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Heavy Isotope Analysis of DNA Replication A researcher switches a culture of \(E\). coli growing in a medium containing \({ }^{15} \mathrm{NH}_{4} \mathrm{Cl}\) to a medium containing \({ }^{14} \mathrm{NH}_{4} \mathrm{Cl}\) for three generations (an eightfold increase in population). What is the molar ratio of hybrid DNA \(\left({ }^{15} \mathrm{~N}^{-14} \mathrm{~N}\right)\) to light DNA \(\left({ }^{14} \mathrm{~N}^{-14} \mathrm{~N}\right)\) at this point?

Short Answer

Expert verified
The molar ratio of hybrid DNA to light DNA is 1:7.

Step by step solution

01

Initial Conditions

Initially, the E. coli population is grown in \(^ {15} N\), so all the DNA is \(^ {15} N - ^ {15} N\). Upon switching to \(^ {14} N\), DNA replication results in hybrid \(^ {15} N - ^ {14} N\) DNA.
02

First Generation

During the first generation in \(^ {14} N\), each \(^ {15} N - ^ {15} N\) DNA molecule will replicate to produce one \(^ {15} N - ^ {14} N\) molecule and one \(^ {14} N - ^ {14} N\) molecule.
03

Second Generation

In the second generation, \(^ {15} N - ^ {14} N\) molecules replicate to produce one hybrid \(^ {15} N - ^ {14} N\) DNA and one light \(^ {14} N - ^ {14} N\) DNA for each existing hybrid molecule. \(^ {14} N - ^ {14} N\) molecules only produce light DNA.
04

Third Generation

In the third generation, each hybrid \(^ {15} N - ^ {14} N\) molecule produces one hybrid and one light DNA molecule. Each light DNA ((^ {14} N - ^ {14} N)) produces two light DNA molecules when replicated.
05

Calculate Molar Ratio

At the end, you have 1/8 hybrid ( (^ {15} N - ^ {14} N )) DNA (1 hybrid initially that doubles each generation) and 7/8 light ( (^ {14} N - ^ {14} N )) DNA. Thus, the molar ratio is 1 part hybrid to 7 parts light.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Heavy Isotope Analysis
Heavy isotope analysis is a crucial technique in understanding DNA replication. It involves using isotopes of different masses to trace biological processes, such as the incorporation of nitrogen isotopes into DNA during bacterial growth. In this specific exercise, the heavier isotope \({ }^{15} \mathrm{N}\) is initially incorporated into E. coli's DNA by growing the bacteria in a medium containing \({ }^{15} \mathrm{NH}_{4} \mathrm{Cl}\). When the medium is switched to \({ }^{14} \mathrm{NH}_{4} \mathrm{Cl}\)—a lighter isotope—the resulting effect on the DNA provides insights into the replication mechanism.
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Heavy isotope analysis allows scientists to distinguish between newly synthesized DNA and the original. This differentiation is possible because heavier isotopes like \({ }^{15} \mathrm{N}\) make the DNA denser, which can then be separated using centrifugation techniques. By analyzing the ratios of hybrid DNA that incorporate both heavy and light isotopes, researchers can determine the dynamics of DNA replication.
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This method is extremely helpful in educational settings to illustrate the semi-conservative nature of DNA replication, where each new DNA helix consists of one old strand and one new strand.
E. coli Growth
Understanding E. coli growth is essential when studying DNA replication and heavy isotope analysis. E. coli is a common bacterium used in molecular biology due to its rapid growth and simple dietary needs. When analyzing DNA replication, E. coli provides a convenient organism for observing generational DNA changes.
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The growth occurs in phases, beginning with a lag phase where the bacteria acclimate to the new environment. Following is an exponential phase where cells divide rapidly, making it an ideal time to observe DNA replication. When the E. coli switch from a \({ }^{15} \mathrm{N}\) medium to a \({ }^{14} \mathrm{N}\) medium, it grows for three generations. Each generation represents a complete cycle of DNA replication, doubling the bacterial population.
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The sequential incorporation of lighter nitrogen from the \({ }^{14} \mathrm{NH}_{4} \mathrm{Cl}\) medium into the DNA allows researchers to separate hybrid and light DNA after several generations. This growth pattern helps demonstrate the transition from heavy to hybrid to light DNA over successive replications.
Hybrid DNA
Hybrid DNA is a key concept in demonstrating DNA replication through heavy isotope analysis. It refers to DNA molecules containing one strand with the heavy isotope \({ }^{15} \mathrm{N}\) and one strand with the lighter isotope \({ }^{14} \mathrm{N}\).
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At the beginning of the experiment, all bacterial DNA consists of \({ }^{15} \mathrm{N}^{-15} \mathrm{N}\) strands. When the medium switches to the lighter isotope, the bacteria reproduce, resulting in new DNA strands incorporating \({ }^{14} \mathrm{N}\).
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After one generation, each original \({ }^{15} \mathrm{N}^{-15} \mathrm{N}\) strand pairs with a \({ }^{14} \mathrm{N}\) strand, forming hybrid \({ }^{15} \mathrm{N}^{-14} \mathrm{N}\) DNA. With every subsequent generation, the amount of hybrid DNA diminishes as light DNA becomes more prevalent. The presence of hybrid DNA at certain stages of bacterial growth is instrumental in confirming the semi-conservative replication model. This model illustrates how each strand serves as a template for a new complementary strand, maintaining genetic continuity.
Molecular Biology Education
Incorporating real-world examples, like heavy isotope analysis and hybrid DNA, significantly enhances molecular biology education. These concepts make abstract ideas like DNA replication more tangible and relatable to students.
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Using methods such as the Meselson-Stahl experiment—integral to understanding DNA replication—educators can vividly demonstrate how heavy and light isotopes help identify the mechanics of DNA synthesis. Through these experiments, students observe how DNA strands separate, replicate, and produce hybrid and light DNA, fostering a deeper understanding.
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Such educational approaches not only clarify complex processes but also equip students with practical scientific skills. By engaging with real experimental techniques, learners gain insights into the experimental design, analysis, and interpretation pivotal to molecular biology research. This hands-on approach encourages critical thinking and a passion for discovery, central to scientific education and career preparation.

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Most popular questions from this chapter

The Ames Test In a nutrient medium that lacks histidine, a thin layer of agar containing \(\sim 10^{9}\) Salmonella typhimurium histidine auxotrophs (mutant cells that require histidine to survive) produces \(\sim 13\) colonies over a two-day incubation period at \(37^{\circ} \mathrm{C}\) (see Eig \(25-19\) ). How do these colonies arise in the absence of histidine? When investigators repeat the experiment in the presence of \(0.4 \mu \mathrm{g}\) of 2 -aminoanthracene, the number of colonies produced over two days exceeds 10,000 . What does this indicate about 2-aminoanthracene? What can you surmise about its carcinogenicity?

Direct Repair Cells normally repair the lesion \(O^{6}\)-meG by directly transferring the methyl group to the protein \(O^{6}\) methylguanine-DNA methyltransferase. For the nucleotide sequence \(\mathrm{AAC}\left(O^{6}-\mathrm{meG}\right) \mathrm{TGCAC}\), with a damaged (methylated) G residue, what would be the sequence of both strands of double- stranded DNA resulting from replication in each of the situations listed? a. Replication occurs before repair. b. Replication occurs after repair. c. Two rounds of replication occur, followed by repair.

Base Composition of DNAs Made from Single-Stranded Templates Predict the base composition of the total DNA synthesized by DNA polymerase on templates provided by an equimolar mixture of the two complementary strands of bacteriophage \(\phi \mathrm{X} 174 \mathrm{DNA}\) (a circular DNA molecule). The base composition of one strand is A, \(24.7 \% ; \mathrm{G}, 24.1 \% ; \mathrm{C}\), \(18.5 \%\); and \(\mathrm{T}, 32.7 \%\). What assumption is necessary to answer this problem?

Strand Invasion in Recombination A key step in many homologous recombination reactions is strand invasion (see step 2 in Fig. 25-29). In almost every case, strand invasion proceeds with a single strand that has a free \(3^{\prime}\) end rather than a \(5^{\prime}\) end. What DNA metabolic advantage is inherent with the use of a free 3 ' end for strand invasion?

Function of DNA Ligase Some \(E\). coli mutants contain defective DNA ligase. When researchers expose these mutants to \({ }^{3} \mathrm{H}\)-labeled thymine and then sediment the DNA produced on an alkaline sucrose density gradient, two radioactive bands appear. One corresponds to a high molecular weight fraction, the other to a low molecular weight fraction. Explain.

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