Biology: The Dynamic Science (MindTap Course List)
4th Edition
ISBN: 9781305389892
Author: Peter J. Russell, Paul E. Hertz, Beverly McMillan
Publisher: Cengage Learning
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Chapter 14, Problem 6TYK
Summary Introduction
Introduction:
Meselson and Stahl experiment explained the replication of DNA (deoxyribonucleic acid) to be semiconservative. The bacteria was grown in a medium containing a heavy isotope of nitrogen and then the bacteria was reproduced in a medium having a light isotope of nitrogen. After each generation, the DNA was centrifuged, in order to separate the DNA molecules on the basis of its density. The result demonstrated a semiconservative replication of the DNA.
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Students have asked these similar questions
The diagram illustrating the polymerase chain reaction (PCR) technique is provided below. How does the number of copies of the DNA region being amplified change at the end of each cycle of the polymerase chain reaction?
Group of answer choices
a. The number of copies triples (or triplicates).
b. The number of copies does not change.
c. The number of copies quadruples (or quadruplicates).
d. The number of copies doubles (or duplicates).
e. The number of copies halves.
Why are Okazaki fragments formed?
A. Okazaki fragments are the result of discontinuous replication due to the supersecondary
structure of the DNA
B. Okazaki fragments are the result of discontinuous replication resulting from the
polymerization that proceeds in the 5'-3' direction.
C. Okazaki fragments are the result of discontinuous replication resulting from the
polymerase having to wait for the primase to add primers so it can proceed in the 5'-3'
direction of synthesis.
D. Okazaki fragments are formed because endonucleases have to correct errors of
replication.
E. Okazaki fragments are formed when the 5'-3' complementary strand is the template for
replication.
In the Meselson-Stahl experiment, what happened after the E. coli was moved to the N14 medium and two rounds of cell division occurred? This result demonstrated that DNA replicated in a semi-conservative manner.
A. The DNA formed one band at the N15 density level and one band at the N14 density level.
B. The DNA formed one band between the N14 and N15 density levels and one band at the N14 density level.
.C. The DNA only formed one band between the N14 and N15 density levels.
Chapter 14 Solutions
Biology: The Dynamic Science (MindTap Course List)
Ch. 14.1 - Prob. 1SBCh. 14.2 - Prob. 1SBCh. 14.2 - Prob. 2SBCh. 14.2 - Prob. 3SBCh. 14.2 - Prob. 4SBCh. 14.3 - What is the importance of complementary base...Ch. 14.3 - Why is a primer needed for DNA replication? How is...Ch. 14.3 - DNA polymerase III and DNA polymerase I are used...Ch. 14.3 - Prob. 4SBCh. 14.4 - Why is a proofreading mechanism important for DNA...
Ch. 14 - Working on the Amazon River, a biologist isolated...Ch. 14 - Prob. 2TYKCh. 14 - Pyrimidines built from a single carbon ring are:...Ch. 14 - Which of the following statements about DNA...Ch. 14 - Which of the following statements about DNA is...Ch. 14 - Prob. 6TYKCh. 14 - Prob. 7TYKCh. 14 - Prob. 8TYKCh. 14 - Prob. 9TYKCh. 14 - Prob. 10TYKCh. 14 - Discuss Concepts Eukaryotic chromosomes can be...Ch. 14 - Prob. 12TYKCh. 14 - Prob. 13TYKCh. 14 - Discuss Concepts During replication, an error...Ch. 14 - Design an Experiment Design an experiment using...Ch. 14 - Prob. 16TYKCh. 14 - Prob. 1ITDCh. 14 - Prob. 2ITD
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- In the procedure for isolating DNA, what is the purpose of adding washing-up liquid? 1.Washing up liquid contains detergents, that help to break down lipids in cell walls, thus freeing the DNA 2.Washing up liquid contains detergents, that help to break down lipids in cell membranes, including the nuclear double envelope, thus freeing the DNA 3.Washing up liquid contains salts, that help to break down lipids and proteins in cell membranes, including the nuclear double envelope, thus freeing the DNA 4.Washing up liquid contains detergents, that help to break down proteins in cell membranes, including the nuclear double envelope, thus freeing the DNAarrow_forwardWhich of the sequences below would serve as a PCR primer that would bind this DNA strand: 5'- AAATTTGGGCCCTTTGGGAAACCC-3, and lead to successful elongation? Select one: O a. 5'-GGGTTTCCC-3" O b.3'-GGGTTTCCC-5" O c. 5'-AAATTTGGG-3 O d.3'-AAATTTGGG-5 O e. 5'-CCCAAAGGG-3"arrow_forwardWhich of the following is the most accurate description of the E.coli replisome? a. Clamp loader positions two Pol III core polymerases on the leading strand and one Pol III core polymerase on the lagging strand b. Single strand template DNA forms two loops on the lagging strand while two Pol III core polymerases synthesize Okazaki fragments c. Okazaki fragments are synthesized in the opposite direction to movement of the replication fork d. Clamp loader only connects with Pol III core polymerases associated with the lagging strand e. Both b. and c. are accuratearrow_forward
- Suppose the experiment of Meselson and Stahl was performed on a sample of 8 cells, each containing one copy of its circular double-stranded DNA genome, and that had been growing on normal 14N medium. You then grew the cells for 3 generations in medium containing 15N. The outcome would be A) 8 cells with single-stranded DNA molecules with 14N, and 24 cells with single-stranded DNA molecules with 15N. B) 16 cells with double-stranded DNA molecules with equal amounts of 14N and 15N, and 48 cells with double-stranded DNA molecules with 15N. C) 8 cells with double-stranded DNA molecules with equal amounts of 14N and 15N, and 24 cells with double-stranded DNA molecules with 15N. D) 8 cells with double-stranded DNA molecules with equal amounts of 14N and 15N, and 32 cells with double-stranded DNA molecules with 15N. E) 65 cells with single-stranded DNA molecules with 15N.arrow_forwardwhat is the difference between human DNA and strawberry DNA?arrow_forwardIn the following drawing, the top strand is the template DNA, and the bottom strand shows the lagging strand prior to the action of DNA polymerase I. The lagging strand contains three Okazaki fragments. The RNA primers have not yet been removed. A. Which Okazaki fragment was made first, the one on the left or the one on the right? B. Which RNA primer will be the first one to be removed by DNA polymerase I, the primer on the left or the primer on the right? For this primer to be removed by DNA polymerase I and for the gap to be filled in, is it necessary for the Okazaki fragment in the middle to have already been synthesized? Explain. C. Let’s consider how DNA ligase connects the left Okazaki fragment with the middle Okazaki fragment. After DNA polymerase I removes the middle RNA primer and fills in the gap with DNA, where does DNA ligase function? See the arrows on either side of the middle RNA primer. Is ligase needed at the left arrow, at the right arrow, or both?arrow_forward
- Which of the following statements BEST DESCRIBES the main findings of the Meselson-Stahl experiment? A. DNA can be separated using centrifugation B. The semiconservative model of DNA replication is more accurate than the dispersive or conservative models of DNA replication C. Using 14N in experiments is an effective way of tracking nitrogen molecules D. Bacteria grown in the presence of a heavier nitrogen isotope (15N) will replicate at a slower rate than those that utilise a lighter nitrogen isotope (14N) E. Both strands of each new DNA double helix are brand new and synthesized from individual nucleotidesarrow_forwardYou have two tubes, each with a pair of DNA fragments inside them. Tube #1 has fragments that are 500bp and 1000 bp in length. Tube #2 has fragments that are 7500bp and 8000bp in size. If you were to perform agarose gel electrophoresis and run the contents of each tube in two separate lanes on the same gel, what would you expect to see? O That the difference between the distances migrated by the two fragments in Tube #1 was much greater than the difference between the distances migrated by the two fragments in Tube #2. O That the difference between the distances migrated by the two fragments in Tube #1 was the same as difference between the distances migrated by the two fragments in Tube #2. O That the difference between the distances migrated by the two fragments in Tube #1 was much less than the difference between the distances migrated by the two fragments in Tube #2. O It is not possible to estimate what we would expect to see.arrow_forwardA researcher sequences the genome of a variety of bacterial and eukaryotic cells. She finds that the bacterial genome is smaller, but that there are more genes for a given number of base pairs in the eukaryotic cells. In other words, there are fewer genes per unit of length of DNA in the eukaryotic cells. What do you predict she will find if she examines the DNA more closely? A. All of the bacterial DNA consists of coding sequences, but this is not true of the eukaryotic DNA. B. There are more repetitive sequences in the eukaryotic DNA than in the bacterial DNA. C. There are densely packed genes in the eukaryotic DNA that were not immediately distinguishable during the first analysis. D. The bacteria have larger quantities of noncoding DNA than the eukaryotic cells.arrow_forward
- After two generations of replication in the Meselson and Stahl experiment, what was the composition of the two bands? One band was all 14N and one band was half 14N and half 15N. One band was all 15N and one band was half 14N and half 15N. One band was all 14N and one band was all 15N. One band was all 15N and one band was one quarter 14N and three quarters 15N. One band was all 14N and one band was one quarter 14N and three quarters 15N.arrow_forwardMeselson-Stahl Experiment showed that DNA replication is semi-conservative. In the experiment, DNA was originally labeled with ¹5N. Then ¹5N-labeled DNA was duplicated in the presence of ¹4N- nucleotides. It was found that the second- generation daughter DNA (after two rounds of replication) consists of all ¹4N-DNA O b. 50% of hybrid 15N-¹4N DNA and 50% ¹5N-DNA all 15N-DNA d. 50% of hybrid ¹5N-¹4N DNA and 50% ¹4N-DNA 15 all hybrid ¹5N-14N DNA a. C. e.arrow_forwardWhat is true of this figure? (can be multiple answers) a. the replication fork is asymmetrical b. the DNA strands are anti-parallel c. one strand runs 3’ to 5’ d. one strand runs 5’ to 3’ e. both strands have identical basesarrow_forward
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