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 strandminus 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

[qdeck style=”min-height: 450px width: 650px !important;” bold_text=”false” random=”false”]

[h]Unit 6 Cumulative Flashcards
[i]

[q json=”true” yy=”4″ unit=”6.Gene_Expression_and_Regulation” dataset_id=”AP_Bio_Flashcards_2022|1948920572110″ question_number=”206″ topic=”6.1-2.DNA_and_RNA,_DNA_Replication”] Compare and contrast how genetic information is stored in bacteria and eukaryotes.

Importance for the AP exam: High 

[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=”AP_Bio_Flashcards_2022|aef94afeae92c” question_number=”207″] Compare and contrast the functions of DNA and RNA

Importance for the AP exam: High 

[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=”AP_Bio_Flashcards_2022|1937d5effb510″ question_number=”208″ topic=”6.1-2.DNA_and_RNA,_DNA_Replication”] What are plasmids? What’s their function, and how are they used in genetic engineering?

Importance for the AP exam: High 

[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=”AP_Bio_Flashcards_2022|192ed041e7d10″ question_number=”209″ 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.

Importance for the AP exam: High 

[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=”AP_Bio_Flashcards_2022|aed4c48425d2c” question_number=”210″] Explain how DNA’s structure allows it to serve as the molecule of heredity.

Importance for the AP exam: High 

[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=”AP_Bio_Flashcards_2022|1925a55316110″ question_number=”211″ 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).

Importance for the AP exam: High 

[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=”AP_Bio_Flashcards_2022|191c9fa502910″ question_number=”212″ 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, primer.

Importance for the AP exam: High 

[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=”AP_Bio_Flashcards_2022|191374b630d10″ question_number=”213″ topic=”6.1-2.DNA_and_RNA,_DNA_Replication”] How is DNA replication at the leading strand different from replication at the lagging strand?

Importance for the AP exam: High 

[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=”AP_Bio_Flashcards_2022|190a49c75f110″ question_number=”214″ topic=”6.1-2.DNA_and_RNA,_DNA_Replication”] Describe the roles of ligase and topoisomerase in DNA replication.

Importance for the AP exam: High 

[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 reseal the break.

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

[q json=”true” yy=”4″ unit=”6.Gene_Expression_and_Regulation” dataset_id=”AP_Bio_Flashcards_2022|18fecacca9510″ question_number=”215″ topic=”6.3.Transcription_and_RNA_Processing”] List the three principal forms of RNA, and describe the function of each one.

Importance for the AP exam: High 

[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=”AP_Bio_Flashcards_2022|18f2470cbfd10″ question_number=”216″ topic=”6.3.Transcription_and_RNA_Processing”] Explain the overall flow of genetic information of cells.

Importance for the AP exam: High 

[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=”AP_Bio_Flashcards_2022|18e7ccd73dd10″ question_number=”217″ topic=”6.3.Transcription_and_RNA_Processing”] What happens during transcription?

Importance for the AP exam: High 

[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=”AP_Bio_Flashcards_2022|af34e1477fbd” question_number=”218″] How is transcription different in prokaryotes and eukaryotes?

Importance for the AP exam: High 

[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=”AP_Bio_Flashcards_2022|18df5c2c23510″ question_number=”219″ topic=”6.3.Transcription_and_RNA_Processing”] Define and describe “template strand,” “minus strand,” “noncoding strand,” or “antisense strand” in relationship to RNA transcription

Importance for the AP exam: High 

[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=”AP_Bio_Flashcards_2022|18d2436940d10″ question_number=”220″ 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.

Importance for the AP exam: High 

[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 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=”AP_Bio_Flashcards_2022|173d636aaee002″ question_number=”221″] What is the function of the 5′ GTP cap and the 3′ poly-A-tail that’s added to mRNA during eukaryotic RNA processing?

Importance for the AP exam: High 

[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=”AP_Bio_Flashcards_2022|18c962fbeb910″ question_number=”222″ topic=”6.3.Transcription_and_RNA_Processing”] Describe, on a big picture level, how transcription is regulated in eukaryotes.

Importance for the AP exam: High 

[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=”AP_Bio_Flashcards_2022|ac507a8247bd” question_number=”223″]Explain how organization of eukaryotic genetic material into introns and exons can increase phenotypic variation.

Importance for the AP exam: High 

[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=”AP_Bio_Flashcards_2022|19537f89990327″ question_number=”224″] Explain how free ribosomes (at 3)  become bound (at 4).

Importance for the AP exam: High 

[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=”AP_Bio_Flashcards_2022|18b2d4c8bad10″ question_number=”225″ topic=”6.4.Translation”] How is translation/protein synthesis different in prokaryotes and eukaryotes?

Importance for the AP exam: High 

[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=”AP_Bio_Flashcards_2022|18a059aa59110″ question_number=”226″ topic=”6.4.Translation”] What is the first phase of translation called? Describe how the process begins.

Importance for the AP exam: High 

[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=”AP_Bio_Flashcards_2022|189753fc45910″ question_number=”227″ topic=”6.4.Translation”] Describe the elongation phase of protein synthesis.

Importance for the AP exam: High 

[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=”AP_Bio_Flashcards_2022|188e290d73d10″ question_number=”228″ topic=”6.4.Translation”] Describe the termination phase of protein synthesis.

Importance for the AP exam: High 

[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=”AP_Bio_Flashcards_2022|1885235f60510″ question_number=”229″ 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.

Importance for the AP exam: High 

[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=”AP_Bio_Flashcards_2022|187bf8708e910″ question_number=”230″ topic=”6.4.Translation”] What is the central dogma of molecular biology?

Importance for the AP exam: High 

[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=”AP_Bio_Flashcards_2022|186e6feb71510″ question_number=”231″ topic=”6.4.Translation”] In the context of the central dogma of molecular biology, why are retroviruses so unique?

Importance for the AP exam: High 

[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=”AP_Bio_Flashcards_2022|18658f7e1c110″ question_number=”232″ topic=”6.5.Regulation_of_Gene_Expression”] What is a promoter?

Importance for the AP exam: High  

[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=”AP_Bio_Flashcards_2022|18658f7e1c110″ question_number=”233″ 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=”AP_Bio_Flashcards_2022|185bf4cd0f910″ question_number=”234″ topic=”6.5.Regulation_of_Gene_Expression”] What’s an operon?

Importance for the AP exam: High 

[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=”AP_Bio_Flashcards_2022|1853a962b3510″ question_number=”235″ 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)

Importance for the AP exam: High 

[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=”AP_Bio_Flashcards_2022|1843f21270510″ question_number=”236″ topic=”6.5.Regulation_of_Gene_Expression”] Explain how the lac operon works when lactose is absent from E. coli’s environment.

Importance for the AP exam: High 

[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=”AP_Bio_Flashcards_2022|a0586d636189″ question_number=”237″] Explain how the lac operon works when lactose is present in E. coli’s environment.

Importance for the AP exam: High 

[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=”AP_Bio_Flashcards_2022|183ac7239e910″ question_number=”238″ 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.

Importance for the AP exam: High 

[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=”AP_Bio_Flashcards_2022|9eb300fb2d89″ question_number=”239″] Explain how the trp operon works when the amino acid tryptophan is available in E. coli’s environment.

Importance for the AP exam: High 

[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=”AP_Bio_Flashcards_2022|9dabe7bb8989″ question_number=”240″]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=”AP_Bio_Flashcards_2022|18319c34ccd10″ question_number=”241″ topic=”6.5.Regulation_of_Gene_Expression”] Using RNA interference as an example, describe the role of small RNA molecules in regulating gene expression.

Importance for the AP exam: Medium 

[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=”AP_Bio_Flashcards_2022|18287145fb110″ question_number=”242″ topic=”6.5.Regulation_of_Gene_Expression”] On a big picture level, define and describe “epigenetics.”

Importance for the AP exam: High 

[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=”AP_Bio_Flashcards_2022|181efbd5acd10″ question_number=”243″ topic=”6.5.Regulation_of_Gene_Expression”] Explain how histone acetylation and DNA methylation can modify DNA expression.

Importance for the AP exam: High 

[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=”AP_Bio_Flashcards_2022|181efbd5acd10″ question_number=”244″ topic=”6.5.Regulation_of_Gene_Expression”] Explain how DNA methylation and acetylation can modify DNA expression.

Importance for the AP exam: High 

[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=”AP_Bio_Flashcards_2022|1814a6e0e9110″ question_number=”245″ topic=”6.5.Regulation_of_Gene_Expression”] Epigenetic modification of DNA is a key process during development. Explain.

Importance for the AP exam: High 

[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=”AP_Bio_Flashcards_2022|16e005051bc040″ question_number=”246″] 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=”AP_Bio_Flashcards_2022|180b7bf217510″ question_number=”247″ 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.

Importance for the AP exam: Medium

[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=”AP_Bio_Flashcards_2022|17f6f7494e110″ question_number=”248″ topic=”6.6.Gene_Expression_and_Cell_Specialization”] What’s the connection between gene expression and phenotypic differences between cells?

Importance for the AP exam: High 

[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=”AP_Bio_Flashcards_2022|17edcc5a7c510″ question_number=”249″ topic=”6.6.Gene_Expression_and_Cell_Specialization”]Explain what induction is and how it works.

Importance for the AP exam: High 

[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=”AP_Bio_Flashcards_2022|17ca4aa527510″ question_number=”250″ topic=”6.6.Gene_Expression_and_Cell_Specialization”] Embryonic development involves sequential gene expression. Explain.

Importance for the AP exam: Medium 

[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=”AP_Bio_Flashcards_2022|17bf3b6cac510″ question_number=”251″ topic=”6.7.Mutation”] 1. Define mutation. 2. Define “point mutation” and describe the three effects of point mutations.

Importance for the AP exam: High 

[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=”AP_Bio_Flashcards_2022|128a585a04d414″ question_number=”252″] Define “silent mutation,” “missense mutation,” and “nonsense mutation.”

Importance for the AP exam: High 

[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=”AP_Bio_Flashcards_2022|17af145a2e910″ question_number=”253″ 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

Importance for the AP exam: High 

[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=”AP_Bio_Flashcards_2022|17a5e96b5cd10″ question_number=”254″ topic=”6.7.Mutation”] What is a frameshift mutation?

Importance for the AP exam: High 

[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=”AP_Bio_Flashcards_2022|179c993bccd10″ question_number=”255″ topic=”6.7.Mutation”] Mutations can be positive, negative, or neutral. Explain.

Importance for the AP exam: High 

[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=”AP_Bio_Flashcards_2022|1287a72c453414″ question_number=”256″] How are mutations important to evolution?

Importance for the AP exam: High 

[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=”AP_Bio_Flashcards_2022|1286cc4fe77c14″ question_number=”257″] How do mutations arise?

Importance for the AP exam: High 

[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=”AP_Bio_Flashcards_2022|1789ae5b30510″ question_number=”258″ topic=”6.7.Mutation”] What is polyploidy?

Importance for the AP exam: Medium 

[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=”AP_Bio_Flashcards_2022|177d4fdc05110″ question_number=”259″ topic=”6.7.Mutation”] Contrast “horizontal gene transfer” with “vertical gene transfer.”

Importance for the AP exam: High 

[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=”AP_Bio_Flashcards_2022|1773ffac75110″ question_number=”260″ topic=”6.7.Mutation”] Describe bacterial conjugation.

Importance for the AP exam: High 

[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=”AP_Bio_Flashcards_2022|128360de709c14″ question_number=”261″] Describe bacterial transformation.

Importance for the AP exam: High 

[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=”AP_Bio_Flashcards_2022|176aaf7ce5110″ question_number=”262″ topic=”6.7.Mutation”] Describe how horizontal gene transfer can occur through viral transduction.

Importance for the AP exam: High 

[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=”AP_Bio_Flashcards_2022|1761a9ced1910″ question_number=”263″ topic=”6.7.Mutation”] Describe how horizontal gene transfer can occur through endosymbiosis.

Importance for the AP exam: Medium 

[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=”AP_Bio_Flashcards_2022|1758345e83510″ question_number=”264″ topic=”6.7.Mutation”] 1. Describe the basic niche of viruses. 2. Describe the size and structure of viruses.

Importance for the AP exam: High 

[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=”AP_Bio_Flashcards_2022|174cdaa48bd10″ question_number=”265″ topic=”6.7.Mutation”] Describe the lytic cycle of a DNA virus.

Importance for the AP exam: Medium 

[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=”AP_Bio_Flashcards_2022|17433ff37f510″ question_number=”266″ topic=”6.7.Mutation”] Describe the lysogenic cycle.

Importance for the AP exam: Medium 

[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=”AP_Bio_Flashcards_2022|17364c715b110″ question_number=”267″ topic=”6.7.Mutation”] Explain how viral life cycles can generate genetic and phenotypic diversity within their own viral populations.

Importance for the AP exam: High 

[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=”AP_Bio_Flashcards_2022|172809a886910″ question_number=”268″ 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

Importance for the AP exam: High

[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=”AP_Bio_Flashcards_2022|171bd06a19910″ question_number=”269″ 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.

Importance for the AP exam: High 

[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=”AP_Bio_Flashcards_2022|170e926678d10″ question_number=”270″ topic=”6.8.Biotechnology”] What is gel electrophoresis? How is it used to analyze DNA?

Importance for the AP exam: High 

[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=”AP_Bio_Flashcards_2022|170e926678d10″ question_number=”271″ topic=”6.8.Biotechnology”] How does gel electrophoresis work?

Importance for the AP exam: High 

[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=”AP_Bio_Flashcards_2022|1703a86ebc110″ question_number=”272″ 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.

Importance for the AP exam: Medium 

[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=”AP_Bio_Flashcards_2022|16f95379f8510″ question_number=”273″ topic=”6.8.Biotechnology”] What is PCR? What is it used for, and how does it work?

Importance for the AP exam: High 

[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 (not shown), 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=”AP_Bio_Flashcards_2022|16f0034a68510″ question_number=”274″ topic=”6.8.Biotechnology”] What is DNA sequencing?

Illustrative example: Genetic Engineering.

Importance for the AP exam: High 

[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. DNA and DNA Replication Click On Challenge

[qwiz quiz_timer=”true” random=”false” dataset_intro=”false” spaced_repetition=”false” use_dataset=”DNA and DNA Replication Click On Challenge Dataset”]

[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]

4. Transcription and Translation Click-On Challenge

[qwiz use_dataset=”Transcription and Translation Click-on Challenge” quiz_timer=”true” random=”false” dataset_intro=”false” spaced_repetition=”false”]

[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]

5. Operons Click-on Challenge

[qwiz quiz_timer=”true” random=”false” dataset_intro=”false” spaced_repetition=”false” use_dataset=”Operons Click-On Challenge Dataset”]

[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]

6. Unit 6 Practice FRQs

[qwiz][h]Unit 6 Practice FRQs
[q json=”true” xx=”1″ multiple_choice=”false” unit=”6.Gene Expression and Regulation” dataset_id=”AP Bio FRQ Dataset 2022|11094d744407″ question_number=”45″ topic=”6.1.DNA_Structure_and_Replication”] The diagram below is a representation of a 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. Identify which of the diagrams below correctly shows the daughter DNA molecules that would result. Justify your answer.

[c]IFNob3cgdGhl IGFuc3dlcg==[Qq]

[f]IElERU5USUZJQ0FUSU9OOiBUaGUgY29ycmVjdCBtb2RlbCBpcyBELg==

Cg==

SlVTVElGSUNBVElPTg==[Qq]: Model D represents the results of semi-conservative DNA replication, which is the model that Meselson and Stahl established in 1958 (Note: this historical information is not required). During DNA replication, a DNA molecule will “unzip” and free nucleotides will move in and pair up according to the base pairing rules (A bonds with T; and C bonds with G). In the scenario described above, the DNA molecule that’s unzipping will consist of 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 be consist of 14N and the paired complementary strand will consist of 15N.

[q json=”true” xx=”1″ multiple_choice=”false” unit=”6.Gene Expression and Regulation” dataset_id=”AP Bio FRQ Dataset 2022|b03f6b4ac07″ question_number=”58″ topic=”6.1.DNA_Structure_and_Replication”] The diagram below shows a short section of a DNA molecule.

Explain the semiconservative model of DNA replication. Add a simple sketch to accompany your explanation.

[c]IFNob3cgdGhl IGFuc3dlcg==[Qq]

[f]IFRoZSBzZW1pY29uc2VydmF0aXZlIG1vZGVsIG9mIEROQSByZXBsaWNhdGlvbiBzdGF0ZXMgdGhhdCB3aGVuIEROQSBpcyByZXBsaWNhdGVkLCBlYWNoIHN0cmFuZCBzZXJ2ZXMgYXMgYSB0ZW1wbGF0ZSBmb3Igc3ludGhlc2lzIG9mIGEgbmV3LCBjb21wbGVtZW50YXJ5IHN0cmFuZC4gQXMgYSByZXN1bHQsIGVhY2ggZGF1Z2h0ZXIgRE5BIG1vbGVjdWxlIHdpbGwgY29uc2lzdCBvZiBvcmlnaW5hbCBzdHJhbmQsIGFuZCBvbmUgbmV3bHkgc3ludGhlc2l6ZWQgc3RyYW5kLg==[Qq]

[q json=”true” xx=”1″ multiple_choice=”false” unit=”6.Gene Expression and Regulation” dataset_id=”AP Bio FRQ Dataset 2022|a7a97f72007″ question_number=”59″ topic=”6.1.DNA_Structure_and_Replication”] The diagram below depicts semi-conservative replication of DNA.

Explain how the Meselson-Stahl experiment proved the semiconservative model to be correct.

[c]IFNob3cgdGhl IGFuc3dlcg==[Qq]

[f]IFRoZSBzZW1pY29uc2VydmF0aXZlIG1vZGVsIG9mIEROQSByZXBsaWNhdGlvbiBzdGF0ZXMgdGhhdCB3aGVuIEROQSBpcyByZXBsaWNhdGVkLCBlYWNoIHN0cmFuZCBzZXJ2ZXMgYXMgYSB0ZW1wbGF0ZSBmb3Igc3ludGhlc2lzIG9mIGEgbmV3LCBjb21wbGVtZW50YXJ5IHN0cmFuZC4gQXMgYSByZXN1bHQsIGVhY2ggZGF1Z2h0ZXIgRE5BIG1vbGVjdWxlIHdpbGwgY29uc2lzdCBvZiB0aGUgb3JpZ2luYWwgc3RyYW5kLCBhbmQgb25lIG5ld2x5IHN5bnRoZXNpemVkIHN0cmFuZC4=

Cg==

Cg==

[Qq]Meselson and Stahl were able to prove the semiconservative model to be correct by using isotopes of nitrogen to label original DNA and newly synthesized DNA in replicating E. coli cells.

They started by growing a culture of E. coli on N15, a heavy isotope of nitrogen so that all of the nitrogen in this population of E. coli’s DNA was composed of this heavier isotope. Then they transferred this E. coli to a medium containing N14, a lighter isotope of nitrogen (and the kind that is much more commonly found in nature). Knowing that E. coli cells can replicate themselves in 20 minutes, they allowed the E. coli to replicate itself once, and then used centrifugation to determine the weight of the nitrogen in the E. coli’s DNA. They found that the DNA’s weight was halfway between N14 and N15. This established that the new DNA was composed of 50% N14 and 50%, N15, which proved an alternative theory (the conservative model of DNA replication) to be incorrect. That’s because if the conservative model were correct, then there would still be some DNA that was all composed of N15 (DNA that was conserved), with other DNA being composed exclusively of N14 (the newly synthesized DNA). The conservative model is represented by image A below, which shows two daughter double helices.

Meselson and Stahl then let the bacteria replicate for a second generation and again sampled the DNA. They found that in the second generation, there continued to be DNA whose weight was halfway between N14 and N15, and new DNA that was completely N14. This proved another model, the dispersive model to be incorrect), and corroborated what one would predict if DNA were semi-conservatively replicated.

[q json=”true” xx=”1″ multiple_choice=”false” unit=”6.Gene Expression and Regulation” dataset_id=”AP Bio FRQ Dataset 2022|96a2e87ec07″ question_number=”61″ topic=”6.1.DNA_Structure_and_Replication”] During their work in determining the structure of DNA, Watson and Crick examined published results from three different laboratories. Each laboratory was reporting on the proportion of nucleotides in skin cells from the same organism.

Which laboratory’s results did Watson and Crick disregard? Explain why they had to disregard the results, and speculate about what might have occured in the lab with the flawed data that caused them to have incorrect results.

[c]IFNob3cgdGhl IGFuc3dlcg==[Qq]

[f]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[Qq]

[q json=”true” xx=”1″ multiple_choice=”false” unit=”6.Gene Expression and Regulation” dataset_id=”AP Bio FRQ Dataset 2022|8551d00f007″ question_number=”63″ topic=”6.1.DNA_Structure_and_Replication”] The diagram below shows the amount of DNA per cell in a dividing cell.

PART 1: Identify what kind of cell division must be occurring.

PART 2: Justify your answer.

PART 3: Explain what’s happening in each numbered stage.

[c]IFNob3cgdGhl IGFuc3dlci4=[Qq]

[f]IFBBUlQgMTo=IFRoZSBkaWFncmFtIHNob3dzIGEgY2VsbCB1bmRlcmdvaW5nIG1laW9zaXMu

Cg==

UEFSVCAyOg==[Qq] It has to be meiosis because the cell starts with a two units of DNA, and ends with 1 unit. The type of cell division that reduces the amount of DNA/cell is meiosis.

PART 3: At I we have a germ cell. II is interphase 1, which doubles the amount of DNA. In III, prophase 1 through telophase 1 are occurring, leading to the first cell division, cytokinesis 1, between III and IV. Meiosis 1 pulls homologous chromosomes apart, Now, in IV, all the steps of meiosis 2 would occur, separating the sister chromatids. In V we have cytokinesis 2, resulting in the haploid gametes in step VI.

[q json=”true” xx=”1″ multiple_choice=”false” unit=”6.Gene Expression and Regulation” dataset_id=”AP Bio FRQ Dataset 2022|212e0be62007″ question_number=”26″ topic=”6.3.Transcription_and_RNA_Processing”] You’re interviewing for a summer internship at a local research lab. During the interview, your graduate student mentor tells you that the lab has just discovered a new eukaryotic protein that’s 197 amino acids long, and that the two of you will be working on finding the gene for this protein. To check your knowledge of biology, the grad student asks, “About how big is this gene going to be?” Provide an answer, and a justification.

[c]IFNob3cgdGhl IGFuc3dlcg==[Qq]

[f]IE5PVEUgRlJPTSBNUi4gVzogdGhlIHRoaXJkIHBhcmFncmFwaCBiZWxvdyBpcyB5b3VyIG1vZGVsIGFuc3dlci4=

Cg==

[Qq]First, you think “The protein is 197 amino acids long. It takes 3 nucleotides to code for one amino acid, and 197 x 3 = 591.” But then you pause, and think some more. You know that during protein synthesis, the first amino acid is methionine, which is coded for by the start codon AUG. However, this first methionine is often snipped off. So you’ll want to add three nucleotides to your total.

But don’t just blurt out “594!” Think about the end of protein synthesis, when the stop codon signals for a release factor. The stop codon is another three nucleotides, so you have to add three more nucleotides, which brings you to 597. But you don’t say that, either. You remember that this is a eukaryotic gene, and might be consist of exons and introns that are excised before translation. So, your final answer is this:

“The minimum number of nucleotides that could code for a 197 amino acid polypeptide would be 597. That’s 591 nucleotides to code for the 197 amino acids, plus three for the start codon and three for the stop codon. But because this is a eukaryotic gene, it might contain introns, and we can’t tell, just from the number of amino acids in the protein, how big it is.”

With a smile, you add, “I’m really looking forward, however, to working with you to figure this out.”

[q json=”true” xx=”1″ multiple_choice=”false” unit=”6.Gene Expression and Regulation” dataset_id=”AP Bio FRQ Dataset 2022|26a7afdc4807″ question_number=”27″ 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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[Qq]

[q json=”true” xx=”1″ multiple_choice=”false” unit=”6.Gene Expression and Regulation” dataset_id=”AP Bio FRQ Dataset 2022|fb0b6947407″ question_number=”48″ topic=”6.4.Translation”] The following image shows a sequence of nucleic acid bases coding for a sequence of amino acids. The third amino acid in the sequence (lysine) has been filled in.

List the remaining amino acids in the sequence. Here’s a genetic code chart to use for reference.

[c]IFNob3cgdGhl IGFuc3dlcg==[Qq]

[f]IG1ldGhpb25pbmUsIGFyZ2luaW5lLCBbbHlzaW5lXSB0aHJlb25pbmUsIHRyeXB0b3BoYW4=[Qq]

[q json=”true” xx=”1″ multiple_choice=”false” unit=”6.Gene Expression and Regulation” dataset_id=”AP Bio FRQ Dataset 2022|237ff6ae8a2006″ question_number=”93″ topic=”6.4.Translation”] The following questions are based on the figure below.

PART 1: Identify whether the figure depicts a eukaryotic cell or a prokaryotic cell. Support your response with evidence from the diagram.

PART 2: Assume that letter B represents a protein, such as insulin, which is exported from cells in the pancreas. Beginning with information in a gene, describe the flow of information and matter that results in the secreted protein at B.

[c]IFNob3cgdGhl IGFuc3dlcg==[Qq]

[f]IFBBUlQgMTogVGhlIGNlbGwgaXMgZXVrYXJ5b3RpYy4gVGhlIGZpcnN0IHBpZWNlIG9mIGV2aWRlbmNlIGlzIHRoZSBwcmVzZW5jZSBvZiBhIG51Y2xldXMuIEluIGFkZGl0aW9uLCBvbmx5IGV1a2FyeW90aWMgY2VsbHMgaGF2ZSBvcmdhbmVsbGVzIHN1Y2ggYXMgbWl0b2Nob25kcmlhIChub3QgbGFiZWxlZCwgYnV0IHR3byBhcmUgdmlzaWJsZSBpbiB0aGUgdG9wIGhhbGYgb2YgdGhlIGNlbGwpLCBhbmQgYW4gZW5kb21lbWJyYW5lIHN5c3RlbSAoaW5jbHVkaW5nIHRoZSByb3VnaCBFUiBhdCBBLCB2ZXNpY2xlcyBhdCBDLCBhbmQgdGhlIEdvbGdpIEFwcGFyYXR1cyBhdCBEKS4=

Cg==

UEFSVCAy[Qq]: The information for a protein like insulin would be encoded in the cell’s DNA. The DNA would be transcribed to RNA, which would be translated into protein. Because insulin is a protein destined for export, it would contain a signal peptide that would bind with a signal recognition particle, which would diffuse toward and bind with the ER membrane. The insulin polypeptide would be synthesized inside the rough ER, and then move to the Golgi for modification. A vesicle would move the insulin to the membrane, and exocytosis would release insulin out of the cell.

[q json=”true” xx=”1″ multiple_choice=”false” unit=”6.Gene Expression and Regulation” dataset_id=”AP Bio FRQ Dataset 2022|1e967aa94c07″ question_number=”28″ topic=”6.5.Regulation_of_Gene_Expression”] Jacques Monod, who co-developed the operon concept (and – I can’t resisting adding – was also leader in the French Resistance to the Nazis during World War II), 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.

[c]IFNob3cgdGhl IGFuc3dlcg==[Qq]

[f]IChOb3RlIGZyb20gTXIuIFc6IEkmIzgyMTc7dmUgaW5jbHVkZWQgZGlhZ3JhbXMgZm9yIHRoZSBwdXJwb3NlcyBvZiByZXZpZXcsIGJ1dCBhIGdvb2QgdGV4dCBleHBsYW5hdGlvbiBpcyBmaW5lKQ==

Cg==

V2hhdCBNb25vZCAoYW5kIGhpcyBjby13b3JrZXIgRnJhbmNvaXMgSmFjb2IpIGRpc2NvdmVyZWQgd2FzIHRoZSA=bGFjIG9wZXJvbi4gQW4gb3Blcm9uIGNvbnNpc3RzIG9mIGEgc2VyaWVzIG9mIHN0cnVjdHVyYWwgZ2VuZXMgKGdlbmVzIHRoYXQgY29kZSBmb3IgZW56eW1lcyBvciBzb21lIG90aGVyIHByb2R1Y3QpIHVuZGVyIHRoZSBjb250cm9sIG9mIGEgc2luZ2xlIHByb21vdGVyLg==

[Qq]

Imagine the promoter (5) as being upstream of the structural genes (7): in other words, 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 structural 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 regulator’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 production of lactose-digesting enzymes.

[q json=”true” xx=”1″ multiple_choice=”false” unit=”6.Gene Expression and Regulation” dataset_id=”AP Bio FRQ Dataset 2022|1a184bb75407″ question_number=”37″ topic=”6.5.Regulation_of_Gene_Expression”] Bacteria require amino acids to produce proteins. For example, bacteria in a human intestine may absorb amino acids from digested food, but at times there may be a deficiency of a particular amino acid. If this is the case, the bacteria will produce the necessary amino acid themselves.

The diagram below depicts a system in bacterial cells that regulates the production of the amino acid tryptophan. This system involves two pathways (X and Y). In both pathways, tryptophan acts as a repressor.

PART 1: Describe the immediate outcome when tryptophan activates pathway X.
PART 2: In the 1950s, biologists Francois Jacob and Jacques Monod developed the concept of the operon to explain how bacterial cells were able to regulate certain biochemical pathways. Explain the operation of the trp (tryptophan) operon.

[c]IFNob3cgdGhl IGFuc3dlcg==[Qq]

[f]IFBBUlQgMTo=IFBhdGh3YXkgWCB3b3VsZCByZXN1bHQgaW4gdGhlIHN1cHByZXNzaW9uIG9mIGdlbmUgdGhyZWUsIHdoaWNoIGluIHR1cm4gd291bGQgYmxvY2sgcHJvZHVjdGlvbiBvZiBlbnp5bWUgMywgcmVzdWx0aW5nIGluIGNlc3NhdGlvbiBvZiB0cnlwdG9waGFuIHN5bnRoZXNpcy4=
Cg==UEFSVCAyOg==IFRoZSA=[Qq]trp operon is a repressible operon, and it works as shown below. (Note from Mr. W: I’ve included diagrams to assist your learning, but they weren’t required in your answer.)

The trp regulatory gene creates a repressor protein (3, below). The repressor protein has two binding sites: one for tryptophan (8), and one for the operator region of the trp operon (6).

When tryptophan is present in the cell’s environment, it binds to the repressor protein, which changes shape so that it can bind to the operator region of the operon. Because the operator is just downstream of the promoter region of the operon (5), the binding of the repressor to the operator blocks RNA polymerase (4), keeping it from transcribing the structural genes (7) that synthesize tryptophan.

However, when tryptophan is absent from the cell’s environment, the repressor protein (3, below) can’t bind with the operator, allowing RNA polymerase to bind with the promoter (5) and transcribe the structural genes (which results in production of the enzymes that synthesize tryptophan (A-E).

[q json=”true” xx=”1″ multiple_choice=”false” unit=”6.Gene Expression and Regulation” dataset_id=”AP Bio FRQ Dataset 2022|1c8f444db007″ question_number=”29″ topic=”6.6.Gene_Expression_and_Cell_Specialization”] The image below is a schematic diagram of a signal transduction pathway within a single cell. Arrows represent activation, while a “T” (which may be upside down: see the one between JNI and Bcl2 below) represents inhibition. The overall result of this pathway is apoptosis (programmed cell death).

Some of the key events that occur within this pathway are as follows: 1) Tumor necrosis factor (TNF) binds to and activates its specific receptor (tumor cell necrosis factor receptor, or TNFR). 2) The activated TNFR activates JNK. 3) The activated JNK activates caspases by causing mitochondria to release cytochrome C, and also by inhibiting BCl-2. BCl-2, when active, promotes cell survival by inhibiting caspases. 4) The active caspases result in DNA damage and cell death.

PART 1: Several elements of the system above have been implicated in the development of cancer. Some of these are loss of function mutations, where a previously functioning component of a cell stops working. Based on the signaling pathway shown, describe the effect of a loss of function mutation of each of the following upon apoptosis (compared with wild-type cells lacking this mutation).

* TNFR * BCl2 * Caspases

PART 2: Gain of function mutations are mutations where a cell component gains a capability that it previously lacked. Based on the schematic above, identify the component(s) of this signaling pathway that might promote cancer cell growth if they had a gain-of-function mutation that increased that component’s activity.

PART 3: Tumor suppressor genes inhibit cell division. List the component(s) of the system above that may be the product of tumor suppressor genes and that would promote cancer cell growth if they had a loss-of-function mutation.

[c]IFNob3cgdGhl IGFuc3dlcg==[Qq]

[f]IFBBUlQgMWE6IA==SWYgVE5GUiBsb3N0IGl0cyBhY3Rpdml0eSwgdGhlIGNlbGwgd291bGQgcmVzcG9uZCBsZXNzIHRvIHRoZSB0dW1vciBuZWNyb3NpcyBmYWN0b3Igc2lnbmFsLiBDZWxsIGRlYXRoIHdvdWxkIERFQ1JFQVNFIGJlY2F1c2UgdGhlIA==Sk4gSyB0byBtaXRvY2hvbmRyaWEgdG8gY3l0b2Nocm9tZSBDIHRvIGNhc3Bhc2VzIHRvIEROQSBkYW1hZ2UgdG8gY2VsbCBkZWF0aA==IHBhdGh3YXkgd291bGQgYmUgYmxvY2tlZC4g[Qq]

PART 1b: If BCl-2 lost its activity, cell death would INCREASE because the inhibitory effect of Bcl2 would be removed, allowing caspases to cause nuclear damage, leading to cell death.

PART 1c: If caspases lost their function then cell death would DECREASE because caspases cause cell death.

PART 2: If a mutation caused BCl-2 to increase its activity (gain of function mutation), it would block caspases, which would decrease cell death/apoptosis. Less cell death means higher survival rates for cancer cells, increasing cancer cell growth.

PART 3: TNFR, JNK, Caspases, and Cytochrome C released from mitochondria are all associated with apoptosis/cell death, which would decrease the spread of cancer cells. Loss of function of any of these could be associated with promoting cancer cell growth.

[q json=”true” xx=”1″ multiple_choice=”false” unit=”6.Gene Expression and Regulation” dataset_id=”AP Bio FRQ Dataset 2022|1219b6e37807″ question_number=”43″ topic=”6.6.Gene_Expression_and_Cell_Specialization”] Proto-oncogenes, when they mutate to become oncogenes, can cause cells to become cancerous.

A lysogenic virus infects a particular cell type and integrates its genome into a site that contains a proto-oncogene. This transforms the cell and increases the level of a protein X. Protein X, in turn, increases cellular proliferation (division and spread).

Tumor suppressor genes code for proteins that prevent cells from becoming cancerous. A compound P is known to increase the level of tumor suppressor proteins in that cell type. A second compound, Q helps in stimulating protein Z. Protein Z has been shown to be capable of binding to X rendering it inactive.

In the graphs below, a minus sign indicates that substance P or Q was not applied to experimental cell cultures, while a plus sign indicates that substance P or Q was applied. Identify which of the following graphs correctly represents the mode of action of P and Q. Justify your response.

[c]IFNob3cgdGhl IGFuc3dlcg==[Qq]

[f]IElERU5USUZJQ0FUSU9OOiBHcmFwaCBDIGlzIGNvcnJlY3Qu

Cg==

SlVTVElGSUNBVElPTjogUCBpbmNyZWFzZXMgdHVtb3Igc3VwcHJlc3NvciBwcm90ZWlucywgd2hpY2ggd291bGQgZGVjcmVhc2UgY2VsbCBwcm9saWZlcmF0aW9uLiA=UQ==IGluZGlyZWN0bHkgaW5hY3RpdmF0ZXMg[Qq]X. Because X increases cell proliferation, inactivating X (which is what substance Q does) should decrease cellular proliferation. Therefore a combination of P and Q should lead to the lowest amount of cellular proliferation, which is shown in the last bar of graph C. P and Q used separately should also decrease cell proliferation, which is what is shown in the second and third bars of graph C. Absence of both substances would lead to an increase in cellular proliferation, which is what is shown in the first bar of graph C.

[q json=”true” xx=”1″ multiple_choice=”false” unit=”6.Gene Expression and Regulation” dataset_id=”AP Bio FRQ Dataset 2022|14d688de8c07″ question_number=”40″ topic=”6.6.Gene_Expression_and_Cell_Specialization”] The bos/seven receptor, shown below, is a receptor tyrosine kinase. This receptor is required for the differentiation of a particular cell, called R7. Like all receptor tyrosine kinases, the protein is inactive as a monomer. Binding of ligand causes the monomeric form of the receptor to form a dimer (shown on the right). Dimerization causes phosphorylation of the receptor’s intracellular domain, activating the protein.
During protein synthesis and processing of the receptor, the extracellular domain is cleaved and a disulfide bridge forms between two cysteine amino acid residues. The disulfide bridge tethers the ligand-binding domain to the rest of the protein.

PART 1: Predict the effect of a mutation in the DNA for the receptor that resulted in changing one of the cysteine amino acid residues into an alanine. Justify your prediction.

PART 2: How would this mutation affect differentiation in cell type R7?

[c]IFNob3cgdGhl IGFuc3dlcg==[Qq]

[f]IFBBUlQgMTo=IEEgbXV0YXRpb24gdGhhdCBjaGFuZ2VzIGEgY3lzdGVpbmUgdG8gYWxhbmluZSB3b3VsZCBlbGltaW5hdGUgdGhlIGRpc3VsZmlkZSBicmlkZ2UgYXR0YWNoaW5nIHRoZSBsaWdhbmQtYmluZGluZyBkb21haW4gdG8gdGhlIHJlc3Qgb2YgdGhlIHByb3RlaW4uIFRoaXMgd291bGQgcmVzdWx0IGluIGFuIGluYWN0aXZlIHJlY2VwdG9yIGJlY2F1c2UgdGhlIGxpZ2FuZCBiaW5kaW5nIGRvbWFpbiBjb3VsZCBubyBsb25nZXIgYXR0YWNoIHRvIHRoZSByZXN0IG9mIHRoZSByZWNlcHRvci4gVGhlIHJlY2VwdG9yIGNvdWxkIG5vIGxvbmdlciBiaW5kIHdpdGggaXRzIGxpZ2FuZCwgZGltZXJpemF0aW9uIHdvdWxkIG5vdCBvY2N1ciwgYW5kIHRoZSBjZWxsdWxhciBwcm9jZXNzZXMgaW5pdGlhdGVkIGJ5IGJpbmRpbmcgb2YgdGhlIGxpZ2FuZCB3b3VsZCBhbHNvIG5vIGxvbmdlciBvY2N1ci4=

Cg==

UEFSVCAyOg==[Qq] This would prevent the differentiation of the R7 cell type.

[q json=”true” xx=”1″ multiple_choice=”false” unit=”6.Gene Expression and Regulation” dataset_id=”AP Bio FRQ Dataset 2022|271c1a2ed007″ question_number=”30″ topic=”6.7.Mutation”] List four types of mutations and describe their effect upon the protein that the mutated DNA codes for.

[c]IFNob3cgdGhl IGFuc3dlcg==[Qq]

[f]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[Qq]

[q json=”true” xx=”1″ multiple_choice=”false” unit=”6.Gene Expression and Regulation” dataset_id=”AP Bio FRQ Dataset 2022|205583944c07″ question_number=”31″ topic=”6.7.Mutation”] Bacteria reproduce asexually (through binary fission). Yet bacterial species can show significant genetic diversity. List and describe 4 processes that generate genetic diversity within bacteria.

[c]IFNob3cgdGhl IGFuc3dlcg==[Qq]

[f]

Cg==

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

Cg==

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[Qq]

[q json=”true” xx=”1″ multiple_choice=”false” unit=”6.Gene Expression and Regulation” dataset_id=”AP Bio FRQ Dataset 2022|12a0c1952007″ question_number=”42″ 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 nucleotide 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” xx=”1″ multiple_choice=”false” unit=”6.Gene Expression and Regulation” dataset_id=”AP Bio FRQ Dataset 2022|9de98da7407″ question_number=”60″ topic=”6.7.Mutation”] Fungi have metabolic pathways that enable them to synthesize their own amino acids. This enables them to grow on a substance called minimal media (a gel that supports fungal growth, but which lacks any amino acids).

A mutant strain of fungus that could not grow on minimal media was discovered. This fungus was able to grow on minimal media to which all of the twenty amino acids were added. These observations are summarized in the diagram below.

In work that was done in the 1940s, George Beadle hypothesized that the abnormal fungus had a fault in a gene coding for the production of one amino acid. To determine which amino acid was involved, Beadle set up the following experiment involving 22 tubes.

Tube 1 had minimal media.
Tube 2 had minimal media, plus all 20 amino acids.
Tubes 3 to 22 had minimal media, plus one of the 20 amino acids, with each tube containing a different amino acid.

Spores from the mutant fungal strain were inoculated onto the media in each test tube

The tubes were incubated under the same conditions for several days and then examined.

Part A: Why were tubes 1 and 2 included in this experiment?

Part B: The amino acid histidine had been added to tube 12. What specific conclusion can be drawn about the mutant fungus based on this result.

[c]IFNob3cgdGhl IGFuc3dlcg==[Qq]
[f]IFBhcnQgQTogVHViZXMgMSBhbmQgMiBzZXJ2ZWQgYXMgY29udHJvbHMuIFR1YmUgMSBzaG93ZWQgdGhhdCB0aGUgbXV0YW50IGZ1bmd1cyBjb3VsZCBub3QgZ3JvdyBvbiBtaW5pbWFsIG1lZGlhLiBUdWJlIDIgc2hvd2VkIHRoYXQgdGhlIG11dGFudCBmdW5ndXMgd2FzIHZpYWJsZSB3aGVuIGdyb3duIG9uIGFsbCBhbWlubyBhY2lkcy4=

Cg==

UGFydCBCOiBUdWJlIDEyIHNob3dlZCB0aGF0IHRoZSBzaW5nbGUgbXV0YXRpb24gdGhhdCB0aGUgbXV0YW50IGZ1bmd1cyBsYWNrZWQgd2FzIGFuIGFiaWxpdHkgdG8gc3ludGhlc2l6ZSBoaXN0aWRpbmUu

Cg==[Qq]

[q json=”true” xx=”1″ multiple_choice=”false” unit=”6.Gene Expression and Regulation” dataset_id=”AP Bio FRQ Dataset 2022|237f5d03795806″ question_number=”94″ topic=”6.7.Mutation”] The diagram below illustrates the effect of a single point mutation in protein P. Protein P is normally an individual protein molecule found in the cytosol of eukaryotic cells. As a result of this mutation, protein P molecules will aggregate together forming chains. The binding that underlies this chain is represented by letters M and S below. The black half-circle S represents a small hydrophobic cavity on one side of protein P. The black ball at M represents a cluster of hydrophobic R groups on the other side of protein P,

Provide a real-life example of this type of mutation and its effects. Identify the human disease with which it is associated.

[c]IFNob3cgdGhl IGFuc3dlcg==[Qq]

[f]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

Cg==

SGVyZSYjODIxNztzIGFuIGltYWdlIHNob3dpbmcgaG93IHRoaXMgd29ya3M6

Cg==

[Qq]

 

[q json=”true” xx=”1″ multiple_choice=”false” unit=”6.Gene Expression and Regulation” dataset_id=”AP Bio FRQ Dataset 2022|24c80e4ad007″ question_number=”32″ topic=”6.8.Biotechnology”] In your new job at the Department of Interspecific Genetic Comparison, you’re told to use biochemical methods to compare two newly discovered bacterial species. What features might you compare?

[c]IFNob3cgdGhl IGFuc3dlcg==[Qq]

[f]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[Qq]

[q json=”true” xx=”1″ multiple_choice=”false” unit=”6.Gene Expression and Regulation” dataset_id=”AP Bio FRQ Dataset 2022|f40f459b407″ question_number=”49″ 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, and 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]IFBBUlQgMTo=IFRoZSBwdXJwb3NlIG9mIFcgaXMgdG8gc2hvdyB0aGF0IHRoZSA=RS4gY29saQ==IGJhY3RlcmlhIGFyZSBhbGl2ZSwgYW5kIHRoZSBwdXJwb3NlIG9mIFggaXMgdG8gZGVtb25zdHJhdGUgdGhhdCBiYWN0ZXJpYSB3aXRob3V0IHRoZSBwbGFzbWlkIGNhbiYjODIxNzt0IGdyb3cgb24gYSBwbGF0ZSB3aXRoIGFtcGljaWxsaW4u

Cg==

[Qq]PART 2: Any bacterium will grow on plate W. Because the nutrient agar has everything that E. coli needs, it grows everywhere it’s deposited (which is all over the plate). By contrast, only transformed bacteria which 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. While the details are somewhat more complicated, you can think of GFP as equivalent to the structural genes in an inducible operon, with arabinose as the inducer). Only plate Y has arabinose on the plate

PART 4: The bacteria on plate X are untransformed, 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]

7. Unit 6 Practice Multiple Choice Questions

[qwiz][h]Unit 6 Cumulative Multiple Choice Questions
[q json=”true” xx=”1″ multiple_choice=”true” unit=”6.Gene Expression and Regulation” dataset_id=”2019 AP Bio Dataset|22d737438c445a” question_number=”21″ 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]IE5vLiBBdCAmIzgyMjA7SUlJJiM4MjIxOyB0aGUgbGluZSBpcyBmbGF0LCBpbmRpY2F0aW5nIHRoYXQgdGhlIGFtb3VudCBvZiBETkEgaXMgc3RheWluZyBjb25zdGFudCBvdmVyIHRpbWUuIFRoZSBlbnp5bWVzIG1lbnRpb25lZCBhYm92ZSBhcmUgYWxsIGludm9sdmVkIGluIEROQSBzeW50aGVzaXMuIFN5bnRoZXNpcyA=aW5jcmVhc2VzIHRoZSBhbW91bnQgb2YgRE5BLiBXaGF0JiM4MjE3O3MgdGhlIG9ubHkgbnVtYmVyIHRoYXQgc2hvd3MgdGhlIGFtb3VudCBvZiBETkEgZ29pbmcgdXA/[Qq]

[f]IE5vLiBBdCAmIzgyMjA7SVYmIzgyMjE7IHRoZSBsaW5lIGlzIGZsYXQsIGluZGljYXRpbmcgdGhhdCB0aGUgYW1vdW50IG9mIEROQSBpcyBzdGF5aW5nIGNvbnN0YW50IG92ZXIgdGltZS4gVGhlIGVuenltZXMgbWVudGlvbmVkIGFib3ZlIGFyZSBhbGwgaW52b2x2ZWQgaW4gRE5BIHN5bnRoZXNpcy4gU3ludGhlc2lzIA==aW5jcmVhc2VzIHRoZSBhbW91bnQgb2YgRE5BLiBXaGF0JiM4MjE3O3MgdGhlIG9ubHkgbnVtYmVyIHRoYXQgc2hvd3MgdGhlIGFtb3VudCBvZiBETkEgZ29pbmcgdXA/[Qq]

[f]IE5vLiBBdCAmIzgyMjA7ViYjODIyMTsgdGhlIGxpbmUgaXMgc2xvcGluZyBkb3dud2FyZCwgaW5kaWNhdGluZyB0aGF0IHRoZSBhbW91bnQgb2YgRE5BIGlzIGRlY3JlYXNpbmcgb3ZlciB0aW1lLiBUaGUgZW56eW1lcyBtZW50aW9uZWQgYWJvdmUgYXJlIGFsbCBpbnZvbHZlZCBpbiBETkEgc3ludGhlc2lzLiBTeW50aGVzaXMgaW5jcmVhc2VzIHRoZSBhbW91bnQgb2YgRE5BLiBXaGF0JiM4MjE3O3MgdGhlIG9ubHkgbnVtYmVyIHRoYXQgc2hvd3MgdGhlIGFtb3VudCBvZiBETkEgZ29pbmcgdXA/[Qq]

[f]IE5vLiBBdCAmIzgyMjA7VkkmIzgyMjE7IHRoZSBsaW5lIGlzIGZsYXQsIGluZGljYXRpbmcgdGhhdCB0aGUgYW1vdW50IG9mIEROQSBpcyBzdGF5aW5nIGNvbnN0YW50IG92ZXIgdGltZS4gVGhlIGVuenltZXMgbWVudGlvbmVkIGFib3ZlIGFyZSBhbGwgaW52b2x2ZWQgaW4gRE5BIHN5bnRoZXNpcy4gU3ludGhlc2lzIA==aW5jcmVhc2VzIHRoZSBhbW91bnQgb2YgRE5BLiBXaGF0JiM4MjE3O3MgdGhlIG9ubHkgbnVtYmVyIHRoYXQgc2hvd3MgdGhlIGFtb3VudCBvZiBETkEgZ29pbmcgdXA/[Qq]

[q json=”true” xx=”1″ multiple_choice=”true” unit=”6.Gene Expression and Regulation” dataset_id=”2019 AP Bio Dataset|21f9e19932045a” question_number=”85″ 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=[Qq]

[q json=”true” xx=”1″ multiple_choice=”true” unit=”6.Gene Expression and Regulation” dataset_id=”2019 AP Bio Dataset|21f6f18e31585a” question_number=”86″ topic=”6.2.Replication”] The diagram below represents a sequence from two DNA molecules from the same species of bacteria. In the experiment’s protocol, the bacteria were first grown for several generations in a medium containing non-radioactive nitrogen. Subsequently, a fraction of the bacteria were transferred to a medium containing radioactive nitrogen.

One of the diagrams below correctly represents the DNA molecules that would be produced after the bacteria were transferred to radioactive nitrogen and then allowed to proceed through one replication cycle.

This model is called

[c]IGNvbnNlcnZhdGl2ZQ==[Qq]

[f]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[Qq]

[c]IGRpc3BlcnNpdmU=[Qq]

[f]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[Qq]

[c]IHNlbWktY29u c2VydmF0aXZl[Qq]

[f]IE5pY2Ugam9iISBETkEgcmVwbGljYXRpb24gaXMgYmVzdCBkZXNjcmliZWQgYXMgc2VtaS1jb25zZXJ2YXRpdmU=LCBhbmQgaXQmIzgyMTc7cyBzaG93biBieSBtb2RlbCBELiBIZXJlJiM4MjE3O3MgYW5vdGhlciByZXByZXNlbnRhdGlvbiBvZiBzZW1pLWNvbnNlcnZhdGl2ZSByZXBsaWNhdGlvbi4=

Cg==

[Qq]

[c]IGludGVyZHVwbGljYXRpdmU=[Qq]

[f]IE5vLiBXaGVuIEROQSByZXBsaWNhdGVzIGl0c2VsZiwgdGhlIEROQSAmIzgyMjA7dW56aXBzJiM4MjIxOyBhbmQgZWFjaCBzdHJhbmQgc3Vic2VxdWVudGx5IGFjdHMgYXMgYSB0ZW1wbGF0ZSBmb3IgdGhlIGZvcm1hdGlvbiBvZiBhIG5ldyBjb21wbGVtZW50YXJ5IHN0cmFuZC4gVGhlIHJlc3VsdGluZyBkYXVnaHRlciBzdHJhbmRzIHdpbGwgZWFjaCBiZSBoYWxmIG9sZCBhbmQgaGFsZiBuZXcgKHdpdGggb25lIHN0cmFuZCBpbiBlYWNoIG1vbGVjdWxlIGVudGlyZWx5IG9sZCwgYW5kIG9uZSBzdHJhbmQgZW50aXJlbHkgbmV3KS4gV2hhdCYjODIxNztzIGEgbG9naWNhbCBuYW1lIGZvciB0aGlzIChjb3JyZWN0KSBtb2RlbCBvZiByZXBsaWNhdGlvbj8=

Cg==

Cg==[Qq]

[q json=”true” xx=”1″ multiple_choice=”true” unit=”6.Gene Expression and Regulation” dataset_id=”2019 AP Bio Dataset|22fc576925cc5a” question_number=”8″ topic=”6.3.Transcription_and_RNA_Processing”] To answer this question, you can use this genetic code table:

The next amino acid that would result from the process below would be

[c]IGd1YW5pbmUu[Qq]

[f]IE5vLiBZb3UgY2FuIGZpbmQgdGhlIDNyZCBhbWlubyBhY2lkIGJ5IGlkZW50aWZ5aW5nIHRoZSBuZXh0IGNvZG9uLiBJdCYjODIxNztzIEFBRy4gV2hlbiB5b3UgbG9vayB1cCBBQUcgaW4gYSBnZW5ldGljIGNvZGUgdGFibGUgd2hhdCBhbWlubyBhY2lkIGRvIHlvdSBnZXQ/[Qq]

[c]IGx5c2 luZS4=[Qq]

[f]IEV4Y2VsbGVudC4gVGhlIG5leHQgY29kb24gaXMgQUFHLCB3aGljaCBpcyB0cmFuc2xhdGVkIGludG8gdGhlIGFtaW5vIGFjaWQgbHlzaW5lLg==[Qq]

[c]IGN5c3RlaW5lLg==[Qq]

[f]IE5vLiBZb3UgY2FuIGZpbmQgdGhlIG5leHQgYW1pbm8gYWNpZCBieSBpZGVudGlmeWluZyB0aGUgbmV4dCBjb2Rvbi4gSXQmIzgyMTc7cyBBQUcuIFdoZW4geW91IGxvb2sgdXAgQUFHIGluIGEgZ2VuZXRpYyBjb2RlIHRhYmxlLCB3aGF0IGFtaW5vIGFjaWQgZG8geW91IGdldD8=[Qq]

[c]IHByb2xpbmUu[Qq]

[f]IE5vLiBZb3UgY2FuIGZpbmQgdGhlIG5leHQgYW1pbm8gYWNpZCBieSBpZGVudGlmeWluZyB0aGUgbmV4dCBjb2Rvbi4gSXQmIzgyMTc7cyBBQUcuIFdoZW4geW91IGxvb2sgdXAgQUFHIGluIGEgZ2VuZXRpYyBjb2RlIHRhYmxlLCB3aGF0IGFtaW5vIGFjaWQgZG8geW91IGdldD8=[Qq]

[q json=”true” xx=”1″ multiple_choice=”true” topic=”6.3.Transcription_and_RNA_Processing” dataset_id=”2019 AP Bio Dataset|22f9b687b9685a” question_number=”9″ unit=”6.Gene Expression and Regulation”] Transcription converts

[c]IFJOQSBpbnRvIHByb3RlaW4u[Qq]

[f]IE5vLiA=VHJhbnNsYXRpb24=IGNvbnZlcnRzIGluZm9ybWF0aW9uIGluIFJOQSBpbnRvIHNlcXVlbmNlcyBvZiBhbWlubyBhY2lkcywgd2hpY2ggbWFrZSB1cCBwcm90ZWluLiBXaGF0IGRvZXMgdHJhbnNjcmlwdGlvbiBkbz8=[Qq]

[c]IHByb3RlaW4gaW50byBSTkEu[Qq]

[f]IE5vLiBUaGVyZSYjODIxNztzIG5vIGJpb2xvZ2ljYWwgcHJvY2VzcyB0aGF0IG1vdmVzIGZyb20gcHJvdGVpbiB0byBSTkEu[Qq]

[c]IEROQSBpbn RvIFJOQS4=[Qq]

[f]IEV4Y2VsbGVudC4gVHJhbnNjcmlwdGlvbiBjb252ZXJ0cyBETkEgaW50byBSTkEu[Qq]

[c]IGNvZG9uIHNlcXVlbmNlcyBpbnRvIGFtaW5vIGFjaWQgc2VxdWVuY2VzLg==[Qq]

[f]IE5vLiBDb2RvbiBzZXF1ZW5jZXMgYXJlIHRyYW5zbGF0ZWQgaW50byBhbWlubyBhY2lkIHNlcXVlbmNlcyBkdXJpbmcgdHJhbnNsYXRpb24uIFdoYXQgaGFwcGVucyBkdXJpbmcgdHJhbnNjcmlwdGlvbj8=[Qq]

[q json=”true” xx=”1″ multiple_choice=”true” topic=”6.3.Transcription_and_RNA_Processing” dataset_id=”2019 AP Bio Dataset|22f764cfe14c5a” question_number=”10″ unit=”6.Gene Expression and Regulation”] In eukaryotic cells, transcription moves information from ________ to _________.

[c]IHRoZSBudWNsZXVzICYjODIz MDsgdGhlIGN5dG9wbGFzbQ==[Qq]

[f]IE5pY2Ugam9iLiBJbiBldWthcnlvdGljIGNlbGxzLCB0cmFuc2NyaXB0aW9uIG1vdmVzIGluZm9ybWF0aW9uIGZyb20gRE5BIGluIHRoZSBudWNsZXVzIHRvIFJOQSwgd2hpY2ggZ29lcyBpbnRvIHRoZSBjeXRvcGxhc20u[Qq]

[c]IHJpYm9zb21lcyAmIzgyMzA7IHByb3RlaW5z[Qq]

[f]IE5vLiBUaGUgY29ubmVjdGlvbiBiZXR3ZWVuIHJpYm9zb21lcyBhbmQgcHJvdGVpbnMgaXMgbW9yZSByZWxhdGVkIHRvIHRyYW5zbGF0aW9uLiBJbiBhbnkgY2FzZSwgcmlib3NvbWVzIGRvbiYjODIxNzt0IGhhdmUgaW5mb3JtYXRpb24sIHRoZXkgdHJhbnNsYXRlIGl0LiBIZXJlJiM4MjE3O3MgYSBoaW50OiB0cmFuc2NyaXB0aW9uIGludm9sdmVzIEROQSwgd2hpY2ggbWFrZXMgdXAgY2hyb21vc29tZXMuIFdoZXJlIGFyZSB0aG9zZSBjaHJvbW9zb21lcyBmb3VuZD8=[Qq]

[c]IHRoZSBjeXRvcGxhc20gJiM4MjMwOy4gdGhlIG51Y2xldXM=[Qq]

[f]IE5vLiBEdXJpbmcgbWFueSBjZWxsIHNpZ25hbGluZyBwcm9jZXNzZXMsIGluZm9ybWF0aW9uIG1vdmVzIGZyb20gdGhlIGN5dG9wbGFzbSB0byB0aGUgbnVjbGV1cy4gQnV0IG5vdCBpbiB0cmFuc2NyaXB0aW9uLiBIZXJlJiM4MjE3O3MgYSBoaW50OiB0cmFuc2NyaXB0aW9uIGludm9sdmVzIEROQSwgd2hpY2ggbWFrZXMgdXAgY2hyb21vc29tZXMuIFdoZXJlIGFyZSB0aG9zZSBjaHJvbW9zb21lcyBmb3VuZD8=[Qq]

[c]IGNvZG9uIHNlcXVlbmNlcyAmIzgyMzA7IGFtaW5vIGFjaWQgc2VxdWVuY2Vz[Qq]

[f]IE5vLiBUaGF0JiM4MjE3O3MgYSBnb29kIGRlZmluaXRpb24gZm9yIHRoZSBpbmZvcm1hdGlvbiB0cmFuc2ZlciB0aGF0IGhhcHBlbnMgZHVyaW5nIA==dHJhbnNsYXRpb24=IChvciBwcm90ZWluIHN5bnRoZXNpcykuIEhlcmUmIzgyMTc7cyBhIGhpbnQ6IHRyYW5zY3JpcHRpb24gaW52b2x2ZXMgRE5BLCB3aGljaCBtYWtlcyB1cCBjaHJvbW9zb21lcy4gV2hlcmUgYXJlIHRob3NlIGNocm9tb3NvbWVzIGZvdW5kPw==[Qq]

[q json=”true” xx=”1″ multiple_choice=”true” topic=”6.3.Transcription_and_RNA_Processing” dataset_id=”2019 AP Bio Dataset|22f272369ccc5a” question_number=”12″ 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]SGVyZSYjODIxNztzIHRoZSBtUk5BIGZyb20gc3RhZ2UgSUku

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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[Qq]

[q json=”true” xx=”1″ multiple_choice=”true” unit=”6.Gene Expression and Regulation” dataset_id=”2019 AP Bio Dataset|22efdaa55ff85a” question_number=”13″ topic=”6.3.Transcription_and_RNA_Processing”] In the diagram below, step E directly results in the production of

[c]IG1lc3NlbmdlciBSTkEu[Qq]

[f]IE5vLiBTdGVwICYjODIyMDtFJiM4MjIxOyA=ZG9lcw==IHByb2R1Y2UgUk5BLCBidXQgdGhhdCBSTkEgaGFzIHRvIGJlIHByb2Nlc3NlZCBiZWZvcmUgaXQmIzgyMTc7cyByZWFkeSB0byBiZSBtZXNzZW5nZXIgUk5BIGFuZCByZWFkIGJ5IGEgcmlib3NvbWUuIFdoYXQgd291bGQgY29tZSA=YmVmb3JlIG1STkE/[Qq]

[c]IHByZS1tZXNzZW 5nZXIgUk5BLg==[Qq]

[f]IEV4Y2VsbGVudC4gV2hhdCB5b3UgaGF2ZSBhZnRlciAmIzgyMjA7RSYjODIyMTsgaXMgYSByYXcgUk5BIHRyYW5zY3JpcHQgY2FsbGVkIHByZS1tUk5BLiBJdCBzdGlsbCBoYXMgdG8gYmUgcHJvY2Vzc2VkIGJlZm9yZSBpdCBjYW4gYmUgbVJOQSwgcmVhZHkgdG8gYmUgcmVhZCBieSBhIHJpYm9zb21lLg==[Qq]

[c]IGEgcG9seXBlcHRpZGUgY2hhaW4u[Qq]

[f]IE5vLiBQb2x5cGVwdGlkZSBjaGFpbnMgYXJlIHNlcXVlbmNlcyBvZiBhbWlubyBhY2lkcywgYW5kIHRoYXQgaGFzIHRvIGRvIHdpdGggcHJvdGVpbnMuIEhlcmUsIHlvdXIgYW5zd2VyIGlzIGdvaW5nIHRvIGJlIHNvbWUgdHlwZSBvZiBSTkEu[Qq]

[c]IGNvbXBsZW1lbnRhcnkgRE5BIChjRE5BKS4=[Qq]

[f]IE5vLiBDb21wbGVtZW50YXJ5IEROQSBpcyB1c2VkIGluIGdlbmV0aWMgZW5naW5lZXJpbmcgdG8gcHJvZHVjZSBldWthcnlvdGljIEROQSB0aGF0IGNhbiBiZSBwbGFjZWQgaW5zaWRlIGJhY3RlcmlhbCBjZWxscyBmb3IgcmVwbGljYXRpb24sIHRyYW5zY3JpcHRpb24sIGFuZCB0cmFuc2xhdGlvbi4gV2hhdCB5b3UmIzgyMTc7cmUgbG9va2luZyBmb3IgaGVyZSBpcyB0aGUgZmlyc3QgcmVzdWx0IG9mIHRyYW5zY3JpcHRpb24u[Qq]

[q json=”true” xx=”1″ multiple_choice=”true” unit=”6.Gene Expression and Regulation” dataset_id=”2019 AP Bio Dataset|224c6afab5705a” question_number=”60″ topic=”6.3.Transcription_and_RNA_Processing”] The following image shows a sequence of nucleic acid bases coding for a sequence of amino acids. The third amino acid in the sequence (lysine) has been filled in.

The sequence of nucleic acid bases on the top line of the image above makes up

[c]IEROQQ==[Qq]

[f]IE5vLiBUaGUgcHJlc2VuY2Ugb2YgdGhlIG51Y2xlb3RpZGUgdXJhY2lsIChVKSB0ZWxscyB5b3UgdGhhdCB0aGlzIGNhbiYjODIxNzt0IGJlIEROQS4gV2hhdCBudWNsZWljIGFjaWQgY29udGFpbnMgdXJhY2lsLCBhbmQgZ2V0cyB0cmFuc2xhdGVkIGludG8gYW1pbm8gYWNpZHM/[Qq]

[c]IHRSTkE=[Qq]

[f]IE5vLiBZb3UmIzgyMTc7cmUgcmlnaHQgYWJvdXQgdGhlIFJOQSBwYXJ0LCBidXQgeW91JiM4MjE3O3ZlIGNob3NlbiB0aGUgd3JvbmcgdHlwZSBvZiBSTkEuIFdoYXQgdHlwZSBvZiBSTkEgZ2V0cyB0cmFuc2xhdGVkIGludG8gYW1pbm8gYWNpZHM/[Qq]

[c]IHJSTkE=[Qq]

[f]IE5vLiBZb3UmIzgyMTc7cmUgcmlnaHQgYWJvdXQgdGhlIFJOQSBwYXJ0LCBidXQgeW91JiM4MjE3O3ZlIGNob3NlbiB0aGUgd3JvbmcgdHlwZSBvZiBSTkEuIFdoYXQgdHlwZSBvZiBSTkEgZ2V0cyB0cmFuc2xhdGVkIGludG8gYW1pbm8gYWNpZHM/[Qq]

[c]IG1S TkE=[Qq]

[f]IEZhbnRhc3RpYy4gbVJOQSBpcyB0aGUgbnVjbGVpYyBhY2lkIHRoYXQgZ2V0cyB0cmFuc2xhdGVkIGludG8gcHJvdGVpbi4=[Qq]

[c]IFJOQWk=[Qq]

[f]IE5vLiBSTkFpIGlzIGFtYXppbmcgc3R1ZmYsIHJlc3BvbnNpYmxlIGZvciBhbGwgdHlwZSBvZiBlcGlnZW5ldGljIGVmZmVjdHMmIzgyMzA7YnV0IGl0JiM4MjE3O3Mgbm90IHRoZSBSTkEgeW91JiM4MjE3O3JlIGxvb2tpbmcgZm9yLiBXaGF0IHR5cGUgb2YgUk5BIGdldHMgdHJhbnNsYXRlZCBpbnRvIGFtaW5vIGFjaWRzPw==[Qq]

[q json=”true” xx=”1″ multiple_choice=”true” unit=”6.Gene Expression and Regulation” dataset_id=”2019 AP Bio Dataset|223283f66cf05a” question_number=”67″ topic=”6.3.Transcription_and_RNA_Processing”] A team of biologists is studying gene expression in rat liver tissue. They find that two different proteins with different structures were translated from two different mRNAs. When they traced the source of the mRNA, however, they found that both mRNAs were transcribed from the same gene in the cell’s nucleus. Which of the following explanation best accounts for what the team has uncovered?

[c]IFRoZSBnZW5lIG1pZ2h0IGhhdmUgYmVlbiBhbHRlcmVkIGJ5IGEgbXV0YXRpb24u[Qq]

[f]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[Qq]

[c]IFRoZSBkaWZmZXJlbnQgZnVuY3Rpb25zIG9mIGVhY2ggcHJvdGVpbiByZXF1aXJlIHRoZW0gdG8gaGF2ZSBkaWZmZXJlbnQgdW5kZXJseWluZyBSTkFzIHdoaWNoIGNvZGUgZm9yIHRoZW0u[Qq]

[f]IE5vLiBSZW1lbWJlciB0aGUgY2VudHJhbCBkb2dtYSBvZiBtb2xlY3VsYXIgYmlvbG9neTogRE5BIG1ha2VzIFJOQSBtYWtlcyBwcm90ZWluLiBUaGlzIGNob2ljZSBpcyBwcm9wb3NpbmcgYSBwcm9jZXNzIHRoYXQgdmlvbGF0ZXMgdGhpcyBrZXkgaWRlYS4gVGhlIG5leHQgdGltZSB5b3Ugc2VlIHRoaXMgcXVlc3Rpb24sIHRoaW5rIGFib3V0IGdlbmUgZXhwcmVzc2lvbiBpbiBldWthcnlvdGVzLCBhbmQgc2VlIGlmIHlvdSBjYW4gcmVtZW1iZXIgYSBtZWNoYW5pc20gYnkgd2hpY2ggdHdvIFJOQXMgY291bGQgcmVzdWx0IGZyb20gb25lIGdlbmUu[Qq]

[c]IERpZmZlcmVudCBzeXN0ZW1zIG9mIGhpc3RvbmUgYWNldHlsYXRpb24gY291bGQgcmVzdWx0IGluIHR3byBkaWZmZXJlbnQgbVJOQXMu[Qq]

[f]IE5vLiBEaWZmZXJlbnQgcGF0dGVybnMgb2YgaGlzdG9uZSBhY2V0eWxhdGlvbiBjYW4gcmVzdWx0IGluIGRpZmZlcmVudCBwYXR0ZXJucyBvZiBnZW5lIGV4cHJlc3Npb24sIGJ1dCBpdCB3b3VsZG4mIzgyMTc7dCBwcm9kdWNlIHR3byBtUk5BcyBmcm9tIG9uZSBnZW5lLiBUaGUgbmV4dCB0aW1lIHlvdSBzZWUgdGhpcyBxdWVzdGlvbiwgdGhpbmsgYWJvdXQgZ2VuZSBleHByZXNzaW9uIGluIGV1a2FyeW90ZXMgYW5kIHNlZSBpZiB5b3UgY2FuIHJlbWVtYmVyIGEgbWVjaGFuaXNtIGJ5IHdoaWNoIHR3byBSTkFzIGNvdWxkIHJlc3VsdCBmcm9tIG9uZSBnZW5lLg==[Qq]

[c]IER1cmluZyBwcmUtbVJOQSBwcm9jZXNzaW5nLCBleG9ucyBjb3VsZCBiZSBzcGxpY2Vk IHRvZ2V0aGVyIGluIHZhcmlvdXMgd2F5cyB0byBtYWtlIGRpZmZlcmVudCBtUk5Bcy4=[Qq]

[f]IFdheSB0byBnbyEgSWYgb25lIGdlbmUgaXMgcmVzdWx0aW5nIGluIHR3byBtUk5BcywgdGhlbiB3aGF0JiM4MjE3O3MgaGFwcGVuaW5nIGlzIGFsdGVybmF0aXZlIHNwbGljaW5nIG9mIFJOQS4gVGhlc2UgZGlmZmVyZW50IFJOQXMgd291bGQgdGhlbiBnbyBvbiB0byBiZSB0cmFuc2xhdGVkIGFzIGRpc3RpbmN0IHByb3RlaW5zLg==[Qq]

[c]IE9uZSBzZXQgb2Ygcmlib3NvbWVzIHJlYWQgdGhlIG1STkEgaW4gdGhlIDUmIzgyNDI7IHRvIDMmIzgyNDI7IGRpcmVjdGlvbi4gQW5vdGhlciBzZXQgcmVhZHMgdGhlbSBpbiB0aGUgMyYjODI0MjsgdG8gNSYjODI0MjsgZGlyZWN0aW9uLg==[Qq]

[f]IE5vLiBBbGwgcmlib3NvbWVzICh3aGV0aGVyIGluIGFuIA==RS4gY29saQ==IGJhY3Rlcml1bSBvciBhbiBvcmNoaWQpIHJlYWQgbVJOQSBpbiB0aGUgNSYjODI0MjsgdG8gMyYjODI0MjsgZGlyZWN0aW9uLiBUaGUgbmV4dCB0aW1lIHlvdSBzZWUgdGhpcyBxdWVzdGlvbiwgdGhpbmsgYWJvdXQgZ2VuZSBleHByZXNzaW9uIGluIGV1a2FyeW90ZXMsIGFuZCBzZWUgaWYgeW91IGNhbiByZW1lbWJlciBhIG1lY2hhbmlzbSBieSB3aGljaCB0d28gUk5BcyBjb3VsZCByZXN1bHQgZnJvbSBvbmUgZ2VuZS4=

Cg==

[Qq]

[q json=”true” xx=”1″ multiple_choice=”true” topic=”6.4.Translation” dataset_id=”2019 AP Bio Dataset|22f4f4d36e9c5a” question_number=”11″ unit=”6.Gene Expression and Regulation”] In eukaryotic cells, translation moves information from ________ to _________.

[c]IHRoZSBudWNsZXVzICYjODIzMDsgdGhlIGN5dG9wbGFzbQ==[Qq]

[f]IE5vLiBUcmFuc2NyaXB0aW9uIGlzIHRoZSBwcm9jZXNzIHRoZSBtb3ZlcyBpbmZvcm1hdGlvbiBmcm9tIHRoZSBudWNsZXVzIHRvIHRoZSBjeXRvcGxhc20u[Qq]

[c]IHJpYm9zb21lcyAmIzgyMzA7IHByb3RlaW5z[Qq]

[f]IE5vLCBidXQgeW91JiM4MjE3O3JlIG9uIHRoZSByaWdodCB0cmFjay4gUmlib3NvbWVzIGFyZSB0aGUgdHJhbnNsYXRvcnMsIGJ1dCB0aGV5IGRvbiYjODIxNzt0IHByb3ZpZGUgdGhlIGluZm9ybWF0aW9uIHRoYXQgZ2V0cyB0cmFuc2xhdGVkLiBUaGF0IGluZm9ybWF0aW9uIGlzIGluIG1STkEsIGFuZCBpdCYjODIxNztzIGluIDMgYmFzZSB1bml0cy4gV2hhdCBhcmUgdGhvc2UgdW5pdHMgY2FsbGVkPw==[Qq]

[c]IHRoZSBjeXRvcGxhc20gJiM4MjMwOy4gdGhlIG51Y2xldXM=[Qq]

[f]IE5vLiBEdXJpbmcgbWFueSBjZWxsIHNpZ25hbGluZyBwcm9jZXNzZXMsIGluZm9ybWF0aW9uIG1vdmVzIGZyb20gdGhlIGN5dG9wbGFzbSB0byB0aGUgbnVjbGV1cy4gQnV0IG5vdCBpbiB0cmFuc2xhdGlvbi4gQ2hvb3NlIGFub3RoZXIgYW5zd2VyIG5leHQgdGltZSB5b3Ugc2VlIHRoaXMgcXVlc3Rpb24u[Qq]

[c]IGNvZG9uIHNlcXVlbmNlcyAmIzgyMz A7IGFtaW5vIGFjaWQgc2VxdWVuY2Vz[Qq]

[f]IEZhYnVsb3VzLiBUcmFuc2xhdGlvbiBtb3ZlcyBpbmZvcm1hdGlvbiBmcm9tIA==Y29kb24gc2VxdWVuY2VzIGluIG1STkEgdG8gYW1pbm8gYWNpZCBzZXF1ZW5jZXMgaW4gcG9seXBlcHRpZGVzLg==[Qq]

[q json=”true” xx=”1″ multiple_choice=”true” unit=”6.Gene Expression and Regulation” dataset_id=”2019 AP Bio Dataset|22ed5a5c9a0c5a” question_number=”14″ topic=”6.4.Translation”] In the diagram below, an anti-codon is indicated by

[c]IGdyb3 VwIEgu[Qq]

[f]IE5pY2UhIEdyb3VwIEggaXMgYW4gYW50aSBjb2Rvbi4=[Qq]

[c]IHByb2R1Y3QgRy4=[Qq]

[f]IE5vLiBHIGlzIG1STkEuIEFuIGFudGljb2RvbiBpcyBhIHNob3J0IHN0cmV0Y2ggb2YgUk5BIG9uIGEgdFJOQS5UaGUgYW50aWNvZG9uIGNvbXBsZW1lbnRzIGEgY29kb24uIEZpbmQgYSB0Uk5BLCBhbmQgeW91JiM4MjE3O2xsIGhhdmUgeW91ciBhbnN3ZXIu[Qq]

[c]IHN0cnVjdHVyZSBLLg==[Qq]

[f]IE5vLiBLIGlzIHRoZSByaWJvc29tZS4gQW4gYW50aWNvZG9uIGlzIGEgc2hvcnQgc3RyZXRjaCBvZiBSTkEgb24gYSB0Uk5BLiBUaGUgYW50aWNvZG9uIGNvbXBsZW1lbnRzIGEgY29kb24uIEZpbmQgYSB0Uk5BLCBhbmQgeW91JiM4MjE3O2xsIGhhdmUgeW91ciBhbnN3ZXIu[Qq]

[c]IGJvbmQgSi4=[Qq]

[f]IE5vLiBCb25kICYjODIyMDtKJiM4MjIxOyBpcyBhIHBlcHRpZGUgYm9uZCwgYSBib25kIHRoYXQgaG9sZHMgcHJvdGVpbnMgdG9nZXRoZXIuIEFuIGFudGljb2RvbiBpcyBhIHNob3J0IHN0cmV0Y2ggb2YgUk5BIG9uIGEgdFJOQS4gVGhlIGFudGljb2RvbiBjb21wbGVtZW50cyBhIGNvZG9uLiBGaW5kIGEgdFJOQSwgYW5kIHlvdSYjODIxNztsbCBoYXZlIHlvdXIgYW5zd2VyLg==[Qq]

[q json=”true” xx=”1″ multiple_choice=”true” topic=”6.4.Translation” dataset_id=”2019 AP Bio Dataset|230429c117bc5a” question_number=”5″ 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?

[c]IENvbWJpbmluZyB3aXRoIGFuIGFtaW5vIGFjaWQgYW5kIHRoZW4gYmluZGluZyB0byBhbiBhbnRpY29kb24u[Qq]

[f]IE5vLiBTdHJ1Y3R1cmUgWCBpcyBjYXJyeWluZyBhbiBhbWlubyBhY2lkLCBidXQgaXQmIzgyMTc7cyBiaW5kaW5nIHdpdGggYSBzZXF1ZW5jZSBvZiAzIGxldHRlcnMgb24gbVJOQS4gV2hhdCBhcmUgdGhvc2UgdGhyZWUgbGV0dGVyIHNlcXVlbmNlcyBjYWxsZWQ/[Qq]

[c]IEJpbmRpbmcgdG8gYW4gYW50aWNvZG9uIGFuZCB0aGVuIGNvbWJpbmluZyB3aXRoIGFuIGFtaW5vIGFjaWQu[Qq]

[f]IE5vLiBTdHJ1Y3R1cmUgWCBpcyA=Y2Fycnlpbmc=IGFuIGFtaW5vIGFjaWQsIGFuZCBpdCYjODIxNztzIGJpbmRpbmcgd2l0aCBhIHNlcXVlbmNlIG9mIDMgbGV0dGVycyBvbiBtUk5BPyBXaGF0IGFyZSB0aG9zZSB0aHJlZSBsZXR0ZXIgc2VxdWVuY2VzIGNhbGxlZD8=[Qq]

[c]IEJpbmRpbmcgdG8gYSBjb2RvbiBhbmQgdGhlbiBjb21iaW5pbmcgd2l0aCBhbiBhbWlubyBhY2lkLg==[Qq]

[f]IE5vLCBidXQgeW91JiM4MjE3O3JlIHZlcnkgY2xvc2UuIFRoZSBzdHJ1Y3R1cmUgYXQgWCBpcyBhIHRSTkEsIGFuZCBpdCBkb2VzIGJpbmQgd2l0aCBib3RoIGFuIGFtaW5vIGFjaWQgYW5kIGEgY29kb24uIFlvdSBqdXN0IG5lZWQgdG8gZ2V0IHRoZSBzZXF1ZW5jZSBvZiBldmVudHMgcmlnaHQu[Qq]

[c]IENvbWJpbmluZyB3aXRoIGFuIGFtaW5vIGFjaWQg YW5kIHRoZW4gYmluZGluZyB0byBhIGNvZG9uLg==[Qq]

[f]IENvcnJlY3QhIFRoZSBzdHJ1Y3R1cmUgYXQgWCBpcyBhIHRSTkEuIEl0IGZpcnN0IGJpbmRzIHdpdGggYW4gYW1pbm8gYWNpZCwgYW5kIHRoZW4gaXQgYmluZHMgd2l0aCBhIGNvZG9uIG9uIHRoZSBtUk5BLg==[Qq]

[q json=”true” xx=”1″ multiple_choice=”true” topic=”6.4.Translation” dataset_id=”2019 AP Bio Dataset|230150fe8df85a” question_number=”6″ 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=[Qq]

[q json=”true” xx=”1″ multiple_choice=”true” topic=”6.4.Translation” dataset_id=”2019 AP Bio Dataset|22fec9b9a4605a” question_number=”7″ 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]IE5vLiBUaGUgbVJOQSBpcyB0aGUgc2luZ2xlIHN0cmFuZGVkIG51Y2xlaWMgYWNpZCB3aXRoIGl0cyA1JiM4MjQyOyBhbmQgMyYjODI0MjsgZW5kcyBsYWJlbGVkLiAmIzgyMjA7WCYjODIyMTsgaXMgd2hhdCBicmluZ3MgYW1pbm8gYWNpZHMgdG8gdGhlIG1STkEgYXQgdGhlIHJpYm9zb21lLg==[Qq]

[c]IHJSTkEu[Qq]

[f]IE5vLiBUaGUgclJOQSBpcyB3aGF0IG1ha2VzIHVwIHRoZSByaWJvc29tZS4gJiM4MjIwO1gmIzgyMjE7IGlzIHdoYXQgYnJpbmdzIGFtaW5vIGFjaWRzIHRvIHRoZSBtUk5BIGF0IHRoZSByaWJvc29tZS4=[Qq]

[q json=”true” xx=”1″ multiple_choice=”true” dataset_id=”2019 AP Bio Dataset|2036a058a4c45a” question_number=”233″ 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. 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=[Qq]

[q json=”true” xx=”1″ multiple_choice=”true” dataset_id=”2019 AP Bio Dataset|211b1c0f810c5a” question_number=”163″ 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=[Qq]

[q json=”true” xx=”1″ multiple_choice=”true” unit=”6.Gene Expression and Regulation” dataset_id=”2019 AP Bio Dataset|22509a0013205a” question_number=”59″ topic=”6.4.Translation”] The following image shows a sequence of nucleic acid bases coding for a sequence of amino acids. The third amino acid in the sequence (lysine) has been filled in.

The processes that would change the nucleic acid information in the top row into the amino acid information in the bottom row occur

[c]IGF0IGEgcm lib3NvbWU=[Qq]

[f]IEV4YWN0bHkhIFRoZSBwcm9jZXNzIGlzIHRyYW5zbGF0aW9uLCBhbmQgaXQgb2NjdXJzIGF0IGEgcmlib3NvbWUu[Qq]

[c]IGluIGVuenltZXMgZW1iZWRkZWQgaW4gdGhlIGNlbGwmIzgyMTc7cyBtZW1icmFuZS4=[Qq]

[f]IE5vLiBUaGUgcHJvY2VzcyB0aGF0IGNvbnZlcnRzIG51Y2xlaWMgYWNpZCBpbmZvcm1hdGlvbiBpbnRvIGFtaW5vIGFjaWQgaW5mb3JtYXRpb24gaXMgdHJhbnNsYXRpb24uIFdoZXJlIGRvZXMgdGhhdCBvY2N1cj8=[Qq]

[c]IGluIHRoZSBudWNsZXVzLg==[Qq]

[f]IE5vLiBUaGUgcHJvY2VzcyB0aGF0IGNvbnZlcnRzIG51Y2xlaWMgYWNpZCBpbmZvcm1hdGlvbiBpbnRvIGFtaW5vIGFjaWQgaW5mb3JtYXRpb24gaXMgdHJhbnNsYXRpb24uIFdoZXJlIGRvZXMgdGhhdCBvY2N1cj8=[Qq]

[c]IGluIHRoZSBzbW9vdGggRW5kb3BsYXNtaWMgcmV0aWN1bHVt[Qq]

[f]IE5vLiBUaGUgcHJvY2VzcyB0aGF0IGNvbnZlcnRzIG51Y2xlaWMgYWNpZCBpbmZvcm1hdGlvbiBpbnRvIGFtaW5vIGFjaWQgaW5mb3JtYXRpb24gaXMgdHJhbnNsYXRpb24uIFdoZXJlIGRvZXMgdGhhdCBvY2N1cj8=

Cg==

Cg==[Qq]

[q json=”true” xx=”1″ multiple_choice=”true” unit=”6.Gene Expression and Regulation” dataset_id=”2019 AP Bio Dataset|2269f9f9a9f85a” question_number=”52″ 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]IEEgcmVndWxhdG9yeSBwcm90ZWluIGJpbmRzIGF0IHRoZSBvcGVyYXRvciwgYm xvY2tpbmcgdHJhbnNjcmlwdGlvbiBvZiB0aGUgc3RydWN0dXJhbCBnZW5lcy4=[Qq]

[f]IENvcnJlY3QhIExhY3Rvc2UgaXMgYW4gaW5kdWNpYmxlIG9wZXJvbiwgbWVhbmluZyB0aGF0IHRoZSBwcmVzZW5jZSBvZiBsYWN0b3NlIGluZHVjZXMgdHJhbnNjcmlwdGlvbiBvZiB0aGUgc3RydWN0dXJhbCBnZW5lcyBmb3IgZGlnZXN0aW5nIGxhY3Rvc2UuIFdoZW4gbGFjdG9zZSBpcyBub3QgaW4gdGhlIGVudmlyb25tZW50LCBhIHJlZ3VsYXRvcnkgcHJvdGVpbiBiaW5kcyBhdCB0aGUgb3BlcmF0b3IgcmVnaW9uLCBibG9ja2luZyB0cmFuc2NyaXB0aW9uIG9mIHRoZSBzdHJ1Y3R1cmFsIGdlbmVzLg==

Cg==

[Qq]

[c]IEEgcmVndWxhdG9yeSBwcm90ZWluIGJpbmRzIHRvIGFuIGFsbG9zdGVyaWMgc2l0ZSBvbiBSTkEgcG9seW1lcmFzZSwgcHJldmVudGluZyBpdCBmcm9tIGJpbmRpbmcgd2l0aCB0aGUgcHJvbW90ZXI=[Qq]

[f]IE5vLiBIZXJlJiM4MjE3O3MgdGhlIHN0cnVjdHVyZSBvZiB0aGUgbGFjIG9wZXJvbiwgc2hvd24gd2hlbiBsYWN0b3NlICg4KSBpcyBwcmVzZW50LiBOb3RlIHRoYXQgbGFjdG9zZSBpcyBiaW5kaW5nIHdpdGggYW4gYWxsb3N0ZXJpYyBzaXRlIG9uIGEgcmVndWxhdG9yeSBwcm90ZWluICgzKSwgY2hhbmdpbmcgaXRzIHNoYXBlIHNvIHRoYXQgaXQgY2FuIG5vIGxvbmdlciBiaW5kIHdpdGggdGhlIG9wZXJhdG9yIHJlZ2lvbiAoNikgYWxsb3dpbmcgdHJhbnNjcmlwdGlvbi4gVGhlIHJlZ3VsYXRvcnkgcHJvdGVpbiBpcyA=bm90IGJpbmRpbmcgd2l0aCBSTkEgcG9seW1lcmFzZSAoNCkuIEtlZXAgdGhhdCBpbiBtaW5kIHRoZSBuZXh0IHRpbWUgeW91IHNlZSB0aGlzIHF1ZXN0aW9uLg==

Cg==

[Qq]

[c]IEEgcmVndWxhdG9yeSBwcm90ZWluIGJpbmRzIHRvIEROQSBwb2x5bWVyYXNlLCBwcmV2ZW50aW5nIHRoZSBnZW5lIGZvciBsYWN0b3NlIHByb2R1Y3Rpb24gdG8gYmUgcmVwbGljYXRlZC4=[Qq]

[f]IE5vLiBETkEgcG9seW1lcmFzZSBpc24mIzgyMTc7dCBpbnZvbHZlZCBpbiB0aGUgZnVuY3Rpb24gb2YgdGhpcyBvcGVyb24uIFRoZSBvcGVyb24gd29ya3MgYXMgc2hvd24gYmVsb3c6IFdoZW4gbGFjdG9zZSAoOCkgaXMgcHJlc2VudCwgbGFjdG9zZSBiaW5kcyB3aXRoIGFuIGFsbG9zdGVyaWMgc2l0ZSBvbiBhIHJlZ3VsYXRvcnkgcHJvdGVpbiAoMyksIGNoYW5naW5nIGl0cyBzaGFwZSBzbyB0aGF0IGl0IGNhbiBubyBsb25nZXIgYmluZCB3aXRoIHRoZSBvcGVyYXRvciByZWdpb24gKDYpIGFsbG93aW5nIHRyYW5zY3JpcHRpb24uIEtlZXAgdGhhdCBpbiBtaW5kIHRoZSBuZXh0IHRpbWUgeW91IHNlZSB0aGlzIHF1ZXN0aW9uLg==

Cg==

[Qq]

[c]IEEgcmVndWxhdG9yeSBwcm90ZWluIGJpbmRzIHRvIHRoZSBzdGFydCBjb2RvbiBvbiBtUk5BLCBwcmV2ZW50aW5nIHRoZSBpbml0aWF0aW9uIG9mIHRyYW5zbGF0aW9uIGJ5IHRoZSByaWJvc29tZS4=[Qq]

[f]IE5vLiBUaGUgb3Blcm9uIGlzIHdvcmtpbmcgYXQgdGhlIGxldmVsIG9mIHRyYW5zY3JpcHRpb24sIG5vdCB0cmFuc2xhdGlvbi4gUmF0aGVyLCB0aGUgb3Blcm9uIHdvcmtzIGFzIHNob3duIGJlbG93OiBXaGVuIGxhY3Rvc2UgKDgpIGlzIHByZXNlbnQsIGxhY3Rvc2UgYmluZHMgd2l0aCBhbiBhbGxvc3RlcmljIHNpdGUgb24gYSByZWd1bGF0b3J5IHByb3RlaW4gKDMpLCBjaGFuZ2luZyBpdHMgc2hhcGUgc28gdGhhdCBpdCBjYW4gbm8gbG9uZ2VyIGJpbmQgd2l0aCB0aGUgb3BlcmF0b3IgcmVnaW9uICg2KSBhbGxvd2luZyB0cmFuc2NyaXB0aW9uLiBLZWVwIHRoYXQgaW4gbWluZCB0aGUgbmV4dCB0aW1lIHlvdSBzZWUgdGhpcyBxdWVzdGlvbi4=

Cg==

[Qq]

[q json=”true” xx=”1″ multiple_choice=”true” unit=”6.Gene Expression and Regulation” dataset_id=”2019 AP Bio Dataset|21a5be6f81d85a” question_number=”114″ topic=”6.5.Regulation_of_Gene_Expression”] In the diagram below, transcription factors will bind with target genes at

[c]IDEg[Qq][c]IDIg[Qq][c]ID Qg[Qq][c]IDU=

Cg==[Qq]

[f]IE5vLiAxIGlzIGEgbGlnYW5kLiBUcmFuc2NyaXB0aW9uIGZhY3RvcnMgYXJlIHN1YnN0YW5jZXMgdGhhdCBlbnRlciBpbnRvIHRoZSBudWNsZXVzIGFuZCBpbmR1Y2UgdHJhbnNjcmlwdGlvbiBvZiBnZW5lcy4gVGFrZSBhbm90aGVyIGxvb2sgYXQgdGhlIGRpYWdyYW0sIGFuZCBpZGVudGlmeSB3aGVyZSB0aGF0IGNvdWxkIGhhcHBlbi4=[Qq]

[f]IE5vLiAyIGlzIGEgcmVjZXB0b3IuIFRyYW5zY3JpcHRpb24gZmFjdG9ycyBhcmUgc3Vic3RhbmNlcyB0aGF0IGVudGVyIGludG8gdGhlIG51Y2xldXMgYW5kIGluZHVjZSB0cmFuc2NyaXB0aW9uIG9mIGdlbmVzLiBUYWtlIGFub3RoZXIgbG9vayBhdCB0aGUgZGlhZ3JhbSwgYW5kIGlkZW50aWZ5IHdoZXJlIHRoYXQgY291bGQgaGFwcGVuLg==[Qq]

[f]IEV4Y2VsbGVudC4gNCByZXByZXNlbnRzIHRoZSBudWNsZXVzLCB3aGljaCBpcyB3aGVyZSB0cmFuc2NyaXB0aW9uIGZhY3RvcnMgKHJlcHJlc2VudGVkIGJ5IDMpIGFyZSBnb2luZyB0byBiaW5kIHdpdGggdGFyZ2V0IGdlbmVzLCBpbmZsdWVuY2luZyB0cmFuc2NyaXB0aW9uLg==[Qq]

[f]IE5vLiA1IHJlcHJlc2VudHMgcHJvdGVpbnMgdGhhdCBoYXZlIGJlZW4gcHJvZHVjZWQgYmVjYXVzZSBvZiB0aGUgYWN0aXZpdHkgb2YgdHJhbnNjcmlwdGlvbiBmYWN0b3JzLiBUcmFuc2NyaXB0aW9uIGZhY3RvcnMgYXJlIHN1YnN0YW5jZXMgdGhhdCBlbnRlciBpbnRvIHRoZSBudWNsZXVzIGFuZCBpbmR1Y2UgdHJhbnNjcmlwdGlvbiBvZiBnZW5lcy4gVGFrZSBhbm90aGVyIGxvb2sgYXQgdGhlIGRpYWdyYW0sIGFuZCBpZGVudGlmeSB3aGVyZSB0aGF0IGNvdWxkIGhhcHBlbi4=[Qq]

[q json=”true” xx=”1″ multiple_choice=”true” unit=”6.Gene Expression and Regulation” dataset_id=”2019 AP Bio Dataset|21edcb4777605a” question_number=”89″ topic=”6.5.Regulation_of_Gene_Expression”] A lysogenic virus infects a particular cell type. and integrates its genome into a site that contains a proto-oncogene. This transforms the cell and increases the level of a protein X. Protein X, in turn, increases cellular proliferation.

Compound P is known to increase the level of tumor suppressor proteins in that same cell type. A second compound, Q helps in stimulating protein Z. Protein Z has been shown to be capable of binding to protein X, rendering it inactive. Which one of the following graphs correctly represents the mode of action of P and Q?

[c]IEEg[Qq][c]IEIg[Qq][c]IE Mg[Qq][c]IEQ=

Cg==[Qq]

[f]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

Cg==

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[q json=”true” xx=”1″ multiple_choice=”true” unit=”6.Gene Expression and Regulation” dataset_id=”2019 AP Bio Dataset|22629e6016745a” question_number=”54″ topic=”6.6.Gene_Expression_and_Cell_Specialization”] The reason why neurons differ from skin cells is that the neurons

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[q json=”true” xx=”1″ multiple_choice=”true” unit=”6.Gene Expression and Regulation” dataset_id=”2019 AP Bio Dataset|2205dbfa5df85a” question_number=”81″ topic=”6.6.Gene_Expression_and_Cell_Specialization”] A team of biologists has started a project to clone the rare blue lobster through somatic cell nuclear transfer. The blue lobster is a variety of the more common purple lobster. The difference in phenotype has been traced to a genetic mutation that results in a blue exoskeleton.

For somatic cell nuclear transfer to be successful, the best choice for a donor cell will be a(n)

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[q json=”true” xx=”1″ multiple_choice=”true” unit=”6.Gene Expression and Regulation” dataset_id=”2019 AP Bio Dataset|2202fc3bb0885a” question_number=”82″ topic=”6.6.Gene_Expression_and_Cell_Specialization”] If you wanted to clone a dog, the least promising tissue to serve as a source of a donor cell would be the

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[c]IGludGVzdGluZXM=[Qq]

[f]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[Qq]

[q json=”true” xx=”1″ multiple_choice=”true” unit=”6.Gene Expression and Regulation” dataset_id=”2019 AP Bio Dataset|21fffbe45ca05a” question_number=”83″ topic=”6.6.Gene_Expression_and_Cell_Specialization”] A team of biologists has started a project to clone the rare blue lobster through somatic cell nuclear transfer. The blue lobster is a variety of the more common purple lobster. The difference in phenotype has been traced to a genetic mutation that results in a blue exoskeleton.

The team has discovered that the rate of production of viable clones is highest when intestinal cells are used. The least likely reason for this would be the difference in

[c]IGhpc3RvbmUgYWNldHlsYXRpb24=[Qq]

[f]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[Qq]

[c]IE NhKys=IGxldmVs[Qq]

[f]IFRlcnJpZmljLiBFdmVyeSBvdGhlciBjaG9pY2UgaXMgY29ubmVjdGVkIGVpdGhlciB3aXRoIEROQSBwYWNrYWdpbmcgb3Igd2l0aCBjb250cm9sIG9mIGdlbmUgZXhwcmVzc2lvbiwgYWxsIG9mIHdoaWNoIGNvdWxkIGFmZmVjdCBob3cgZ2VuZXMgYXJlIGV4cHJlc3NlZCBpbiBhIGRldmVsb3BpbmcgZW1icnlvLCB3aGljaCBjb3VsZCBpbiB0dXJuIGFmZmVjdCB0aGUgdmlhYmlsaXR5IG9mIHRob3NlIGVtYnJ5b3MuIENhKys=IGxldmVscyBzZWVtIGxpa2UgdGhlIGZhY3RvciB0aGF0IHdvdWxkIGhhdmUgdGhlIGxlYXN0IGltcGFjdC4=[Qq]

[c]IEROQSBtZXRoeWxhdGlvbg==[Qq]

[f]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[Qq]

[c]IGNocm9tYXRpbiBzdHJ1Y3R1cmU=[Qq]

[f]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[Qq]

[q json=”true” xx=”1″ multiple_choice=”true” unit=”6.Gene Expression and Regulation” dataset_id=”2019 AP Bio Dataset|21fccaa80f045a” question_number=”84″ topic=”6.6.Gene_Expression_and_Cell_Specialization”] A team of biologists is involved in a project to create a population of cloned, genetically engineered sheep that secrete, in their milk, a protein that inhibits the growth of a specific type of tumor. In this new population of clones, many have behavioral defects. From the list below, the most likely reason for this would be

[c]IGNvbXBsZXRlIHJlcHJvZ3JhbW1pbmcgb2YgdGhlIEROQSBpbiB0aGUgbmV3bHkgZm9ybWVkIHp5Z290ZSB0byBhbiBlbWJyeW9uaWMgc3RhdGUgYWZ0ZXIgbnVjbGVhciB0cmFuc2Zlci4=[Qq]

[f]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[Qq]

[c]IG1pdG9jaG9uZHJpYWwgaW5jb21wYXRpYmlsaXR5IHdpdGhpbiB0aGUgZWdn[Qq]

[f]IE5vLiBJbiBzb21hdGljIGNlbGwgbnVjbGVhciB0cmFuc2ZlciwganVzdCB0aGUgbnVjbGV1cyBvZiB0aGUgZG9ub3IgY2VsbCBpcyBpbXBsYW50ZWQgaW50byBhbiBlZ2cgY2VsbCB0aGF0IGhhcyBoYWQgaXRzIG51Y2xldXMgcmVtb3ZlZC4gSW4gb3RoZXIgd29yZHMsIGFsbCB0aGUgbWl0b2Nob25kcmlhIHNob3VsZCBiZSBjb21pbmcgZnJvbSB0aGUgZWdnICh3aGljaCBpcyB3aGF0IGhhcHBlbnMgZHVyaW5nIHRoZSBub3JtYWwgZmVydGlsaXphdGlvbiBwcm9jZXNzKS4gQXMgYSByZXN1bHQsIHRoZXJlJiM4MjE3O3Mgbm8gY2hhbmNlIG9mIG1pdG9jaG9uZHJpYWwgaW5jb21wYXRpYmlsaXR5LiBMb29rIGF0IHRoZSBsaXN0IGFnYWluLCBhbmQgc2VlIGlmIHlvdSBjYW4gZmluZCBzb21ldGhpbmcgdGhhdCBjb3VsZCBsZWFkIHRvIGdlbmV0aWMgcHJvYmxlbXMgZHVyaW5nIGRldmVsb3BtZW50Lg==[Qq]

[c]IGRvbm9yIGNlbGwgdGVsb21lcmVz IHRoYXQgd2VyZSB0b28gc2hvcnQ=[Qq]

[f]IEdyZWF0IGpvYi4gQXMgYSBtdWx0aWNlbGx1bGFyIGV1a2FyeW90aWMgb3JnYW5pc20gYWdlcywgdGhlIHRlbG9tZXJlcyBpbiBpdHMgY2VsbHMgc2hvcnRlbi4gQXQgYSBjZXJ0YWluIHBvaW50LCB0aGUgdGVsb21lcmVzIGFyZSBnb25lLCBhbmQgZnVydGhlciBjZWxsIGRpdmlzaW9uIHJlc3VsdHMgaW4gbG9zcyBvZiBnZW5ldGljIEROQS4gVGhpcyBpcyBhIGtub3duIHByb2JsZW0gaW4gY2xvbmVkIG9yZ2FuaXNtcywgYW5kIGNvdWxkIGFjY291bnQgZm9yIHRoZSBiZWhhdmlvcmFsIHByb2JsZW1zIGRlc2NyaWJlZCBhYm92ZS4=[Qq]

[c]IGFjdGluIHBvbHltZXJpemF0aW9uIGxldmVscyBpbiB0aGUgZWdnIHRoYXQgd2VyZSB0b28gaGlnaA==[Qq]

[f]IE5vLiBXaGlsZSBhY3RpbiBwb2x5bWVyaXphdGlvbiBsZXZlbHMgdGhhdCB3ZXJlIHRvbyBoaWdoIGNvdWxkIGNhdXNlIHByb2JsZW1zIChvciBiZSBhIHN5bXB0b20gb2YgYSBwcm9ibGVtKSwgdGhlcmUmIzgyMTc7cyBhIG1vcmUgb2J2aW91cyBjaG9pY2UgdGhhdCBjb3VsZCBsZWFkIHRvIGdlbmV0aWMgcHJvYmxlbXMgZHVyaW5nIGRldmVsb3BtZW50Lg==

Cg==

Cg==[Qq]

[q json=”true” xx=”1″ multiple_choice=”true” dataset_id=”2019 AP Bio Dataset|20593de16c7c5a” question_number=”222″ 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]IEZld2VyIGdlbmUgY29waWVzIGxlYWRzIHRvIGZhc3RlciBETkEgcmVwbGljYXRpb24sIGFuZCBmYXN0ZXIgRE5BIHJlcGxpY2F0aW9uIGFsbG93cyBhIHBvcHVsYXRpb24gdG8gZXZvbHZlIG1vdmUgcXVpY2tseS4=[Qq]

[f]IE5vLiBBcyB5b3UgY2FuIHNlZSBiZWxvdywgZ2VuZSBkdXBsaWNhdGlvbiBsZWFkcyB0byA=bW9yZQ==IGdlbmUgY29waWVzLCBub3QgZmV3ZXIu

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]IE5vLCBidXQgeW91IGNob3NlIHRoZSBzZWNvbmQgYmVzdCBjaG9pY2UuIEl0IHNvdW5kcyBsaWtlIHlvdSB3ZXJlIHRoaW5raW5nIG9mIHdoYXQgaGFwcGVucyB3aXRoIHJlcGxpY2F0aXZlIHRyYW5zcG9zaXRpb24sIHdoZXJlIGEgdHJhbnNwb3NvbiAoYSBtb2JpbGUgRE5BIGVsZW1lbnQpIG1ha2VzIGEgY29weSBvZiBpdHNlbGYgdGhhdCBpbnNlcnRzIHNvbWV3aGVyZSBlbHNlIGluIHRoZSBnZW5vbWUuIFRoaXMgY2FuIGVmZmVjdCBleHByZXNzaW9uIG9mIGdlbmVzLCBlc3BlY2lhbGx5IGlmIHRoZSB0cmFuc3Bvc29uIGRpc3J1cHRzIGFub3RoZXIgZ2VuZSYjODIxNztzIHByb21vdGVyLg==

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==

[Qq]

[q json=”true” xx=”1″ multiple_choice=”true” unit=”6.Gene Expression and Regulation” dataset_id=”2019 AP Bio Dataset|2236274901305a” question_number=”66″ 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” xx=”1″ multiple_choice=”true” unit=”6.Gene Expression and Regulation” dataset_id=”2019 AP Bio Dataset|22a3bf60b3385a” question_number=”36″ 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 separation of these fragments by gel electrophoresis?

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

Cg==[Qq]

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[q json=”true” xx=”1″ multiple_choice=”true” unit=”6.Gene Expression and Regulation” dataset_id=”2019 AP Bio Dataset|22b1662a6b0c5a” question_number=”32″ topic=”6.8.Biotechnology”] The diagram below shows some of the steps involved in cloning. Which of the statements below is true about processes X and Y?

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Cg==

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Cg==

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Cg==

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[f]IE5pY2UgSm9iLiBZb3UgaGF2ZSBhIGdvb2QgdW5kZXJzdGFuZGluZyBvZiBob3cgY2xvbmluZyB3b3JrcyE=[Qq]

[q json=”true” xx=”1″ multiple_choice=”true” unit=”6.Gene Expression and Regulation” dataset_id=”2019 AP Bio Dataset|2248b75bcdf45a” question_number=”61″ 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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Cg==

[Qq]

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Cg==

[Qq]

[c]IHJlcGxpY2F0aW9uIG9mIG1lc3NlbmdlciBSTkEgd2l0aGluIHRoZSBiYWN0ZXJpYWwgY2VsbA==[Qq]

[f]IE5vLiBUYWtlIGEgbG9vayBhdCB0aGlzIHBpY3R1cmUgb2YgRE5BIHRoYXQmIzgyMTc7cyBiZWVuIGN1dCBpbiBhIHdheSB0aGF0IGxlYXZlcyBpdCB3aXRoIHN0aWNreSBlbmRzLiBOb3csIGxvb2sgYWdhaW4gYXQgdGhlIGNob2ljZXMgYW5kIHNlZSBpZiB5b3UgY2FuIGZpbmQgYSBiZXR0ZXIgYW5zd2VyLg==

Cg==

[Qq]

[q json=”true” xx=”1″ multiple_choice=”true” unit=”6.Gene Expression and Regulation” dataset_id=”2019 AP Bio Dataset|225ee3c50b4c5a” question_number=”55″ topic=”6.8.Biotechnology”] In order to clone genes, bacteria are frequently used. This is because bacteria

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[c]IHVzZSBtaXRvc2lzIHRvIGV4cG9uZW50aWFsbHkgaW5jcmVhc2UgdGhlaXIgcG9wdWxhdGlvbi4=[Qq]

[f]IE5vLiBNaXRvc2lzIGlzIHNvbWV0aGluZyB0aGF0JiM4MjE3O3MgZG9uZSBleGNsdXNpdmVseSBieSBldWthcnlvdGVzIChhbmQgbm90IGJ5IGJhY3RlcmlhLCB3aGljaCBhcmUgcHJva2FyeW90ZXMpLiBTZWUgaWYgdGhlcmUmIzgyMTc7cyBhIGNob2ljZSB0aGF0IGxvb2tzIGxpa2UgaXQgd291bGQgYmUgdXNlZnVsIGZvciBtYWtpbmcgbWFueSBjb3BpZXMgb2YgYSBnZW5lLg==[Qq]

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[q json=”true” xx=”1″ multiple_choice=”true” unit=”6.Gene Expression and Regulation” dataset_id=”2019 AP Bio Dataset|225b6f0364dc5a” question_number=”56″ 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 disaccharide, 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]

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[c]IHBsYXRlIFg=[Qq]

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[c]IHBsYX RlIFk=[Qq]

[f]IEZhYnVsb3VzLiBUaGVzZSBiYWN0ZXJpYSBoYXZlIHRha2VuIHVwIHRoZSBwbGFzbWlkLCB3aGljaCBwcm92aWRlcyB0aGVtIHdpdGggdGhlIGdlbmUgdG8gc3ludGhlc2l6ZSBHRlAuIE9uIHRvcCBvZiB0aGF0LCB0aGV5IGhhdmUgYmVlbiBwcm92aWRlZCB3aXRoIGFyYWJpbm9zZSwgd2hpY2ggaW5kdWNlcyB0aGUgY2VsbCB0byBzeW50aGVzaXplIEdGUC4=[Qq]

[c]IHBsYXRlIFo=[Qq]

[f]IE5vLiBXaGlsZSB0aGVzZSBiYWN0ZXJpYSBoYXZlIGJlZW4gdHJhbnNmb3JtZWQgKGFzIGRlbW9uc3RyYXRlZCBieSB0aGUgZmFjdCB0aGF0IHRoZXkmIzgyMTc7cmUgZ3Jvd2luZyBpbiB0aGUgcHJlc2VuY2Ugb2YgdGhlIGFudGliaW90aWMgYW1waWNpbGxpbiksIHRoZXkgaGF2ZSBub3QgYmVlbiBmZWQgYXJhYmlub3NlLCB3aGljaCBpcyByZXF1aXJlZCB0byBpbmR1Y2UgdGhlIGNlbGwgdG8gc3ludGhlc2l6ZSBHRlAuIFJlbWVtYmVyIHRoYXQgbmV4dCB0aW1lIHlvdSBzZWUgdGhpcyBxdWVzdGlvbiwgYW5kIG1ha2UgYW5vdGhlciBjaG9pY2Uu[Qq]

[q json=”true” xx=”1″ multiple_choice=”true” unit=”6.Gene Expression and Regulation” dataset_id=”2019 AP Bio Dataset|2257cbb0d09c5a” question_number=”57″ 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 disaccharide, 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?

[c]IFBsYXRlIFcgc2hvd2VkIHN1Y2Nlc3NmdWwgZGVzdHJ1Y3Rpb24gb2YgdGhlIHBsYXNtaWQgYnkgYmFjdGVyaWFsIHJlc3RyaWN0aW9uIGVuenltZXMu[Qq]

[f]IE5vLiBQbGF0ZSBXIGhhcyB1bnRyYW5zZm9ybWVkIGJhY3RlcmlhIHRoYXQgbmV2ZXIgdG9vayB1cCB0aGUgcGxhc21pZCBpbiB0aGUgZmlyc3QgcGxhY2UuIE5leHQgdGltZSB5b3Ugc2VlIHRoaXMgcXVlc3Rpb24sIHRoaW5rIGFib3V0IHRoZSBpZGVhIA==ZXhwZXJpbWVudGFsIGNvbnRyb2xzLCBhbmQgbWFrZSBhIGRpZmZlcmVudCBjaG9pY2Uu[Qq]

[c]IFBsYXRlIFcgc2hvd2VkIHRoYXQgdGhlcmUgd2FzIGV4dHJlbWVseSBoaWdoIHRyYW5zZm9ybWF0aW9uIGVmZmljaWVuY3ku[Qq]

[f]IE5vLiBUaGUgZ3Jvd3RoIG9uIHBsYXRlIFcgaXMgbm90IGFib3V0IHRyYW5zZm9ybWF0aW9uIGVmZmljaWVuY3ksIGJlY2F1c2UgdGhlIGJhY3RlcmlhIG9uIHRoYXQgcGxhdGUgd2VyZSBub3QgdHJhbnNmb3JtZWQuIE5leHQgdGltZSB5b3Ugc2VlIHRoaXMgcXVlc3Rpb24sIHRoaW5rIGFib3V0IHRoZSBpZGVhIA==ZXhwZXJpbWVudGFsIGNvbnRyb2xzLCBhbmQgbWFrZSBhIGRpZmZlcmVudCBjaG9pY2Uu[Qq]

[c]IFBsYXRlIFggc2hvd2VkIGhvdyBudXRyaWVudCBhZ2FyIGNhbiBlbmhhbmNlIHRoZSBncm93dGggb2YgdmlhYmxlIGJhY3RlcmlhLg==[Qq]

[f]IE5vLiBUaGVyZSYjODIxNztzIG5vIGdyb3d0aCBvbiBwbGF0ZSBYLCBhbmQgdGhhdCYjODIxNztzIGR1ZSB0byB0aGUgcHJlc2VuY2Ugb2YgdGhlIGFudGliaW90aWMgYW1waWNpbGxpbi4gTmV4dCB0aW1lIHlvdSBzZWUgdGhpcyBxdWVzdGlvbiwgdGhpbmsgYWJvdXQgdGhlIGlkZWEgZXhwZXJpbWVudGFsIGNvbnRyb2xzLCBhbmQgbWFrZSBhIGRpZmZlcmVudCBjaG9pY2Uu[Qq]

[c]IFBsYXRlIFggc2hvd2VkIHRoYXQgYW1waWNpbGxpbiB3YXMgZWZmZWN0aXZlIGluIGluaGliaXRpbm cgdGhlIGdyb3d0aCBvZiBiYWN0ZXJpYSB0aGF0IGhhZCBub3QgdGFrZW4gdXAgdGhlIHBsYXNtaWQu[Qq]

[f]IFRoYXQmIzgyMTc7cyBjb3JyZWN0ISBQbGF0ZSBYIHNob3dzIHRoYXQgdGhlcmUgd29uJiM4MjE3O3QgYmUgYW55IGdyb3d0aCBpbiB1bnRyYW5zZm9ybWVkIGJhY3RlcmlhLCB3aGljaCBoYXZlbiYjODIxNzt0IHRha2VuIHVwIHRoZSBwbGFzbWlkIHRoYXQgcHJvdmlkZXMgdGhlbSB3aXRoIGFudGliaW90aWMgcmVzaXN0YW5jZSAoaW4gYWRkaXRpb24gdG8gdGhlIGdlbmUgZm9yIEdGUCku[Qq]

[q json=”true” xx=”1″ multiple_choice=”true” unit=”6.Gene Expression and Regulation” dataset_id=”2019 AP Bio Dataset|22546e37a1145a” question_number=”58″ topic=”6.8.Biotechnology”] The following image shows a sequence of nucleic acid bases coding for a sequence of amino acids. The third amino acid in the sequence (lysine) has been filled in.

In a very schematic form, the above process represents

[c]IEROQSByZXBsaWNhdGlvbg==[Qq]

[f]IE5vLiBUaGUgJiM4MjIwO1UmIzgyMjE7IG51Y2xlb3RpZGUgaW5kaWNhdGVzIHVyYWNpbCwgZm91bmQgaW4gUk5BIGJ1dCBub3QgRE5BLiBTbywgeW91IGNhbiBydWxlIG91dCBETkEgcmVwbGljYXRpb24uIFdoYXQgcHJvY2VzcyBjb252ZXJ0cyBSTkEgbnVjbGVvdGlkZXMgaW50byBhbWlubyBhY2lkcz8=[Qq]

[c]IHRyYW5zY3JpcHRpb24=[Qq]

[f]IE5vLiBUcmFuc2NyaXB0aW9uIGNoYW5nZXMgaW5mb3JtYXRpb24gaW4gRE5BIGludG8gUk5BLiBIZXJlLCBpbmZvcm1hdGlvbiBpbiBSTkEgY29kb25zIGlzIGJlaW5nIGNoYW5nZWQgaW50byBzZXF1ZW5jZXMgb2YgYW1pbm8gYWNpZHMuIEl0JiM4MjE3O3MgYW5hbG9nb3VzIHRvIHdoYXQgaGFwcGVucyB3aGVuIHlvdSBjaGFuZ2UgdGhlIHdvcmRzIGluIG9uZSBsYW5ndWFnZSBpbnRvIGVxdWl2YWxlbnQgd29yZHMgaW4gYW5vdGhlciBsYW5ndWFnZS4gV2hhdCBwcm9jZXNzIGlzIHRoYXQ/[Qq]

[c]IHRyYW5zZm9ybWF0aW9u[Qq]

[f]IE5vLiBUaGUgdGVybSA=dHJhbnNmb3JtYXRpb24=IHJlZmVycyB0byBhIHdoZW4gYSBiYWN0ZXJpYWwgY2VsbCB0YWtlcyB1cCBETkEgZnJvbSBhIHBsYXNtaWQgKG9yIGZyb20gdGhlIGVudmlyb25tZW50KSwgdGhlcmVieSBhY3F1aXJpbmcgbmV3IGdlbmVzLiBIZXJlLCBpbmZvcm1hdGlvbiBpbiBSTkEgaXMgYmVpbmcgY2hhbmdlZCBpbnRvIHByb3RlaW4uIEl0JiM4MjE3O3MgYW5hbG9nb3VzIHRvIHdoYXQgaGFwcGVucyB3aGVuIHlvdSBjaGFuZ2UgdGhlIHdvcmRzIGluIG9uZSBsYW5ndWFnZSBpbnRvIGVxdWl2YWxlbnQgd29yZHMgaW4gYW5vdGhlciBsYW5ndWFnZS4gV2hhdCBwcm9jZXNzIGlzIHRoYXQ/[Qq]

[c]IHRyYW5z bGF0aW9u[Qq]

[f]IEZhYnVsb3NvISBUaGUgcHJvY2VzcyBhdCB3b3JrIGFib3ZlIGlzIHRyYW5zbGF0aW9uLCB3aGljaCB0cmFuc2xhdGVzIFJOQSBzZXF1ZW5jZXMgaW50byBhbWlubyBhY2lkIHNlcXVlbmNlcy4=[Qq]

[c]IHRyYW5zZHVjdGlvbg==[Qq]

[f]IE5vLiBUcmFuc2R1Y3Rpb24gcmVmZXJzIHRvIHNldmVyYWwgcHJvY2Vzc2VzIGluIGJpb2xvZ3ksIHN1Y2ggYXMgd2hlbiBhIHZpcnVzIGJyaW5ncyBETkEgZnJvbSBvbmUgYmFjdGVyaWFsIGNlbGwgdG8gYW5vdGhlciBkdXJpbmcgYSB2aXJhbCByZXBsaWNhdGlvbiBjeWNsZS4gSGVyZSwgaW5mb3JtYXRpb24gaW4gUk5BIGlzIGJlaW5nIGNoYW5nZWQgaW50byBwcm90ZWluLiBJdCYjODIxNztzIGFuYWxvZ291cyB0byB3aGF0IGhhcHBlbnMgd2hlbiB5b3UgY2hhbmdlIHRoZSB3b3JkcyBpbiBvbmUgbGFuZ3VhZ2UgaW50byBlcXVpdmFsZW50IHdvcmRzIGluIGFub3RoZXIgbGFuZ3VhZ2UuIFdoYXQgcHJvY2VzcyBpcyB0aGF0Pw==[Qq]

[q json=”true” xx=”1″ multiple_choice=”true” unit=”6.Gene Expression and Regulation” dataset_id=”2019 AP Bio Dataset|22416e6299905a” question_number=”63″ 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]IE5vLiBUaGUga2V5IHRoaW5nIHRvIG5vdGljZSBpcyB0aGF0IHRoZSBsb25nIGFsbGVsZSBETkEgaGFzIGZvdXIgZnJhZ21lbnRzLCBhbmQgdGhlIHNob3J0IGFsbGVsZSBETkEgaGFzIGZpdmUuIEEgbXV0YXRpb24gdGhhdCBlbGltaW5hdGVkIGEgcmVzdHJpY3Rpb24gc2l0ZSB3b3VsZCBwcm9kdWNlIGZld2VyIGZyYWdtZW50cyBpbiB0aGUgc2hvcnQgYWxsZWxlLiBZb3VyIGpvYiBpcyB0byBleHBsYWluIHdoeSB0aGUgc2hvcnQgYWxsZWxlIGhhcyA=bW9yZQ==IGZyYWdtZW50cy4gSG93IGNvdWxkIHRoYXQgaGFwcGVuPw==[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]

[q json=”true” xx=”1″ multiple_choice=”true” unit=”6.Gene Expression and Regulation” dataset_id=”2019 AP Bio Dataset|21eadfe48e7c5a” question_number=”90″ topic=”6.8.Biotechnology”] Two cloning vectors used in genetic engineering are bacterial viruses and plasmids. Which of the following statements about viruses and plasmids is FALSE?

[c]IFBsYXNtaWRzIHVzZSB0aGUgdHJhbnNsYXRpb24gbWFjaGluZXJ5IG9mIHRoZSBjZWxsLg==[Qq]

[f]IE5vLiBSZW1lbWJlciB0aGF0IHlvdSYjODIxNztyZSBsb29raW5nIGZvciBhIGZhbHNlIHN0YXRlbWVudCwgYW5kIHRoaXMgb25lIGlzIHRydWUuIE9uY2UgeW91IGludHJvZHVjZSBhIHBsYXNtaWQgaW50byBhIGhvc3QgY2VsbCwgaXQgY2FuIGJlIHJlcGxpY2F0ZWQsIHRyYW5zY3JpYmVkLCBhbmQgdHJhbnNsYXRlZCB1c2luZyB0aGUgaG9zdCBjZWxsJiM4MjE3O3MgbW9sZWN1bGFyIG1hY2hpbmVyeS4gUmVtZW1iZXIgdGhhdCwgYW5kIG5leHQgdGltZSBjaG9vc2UgYW5vdGhlciBhbnN3ZXIu[Qq]

[c]IFZpcnVzZXMgaGF2ZSBhIHByb3RlaW4gY2Fwc2lkLg==[Qq]

[f]IE5vLiBSZW1lbWJlciB0aGF0IHlvdSYjODIxNztyZSBsb29raW5nIGZvciBhIGZhbHNlIHN0YXRlbWVudCwgYW5kIHRoaXMgb25lIGlzIHRydWUuIEEgdHlwaWNhbCB2aXJ1cyBoYXMgYSBudWNsZWljIGFjaWQgY29yZSAoY29tcG9zZWQgb2YgRE5BIG9yIFJOQSkgc3Vycm91bmRlZCBieSBhIHByb3RlaW4gY29hdCAob3IgY2Fwc2lkKS4gUmVtZW1iZXIgdGhhdCwgYW5kIG5leHQgdGltZSBjaG9vc2UgYW5vdGhlciBhbnN3ZXIu[Qq]

[c]IFZpcnVzZXMgY2FuIHJlcGxpY2F0ZSBpbiB0aG UgYWJzZW5jZSBvZiBhIGNlbGx1bGFyIGhvc3Qu[Qq]

[f]IEV4Y2VsbGVudC4gWW91JiM4MjE3O3ZlIG9idmlvdXNseSByZW1lbWJlcmVkIHRoYXQgdmlydXNlcyBhcmUgb2JsaWdhdGUgaW50cmFjZWxsdWxhciBwYXJhc2l0ZXM=LCB0aGF0IGRlcGVuZCBvbiB0aGVpciBob3N0IGluIG9yZGVyIHRvIHJlcGxpY2F0ZS4=[Qq]

[c]IFBsYXNtaWRzIGNhcnJ5IGdlbmVzLg==[Qq]

[f]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[Qq]

[c]IFBsYXNtaWRzIGhhdmUgcmVzdHJpY3Rpb24gc2l0ZXMu[Qq]

[f]IE5vLiBSZW1lbWJlciB0aGF0IHlvdSYjODIxNztyZSBsb29raW5nIGZvciBhIGZhbHNlIHN0YXRlbWVudCwgYW5kIHRoaXMgb25lIGlzIHRydWUuIFBsYXNtaWRzIGRvIGhhdmUgcmVzdHJpY3Rpb24gc2l0ZXMgKHNpdGVzIHdoZXJlIHRoZXkgY2FuIGJlIGN1dCBvcGVuIGJ5IGVuenltZXMpLiBUaGF0JiM4MjE3O3MgaG93IHBsYXNtaWRzIGNhbiBiZSBlbmdpbmVlcmVkIHRvIGNvbWJpbmUgd2l0aCBvdGhlciBETkEsIG1ha2luZyByZWNvbWJpbmFudCBETkEgdGVjaG5vbG9neSBwb3NzaWJsZS4gUmVtZW1iZXIgdGhhdCBuZXh0IHRpbWUsIGFuZCBjaG9vc2UgYW5vdGhlciBhbnN3ZXIu[Qq]

[q json=”true” xx=”1″ multiple_choice=”true” unit=”6.Gene Expression and Regulation” dataset_id=”2019 AP Bio Dataset|22c982818d185a” question_number=”25″ topic=”6.8.Biotechnology”] A family pedigree and a series of RFLPs for a rare genetic disease is shown below.

Which individual is a heterozygote?

[c]IDQg[Qq][c]ID Ug[Qq][c]IDg=

Cg==[Qq]

[f]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[Qq]

[f]IEZhYnVsb3VzLiA1IGRvZXNuJiM4MjE3O3QgaGF2ZSB0aGlzIGNvbmRpdGlvbiwgd2hpY2ggaXMgY2F1c2VkIGJ5IGFuIGF1dG9zb21hbCByZWNlc3NpdmUgYWxsZWxlLiA1IG11c3QgYmUgYSBoZXRlcm96eWdvdGUgYmVjYXVzZSBoaXMgUkZMUCBwYXR0ZXJuIHNob3dzIHR3byBiYW5kcyAod2hpY2ggaXMgd2hhdCB5b3UmIzgyMTc7ZCBleHBlY3QgaW4gYSBoZXRlcm96eWdvdGUsIHdpdGggYSBkb21pbmFudCBhbmQgYSByZWNlc3NpdmUgYWxsZWxlKS4=[Qq]

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[q json=”true” xx=”1″ multiple_choice=”true” unit=”6.Gene Expression and Regulation” dataset_id=”2019 AP Bio Dataset|22c651453f7c5a” question_number=”26″ topic=”6.8.Biotechnology”] A family pedigree and a series of RFLPs for a rare genetic disease is shown below.

Which individual is homozygous dominant?

[c]IDEg[Qq][c]IDIg[Qq][c]ID Mg[Qq][c]IDQg[Qq][c]IDU=

Cg==[Qq]

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[f]IE5vLiA1IGlzIHNob3dpbmcgdGhlIGRvbWluYW50IHBoZW5vdHlwZSwgYW5kIHRoZSBjb25kaXRpb24gaXMgY2xlYXJseSByZWNlc3NpdmUgKHBhc3NlZCBvbnRvIG9mZnNwcmluZyBieSBwYXJlbnRzIHdobyBkb24mIzgyMTc7dCBoYXZlIHRoZSBjb25kaXRpb24pLiBZb3UgY2FuIHRlbGwgdGhhdCA1IGlzIGEgaGV0ZXJvenlnb3RlIGJhc2VkIG9uIGhpcyBSRkxQIHBhdHRlcm4uIFRoZSBwYXR0ZXJuIGhhcyB0d28gYmFuZHMsIHdoaWNoIGlzIHdoYXQgeW91JiM4MjE3O2Qgc2VlIGluIGEgaGV0ZXJvenlnb3RlICh0d28gYmFuZHMsIG9uZSBmb3IgZWFjaCBhbGxlbGUpLg==[Qq]
[x][restart]
[/qwiz]