AP Bio unit 12-13

110-Some viruses can be crystallized and their structures analyzed. One such virus is yellow mottle virus, which infects beans. This virus has a single-stranded RNA genome containing about 6300 nucleotides. Its capsid is 25-30 nm in diameter and contains 180 identical capsomeres. If the yellow mottle virus begins its infection of a cell by using its genome as mRNA, which of the following would you expect to be able to measure? • translation rate • transcription rate • formation of new transcription factors • replication rate

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11H-The bulldog ant has a diploid number of two chromosomes. Therefore, following meiosis, each daughter cell will have a single chromosome. Diversity in this species may be generated by mutations and_____ 1. crossing over and independent assortment 2. crossing over 3. nothing else 4. independent assortment

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11J-A man who is an achondroplastic dwarf with normal vision marries a color-blind woman of normal height. The man's father was six feet tall, and both the woman's parents were of average height. Achondroplastic dwarfism is autosomal dominant, and red-green color blindness is X-linked recessive. How many of their daughters might be expected to be color-blind dwarfs? 1. none 2. one out of four 3. half 4. three out of four

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11J-The pedigree in the figure above shows the transmission of a trait in a particular family. Based on this pattern of transmission, the trait is most likely 1. mitochondrial 2. sex-linked dominant 3. autosomal dominant 4. sex-linked recessive

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11K-Hershey and Chase used a DNA-based virus for their work. What would the results have been if they had used an RNA virus? With an RNA virus radioactive RNA would have been in the final pellet. With an RNA virus neither sample would have had a radioactive pellet. With an RNA virus the protein shell would have been radioactive in both samples. With an RNA virus radioactive protein would have been in the final pellet.

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11K-What is a major difference between eukaryotic DNA replication and prokaryotic DNA replication? 1. Prokaryotic chromosomes have a single origin of replication, while eukaryotic chromosomes have multiple origins of replication. 2. DNA polymerases of prokaryotes can add nucleotides to both 3' and 5' ends of DNA strands, while those of eukaryotes function only in the 5' ® 3' direction. 3. Prokaryotic replication does not require a primer. 4. DNA replication in prokaryotic cells is conservative. DNA replication in eukaryotic cells is semi-conservative.

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11L-In an analysis of the nucleotide composition of DNA, which of the following will be found? A+ C = G+ T A = G and C = T A = C G + C = T + A

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11N-Once researchers identified DNA as the unit of inheritance, they asked how information was transferred from the DNA in the nucleus to the site of protein synthesis in the cytoplasm. What is the mechanism of information transfer in eukarotes? Messenger RNA is transcribed from a single gene and transfers information from the DNA in the nucleus to the cytoplasm, where protein synthesis takes place. Transfer RNA takes information from DNA directly to a ribosome, where protein synthesis takes place. Proteins transfer information from the nucleus to the ribosome, where protein synthesis takes place. DNA from a single gene is replicated and transferred to the cytoplasm, where it serves as a template for protein synthesis.

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11N-Poliovirus is an RNA virus of the picornavirus group, which uses its RNA as mRNA. At its 5' end, the RNA genome has a viral protein (VPg) instead of a 5' cap. This is followed by a nontranslated leader sequence, and then a single long protein-coding region (~7000 nucleotides), followed by a poly-A tail. Observations were made that used radioactive amino acid analogues. Short period use of the radioactive amino acids result in labeling of only very long proteins, while longer periods of labeling result in several different short polypeptides. What conclusion is most consistent with the results of the radioactive labeling experiment? The RNA is only translated into a single long polypeptide, which is then cleaved into shorter ones. The RNA is translated into short polypeptides, which are subsequently assembled into large ones. Host cell ribosomes only translate the viral code into short polypeptides. The large radioactive polypeptides are coded by the host, whereas the short ones are coded for by the virus

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11N-Which of the following contradicts the one-gene, one-enzyme hypothesis? A single antibody gene can code for different related proteins, depending on the splicing that takes place post-transcriptionally. A mutation in a single gene can result in a defective protein. Sickle-cell anemia results in defective hemoglobin. Alkaptonuria results when individuals lack a single enzyme involved in the catalysis of homogentisic acid.

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12B-The lactose operon is likely to be transcribed when the cyclic AMP and lactose levels are both high within the cell the cAMP level is high and the lactose level is low there is more glucose in the cell than lactose C there is glucose but no lactose in the cell

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12C-A lack of which molecule would result in a cell's inability to "turn off" genes? •corepressor • promoter •operon • inducer

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12D-The reason for differences in the sets of proteins expressed in a nerve and a pancreatic cell of the same individual is that nerve and pancreatic cells contain different • sets of regulatory proteins •promoters •genes •regulatory sequences

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12D-Using retroviral vectors for gene therapy might increase the patient's risk of developing cancer because they might •integrate recombinant DNA into the genome in ways that misregulate the expression of genes at or near the site of integration • introduce proteins from the virus •not express the genes that were introduced into a patient's cells •not integrate their recombinant DNA into the patient's genome

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12E-One of two major forms of a human condition called neurofibromatosis (NF 1) is inherited as a dominant gene, although it may range from mildly to very severely expressed. Which of the following is the best explanation for why a young, affected child is the first in her family to be diagnosed? 1. One of the parents has a mild expression of the gene 2. The condition skipped a generation in the family. 3. The child has one more chromosome than either of the parents. 4. The mother carries the gene but does not express it.

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Mutant fruit flies that make only one antimicrobial peptide were tested for survival after infection with Neurospora crassa fungi or with Micrococcus lutes bacteria. 12E-The results shown in the graphs support the hypothesis that 1. adding the drosomycin gene to such mutants protects them from death by fungal infection 2. wild-type flies with the full set of genes for antimicrobial peptides are highly susceptible to these infective agents 3. adding the defensin gene to such mutants protects them from death by fungal infection

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S3C-Inheritance patterns cannot always be explained by Mendel's models of inheritance. If a pair of homologous chromosomes fails to separate during meiosis I, select the choice that shows the chromosome number of the four resulting gametes with respect to the normal haploid number (n)? n+1;n+1;n-1; n-1 n+1; n-1; n-1; n-1 n+1; n+1; n; n n+1; n-1; n; n

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111-A black guinea pig crossed with an albino guinea pig produced twelve black offspring. When the albino was crossed with a second black animal, six blacks and six albinos were obtained. What is the best explanation for this genetic situation? 1. Albino is dominant; black is incompletely dominant. 2. Albino is recessive; black is dominant. 3. Albino is recessive; black is codominant. 4. Albino and black are codominant.

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111-Skin color in a certain species of fish is inherited via a single gene with four different alleles. One fish of this type has alleles 1 and 3 (SS3) and its mate has alleles 2 and 4 (S,S4). If each allele confers a unit of color darkness such that S1 has one unit, S2 has two units, and so on, then what proportion of their offspring would be expected to have five units of color? 1. 1/8 2. 1/2 3. 1/4 4. 0

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11F-Somatic cells of roundworms have four individual chromosomes per cell. How many chromosomes would you expect to find in an ovum from a roundworm? 1. a diploid number 2. two 3. eight 4. four

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11G-What is a major difference between mitosis and meiosis I in a diploid organism? 1. Sister chromatids separate in mitosis, while homologous pairs of chromosomes separate in meiosis II. 2. Sister chromatids separate in mitosis, while homologous pairs of chromosomes separate in meiosis I. 3. DNA replication takes place prior to mitosis, but not before meiosis I. 4. Only meiosis I results in daughter cells that contain identical genetic information.

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11G-Which of the following can occur by the process of meiosis but not mitosis? 1. Haploid cells multiply into more haploid cells. 2. Diploid cells form haploid cells. 3. A diploid cell combines with a haploid cell. 4. Haploid cells fuse to form diploid cells.

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11H-Independent assortment of chromosomes is a result of 1. the diverse combination of alleles that may be found within any given chromosome 2. the random way each pair of homologous chromosomes lines up at the metaphase plate during meiosis I 3. the random combinations of eggs and sperm during fertilization 4. the random distribution of the sister chromatids to the two daughter cells during anaphase II

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11H-Which of the following happens at the conclusion of meiosis I? 1. Sister chromatids are separated. 2. Homologous chromosomes of a pair are separated from each other. 3. The chromosome number per cell remains the same. 4. Four daughter cells are formed.

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11J-Gene S controls the sharpness of spines in a type of cactus. Cactuses with the dominant allele, S, have sharp spines, whereas homozygous recessive ss cactuses have dull spines. At the same time, a second gene, N, determines whether or not cactuses have spines. Homozygous recessive n cactuses have no spines at all. If doubly heterozygous SsNn cactuses were allowed to self-pollinate, the F2 would segregate in which of the following ratios? 1. 1 sharp-spined:2 dull-spined:1 spineless 2. 9 sharp-spined:3 dull-spined:4 spineless 3. 1 sharp-spined:1 dull-spined:1 spineless 4. 3 sharp-spined: 1 spineless

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12B-Suppose an experimenter becomes proficient with a technique that allows her to move DNA sequences within a prokaryotic genome. If a researcher moves the repressor gene (lac), along with its promoter, to a position at some several thousand base pairs away from its normal position, which of the following results would be expected? The lac operon will be expressed continuously. The lac operon will function normally. The repressor will no longer bind to the operator. The repressor will no longer bind to the inducer.

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12C - Lactase is the enzyme needed to digest lactose, the sugar found in milk. Most mammals produce lactase when they are young but stop once nursing ends. In humans however, many people continue to produce lactase into adulthood and are referred to as lactase-persistent. Which of the following mutations is most likely to cause lactase persistence in humans? A mutation that turns off the expression of transcription factors that activate the expression of lactase A mutation that increases the binding of transcription factors to the promoter of the lactase gene A nucleotide substitution in the coding region of the lactase gene that interferes with the interaction between lactase and lactose The insertion of a single nucleotide into the lactase gene that results in the formation of a stop codon

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12C-For a repressible operon to be transcribed, which of the following must occur? RNA polymerase must not occupy the promoter, and the repressor must be inactive. RNA polymerase must bind to the promoter, and the repressor must be inactive. RNA polymerase and the active repressor must be present. A corepressor must be present.

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12C-If a researcher moves the promoter for the lac operon to the region between the beta galactosidase (lacZ) gene and the permease (lacY) gene, which of the following would be likely? 1. RNA polymerase will no longer transcribe permease. 2. Beta galactosidase will not be produced. 3. The operon will still transcribe the lacZ and lacY genes, but the mRNA will not be translated. 4. The three structural genes will be expressed normally.

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12D-A research introduces double-stranded RNA into a culture of mammalian cells and can identify its location or that of its smaller subsection experimentally, using a fluorescent probe. When she finds that the introduced strand separates into single-stranded RNAs, what other evidence of this single-stranded RNA piece's activity can she find? The researcher can measure the degradation rate of the remaining single strand. The rate of accumulation of the polypeptide encoded by the target mRNA is reduced. The cell's translation ability is entirely shut down. The amount of miRNA is multiplied by its replication.

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12E-The most commonly occurring mutation in people with cystic fibrosis is a deletion of a single codon. This results in •a nonsense mutation •a base-pair substitution •a polypeptide missing an amino acid •a frameshift mutation

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14A-Which of the following might result in a human zygote with 45 chromosomes? 1. failure of an egg to complete meiosis II 2. an error in either egg or sperm meiotic anaphase 3. incomplete cytokinesis during spermatogenesis after meiosis | 4. failure of the egg nucleus to be fertilized by the sperm

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14B-Imagine that there are twenty-five different species of protists living in a tide pool. Some of these species reproduce both sexually and asexually, and some of them can reproduce only asexually. The pool gradually becomes infested with disease-causing viruses and bacteria. Which species are more likely to thrive in the changing environment? 1. Sexually and asexually reproducing species are equally likely to thrive. 2. the sexually reproducing species 3. the asexually reproducing species

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14B-This is a figure showing transduction, in which viruses that infect bacteria carry genes from one host cell to another. There is a process in sexually-reproducing eukaryotes that uses enzymes similar to the ones which result in the incorporation of the At gene into the host genome shown in the figure. This analogous process in eukaryotes is 1. Fertilization during sexual reproduction. 2. Crossing over during meiosis. 3. Independent assortment during mitosis. 4 Transduction during nondisjunction.

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A B C 11P-The segment of DNA shown in the figure above has restriction sites I and II, which create restriction fragments A, B, and C. Which of the gels produced by electrophoresis shown below best represents the separation and identity of these fragments? 1. C A B 2. C A B 3. A B C 4. B A C

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EV2C-A principal problem with inserting an unmodified mammalian gene into a plasmid and then getting that gene expressed in bacteria is that 1. bacterial RNA polymerase cannot make RNA complementary to mammalian DNA 2. bacteria cannot remove eukaryotic introns 3. prokaryotes use a different genetic code from that of eukaryotes 4. bacteria translate only mRNAs that have multiple messages

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In an experiment, DNA from the linear form of the bacteriophage Lambda was cut into fragments using the restriction enzyme Hind III. Restriction enzymes are isolated from bacteria and cut DNA in specific locations. Hind III cuts the Lambda DNA between the adenine nucleotides on the complimentary strands in a specific sequence, as indicated in the diagram, producing eight different size fragments. These fragments are then separated with an electrical current based on size after the DNA fragments are placed in a porous gel, a process called gel electrophoresis. cut site - S'AAGCTT3' 3'TTCGAA 5' cut site 11P-Select an observation that best describes a correct aspect of the two processes of restriction digest and gel electrophoresis: 1. If an electrical current is not used, eight separate DNA bands would be visible, but they would not be separated as much as when an electrical current is used. 2. When separated on a gel, the pattern of DNA bands will be characteristic of those cut with Hind lIl, different restriction enzymes will not produce these same fragments. 3. Only the restriction enzyme Hind Ill can be used to cut Lambda DNA since restriction enzymes are specific to the type of DNA they can cut. 4.The sequence AAGCTT is found eight times in the Lambda genome and the restriction enzyme Hind Ill finds each location.

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S3B-Human melanocytes produce two types of melanin. Eumelanin produces brown or black pigmentation while pheomelanin produces red or blonde pigmentation. When exposed to ultraviolet light, 1. an increase in pheomelanin produces darker skin and greater protection against the effects of increasing ultraviolet light. 2. the ratio of eumelanin to pheomelanin in melanocytes increases. 3. a decrease in eumelanin produces lighter skin and absorbs the energy from red and yellow wavelengths of sunlight. 4. the concentration of eumelanin melanocytes increases but the concentration of pheomelanin melanocytes decreases.

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S3B-Refer to the accompanying figure. In every case, caterpillars that feed on oak flowers look like oak flowers. In every case, caterpillars that were raised on oak leaves looked like twigs. These results support which of the following hypotheses? 1. The longer day lengths of summer trigger the development of twig-like caterpillars. 2. Differences in diet trigger the development of different types of caterpillars.

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S3E-Several scientific laboratories across the globe are performing research concerning the origin of life on Earth. Which of the following are the most likely results of abiotic experiments testing the potential for hydrogen bonding between various nucleic acids and amino acids? Base your response on our current understanding of Earth's first genetic systems. graph with more DNA-DNA graph with more RNA- amino acid

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SP3B-Acidity in the soil affects the color range of flowers of the same genetic variety. The color ranges from yellow (high pH soil) to blue (low pH soil). Predict the effect of acid rain on the flower color. 1. All green flowers. 2. More blue than yellow flowers. 3. More yellow than blue flowers. 4. Yellow with blue spots and/or blue with yellow spots flowers.

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SP3B-Scientists expose aneuploidy and polyploidy zygotes to radiation that introduces random mutations. Predict which type is more likely to possess at least one copy in its genome of functional genes needed for metabolism. 1. Both have the same chance. 2. Polyploidy 3. Aneuploidy

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SP5D-Plants that are homozygous dominant (BB) and heterozygous (Bb) for the flower color gene are blue. A true-breeding blue plant is crossed with a purple plant. The F1 generation are allowed to self-pollinate. The data table represents the F2 generation. blue: 228 purple: 75 Determine if the null hypothesis should be rejected and if the data supports the alternative hypothesis. 1. The null hypothesis should be rejected and the data does not support the alternative hypothesis. 2. The null hypothesis should be rejected and the data does support the alternative hypothesis. 3. The null hypothesis should not be rejected and the data does support the alternative hypothesis. 4. The null hypothesis should not be rejected and the data does not support the alternative hypothesis.

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111-Suppose two AaBbCc individuals are mated. Assuming that the genes are not linked, what fraction of the offspring are expected to be homozygous recessive for the three traits? 1. 1/16 2. 1/4 3. 1/64 4. 1/8

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11F-After telophase I of meiosis, the chromosomal makeup of each daughter cell is _______. 1. diploid, and the chromosomes are each composed of two chromatids 2. haploid, and the chromosomes are each composed of a single chromatid 3. haploid, and the chromosomes are each composed of two chromatids 4. diploid, and the chromosomes are each composed of a single chromatid

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11F-If a cell has completed meiosis I and is just beginning meiosis Il, which of the following is an appropriate description of its contents? 1. It has half the chromosomes but twice the DNA of the originating cell. 2. It is identical in content to another cell formed from the same meiosis I event. 3. It has half the amount of DNA as the cell that began meiosis. 4. It has one-fourth the DNA and one-half the chromosomes as the originating cell.

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11G-In eukaryotes, genetic information is passed to the next generation by processes that include mitosis or meiosis. Which of the explanations identifies the correct process and supports the claim that heritable information is passed from one generation to another? 1. During mitosis, DNA replication occurs twice within the cell cycle to insure a full set of chromosomes within each of the daughter cells produced. 2. Mitosis, followed by cytokinesis, produces daughter cells that are genetically different from the parent cell, thus insuring variation within the population. 3. In asexual reproduction, a single individual is the sole parent and passes copies of its genes to its offspring without the fusion of gametes. 4. Single-celled organisms can fuse their cells, reproducing asexually through mitosis to form new cells that are not identical to the parent cell.

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11G-What is a major difference between meiosis I and mitosis in a diploid animal? 1. Crossover takes place in meiosis II. 2. Sister chromatids separate in mitosis, and homologues separate in meiosis I. 3. Meiosis Il occurs in a haploid cell, while mitosis occurs in diploid cells. 5. Homologues align on the metaphase plate in meiosis I.

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11G-Which of the following is a true statement about sexual vs. asexual reproduction? 1. In asexual reproduction, offspring are produced by fertilization without meiosis. 2. Asexual reproduction, but not sexual reproduction, is characteristic of plants and fungi. 3. In sexual reproduction, individuals transmit half of their nuclear genes to each of their offspring. 4. Asexual reproduction produces only haploid offspring.

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11H-When homologous chromosomes cross over, what occurs? 1. Maternal alleles are "corrected" to be like paternal alleles and vice versa. 2. Two chromatids get tangled, resulting in one re-sequencing its DNA. 3. Corresponding segments of non-sister chromatids are exchanged. 4. Two sister chromatids exchange identical pieces of DNA.

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11J-What does a frequency of recombination of 50% indicate? 1. All of the offspring have combinations of traits that match one of the two parents. 2. Abnormal meiosis has occurred. 3. The two genes are likely to be located on different chromosomes. 4. The genes are located on sex chromosomes.

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11L-Within a double-stranded DNA molecule, adenine forms hydrogen bonds with thymine and cytosine forms hydrogen bonds with guanine. This arrangement 1. determines the tertiary structure of a DNA molecule 2. allows variable width of the double helix 3. permits complementary base pairing 4. determines the type of protein produced

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11M-Suppose you are provided with an actively dividing culture of E. coli bacteria to which radioactive thymine has been added. What would happen if a cell replicates once in the presence of this radioactive base? Neither of the two daughter cells would be radioactive. All four bases of the DNA would be radioactive. DNA in both daughter cells would be radioactive. One of the daughter cells, but not the other, would have radioactive DNA.

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12A-Which of the following problems with animal cloning might result in premature death of the clones? use of pluripotent instead of totipotent stem cells use of nuclear DNA as well as mtDNA abnormal gene regulation due to variant methylation the indefinite replication of totipotent stem cells

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12B-Suppose an experimenter becomes proficient with a technique that allows her to move DNA sequences within a prokaryotic genome. If the operator is moved to the far end of the operon, past the transacetylase (lacA) gene, which of the following would likely occur when the cell is exposed to lactose? The inducer will no longer bind to the repressor. The operon will never be transcribed. The structural genes will be transcribed continuously. The repressor will no longer bind to the operator.

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12B-Transcription of structural genes in an inducible operon stops when the pathway's product is present starts when the pathway's product is present starts when the pathway's substrate is present occurs continuously in the cell

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12D-Which of the following is one of the technical reasons why gene therapy is problematic? 1. Most cells with an engineered gene do not produce gene product. 2. Cells with transferred genes are unlikely to replicate. 3. Transferred genes may not have appropriately controlled activity. 4. mRNA from transferred genes cannot be translated.

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12E-Hutchinson-Gilford progeria is an exceedingly rare human genetic disorder in which there is very early senility and death, usually from coronary artery disease, at an average age of 13 years. Patients, who look very old even as children, do not live to reproduce. Which of the following represents the most likely assumption? 1. The disorder will increase in frequency in successive generations within a family. 2. Each patient will have had at least one affected grandparent or parent. 3. The disorder may be due to mutation in a single protein-coding gene. 4. The disease is autosomal dominant.

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EV2C-Which of the following statements describes a eukaryotic chromosome? a chromosome with different numbers of genes in different cell types of an organism a single strand of DNA a single linear molecule of double-stranded DNA plus proteins a series of nucleosomes wrapped around two DNA molecules

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S3B-Hydrangea plants of the same genotype are planted in a large flower garden. Some of the plants produce blue flowers and others pink flowers. This can be best explained by which of the following? 1. the allele for blue hydrangea being completely dominant 2. the knowledge that multiple alleles are involved 3. environmental factors such as soil pH 4. the alleles being codominant

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S3C-If cell X enters meiosis, and nondisjunction of one chromosome occurs in one of its daughter cells during meiosis II, what will be the result at the completion of meiosis? 1. Two of the four gametes descended from cell X will be haploid, and two will be diploid. 2. Half of the gametes descended from cell X will be n + 1, and half will be n - 1. 3. 1/4 of the gametes descended from cell X will be n+ 1, 1/4 will be n - 1, and 1/2 will be n. 4. All the gametes descended from cell X will be diploid.

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S3C-The individual with genotype AaBbCCDdEE can make many kinds of gametes. Which of the following is the major reason? 1. the tendency for dominant alleles to segregate together 2. crossing over during prophase I 3. different possible assortment of chromosomes into gametes 4. recurrent mutations forming new alleles

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S3E-Stanley Miller applied heat and electrical sparks to a mixture of simple inorganic compounds such as methane, hydrogen gas, ammonia, and water vapor. What was the main purpose of the available free energy in this experiment? Remove nonreactive elements in the atmosphere. Provide energy for the formation of oxygen gas. Provide energy for the breakdown of the inorganic compounds to form new organic compounds. Evaporate water to produce oxygen gas.

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SP3B-The ability to taste PTC is due to a dominant allele (T). An offspring of a non-taster can taste PTC. Predict the genotype of the other parent if another child is a non-taster. Predict the genotype of the other parent if another child is a taster. First prediction is Tt. Second prediction is tt. First prediction is tt. Second prediction is Tt. First prediction is Tt. Second prediction is Tt. First prediction is TT. Second prediction is t

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110-The first class of drugs developed to treat AIDS, such as AZT, were known as reverse transcriptase inhibitors. They worked because they bonded to the dsDNA genome of the virus in such a way that it could not separate for replication to Occur prevented host cells from producing the enzymes used by the virus to replicate its genome targeted and destroyed the viral genome before it could be reverse transcribed into DNA bonded to the viral reverse transcriptase enzyme, thus preventing the virus from making a DNA copy of its RNA genome

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111-A sexually reproducing animal has two unlinked genes, one for head shape (H) and one for tail length (T). Its genotype is HhTt. Which of the following genotypes is possible in a gamete from this organism? 1. T 2. Hh 3. HhTt 4. НТ

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11J-A woman who has blood type A positive has a daughter who is type O positive and a son who is type B negative. Rh positive is a trait that shows simple dominance over Rh negative. Which of the following is a possible phenotype for the father? 1. AB negative 2. O negative 3. A negative 4. B positive

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11M-DNA contains the template needed to copy itself, but it has no catalytic activity in cells. What catalyzes the formation of phosphodiester bonds between adjacent nucleotides in the DNA polymer being formed? deoxyribonucleotide triphosphates ribozymes ATP DNA polymerase

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11M-In E. coli replication the enzyme primase is used to attach a 5 to 10 base ribonucleotide strand complementary to the parental DNA strand. The RNA strand serves as a starting point for the DNA polymerase that replicates the DNA. If a mutation occurred in the primase gene, which of the following would you expect? Replication would only occur on the leading strand. Replication would only occur on the lagging strand. Replication would not be affected as the enzyme primase in involved with RNA synthesis. Replication would not occur on either the leading or lagging strand.

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11P-Imagine that you compare two DNA sequences found in the same location on homologous chromosomes. On one of the homologs, the sequence is AACTACGA. On the other homolog, the sequence is AACTTCGA. Within a population, you discover that each of these sequences is common. These sequences 1. identify a protein-coding region of a gene 2. do none of the listed actions 3. cause disease 4. contain an SNP that may be useful for genetic mapping

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11P-Sequencing an entire genome, such as that of C. elegans, a nematode, is most important because 1. it allows researchers to use the sequence to build a "better" nematode, which is resistant to disease 2. it allows research on a group of organisms we do not usually care much about 3. a sequence that is found to have no introns in the nematode genome is likely to have acquired the introns from higher organisms 4. a sequence that is found to have a particular function in the nematode is likely to have a closely related function in vertebrates

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11P-Why is it so important to be able to amplify DNA fragments when studying genes? 1. Before amplification, DNA fragments are likely to bind to RNA and no longer be able to be analyzed. 2. Restriction enzymes (endonucleases) cut DNA into fragments that are too small. 3. A clone requires multiple copies of each gene per clone. 4. A gene may represent only a millionth of the cell's DNA.

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12A-Use the accompanying diagram of the key developmental events in the life cycle of the fruit fly Drosophila to select the statement that best describes the timing and coordination of events necessary for normal fruit fly development. Nurse cells in the fertilized egg trigger segmentation in the three different larval stages that occur in the fruit fly's development. Fruit flies have an ordered series of segments in their bodies that are switched during metamorphosis. The male gamete contains cytoplasmic determinants that determine the fate of the embryo and regulate gene expression for metamorphosis. Cytoplasmic determinants localized in the unfertilized egg provide positional information for placement of the anterior-posterior axis in the embryo.

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12E-Refer to the metabolic pathway illustrated above. If A, B, and C are all required for growth, a strain mutant for the gene-encoding enzyme B would be able to grow on medium supplemented with 1. nutrient B only 2. nutrients A and C 3. nutrient A only 4. nutrient C only

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12E-Sex determination in mammals is due to the SRY gene. Which of the following could allow a person with an XX karyotype to develop a male phenotype? 1. a person with an extra autosomal chromosome 2. a person with one normal and one shortened (deleted) X 3. the loss of the SRY gene from an autosome 4. translocation of SRY to a X chromosome

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14A-Choose the best evidence for why the karyotype(s) above results in sterility due to an alteration of chromosome numbers. 1. Karyotype 2 contains one X and one Y sex chromosome. 2. Karyotype 1 contains a missing Y sex chromosome. 3. Karyotypes 1 contains a missing chromosome and 3 contains an odd number of chromosomes. 4. Karyotype 3 contains an odd number of chromosome 12

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14A-Explain how changes in chromosome number results in a new phenotype. 1. Turner's Syndrome results in reduced fertility because some haploid gametes contain 3 sex chromosomes. 2. Triploidy (3n) increases vigor because the additional set of chromosomes contains functional copies of all genes involved in reproduction. 3. Any change in chromosome number results in death of the zygote. 4. Individuals with Trisomy 21 fail to produce haploid gametes, resulting in sterility.

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14A-Of the following human aneuploidies, which is the one that generally has the most severe impact on the health of the individual? 1. 47, XXX 2. 45, X 3. 47, XXY 4. 47, trisomy 21

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14B-How is natural selection related to sexual reproduction as opposed to asexual reproduction? 1. Sexual reproduction allows the greatest number of offspring to be produced. 2. Sexual reproduction utilizes far less energy than asexual reproduction. 3. Sexual reproduction results in the greatest number of new mutations. 4. Sexual reproduction results in many new gene combinations, some of which will lead to differential reproduction.

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14B-Swine are vulnerable to infection by bird flu virus and human flu virus, which can both be present in an individual pig at the same time. When this occurs, it is possible for genes from bird flu virus and human flu virus to be combined. If the human flu virus contributes a gene for Tamiflu resistance (Tamiflu is an antiviral drug) to the new virus, and if the new virus is introduced to an environment lacking Tamiflu, then what is most likely to occur? 1. The Tamiflu-resistance gene will undergo mutations that convert it into a gene that has a useful function in this environment. 2. If the Tamiflu-resistance gene confers no benefit in the current environment, and has no cost, the virus will increase in frequency. 3. The new virus will maintain its Tamiflu-resistance gene, in case of future exposure to Tamiflu. 4. If the Tamiflu-resistance gene involves a cost, it will experience directional selection leading to reduction in its frequency.

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A certain (hypothetical) organism is diploid, has either blue or orange wings as the consequence of one of its genes on chromosome 12, and has either long or short antennae as the result of a second gene on chromosome 19, as shown in the figure. 11F-If a female of this species has one chromosome 12 with a blue gene and another chromosome 12 with an orange gene, and has both number 19 chromosomes with short genes, she will produce which of the following egg types? 1. three-fourths blue short and one-fourth orange short gene eggs 2. only blue short gene eggs 3. only orange short gene eggs 4. one-half blue short and one-half orange short gene eggs

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A researcher found a method she could use to manipulate and quantify phosphorylation and methylation in embryonic cells in culture. 12A-One of her colleagues suggested she try increased methylation of C nucleotides in the DNA of promoters of a mammalian system. Which of the following results would she most likely see? activation of histone tails for enzymatic function decreased chromatin condensation higher levels of transcription of certain genes inactivation of the selected genes

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EV2C-A primary transcript in the nucleus of a eukaryotic cell is _ the functional mRNA, while a primary transcript in a prokaryotic cell is the functional mRNA. the same size as; larger than the same size as; smaller than larger than; smaller than larger than; the same size as

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S3C-A phenotypically normal prospective couple seeks genetic counseling because the man knows that he has a translocation of a portion of his chromosome 4 that has been exchanged with a portion of his chromosome 12. Although his translocation is balanced, he and his wife want to know the probability that his sperm will be abnormal. What is your prognosis regarding his sperm? 1. None will carry the translocation. 2. 1/2 will be normal and the rest will have the father's translocation. 3. All will carry the same translocation as the father. 4. 1/4 will carry the two normal chromosomes, 4 and 12, 1/4 will have only the two translocation chromosomes and no normal chromosomes 4 and 12, and 1/2 will have both normal and translocated chromosomes.

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S3E-Several scientific laboratories across the globe are involved in research concerning the origin of life on Earth. Which of these questions is currently the most problematic and would have the greatest impact on our understanding if we were able to answer it? How can amino acids, simple sugars, and nucleotides be synthesized abiotically? How could polymers involving lipids and/or proteins form membranes in aqueous environments? How can RNA molecules act as templates for the synthesis of complementary RNA molecules? How did RNA sequences come to carry the code for amino acid sequences?

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S3E-The first genetic material on Earth was probably DNA produced by reverse transcriptase from abiotically produced RNA DNA molecules whose information was transcribed to RNA and later translated in polypeptides oligopeptides located within protobionts self-replicating RNA molecules

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SP5D-The dominant allele for stem color in plants is yellow (Y). The recessive allele for stem color is blue (y). You cross a true-breeding yellow stem plant with a blue stem plant. If stem color is an example of incomplete dominance, you would predict a phenotypic ratio of 1. All green in the F2 generation 2. 1 yellow:2 yellow with blue stripes:1 blue in the F2 generation 3. 75% yellow and 25% blue in the F2 generation 4. 1 yellow:2 green:1 blue in the F2 generation

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EV2A-Codons are three-base sequences that specify the addition of a single amino acid. How do eukaryotic codons and prokaryotic codons compare? 1. The translation of codons is mediated by tRNAs in eukaryotes, but translation requires no intermediate molecules such as tRNAs in prokaryotes. 2. Prokaryotic codons usually contain different bases than those of eukaryotes. 4. Prokaryotic codons usually specify different amino acids than those of eukaryotes. 5. Codons are a nearly universal language among all organisms.

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