AP®︎ Biology: Unit 6 Practice Test
Prepare for your quiz, test, or the AP exam with focused practice questions on Unit 6 of AP Biology – Gene Expression and Regulation.
Questions List
Unit 6 (All Topics)
Question 1 Topic 6.1Easy
This question tests the following: IST-1.L
What is the basic structure of DNA?
What You’re Being Tested On:
Explore the learning objectives taken directly from the College Board’s AP® Biology Curriculum. Ensure you’re prepared for the exact topics covered on the AP® exam, in-class tests, and quizzes, and gain confidence in your mastery of the material.
Unit 6: Gene Expression and Regulation
How genetic information flows from DNA to RNA to protein, and how cells control which genes are turned on or off.
Topic 6.1: DNA and RNA Structure
Learning Objective: 6.1.A
Describe the structures involved in passing hereditary information from one generation to the next.
Essential Knowledge: 6.1.A.1
Genetic information is stored in and passed to subsequent generations through DNA molecules and, in some cases, RNA molecules. i. Prokaryotic organisms typically have circular chromosomes. ii. Eukaryotic organisms typically have multiple linear chromosomes that are comprised of DNA. These chromosomes are condensed using histones and associated proteins.
Essential Knowledge: 6.1.A.2
Prokaryotes and eukaryotes can contain plasmids, which are extra-chromosomal circular molecules of DNA.
Learning Objective: 6.1.B
Describe the characteristics of DNA that allow it to be used as hereditary material.
Essential Knowledge: 6.1.B.1
Nucleic acids exhibit specific nucleotide base pairing that is conserved through evolution. i. Purines (guanine and adenine) have a double ring structure. ii. Pyrimidines (cytosine, thymine, and uracil) have a single ring structure. iii. Purines pair with pyrimidines: adenine with thymine (or uracil in RNA) and guanine with cytosine.
Topic 6.2: DNA Replication
Learning Objective: 6.2.A
Describe the mechanisms by which genetic information is copied for transmission between generations.
Essential Knowledge: 6.2.A.1
DNA replication ensures continuity of hereditary information. i. DNA is synthesized in the 5’ to 3’ direction. ii. Replication is a semiconservative process, meaning one strand of DNA serves as the template for a new strand of complementary DNA. iii. Helicase unwinds the DNA strands. iv. Topoisomerase relaxes supercoiling in front of the replication fork. v. DNA polymerase requires RNA primers to initiate DNA synthesis. vi. DNA polymerase synthesizes new strands of DNA continuously on the leading strand and discontinuously on the lagging strand. vii. Ligase joins the fragments on the lagging strand. Exclusion: The names of the steps and particular enzymes involved, excluding DNA polymerase, ligase, RNA polymerase, helicase, and topoisomerase, are beyond the scope of the AP Exam.
Topic 6.3: Transcription and RNA Processing
Learning Objective: 6.3.A
Describe the mechanisms by which genetic information flows from DNA to RNA to protein.
Essential Knowledge: 6.3.A.1
The sequence of the RNA bases, together with the structure of the RNA molecule, determines RNA function. i. Messenger RNA (mRNA) molecules carry information from DNA in the nucleus to the ribosome in the cytoplasm. ii. Distinct transfer RNA (tRNA) molecules bind specific amino acids and have anticodon sequences that base pair with the codons of mRNA. tRNA is recruited to the ribosome during translation to generate the primary peptide sequence based on the mRNA sequence. iii. Ribosomal RNA (rRNA) molecules are functional building blocks of ribosomes.
Essential Knowledge: 6.3.A.2
RNA polymerases use a single template strand of DNA to direct the inclusion of bases in the newly formed RNA molecule. This process is known as transcription.
Essential Knowledge: 6.3.A.3
The enzyme RNA polymerase synthesizes mRNA molecules in the 5’ to 3’ direction by reading the template DNA strand in the 3’ to 5’ direction.
Essential Knowledge: 6.3.A.4
In eukaryotic cells the mRNA transcript undergoes a series of enzyme-mediated modifications. i. The addition of a poly-A tail makes mRNA more stable. ii. The addition of a GTP cap helps with ribosomal recognition. iii. The excision of introns, along with the splicing and retention of exons, generates different versions of the resulting mature mRNA molecule. This process is known as alternative splicing.
Topic 6.4: Translation
Learning Objective: 6.4.A
Explain how the phenotype of an organism is determined by its genotype.
Essential Knowledge: 6.4.A.1
Translation of the mRNA to generate a polypeptide occurs on ribosomes that are present in the cytoplasm of both prokaryotic and eukaryotic cells, as well as the cytoplasmic surface of the rough ER of eukaryotic cells.
Essential Knowledge: 6.4.A.2
In prokaryotic organisms, translation of the mRNA molecule occurs while it is being transcribed.
Essential Knowledge: 6.4.A.3
Translation involves many sequential steps, including initiation, elongation, and termination. The salient features of translation include: i. Translation is initiated when the rRNA in the ribosome interacts with the mRNA at the start codon (AUG, coding for the amino acid methionine). ii. The sequence of nucleotides on the mRNA is read in triplets, called codons. iii. Each codon encodes a specific amino acid, which can be deduced by using a genetic code chart. Many amino acids are encoded by more than one codon. iv. Nearly all living organisms use the same genetic code, which is evidence for the common ancestry of all living organisms. v. tRNA brings the correct amino acid to the place specified by the codon on the mRNA. vi. The amino acid is transferred to the growing polypeptide chain. vii. The process continues along the mRNA until a stop codon is reached. viii. Translation terminates with the release of the newly synthesized protein. Exclusion: The details and names of the enzymes and factors involved in each of these steps are beyond the scope of the AP Exam. Exclusion: Memorization of the genetic code, with the exception of the start codon AUG, is beyond the scope of the AP Exam.
Essential Knowledge: 6.4.A.4
Genetic information in retroviruses is a special case and has an alternate flow of information: from RNA to DNA, made possible by reverse transcriptase, an enzyme that copies the viral RNA genome into DNA. This DNA integrates into the host genome and is transcribed and translated for the assembly of new viral progeny.
Topic 6.5: Regulation of Gene Expression
Learning Objective: 6.5.A
Describe the types of interactions that regulate gene expression.
Essential Knowledge: 6.5.A.1
Regulatory sequences are stretches of DNA that interact with regulatory proteins to control transcription. Some genes are constitutively expressed, and others are inducible.
Essential Knowledge: 6.5.A.2
Epigenetic changes can affect gene expression through reversible modifications of DNA or histones.
Essential Knowledge: 6.5.A.3
The phenotype of a cell or an organism is determined by the combination of genes that are expressed and the levels at which they are expressed. i. Observable cell differentiation results from the expression of genes for tissue-specific proteins. ii. Induction of transcription factors during development results in sequential gene expression. iii. The function and amount of gene products determine the phenotype of organisms.
Learning Objective: 6.5.B
Explain how the location of regulatory sequences relates to their function.
Essential Knowledge: 6.5.B.1
Both prokaryotes and eukaryotes have groups of genes that are coordinately regulated. i. Prokaryotes regulate operons in an inducible or repressible system. ii. In eukaryotes, groups of genes may be influenced by the same transcription factors to coordinately regulate expression.
Topic 6.6: Gene Expression and Cell Specialization
Learning Objective: 6.6.A
Explain how the binding of transcription factors to promoter regions affects gene expression and the phenotype of the organism.
Essential Knowledge: 6.6.A.1
RNA polymerase and transcription factors bind to promoter or enhancer DNA sequences to initiate transcription. These sequences can be upstream or downstream of the transcription start site.
Essential Knowledge: 6.6.A.2
Negative regulatory molecules inhibit gene expression by binding to DNA and blocking transcription.
Learning Objective: 6.6.B
Explain the connection between the regulation of gene expression and phenotypic differences in cells and organisms.
Essential Knowledge: 6.6.B.1
Gene regulation results in differential gene expression and influences cell products and functions.
Essential Knowledge: 6.6.B.2
Certain small RNA molecules have roles in regulating gene expression.
Topic 6.7: Mutations
Learning Objective: 6.7.A
Describe the various types of mutation.
Essential Knowledge: 6.7.A.1
Alterations in a DNA sequence are mutations that can cause changes in the type or amount of the protein produced and the consequent phenotype. DNA mutations can be beneficial, detrimental, or neutral based on the effect or the lack of effect they have on the resulting nucleic acid or protein and the phenotypes that are conferred by the protein. i. Point mutations occur when one nucleotide has been substituted for a different nucleotide. ii. Frameshift mutations occur when one or more nucleotides are inserted or deleted, causing the reading frame to be shifted. iii. Nonsense mutations occur when there is a point mutation that causes a premature stop. iv. Silent mutations occur when the change in the nucleotide sequence has no effect on the amino acid sequence. Exclusion: Knowledge of specific mutations and their effects is beyond the scope of the AP Exam. Illustrative examples: Mutations in the CFTR gene disrupt ion transport and result in cystic fibrosis; Mutations in the MC1R gene give adaptive melanism in pocket mice.
Learning Objective: 6.7.B
Explain how changes in genotype may result in changes in phenotype.
Essential Knowledge: 6.7.B.1
Errors in DNA replication or DNA repair mechanisms as well as external factors, including radiation and reactive chemicals, can cause random mutations in the DNA. i. Whether a mutation is beneficial, detrimental, or neutral depends on the environmental context. ii. Mutations are a source of genetic variation.
Essential Knowledge: 6.7.B.2
Errors in mitosis or meiosis can result in changes in phenotype. i. Changes in chromosome number resulting from nondisjunction often result in new phenotypes caused by triploidy (aneuploidy). ii. Changes in chromosome number often result in disorders with developmental limitations. iii. Alterations in chromosome structure lead to genetic disorders. Exclusion: Knowledge of specific disorders related to changes in chromosome number is beyond the scope of the AP Exam.
Learning Objective: 6.7.C
Explain how alterations in DNA sequences contribute to variation that can be subject to natural selection.
Essential Knowledge: 6.7.C.1
Changes in genotype may affect phenotypes that are subject to natural selection. Genetic changes that enhance survival and reproduction can be selected for by environmental conditions. i. The horizontal acquisitions of genetic information in prokaryotes via transformation (uptake of DNA), transduction (viral transmission of genetic information), conjugation (cell-to-cell transfer of DNA), and transposition (movement of DNA segments within and between DNA molecules) increase genetic variation. ii. Related viruses can recombine genetic information if they infect the same host cell. iii. Reproductive processes that increase genetic variation are evolutionarily conserved and are shared by various organisms. Illustrative examples: Sickle cell anemia.
Topic 6.8: Biotechnology
Learning Objective: 6.8.A
Explain the use of genetic engineering techniques in analyzing or manipulating DNA.
Essential Knowledge: 6.8.A.1
Genetic engineering techniques can be used to analyze and manipulate DNA and RNA. i. Gel electrophoresis is a process that separates DNA fragments by size and charge. ii. During polymerase chain reaction (PCR), DNA fragments are amplified by denaturing DNA, annealing primers to the original strand, and extending the new DNA molecule. iii. Bacterial transformation introduces foreign DNA into bacterial cells. iv. DNA sequencing technology determines the order of nucleotides in a DNA molecule. Typically, these techniques result in a DNA fingerprint that allows for the comparison of DNA sequences from various samples. Exclusion: Knowledge of the details of each of these genetic engineering techniques is beyond the scope of the AP Exam. Illustrative examples: Amplified DNA fragments can be used to identify organisms and perform phylogenetic analysis; Analysis of DNA can be used for forensic identification; Genetically modified organisms include transgenic animals; Gene cloning allows propagation of DNA fragments.
