Exam 2015, questions and answers - question bank for professor amro chapter 7-8 PDF

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Question Bank for Professor Amro Chapter 7-8...

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TestBanks Chapter 7: DNA: Structure and Replication

Description Question pool for TestBanks Chapter 7: DNA: Structure and Replication Instructions Add Question Here Multiple Choice

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Question In the classic experiment by Griffith, evidence of the action of a hereditary biomolecule was identified by: Answer

growth of the bacterial cells. transformation (phenotypic change) of the “R strain” by “S strain” biomolecules. rapid death of mice following injection with bacteria. isolation of pure DNA from the “S strain” cells. the ability of heat to destroy deadly biomolecules. Add Question Here

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Question Mendel identified evidence for the location of genes on separate structures (chromosomes) through his: Answer

electrophoresis experiments. purification of chromosomes from pea plants. identification of dominant and recessive alleles of genes. identification of independent assortment of distinct genes during meiosis. use of high resolution microscopic techniques. Add Question Here

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Question Smooth (S) and rough (R) strains of Streptococcus pneumonia are distinct because of their ability to cause illness (death in rodents). This trait is controlled by genes that regulate: Answer

the production of a protective polysaccharide coat (capsule) around each cell. the production of lethal toxins. the speed of bacterial growth. the metabolic rate of microbial glycolysis. the bacterial ability to penetrate through the membrane of animal cells. Add Question Here

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Multiple Answer Question Evidence of transformation of ¿R strain¿ cells by biomolecules from ¿S strain¿ cells includes: Answer

the ¿R strain¿ cells begin to produce high levels of neurotoxin. the transformed ¿R strain¿ cells gain the ability to kill mice. the transformed ¿R strain¿ cells become genetically identical to ¿S strain¿ cells. radioactively labeled DNA is transferred from ¿S¿ to ¿R¿ strains. isolation of smooth-looking cells from animals infected with “R strain” cells that had been incubated with heat-killed “S strain” cells. Add Question Here

Multiple Answer Question Oswald Avery and colleagues strengthened scientific support that DNA was the transforming factor by replicating the Griffith experiment with some important differences in experimental design. The key difference was: Answer

purifying “S strain” DNA and RNA from other biomolecules. systematically eliminating the impact of classes of “S strain” biomolecules using enzymatic digestion before mixing with “R strain” live cells. injecting “S strain” DNA directly into “R strain” cells and observing the change in phenotype. Avery used Escherichia coli rather than Streptococcus as it is easier to transform. all of the above. Add Question Here

Multiple Choice Question Oswald Avery and colleagues strengthened scientific support that DNA was the transforming factor by replicating the Griffith experiment with some important differences in experimental design. They found that the enzyme ______________ was effective at destroying the transforming capacity of ¿S strain¿ biomolecules. Answer

protease (protein destruction) RNase (RNA destruction) DNase (DNA destruction) polysaccharide-destroying enzymes lipase (lipid destruction) Add Question Here

Multiple Choice Question Alfred Hershey and Martha Chase examined transformation using bacteriophage (bacterial DNA virus) and bacterial cells. In this experiment, how is transformation (altered phenotype) is displayed? Answer

The bacteria are transformed (by phage DNA) into virus synthesizing cells. The bacteria are transformed from live to dead by the virus. The bacteria are transformed from rough to smooth by virus. The bacteria are transformed into a virus. Transformation is not exhibited by a virus-infected bacterial cell.

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file:///C:/Users/moh/AppData/Local/Temp/Rar$EX96.960/ch7/CourseC Add Question Here Multiple Choice Question Alfred Hershey and Martha Chase examined transformation using bacteriophage (bacterial DNA virus) and bacterial cells. In these experiments, they used radioactivity to label nucleic acids (in this case DNA) proteins. What do radioactive sulfur and phosphate specifically label? Answer

Sulfur labels DNA, and phosphate labels protein. Sulfur labels both protein and DNA. Phosphate labels both protein and DNA. Sulfur labels protein, and phosphate labels DNA. None of the above. Add Question Here

Multiple Answer Question Alfred Hershey and Martha Chase examined transformation using bacteriophage (bacterial DNA virus) and bacterial cells. Sulfur was used to label cellular components in this experiment because of its presence in the amino acid(s) Answer

histidine. methionine and cysteine. tyrosine. glycine. serine. Add Question Here

Multiple Answer Question Alfred Hershey and Martha Chase examined transformation using bacteriophage (bacterial DNA virus) and bacterial cells. If phage are labeled with radioactive sulfur and allowed to infect bacterial cells, the radioactive sulfur will be localized to: Answer

the inside of infected cells (in phage DNA). the outside of infected cells (in phage ghosts). in newly synthesized phage viruses in the host bacterial cell. No radioactivity will remain after infection. The sulfur will be metabolically consumed, and therefore the radioactivity will be destroyed. Add Question Here

Multiple Answer Question Alfred Hershey and Martha Chase examined transformation using bacteriophage (bacterial DNA virus) and bacterial cells. If phage are labeled with radioactive phosphate and allowed to infect bacterial cells, the radioactive phosphate will be localized to: Answer

the inside of infected cells (in phage DNA). the outside of infected cells (in phage ghosts). in newly synthesized phage viruses in the host bacterial cell. No radioactivity will remain after infection.

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The phosphate will be metabolically consumed, and therefore the radioactivity will be destroyed. Add Question Here Multiple Answer Question Which of the following is/are NOT a key structural property displayed by the heritable (transforming) material? Answer

The structure must be accurately copied before cell division. The structure must contain complex information (genetic code). The structure must be composed of rare/unusual elements. The structure must allow or tolerate small changes that faciliate non-lethal variation. The structure must be enzymatically active, able to mediate cellular metabolism. Add Question Here

Multiple Answer Question Chargaff's rules do not hold for which of the following genome types? Answer

Yeast Bacteria Single-stranded DNA virus Invertebrates

Archaeobacteria Add Question Here Multiple Answer Question Of the three key building blocks of DNA, which type(s) of building block stabilizes the structure via weak van der Waals interactions (through stacking), and regular hydrogen bonds? Answer

Phosphate Deoxyribose Nitrogenouse bases Both phosphate and nitrogenous bases All of the above Add Question Here

Multiple Answer Question Of the three key building blocks of DNA, which type(s) of building block is/are negatively charged and oriented on the outside of the double helical structure? Answer

Phosphate Deoxyribose Nitrogenouse bases Both phosphate and nitrogenous bases All of the above Add Question Here

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Multiple Answer Question Of the three key building blocks of DNA, which type(s) of building block is/are structurally different in RNA molecules? Answer

Phosphate Deoxyribose Nitrogenouse bases Both deoxyribose and nitrogenous bases All of the above Add Question Here

Multiple Answer Question When comparing the three key models of DNA replication, the model that included the synthesis of a brand new double stranded DNA molecule from an original molecule was named: Answer

dispersive replication. conservative replication. semiconservative replication. liberal replication. none of the above. Add Question Here

Multiple Answer Question When comparing the three key models of DNA replication, the model that included the separation of the two strands of the original DNA (template) and using those strands as templates to synthesize two new DNA strands is called: Answer

dispersive replication. conservative replication. semiconservative replication. liberal replication. none of the above. Add Question Here

Multiple Answer Question The Meselson-Stahl experiment made clear predictions regarding experimental outcomes if dispersive, conservative, or semiconservative DNA replication was occurring in their Escherichia coli cells. This experiment enables the detection of ¿new¿ and ¿old¿ DNA by assessing the _______________ of DNA molecules in the cells. Answer

intensity of radioactive labeling density charge distribution along the double helix number of mutations size (length) Add Question Here

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Question The Meselson-Stahl experiment made clear predictions regarding experimental outcomes if dispersive, conservative, or semiconservative DNA replication was occurring in their Escherichia coli cells. To begin the experiment, cells were grown for a period of time in media containing __________ to label the chromosome. Answer

15

heavy nitrogen - N 32

radioactive phosphate - P chemically modified nitrogenous bases only sulfur and no nitrogen none of the above Add Question Here Multiple Answer Question If Escherichia coli, grown for a period of time in

15

N, is transferred to

14

N for one

14

generation of DNA replication, the resulting DNA should have N added to all ¿new¿ DNA. If 14 conservative replication is occurring, the N-containing ¿new¿ DNA will compose _________________. Answer

one strand of all bacterial chromosomes both strands of DNA in half of all bacterial chromosomes regions dispersed throughout all bacterial DNA none of the new DNA, it will only be found in the old DNA all of the above Add Question Here

Multiple Answer Question If Escherichia coli, grown for a period of time in 15N, is transferred to 14N for one 14 generation of DNA replication, the resulting DNA should have N added to all ¿new¿ DNA. If 14 semiconservative replication is occurring, the N-containing ¿new¿ DNA will compose

_________________. Answer

one strand of all bacterial chromosomes both strands of DNA in half of all bacterial chromosomes regions dispersed throughout all bacterial DNA none of the new DNA, it will only be found in the old DNA none of the above Add Question Here

Multiple Answer Question In 1959, Arthur Kornberg isolated DNA polymerase activity fromEscherichia coli cells, revealing its ability to copy DNA by mixing ___________________ in a tube and measuring synthesis of a DNA strand. Answer

dATP, dGTP, dUTP, dCTP, template DNA, and pure DNA polymerase enzyme ATP, GTP, TTP, CTP, template DNA, and pure DNA polymerase enzyme dATP, dGTP, dTTP, dCTP, template RNA, and pure RNA polymerase enzyme dATP, dGTP, dTTP, dCTP, template DNA, and pure DNA polymerase enzyme dATP, dGTP, dTTP, dCTP, template DNA, and cell enzymatic extract Add Question Here

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Multiple Answer Question Examine Figure 7-16, recalling that DNA synthesis by DNA polymerases always occurs in the 5' to 3¿ direction. The predominant location of small Okazaki fragments during DNA replication occurs at the: Answer

replication fork. leading strand of replication. lagging strand of replication. site of primer synthesis by RNA primase. major groove. Add Question Here

Multiple Answer Question The spontaneous (though infrequent) change of a nitrogenous base to a slightly distinct form is called Answer

ketomerization. tautomerization. methylation. mutagenesis. proton exchange. Add Question Here

Multiple Answer Question The replisome contains a protein subunit responsible for unwinding the double helix to enable DNA replication. This subunit/enzyme is named: Answer

pol III holoenzyme. the beta clamp. primase. ligase. helicase. Add Question Here

Multiple Answer Question The replisome contains a protein subunit responsible for attaching free ends of DNA on the newly formed strand. This subunit/enzyme is named: Answer

pol III holoenzyme. the beta clamp. primase. ligase. helicase. Add Question Here

Multiple Answer

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Question Topoisomerase and helicase have distinct functions that include which of the following? Answer

Topoisomerase is responsible for unwinding the double helix (separating strands). Helicase is responsible for unwinding the double helix (separating strands). Helicase relieves supercoiling that occurs in front of the replication fork. The enzymes have nearly identical activities but are located at different sites during DNA replication. These enzymes are nearly identical but have been given distinct names. Add Question Here

Multiple Choice Question The complexity of lagging strand replication is necessary because: Answer

there is room for only a single DNA polymerase III enzyme in the replisome. the helicase can only unwind double helical DNA slowly. as polymerization occurs only in the 5' to 3' direction, the lagging strand must be synthesized in consecutive small fragments. the RNA primase works better on the leading strand DNA. DNA polymerase I is more often associated with the lagging strand. Add Question Here

Multiple Choice Question Initiation of replication occurs at an ¿origin of replication¿ site that typically includes an AT-rich region. Initiation benefits from these AT-rich regions because Answer

adenine-thymine pairs are held together by two H-bonds, making them easier to separate during unwinding. the “AT-rich” region recruits DNA polymerase to begin the process of DNA replication. the “AT-rich” region recruits topoisomerase to begin the process of DNA replication. GC-rich regions are impossible to replicate because of their strong H-bonds. none of the above. Add Question Here

Multiple Choice Question Experiments on chromosome structure and function have shown that in eukaryotic chromatids: Answer

there are at least 50 DNA molecules per chromatid. there is more than one DNA replicating point per chromatid. DNA molecules duplicate conservatively rather than semiconservatively. DNA segragation to daughter chromatids occurs in a dispersive pattern. most DNA synthesis occurs during the M (mitosis) phase of the cell cycle. Add Question Here

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Question In eukaryotic DNA replication, re-association of histones with newly formed DNA is accomplished by: Answer

spontaneous association of histones with DNA. chromatin assembly factor (CAF-1) attaching to the primase enzyme. histones only associating with new DNA after the chromosome is completely replicated. histones not releasing DNA during replication. chromatin assembly factor 1 (CAF-1) and histones binding to the sliding clamp structure. Add Question Here

Multiple Choice Question Cell cycle progress enables the initiation of genome replication by: Answer

activating DNA polymerase III activity directly. providing Cdc6 and Cdt1, which aid assembly of initiation components at the origin of replication. activating histone degradation near the origin of replication. activating expression of the ORC protein and helicase proteins. all of the above. Add Question Here

Multiple Choice Question When replicating the end of a chromosome, the lagging strand cannot copy the last ~10 nucleotides at the end of the chromosome. As a result, chromosomes contain telomere sequences at their ends, which are defined as: Answer

special DNA sequences that do not require priming for replication. non-coding, repetitive sequences that can be copied independent of the replisome. non-coding, repetitive sequences that are not replicated during DNA replication. protein-based structures at the ends of chromosomes that protect the chromosome end. all of the above. Add Question Here

Multiple Choice Question Telomerase activity relies upon ________________ for appropriate priming. Answer

RNA primase a short, telomeric RNA sequence that is carried within its structure a primosome subunit associated with the telomere structure a unique protein-based primer DNA polymerase III holoenzyme Add Question Here

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Question Even before the structure of DNA was elucidated, experimental evidence indicated that the hereditary material had to have three key properties. Name them. Answer 1. Because essentially every cell in the body of an organism has the same genetic makeup, it is crucial that the genetic material be faithfully replicated at every cell division. Thus, the structural features of DNAmust allow faithful replication. 2. Because it must encode the constellation of proteins expressed by an organism, the genetic materialmust have informational content . 3. Because hereditary changes, called mutations, provide the raw material on which evolutionary selection operates, the genetic materialmust be able to change on rare occasions. Add Question Here Essay

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Question If the structure of DNA was, in fact, as proposed by Linus Pauling (Pauling and Corey, 1953), it would be found as a triple helix. Would this arrangement provide the simplest method for genetic material to be passed to daughter cells during mitotic cell division? Explain your answer. Answer A triple helix implies that there are three strands of genetic material. In order for the cell to divide, one strand would have to go to one cell, and two would go to the other. This might be a tricky method of separating the material. The asymmetry of a triple helix does not preclude cell division, but the double-helix structure lends itself to an easier separation of molecules to the daughter cell destinations. Also, base pairing is easier to support when only two bases are involved (though Pauling's model put them toward the outside, so no base pairing was possible). Add Question Here Essay

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Question Imagine it is the early 1900s and the nature of genetic material is not yet known. You believe that proteins (polypeptides) are the most likely candidates for storing heritable information, rather than...