Unit 6 Review

AP Bio Review Unit 6: Gene Expression and Regulation Multiple Choice Practice Total: 50 /57 1. When DNA replicates, each strand of the original DNA molecule is used as a template for the synthesis of a second, complementary strand. Which of the following figures most accurately illustrates enzyme-mediated synthesis of new DNA at a replication fork? a. b. c. d. 2. The human TPM1 gene encodes members of the tropomyosin family of cytoskeletal proteins. Which of the following best explains how different proteins can be made in different cell types from the one TPM1 gene? a. Different introns are selectively converted to exons. b. Different exons are retained or spliced out of the primary transcript. c. The GTP cap is selectively added to and activates different exons. d. Different portions of the primary transcript remain bound to the template DNA. 3. The first diagram below shows the levels of mRNA from two different genes (bicoid and caudal) at different positions along the anterior-posterior axis of a Drosophila egg immediately before fertilization. The second diagram shows the levels of the two corresponding proteins along the anterior-posterior axis shortly after fertilization. Which of the following conclusions is best supported by the data? a. Bicoid protein inhibits translation of caudal mRNA. b. Bicoid protein stabilizes caudal mRNA. c. Translation of bicoid mRNA produces caudal protein.

d. Caudal protein stimulates development of anterior structures. Use the following information to answer questions 4 & 5: The following figures display data collected while studying a family, some members of which have sickle-cell disease—a rare genetic disorder caused by a mutation in the hemoglobin beta gene (HBB). There are at least two alleles of the HBB gene: the HbA allele encodes wild-type hemoglobin and the HbS allele encodes the sickle-cell form of hemoglobin. Genetic testing provided insight into the inheritance pattern for sickle-cell disease. Figure 1. Pedigree of a family with affected individuals. Squares represent males, circles represent females, shaded symbols represent individuals with sickle-cell disease. 5' CTG ACT CCT GAG GAG AAG TCT 3' Non-template Strand 3' GAC TGA GGA CTC CTC TTC AGA 5' Template Strand Figure 2. A portion of the DNA sequence from the wild-type hemoglobin allele (HbA) that codes for normal hemoglobin. Figure 3. Codon table showing nucleotide sequences for each amino acid. Figure 4. Image of a gel following electrophoretic separation of DNA fragments of the HBB gene from three individuals in the pedigree in Figure 1.

a. b. c. d. 8. Mutations in the MYO6 and POU4F3 genes have been associated with a form of hereditary hearing loss in humans. Researchers studying the genes have proposed that POU4F3 encodes a transcription factor that influences the regulation of MYO6. Which of the following questions will best help guide the researchers toward a direct test of their proposal? a. Have mutations in other genes also been associated with hearing loss? b. In what types of cells are the mutant forms of the POU4F3 gene expressed? c. Are mutations in the MYO6 and POU4F3 genes also found in mice? d. Do mutations in the POU4F3 gene affect MYO6 mRNA levels in cells? 9. Sickle-cell anemia results from a point mutation in the HBB gene. The mutation results in the replacement of an amino acid that has a hydrophilic R-group with an amino acid that has a hydrophobic R-group on the exterior of the hemoglobin protein. Such a mutation would most likely result in altered a. properties of the molecule as a result of abnormal interactions between adjacent hemoglobin molecules b. DNA structure as a result of abnormal hydrogen bonding between nitrogenous bases c. fatty acid structure as a result of changes in ionic interactions between adjacent fatty acid chains d. protein secondary structure as a result of abnormal hydrophobic interactions between R-groups in the backbone of the protein 10.Cystic fibrosis is a recessively inherited disorder that results from a mutation in the gene encoding CFTR chloride ion channels located on the surface of many epithelial cells. As shown in the figure, the mutation prevents the normal movement of chloride ions from the cytosol of the cell to the extracellular fluid. As a consequence of the mutation, the mucus layer that is normally present on the surface of the cells becomes exceptionally dehydrated and viscous.

An answer to which of the following questions would provide the most information about the association between the CFTR mutation and the viscous mucus? a. Is the mucus also secreted from the cells through the CFTR proteins? b. How does the disrupted chloride movement affect the movement of sodium ions and water by the cell? c. How does the mutation alter the structure of the CFTR proteins? d. What is the change in nucleotide sequence that results in the CFTR mutation? 11.The processes illustrated in the models depicted below all result in which of the following? a. Transcription b. An increase in genetic variation c. An increase in the chromosome number d. Horizontal gene transfer transduction, conjugation, and transformation 12.A new mutation that arose in one copy of gene X in a somatic cell resulted in the formation of a tumor. Which of the following pieces of evidence best describes how the new mutation directly caused the tumor? a. Protein X normally stimulates cell division, and the mutation created an overactive version of protein X. b. Protein X normally activates a growth hormone receptor, and the mutation decreased the stability of protein X. c. Protein X normally prevents passage through the cell cycle, and the mutation created an overactive version of protein X. d. Protein X normally regulates gene expression, and the mutation created an underactive version of protein X that blocked the cell cycle. Use the following information for questions 13-17: In a transformation experiment, a sample of E. coli bacteria was mixed with a plasmid containing the gene for resistance to the antibiotic ampicillin (amp r ). Plasmid was not added to a second sample. Samples were plated on nutrient agar plates, some of which were supplemented with the antibiotic ampicillin. The results of E. coli growth are summarized below. The shaded area represents extensive growth of bacteria; dots represent individual colonies of bacteria.

In a classic experiment from the 1970s investigating gene expression, a solution containing equal amounts of rabbit a-hemoglobin mRNA and b-hemoglobin mRNA, which encode subunits of a protein found in red blood cells, was injected into newly fertilized frog eggs. The injected mRNA was not degraded during the course of the experiment. Tadpoles that developed from the injected eggs were dissected into two fragments, one containing predominantly the notochord, muscle tissue, and nerve tissue and the other containing predominantly the other tissue types. Equal amounts of total protein were analyzed after separation by electrophoresis to identify the relative amounts of the different proteins present in each sample. The thickness of the bands indicates the relative amounts of rabbit α -hemoglobin, rabbit β -hemoglobin, and frog tubulin (a cytoskeletal protein that is expressed at relatively constant levels in all tissues) present in each tadpole sample. The experimental protocol and results are summarized in the figure below. 18.The electrophoresis results best support which of the following conclusions? a. Cell specialization during development results in some cells losing the ability to synthesize proteins. b. Cells from different tissues share a common ability to use genetic material from a foreign source to produce protein. c. In comparison with other cells, nerve cells have a superior ability to produce cytoskeletal proteins. d. Muscle cells produce more b-hemoglobin than do cells from the other tissues in a tadpole. 19.Which of the following conclusions is most consistent with the results of the experiment? a. Rabbit mRNA is composed of nucleotides that are absent from frog mRNA. b. A larger volume of blood circulates through a rabbit than through a frog. c. The subunits of hemoglobin differ in size, shape, or charge. d. Synthesis of β -hemoglobin occurs at a faster rate in muscle cells than in other body cells. 20.Given that equal amounts of the different mRNAs were injected into fertilized frog eggs, which of the following conclusions is most consistent with the electrophoresis results? (the beta band is larger, like thicker, which means that it was translated more efficiently) a. β -hemoglobin mRNA is translated more efficiently than is α -hemoglobin mRNA. b. α -hemoglobin is present only in cells where β -hemoglobin is absent. c. α -hemoglobin mRNA is more stable than β -hemoglobin mRNA. d. Tubulin inhibits translation of hemoglobin mRNA. 21.How does the DNA in a prokaryote differ from a eukaryote?

a. Prokaryote has circular DNA, Eukaryotes has linear DNA b. Prokaryote has single stranded DNA, Eukaryotes has double stranded DNA c. Prokaryote has linear DNA, Eukaryotes has prokaryote DNA d. Prokaryote has double stranded DNA, Eukaryotes has single stranded DNA 22.Both eukaryotes and prokaryotes have plasmids. What are plasmids? a. large extra-chromosomal, double-stranded, linear DNA molecules b. large extra-chromosomal, double-stranded, circular DNA molecules c. small extra-chromosomal, double-stranded, circular DNA molecules d. small extra-chromosomal, double-stranded, linear DNA molecules 23.Identify which of the following best describes purines and pyrimidines. a. Purines (C, T, & U) have single ring, Pyrimidines (A & G) have double ring b. Purines (C, T, & U) have double ring, Pyrimidines (A & G) have single ring c. Purines (A & G) have single ring, Pyrimidines (C, T, & U) have double ring d. Purines (A & G) have double ring, Pyrimidines (C, T, & U) have single ring 24.Which of the following demonstrates base pairing rules to 5' - CAGGT - 3' a. 3' - TGGAC - 5' b. 3' - GTCCA - 5' c. 5' - GTCCA - 3' d. 5' - CAGGT - 3' 25.Which enzyme is responsible for relaxing supercoiling in front of the replication fork? a. helicase b. DNA polymerase c. topoisomerase d. ligase 26.Which enzyme is responsible for unwinding the DNA strands? a. helicase b. DNA polymerase c. topoisomerase d. ligase 27.Which enzyme is requires RNA primers to initiate DNA synthesis? a. helicase b. DNA polymerase c. topoisomerase d. ligase 28.Which enzyme is responsible for joining the fragments on the lagging strand? a. helicase b. DNA polymerase c. topoisomerase d. ligase 29.Describe the difference between the synthesis of the leading and lagging strands. a. directionality: leading = 3' -> 5', lagging = 5' -> 3' b. synthesis continuity: leading = discontinuous & lagging = continuous c. directionality: leading = 5' -> 3', lagging = 3' -> 5' d. synthesis continuity: leading = continuous & lagging = discontinuous 30.How do the three types of RNA work together? a. mRNA carries message, tRNA carries amino acids, and rRNA makes ribosome b. mRNA makes ribosomes, tRNA carries amino acids, and rRNA relays information c. mRNA carries amino acids, tRNA transfer message, and rRNA makes ribosome d. mRNA is made, tRNA holds the DNA in place, and rRNA removes the deoxyribose 31.Which of the following are NOT post-transcriptional modifications? a. 5' cap b. intron removal c. poly-A tail d. TATA box ( a type of promoter sequence where transcription begins)

41. Which of the following genetic engineering techniques is incorrectly described? a. Electrophoresis separates molecules based on size & charge b. Bacterial transformation introduces DNA to bacterial cells c. PCR allows for DNA fragments to be stabilized for engineering processes d. DNA sequencing determines the order of nucleotides in a DNA molecule Free Response Practice 1. The yeast Saccharomyces cerevisiae is a single-celled organism. Amino acid synthesis in yeast cells occurs through metabolic pathways, and enzymes in the synthesis pathways are encoded by different genes. The synthesis of a particular amino acid can be prevented by mutation of a gene encoding an enzyme in the required pathway. A researcher conducted an experiment to determine the ability of yeast to grow on media that differ in amino acid content. Yeast can grow as both haploid and diploid cells. The researcher tested two different haploid yeast strains (Mutant 1 and Mutant 2), each of which has a single recessive mutation, and a haploid wild-type strain. The resulting data are shown in Table 1. (a) Identify the role of treatment I in the experiment. It is the control group to test the viability of the yeast strains so that the scientists can know that the changes in the medium are what affects the growth in the strains.. (b) Provide reasoning to explain how Mutant 1 can grow on treatment I medium but cannot grow on treatment III medium. Mutant one uses methionine while it is present in treatment one because all the amino acids are present but it cannot grow on treatment 3 because of the lack of methionine. There is a mutation that prevents it from synthesizing the methionine.

(c) Yeast mate by fusing two haploid cells to make a diploid cell. In a second experiment, the researcher mates the Mutant 1 and Mutant 2 haploid strains to produce diploid cells. Using the table provided, predict whether the diploid cells will grow on each of the four media. Use a plus sign (+) to indicate growth and a minus sign (-) to indicate no growth. There is growth in all four cells (+) . Or original answer might be correct too, (+, -, +, +) 2. A researcher is studying patterns of gene expression in mice. The researcher collected samples from six different tissues in a healthy mouse and measured the amount of mRNA from six genes. The data are shown in Figure 1. (a) Based on the data provided, identify the gene that is most likely to encode a protein that is an essential component of glycolysis. Provide reasoning to support your identification. Gene G because it is the only gene that is expressed in all six tissues and glycolysis occurs in all of the six tissues.

P450 and Abc8 deletion and only Cps present means that the insecticide will not be detoxified and it will remain inside the cell which lowers the chances of survival for bedbugs. However if only Cps is deleted and P450 and Abc8 are present then, the insecticide will be detoxified and moved out of the cell which will result in a higher chance of survival of the bedbugs. Based on the data, strain 5 has a very low percent of survival while strain 4, compared to strain 5 has a significantly higher percent survival. 4. Gibberellin is the primary plant hormone that promotes stem elongation. GA 3-beta-hydrozylase ( GA3H ) is the enzyme that catalyzes the reaction that converts a precursor of gibberellin to the active form of gibberellin. A mutation in the GA3H gene results in a short plant phenotype. When a pure-breeding tall plant is crossed with a pure-breeding short plant, all offspring in the F 1 generation are tall. When the F 1 plants are crossed with each other, 75 percent of the plants in the F 2 generation are tall and 25 percent of the plants are short. (a) The wild-type allele encodes a GA3H enzyme with alanine (Ala), a nonpolar amino acid, at position 229. The mutant allele encodes a GA3H enzyme with threonine (Thr), a polar amino acid, at position 229. Describe the effect of the mutation on the enzyme and provide reasoning to support how this mutation results in a short plant phenotype in homozygous recessive plants. The mutation will result in a change in shape and structure of the protein. The mutation eliminates the gibberellin production, which promotes stem elongation, so it results in short plant phenotypes. (b) Using the codon chart provided, predict the change in the codon sequence that resulted in the substitution of alanine for threonine at amino acid position 229. The A nucleotide was substituted in for the G nucleotide. (c) Describe how individuals with one (heterozygous) or two (homozygous) copies of the wild-type GA3H allele can have the same phenotype. A lot of gibberellin is produced from just one dominant allele of the wild type GA3H. The dominant allele suppresses the recessive allele in the heterozygous copy.

5. A comet assay is a technique used to determine the amount of double-stranded breaks in DNA (DNA damage) in cells. The nucleus of an individual cell is placed on a microscope slide coated with an agarose gel. An electric current is applied to the gel that causes DNA to move (electrophoresis), and the DNA is stained with a fluorescent dye. When viewed using a microscope, undamaged DNA from the nucleus appears as a round shape (the head), and the fragments of damaged DNA extend out from the head (the tail). The length of the tail corresponds to the amount of the damage in the DNA (see Figure 1). (a) To explain the movement of DNA fragments in the comet assay, identify one property of DNA and provide reasoning to support how the property contributes to the movement during the comet assay technique. DNA has a negative charge so in gel electrophoresis it moves towards the positive end. The negative charge is attracted to the positive end. (b) In a different experiment, cells are treated with a chemical mutagen that causes only nucleotide substitutions in DNA. Predict the likely results of a comet assay from this treatment. There will be only a head and no tail.