sites, E, P, and A. The first tRNA is sitting in the P site. The ribosome waits until a tRNA with an anticodon that complements the second codon binds at the A site, bringing a second amino acid. Once that amino acid is in place, the ribosome catalyzes a peptide bond between the P site and A site amino acids. Next, the ribosome translocates, moving over by one codon. The first tRNA, now in the E site, breaks away, leaving its amino acid behind. The second tRNA is now in the P site, and the A site is empty. Another tRNA with an anticodon matching the third codon binds at the A site, and elongation continues, codon after codon.

[q json=”true” yy=”4″ unit=”6.Gene_Expression_and_Regulation” dataset_id=”Unit 6 Cumulative Flashcards|4195ce7d9378″ question_number=”23″ topic=”6.4.Translation”] Describe the termination phase of protein synthesis.

[a] Termination follows elongation, which ends when the ribosome reaches a stop codon. The stop codon has no corresponding amino acid. Instead of another tRNA, a protein called a release factor enters the A site. This causes the polypeptide to dissociate from the last tRNA. The ribosome itself dissociates, breaking into its large and small subunits. Depending on the cell’s needs, the mRNA might be broken down, or it might be translated again.

[q json=”true” yy=”4″ unit=”6.Gene_Expression_and_Regulation” dataset_id=”Unit 6 Cumulative Flashcards|417e8606ab78″ question_number=”24″ topic=”6.4.Translation”] The genetic code is said to be universal, specific, and redundant. Explain what the genetic code is, and why it has these properties.

[a] The genetic code specifies which three-letter codon sequence in mRNA is translated into which amino acid. It’s universal in that it’s used by all organisms in all three domains of life (bacteria, archaea, and eukarya). The code is nearly identical throughout nature (though there are a few differences where, for example, a stop codon in one group of organisms is translated into a specific amino acid in others). It’s specific in that each codon codes for only one amino acid. It’s redundant in that there are synonyms, with multiple codons coding for the same amino acid.

[q json=”true” yy=”4″ unit=”6.Gene_Expression_and_Regulation” dataset_id=”Unit 6 Cumulative Flashcards|4169919ba778″ question_number=”25″ topic=”6.4.Translation”] What is the central dogma of molecular biology?

[a] The central dogma is the idea that genetic information in living organisms flows from DNA to RNA to Protein. DNA is transcribed into RNA, which is translated into protein.

[q json=”true” yy=”4″ unit=”6.Gene_Expression_and_Regulation” dataset_id=”Unit 6 Cumulative Flashcards|41549d30a378″ question_number=”26″ topic=”6.4.Translation”] In the context of the central dogma of molecular biology, why are retroviruses so unique?

[a] Retroviruses, which include HIV (a), are RNA viruses that use an enzyme called reverse transcriptase (b) to reverse part of the central dogma. Retroviruses inject their RNA genome (c) into eukaryotic cells. Then, reverse transcriptase reverse-transcribes the injected RNA into DNA (f), which can then incorporate into the host cell genome as a provirus (h). Embedded in the cell’s chromosomes, the provirus can exploit the infected cell’s replication cycle, and be replicated along with the rest of the cell each time its host cell divides.

[!]6.5.Regulation of Gene Expression[/!]

[q json=”true” yy=”4″ unit=”6.Gene_Expression_and_Regulation” dataset_id=”Unit 6 Cumulative Flashcards|413fa8c59f78″ question_number=”27″ topic=”6.5.Regulation_of_Gene_Expression”] What is a promoter? 

[a] A promoter is a DNA sequence where RNA polymerase (RNAP) can bind to initiate transcription. The promoter is always upstream from the gene.

[q json=”true” yy=”4″ unit=”6.Gene_Expression_and_Regulation” dataset_id=”Unit 6 Cumulative Flashcards|412ab45a9b78″ question_number=”28″ topic=”6.5.Regulation_of_Gene_Expression”] What are transcription factors?

[a]

Transcription factors (5) are proteins that bind to DNA at or near a gene’s promoter (1). Transcription factors play a regulatory role, enhancing or inhibiting gene expression by promoting or inhibiting the binding of RNA polymerase (6)

[q json=”true” yy=”4″ unit=”6.Gene_Expression_and_Regulation” dataset_id=”Unit 6 Cumulative Flashcards|411813fb7b78″ question_number=”29″ topic=”6.5.Regulation_of_Gene_Expression”] What’s an operon?

[a] An operon is a group of genes transcribed in a single mRNA molecule, which can then be translated into multiple proteins, all of which are part of the same metabolic pathway or process. Operons generally involve a regulatory system in which the expression of the operon is under the control of a regulatory protein that interacts with an operator region near the operon’s promoter (the details of which are covered in several other cards).

[q json=”true” yy=”4″ unit=”6.Gene_Expression_and_Regulation” dataset_id=”Unit 6 Cumulative Flashcards|4105739c5b78″ question_number=”30″ topic=”6.5.Regulation_of_Gene_Expression”] Explain the  difference between an inducible operon (like the lac operon) and a repressible operon (like the trp operon)

[a]  Inducible operons are turned “on” (induced) by the presence of a metabolite. For example, the lac operon produces enzymes for the digestion of the disaccharide lactose.  Expression of the operon is induced by the presence of lactose. When all the lactose is digested, the operon switches off (and no longer produces enzymes for lactose digestion).

Repressible operons are turned “off” (repressed) by the presence of a metabolite. For example, the trp operon produces enzymes for the synthesis of tryptophan, an amino acid.  When there’s no tryptophan in the environment, the operon turns “on,” and the cell produces enzymes needed for tryptophan synthesis. When tryptophan is present it represses the operon.

[q json=”true” yy=”4″ unit=”6.Gene_Expression_and_Regulation” dataset_id=”Unit 6 Cumulative Flashcards|40f07f315778″ question_number=”31″ topic=”6.5.Regulation_of_Gene_Expression”] Explain how the lac operon works when lactose is absent from E. coli’s environment.

[a] The lac operon is an inducible operon, consisting of a regulatory gene (1), a promoter (5), an operator (6), and structural genes (7). The structural genes code for enzymes that digest lactose into monosaccharides. When lactose is absent, the regulatory protein (3) binds to the operator region (6), which blocks RNA polymerase (4), preventing it from transcribing the structural genes (7).

[q json=”true” yy=”4″ unit=”6.Gene_Expression_and_Regulation” topic=”6.5.Regulation_of_Gene_Expression” dataset_id=”Unit 6 Cumulative Flashcards|40d936ba6f78″ question_number=”32″] Explain how the lac operon works when lactose is present in E. coli’s environment.

[a] When lactose (8) is present in the cell’s environment, it diffuses into the cell and binds with the lac regulatory protein (3). Binding causes an allosteric change in the protein (3a), making it unable to bind with the operator region (6). With the operator unblocked, RNA polymerase can transcribe the genes for lactose digestion (7a – 7c).

[q json=”true” yy=”4″ unit=”6.Gene_Expression_and_Regulation” dataset_id=”Unit 6 Cumulative Flashcards|40c6965b4f78″ question_number=”33″ topic=”6.5.Regulation_of_Gene_Expression”] Explain how the trp operon works when the amino acid tryptophan is not present in E. coli’s environment.

[a] The trp operon is repressible, consisting of a regulatory gene (1), a promoter (5), an operator (6), and structural genes (7a – 7e), which produce enzymes that synthesize the amino acid tryptophan. When tryptophan is absent, the regulatory protein cannot bind with the operator. Because the operator isn’t blocked, RNA polymerase (4), can transcribe the structural genes.

[q json=”true” yy=”4″ unit=”6.Gene_Expression_and_Regulation” topic=”6.5.Regulation_of_Gene_Expression” dataset_id=”Unit 6 Cumulative Flashcards|40af4de46778″ question_number=”34″] Explain how the trp operon works when the amino acid tryptophan is available in E. coli’s environment.

[a] When tryptophan (8) is available in the cell’s environment, it diffuses into the cell and binds with the trp regulatory protein (3). This causes an allosteric change in the trp regulator so that it binds with the operator region (6), blocking RNA polymerase from transcribing the genes for tryptophan synthesis.

[q json=”true” yy=”4″ unit=”6.Gene_Expression_and_Regulation” topic=”6.5.Regulation_of_Gene_Expression” dataset_id=”Unit 6 Cumulative Flashcards|4098056d7f78″ question_number=”35″] The lac operon system works through negative feedback. Explain.

[a] The lac operon system is an inducible operon. When lactose is present it binds with the regulatory protein, causing an allosteric shift that keeps that protein from binding with the operator, permitting transcription to occur. When all the lactose is digested, the regulatory protein reverts to its unbound shape, enabling it to bind with the operator again, blocking transcription. Thus, the entire system is a negative feedback system in which the activity of the system has the effect of turning the system off.

[q json=”true” yy=”4″ unit=”6.Gene_Expression_and_Regulation” dataset_id=”Unit 6 Cumulative Flashcards|408311027b78″ question_number=”36″ topic=”6.5.Regulation_of_Gene_Expression”] Using RNA interference as an example, describe the role of small RNA molecules in regulating gene expression.

[a] One of the main functions of small RNA molecules (typically about 22 nucleotides long) is RNA interference. In RNA interference, small RNAs (E) bind with messenger RNA (I), often blocking the translation of the mRNA into protein (J), and sometimes signaling for that mRNA’s enzymatic destruction (K).

[q json=”true” yy=”4″ unit=”6.Gene_Expression_and_Regulation” dataset_id=”Unit 6 Cumulative Flashcards|406bc88b9378″ question_number=”37″ topic=”6.5.Regulation_of_Gene_Expression”] On a big picture level, define and describe “epigenetics.”

[a] “Epigenetics” refers to changes in DNA expression that involve reversible chemical modifications of DNA or modifications in DNA packaging, but not changes in the sequence of nucleotides. In addition to changing DNA expression in cells, epigenetic changes are responsible for the differentiation of tissues during development. There’s also increasing evidence that these changes can be transmitted from one generation to the next.

[q json=”true” yy=”4″ unit=”6.Gene_Expression_and_Regulation” dataset_id=”Unit 6 Cumulative Flashcards|4059282c7378″ question_number=”38″ topic=”6.5.Regulation_of_Gene_Expression”] Explain how histone acetylation and DNA methylation can modify DNA expression.

[a] DNA in eukaryotes is packaged by wrapping the DNA around proteins called histones (3). DNA that is tightly wrapped up (2) is unavailable for transcription, while DNA that is more loosely packed (7) is available for transcription. This association with histones can vary throughout the life of the organism, varying based on developmental timing and influences from the environment.

[q json=”true” yy=”4″ unit=”6.Gene_Expression_and_Regulation” dataset_id=”Unit 6 Cumulative Flashcards|403f8ba9a778″ question_number=”39″ topic=”6.5.Regulation_of_Gene_Expression”] Explain how DNA methylation and acetylation can modify DNA expression.

[a] When DNA is methylated (6), it becomes unavailable for transcription. Modification of histone proteins with acetyl groups (7) makes DNA more available for transcription.

[q json=”true” yy=”4″ unit=”6.Gene_Expression_and_Regulation” dataset_id=”Unit 6 Cumulative Flashcards|40284332bf78″ question_number=”40″ topic=”6.5.Regulation_of_Gene_Expression”] Epigenetic modification of DNA is a key process during development. Explain.

[a] Epigenetics explains why, even though all cells in a multicellular eukaryote have the same DNA, they can be functionally and morphologically different. The differences are a function of epigenetic packaging, acetylation, methylation, and other types of transcriptional control. For example, as a result of epigenetic modification, muscle tissue is expressing muscle-tissue-specific proteins (with most other genes shut down), while nerve cells are expressing nerve-cell-specific proteins (with most other genes shut down).

[q json=”true” yy=”4″ unit=”6.Gene_Expression_and_Regulation” topic=”6.5.Regulation_of_Gene_Expression” dataset_id=”Unit 6 Cumulative Flashcards|400ea6aff378″ question_number=”41″] Epigenetic modifications of DNA can be transmitted from one generation to the next. Explain.

[a] Some epigenetic modifications seem to be transmissible from one generation to the next. This results in a kind of Lamarckian inheritance, where the environment experienced by one generation can epigenetically modify DNA in a way that gets transmitted to the next generation. An example of this has been documented in the children born to women who experienced starvation during their pregnancies during the “Dutch Hunger Winter,” which occurred during World War II in Holland. These children experienced lifelong metabolic changes, depending on when during the pregnancy their mothers experienced starvation. Some of these metabolic changes were even transmissible to the grandchildren of the mothers who were pregnant during WW II.

[q json=”true” yy=”4″ unit=”6.Gene_Expression_and_Regulation” dataset_id=”Unit 6 Cumulative Flashcards|3ff9b244ef78″ question_number=”42″ topic=”6.5.Regulation_of_Gene_Expression”] Using estrogen as an example, explain how gene expression can be coordinated in a variety of tissues.

Illustrative example: Coordinated gene expression in eukaryotes.

[a] In eukaryotes, gene expression can be coordinated by having separate genes share regulatory sequences that respond to the same transcription factors. For example, cells that respond to estrogen develop in a way so that in addition to the tissue-specific genes that they express, they also express the gene for the estrogen receptor. As a result, when estrogen (F) is released from the ovaries, it will bind with these cells’ cytoplasmic estrogen receptors (G) and induce transcription (I) of estrogen-related genes.

 

[!]6.6.Gene Expression and Cell Specialization[/!]

[q json=”true” yy=”4″ unit=”6.Gene_Expression_and_Regulation” dataset_id=”Unit 6 Cumulative Flashcards|3fe4bdd9eb78″ question_number=”43″ topic=”6.6.Gene_Expression_and_Cell_Specialization”] What’s the connection between gene expression and phenotypic differences between cells?

[a] All cells in the same organism are genomically equivalent. They have the same DNA, derived from the zygote (fertilized egg). During development, different genes get expressed. That difference in gene expression is what accounts for the differences among the cells of an organism. For example, the cells making up the lens of your eye possess the same DNA as the cells making up the muscle tissue in your biceps. The difference is that the eye lens cells express only eye-lens proteins, while muscle cells express an array of muscle tissue proteins.

[q json=”true” yy=”4″ unit=”6.Gene_Expression_and_Regulation” dataset_id=”Unit 6 Cumulative Flashcards|3fcd75630378″ question_number=”44″ topic=”6.6.Gene_Expression_and_Cell_Specialization”] Explain what induction is and how it works.

[a] Induction is a key process that occurs during embryonic development when certain cells (indicated by “A”) release morphogens: substances that affect the development of form. Morphogens (“1″) work by activating transcription factors that can influence genes controlling cell division, activation of tissue-specific genes leading to cellular differentiation, or apoptosis (programmed cell death).

[q json=”true” yy=”4″ unit=”6.Gene_Expression_and_Regulation” dataset_id=”Unit 6 Cumulative Flashcards|3fb3d8e03778″ question_number=”45″ topic=”6.6.Gene_Expression_and_Cell_Specialization”] Embryonic development involves sequential gene expression. Explain.

[a] During embryonic development, genes are expressed in a way in which a general form is expressed before a more specific form. For example, in the development of the fruit fly Drosophila melanogaster, among the first genes expressed are those that lay down the head-to-tail axis (1). The next set of genes to be activated divide the embryo into a few major regions (2). The next divide these regions into specific segments (3 and 4), each of which has an anterior end and a posterior end. Finally, homeotic genes are expressed, determining which appendages are expressed on each segment.

[!]6.7.Mutation[/!]

[q json=”true” yy=”4″ unit=”6.Gene_Expression_and_Regulation” dataset_id=”Unit 6 Cumulative Flashcards|3f9ee4753378″ question_number=”46″ topic=”6.7.Mutation”] 1. Define mutation. 2. Define “point mutation” and describe the three effects of point mutations.

[a] A mutation is a change in a DNA sequence (or the entire structure of a chromosome, though that will be addressed in another card). A point mutation involves a change in a single base pair. This change can result in a substitution (for example, substituting an adenine for a guanine), a deletion of a single base, or an insertion of a single base.

[q json=”true” yy=”4″ unit=”6.Gene_Expression_and_Regulation” topic=”6.7.Mutation” dataset_id=”Unit 6 Cumulative Flashcards|3f89f00a2f78″ question_number=”47″] Define “silent mutation,” “missense mutation,” and “nonsense mutation.”

[a]

  • Silent mutations don’t change the amino acid that the gene is coding for. That’s because the genetic code is redundant, with many codons coding for the same amino acid. As a result, substitution mutations often don’t result in an amino acid change, and this is especially common when the substitution is in the 3rd base of a triplet.
  • A missense mutation changes the amino acid in a protein (for example, from glycine to alanine).
  • A nonsense mutation changes a codon coding for an amino acid to one that codes for a stop codon.

[q json=”true” yy=”4″ unit=”6.Gene_Expression_and_Regulation” dataset_id=”Unit 6 Cumulative Flashcards|3f74fb9f2b78″ question_number=”48″ topic=”6.7.Mutation”] Sickle cell disease is caused by single substitution mutation. Explain how one such substitution can cause sickle cell disease.

Illustrative example: Mutation

[a] In sickle cell anemia, a substitution mutation in the gene for hemoglobin changes the DNA triplet CAG to CTG. This causes an acidic amino acid (glutamic acid) to be replaced by a hydrophobic amino acid (valine). This change occurs on the outer surface of hemoglobin, and, in low oxygen situations (such as that caused by physical exertion) it leads hemoglobin molecules to form polymer-like chains (below). These chains distort the shape of red blood cells, causing the sickling that is associated with this disease, and which also causes the disease’s negative effects by blocking capillaries, causing pain crises and tissue damage.

[q json=”true” yy=”4″ unit=”6.Gene_Expression_and_Regulation” dataset_id=”Unit 6 Cumulative Flashcards|3f5db3284378″ question_number=”49″ topic=”6.7.Mutation”] What is a frameshift mutation?

[a] Protein-coding genes have a reading frame. That’s because codons are sequences of 3 RNA bases, and starting with the start codon on a stretch of mRNA, the codons are read one at a time, with each codon being converted into an amino acid. A frameshift mutation involves the insertion or deletion of a nucleotide in DNA in a way that disrupts the reading frame in mRNA, causing all of the codons following the mutation to be misread by the ribosome. The most common result of this is extensive missense (codons coding for the wrong amino acids), though nonsense and silent mutations (coding for a stop codon) are also possible.

[q json=”true” yy=”4″ unit=”6.Gene_Expression_and_Regulation” dataset_id=”Unit 6 Cumulative Flashcards|3f3f6e8daf78″ question_number=”50″ topic=”6.7.Mutation”] Mutations can be positive, negative, or neutral. Explain.

[a]

  • A positive mutation improves a protein (or regulatory pathway, or any other phenotype) in a way that increases evolutionary fitness (an organism’s chance of survival and reproduction).
  • A negative mutation does the opposite (reducing fitness).
  • A neutral mutation does not affect phenotype (because it’s in a non-coding or non-regulatory part of the genome,) or if it’s a mutation that doesn’t impact protein or RNA structure, such as a silent mutation).

[q json=”true” yy=”4″ unit=”6.Gene_Expression_and_Regulation” topic=”6.7.Mutation” dataset_id=”Unit 6 Cumulative Flashcards|3f2a7a22ab78″ question_number=”51″] How are mutations important to evolution?

[a] Mutations provide the raw material upon which natural selection acts, and makes evolution a creative process that results in adaptation. Without mutation, natural selection could only cull harmful variants from a population. With mutation, new variants can arise that increase a population’s fitness.

[q json=”true” yy=”4″ unit=”6.Gene_Expression_and_Regulation” topic=”6.7.Mutation” dataset_id=”Unit 6 Cumulative Flashcards|3f1585b7a778″ question_number=”52″] How do mutations arise?

[a] Mutations can come about through errors in DNA replication or repair. In addition, mutations can be caused by environmental factors such as ionizing radiation or chemicals that interact with and alter DNA.

[q json=”true” yy=”4″ unit=”6.Gene_Expression_and_Regulation” dataset_id=”Unit 6 Cumulative Flashcards|3f00914ca378″ question_number=”53″ topic=”6.7.Mutation”] What is polyploidy? Note: we’ll cover this more in Unit 7, when we learn about speciation

[a] Polyploidy is a doubling (or more) of chromosome number. If it occurs in a gamete during sexual reproduction that produces a viable zygote, it can result in sympatric speciation that occurs, essentially, within one generation. This can also happen if polyploidy occurs during mitosis, followed by vegetative propagation (a form of asexual reproduction that occurs in plants like strawberries or aspen trees). Polyploidy is rare in animals but is a common mechanism for speciation in plants, where between 30% to 80% of species arose through some type of polyploidy.

[q json=”true” yy=”4″ unit=”6.Gene_Expression_and_Regulation” dataset_id=”Unit 6 Cumulative Flashcards|3eeb9ce19f78″ question_number=”54″ topic=”6.7.Mutation”] Contrast “horizontal gene transfer” with “vertical gene transfer.”

[a] In vertical gene transfer, parents transmit all or half of their genome to their offspring (depending on whether reproduction is asexual or sexual, respectively). In horizontal gene transfer, one organism transfers genes to another organism that is not its offspring. If the recipient is unicellular, then these newly acquired genes become part of the recipient’s genome, and will then get passed to its offspring. In a multicellular organism, horizontal transfer only has long-lasting results if the genes are transferred into the germline (the cells that create an egg or sperm cells).

[q json=”true” yy=”4″ unit=”6.Gene_Expression_and_Regulation” dataset_id=”Unit 6 Cumulative Flashcards|3ed6a8769b78″ question_number=”55″ topic=”6.7.Mutation”] Describe bacterial conjugation.

[a] Bacteria have, in addition to their main chromosome, small circles of DNA called plasmids (b). These plasmids can be copied in one bacterial cell, with the copy transmitted to a second bacterial cell via a membrane extension called a pilus (c). This transfer of plasmids transfers whatever genes are encoded on the plasmid to the second cell, which can, in turn, transmit this plasmid to another bacterial cell (or its offspring).

[q json=”true” yy=”4″ unit=”6.Gene_Expression_and_Regulation” topic=”6.7.Mutation” dataset_id=”Unit 6 Cumulative Flashcards|3eb60fd02378″ question_number=”56″] Describe bacterial transformation.

[a] In bacterial transformation, bacteria pick up DNA fragments (a) from the environment, which become incorporated into the genome (c). In genetic engineering, small loops of DNA called plasmids can be forced into bacterial cells, and this forced uptake of plasmids is also referred to as transformation.

[q json=”true” yy=”4″ unit=”6.Gene_Expression_and_Regulation” dataset_id=”Unit 6 Cumulative Flashcards|3ea11b651f78″ question_number=”57″ topic=”6.7.Mutation”] Describe how horizontal gene transfer can occur through viral transduction.

[a]

Transduction is a type of horizontal gene transfer that can occur through viruses. During viral infections, the virus breaks apart the host’s genome. Sometimes, DNA fragments from the host are mistakenly incorporated into a virus. As a result, when that virus infects a cell in another organism, it can bring in that other organism’s DNA. If the virus infects a germ-line cell, then new genes can be incorporated into the gene pool of the recipient.

[q json=”true” yy=”4″ unit=”6.Gene_Expression_and_Regulation” dataset_id=”Unit 6 Cumulative Flashcards|3e89d2ee3778″ question_number=”58″ topic=”6.7.Mutation”] Describe how horizontal gene transfer can occur through endosymbiosis.

[a] Genes can be transferred horizontally between host cells and endosymbionts, such as mitochondria or chloroplasts. Throughout eukaryotic evolution, mitochondria and chloroplasts have transferred up to 99% of their genes to their host cells (with these genes being incorporated into the genome of the host species). Since both mitochondria and chloroplasts are of bacterial origin, this amounts to a significant horizontal transfer of genetic material from bacteria to eukaryotes.

[q json=”true” yy=”4″ unit=”6.Gene_Expression_and_Regulation” dataset_id=”Unit 6 Cumulative Flashcards|3e70366b6b78″ question_number=”59″ topic=”6.7.Mutation”] 1. Describe the basic niche of viruses. 2. Describe the size and structure of viruses.

[a] 1) Viruses are tiny, obligate, intracellular parasites that reproduce themselves by hijacking a host cell’s metabolic machinery and using it to replicate more viruses.

2) Viruses can be as small as 30 nanometers across. The structure of a virus consists of a core of genetic material (DNA or RNA, at “1”) surrounded by proteins that form an outer protein coat (at “2”). Some viruses have an additional phospholipid bilayer, usually stolen from the host cell, and this phospholipid bilayer can contain embedded membrane proteins.

[q json=”true” yy=”4″ unit=”6.Gene_Expression_and_Regulation” dataset_id=”Unit 6 Cumulative Flashcards|3e5d960c4b78″ question_number=”60″ topic=”6.7.Mutation”] Describe the lytic cycle of a DNA virus.

[a] In the lytic cycle, a virus lands on a cell’s outer surface (the cell wall or membrane, at”d”). Next, the virus injects its DNA (“e”) into the cell’s cytoplasm. The  DNA is replicated using the cell’s replication machinery (DNA polymerases and other enzymes). The host cell’s RNA polymerases are used to transcribe and translate the viral DNA into viral proteins. The viral proteins and genetic material self-assemble into new viruses (5). At a certain point, the host bursts open (6), releasing new viruses into the environment.

[q json=”true” yy=”4″ unit=”6.Gene_Expression_and_Regulation” dataset_id=”Unit 6 Cumulative Flashcards|3e464d956378″ question_number=”61″ topic=”6.7.Mutation”] Describe the lysogenic cycle.

[a]  In the lysogenic cycle, a virus injects its DNA into its host. Then the viral DNA  incorporates itself into the host cell’s chromosome as a prophage (in a bacterial host) or a provirus (in a eukaryotic host). Each time the host cell replicates, it also replicates the viral DNA. At a certain point, the prophage or provirus breaks out and initiates a lytic cycle, replicating many new viruses and destroying its host in the process.

[q json=”true” yy=”4″ unit=”6.Gene_Expression_and_Regulation” dataset_id=”Unit 6 Cumulative Flashcards|3e31592a5f78″ question_number=”62″ topic=”6.7.Mutation”] Explain how viral life cycles can generate genetic and phenotypic diversity within their own viral populations.

[a] Genetic and phenotypic diversity within viral populations can arise in two ways. The first is mutation, which occurs during viral replication. The second is the mixing of viral genomes. When two viruses invade the same host cell, the resulting viral offspring can wind up with recombined DNA from the two invading viruses. This type of recombination in the influenza virus, which can infect humans, birds, and pigs, explains the virus’s variability from year to year (and the resulting need for the development of a new influenza vaccine each year).

[!]6.8.Biotechnology[/!]

[q json=”true” yy=”4″ unit=”6.Gene_Expression_and_Regulation” dataset_id=”Unit 6 Cumulative Flashcards|3e1c64bf5b78″ question_number=”63″ topic=”6.8.Biotechnology”] It takes several steps to transfer human genes to bacterial cells that can subsequently produce human gene products (such as the human protein insulin). One step involves dealing with introns. Explain what introns are, why they’re a problem, and how they can be removed.

Illustrative example: Genetic Engineering

[a] Introns are non-coding sequences of DNA within eukaryotic genes that have to be spliced out before the gene’s RNA can be translated into protein. So, to transfer a human gene to a bacterium to create a gene product, you have to use DNA from which the introns have been removed. You can get this intron-free in two ways. First, you can find cells that produce the desired protein, extract the mRNA that codes for this protein from those cells, and then use reverse transcriptase to create cDNA (complementary DNA) from the mRNA. Alternatively, if you know the amino acid sequence for the protein, you can reverse-engineer DNA that codes for that amino acid sequence.

[q json=”true” yy=”4″ unit=”6.Gene_Expression_and_Regulation” dataset_id=”Unit 6 Cumulative Flashcards|3e0770545778″ question_number=”64″ topic=”6.8.Biotechnology”] Using restriction enzymes and DNA ligase, you can create a recombinant plasmid with a human gene that can be used in genetic engineering. Explain how. Note: you can assume that introns have already been removed from the human DNA.

[a] To create a recombinant plasmid, first cut the target human DNA with a restriction enzyme (2) so that its ends are “sticky,” with single-stranded overhangs that can form hydrogen bonds with complementary sequences. Second, extract a plasmid (3) from a bacterial cell, and then cut open that plasmid with the same restriction enzyme that you used to cut the sticky ends on the human DNA. Third, combine the human DNA fragment and the plasmid DNA. Because they’ve been cut with the same restriction enzyme, they’ll combine by forming hydrogen bonds between their complementary sticky ends. Finally, use DNA ligase (4) to create sugar-phosphate bonds between adjacent, unbonded nucleotides. The result is a recombinant plasmid (g) that contains a human gene.

[q json=”true” yy=”4″ unit=”6.Gene_Expression_and_Regulation” dataset_id=”Unit 6 Cumulative Flashcards|3df027dd6f78″ question_number=”65″ topic=”6.8.Biotechnology”] What is gel electrophoresis? How is it used to analyze DNA?

[a]

Gel electrophoresis is used to sort molecules by size and/or electrical charge. It’s the basis of restriction fragment analysis, commonly referred to as DNA fingerprinting (though it can be used to analyze other types of molecules as well, such as proteins.

[q json=”true” yy=”4″ unit=”6.Gene_Expression_and_Regulation” dataset_id=”Unit 6 Cumulative Flashcards|3dd4374ebf78″ question_number=”66″ topic=”6.8.Biotechnology”] How does gel electrophoresis work?

[a] Gel electrophoresis involves placing molecules in a porous gel (5) and then running an electrical current through the gel. Because DNA’s phosphate groups are negatively charged, DNA fragments will move away from the negatively charged side of an electrophoresis chamber (2) and toward the positive side (4). As this occurs, small fragments will be less impeded by the gel than large fragments, meaning that during the same period, smaller fragments will move more than larger ones, enabling the fragments to be sorted by size.

[q json=”true” yy=”4″ unit=”6.Gene_Expression_and_Regulation” dataset_id=”Unit 6 Cumulative Flashcards|3db846c00f78″ question_number=”67″ topic=”6.8.Biotechnology”] Restriction enzymes and gel electrophoresis can be used to create unique patterns (often called “DNA fingerprints”) that are widely used in biological research and genetic engineering. Explain how this process could be used to distinguish between the DNA of any two organisms.

[a] Electrophoresis can be used to create “DNA fingerprints.” Because the DNA of any two organisms (who are not clones) is different, applying the same restriction enzymes to different DNA will result in differences in the restriction fragments that result. That’s because some of the restriction sites in one organism’s DNA won’t be present in another organism’s DNA, leading to restriction fragment length polymorphisms (RFLPs). Subsequent electrophoresis (which sorts these fragments by size), followed by using probes that enable visualization of a small percentage of the fragments, creates a unique pattern.

[q json=”true” yy=”4″ unit=”6.Gene_Expression_and_Regulation” dataset_id=”Unit 6 Cumulative Flashcards|3da0fe492778″ question_number=”68″ topic=”6.8.Biotechnology”] What is PCR? What is it used for, and how does it work?

[a] PCR stands for “polymerase chain reaction.” It’s a cell-free technique for cloning DNA. PCR requires the DNA sample that you want to clone (a), primers—short strands of single-stranded DNA that bind to sequences at the start of the DNA that you want to amplify (c), heat-resistant DNA polymerase (g), and free nucleotides for DNA synthesis (b). PCR involves repeated cycles of heating the DNA to separate it into single strands (1); then cooling the DNA so that primers can bind and so that DNA polymerase can synthesize new DNA (2 and 3)) Every heating and cooling cycle doubles the amount of DNA. After 10 cycles you have 1000 times more DNA than you started with. After  30 cycles, you’ve amplified your DNA a billionfold.

[q json=”true” yy=”4″ unit=”6.Gene_Expression_and_Regulation” dataset_id=”Unit 6 Cumulative Flashcards|3d6b7137ab78″ question_number=”69″ topic=”6.8.Biotechnology”] What is DNA sequencing?

Illustrative example: Genetic Engineering.

[a] Sequencing is taking a sample of DNA – anything from a small fragment to the entire genome of an organism – and figuring out the specific sequence of A, T, C, and G nucleotides that make it up.

[x] [restart]

[/qdeck]

3. Unit 6 Practice Multiple Choice Quiz 1

[qwiz qrecord_id=”sciencemusicvideosMeister1961-Unit 6 Cumulative Practice Multiple Choice” dataset=”Unit 6 Cumulative Multiple Choice” style=”width: 600px !important; min-height: 450px !important;”]

[h] Unit 6 Cumulative Multiple Choice Quiz 1

[i]

[q json=”true” multiple_choice=”true” unit=”6.Gene Expression and Regulation” topic=”6.1.DNA_and_RNA” dataset_id=”Unit 6 Cumulative Multiple Choice|13bb16ec9be6e2″ question_number=”1″] Infectious Bursal Disease Virus is a double-stranded RNA virus. If the percentage of guanine in this virus is about 23%, then what is the percentage of uracil?

[c]IDQ2JSA=[Qq][c]IDIzJSA=[Qq][c]IDI3 JSA=[Qq][c]IDU0JQ==

Cg==[Qq]

[f]IE5vLiBIZXJlJiM4MjE3O3MgYSBoaW50LiBJbiBhIGRvdWJsZS1zdHJhbmRlZCBSTkEgdmlydXMsIHRoZSBwZXJjZW50YWdlIG9mIEd1YW5pbmUgd2lsbCBtYXRjaCB0aGF0IG9mIEN5dG9zaW5lLCBhbmQgdGhlIHBlcmNlbnRhZ2Ugb2YgQWRlbmluZSB3aWxsIG1hdGNoIHRoYXQgb2YgVXJhY2lsLiBUaGVyZWZvcmUgaWYgR3VhbmluZSBpcyAyMyUsIHRoZW4gY3l0b3NpbmUgYWxzbyBoYXMgdG8gYmUgMjMlLiBBZGQgdGhvc2UgdHdvIHBlcmNlbnRhZ2VzIHRvZ2V0aGVyLMKgc3VidHJhY3QgZnJvbSAxMDAsIGFuZCBkaXZpZGUgdGhhdCBpbiBoYWxmIHRvIGdldCB5b3VyIGFuc3dlci4=[Qq]

[f]IE5vLiBIZXJlJiM4MjE3O3MgYSBoaW50LiBJbiBhIGRvdWJsZS1zdHJhbmRlZCBSTkEgdmlydXMsIHRoZSBwZXJjZW50YWdlIG9mIEd1YW5pbmUgd2lsbCBtYXRjaCB0aGF0IG9mIEN5dG9zaW5lLCBhbmQgdGhlIHBlcmNlbnRhZ2Ugb2YgQWRlbmluZSB3aWxsIG1hdGNoIHRoYXQgb2YgVXJhY2lsLiBUaGVyZWZvcmUgaWYgR3VhbmluZSBpcyAyMyUsIHRoZW4gY3l0b3NpbmUgYWxzbyBoYXMgdG8gYmUgMjMlLiBBZGQgdGhvc2UgdHdvIHBlcmNlbnRhZ2VzIHRvZ2V0aGVyLMKgc3VidHJhY3QgZnJvbSAxMDAsIGFuZCBkaXZpZGUgdGhhdCBpbiBoYWxmIHRvIGdldCB5b3VyIGFuc3dlci4=[Qq]

[f]IEZhYnVsb3VzLiBJbiBhIGRvdWJsZS1zdHJhbmRlZCBSTkEgdmlydXMsIHRoZSBwZXJjZW50YWdlIG9mIEd1YW5pbmUgd2lsbCBtYXRjaCB0aGF0IG9mIEN5dG9zaW5lLCBhbmQgdGhlIHBlcmNlbnRhZ2Ugb2YgQWRlbmluZSB3aWxsIG1hdGNoIHRoYXQgb2YgVXJhY2lsLiBUaGVyZWZvcmUgaWYgR3VhbmluZSBpcyAyMyUsIHRoZW4gY3l0b3NpbmUgYWxzbyBoYXMgdG8gYmUgMjMlLiBJZiBHICsgQyA9IDQ2JSwgdGhlbiBBICsgVSBoYXMgdG8gYmUgNTQlLiBIYWxmIG9mIDU0IGlzIDI3Lg==[Qq]

[f]IE5vLiBIZXJlJiM4MjE3O3MgYSBoaW50LiBJbiBhIGRvdWJsZS1zdHJhbmRlZCBSTkEgdmlydXMsIHRoZSBwZXJjZW50YWdlIG9mIEd1YW5pbmUgd2lsbCBtYXRjaCB0aGF0IG9mIEN5dG9zaW5lLCBhbmQgdGhlIHBlcmNlbnRhZ2Ugb2YgQWRlbmluZSB3aWxsIG1hdGNoIHRoYXQgb2YgVXJhY2lsLiBUaGVyZWZvcmUgaWYgR3VhbmluZSBpcyAyMyUsIHRoZW4gY3l0b3NpbmUgYWxzbyBoYXMgdG8gYmUgMjMlLiBBZGQgdGhvc2UgdHdvIHBlcmNlbnRhZ2VzIHRvZ2V0aGVyLMKgc3VidHJhY3QgZnJvbSAxMDAsIGFuZCBkaXZpZGUgdGhhdCBpbiBoYWxmIHRvIGdldCB5b3VyIGFuc3dlci4=

Cg==

[Qq]

[q json=”true” multiple_choice=”true” unit=”6.Gene Expression and Regulation” topic=”6.1.DNA_and_RNA” dataset_id=”Unit 6 Cumulative Multiple Choice|13bb01f830e2e2″ question_number=”2″] The genetic material of a newly discovered virus is analyzed. Its base composition is

  • 11% A,
  • 32% G,
  • 18% U, and
  • 39% C.

What is the best description of this virus’s genetic material?

[c]IHNpbmdsZS1zdH JhbmRlZCBSTkE=[Qq]

[f]IEV4YWN0bHkuIFRoZSBwcmVzZW5jZSBvZiBVICh1cmFjaWwpIHNob3dzIHRoYXQgaXQmIzgyMTc7cyBhbiBSTkEgdmlydXMuIElmIGl0IHdlcmUgZG91YmxlLXN0cmFuZGVkLCB0aGVuIHRoZSBwZXJjZW50YWdlIG9mIEEgd291bGQgbWF0Y2ggVSBhbmQgQyB3b3VsZCBtYXRjaCBHLiBCZWNhdXNlIGl0IGRvZXNuJiM4MjE3O3QgbWF0Y2gsIHdlIGtub3cgdGhlIFJOQSBpcyBzaW5nbGUtc3RyYW5kZWQu[Qq]

[c]IGRvdWJsZS1zdHJhbmRlZCBSTkE=[Qq]

[f]IE5vLiBJbiBkb3VibGUtc3RyYW5kZWQgUk5BLCB0aGUgcGVyY2VudGFnZSBvZiBBIHdvdWxkIG1hdGNoIFUgYW5kIEMgd291bGQgbWF0Y2ggRy4=[Qq]

[c]IHNpbmdsZS1zdHJhbmRlZCBETkE=[Qq]

[f]IE5vLiBUaGUgYmFzZSBVICh1cmFjaWwpIGlzIG5ldmVyIHByZXNlbnQgaW4gRE5BLg==[Qq]

[c]IGRvdWJsZS1zdHJhbmRlZCBETkE=[Qq]

[f]IE5vLiBUaGUgYmFzZSBVICh1cmFjaWwpIGlzIG5ldmVyIHByZXNlbnQgaW4gRE5BLg==

Cg==

Cg==

[Qq]

[q json=”true” multiple_choice=”true” unit=”6.Gene Expression and Regulation” topic=”6.2.Replication” dataset_id=”Unit 6 Cumulative Multiple Choice|13baeaafb9fae2″ question_number=”3″] The images below show various enzymes involved in DNA replication. Which of the following correctly identifies the enzymes?

[c]IELigJRoZWxpY2FzZTsgR+KAlEROQSBQb2x5bWVyYXNlIElJ SSwgS+KAlEROQSBQb2x5bWVyYXNlIEksIE7igJRsaWdhc2U=[Qq]

[f]IEV4Y2VsbGVudCEgSGVsaWNhc2UgKEIpIGlzIG9wZW5pbmcgdXAgdGhlIGhlbGl4OyBETkEgcG9seW1lcmFzZSBJSUkgKEcpwqAgaXMgYWRkaW5nIG5ldyBudWNsZW90aWRlcyB0byB0aGUgZ3Jvd2luZyBkYXVnaHRlciBzdHJhbmQ7IEROQSBwb2x5bWVyYXNlIEkgKEspIGlzIHJlcGxhY2luZyB0aGUgUk5BIHByaW1lciB3aXRoIEROQTsgRE5BIGxpZ2FzZSAoTikgaXMgc2VhbGluZyB0aGUgZ2FwcyBiZXR3ZWVuIGZyYWdtZW50cy4=[Qq]

[c]IELigJRETkEgcG9seW1lcmFzZSBJOyBH4oCUaGVsaWNhc2UsIEvigJRETkEgUG9seW1lcmFzZSBJSUksIE7igJRsaWdhc2U=[Qq]

[f]IE5vLiBZb3Ugd2VyZSByaWdodCBhYm91dCBoZWxpY2FzZSBhbmQgbGlnYXNlLiBOZXh0IHRpbWUsIHJlbWVtYmVyIHRoYXQgRE5BIHBvbHltZXJhc2UgMSByZXBsYWNlcyB0aGUgUk5BIHByaW1lciB3aXRoIEROQS4=[Qq]

[c]IELigJRsaWdhc2U7IEfigJRoZWxpY2FzZSwgS+KAlEROQSBQb2x5bWVyYXNlIElJSSwgTuKAlEROQSBQb2x5bWVyYXNlIDE=[Qq]

[f]IE5vLiBZb3Ugd2VyZSByaWdodCBhYm91dCBoZWxpY2FzZS4gTmV4dCB0aW1lLCByZW1lbWJlciB0aGF0IGxpZ2FzZSBzZWFscyB0aGUgZ2FwcyBiZXR3ZWVuIGZyYWdtZW50cyB3aXRoIGEgc3VnYXItcGhvc3BoYXRlIGJvbmQsIGFuZCBETkEgcG9seW1lcmFzZSAxIHJlcGxhY2VzIHRoZSBSTkEgcHJpbWVyIHdpdGggRE5BLg==[Qq]

[c]IELigJRoZWxpY2FzZTsgR+KAlEROQSBsaWdhc2UsIEvigJRETkEgUG9seW1lcmFzZSBJSUksIE7igJRETkEgUG9seW1lcmFzZSAx[Qq]

[f]IE5vLiBZb3Ugd2VyZSByaWdodCBhYm91dCBoZWxpY2FzZS4gTmV4dCB0aW1lLCByZW1lbWJlciB0aGF0IGxpZ2FzZSBzZWFscyB0aGUgZ2FwcyBiZXR3ZWVuIGZyYWdtZW50cyB3aXRoIGEgc3VnYXItcGhvc3BoYXRlIGJvbmQsIGFuZCBETkEgcG9seW1lcmFzZSAxIHJlcGxhY2VzIHRoZSBSTkEgcHJpbWVyIHdpdGggRE5BLg==

Cg==

[Qq]

[q json=”true” xyz=”2″ multiple_choice=”true” unit=”6.Gene Expression and Regulation” dataset_id=”Unit 6 Cumulative Multiple Choice|13bad3674312e2″ question_number=”4″ topic=”6.2.Replication”] The diagram below shows the amount of DNA per cell in a dividing cell. Which roman numeral would require the involvement of enzymes like DNA polymerase, helicase, and ligase?

[c]IElJ IA==[Qq][c]IElJSSA=[Qq][c]IElWIA==[Qq][c]IFYg[Qq][c]IFZJ

Cg==[Qq]

[f]IEV4Y2VsbGVudCEgRW56eW1lcyBsaWtlIEROQSBwb2x5bWVyYXNlLCBoZWxpY2FzZSwgYW5kIGxpZ2FzZSB3b3VsZCBiZSBpbnZvbHZlZCBpbiB0aGUgc3ludGhlc2lzIG9mIEROQSB0aGF0IGhhcHBlbnMgZHVyaW5nIGludGVycGhhc2UgMSwgd2hpY2ggaXMgc2hvd24gaW4gc3RlcCBJSS4=[Qq]

[f]IE5vLiBBdCAmIzgyMjA7SUlJJiM4MjIxOyB0aGUgbGluZSBpcyBmbGF0LCBpbmRpY2F0aW5nIHRoYXQgdGhlIGFtb3VudCBvZiBETkEgaXMgc3RheWluZyBjb25zdGFudCBvdmVyIHRpbWUuIFRoZSBlbnp5bWVzIG1lbnRpb25lZCBhYm92ZSBhcmUgYWxsIGludm9sdmVkIGluIEROQSBzeW50aGVzaXMuIFN5bnRoZXNpcyBpbmNyZWFzZXMgdGhlIGFtb3VudCBvZiBETkEuIFdoYXQmIzgyMTc7cyB0aGUgb25seSBudW1iZXIgdGhhdCBzaG93cyB0aGUgYW1vdW50IG9mIEROQSBnb2luZyB1cD8=[Qq]

[f]IE5vLiBBdCAmIzgyMjA7SVYmIzgyMjE7IHRoZSBsaW5lIGlzIGZsYXQsIGluZGljYXRpbmcgdGhhdCB0aGUgYW1vdW50IG9mIEROQSBpcyBzdGF5aW5nIGNvbnN0YW50IG92ZXIgdGltZS4gVGhlIGVuenltZXMgbWVudGlvbmVkIGFib3ZlIGFyZSBhbGwgaW52b2x2ZWQgaW4gRE5BIHN5bnRoZXNpcy4gU3ludGhlc2lzIGluY3JlYXNlcyB0aGUgYW1vdW50IG9mIEROQS4gV2hhdCYjODIxNztzIHRoZSBvbmx5IG51bWJlciB0aGF0IHNob3dzIHRoZSBhbW91bnQgb2YgRE5BIGdvaW5nIHVwPw==[Qq]

[f]IE5vLiBBdCAmIzgyMjA7ViYjODIyMTsgdGhlIGxpbmUgaXMgc2xvcGluZyBkb3dud2FyZCwgaW5kaWNhdGluZyB0aGF0IHRoZSBhbW91bnQgb2YgRE5BIGlzIGRlY3JlYXNpbmcgb3ZlciB0aW1lLiBUaGUgZW56eW1lcyBtZW50aW9uZWQgYWJvdmUgYXJlIGFsbCBpbnZvbHZlZCBpbiBETkEgc3ludGhlc2lzLiBTeW50aGVzaXMgaW5jcmVhc2VzIHRoZSBhbW91bnQgb2YgRE5BLiBXaGF0JiM4MjE3O3MgdGhlIG9ubHkgbnVtYmVyIHRoYXQgc2hvd3MgdGhlIGFtb3VudCBvZiBETkEgZ29pbmcgdXA/[Qq]

[f]IE5vLiBBdCAmIzgyMjA7VkkmIzgyMjE7IHRoZSBsaW5lIGlzIGZsYXQsIGluZGljYXRpbmcgdGhhdCB0aGUgYW1vdW50IG9mIEROQSBpcyBzdGF5aW5nIGNvbnN0YW50IG92ZXIgdGltZS4gVGhlIGVuenltZXMgbWVudGlvbmVkIGFib3ZlIGFyZSBhbGwgaW52b2x2ZWQgaW4gRE5BIHN5bnRoZXNpcy4gU3ludGhlc2lzIGluY3JlYXNlcyB0aGUgYW1vdW50IG9mIEROQS4gV2hhdCYjODIxNztzIHRoZSBvbmx5IG51bWJlciB0aGF0IHNob3dzIHRoZSBhbW91bnQgb2YgRE5BIGdvaW5nIHVwPw==

Cg==

[Qq]

[q json=”true” multiple_choice=”true” unit=”6.Gene Expression and Regulation” topic=”6.2.Replication” dataset_id=”Unit 6 Cumulative Multiple Choice|13babe72d80ee2″ question_number=”5″] The image below is a model of two steps that occur in sequence during DNA replication.

Enzyme “C” is

[c]IEROQSBoZWxpY2FzZQ==[Qq]

[f]IE5vLiBIZWxpY2FzZSBpcyB0aGUgZW56eW1lIHRoYXQgc2VwYXJhdGVzIHRoZSBzdHJhbmRzIGF0IHRoZSBvcmlnaW4gb2YgcmVwbGljYXRpb24gKHdoaWNoIGlzIHRoZSAxc3Qgc3RlcCBpbiB0aGUgcmVwbGljYXRpb24gcHJvY2VzcykuIEhlcmUmIzgyMTc7cyBhIGhpbnQ6IEVuenltZSAmIzgyMjA7QyYjODIyMTsgaXMgaW5jb3Jwb3JhdGluZyBmcmVlIG51Y2xlb3RpZGVzIGludG8gYSBuZXcgZGF1Z2h0ZXIgc3RyYW5kIG9mIEROQS4gSW4gb3RoZXIgd29yZHMsIGl0JiM4MjE3O3MgY29udmVydGluZyBtb25vbWVycyBpbnRvIGEgcG9seW1lci4=[Qq]

[c]IEROQSBsaWdhc2U=[Qq]

[f]IE5vLiBETkEgbGlnYXNlJiM4MjE3O3MgZnVuY3Rpb24gaXMgdG8gY3JlYXRlIHN1Z2FyLXBob3NwaGF0ZSBib25kcyBiZXR3ZWVuIGFkamFjZW50IHNlZ21lbnRzIG9mIEROQS4gSGVyZSYjODIxNztzIGEgaGludDogRW56eW1lICYjODIyMDtDJiM4MjIxOyBpcyBpbmNvcnBvcmF0aW5nIGZyZWUgbnVjbGVvdGlkZXMgaW50byBhIG5ldyBkYXVnaHRlciBzdHJhbmQgb2YgRE5BLiBJbiBvdGhlciB3b3JkcywgaXQmIzgyMTc7cyBjb252ZXJ0aW5nIG1vbm9tZXJzIGludG8gYSBwb2x5bWVyLg==[Qq]

[c]IEROQSBwb2x5bW VyYXNlIElJSQ==[Qq]

[f]IE5pY2VseSBkb25lLiBFbnp5bWUgJiM4MjIwO0MmIzgyMjE7IGlzIEROQSBwb2x5bWVyYXNlIElJSS4=[Qq]

[c]IHByaW1hc2U=[Qq]

[f]IE5vLiBQcmltYXNlJiM4MjE3O3MgZnVuY3Rpb24gaXMgdG8gbGF5IGRvd24gYSBzaG9ydCBzdHJldGNoIG9mIFJOQSAoc2hvd24gYXQgJiM4MjIwO0ImIzgyMjE7IGluIHN0ZXAgMSBhYm92ZSkgYXQgdGhlIG9yaWdpbiBvZiByZXBsaWNhdGlvbi4gSGVyZSYjODIxNztzIGEgaGludDogRW56eW1lICYjODIyMDtDJiM4MjIxOyBpcyBpbmNvcnBvcmF0aW5nIGZyZWUgbnVjbGVvdGlkZXMgaW50byBhIG5ldyBkYXVnaHRlciBzdHJhbmQgb2YgRE5BLiBJbiBvdGhlciB3b3JkcywgaXQmIzgyMTc7cyBjb252ZXJ0aW5nIG1vbm9tZXJzIGludG8gYSBwb2x5bWVyLg==

Cg==

Cg==

[Qq]

[q json=”true” multiple_choice=”true” unit=”6.Gene Expression and Regulation” topic=”6.3.Transcription_and_RNA_Processing” dataset_id=”Unit 6 Cumulative Multiple Choice|13baa72a6126e2″ question_number=”6″] In the model below

[c]IE51bWJlciA1IGlzIEROQSBwb2x5bWVyYXNlLCBhbmQgdGhlIHByb2Nlc3MgaXMgRE5BIHJlcGxpY2F0aW9uLg==[Qq]

[f]IE5vLiBIZXJlJiM4MjE3O3MgYSBoaW50LiBOb3RpY2UgdGhhdCBzdHJhbmQgdGhhdCYjODIxNztzIG51bWJlcmVkICYjODIyMDsyJiM4MjIxOyBpcyBwYXJ0IG9mIGEgZG91YmxlIGhlbGl4IHRoYXQmIzgyMTc7cyBjb21wb3NlZCBvZiB0aGUgbW9ub21lcnMgQSwgRywgQywgYW5kIFQuIFRoZSBzdHJhbmQgdGhhdCYjODIxNztzIG51bWJlcmVkICYjODIyMDsxJiM4MjIxOyBpcyBzaW5nbGUtc3RyYW5kZWQgYW5kIGNvbXBvc2VkIG9mIEEsIEcsIEMsIGFuZCBVLiBXaGF0IGlzIG1vbGVjdWxlIDIsIHdoYXQgaXMgbW9sZWN1bGUgMSwgYW5kIHdoYXQgZW56eW1lIGFuZCBwcm9jZXNzIGNvbnZlcnRzIHRoZSBmb3JtZXIgdG8gdGhlIGxhdHRlcj8=[Qq]

[c]IE51bWJlciA1IGlzIGEgcmlib3NvbWUsIGFuZCB0aGUgcHJvY2VzcyBpcyB0cmFuc2xhdGlvbi4=[Qq]

[f]IE5vLiBUcmFuc2xhdGlvbiBjb252ZXJ0cyBtUk5BIGludG8gYSBzZXF1ZW5jZSBvZiBhbWlubyBhY2lkcywgYXMgc2hvd24gYmVsb3cu

Cg==

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

Cg==

[Qq]

[c]IE51bWJlciA1IGlzIGEgcmlib3NvbWUsIGFuZCB0aGUgcHJvY2VzcyBpcyB0cmFuc2NyaXB0aW9uLg==[Qq]

[f]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

Cg==

[Qq]

[c]IE51bWJlciA1IGlzIFJOQSBwb2x5bWVyYXNlLCBhbm QgdGhlIHByb2Nlc3MgaXMgdHJhbnNjcmlwdGlvbi4=[Qq]

[f]IEV4Y2VsbGVudC4gUk5BIHBvbHltZXJhc2UgKDUpIHRyYW5zY3JpYmVzIFJOQSAoMSkgZnJvbSBhIEROQSB0ZW1wbGF0ZSAoMik=

Cg==

[Qq]

[q json=”true” xyz=”2″ multiple_choice=”true” topic=”6.3.Transcription_and_RNA_Processing” dataset_id=”Unit 6 Cumulative Multiple Choice|13ba8b39d276e2″ question_number=”7″ unit=”6.Gene Expression and Regulation”] If you wanted to genetically engineer bacteria that could synthesize the same protein that’s being synthesized below, the sequence you would insert in these bacteria would be

 

[c]IFRBQyBUQUMgR0dDIENUQyBDVEcgQ0NUIEdHRyBHR0EgQ1RBIEFDVA==[Qq]

[f]IE5vLiBIZXJlJiM4MjE3O3MgaG93IHRvIGRvIHRoaXMu

Cg==

SGVyZSYjODIxNztzIHRoZSBtUk5BIGZyb20gc3RhZ2UgSUku

Cg==

[Qq]

Now use the base pairing rules between RNA and DNA to figure out the DNA: A bonds with T, U with A, C with G, and G with C. The first DNA bases (organized into triplets) will be

TAC TAC GGC …

[c]IFRBQyBUQUMgR0dDIENUVCBDVEcgQ0NUIENDQyBHR0EgQ1RBIEFDVA==[Qq]

[f]IEhlcmUmIzgyMTc7cyB0aGUgbVJOQSBmcm9tIHN0YWdlIElJLg==

Cg==

Cg==

[Qq]Now use the base pairing rules between RNA and DNA to figure out the DNA: A bonds with T, U with A, C with G, and G with C. The first DNA bases (organized into triplets) will be

TAC TAC GGC …

[c]IFRBQyBUQUMgR0dDIENUVCBDVEcg Q0NUIEdHRyBHR1QgQ1RBIEFDVA==[Qq]

[f]IEV4Y2VsbGVudC4gWW91JiM4MjE3O3ZlIHVzZWQgdGhlIGJhc2UtcGFpcmluZyBydWxlcyBiZXR3ZWVuIFJOQSBhbmQgRE5BIHRvIGZpZ3VyZSBvdXQgdGhlIEROQSAoIEEgYm9uZHMgd2l0aCBULCBVIHdpdGggQSwgQyB3aXRoIEcsIGFuZCBHIHdpdGggQyk=[Qq]

[c]IEl0IGNhbiYjODIxNzt0IGJlIGRldGVybWluZWQgYmVjYXVzZSB0aGUgZGlhZ3JhbSBkb2VzbiYjODIxNzt0IHNob3cgdGhlIHNlcXVlbmNlIG9mIHRoZSBpbnRyb24gRE5BIHRoYXQgd2FzIHJlbW92ZWQgYXQgc3RlcCBGIGluIHN0YWdlIEku[Qq]

[f]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

Cg==

Cg==

[Qq]

[q json=”true” xyz=”2″ multiple_choice=”true” unit=”6.Gene Expression and Regulation” dataset_id=”Unit 6 Cumulative Multiple Choice|13ba76456772e2″ question_number=”8″ topic=”6.4.Translation”] In the diagram below, an anti-codon is indicated by

[c]IGdyb3 VwIEgu[Qq]

[f]IE5pY2UhIEdyb3VwIEggaXMgYW4gYW50aWNvZG9uLg==[Qq]

[c]IHByb2R1Y3QgRy4=[Qq]

[f]IE5vLiBHIGlzIG1STkEuIEFuIGFudGljb2RvbiBpcyBhIHNob3J0IHN0cmV0Y2ggb2YgUk5BIG9uIGEgdFJOQS5UaGUgYW50aWNvZG9uIGNvbXBsZW1lbnRzIGEgY29kb24uIEZpbmQgYSB0Uk5BLCBhbmQgeW91JiM4MjE3O2xsIGhhdmUgeW91ciBhbnN3ZXIu[Qq]

[c]IHN0cnVjdHVyZSBLLg==[Qq]

[f]IE5vLiBLIGlzIHRoZSByaWJvc29tZS4gQW4gYW50aWNvZG9uIGlzIGEgc2hvcnQgc3RyZXRjaCBvZiBSTkEgb24gYSB0Uk5BLiBUaGUgYW50aWNvZG9uIGNvbXBsZW1lbnRzIGEgY29kb24uIEZpbmQgYSB0Uk5BLCBhbmQgeW91JiM4MjE3O2xsIGhhdmUgeW91ciBhbnN3ZXIu[Qq]

[c]IGJvbmQgSi4=[Qq]

[f]IE5vLiBCb25kICYjODIyMDtKJiM4MjIxOyBpcyBhIHBlcHRpZGUgYm9uZCwgYSBib25kIHRoYXQgaG9sZHMgcHJvdGVpbnMgdG9nZXRoZXIuIEFuIGFudGljb2RvbiBpcyBhIHNob3J0IHN0cmV0Y2ggb2YgUk5BIG9uIGEgdFJOQS4gVGhlIGFudGljb2RvbiBjb21wbGVtZW50cyBhIGNvZG9uLiBGaW5kIGEgdFJOQSwgYW5kIHlvdSYjODIxNztsbCBoYXZlIHlvdXIgYW5zd2VyLg==

Cg==

[Qq]

[q json=”true” xyz=”2″ multiple_choice=”true” topic=”6.4.Translation” dataset_id=”Unit 6 Cumulative Multiple Choice|13ba5efcf08ae2″ question_number=”9″ unit=”6.Gene Expression and Regulation”] Which process is occurring below?

[c]IEROQSByZXBsaWNhdGlvbg==[Qq]

[f]IE5vLiBJZiB0aGlzIHdlcmUgRE5BIHJlcGxpY2F0aW9uLCB5b3UmIzgyMTc7ZCBzZWUgYSByZXBsaWNhdGlvbiBmb3JrLCBudWNsZW90aWRlcywgYW5kIGFsbCB0eXBlcyBvZiBlbnp5bWVzLiBIZXJlIHdoYXQgeW91IHNlZSBpcyBhIHJpYm9zb21lLCBtUk5BLCBhbmQgYSB0Uk5BIGNhcnJ5aW5nIGFuIGFtaW5vIGFjaWQuIFdoYXQgcHJvY2VzcyBhcmUgdGhlc2UgaW52b2x2ZWQgd2l0aD8=[Qq]

[c]IHRyYW5zY3JpcHRpb24gb2YgRE5BIGludG8gUk5B[Qq]

[f]IE5vLiBJZiB0aGlzIHdlcmUgdHJhbnNjcmlwdGlvbiBvZiBETkEgaW50byBSTkEsIHlvdSYjODIxNztkIHNlZSBETkEsIFJOQSwgYW5kIGVuenltZXMgbGlrZSBSTkEgcG9seW1lcmFzZS4gSGVyZSB3aGF0IHlvdSBzZWUgaXMgYSByaWJvc29tZSwgbVJOQSwgYW5kIGEgdFJOQSBjYXJyeWluZyBhbiBhbWlubyBhY2lkLiBXaGF0IHByb2Nlc3MgYXJlIHRoZXNlIGludm9sdmVkIHdpdGg/[Qq]

[c]IHRyYW5zY3JpcHRpb24gb2YgUk5BIGludG8gcHJvdGVpbg==[Qq]

[f]IE5vLiBUcmFuc2NyaXB0aW9uIGludm9sdmVzIGNoYW5naW5nIEROQSBpbmZvcm1hdGlvbiBpbnRvIFJOQSBpbmZvcm1hdGlvbi4gSGVyZSB3aGF0IHlvdSBzZWUgaXMgYSByaWJvc29tZSwgbVJOQSwgYW5kIGEgdFJOQSBjYXJyeWluZyBhbiBhbWlubyBhY2lkLiBXaGF0IHByb2Nlc3MgYXJlIHRoZXNlIGludm9sdmVkIHdpdGg/[Qq]

[c]IHRyYW5zbGF0aW9uIG9mIF JOQSBpbnRvIHByb3RlaW4=[Qq]

[f]IEV4Y2VsbGVudCEgVGhlIGltYWdlIGFib3ZlIHNob3dzIGEgcmlib3NvbWUgdHJhbnNsYXRpbmcgUk5BIGluZm9ybWF0aW9uIGludG8gcHJvdGVpbi4=

Cg==

[Qq]

[q json=”true” xyz=”2″ multiple_choice=”true” dataset_id=”Unit 6 Cumulative Multiple Choice|13ba45606dbee2″ question_number=”10″ unit=”6.Gene Expression and Regulation” topic=”6.4.Translation”] The nucleotide sequence below is part of the template strand for the gene coding protein Q.

The “G” on the left is at the 5′ end, and the A on the right is at the 3′ end. If the first amino acid is coded from “GTT”, then what is the second amino acid that this DNA is coding for? You can use the genetic code below to find the answer.

[c]IEdsbiAoZ2x1dGFtaW5lKQ==[Qq]

[f]IE5vLiBUaGUgc2Vjb25kIGFtaW5vIGFjaWQgd2lsbCBiZSB0aGUgb25lIGNvZGVkIGZvciBmcm9tIHRoZSBzZWNvbmQgdHJpcGxldC4gQ0dULiBGaW5kIHRoZSBtUk5BIGNvbXBsZW1lbnQgb2YgQ0dULCBhbmQgdGhlbiBsb29rIHRoYXQgdXAgaW4gdGhlIGdlbmV0aWMgY29kZSBjaGFydC4=[Qq]

[c]IEFsYSAoYW xhbmluZSk=[Qq]

[f]IEV4Y2VsbGVudC4gQ0dUIGluIEROQSBnZXRzIHRyYW5zY3JpYmVkIGludG8gR0NBIGluIG1STkEsIHdoaWNoIGNvZGVzIGZvciBBbGFuaW5lLg==[Qq]

[c]IEdseSAoZ2x5Y2luZSku[Qq]

[f]IE5vLiBUaGUgc2Vjb25kIGFtaW5vIGFjaWQgd2lsbCBiZSB0aGUgb25lIGNvZGVkIGZvciBmcm9tIHRoZSBzZWNvbmQgdHJpcGxldC4gQ0dULiBGaW5kIHRoZSBtUk5BIGNvbXBsZW1lbnQgb2YgQ0dULCBhbmQgdGhlbiBsb29rIHRoYXQgdXAgaW4gdGhlIGdlbmV0aWMgY29kZSBjaGFydC4=[Qq]

[c]IEdsbiAoZ2x1dGFtaW5lKQ==[Qq]

[f]IE5vLiBUaGUgc2Vjb25kIGFtaW5vIGFjaWQgd2lsbCBiZSB0aGUgb25lIGNvZGVkIGZvciBmcm9tIHRoZSBzZWNvbmQgdHJpcGxldC4gQ0dULiBGaW5kIHRoZSBtUk5BIGNvbXBsZW1lbnQgb2YgQ0dULCBhbmQgdGhlbiBsb29rIHRoYXQgdXAgaW4gdGhlIGdlbmV0aWMgY29kZSBjaGFydC4=[Qq]

[c]IFRociAodGhyZW9uaW5lKQ==[Qq]

[f]IE5vLiBUaGUgc2Vjb25kIGFtaW5vIGFjaWQgd2lsbCBiZSB0aGUgb25lIGNvZGVkIGZvciBmcm9tIHRoZSBzZWNvbmQgdHJpcGxldC4gQ0dULiBGaW5kIHRoZSBtUk5BIGNvbXBsZW1lbnQgb2YgQ0dULCBhbmQgdGhlbiBsb29rIHRoYXQgdXAgaW4gdGhlIGdlbmV0aWMgY29kZSBjaGFydC4=

Cg==

Cg==

[Qq]

[q json=”true” multiple_choice=”true” unit=”6.Gene Expression and Regulation” topic=”6.5.Regulation_of_Gene_Expression” dataset_id=”Unit 6 Cumulative Multiple Choice|13ba2e17f6d6e2″ question_number=”11″] The diagram below is a model of the trp operon.

Which of the following is true of the repressor molecule?

[c]IGl0JiM4MjE3O3MgYW4gYW1pbm8gYWNpZA==[Qq]

[f]IE5vLiBUaGVyZSA=aXM=IGFuIGFtaW5vIGFjaWQgc2hvd24gaW4gdGhpcyBkaWFncmFtLiBJdCYjODIxNztzIHRyeXB0b3BoYW4gKDgpLiBIb3dldmVyLCB0cnlwdG9waGFuIGlzIA==bm90IHRoZSByZXByZXNzb3IuIEhlcmUmIzgyMTc7cyBhIGhpbnQ6IHRoZSByZXByZXNzb3IgYmluZHMgd2l0aCBETkEgaW4gYSB3YXkgdGhhdCBtYWtlcyBpdCBpbXBvc3NpYmxlIGZvciBSTkEgcG9seW1lcmFzZSB0byB0cmFuc2NyaWJlIHRoZSBvcGVyb24mIzgyMTc7cyBzdHJ1Y3R1cmFsIGdlbmVzLg==[Qq]

[c]IGl0JiM4MjE3O3MgYSBzbWFsbCBSTkE=[Qq]

[f]IE5vLiBUaGUgUk5BIHNob3duIGluIHRoaXMgZGlhZ3JhbSAoMikgaXMgbm90IHRoZSByZXByZXNzb3IuIEhlcmUmIzgyMTc7cyBhIGhpbnQ6IHRoZSByZXByZXNzb3IgYmluZHMgd2l0aCBETkEgaW4gYSB3YXkgdGhhdCBtYWtlcyBpdCBpbXBvc3NpYmxlIGZvciBSTkEgcG9seW1lcmFzZSB0byB0cmFuc2NyaWJlIHRoZSBvcGVyb24mIzgyMTc7cyBzdHJ1Y3R1cmFsIGdlbmVzLg==[Qq]

[c]IGl0IGJpbmRzIHRvIHRoZS BvcGVyYXRvciByZWdpb24=[Qq]

[f]IENvcnJlY3QuIEluIHRoZSBwcmVzZW5jZSBvZiB0cnlwdG9waGFuICg4KSwgdGhlIHJlcHJlc3NvciBtb2xlY3VsZSAoMykgYmluZHMgd2l0aCB0aGUgb3BlcmF0b3IgcmVnaW9uICg2KSwgcHJldmVudGluZyB0aGUgdHJhbnNjcmlwdGlvbiBvZiB0aGUgc3RydWN0dXJhbCBnZW5lcy4=[Qq]

[c]IEl0JiM4MjE3O3MgY29kZWQgZm9yIGJ5IGEgc3RydWN0dXJhbCBnZW5lLg==[Qq]

[f]IE5vLiBUaGUgc3RydWN0dXJhbCBnZW5lcyBvZiB0aGUgb3Blcm9uIGFyZSBzaG93biBhdCAmIzgyMjA7NyYjODIyMTsgYWJvdmUgYW5kICYjODIyMDs3YSAmIzgyMTE7IDdlJiM4MjIxOyBiZWxvdywgYW5kIHRoZXkgY29kZSBmb3IgZW56eW1lcyAoJiM4MjIwOzkmIzgyMjE7IGJlbG93KS4=

Cg==

Cg==

[Qq]

[q json=”true” multiple_choice=”true” unit=”6.Gene Expression and Regulation” topic=”6.5.Regulation_of_Gene_Expression” dataset_id=”Unit 6 Cumulative Multiple Choice|13ba19238bd2e2″ question_number=”12″] The trp operon, shown below, is a system for gene regulation found in E. coli and many other bacterial species.

The diagram below shows the system when the amino acid tryptophan is present in the environment.

This diagram shows the system when tryptophan is absent from the environment

The adaptive value of the trp operon to E. coli (and other species) is that it

[c]IGFsbG93cyA=RS4gY29saQ==IHRvIHByb2R1Y2UgdGhlIGVuenltZXMgZm9yIGRpZ2VzdGluZyB0cnlwdG9waGFuIHdoZW4gdHJ5cHRvcGhhbiBpcyBwcmVzZW50IGluIHRoZSBlbnZpcm9ubWVudC4=[Qq]

[f]IE5vLCBmb3IgdHdvIHJlYXNvbnMuIDEpIE5vdGUgdGhhdCB3aGVuIHRyeXB0b3BoYW4gaXMgcHJlc2VudCBpbiB0aGUgZW52aXJvbm1lbnQsIHRoZSBzeXN0ZW0gdHVybnMgb2ZmLiAyKSBUaGUgdHJwIG9wZXJvbiBwcm9kdWNlcyBlbnp5bWVzIGZvciBzeW50aGVzaXppbmcgdHJ5cHRvcGhhbiwgbm90IGZvciBkaWdlc3RpbmcgaXQu[Qq]

[c]IHByb2R1Y2VzIGVuenltZXMgdGhhdCBpbmNyZWFzZSBSTkEgcG9seW1lcmFzZSYjODIxNztzIGFiaWxpdHkgdG8gYmluZCBhdCBpdHMgcHJvbW90ZXIu[Qq]

[f]IE5vLiBXaGlsZSBSTkEgcG9seW1lcmFzZSBkb2VzIGJpbmQgYXQgdGhlIG9wZXJvbiYjODIxNztzIHByb21vdGVyIHJlZ2lvbiwgdGhhdCBiaW5kaW5nIGlzbiYjODIxNzt0IGFzc29jaWF0ZWQgd2l0aCB0aGUgZW56eW1lcyB0aGF0IHRoaXMgc3lzdGVtIHByb2R1Y2VzLg==[Qq]

[c]IGFsbG 93cyA=RS4gY29saQ==IHRvIG9ubHkgcHJvZHVjZSBlbnp5bWVzIGZvciBzeW50aGVzaXppbmcgdHJ5cHRvcGhhbiB3aGVuIHRyeXB0b3BoYW4gaXMgYWJzZW50IGZyb20gdGhlIGVudmlyb25tZW50Lg==[Qq]

[f]IE5pY2VseSBkb25lLiBCeSBvbmx5IHByb2R1Y2luZyBlbnp5bWVzIGZvciB0cnlwdG9waGFuIHN5bnRoZXNpcyB3aGVuIHRyeXB0b3BoYW4gaXMgYWJzZW50LCB0aGUgb3Blcm9uIGhlbHBzIGNvbnNlcnZlIG1hdHRlciBhbmQgZW5lcmd5Lg==[Qq]

[c]IGVuYWJsZXMgcmVndWxhdG9yeSBnZW5lcyB0byBwcm9kdWNlIHJlZ3VsYXRvcnkgcHJvdGVpbnMgdGhhdCBjYW4gYmluZCB3aXRoIGFuIG9wZXJhdG9yIHJlZ2lvbi4=[Qq]

[f]IE5vLiBXaGlsZSB0aGUgYmluZGluZyBvZiByZWd1bGF0b3J5IGdlbmVzIHRvIG9wZXJhdG9ycyBpcyBhIGtleSBwYXJ0IG9mIGhvdyBvcGVyb25zIHdvcmssIHRoYXQgZG9lc24mIzgyMTc7dCBleHBsYWluIHdoeSBvcGVyb25zIGxpa2UgdHJwIGhlbHAgYmFjdGVyaWEgdG8gc3Vydml2ZS4=

Cg==

[Qq]

[q json=”true” xyz=”2″ unit=”6.Gene Expression and Regulation” topic=”6.5.Regulation_of_Gene_Expression” multiple_choice=”true” dataset_id=”Unit 6 Cumulative Multiple Choice|13ba01db14eae2″ question_number=”13″] The diagram below shows the lac operon in the presence and absence of lactose. The genes lacZ, lacY, and lacA allow the bacterium to use lactose as an energy source.

A researcher deletes the operator sequence. Which of the following would be the result?

[c]IFJOQSBwb2x5bWVyYXNlIHdvdWxkIGJlIHVuYWJsZSB0byB0cmFuc2NyaWJlIHRoZSBnZW5lcyBiZWNhdXNlIHRoZSBvcGVyYXRvciBzZXF1ZW5jZSBpcyBuZWNlc3NhcnkgZm9yIHRyYW5zY3JpcHRpb24u[Qq]

[f]IE5vLiBSTkEgcG9seW1lcmFzZSBjb3VsZCBzdGlsbCBhdHRhY2ggdG8gdGhlIHByb21vdGVyIGFuZCB0cmFuc2NyaWJlIHRoZSBnZW5lcyBmb3IgbGFjdG9zZSBkaWdlc3Rpb24uIExvb2sgYXQgdGhlIGRpYWdyYW0gYWdhaW4sIGFuZCBmaWd1cmUgb3V0IHdoYXQgd291bGQgaGFwcGVuIGlmIHRoZSBvcGVyYXRvciB3YXMgZGVsZXRlZC4=[Qq]

[c]IFRoZSBnZW5lcyBsYWNaLCBsYWNZLCBhbmQgbGFjQSB3b3VsZCBvbmx5IGJlIGV4cHJlc3NlZCBpbiB0aGUgcHJlc2VuY2Ugb2YgbGFjdG9zZS4=[Qq]

[f]IE5vLiBXaXRob3V0IHRoZSBvcGVyYXRvciwgdGhlcmUmIzgyMTc7cyBubyB3YXkgZm9yIHRoZSByZXByZXNzb3IgdG8gYmluZCB0byB0aGUgRE5BLCBhbmQgdGhlIHByZXNlbmNlIG9yIGFic2VuY2Ugb2YgbGFjdG9zZSB3b3VsZG4mIzgyMTc7dCBoYXZlIGFueSBlZmZlY3Qgb24gdHJhbnNjcmlwdGlvbi4gTG9vayBhdCB0aGUgZGlhZ3JhbSBhZ2FpbiwgYW5kIGZpZ3VyZSBvdXQgd2hhdCB3b3VsZCBoYXBwZW4gaWYgdGhlIG9wZXJhdG9yIHdhcyBkZWxldGVkLg==[Qq]

[c]IFRoZSBnZW5lcyBsYWNaLCBsYWNZLCBhbmQgbGFj QSB3b3VsZCBhbHdheXMgYmUgZXhwcmVzc2VkLg==[Qq]

[f]IEV4YWN0bHkuIFdpdGhvdXQgYW4gb3BlcmF0b3Igc2VxdWVuY2UgZm9yIHRoZSByZXByZXNzb3IgdG8gYmluZCB0bywgd2l0aG91dCBhbiBvcGVyYXRvciBzZXF1ZW5jZSwgdGhlIHJlcHJlc3NvciBjb3VsZCBub3QgYXR0YWNoIHRvIHRoZSBETkEsIGFuZCBzbyBjb3VsZCBub3QgYmxvY2sgdGhlIFJOQSBwb2x5bWVyYXNlIGZyb20gdHJhbnNjcmliaW5nIHRoZSBnZW5lcy4gQXMgYSByZXN1bHQsIHRoZSBzdHJ1Y3R1cmFsIGdlbmVzICh0aGUgZ2VuZXMgZm9yIHRoZSB0aHJlZSBlbnp5bWVzKSB3b3VsZCBhbHdheXMgYmUgZXhwcmVzc2VkLg==[Qq]

[c]IFRoZSBnZW5lcyBsYWNaLCBsYWNZLCBhbmQgbGFjQSB3b3VsZCBvbmx5IGJlIGV4cHJlc3NlZCBpbiB0aGUgYWJzZW5jZSBvZiBsYWN0b3NlLg==[Qq]

[f]IE5vLiBXaXRob3V0IHRoZSBvcGVyYXRvciwgdGhlcmUmIzgyMTc7cyBubyB3YXkgZm9yIHRoZSByZXByZXNzb3IgdG8gYmluZCB0byB0aGUgRE5BLCBhbmQgdGhlIHByZXNlbmNlIG9yIGFic2VuY2Ugb2YgbGFjdG9zZSB3b3VsZG4mIzgyMTc7dCBoYXZlIGFueSBlZmZlY3Qgb24gdHJhbnNjcmlwdGlvbi4gTG9vayBhdCB0aGUgZGlhZ3JhbSBhZ2FpbiwgYW5kIGZpZ3VyZSBvdXQgd2hhdCB3b3VsZCBoYXBwZW4gaWYgdGhlIG9wZXJhdG9yIHdhcyBkZWxldGVkLg==

Cg==

[Qq]

[q json=”true” multiple_choice=”true” unit=”6.Gene Expression and Regulation” topic=”6.6.Gene_Expression_and_Cell_Specialization” dataset_id=”Unit 6 Cumulative Multiple Choice|13b9ea929e02e2″ question_number=”14″] Early in the development of a female mammal one of her two X-chromosomes is randomly inactivated. As shown below, the pattern of X-inactivation is maintained in all of the descendant cells, leading to patches of tissues with either the paternal X-chromosome activated (Xp) or the maternal chromosome activated (Xm).

The image below shows two forms of DNA packaging.

A biochemical examination of the DNA in an X-chromosome that’s been inactivated would

[c]IGJlIGxpa2UgZm9ybSAxLCB3aXRoIGEgaGlnaC BkZWdyZWUgb2YgRE5BIG1ldGh5bGF0aW9uLg==[Qq]

[f]IE5pY2Ugam9iLiBYLWluYWN0aXZhdGlvbiBvY2N1cnMgdGhyb3VnaCB0aWdodGx5IHBhY2thZ2luZyB0aGUgRE5BLCBhbmQgdGhyb3VnaCBETkEgbWV0aHlsYXRpb24sIGJvdGggb2Ygd2hpY2ggbWFrZSB0aGUgRE5BIHVuYXZhaWxhYmxlIGZvciB0cmFuc2NyaXB0aW9uLg==[Qq]

[c]IGJlIGxpa2UgZm9ybSAyLCB3aXRoIGEgaGlnaCBkZWdyZWUgb2YgRE5BIG1ldGh5bGF0aW9uLg==[Qq]

[f]IE5vLiBIZXJlJiM4MjE3O3MgYSBoaW50LiBJbmFjdGl2YXRlZCBETkEgaXMgbm90IGF2YWlsYWJsZSBmb3IgdHJhbnNjcmlwdGlvbi4gRm9yIEROQSB0byBiZSB0cmFuc2NyaWJlZCwgUk5BIHBvbHltZXJhc2UgbmVlZHMgdG8gYmUgYWJsZSB0byBhY2Nlc3MgdGhlIEROQS4gV2hpY2ggZm9ybSBvZiBETkEgYWJvdmUgd291bGQgYmUgbW9yZSBkaWZmaWN1bHQgdG8gdHJhbnNjcmliZT8=[Qq]

[c]IGJlIGxpa2UgZm9ybSAyLCB3aXRoIGEgaGlnaCBkZWdyZWUgb2YgaGlzdG9uZSBhY2V0eWxhdGlvbi4=[Qq]

[f]IE5vLiBBY2V0eWxhdGlvbiBvZiBoaXN0b25lIHByb3RlaW5zIGFuZCBETkEgbWFrZXMgdGhlIEROQSBsb29zZWx5IHBhY2tlZCAobGlrZSBmb3JtIDIpIGFuZCBtb3JlIGF2YWlsYWJsZSBmb3IgdHJhbnNjcmlwdGlvbi4gV2hpY2ggZm9ybSBvZiBETkEgYWJvdmUgd291bGQgYmUgbW9yZSBkaWZmaWN1bHQgdG8gdHJhbnNjcmliZT8=[Qq]

[c]IGJlIGxpa2UgZm9ybSAxLCB3aXRoIGxhcmdlIHJlZ2lvbnMgaW4gd2hpY2ggdGhlIGdlbmVzIG9uIHRoZSBpbmFjdGl2YXRlZCBjaHJvbW9zb21lIGhhdmUgYmVlbiBkZWxldGVkLg==[Qq]

[f]IE5vLiBBIGtleSBwcmluY2lwbGUgb2YgZXVrYXJ5b3RpYyBnZW5lIGV4cHJlc3Npb24gaXMgdGhhdCBhbGwgY2VsbHMgaGF2ZSB0aGUgc2FtZSBETkEuIEl0JiM4MjE3O3MgdGhlIHBhY2thZ2luZyBvZiB0aGUgRE5BIChhbG9uZyB3aXRoIG90aGVyIGZhY3RvcnMpIHRoYXQgZGV0ZXJtaW5lIHdoZXRoZXIgdGhlIGdlbmVzIGFyZSBleHByZXNzZWQu

Cg==

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Cg==

[Qq]

[q json=”true” multiple_choice=”true” unit=”6.Gene Expression and Regulation” topic=”6.6.Gene_Expression_and_Cell_Specialization” dataset_id=”Unit 6 Cumulative Multiple Choice|13b9d34a271ae2″ question_number=”15″] The diagram below shows some of the events that occur during fertilization and early development. Which of the following is the best explanation for the differences between tissue type A and tissue type B?

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[f]IE5vLiBGZXJ0aWxpemF0aW9uIHJlc3VsdHMgaW4gYSBkaXBsb2lkIG51Y2xldXMgd2l0aCBvbmUgc2V0IGZyb20gdGhlIGZhdGhlciYjODIxNztzIHNwZXJtIGFuZCB0aGUgc2Vjb25kIGZyb20gdGhlIG1vdGhlciYjODIxNztzIGVnZy4gV2hhdCBlbHNlIGNvdWxkIGV4cGxhaW4gdGhlIGRpZmZlcmVuY2VzIHRoYXQgZGV2ZWxvcCBiZXR3ZWVuIHRoZSB0d28gbGluZXMgb2YgY2VsbHM/[Qq]

[c]IE9uZSBkYXVnaHRlciBjZWxsIGhhcyBhbGwgb2YgaXRzIG1pdG9jaG9uZHJpYSBmcm9tIHRoZSB1bmZlcnRpbGl6ZWQgZWdnIGNlbGwsIGFuZCB0aGUgb3RoZXIgaGFzIGFsbCBvZiBpdHMgbWl0b2Nob25kcmlhIGZyb20gdGhlIHNwZXJtIGNlbGwu[Qq]

[f]IE5vLiBBbGwgb2YgdGhlIG1pdG9jaG9uZHJpYSBjb21lIGZyb20gdGhlIGVnZy4gV2hhdCBlbHNlIGNvdWxkIGV4cGxhaW4gdGhlIGRpZmZlcmVuY2VzIHRoYXQgZGV2ZWxvcCBiZXR3ZWVuIHRoZSB0d28gbGluZXMgb2YgY2VsbHM/[Qq]

[c]IEVhY2ggZGF1Z2h0ZXIgY2VsbCBoYXMgZGlmZmVyZW50IG1vbGVjdWxlcyBwcmVzZW50IGluIGl0cyBjeX RvcGxhc20sIHJlc3VsdGluZyBpbiB0aGUgYWN0aXZhdGlvbiBvZiB0aXNzdWUtc3BlY2lmaWMgZ2VuZXMu[Qq]

[f]IE5pY2Ugam9iLiBUaGUgZGlmZmVyZW50IG1vbGVjdWxlcyBwcmVzZW50IGluIGRpZmZlcmVudCBwYXJ0cyBvZiB0aGUgZWdnIGFjdCBhcyB0cmFuc2NyaXB0aW9uIGZhY3RvcnMuIElmIHRoZXNlIG1vbGVjdWxlcyBhcmUgdW5lcXVhbGx5IGRpc3RyaWJ1dGVkIChhcyB0aGV5IGFyZSBhYm92ZSksIHRoZW4gZGlmZmVyZW50IGdlbmVzIHdpbGwgYmUgYWN0aXZhdGVkIGluIGVhY2ggY2VsbCBsaW5lLCBsZWFkaW5nIGVhY2ggdG8gZGV2ZWxvcCBpbnRvIGRpc3RpbmN0IHRpc3N1ZXMu[Qq]

[c]IFNlbGVjdGl2ZSBnZW5lIGRlbGV0aW9uIGNhdXNlZCBieSBjeXRvcGxhc20gZGV0ZXJtaW5hbnRzIGluIGVhY2ggZGF1Z2h0ZXIgY2VsbCBsZWFkcyB0byBnZW5ldGljIGRpZmZlcmVuY2VzLCByZXN1bHRpbmcgaW4gdGhlIGV4cHJlc3Npb24gb2YgZGlmZmVyZW50IGdlbmVzLg==[Qq]

[f]IE5vLiBBIGtleSBwcmluY2lwbGUgb2YgZGV2ZWxvcG1lbnQgaXMgdGhhdCBhbGwgY2VsbHMgaGF2ZSB0aGUgc2FtZSBnZW5lcy4gV2hhdCYjODIxNztzIGRpZmZlcmVudCBpcyB3aGljaCBnZW5lcyBnZXQgZXhwcmVzc2VkLiBXaGF0IGVsc2UgY291bGQgZXhwbGFpbiB0aGUgZGlmZmVyZW5jZXMgdGhhdCBkZXZlbG9wIGJldHdlZW4gdGhlIHR3byBsaW5lcyBvZiBjZWxscz8=

Cg==

Cg==

[Qq]

[q json=”true” xyz=”2″ multiple_choice=”true” unit=”6.Gene Expression and Regulation” topic=”6.7.Mutation” dataset_id=”Unit 6 Cumulative Multiple Choice|13b9b2b180a2e2″ question_number=”16″]

Self-incompatibility in plants is one mechanism by which plants ensure fertilization by a genetically different plant. The S-locus controls self-incompatibility.  The S-locus includes a ligand, which is expressed in the stigma (the female part of the flower which accepts a pollen grain), and a matching receptor which is expressed in the pollen (which is analogous to a male gamete).  The diagram below shows the interactions between two different types of pollen and the surface of the stigma.

A researcher introduces a mutation that deletes the fourth base pair in the coding sequence of the S-receptor gene.  Which of the following will be the result?

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[f]IE5vLiBUaGluayBhYm91dCB0aGUgY29uc2VxdWVuY2VzIG9mIGEgZnJhbWVzaGlmdCBtdXRhdGlvbi4gSG93IGl0IHdvdWxkIGFmZmVjdCB0aGUgYW1pbm8gYWNpZCBzZXF1ZW5jZSBpbiB0aGUgcmVzdWx0aW5nIFMtcmVjZXB0b3IgcHJvdGVpbiwgYW5kIGhvdyB0aGF0IHdvdWxkIGFmZmVjdCB0aGUgc2hhcGUgb2YgdGhlIHJlY2VwdG9yPw==[Qq]

[c]IFRoZSByZWNlcHRvciB3aWxsIG5vIGxvbmdlciBiaW5kIHRvIHRoZSBsaWdhbmQuwqAgQXMgYSByZXN1bHQsIHRoZS BwbGFudCB3aWxsIGJlIHVuYWJsZSB0byBkaXN0aW5ndWlzaCBiZXR3ZWVuIHNlbGYgYW5kIGNyb3NzIHBvbGxlbi4=[Qq]

[f]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[Qq]

[c]IFRoZSByZWNlcHRvciB3aWxsIG5vIGxvbmdlciBiaW5kIHRvIHRoZSBsaWdhbmQuwqAgQXMgYSByZXN1bHQsIG5laXRoZXIgc2VsZiBub3IgY3Jvc3MgcG9sbGVuIHdpbGwgZ2VybWluYXRlLg==[Qq]

[f]IE5vLiBEZWxldGluZyB0aGUgZm91cnRoIGJhc2UgcGFpciBvZiB0aGUgY29kaW5nIHNlcXVlbmNlIG9mIGEgZ2VuZSB3aWxsIGNhdXNlIGEgZnJhbWVzaGlmdCBtdXRhdGlvbiwgY2hhbmdpbmcgZXZlcnkgYW1pbm8gYWNpZCBwYXN0IHRoZSBpbml0aWFsIG1ldGhpb25pbmUuIFRoYXQgd2lsbCBjaGFuZ2UgdGhlIHNoYXBlIG9mIHRoZSByZWNlcHRvci4gSW4gdGVybXMgb2YgdGhlIHJlY2VwdG9yJiM4MjE3O3MgYWJpbGl0eSB0byBiaW5kIHdpdGggdGhlIGxpZ2FuZCwgd2hhdCB3aWxsIGJlIHRoZSBjb25zZXF1ZW5jZT8=[Qq]

[c]IFdpdGggdGhlIGluZm9ybWF0aW9uIHRoYXQmIzgyMTc7cyBnaXZlbiwgaXQmIzgyMTc7cyBub3QgcG9zc2libGUgdG8gZGV0ZXJtaW5lIHRoZSBjb25zZXF1ZW5jZXMu[Qq]

[f]IE5vLiBUaGluayBhYm91dCB0aGUgY29uc2VxdWVuY2VzIG9mIGEgZnJhbWVzaGlmdCBtdXRhdGlvbi4gSG93IHdvdWxkIGl0IGFmZmVjdCB0aGUgYW1pbm8gYWNpZCBzZXF1ZW5jZSBpbiB0aGUgcmVzdWx0aW5nIHByb3RlaW4sIGFuZCBob3cgdGhhdCB3b3VsZCBhZmZlY3QgdGhlIHNoYXBlIG9mIHRoZSByZWNlcHRvcj8=

Cg==

[Qq]

[q json=”true” xyz=”2″ multiple_choice=”true” unit=”6.Gene Expression and Regulation” dataset_id=”Unit 6 Cumulative Multiple Choice|13b99914fdd6e2″ question_number=”17″ topic=”6.7.Mutation”] Imagine that an error has occurred during DNA replication in a cell. The error substitutes a G in place of a T in one of the cell’s genes. Which of the following is the most likely effect that this will have on the cell?

[c]IEVhY2ggb2YgdGhlIGNlbGwmIzgyMTc7cyBwcm90ZWlucyB3aWxsIGNvbnRhaW4gYW4gaW5jb3JyZWN0IGFtaW5vIGFjaWQu[Qq]

[f]IE5vLiBUaGUgbWFpbiBwcm9ibGVtIHdpdGggdGhpcyBjaG9pY2UgaXMgdGhhdCB0aGUgbXV0YXRpb24gd2lsbCBhZmZlY3Qgb25seSBhIHNpbmdsZSBwcm90ZWluIChub3QgZWFjaCBvZiB0aGUgY2VsbCYjODIxNztzIHByb3RlaW5zKS4gV2hlbiB5b3UgbmV4dCBzZWUgdGhpcyBxdWVzdGlvbiwgdGhpbmsgb2YgaG93IGluZm9ybWF0aW9uIGluIEROQSBnZXRzIHRyYW5zbGF0ZWQgaW50byBwcm90ZWluLCBhbmQgY29uc2lkZXIgdGhlIGxpa2VseSBlZmZlY3Qgb2YgYSBzaW5nbGUgYmFzZSBwYWlyIHN1YnN0aXR1dGlvbi4=[Qq]

[c]IFRoZSBhbWlubyBhY2lkIHNlcXVlbmNlIG9mIG9uZSBvZiB0aGUgY2VsbCYjODIxNztzIHByb3RlaW5zIHdpbGwgYmUgY29tcGxldGVseSBhbHRlcmVkLg==[Qq]

[f]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[Qq]

[c]IE9uZSBhbWlubyBhY2lkIHdpbGwgYmUgbWlzc2luZyBmcm9tIG9uZSBvZiB0aGUgY2VsbCYjODIxNztzIHByb3RlaW5zLA==[Qq]

[f]IE5vLiBBbGwgdGhhdCB0aGlzIG11dGF0aW9uIGRvZXMgaXMgc3Vic3RpdHV0ZSBvbmUgbnVjbGVvdGlkZSBmb3IgYW5vdGhlci4gVGhhdCBtaWdodCByZXN1bHQgaW4gYW4gaW5jb3JyZWN0IGFtaW5vIGFjaWQsIGJ1dCBpdCB3b3VsZG4mIzgyMTc7dCByZXN1bHQgaW4gYSBtaXNzaW5nIGFtaW5vIGFjaWQu[Qq]

[c]IE9uZSBvZiB0aGUgY2VsbCYjODIxNztzIHByb3RlaW5zIG1p Z2h0IGNvbnRhaW4gYW4gaW5jb3JyZWN0IGFtaW5vIGFjaWQu[Qq]

[f]IENvcnJlY3QuIEEgbGlrZWx5IGVmZmVjdCBvZiBhIHNpbmdsZSBudWNsZW90aWRlIGNoYW5nZSBpcyB0aGF0IG9uZSBvZiB0aGUgY2VsbCYjODIxNztzIHByb3RlaW5zIG1pZ2h0IGNvbnRhaW4gYW4gaW5jb3JyZWN0IGFtaW5vIGFjaWQu

Cg==

Cg==

[Qq]

[q json=”true” xyz=”2″ multiple_choice=”true” unit=”6.Gene Expression and Regulation” dataset_id=”Unit 6 Cumulative Multiple Choice|13b9787c575ee2″ question_number=”18″ topic=”6.8.Biotechnology”] As shown below, a DNA fragment from a plasmid in E. coli has two restriction sites for the restriction endonuclease Hind III, which creates fragments X, Y, and Z.

Which of the following patterns would result from the separation of these fragments by gel electrophoresis?

[c]IEEg[Qq][c]IE Ig[Qq][c]IEMg[Qq][c]IEQ=

Cg==[Qq]

[f]IE5vLiBUaGUgcmVzdHJpY3Rpb24gZW5kb251Y2xlYXNlIGRlc2NyaWJlZCBpbiB0aGUgcXVlc3Rpb24gd2lsbCBjdXQgdGhlIEROQSBpbnRvIHRocmVlIHBpZWNlcy4gSnVzdCBieSBsb29raW5nIGF0IHRoZSBkaWFncmFtIG9mIHRoZSBETkEsIHlvdSBjYW4gdGVsbCB0aGF0IFogaXMgdGhlIGxvbmdlc3QgZnJhZ21lbnQsIGZvbGxvd2VkIGJ5IFgsIGZvbGxvd2VkIGJ5IFkuIE5vdywgY29ubmVjdCB0aGlzIHdpdGggdGhlIGlkZWEgdGhhdCBlbGVjdHJvcGhvcmVzaXMgb2YgRE5BIGZyYWdtZW50cyBzZXBhcmF0ZXMgdGhlbSBiYXNlZCBvbiBzaXplLCB3aXRoIGxhcmdlciBmcmFnbWVudHMgbW92aW5nIGxlc3MsIGFuZCBzbWFsbGVyIGZyYWdtZW50cyBtb3ZpbmcgbW9yZS4gSWYgYWxsIG9mIHRoZXNlIGZyYWdtZW50cyBzdGFydCBhdCB0aGUgbmVnYXRpdmUgcG9sZSBhdCB0aGUgdG9wLCB0aGVuIHdoaWNoIGZyYWdtZW50IHdpbGwgbW92ZSB0aGUgbW9zdCwgd2hpY2ggdGhlIHNlY29uZCBtb3N0LCBhbmQgd2hpY2ggdGhlIGxlYXN0PyBGaWd1cmUgdGhhdCBvdXQsIGFuZCB5b3UmIzgyMTc7bGwgaGF2ZSB0aGUgYW5zd2VyIHRoZSBuZXh0IHRpbWUgeW91IHNlZSB0aGlzIHF1ZXN0aW9uLg==[Qq]

[f]IFllcy4gQmFzZWQgb24gdGhlIGRpYWdyYW0sIFogaXMgdGhlIGxvbmdlc3QgZnJhZ21lbnQsIGZvbGxvd2VkIGJ5IFgsIGZvbGxvd2VkIGJ5IFkuIEFzIGEgcmVzdWx0LCB0aGUgcGF0dGVybiB0aGF0IHdpbGwgcmVzdWx0IGZyb20gZWxlY3Ryb3Bob3Jlc2lzIGlzIEIu[Qq]

[f]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[Qq]

[f]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

Cg==

[Qq]

[q json=”true” xyz=”2″ multiple_choice=”true” unit=”6.Gene Expression and Regulation” dataset_id=”Unit 6 Cumulative Multiple Choice|13b95a37bccae2″ question_number=”19″ topic=”6.8.Biotechnology”] The image below represents the pGLO plasmid, which is used both in research and in education.

As indicated, the plasmid contains a restriction site and three genes:

  • ampR: encodes for a protein that provides resistance to ampicillin, an antibiotic.
  • gfp: encodes the green fluorescent protein (GFP), which glows green (fluoresces) under ultraviolet (UV) light.
  • araC: encodes a protein required for the expression of GFP when arabinose, a monosaccharide, is present.

The results of a bacterial transformation experiment using the pGLO plasmid are shown in the table below.

Which plate would contain bacteria that fluoresce (glow green) under UV light?

[c]IHBsYXRlIFc=[Qq]

[f]IE5vLiBUaGVzZSBiYWN0ZXJpYSBoYXZlIG5vdCBiZWVuIHRyYW5zZm9ybWVkIChtZWFuaW5nIHRoYXQgdGhleSBoYXZlbiYjODIxNzt0IHRha2VuIHVwIHRoZSBwbGFzbWlkKS4gV2l0aG91dCB0aGUgcGxhc21pZCwgdGhleSBsYWNrIHRoZSBHRlAgZ2VuZSwgYW5kIHRoZXJlZm9yZSB3b3VsZCBoYXZlIG5vIHdheSB0byBzeW50aGVzaXplIHRoZSBwcm90ZWluIHRoYXQgd291bGQgZW5hYmxlIHRoZW0gdG8gZ2xvdy4gTmV4dCB0aW1lLCBjaG9zZSBmcm9tIG9uZSBvZiB0aGUgdHdvIHBsYXRlcyB3aXRoIHRyYW5zZm9ybWVkIGJhY3RlcmlhLg==[Qq]

[c]IHBsYXRlIFg=[Qq]

[f]IE5vLiBUaGVzZSBiYWN0ZXJpYSBoYXZlIG5vdCBiZWVuIHRyYW5zZm9ybWVkIChtZWFuaW5nIHRoYXQgdGhleSBoYXZlbiYjODIxNzt0IHRha2VuIHVwIHRoZSBwbGFzbWlkKS4gT24gdG9wIG9mIHRoYXQsIHRoZXkmIzgyMTc7dmUgYmVlbiBwbGFjZWQgdXBvbiB0aGUgYW50aWJpb3RpYyBhbXBpY2lsbGluLCB3aGljaCBpcyBwcmV2ZW50aW5nIGFueSBiYWN0ZXJpYWwgZ3Jvd3RoLiBOZXh0IHRpbWUsIGNob3NlIGZyb20gb25lIG9mIHRoZSB0d28gcGxhdGVzIHdpdGggdHJhbnNmb3JtZWQgYmFjdGVyaWEu[Qq]

[c]IHBsYX RlIFk=[Qq]

[f]IEZhYnVsb3VzLiBUaGVzZSBiYWN0ZXJpYSBoYXZlIHRha2VuIHVwIHRoZSBwbGFzbWlkLCB3aGljaCBwcm92aWRlcyB0aGVtIHdpdGggdGhlIGdlbmUgdG8gc3ludGhlc2l6ZSBHRlAuIE9uIHRvcCBvZiB0aGF0LCB0aGV5IGhhdmUgYmVlbiBwcm92aWRlZCB3aXRoIGFyYWJpbm9zZSwgd2hpY2ggaW5kdWNlcyB0aGUgY2VsbCB0byBzeW50aGVzaXplIEdGUC4=[Qq]

[c]IHBsYXRlIFo=[Qq]

[f]IE5vLiBXaGlsZSB0aGVzZSBiYWN0ZXJpYSBoYXZlIGJlZW4gdHJhbnNmb3JtZWQgKGFzIGRlbW9uc3RyYXRlZCBieSB0aGUgZmFjdCB0aGF0IHRoZXkmIzgyMTc7cmUgZ3Jvd2luZyBpbiB0aGUgcHJlc2VuY2Ugb2YgdGhlIGFudGliaW90aWMgYW1waWNpbGxpbiksIHRoZXkgaGF2ZSBub3QgYmVlbiBmZWQgYXJhYmlub3NlLCB3aGljaCBpcyByZXF1aXJlZCB0byBpbmR1Y2UgdGhlIGNlbGwgdG8gc3ludGhlc2l6ZSBHRlAuIFJlbWVtYmVyIHRoYXQgbmV4dCB0aW1lIHlvdSBzZWUgdGhpcyBxdWVzdGlvbiwgYW5kIG1ha2UgYW5vdGhlciBjaG9pY2Uu

Cg==

[Qq]

[q json=”true” xyz=”2″ multiple_choice=”true” unit=”6.Gene Expression and Regulation” dataset_id=”Unit 6 Cumulative Multiple Choice|13b93bf32236e2″ question_number=”20″ topic=”6.8.Biotechnology”] A research team is studying the genetics of beak length in house sparrows, where beak length has two phenotypes, long and short. Pedigree studies have shown that this phenotype is under the control of a single autosomal gene with two alleles. The DNA for each allele is isolated and sequenced, and it’s determined that a single nucleotide difference accounts for the short mutation.

When the DNA of both alleles is subjected to restriction fragment analysis, the pattern below is produced.

Based on the diagram, the most reasonable explanation for the different RFLPs associated with the long and short-beaked alleles is that a mutation in the short allele DNA

[c]IGVsaW1pbmF0ZWQgYSByZXN0cmljdGlvbiBzaXRlLg==[Qq]

[f]IE5vLiBUaGUga2V5IHRoaW5nIHRvIG5vdGljZSBpcyB0aGF0IHRoZSBsb25nIGFsbGVsZSBETkEgaGFzIGZvdXIgZnJhZ21lbnRzLCBhbmQgdGhlIHNob3J0IGFsbGVsZSBETkEgaGFzIGZpdmUuIEEgbXV0YXRpb24gdGhhdCBlbGltaW5hdGVkIGEgcmVzdHJpY3Rpb24gc2l0ZSB3b3VsZCBwcm9kdWNlIGZld2VyIGZyYWdtZW50cyBpbiB0aGUgc2hvcnQgYWxsZWxlLiBZb3VyIGpvYiBpcyB0byBleHBsYWluIHdoeSB0aGUgc2hvcnQgYWxsZWxlIGhhcyBtb3JlIGZyYWdtZW50cy4gSG93IGNvdWxkIHRoYXQgaGFwcGVuPw==[Qq]

[c]IGFjdGl2YXRlZCBhbiBpbmFjdGl2ZSBnZW5lLCBjcmVhdGluZyBuZXcgRE5BIGZyYWdtZW50cy4=[Qq]

[f]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[Qq]

[c]IGNyZWF0ZWQgYSBuZXcgcm VzdHJpY3Rpb24gc2l0ZS4=[Qq]

[f]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[Qq]

[c]IGRpc2FibGVkIGFuIGFjdGl2ZSBnZW5lLg==[Qq]

[f]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[Qq]

[x]

[restart]
[/qwiz]

4. Unit 6 Multiple Choice Quiz 2

[qwiz style=”width: 600px !important; min-height: 450px !important;” dataset=”Unit 6 Multiple Choice Quiz 2″ qrecord_id=”sciencemusicvideosMeister1961-Unit 6 Cumulative Multiple Choice Quiz 2″]

[h]Unit 6 Multiple Choice Quiz 2

[i]

[q json=”true” multiple_choice=”true” unit=”6.Gene Expression and Regulation” topic=”6.1.DNA_and_RNA” dataset_id=”Unit 6 Multiple Choice Quiz 2|ef9ee54995e3a” question_number=”1″] If adenine makes up 19% of the nucleotide bases in a sample of DNA from a bacterial cell, then the same sample will have what percentage of guanine?

[c]IDE5JSA=[Qq][c]IDM4JSA=[Qq][c]IDYyJSA=[Qq][c]IDMx JQ==

Cg==[Qq]

[f]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[Qq]

[f]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[Qq]

[f]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[Qq]

[f]IFBlcmZlY3QuwqAgSWYgQSYjODIxNztzIHBlcmNlbnRhZ2UgaXMgMTklLCB0aGVuIFQmIzgyMTc7cyBwZXJjZW50YWdlIHdpbGwgYWxzbyBiZSAxOSUuIFRoZXJlZm9yZSBndWFuaW5lJiM4MjE3O3MgcGVyY2VudGFnZSBoYXMgdG8gYmUgMzElLg==

Cg==

[Qq]

[q json=”true” multiple_choice=”true” unit=”6.Gene Expression and Regulation” topic=”6.1.DNA_and_RNA” dataset_id=”Unit 6 Multiple Choice Quiz 2|ef9d9602e5a3a” question_number=”2″] The genetic material of a newly discovered virus is analyzed. Its base composition is

  • 18% A,
  • 32% G,
  • 0% U,
  • 18% T, and
  • 32% C.

What is the best description of this virus’s genetic material?

[c]IHNpbmdsZS1zdHJhbmRlZCBSTkE=[Qq]

[f]IE5vLiBSTkEgdmlydXNlcyB3b3VsZCBoYXZlIFUgKHVyYWNpbCkgYXMgcGFydCBvZiB0aGVpciBnZW5ldGljIG1hdGVyaWFsLg==[Qq]

[c]IGRvdWJsZS1zdHJhbmRlZCBSTkE=[Qq]

[f]IE5vLiBSTkEgdmlydXNlcyB3b3VsZCBoYXZlIFUgKHVyYWNpbCkgYXMgcGFydCBvZiB0aGVpciBnZW5ldGljIG1hdGVyaWFsLg==[Qq]

[c]IFNpbmdsZS1zdHJhbmRlZCBETkE=[Qq]

[f]IE5vLiBOb3RlIGhvdyB0aGUgcGVyY2VudGFnZSBvZiBBIGlzIGVxdWFsIHRvIHRoYXQgb2YgVCwgYW5kIHRoYXQgdGhlIHBlcmNlbnRhZ2Ugb2YgQyBlcXVhbHMgdGhhdCBvZiBHLiBXaGF0IGRvZXMgdGhhdCB0ZWxsIHlvdSBhYm91dCB0aGUgc3RydWN0dXJlPw==[Qq]

[c]IGRvdWJsZS1zdH JhbmRlZCBETkE=[Qq]

[f]IFdheSB0byBnby4gQmVjYXVzZSB0aGUgcGVyY2VudGFnZSBvZiBBIGlzIGVxdWFsIHRvIHRoYXQgb2YgVCBhbmQgdGhlIHBlcmNlbnRhZ2Ugb2YgQyBlcXVhbHMgdGhhdCBvZiBHLCB5b3Uga25vdyB0aGF0IHRoZSB2aXJ1cyYjODIxNztzIGdlbmV0aWMgbWF0ZXJpYWwgaGFzIHRvIGJlIGRvdWJsZS1zdHJhbmRlZCBETkEu

Cg==

Cg==

[Qq]

[q json=”true” xyz=”2″ multiple_choice=”true” dataset_id=”Unit 6 Multiple Choice Quiz 2|ef9c46bc3563a” question_number=”3″ unit=”6.Gene Expression and Regulation” topic=”6.2.Replication”] The nucleotide sequence below is part of the template strand for the gene coding protein Q. If during DNA replication a complementary strand was synthesized from this sequence, which nucleotide would be inserted at the point indicated by “*”?

[c]IEEg[Qq][c]IEMg[Qq][c]IE cg[Qq][c]IFQg[Qq][c]IFU=

Cg==[Qq]

[f]IE5vLiBUaGlzIHF1ZXN0aW9uIGlzIGNoZWNraW5nIHlvdXIgdW5kZXJzdGFuZGluZyBvZiB0aGUgYmFzZS1wYWlyaW5nIHJ1bGVzIGZvciBETkE6IEEgKGFkZW5pbmUpIGJvbmRzIHdpdGggVCAodGh5bWluZSk7IEMgKGN5dG9zaW5lKSBib25kcyB3aXRoIEcgKGd1YW5pbmUpLg==[Qq]

[f]IE5vLiBUaGlzIHF1ZXN0aW9uIGlzIGNoZWNraW5nIHlvdXIgdW5kZXJzdGFuZGluZyBvZiB0aGUgYmFzZS1wYWlyaW5nIHJ1bGVzIGZvciBETkE6IEEgKGFkZW5pbmUpIGJvbmRzIHdpdGggVCAodGh5bWluZSk7IEMgKGN5dG9zaW5lKSBib25kcyB3aXRoIEcgKGd1YW5pbmUpLg==[Qq]

[f]IFllcy4gVGhlIG51Y2xlb3RpZGUgQyAoY3l0b3NpbmUpIGFsd2F5cyBib25kcyB3aXRoIEcgKGd1YW5pbmUp[Qq]

[f]IE5vLiBUaGlzIHF1ZXN0aW9uIGlzIGNoZWNraW5nIHlvdXIgdW5kZXJzdGFuZGluZyBvZiB0aGUgYmFzZS1wYWlyaW5nIHJ1bGVzIGZvciBETkE6IEEgKGFkZW5pbmUpIGJvbmRzIHdpdGggVCAodGh5bWluZSk7IEMgKGN5dG9zaW5lKSBib25kcyB3aXRoIEcgKGd1YW5pbmUpLg==[Qq]

[f]IE5vLiBUaGlzIHF1ZXN0aW9uIGlzIGFib3V0IEROQSByZXBsaWNhdGlvbiwgbm90IHRyYW5zY3JpcHRpb24uIFRoZSBSTkEgYmFzZSBVIChVcmFjaWwpIGlzbiYjODIxNzt0IGludm9sdmVkLiBUcnkgdGhpcyBhZ2FpbiwgZm9jdXNpbmcgb24gRE5BIHJlcGxpY2F0aW9uIGFuZCB0aGUgYmFzZS1wYWlyaW5nIHJ1bGVzIGZvciBETkEu

Cg==

[Qq]

[q json=”true” xyz=”2″ multiple_choice=”true” unit=”6.Gene Expression and Regulation” dataset_id=”Unit 6 Multiple Choice Quiz 2|ef9af7758523a” question_number=”4″ topic=”6.2.Replication”] The diagram below represents a sequence from two DNA molecules from the same species of bacteria. In an experimental protocol, the bacteria were first grown for several generations in a medium containing non-radioactive nitrogen (14N). Subsequently, a fraction of the bacteria were transferred to a medium containing radioactive nitrogen(15N).

Which of the diagrams below best represents the DNA molecules produced after the bacteria were transferred to radioactive nitrogen and then allowed to proceed through one replication cycle.

[c]IEEg[Qq][c]IEIg[Qq][c]IEMg[Qq][c]IE Q=

Cg==[Qq]

[f]IE5vLiBNb2RlbCBBIHNob3dzIHRoZSB0d28gZGF1Z2h0ZXIgc3RyYW5kcyB0aGF0IHJlc3VsdCBmcm9tIEROQSByZXBsaWNhdGlvbiBhcyBjb25zaXN0aW5nIG9mIG9uZSBkb3VibGUtc3RyYW5kZWQgZGF1Z2h0ZXIgY29uc2lzdGluZyBlbnRpcmVseSBvZiBETkEgbWFkZSBmcm9tIA==MTQ=TiAodGhlIGxpZ2h0ZXIgb25lIG9uIHRoZSBsZWZ0KSwgYW5kIHRoZSBvdGhlciBvbmUgZW50aXJlbHkgbWFkZSBvZiA=MTU=TiAodGhlIGRhcmtlciBvbmUgb24gdGhlIHJpZ2h0KS4gQnV0IHRoYXQmIzgyMTc7cyBub3QgaG93IEROQSByZXBsaWNhdGlvbiB3b3Jrcy4gV2hlbiBETkEgcmVwbGljYXRlcyBpdHNlbGYsIHRoZSBETkEgJiM4MjIwO3VuemlwcyYjODIyMTsgYW5kIGVhY2ggc3RyYW5kIHN1YnNlcXVlbnRseSBhY3RzIGFzIGEgdGVtcGxhdGUgZm9yIHRoZSBmb3JtYXRpb24gb2YgYSBuZXcgY29tcGxlbWVudGFyeSBzdHJhbmQuIFRoZSByZXN1bHRpbmcgZGF1Z2h0ZXIgc3RyYW5kcyB3aWxsIGVhY2ggYmUgaGFsZiBvbGQgYW5kIGhhbGYgbmV3LiBTZWUgaWYgeW91IGNhbiBmaWd1cmUgb3V0IHdoaWNoIGRpYWdyYW0gcmVwcmVzZW50cyB0aGlzIGNvcnJlY3QgbW9kZWwu[Qq]

[f]IE5vLiBNb2RlbCBCIHNob3dzIHRoZSB0d28gZGF1Z2h0ZXIgc3RyYW5kcyB0aGF0IHJlc3VsdCBmcm9tIEROQSByZXBsaWNhdGlvbiBhcyBjb25zaXN0aW5nIGVudGlyZWx5IG9mIEROQSBtYWRlIGZyb20gMTU=Ti4gQnV0IHRoYXQmIzgyMTc7cyBub3QgaG93IEROQSByZXBsaWNhdGlvbiB3b3Jrcy4gV2hlbiBETkEgcmVwbGljYXRlcyBpdHNlbGYsIHRoZSBETkEgJiM4MjIwO3VuemlwcyYjODIyMTsgYW5kIGVhY2ggc3RyYW5kIHN1YnNlcXVlbnRseSBhY3RzIGFzIGEgdGVtcGxhdGUgZm9yIHRoZSBmb3JtYXRpb24gb2YgYSBuZXcgY29tcGxlbWVudGFyeSBzdHJhbmQuIFRoZSByZXN1bHRpbmcgZGF1Z2h0ZXIgc3RyYW5kcyB3aWxsIGVhY2ggYmUgaGFsZiBvbGQgYW5kIGhhbGYgbmV3LiBTZWUgaWYgeW91IGNhbiBmaWd1cmUgb3V0IHdoaWNoIGRpYWdyYW0gcmVwcmVzZW50cyB0aGlzIGNvcnJlY3QgbW9kZWwu[Qq]

[f]IE5vLiBNb2RlbCBDIGhhcyB0d28ga2V5IGZlYXR1cmVzOiAxKSBJdCBzaG93cyB0aGUgdHdvIGRhdWdodGVyIHN0cmFuZHMgdGhhdCByZXN1bHQgZnJvbSBETkEgcmVwbGljYXRpb24gYXMgY29uc2lzdGluZyBvZiBhIG1peHR1cmUgb2YgRE5BIG1hZGUgZnJvbSA=MTU=TiBhbmQgMTQ=TiwgYW5kIGl0IHNob3dzIGVhY2ggaW5kaXZpZHVhbCBzdHJhbmQgY29uc2lzdGluZyBvZiBib3RoIA==[Qq]15N and 14N. But that’s not how DNA replication works. When DNA replicates itself, the DNA “unzips” and each strand subsequently acts as a template for the formation of a new complementary strand. The resulting daughter strands will each be half old and half new. See if you can figure out which diagram represents this correct model.

[f]IEZhYnVsb3VzLiBNb2RlbCBEIGNvcnJlY3RseSByZXByZXNlbnRzIHdoYXQgaGFwcGVucyBkdXJpbmcgRE5BIHJlcGxpY2F0aW9uLCB3aXRoIGVhY2ggZGF1Z2h0ZXIgc3RyYW5kIGNvbnNpc3Rpbmcgb2Ygb25lIG9mIHRoZSBvcmlnaW5hbCBwYXJlbnQgc3RyYW5kcyBhbmQgb25lIG5ldyBjb21wbGVtZW50YXJ5IHN0cmFuZC4=

Cg==

[Qq]

[q json=”true” multiple_choice=”true” unit=”6.Gene Expression and Regulation” topic=”6.2.Replication” dataset_id=”Unit 6 Multiple Choice Quiz 2|ef99a82ed4e3a” question_number=”5″] The image below is a model of two steps that occur in sequence during DNA replication.

“A” represents an enzyme that creates structure “B.” The creation of structure “B” is required because

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[f]IE5vLiBJdCBsb29rcyBsaWtlIHlvdSB3ZXJlIHRoaW5raW5nIG9mIHNpbmdsZS1zdHJhbmQgYmluZGluZyBwcm90ZWlucy4=[Qq]

[c]IEROQSBwb2x5bWVyYXNlIElJSSBjYW4gb25seSBhZGQgbmV3 IG51Y2xlb3RpZGVzIHRvIGFuIGV4aXN0aW5nIHN0cmFuZC4=[Qq]

[f]IENvcnJlY3QhIFRoZSByZWQgc2VnbWVudCAobGFiZWxlZCAmIzgyMjA7QiYjODIyMTsgaW4gc3RlcCAxKSBpcyBhbiBSTkEgcHJpbWVyLiBCZWNhdXNlIEROQSBwb2x5bWVyYXNlIElJSSBjYW4gb25seSBhZGQgbmV3IG51Y2xlb3RpZGVzIHRvIGFuIGV4aXN0aW5nIHN0cmFuZCwgYW5vdGhlciBlbnp5bWUgKFByaW1hc2UpIGhhcyB0byBiZWdpbiB0aGUgcmVwbGljYXRpb24gcHJvY2VzcyBieSBsYXlpbmcgZG93biBhIHNob3J0IHNlcXVlbmNlIG9mIFJOQSwgYWxzbyBrbm93biBhcyBhIHByaW1lci4=[Qq]

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[f]IE5vLiBXaGlsZSBpdCYjODIxNztzIHRydWUgdGhhdCBETkEgcG9seW1lcmFzZSBjYW4gb25seSBzeW50aGVzaXplIGluIGEgNSYjODI0MjsgdG8gMyYjODI0MjsgZGlyZWN0aW9uLCB0aGF0IGRvZXNuJiM4MjE3O3QgZXhwbGFpbiB3aHkgJiM4MjIwO0ImIzgyMjE7IGlzIHJlcXVpcmVkLg==[Qq]

[c]IEROQSByZXBsaWNhdGlvbiBvbiB0aGUgbGFnZ2luZyBzdHJhbmQgY2FuIG9ubHkgYmUgYWNjb21wbGlzaGVkIGluIHNob3J0IHN0cmV0Y2hlcyB1c2luZyBPa2F6YWtpIGZyYWdtZW50cy4=[Qq]

[f]IE5vLiBXaGlsZSBpdCYjODIxNztzIHRydWUgdGhhdCBETkEgcmVwbGljYXRpb24gb24gdGhlIGxhZ2dpbmcgc3RyYW5kIGNhbiBvbmx5IGJlIGFjY29tcGxpc2hlZCBpbiBzaG9ydCBzdHJldGNoZXMgdXNpbmcgT2themFraSBmcmFnbWVudHMsIHRoYXQgZG9lc24mIzgyMTc7dCBleHBsYWluIHdoeSAmIzgyMjA7QiYjODIyMTsgaXMgcmVxdWlyZWQu

Cg==

Cg==

[Qq]

[q json=”true” xyz=”2″ multiple_choice=”true” unit=”6.Gene Expression and Regulation” dataset_id=”Unit 6 Multiple Choice Quiz 2|ef9858e824a3a” question_number=”6″ topic=”6.3.Transcription_and_RNA_Processing”] To answer this question, you can use this genetic code table:

In the diagram below, structure “X” is attached to the bases “G-U-C.” The next three bases are “A-A-G.” The third amino acid that would result from the process below would be

[c]IGd1YW5pbmUu[Qq]

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[c]IGx5c2 luZS4=[Qq]

[f]IEV4Y2VsbGVudC4gVGhlIG5leHQgY29kb24gaXMgQUFHLCB3aGljaCBpcyB0cmFuc2xhdGVkIGludG8gdGhlIGFtaW5vIGFjaWQgbHlzaW5lLg==[Qq]

[c]IGN5c3RlaW5lLg==[Qq]

[f]IE5vLiBZb3UgY2FuIGZpbmQgdGhlIG5leHQgYW1pbm8gYWNpZCBieSBpZGVudGlmeWluZyB0aGUgbmV4dCBjb2Rvbi4gSXQmIzgyMTc7cyBBQUcuIFdoZW4geW91IGxvb2sgdXAgQUFHIGluIGEgZ2VuZXRpYyBjb2RlIHRhYmxlLCB3aGF0IGFtaW5vIGFjaWQgZG8geW91IGdldD8=[Qq]

[c]IHByb2xpbmUu[Qq]

[f]IE5vLiBZb3UgY2FuIGZpbmQgdGhlIG5leHQgYW1pbm8gYWNpZCBieSBpZGVudGlmeWluZyB0aGUgbmV4dCBjb2Rvbi4gSXQmIzgyMTc7cyBBQUcuIFdoZW4geW91IGxvb2sgdXAgQUFHIGluIGEgZ2VuZXRpYyBjb2RlIHRhYmxlLCB3aGF0IGFtaW5vIGFjaWQgZG8geW91IGdldD8=

Cg==

[Qq]

[q json=”true” xyz=”2″ multiple_choice=”true” unit=”6.Gene Expression and Regulation” dataset_id=”Unit 6 Multiple Choice Quiz 2|ef9709a17463a” question_number=”7″ topic=”6.3.Transcription_and_RNA_Processing”] In the diagram below, step E directly results in the production of

[c]IG1lc3NlbmdlciBSTkEu[Qq]

[f]IE5vLiBTdGVwICYjODIyMDtFJiM4MjIxOyBkb2VzIHByb2R1Y2UgUk5BLCBidXQgdGhhdCBSTkEgaGFzIHRvIGJlIHByb2Nlc3NlZCBiZWZvcmUgaXQmIzgyMTc7cyByZWFkeSB0byBiZSBtZXNzZW5nZXIgUk5BIGFuZCByZWFkIGJ5IGEgcmlib3NvbWUuIFdoYXQgd291bGQgY29tZSBiZWZvcmUgbVJOQT8=[Qq]

[c]IHByZS1tZXNzZW 5nZXIgUk5BLg==[Qq]

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[c]IGEgcG9seXBlcHRpZGUgY2hhaW4u[Qq]

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[c]IGNvbXBsZW1lbnRhcnkgRE5BIChjRE5BKS4=[Qq]

[f]IE5vLiBDb21wbGVtZW50YXJ5IEROQSBpcyB1c2VkIGluIGdlbmV0aWMgZW5naW5lZXJpbmcgdG8gcHJvZHVjZSBldWthcnlvdGljIEROQSB0aGF0IGNhbiBiZSBwbGFjZWQgaW5zaWRlIGJhY3RlcmlhbCBjZWxscyBmb3IgcmVwbGljYXRpb24sIHRyYW5zY3JpcHRpb24sIGFuZCB0cmFuc2xhdGlvbi4gV2hhdCB5b3UmIzgyMTc7cmUgbG9va2luZyBmb3IgaGVyZSBpcyB0aGUgZmlyc3QgcmVzdWx0IG9mIHRyYW5zY3JpcHRpb24u

Cg==

Cg==

[Qq]

[q json=”true” xyz=”2″ multiple_choice=”true” topic=”6.4.Translation” dataset_id=”Unit 6 Multiple Choice Quiz 2|ef95ba5ac423a” question_number=”8″ unit=”6.Gene Expression and Regulation”] Which of the choices below correctly shows the sequence of processes that was carried out by the structure labeled “X” before it arrived at its current position?

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[f]IE5vLiBTdHJ1Y3R1cmUgWCBpcyBjYXJyeWluZyBhbiBhbWlubyBhY2lkLCBidXQgaXQmIzgyMTc7cyBiaW5kaW5nIHdpdGggYSBzZXF1ZW5jZSBvZiAzIGxldHRlcnMgb24gbVJOQS4gV2hhdCBhcmUgdGhvc2UgdGhyZWUtbGV0dGVyIHNlcXVlbmNlcyBjYWxsZWQ/[Qq]

[c]IEJpbmRpbmcgdG8gYW4gYW50aWNvZG9uIGFuZCB0aGVuIGNvbWJpbmluZyB3aXRoIGFuIGFtaW5vIGFjaWQu[Qq]

[f]IE5vLiBTdHJ1Y3R1cmUgWCBpcyBjYXJyeWluZyBhbiBhbWlubyBhY2lkLCBhbmQgaXQmIzgyMTc7cyBiaW5kaW5nIHdpdGggYSBzZXF1ZW5jZSBvZiAzIGxldHRlcnMgb24gbVJOQS4gV2hhdCBhcmUgdGhvc2UgdGhyZWUtbGV0dGVyIHNlcXVlbmNlcyBjYWxsZWQ/[Qq]

[c]IEJpbmRpbmcgdG8gYSBjb2RvbiBhbmQgdGhlbiBjb21iaW5pbmcgd2l0aCBhbiBhbWlubyBhY2lkLg==[Qq]

[f]IE5vLCBidXQgeW91JiM4MjE3O3JlIHZlcnkgY2xvc2UuIFRoZSBzdHJ1Y3R1cmUgYXQgWCBpcyBhIHRSTkEsIGFuZCBpdCBkb2VzIGJpbmQgd2l0aCBib3RoIGFuIGFtaW5vIGFjaWQgYW5kIGEgY29kb24uIFlvdSBqdXN0IG5lZWQgdG8gZ2V0IHRoZSBzZXF1ZW5jZSBvZiBldmVudHMgcmlnaHQu[Qq]

[c]IENvbWJpbmluZyB3aXRoIGFuIGFtaW5vIGFjaWQg YW5kIHRoZW4gYmluZGluZyB0byBhIGNvZG9uLg==[Qq]

[f]IENvcnJlY3QhIFRoZSBzdHJ1Y3R1cmUgYXQgWCBpcyBhIHRSTkEuIEl0IGZpcnN0IGJpbmRzIHdpdGggYW4gYW1pbm8gYWNpZCwgYW5kIHRoZW4gaXQgYmluZHMgd2l0aCBhIGNvZG9uIG9uIHRoZSBtUk5BLg==

Cg==

[Qq]

[q json=”true” xyz=”2″ multiple_choice=”true” topic=”6.4.Translation” dataset_id=”Unit 6 Multiple Choice Quiz 2|ef9445d355a3a” question_number=”9″ unit=”6.Gene Expression and Regulation”] The structure labeled “X” below is a(n)

[c]IHBvbHlwZXB0aWRlLg==[Qq]

[f]IE5vLiBBIHBvbHlwZXB0aWRlIGlzIGEgY2hhaW4gb2YgYW1pbm8gYWNpZHMsIGNvbm5lY3RlZCBieSBwZXB0aWRlIGJvbmRzLiBBYm92ZSwgYWxsIHlvdSBzZWUgaXMgb25lIGFtaW5vIGFjaWQgYXR0YWNoZWQgdG8gc3RydWN0dXJlIFguIER1cmluZyBwcm90ZWluIHN5bnRoZXNpcywgd2hhdCBtb2xlY3VsZSBicmluZ3MgYW1pbm8gYWNpZHMgdG8gdGhlIHJpYm9zb21lPw==[Qq]

[c]IHRS TkEu[Qq]

[f]IFRlcnJpZmljISAmIzgyMjA7WCYjODIyMTsgaXMgYSB0Uk5BLg==[Qq]

[c]IG1STkEu[Qq]

[f]IE5vLiBUaGUgbVJOQSBpcyB0aGUgc2luZ2xlLXN0cmFuZGVkIG51Y2xlaWMgYWNpZCB3aXRoIGl0cyA1JiM4MjQyOyBhbmQgMyYjODI0MjsgZW5kcyBsYWJlbGVkLiAmIzgyMjA7WCYjODIyMTsgaXMgd2hhdCBicmluZ3MgYW1pbm8gYWNpZHMgdG8gdGhlIG1STkEgYXQgdGhlIHJpYm9zb21lLg==[Qq]

[c]IHJSTkEu[Qq]

[f]IE5vLiBUaGUgclJOQSBpcyB3aGF0IG1ha2VzIHVwIHRoZSByaWJvc29tZS4gJiM4MjIwO1gmIzgyMjE7IGlzIHdoYXQgYnJpbmdzIGFtaW5vIGFjaWRzIHRvIHRoZSBtUk5BIGF0IHRoZSByaWJvc29tZS4=

Cg==

[Qq]

[q json=”true” xyz=”2″ multiple_choice=”true” dataset_id=”Unit 6 Multiple Choice Quiz 2|ef92f68ca563a” question_number=”10″ unit=”6.Gene Expression and Regulation” topic=”6.4.Translation”] In the figure below, what structure is labeled X?

[c]IEROQQ==[Qq]

[f]IE5vLiBIZXJlJiM4MjE3O3MgYSBoaW50LiBUaGUgc3RydWN0dXJlIGlzIGEgdFJOQSAodHJhbnNmZXIgUk5BKS4gV2hhdCBkbyB0Uk5BcyB0cmFuc2Zlcj8=[Qq]

[c]IFZlc2ljbGU=[Qq]

[f]IE5vLiBWZXNpY2xlcyBjYW4gYmUgY2lyY3VsYXIsIHNvIHRoYXQmIzgyMTc7cyBwb3NzaWJseSB3aGF0IHN0ZWVyZWQgeW91IHRvd2FyZCB0aGlzIGNob2ljZS4gSGVyZSYjODIxNztzIGEgaGludC4gVGhlIHN0cnVjdHVyZSBpcyBhIHRSTkEgKHRyYW5zZmVyIFJOQSkuIFdoYXQgZG8gdFJOQXMgdHJhbnNmZXI/[Qq]

[c]IEFtaW5v IEFjaWQ=[Qq]

[f]IEV4Y2VsbGVudC4gVGhlIHN0cnVjdHVyZSBpcyBhIHRSTkEsIGFuZCBYIGlzIGFuIGFtaW5vIGFjaWQgdGhhdCB0aGUgdFJOQSBpcyB0cmFuc2ZlcnJpbmcgdG8gYSByaWJvc29tZSBmb3IgcHJvdGVpbiBzeW50aGVzaXMu[Qq]

[c]IFRocmVlIG51Y2xlb3RpZGVz[Qq]

[f]IE5vLiBUaGUgc3RydWN0dXJlIGlzIGEgdFJOQS4gV2hpbGUgdFJOQXMgYXJlIG1hZGUgb2YgbnVjbGVvdGlkZXMsIHBhcnQgJiM4MjIwO1gmIzgyMjE7IGlzIG5vdCByZWFsbHkgcGFydCBvZiB0aGUgdFJOQSwgYnV0IHJhdGhlciBzb21ldGhpbmcgdGhhdCB0aGUgdFJOQSBpcyBjYXJyeWluZy4gSGVyZSYjODIxNztzIGEgaGludC4gdFJOQXMgYXJlIGludm9sdmVkIGluIHByb3RlaW4gc3ludGhlc2lzLiBXaGF0IGFyZSBwcm90ZWlucyBtYWRlIG9mLCBhbmQgd2hhdCBkbyB0Uk5BcyB0cmFuc2Zlcj8=

Cg==

Cg==

[Qq]

[q json=”true” multiple_choice=”true” unit=”6.Gene Expression and Regulation” topic=”6.5.Regulation_of_Gene_Expression” dataset_id=”Unit 6 Multiple Choice Quiz 2|ef91820536e3a” question_number=”11″] The diagram below is a model of the lac operon.

A key feature of this system is

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[f]IEZhYnVsb3VzLiBXaGVuIGxhY3Rvc2UgKDgpIGJpbmRzIHdpdGggdGhlIHJlcHJlc3NvciAoM2EpLCB0aGUgcmVwcmVzc29yIGNoYW5nZXMgc2hhcGUgYW5kIGNhbiBubyBsb25nZXIgYmluZCB3aXRoIEROQSBhdCB0aGUgb3BlcmF0b3IgcmVnaW9uICg2KQ==[Qq]

[c]IGxhY3Rvc2UmIzgyMTc7cyBhbGxvc3RlcmljIGFjdGl2YXRpb24gb2YgdGhlIHJlcHJlc3NvciBtb2xlY3VsZS4=[Qq]

[f]IE5vLiBOb3RlIHRoYXQgd2hlbiB0aGUgcmVwcmVzc29yIGlzIHVuYm91bmQgKGFzIGluIDMgYmVsb3cpKSwgaXRzIHNoYXBlIGVuYWJsZXMgaXQgdG8gYmluZCB3aXRoIHRoZSBvcGVyYXRvci4gV2hlbiBsYWN0b3NlIGJpbmRzIHdpdGggdGhlIHJlcHJlc3NvciAoM2EsIGFib3ZlKSwgdGhlIHJlcHJlc3NvciBjaGFuZ2VzIHNoYXBlIHNvIHRoYXQgaXQgY2FuIG5vIGxvbmdlciBiaW5kICh3aGljaCBpcyBpbmhpYml0aW9uKS4=

Cg==

[Qq]

[c]IGxhY3Rvc2UmIzgyMTc7cyBjb21wZXRpdGl2ZSBpbmhpYml0aW9uIG9mIHRoZSByZXByZXNzb3IgbW9sZWN1bGUu[Qq]

[f]IE5vLiBJZiBsYWN0b3NlIHdlcmUgYSBjb21wZXRpdGl2ZSBpbmhpYml0b3Igb2YgdGhlIHJlcHJlc3NvciwgdGhlbiBpdCB3b3VsZCBhbHNvIGJlIGJpbmRpbmcgd2l0aCB0aGUgb3BlcmF0b3IgKDYpLiBJbnN0ZWFkLCBsYWN0b3NlIGlzIGJpbmRpbmcgd2l0aCB0aGUgcmVwcmVzc29yIG1vbGVjdWxlIGl0c2VsZiwgYW5kIGNoYW5naW5nIGl0cyBzaGFwZS4=

Cg==

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Cg==

[Qq]

[c]IFJOQSBpbnRlcmZlcmVuY2UgdGhhdCBwcmV2ZW50cyB0aGUgdHJhbnNsYXRpb24gb2Ygc3RydWN0dXJhbCBSTkEu[Qq]

[f]IE5vLiBSTkEgaW50ZXJmZXJlbmNlIGlzIG5vdCBhdCB3b3JrIGluIHRoaXMgc3lzdGVtLg==

Cg==

SGVyZSYjODIxNztzIGEgaGludC4gVGhlIHN5c3RlbSBpbnZvbHZlcyB0aGUga2luZCBvZiBpbmhpYml0aW9uIHNob3duIGJlbG93IGJ5IG1vbGVjdWxlICYjODIyMDs0LiYjODIyMTs=

Cg==

[Qq]

[q json=”true” xyz=”2″ multiple_choice=”true” unit=”6.Gene Expression and Regulation” dataset_id=”Unit 6 Multiple Choice Quiz 2|ef900d7dc863a” question_number=”12″ topic=”6.5.Regulation_of_Gene_Expression”] In the lac operon, which of the following events takes place when lactose is not available in the environment?

[c]IEEgcmVndWxhdG9yeSBwcm90ZWluIGJpbmRzIGF0IHRoZSBvcGVyYXRvciwgYmxv Y2tpbmcgdGhlIHRyYW5zY3JpcHRpb24gb2YgdGhlIHN0cnVjdHVyYWwgZ2VuZXMu[Qq]

[f]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

Cg==

[Qq]

[c]IEEgcmVndWxhdG9yeSBwcm90ZWluIGJpbmRzIHRvIGFuIGFsbG9zdGVyaWMgc2l0ZSBvbiBSTkEgcG9seW1lcmFzZSwgcHJldmVudGluZyBpdCBmcm9tIGJpbmRpbmcgd2l0aCB0aGUgcHJvbW90ZXI=[Qq]

[f]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

Cg==

[Qq]

[c]IEEgcmVndWxhdG9yeSBwcm90ZWluIGJpbmRzIHRvIEROQSBwb2x5bWVyYXNlLCBwcmV2ZW50aW5nIHRoZSBnZW5lIGZvciBsYWN0b3NlIHByb2R1Y3Rpb24gdG8gYmUgcmVwbGljYXRlZC4=[Qq]

[f]IE5vLiBETkEgcG9seW1lcmFzZSBpc24mIzgyMTc7dCBpbnZvbHZlZCBpbiB0aGUgZnVuY3Rpb24gb2YgdGhpcyBvcGVyb24uIFRoZSBvcGVyb24gd29ya3MgYXMgc2hvd24gYmVsb3c6IFdoZW4gbGFjdG9zZSAoOCkgaXMgcHJlc2VudCwgbGFjdG9zZSBiaW5kcyB3aXRoIGFuIGFsbG9zdGVyaWMgc2l0ZSBvbiBhIHJlZ3VsYXRvcnkgcHJvdGVpbiAoMyksIGNoYW5naW5nIGl0cyBzaGFwZSBzbyB0aGF0IGl0IGNhbiBubyBsb25nZXIgYmluZCB3aXRoIHRoZSBvcGVyYXRvciByZWdpb24gKDYpIGFsbG93aW5nIHRyYW5zY3JpcHRpb24uIEtlZXAgdGhhdCBpbiBtaW5kIHRoZSBuZXh0IHRpbWUgeW91IHNlZSB0aGlzIHF1ZXN0aW9uLg==

Cg==

[Qq]

[c]IEEgcmVndWxhdG9yeSBwcm90ZWluIGJpbmRzIHRvIHRoZSBzdGFydCBjb2RvbiBvbiBtUk5BLCBwcmV2ZW50aW5nIHRoZSBpbml0aWF0aW9uIG9mIHRyYW5zbGF0aW9uIGJ5IHRoZSByaWJvc29tZS4=[Qq]

[f]IE5vLiBUaGUgb3Blcm9uIGlzIHdvcmtpbmcgYXQgdGhlIGxldmVsIG9mIHRyYW5zY3JpcHRpb24sIG5vdCB0cmFuc2xhdGlvbi4gUmF0aGVyLCB0aGUgb3Blcm9uIHdvcmtzIGFzIHNob3duIGJlbG93OiBXaGVuIGxhY3Rvc2UgKDgpIGlzIHByZXNlbnQsIGxhY3Rvc2UgYmluZHMgd2l0aCBhbiBhbGxvc3RlcmljIHNpdGUgb24gYSByZWd1bGF0b3J5IHByb3RlaW4gKDMpLCBjaGFuZ2luZyBpdHMgc2hhcGUgc28gdGhhdCBpdCBjYW4gbm8gbG9uZ2VyIGJpbmQgd2l0aCB0aGUgb3BlcmF0b3IgcmVnaW9uICg2KSBhbGxvd2luZyB0cmFuc2NyaXB0aW9uLiBLZWVwIHRoYXQgaW4gbWluZCB0aGUgbmV4dCB0aW1lIHlvdSBzZWUgdGhpcyBxdWVzdGlvbi4=

Cg==

Cg==

[Qq] 

[q json=”true” xyz=”2″ unit=”6.Gene Expression and Regulation” topic=”6.5.Regulation_of_Gene_Expression” multiple_choice=”true” dataset_id=”Unit 6 Multiple Choice Quiz 2|ef8ebe371823a” question_number=”13″] The diagram below shows the lac operon in the presence and absence of lactose. The genes lacZ, lacY, and lacA allow the bacterium to use lactose as an energy source.

Which of the following accurately connects the structure and function of the repressor molecule?

[c]IFdoZW4gdGhlIGJhY3Rlcml1bSBuZWVkcyBlbmVyZ3ksIEFUUCBiaW5kcyB0aGUgcmVwcmVzc29yLCBhbGxvd2luZyBSTkEgcG9seW1lcmFzZSB0byB0cmFuc2NyaWJlIHRoZSBnZW5lcyBuZWVkZWQgdG8gYnJlYWsgZG93biBsYWN0b3NlIGFuZCBtYWtlIEFUUC4=[Qq]

[f]IE5vLiBMb29rIGFnYWluIGF0IHRoZSBkaWFncmFtLiBXaGF0IGJpbmRzIHdpdGggdGhlIHJlcHJlc3NvciwgY2F1c2luZyBhbiBhbGxvc3RlcmljIHNoaWZ0IHRoYXQga2VlcHMgdGhlIHJlcHJlc3NvciBmcm9tIGJpbmRpbmcgd2l0aCB0aGUgb3BlcmF0b3IgYW5kIGFsbG93aW5nIHRoZSBnZW5lcyBmb3IgbGFjdG9zZSBkaWdlc3Rpb24gdG8gYmUgdHJhbnNjcmliZWQ/[Qq]

[c]IFRoZSByZXByZXNzb3IgbW9sZWN1bGUgaGFzIHR3byBiaW5kaW5nIHNpdGVzIGVhY2ggd2l0aCB0aGUgY29ycmVjdCBzaGFwZSBhbmQgY2hhcmdlIHRvIGJpbmQgZWl0aGVyIEROQSBvciBsYWN0b3NlLsKgIFdoZW4gbGFjdG9zZSBpcy BwcmVzZW50IGFuZCBiaW5kcyB0aGUgcmVwcmVzc29yLCB0aGUgcmVwcmVzc29yIGNoYW5nZXMgc2hhcGUgc28gdGhhdCBpdCBjYW4gbm8gbG9uZ2VyIGJpbmQgRE5BLCBhbGxvd2luZyB0cmFuc2NyaXB0aW9uIHRvIHRha2UgcGxhY2Uu[Qq]

[f]IEV4Y2VsbGVudC4gVGhlIGZ1bmN0aW9uIG9mIHRoZSBsYWMgb3Blcm9uIGlzIGFib3V0IHRoZSByZXByZXNzb3IgaGF2aW5nIGJpbmRpbmcgc2l0ZXMgZm9yIGxhY3Rvc2UgQU5EIEROQS4gV2hlbiBsYWN0b3NlIGJpbmRzIHdpdGggdGhlIHJlcHJlc3NvciwgaXQgY2F1c2VzIGFuIGFsbG9zdGVyaWMgc2hpZnQgdGhhdCBwcmV2ZW50cyB0aGUgcmVwcmVzc29yIGZyb20gYmluZGluZyB3aXRoIHRoZSBvcGVyYXRvciwgYWxsb3dpbmcgUk5BIHBvbHltZXJhc2UgdG8gdHJhbnNjcmliZSB0aGUgZ2VuZXMgZm9yIGxhY3Rvc2UgZGlnZXN0aW9uLg==[Qq]

[c]IFRoZSByZXByZXNzb3IgbW9sZWN1bGUgaGFzIGEgc2l0ZSB3aXRoIHRoZSBjb3JyZWN0IHNoYXBlIGFuZCBjaGFyZ2UgZm9yIGJpbmRpbmcgRE5BIGFuZCBhIHNlY29uZCBzaXRlIHdpdGggdGhlIGNvcnJlY3Qgc2hhcGUgYW5kIGNoYXJnZSB0byBiaW5kIHRvIGxhY3Rvc2UuwqAgV2hlbiBsYWN0b3NlIGlzIE5PVCBwcmVzZW50IGFuZCBpcyBOT1QgYm91bmQgdG8gdGhlIHJlcHJlc3NvciwgdGhlIHJlcHJlc3NvciBjaGFuZ2VzIHNoYXBlIHNvIHRoYXQgaXQgYmluZHMgRE5BLCBhbGxvd2luZyB0cmFuc2NyaXB0aW9uIHRvIHRha2UgcGxhY2Uu[Qq]

[f]IE5vLiBJdCYjODIxNztzIHRoZSBvcHBvc2l0ZS4gV2hlbiB0aGUgcmVwcmVzc29yIGJpbmRzIHRvIHRoZSBvcGVyYXRvciwgdHJhbnNjcmlwdGlvbiBpcyBibG9ja2VkLiBMb29rIGFnYWluIGF0IHRoZSBkaWFncmFtIGFuZCBjaG9vc2UgYSBiZXR0ZXIgcmVzcG9uc2Uu[Qq]

[c]IFdoZW4gdGhlIGJhY3Rlcml1bSBkb2VzbuKAmXQgbmVlZCBlbmVyZ3ksIGxhY3Rvc2UgY2hhbmdlcyBzaGFwZSBhbmQgY2hhcmdlIHNvIHRoYXQgaXQgYmluZHMgdG8gdGhlIG9wZXJhdG9yLsKgIFNpbmNlIHRoZSBvcGVyYXRvciBpcyBib3VuZCBieSB0aGUgbGFjdG9zZSwgdGhlIHJlcHJlc3NvciBiaW5kcyB0byB0aGUgUk5BIHBvbHltZXJhc2UsIHByZXZlbnRpbmcgaXQgZnJvbSBhdHRhY2hpbmcgdG8gdGhlIEROQS7CoCBBcyBhIHJlc3VsdCwgdHJhbnNjcmlwdGlvbiBkb2VzIG5vdCBvY2N1ci4=[Qq]

[f]IE5vLiBMYWN0b3NlIGRvZXNuJiM4MjE3O3QgYmluZCB3aXRoIFJOQSBwb2x5bWVyYXNlLiBMb29rIGFnYWluIGF0IHRoZSBkaWFncmFtLCBhbmQgbWFrZSBhbm90aGVyIGNob2ljZSB0aGUgbmV4dCB0aW1lIHlvdSBzZWUgdGhpcyBxdWVzdGlvbi4=

Cg==

[Qq]

[q json=”true” multiple_choice=”true” unit=”6.Gene Expression and Regulation” topic=”6.6.Gene_Expression_and_Cell_Specialization” dataset_id=”Unit 6 Multiple Choice Quiz 2|ef8d94312623a” question_number=”14″] Which of the following statements most closely matches what’s being depicted in the model of gene expression shown below?

[c]IEV2ZXJ5IEROQSBnZW5lIGhhcyBvbmUgY29ycmVzcG9uZGluZyBwcm90ZWluIHByb2R1Y3Qu[Qq]

[f]IE5vLiBMb29rIGF0ICYjODIyMDtELCYjODIyMTsgJiM4MjIwO0UsJiM4MjIxOyBhbmQgJiM4MjIwO0YsJiM4MjIxOyBhYm92ZS4gVGhyZWUgcHJvdGVpbnMgY29tZSBmcm9tIG9uZSBnZW5lLg==[Qq]

[c]IE11bHRpcGxlIHByb3RlaW5zIGNhbiBiZS Bwcm9kdWNlZCBmcm9tIG9uZSBnZW5lLg==[Qq]

[f]IEV4YWN0bHkuIEJ5IHNodWZmbGluZyBvciBkZWxldGluZyBleG9ucywgYWx0ZXJuYXRpdmUgc3BsaWNpbmcgYWxsb3dzIG11bHRpcGxlIHByb3RlaW5zIHRvIGJlIGNyZWF0ZWQgZnJvbSBvbmUgY29kaW5nIEROQS4=[Qq]

[c]IFByb2thcnlvdGVzIGhhdmUgbW9yZSBmbGV4aWJpbGl0eSBpbiBob3cgZ2VuZXMgYXJlIGV4cHJlc3NlZCB0aGFuIGRvIGV1a2FyeW90ZXM=[Qq]

[f]IE5vLiBUaGF0JiM4MjE3O3MgcHJvYmFibHkgbm90IHRydWUgKGp1c3QgY29uc2lkZXIgdGhlIGFsdGVybmF0aXZlIHNwbGljaW5nIHNob3duIGFib3ZlKSwgYW5kIGl0JiM4MjE3O3Mgbm90IGRlcGljdGVkIGluIHRoZSBpbWFnZSBhYm92ZS4=[Qq]

[c]IE1vZGlmaWNhdGlvbiBvZiBETkEgZnJvbSBjZWxsIHRvIGNlbGwgd2l0aGluIGEgZXVrYXJ5b3RpYyBvcmdhbmlzbSBjYW4gZW5oYW5jZSBwaGVub3R5cGljIGRpdmVyc2l0eS4=[Qq]

[f]IE5vLiBOb3RlIHRoYXQgaXQmIzgyMTc7cyBub3QgdGhlIEROQSB0aGF0JiM4MjE3O3MgYmVpbmcgbW9kaWZpZWQuIFRoZSBtb2RpZmljYXRpb24gb2NjdXJzIGFmdGVyIHRyYW5zY3JpcHRpb24sIGFuZCBpdCBpbnZvbHZlcyBSTkEu

Cg==

[Qq]

[q json=”true” multiple_choice=”true” unit=”6.Gene Expression and Regulation” topic=”6.6.Gene_Expression_and_Cell_Specialization” dataset_id=”Unit 6 Multiple Choice Quiz 2|ef8c1fa9b7a3a” question_number=”15″] The Pitx1 gene plays a role in gene expression in various tissues in the vertebrate body, including the pelvis, the jaw, and the pituitary gland within the brain.

The diagram below is a model of gene expression in the three-spine stickleback. The enhancer sequences control expression of the Pitx1 gene in a variety of tissues. The difference between the form on the left (with a pelvic spine) and the form on the right (without a pelvic spine) is most likely connected to

[c]IEEgbXV0YXRpb24gdGhhdCBwcmV2ZW50cyBSTkEgcG9seW1lcmFzZSBmcm9tIGJpbmRpbmcgd2l0aCB0aGUgUGl0eDEgZ2VuZSBwcm9tb3Rlci4=[Qq]

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[c]IEFjdGl2YXRpb24gb2YgdGhlIHBlbHZpYyBlbmhhbmNlciBpbiB0aGUgc3BpbmVsZXNzIGZvcm0u[Qq]

[f]IE5vLiBBY3RpdmF0aW9uIG9mIHRoZSBwZWx2aWMgZW5oYW5jZXIgd291bGQgcHJvbW90ZSBwZWx2aWMgc3BpbmUgZGV2ZWxvcG1lbnQuIFdoaWNoIG9mIHRoZSBjaG9pY2VzIHdvdWxkIHByZXZlbnQgcGVsdmljIHNwaW5lIGRldmVsb3BtZW50Pw==[Qq]

[c]IEEgZGVsZXRpb24gbXV0YXRpb24gaW4gdGhlIHBlbHZp YyBlbmhhbmNlciBpbiB0aGUgc3BpbmVsZXNzIGZvcm0u[Qq]

[f]IEV4Y2VsbGVudC4gQSBkZWxldGlvbiBpbiB0aGUgcGVsdmljIGVuaGFuY2VyIHdvdWxkIGFmZmVjdCB0aGUgZXhwcmVzc2lvbiBvZiB0aGUgUGl0eDEgZ2VuZSBpbiB0aGUgcGVsdmlzLCBidXQgbm90IGluIG90aGVyIHRpc3N1ZXMu[Qq]

[c]IGNocm9tb3NvbWFsIHRyYW5zbG9jYXRpb25zIHRoYXQgbW92ZSBQaXR4MSB0byBhIGRpZmZlcmVudCBjaHJvbW9zb21lIGluIHRoZSBzcGluZWxlc3MgZm9ybS4=[Qq]

[f]IE5vLiBUaGUgbW9kZWwgYWJvdmUgc2hvd3Mgbm8gZXZpZGVuY2Ugb2YgYSBjaHJvbW9zb21hbCB0cmFuc2xvY2F0aW9uLiBMb29rIG92ZXIgdGhlIGNob2ljZXMgYWdhaW4sIGFuZCBkZXRlcm1pbmUgd2hpY2ggb25lIHdvdWxkIHByZXZlbnQgcGVsdmljIHNwaW5lIGRldmVsb3BtZW50Lg==

Cg==

Cg==

[Qq]

[q json=”true” xyz=”2″ multiple_choice=”true” dataset_id=”Unit 6 Multiple Choice Quiz 2|ef8a85e18ae3a” question_number=”16″ unit=”6.Gene Expression and Regulation” topic=”6.7.Mutation”] Which of the following best describes the evolutionary role of gene duplication?

[c]IEJlY2F1c2Ugb2YgZ2VuZSBkdXBsaWNhdGlvbiwgdGhlIGdlbmV0aWMgY29kZSBjYW4gYmUgbW9yZSBmbGV4aWJsZS4=[Qq]

[f]IE5vLiBHZW5lIGR1cGxpY2F0aW9uIGlzIGV4YWN0bHkgd2hhdCBpdCBzb3VuZHMgbGlrZTogdGhlIGNvcHlpbmcgb2YgZ2VuZSBzbyB0aGF0IGluc3RlYWQgb2YgdGhlcmUgYmVpbmcgb25lIGNvcHksIHRoZXJlIGFyZSBtdWx0aXBsZSBjb3BpZXMu

Cg==

Cg==

[Qq]Gene duplication has no effect on the genetic code (the way that codon sequences are translated into protein). But they do play a big role in evolution. Take a look at the other choices, and see if one of them looks like it could help a population evolve.

[c]IEZld2VyIGdlbmUgY29waWVzIGxlYWRzIHRvIGZhc3RlciBETkEgcmVwbGljYXRpb24sIGFuZCBmYXN0ZXIgRE5BIHJlcGxpY2F0aW9uIGFsbG93cyBhIHBvcHVsYXRpb24gdG8gZXZvbHZlIG1vcmUgcXVpY2tseS4=[Qq]

[f]IE5vLiBBcyB5b3UgY2FuIHNlZSBiZWxvdywgZ2VuZSBkdXBsaWNhdGlvbiBsZWFkcyB0byBtb3JlIGdlbmUgY29waWVzLCBub3QgZmV3ZXIu

Cg==

Cg==

[Qq]Take a look at the other choices, and see if one of them indicates a process that could help a population evolve.

[c]IER1cGxpY2F0aW9uIGNyZWF0ZXMgaW50cm9ucywgYW5kIGludHJvbnMgYWxsb3cgZm9yIHRoZSBnZW5lIHNodWZmbGluZyB0aGF0IHByb21vdGVzIHBoZW5vdHlwaWMgdmFyaWF0aW9uLg==[Qq]

[f]IE5vLiBXaGlsZSB0aGUgc2Vjb25kIHBhcnQgb2YgdGhpcyBzdGF0ZW1lbnQgaXMgdHJ1ZSwgdGhlIGZpcnN0IGlzIGluY29ycmVjdC4gR2VuZSBkdXBsaWNhdGlvbiBpcyBleGFjdGx5IHdoYXQgaXQgc291bmRzIGxpa2U6IHRoZSBjb3B5aW5nIG9mIGdlbmUgc28gdGhhdCBpbnN0ZWFkIG9mIHRoZXJlIGJlaW5nIG9uZSBjb3B5LCB0aGVyZSBhcmUgbXVsdGlwbGUgY29waWVzLg==

Cg==

Cg==

[Qq]Take a look at the other choices, and see if one of them looks like it could help a population evolve.

[c]IER1cGxpY2F0aW9uIG9mIGdlbmVzIGNoYW5nZXMgdGhlaXIgcG9zaXRpb24gaW4gdGhlIGdlbm9tZSwgYW5kIHRoYXQgbGVhZHMgdG8gY2hhbmdlcyBpbiBnZW5lIGV4cHJlc3Npb24u[Qq]

[f]IE5vLCBidXQgeW91IGNob3NlIHRoZSBzZWNvbmQtYmVzdCBjaG9pY2UuIEl0IHNvdW5kcyBsaWtlIHlvdSB3ZXJlIHRoaW5raW5nIG9mIHdoYXQgaGFwcGVucyB3aXRoIHJlcGxpY2F0aXZlIHRyYW5zcG9zaXRpb24sIHdoZXJlIGEgdHJhbnNwb3NvbiAoYSBtb2JpbGUgRE5BIGVsZW1lbnQpIG1ha2VzIGEgY29weSBvZiBpdHNlbGYgdGhhdCBpbnNlcnRzIHNvbWV3aGVyZSBlbHNlIGluIHRoZSBnZW5vbWUuIFRoaXMgY2FuIGFmZmVjdCBleHByZXNzaW9uIG9mIGdlbmVzLCBlc3BlY2lhbGx5IGlmIHRoZSB0cmFuc3Bvc29uIGRpc3J1cHRzIGFub3RoZXIgZ2VuZSYjODIxNztzIHByb21vdGVyLg==

Cg==

R2VuZSBkdXBsaWNhdGlvbiBpcyBleGFjdGx5IHdoYXQgaXQgc291bmRzIGxpa2U6IHRoZSBjb3B5aW5nIG9mIGdlbmUgc28gdGhhdCBpbnN0ZWFkIG9mIHRoZXJlIGJlaW5nIG9uZSBjb3B5LCB0aGVyZSBhcmUgbXVsdGlwbGUgY29waWVzLg==

Cg==

[Qq]

Take a look at the other choices, and see if one of them looks like it could help a population evolve.

[c]IEFuIGV4dHJhIGdlbmUgY29weSBjYW4gdW5kZXJnbyBhZGFwdGl2ZSBjaGFuZ2VzIHdo aWxlIHRoZSBmaXJzdCBjb3B5IHBlcmZvcm1zIGl0cyBvcmlnaW5hbCBmdW5jdGlvbi4=[Qq]

[f]IE5pY2Ugam9iLiBUaGUgYmVuZWZpdCBvZiBnZW5lIGR1cGxpY2F0aW9uIGlzIHRoYXQgaXQgcHJvdmlkZXMgYSBtZWNoYW5pc20gYnkgd2hpY2ggYSBnZW5lIHBvb2wgY2FuIGNyZWF0ZSBnZW5lcyB0aGF0IGNvZGUgZm9yIG5ldyBhZGFwdGl2ZSB0cmFpdHMsIHdoaWxlIG1haW50YWluaW5nIHRoZSBvcmlnaW5hbCBmdW5jdGlvbiBvZiB0aGUgZHVwbGljYXRlZCBnZW5lLg==

Cg==

Cg==

[Qq]

[q json=”true” xyz=”2″ multiple_choice=”true” unit=”6.Gene Expression and Regulation” topic=”6.7.Mutation” dataset_id=”Unit 6 Multiple Choice Quiz 2|ef89115a1c63a” question_number=”17″] In the diagram below, a nonsense mutation is shown at

[c]IEEg[Qq][c]IE Ig[Qq][c]IEMg[Qq][c]IEQ=

Cg==[Qq]

[f]IE5vLiBNdXRhdGlvbiAmIzgyMjA7QSYjODIyMTsgY2hhbmdlcyB0aGUgRE5BIGFuZCBSTkEsIGJ1dCBub3QgdGhlIHJlc3VsdGluZyBhbWlubyBhY2lkLiBBIG5vbnNlbnNlIG11dGF0aW9uIGhhcyBhIG11Y2ggYmlnZ2VyIGltcGFjdCBvbiB0aGUgcmVzdWx0aW5nIHByb3RlaW4u[Qq]

[f]IEV4Y2VsbGVudC4gTXV0YXRpb24gJiM4MjIwO0ImIzgyMjE7IGluc2VydHMgYSBzdG9wIGNvZG9uIGluc3RlYWQgb2YgYW4gYW1pbm8gYWNpZC4gVGhhdCYjODIxNztzIGEgbm9uc2Vuc2UgbXV0YXRpb24u[Qq]

[f]IE5vLiBNdXRhdGlvbiAmIzgyMjA7QyYjODIyMTsgY2hhbmdlcyB0aGUgRE5BLCB0aGUgUk5BLCBhbmQgdGhlIGFtaW5vIGFjaWQsIG1ha2luZyBpdCBhICYjODIyMDttaXNzZW5zZSYjODIyMTsgbXV0YXRpb24uIEEgbm9uc2Vuc2UgbXV0YXRpb24gaGFzIGEgbXVjaCBiaWdnZXIgaW1wYWN0IG9uIHRoZSByZXN1bHRpbmcgcHJvdGVpbi4=[Qq]

[f]IE5vLiBNdXRhdGlvbiAmIzgyMjA7RCYjODIyMTsgY2hhbmdlcyB0aGUgRE5BLCB0aGUgUk5BLCBhbmQgdGhlIGFtaW5vIGFjaWQsIG1ha2luZyBpdCBhICYjODIyMDttaXNzZW5zZSYjODIyMTsgbXV0YXRpb24uIEEgbm9uc2Vuc2UgbXV0YXRpb24gaGFzIGEgbXVjaCBiaWdnZXIgaW1wYWN0IG9uIHRoZSByZXN1bHRpbmcgcHJvdGVpbi4=

Cg==

Cg==

[Qq]

[q json=”true” xyz=”2″ multiple_choice=”true” unit=”6.Gene Expression and Regulation” dataset_id=”Unit 6 Multiple Choice Quiz 2|ef879cd2ade3a” question_number=”18″ topic=”6.8.Biotechnology”] Certain restriction enzymes cut DNA in a way that leaves the fragments with “sticky ends.” These “sticky ends” are useful in genetic engineering because they allow for

[c]IHNjcmVlbmluZyBvZiBwbGFzbWlkcyB3aXRoIGFudGliaW90aWMgcmVzaXN0YW5jZSBnZW5lcy4=[Qq]

[f]IE5vLiBUYWtlIGEgbG9vayBhdCB0aGlzIHBpY3R1cmUgb2YgRE5BIHRoYXQmIzgyMTc7cyBiZWVuIGN1dCBpbiBhIHdheSB0aGF0IGxlYXZlcyBpdCB3aXRoIHN0aWNreSBlbmRzLiBOb3csIGxvb2sgYWdhaW4gYXQgdGhlIGNob2ljZXMgYW5kIHNlZSBpZiB5b3UgY2FuIGZpbmQgYSBiZXR0ZXIgYW5zd2VyLg==

Cg==

[Qq]

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[q json=”true” xyz=”2″ multiple_choice=”true” unit=”6.Gene Expression and Regulation” dataset_id=”Unit 6 Multiple Choice Quiz 2|ef86284b3f63a” question_number=”19″ topic=”6.8.Biotechnology”] The image below represents the pGLO plasmid, which is used both in research and in education.

As indicated, the plasmid contains a restriction site and three genes:

  • ampR—confers resistance to ampicillin, an antibiotic.
  • gfp—encodes the green fluorescent protein (GFP), which glows green (fluoresces) under ultraviolet (UV) light
  • araC—encodes a protein required to promote the expression of GFP when arabinose, a monosaccharide, is present

The results of a bacterial transformation experiment using the pGLO plasmid are shown in the table below.

Which of the following statements about the purpose of plates W or X is correct?

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[Qq]

[q json=”true” multiple_choice=”true” unit=”6.Gene Expression and Regulation” question_number=”20″ topic=”6.8.Biotechnology” dataset_id=”Unit 6 Multiple Choice Quiz 2|ef8110713ca3a”] The diagram below shows the lac operon in the presence and absence of lactose. The genes lacZ, lacY, and lacA allow the bacterium to use lactose as an energy source.

A researcher adds a gene for ampicillin resistance (AMPR)between the lacZ and lacY genes as shown below.

Then the researcher plates the bacteria on the following media. All media contain fatty acids as a secondary energy source. Bacteria, when given the choice, metabolize sugars before fatty acids.

  • Plate 1: No ampicillin or lactose
  • Plate 2: Ampicillin and no lactose
  • Plate 3: Lactose and no ampicillin
  • Plate 4: Ampicillin and lactose

Which plates would there be growth on?

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[/qwiz]

5. Unit 6 Practice FRQs

[qwiz qrecord_id=”sciencemusicvideosMeister1961-Unit 6 Cumulative Practice FRQs (v2.0)” dataset=”Unit 6 Cumulative Practice FRQs” style=”width: 600px !important; min-height: 450px !important;”][h]Unit 6 Practice FRQs

[i]

[q json=”true” xx=”1″ multiple_choice=”false” unit=”6.Gene Expression and Regulation” dataset_id=”Unit 6 Cumulative Practice FRQs|13b77cea3736e2″ question_number=”1″ topic=”6.1.DNA_Structure_and_Replication”] The diagram below is a representation of two fragments of DNA, each with the same nucleotide sequence, from the same species of bacterium. The fragment of DNA on the left was grown in a medium containing 14N, the most common isotope of nitrogen. The fragment on the right was grown in a medium containing 15N, a less common, radioactive isotope of nitrogen that contains an additional neutron.

Bacteria with DNA consisting of 14N were transferred to a medium consisting of 15N, and allowed to replicate once.

PART 1: Identify which of the diagrams below correctly shows the daughter DNA molecules that would result.

PART 2: Justify your answer.

[c]IFNob3cgdGhl IGFuc3dlcg==[Qq]

[f]IFBBUlQgMTrCoFRoZSBjb3JyZWN0IG1vZGVsIGlzIEQu

Cg==

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[Qq]14N, but the free nucleotides coming in will contain the 15N nitrogen in their nucleotide bases. As a result, in each of the duplicated DNA molecules, one strand will consist of nucleotides with 14N, and the paired complementary strand will consist of nucleotides with 15N.

[q json=”true” xx=”1″ multiple_choice=”false” unit=”6.Gene Expression and Regulation” dataset_id=”Unit 6 Cumulative Practice FRQs|13b767f5cc32e2″ question_number=”2″ topic=”6.3.Transcription_and_RNA_Processing”] A student is interviewing for a summer internship at a local research lab. During the interview, a graduate student mentor tells the student that the lab has just discovered a new eukaryotic protein that’s 197 amino acids long and that the summer project will be identifying the gene for this protein.

PART 1: Calculate, in terms of nucleotides, the minimum size of the gene for this protein. Justify your response.

PART 2: Explain why the number of nucleotides in the gene might be much larger than the minimum number.

PART 3: The research team wants to insert the gene for this protein into a bacterial plasmid so that a bacterium can synthesize the protein. Describe two techniques by which the research team could create DNA for this protein that could be correctly translated by a bacterium.

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Cg==

[Qq]PART 2: Because this is a eukaryotic gene, it probably contains introns, which would make the number of nucleotides larger than the minimum required.  

PART 3: For a bacterium to properly translate DNA for a eukaryotic protein, the DNA that codes for this protein must be intron free. To create intron-free DNA, the research team could isolate cells that are producing this protein, and then isolate the mRNA for this protein. Using reverse transcriptase, they could convert the mRNA into DNA, which could then be inserted into a bacterial plasmid. Alternatively, the research team could reverse engineer DNA from the amino acid sequence of the protein.

[q json=”true” multiple_choice=”false” unit=”6.Gene Expression and Regulation” topic=”6.4.Translation” dataset_id=”Unit 6 Cumulative Practice FRQs|13b750ad554ae2″ question_number=”3″] Scientists are studying the effectiveness of a drug, Evernimicin, on inhibiting ribosome function in the bacteria Escherichia coli. To do this, they measured the synthesis of luciferase, a genetically engineered protein introduced into E. coli to make them luminescent (glow). In their experiment, scientists tested different concentrations of Evernimicin at both 37° Celsius (white boxes) and 25° Celsius (black boxes).

PART 1: Identify the control group in this experiment and describe the difference in ribosome function in the control group in terms of relative luminescence at 37° and 25° Celsius.

PART 2: Explain the relationship between ribosome function and luciferase production in E. coli.

PART 3: A scientist proposes that Evernimicin’s effect is mostly on the cell membrane of E. coli. Identify one part of the plasma membrane that could be affected by an elevated dose of Evernimicin in E. coli.

[c]IFNob3cgdGhl IGFuc3dlcg==[Qq]

[f]

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[Qq]

PART 3: Proteins in the plasma membrane (receptors, channels, etc.) could be affected by an increased dose of Evernimicin. The data shows that Evernimicin inhibits ribosome function. Since some of the proteins created by ribosomes are membrane proteins, then ribosome inhibition will affect these membrane components.

[q json=”true” xx=”1″ multiple_choice=”false” unit=”6.Gene Expression and Regulation” dataset_id=”Unit 6 Cumulative Practice FRQs|13b73bb8ea46e2″ question_number=”4″ topic=”6.4.Translation”] Starting from its synthesis at a ribosome, describe the pathway of a protein (such as a protein hormone or an antibody) that is going to be secreted from the cell.

[c]IFNob3cgdGhl IGFuc3dlcg==[Qq]

[f]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

Cg==

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[Qq]For instructional purposes, the process is shown above. 1 is mRNA. 2 is a ribosome. 3 is the signal polypeptide. 4 is the signal recognition particle. 6 is a receptor in the rough ER, and 5 is a channel through which the protein (8) is extruded into the ER lumen.

[q json=”true” xx=”1″ multiple_choice=”false” unit=”6.Gene Expression and Regulation” dataset_id=”Unit 6 Cumulative Practice FRQs|13b726c47f42e2″ question_number=”5″ topic=”6.5.Regulation_of_Gene_Expression”] Jacques Monod, who co-developed the operon concept, noticed that when E. coli bacteria were fed lactose, they were able to activate a regulatory pathway that produced lactose-digesting enzymes. When the lactose was digested, this regulatory pathway was turned off, and the enzymes for digesting lactose were no longer produced.

Explain the biology behind what Monod discovered.

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[Qq]

The promoter (5) is upstream of the structural genes (7). To transcribe the genes for lactose-digesting enzymes, RNA polymerase (4) will bind with the promoter, and then proceed on to the structural genes. However, between the promoter and the structural genes is an operator (6): a DNA region that allows for regulatory control. In the case of the lac operon, there’s a regulatory gene (1) that produces a regulatory protein (3). In the absence of lactose, the regulatory protein can bind with the operator in a way that blocks RNA polymerase, preventing it from transcribing the genes that code for lactose-digesting enzymes. In other words, when lactose is not present, the cell won’t transcribe the structural genes or translate the resulting mRNA into protein (both of which would be a waste of energy in lactose’s absence).

However, the regulatory protein, in addition to having a site for binding with the operator, also has a site for binding with lactose (8, below).

When lactose (8) binds to this regulatory protein, it changes the regulatory protein’s shape so that it can no longer bind with the operator. This allows RNA polymerase to move along from the promoter to transcribe the structural genes. In other words, the presence of lactose induces changes that allow for the production of lactose-digesting enzymes.

[q json=”true” xx=”1″ multiple_choice=”false” unit=”6.Gene Expression and Regulation” dataset_id=”Unit 6 Cumulative Practice FRQs|13b70d27fc76e2″ question_number=”6″ topic=”6.5.Regulation_of_Gene_Expression”] The diagram below depicts a system in bacterial cells that regulates the production of a protein product A. This system involves two regulatory pathways, X and Y. In both pathways, product A acts as an inhibitor.

PART 1: Propose a mechanism by which regulatory pathway X would inhibit the production of product A. Identify a system in bacterial cells that works by the same mechanism as regulatory pathway X.

PART 2: Explain how this regulatory pathway is beneficial for the cell.

PART 3: Propose two mechanisms by which regulatory pathway Y could inhibit the production of product A.

[c]IFNob3cgdGhl IGFuc3dlcg==[Qq]

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Cg==

Cg==

[Qq]PART 2: Production of product A requires the cell to use energy for the production of enzymes 1 through 5 and for the chemical reactions carried out by each enzyme in the biochemical pathway by which the precursor molecule is converted into product A. When product A is present in high concentration, it’s of benefit to the cell to turn off production of product A until its concentration falls to a level where the cell once again needs to synthesize it.

PART 3: Regulatory pathway Y could work through competitive inhibition. In that case, product A could fit into the active site of enzyme 3 and block compound 2 from binding with enzyme 3’s active site. Alternatively, product A could work through allosteric inhibition. In that case, product A would bind away from the active site of enzyme 3 at an allosteric site. The binding of compound A with the allosteric site would change the shape of enzyme 3’s active site so that compound 2 could no longer bind with enzyme 3.

In both cases, preventing enzyme 3 from converting compound 2 into compound 3 would shut down the pathway leading to the production of product A.

In the diagram below (included for instructional purposes), number 1 represents compound 2 (above). Diagram (b) shows competitive inhibition, with compound A represented by number 4. Diagram (c) shows allosteric regulation. In this scenario, compound A would be number 6 in the diagram, binding at an allosteric site.

[q json=”true” multiple_choice=”false” unit=”6.Gene Expression and Regulation” topic=”6.6.Gene_Expression_and_Cell_Specialization” dataset_id=”Unit 6 Cumulative Practice FRQs|13b6f5df858ee2″ question_number=”7″] One important class of proteins in eukaryotic cells is called ATP-dependent chromatin remodelers. Typically, DNA in chromatin form is wound around groups of histone proteins called the nucleosome, forming a histone-chromatin complex. ATP-dependent chromatin remodelers change the structure of the histone-chromatin complex, as shown in the model below.

 

PART 1: Identify how ATP-dependent chromatin remodelers change the structure of the histone-chromatin complex, using evidence from the model above.

PART 2: Make a claim about how the abundance of ATP-dependent chromatin remodelers would change the rate of production of an mRNA transcript.

PART 3: Predict the effects of inhibiting the supply of ATP on the structure of the histone-chromatin complex

[c]IFNob3cgdGhl IGFuc3dlcg==[Qq]

[f]

Cg==

UEFSVCAxOiDCoEFjY29yZGluZyB0byB0aGUgbW9kZWwsIEFUUC1kZXBlbmRlbnQgY2hyb21hdGluIHJlbW9kZWxlcnMgbG9vc2VuIHRoZSBoaXN0b25lLWNocm9tYXRpbiBjb21wbGV4IGJ5IHNsaWRpbmcgbnVjbGVvc29tZXMgYXBhcnQsIHVud3JhcHBpbmcgY2hyb21hdGluIGZyb20gbnVjbGVvc29tZXMsIG9yIGVqZWN0aW5nIG51Y2xlb3NvbWVzIGNvbXBsZXRlbHku

Cg==

UEFSVCAyOiBQcm9kdWN0aW9uIG9mIG1STkEgdHJhbnNjcmlwdHMgd291bGQgaW5jcmVhc2UgaW4gdGhlIHByZXNlbmNlIG9mIGNocm9tYXRpbiByZW1vZGVsZXJzIGJlY2F1c2UgdGhleSBtYWtlIEROQSBhY2Nlc3NpYmxlIHRvIFJOQSBwb2x5bWVyYXNlLCB3aGljaCBpcyB0aGUgZW56eW1lIHRoYXQgcHJvZHVjZXMgbVJOQS4gV2l0aG91dCB0aGVzZSByZW1vZGVsZXJzLCBSTkEgcG9seW1lcmFzZSB3b3VsZCBoYXZlIGRpZmZpY3VsdHkgYWNjZXNzaW5nIGFuZCBiaW5kaW5nIHRvIHRoZSBETkEgYXQgdGhhdCBsb2N1cy4=

[Qq]

PART 3: If the supply of ATP was decreased, we would expect the histone-chromatin complex to remain intact and bound tightly. Because the chromatin-remodeler is ATP-dependent, without ATP there would be no energy to modify the packaging of DNA.

[q json=”true” xx=”1″ multiple_choice=”false” unit=”6.Gene Expression and Regulation” dataset_id=”Unit 6 Cumulative Practice FRQs|13b6c2a67ff6e2″ question_number=”8″ topic=”6.7.Mutation”] In Tay-Sachs disease, a genetic mutation in an enzyme called beta-hexosaminidase A results in the build-up of a molecule called GM2 ganglioside. GM2 ganglioside buildup is toxic and ultimately fatal. Victims usually die before the age of five, and there is no treatment.

The table below shows a portion of a DNA sequence for a person with a normal version of beta-hexosaminidase A and for a person with the Tay-Sachs mutation.

DNA Sequence
Normal Individual …GAGAGGTTT
A person with Tay Sachs …GAGGGGTTT

PART 1: Based on the DNA sequence above, list the amino acid sequence for an individual with Tay-Sachs. Use the genetic code chart below.

PART 2: With reference to protein structure and amino acid chemistry, explain how the Tay Sachs mutation affects the hexosaminidase A enzyme. Use the same chart listed above for reference.

PART 3: Part of the metabolic pathway connected with Tay-Sachs is shown below. Using this diagram, describe some of the effects of the Tay-Sachs mutation.

[c]IFNob3cgdGhl IGFuc3dlcg==[Qq]

[f]IE5vdGUgdGhhdCB0aGUgdmFyaW91cyBwYXJ0cyBvZiB0aGUgYW5zd2VyIGNhbiBvY2N1ciBpbiBhbnkgb2YgdGhlIHNlY3Rpb25zIGJlbG93Lg==

Cg==

UEFSVCAxOg==[Qq] Leu-Pro-Lys

PART 2: The mutation changes only one amino acid: Serine to Proline. However, Serine (ser) has a polar side chain, while proline has a nonpolar side chain. Because a protein’s 3-D conformation (shape) is based on interactions between side chains, this single shift could change the shape of the enzyme. If this changes the enzyme’s active site, then the enzyme’s ability to break down its substrate could be affected. Since Tay-Sachs results in the buildup of a toxic substance that is normally broken down by the healthy version of the enzyme, then that is probably the molecular cause of Tay-Sachs disease.

PART 3: If the Tay-Sachs mutation has occurred and the Hex A enzyme has been affected, then it will not be able to break down the glycolipid into compound Q. Compound Q, in turn, will not be able to be converted into compounds S and T. Instead, compound Q will build up to toxic levels.

[q json=”true” multiple_choice=”false” unit=”6.Gene Expression and Regulation” topic=”6.8.Biotechnology” dataset_id=”Unit 6 Cumulative Practice FRQs|13b683c93eeae2″ question_number=”9″]

E.coli is a bacteria that can cause food poisoning when it produces two key toxins: Shiga toxin and intimin toxin.

When the genes for these toxins are digested with certain restriction enzymes, the gene for the Shiga toxin produces fragments of about 210 base pairs of DNA, and the gene for intimin produces fragments of about 280 base pairs.

Food scientists wished to identify which, if any, samples of food were contaminated with toxin-producing E. coli. They amplified the DNA of three different samples via PCR, loaded them into an agarose gel, and then ran an electrical current through the samples to separate the fragments of DNA (shown below).

PART 1: Rank the 3 samples in terms of their likeliness to cause food poisoning and justify your answer.

PART 2: Explain why running an electrical current can separate the fragments of DNA.

[c]IFNob3cgdGhl IGFuc3dlcg==[Qq]

[f]

Cg==

UEFSVCAxOiBTYW1wbGUgMiBpcyB0aGUgbW9zdCBsaWtlbHkgdG8gY2F1c2UgZm9vZCBwb2lzb25pbmcsIGFzIGl0IGNvbnRhaW5zIGJhbmRzIHRoYXQgYXJlIGFzc29jaWF0ZWQgd2l0aCBib3RoIFNoaWdhIHRveGluICgyMTAgYnApIGFuZCBpbnRpbWluIHRveGluICgyODAgYnApLiBTYW1wbGUgMyBwcm9kdWNlcyBvbmUgb2YgdGhlIHR3byB0b3hpbnMsIGFuZCBzYW1wbGUgMSBwcm9kdWNlcyBub25lIG9mIHRoZSB0b3hpbnMu

Cg==

UEFSVCAyOiBSdW5uaW5nIGFuIGVsZWN0cmljYWwgY3VycmVudCB0aHJvdWdoIHRoZSBETkEgY2FuIHNlcGFyYXRlIGZyYWdtZW50cyBiZWNhdXNlIEROQSBpcyBuZWdhdGl2ZWx5IGNoYXJnZWQuIFRoZXJlZm9yZSwgRE5BIHdpbGwgbWlncmF0ZSBmcm9tIHRoZSBuZWdhdGl2ZSB0byB0aGUgcG9zaXRpdmVseSBjaGFyZ2VkIGVuZCBvZiB0aGUgZ2VsLCB3aXRoIGxhcmdlciBzdHJhbmRzIHRha2luZyBsb25nZXIgdG8gbW92ZSB0aGFuIHNtYWxsZXIgc3RyYW5kcy4=

[Qq]

[q json=”true” xx=”1″ multiple_choice=”false” unit=”6.Gene Expression and Regulation” dataset_id=”Unit 6 Cumulative Practice FRQs|13b5e576165ae2″ question_number=”10″ topic=”6.8.Biotechnology”] The image below represents the pGLO plasmid, which is used both in research and education.

As indicated, the plasmid contains a restriction site and three genes:

  • ampR: confers resistance to ampicillin, an antibiotic.
  • gfp: encodes the green fluorescent protein (GFP), which glows green (fluoresces) under ultraviolet (UV) light
  • araC: encodes a protein required to promote the expression of GFP when arabinose, a disaccharide, is present

The results of a bacterial transformation experiment using the pGLO plasmid are shown in the table below.

PART 1: Explain the purpose of plates W and X.

PART 2: Explain why there is a lawn of bacteria on plate W, as opposed to only discrete colonies on plates Y and Z.

PART 3: Explain why only the bacteria on plate Z glow under ultraviolet light.

PART 4: Explain why there is no growth on plate X.

[c]IFNob3cgdGhl IGFuc3dlcg==[Qq]

[f]IFBBUlQgMTogUGxhdGVzIFcgYW5kIFggYXJlIGNvbnRyb2xzLiBUaGUgcHVycG9zZSBvZiBXIGlzIHRvIHNob3cgdGhhdCB0aGUgRS4gY29saQ==IGJhY3RlcmlhIGFyZSBhbGl2ZSwgYW5kIHRoZSBwdXJwb3NlIG9mIFggaXMgdG8gZGVtb25zdHJhdGUgdGhhdCBiYWN0ZXJpYSB3aXRob3V0IHRoZSBwbGFzbWlkIGNhbiYjODIxNzt0IGdyb3cgb24gYSBwbGF0ZSB3aXRoIGFtcGljaWxsaW4u

Cg==

UEFSVCAyOiBBbnkgYmFjdGVyaXVtIHdpbGwgZ3JvdyBvbiBwbGF0ZSBXLiBCZWNhdXNlIHRoZSBudXRyaWVudCBhZ2FyIGhhcyBldmVyeXRoaW5nIHRoYXQg[Qq]E. coli needs, it grows everywhere it’s deposited (which is all over the plate). By contrast, only transformed bacteria that took up the plasmid (which contains an antibiotic resistance gene)will grow on plates with ampicillin. Because only a small proportion of the cells exposed to the plasmid will be transformed, the resulting growth will show up as discrete colonies (with every colony representing a successfully transformed cell).

PART 3: GFP is only expressed when the araC gene produces a regulatory protein that permits transcription of the GFP genes. araC, in turn, will only produce this regulatory protein in the presence of arabinose. Only plate Y has arabinose on the plate

PART 4: The bacteria on plate X have not been transformed, which means they haven’t taken up the plasmid, which means that they lack the ampicillin resistance gene, and won’t grow in the presence of ampicillin.

[x][restart]
[/qwiz]

6. DNA and DNA Replication Click On Challenge

Note: if you’re logged in on the Biomania phone app, you can get the same credit for completing this click-on challenge, and the ones below.. Your progress on the app will be remembered on the website, and vice versa.

[qwiz style=”width: 650px !important; min-height: 450px !important;” quiz_timer=”true” random=”false” dataset_intro=”false” spaced_repetition=”false” use_dataset=”DNA and DNA Replication Click On Challenge Dataset” qrecord_id=”sciencemusicvideosMeister1961-Unit 6 DNA Click-on Challenge (v2.0)”]

[h] DNA and DNA Replication Click-On Challenge

[i] Note the timer in the top right. Your goal is accuracy and speed. A good strategy: once through slowly, then additional tries for improvement.

[/qwiz]

7. Transcription and Translation Click-On Challenge

[qwiz style=”width: 650px !important; min-height: 450px !important;” use_dataset=”Transcription and Translation Click-on Challenge” quiz_timer=”true” random=”false” dataset_intro=”false” spaced_repetition=”false” qrecord_id=”sciencemusicvideosMeister1961-Unit 6 Transcription and Translation Click-on Challenge (v2.0)”]

[h] Transcription and Translation Click-On Challenge

[i] Notice the timer at the top right. Your goal is to build speed and accuracy. A good strategy: once through slowly, then additional trials to climb higher in the leaderboard.

[/qwiz]

8. Operons Click-on Challenge

[qwiz style=”width: 650px !important; min-height: 450px !important;” quiz_timer=”true” random=”false” dataset_intro=”false” spaced_repetition=”false” use_dataset=”Operons Click-On Challenge Dataset” qrecord_id=”sciencemusicvideosMeister1961-Unit 6 Operons Click-on Challenge (v2.0)”]

[h] Operons Click-On Challenge

[i] Note the timer in the top right. Your goal is accuracy and speed. A good strategy: once through slowly, then additional trials for improvement.

[/qwiz]

9. Eukaryotic Gene Regulation Click-on Challenge

[qwiz style=”width: 650px !important; min-height: 450px !important;” quiz_timer=”true” random=”true” spaced_repetition=”false” use_dataset=”Eukaryotic Gene Regulation Click-on Challenge Dataset” dataset_intro=”false” qrecord_id=”sciencemusicvideosMeister1961-Unit 6 Eukaryotic Gene Regulation Click-on Challenge”]

[h] Eukaryotic Gene Regulation Click-on Challenge

[i]

[x][restart]

[/qwiz]